From ea6d2c80d7fd3cc5030417b0d21a9d60988bfa7d Mon Sep 17 00:00:00 2001 From: Morten Hjorth-Jensen Date: Wed, 27 Oct 2021 15:38:28 +0200 Subject: [PATCH] added codes and more --- doc/src/week36/{ => programs}/franke.py | 0 doc/src/week36/{ => programs}/scale.py | 0 doc/src/week36/{ => programs}/scale2.py | 0 doc/src/week36/{ => programs}/scale3.py | 0 doc/src/week36/{ => programs}/scaler.py | 0 doc/src/week36/{ => programs}/test.py | 0 doc/src/week36/{ => programs}/test2.py | 0 doc/src/week39/codes/test10.py | 74 +- doc/src/week39/codes/test11.py | 154 ++ doc/src/week39/codes/test12.py | 741 ++++++ doc/src/week43/LatexFigures/Graph | 12 + doc/src/week43/LatexFigures/fig1.pdf | Bin 0 -> 41515 bytes doc/src/week43/LatexFigures/fig1.tex | 28 + doc/src/week43/LatexFigures/fig2.pdf | Bin 0 -> 39214 bytes doc/src/week43/LatexFigures/fig2.tex | 64 + doc/src/week43/LatexFigures/fig3.tex | 55 + doc/src/week43/LatexFigures/nn.png | Bin 0 -> 173946 bytes doc/src/week43/LatexFigures/nn.py | 51 + 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doc/src/week43/LatexFigures/nn.png create mode 100644 doc/src/week43/LatexFigures/nn.py delete mode 100644 doc/src/week43/chapter10.do.txt delete mode 100644 doc/src/week43/odenn.do.txt diff --git a/doc/src/week36/franke.py b/doc/src/week36/programs/franke.py similarity index 100% rename from doc/src/week36/franke.py rename to doc/src/week36/programs/franke.py diff --git a/doc/src/week36/scale.py b/doc/src/week36/programs/scale.py similarity index 100% rename from doc/src/week36/scale.py rename to doc/src/week36/programs/scale.py diff --git a/doc/src/week36/scale2.py b/doc/src/week36/programs/scale2.py similarity index 100% rename from doc/src/week36/scale2.py rename to doc/src/week36/programs/scale2.py diff --git a/doc/src/week36/scale3.py b/doc/src/week36/programs/scale3.py similarity index 100% rename from doc/src/week36/scale3.py rename to doc/src/week36/programs/scale3.py diff --git a/doc/src/week36/scaler.py b/doc/src/week36/programs/scaler.py similarity index 100% rename from doc/src/week36/scaler.py rename to doc/src/week36/programs/scaler.py diff --git a/doc/src/week36/test.py b/doc/src/week36/programs/test.py similarity index 100% rename from doc/src/week36/test.py rename to doc/src/week36/programs/test.py diff --git a/doc/src/week36/test2.py b/doc/src/week36/programs/test2.py similarity index 100% rename from doc/src/week36/test2.py rename to doc/src/week36/programs/test2.py diff --git a/doc/src/week39/codes/test10.py b/doc/src/week39/codes/test10.py index 946a8edd2..6de1794db 100644 --- a/doc/src/week39/codes/test10.py +++ b/doc/src/week39/codes/test10.py @@ -1,44 +1,52 @@ +""" +Code to test Ridge with own gradient descent and SGD +""" -from random import random, seed import numpy as np +import pandas as pd import matplotlib.pyplot as plt -from mpl_toolkits.mplot3d import Axes3D -from matplotlib import cm -from matplotlib.ticker import LinearLocator, FormatStrFormatter -import sys +from sklearn.model_selection import train_test_split +from sklearn import linear_model +from sklearn.neural_network import MLPRegressor +from sklearn.metrics import accuracy_score +import seaborn as sns +import autograd.numpy as np +from autograd import grad + + +def MSE(y_data,y_model): + n = np.size(y_model) + return np.sum((y_data-y_model)**2)/n +# A seed just to ensure that the random numbers are the same for every run. +# Useful for eventual debugging. +np.random.seed(315) -# the number of datapoints n = 100 -x = 2*np.random.rand(n,1) -y = 4+3*x*x+np.random.randn(n,1) +x = np.random.rand(n) +y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2) -X = np.c_[np.ones((n,1)), x, x*x] -XT_X = X.T @ X +Maxpolydegree = 5 +X = np.zeros((n,Maxpolydegree-1)) -#Ridge parameter lambda -lmbda = 0.001 -Id = lmbda* np.eye(XT_X.shape[0]) +for degree in range(1,Maxpolydegree): #No intercept column + X[:,degree-1] = x**(degree) -beta_linreg = np.linalg.inv(XT_X+Id) @ X.T @ y -print(beta_linreg) -# Start plain gradient descent -beta = np.random.randn(2,1) +# We split the data in test and training data +X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2) -eta = 0.1 -Niterations = 100 -for iter in range(Niterations): - gradients = 2.0/n*X.T @ (X @ (beta)-y)+2*lmbda*beta - beta -= eta*gradients +nlambdas = 10 +lmbd_vals = np.logspace(-4, 0, nlambdas) +MSERidgePredict = np.zeros(nlambdas) +for i in range(nlambdas): + lmb = lmbd_vals[i] + RegRidge = linear_model.Ridge(lmb,fit_intercept=False) + RegRidge.fit(X_train,y_train) + ypredictRidge = RegRidge.predict(X_test) + MSERidgePredict[i] = MSE(y_test,ypredictRidge) -print(beta) -ypredict = X @ beta -ypredict2 = X @ beta_linreg -plt.plot(x, ypredict, "r-") -plt.plot(x, ypredict2, "b-") -plt.plot(x, y ,'ro') -plt.axis([0,2.0,0, 15.0]) -plt.xlabel(r'$x$') -plt.ylabel(r'$y$') -plt.title(r'Gradient descent example for Ridge') -plt.show() +beta = np.random.randn(X_train.shape[1],1) +loss = np.mean((y_train.reshape(-1,1) - X_train@beta)**2) +print(loss) +get_grad = grad(loss,argnum=2) +grad_beta = get_grad(X_train,y_train,beta) diff --git a/doc/src/week39/codes/test11.py b/doc/src/week39/codes/test11.py new file mode 100644 index 000000000..1dd549d09 --- /dev/null +++ b/doc/src/week39/codes/test11.py @@ -0,0 +1,154 @@ +##Make synthetic data +n = 1000 +np.random.seed(20) +x1 = np.random.rand(n) +x2 = np.random.rand(n) +X = designMatrix(x1, x2, 4) +y = franke(x1, x2) + +##Train-validation-test samples. +# We choose / play with hyper-parameters on the validation data and then test predictions on the test data + +X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=1) +X_train, X_val, y_train, y_val = train_test_split(X_train, y_train, test_size=0.25, random_state=1) # 0.25 x 0.8 = 0.2 + +scaler = StandardScaler() +scaler.fit(X_train) +X_train = scaler.transform(X_train) +X_test = scaler.transform(X_test) +X_val = scaler.transform(X_val) + +X_train[:, 0] = 1 +X_test[:, 0] = 1 +X_val[:, 0] = 1 + + + +linreg = linregOwn(method='ols') +#print('Invert OLS:', linreg.fit(X_train, y_train)) +beta = SGD(X_train, y_train, learning_rate=0.07) +#print('SGD OLS:', beta) + + +linreg = linregOwn(method='ridge') +#print('Invert Ridge:', linreg.fit(X_train, y_train, lambda_= 0.01)) +beta = SGD(X_train, y_train, learning_rate=0.0004, method='ridge') +#print('SGD Ridge:', beta) + + +sgdreg = SGDRegressor(max_iter = 100, penalty=None, eta0=0.1) +sgdreg.fit(X_train[:, 1:],y_train.ravel()) +#print('sklearn:', sgdreg.coef_) +#print('sklearn intercept:', sgdreg.intercept_) + + +def plot_MSE(method = 'ridge', scheme = None): + eta = np.logspace(-5, -3, 10) + lambda_ = np.logspace(-5, -1, 10) + MSE_ols = [] + MSE_ridge = [] + + if scheme == 'joint': + + if method == 'ridge': + + for lmbd in lambda_: + + for i in eta: + + beta = SGD(X_train, y_train, learning_rate=i, lambda_ = lmbd, method = method) + mse_ols_test, mse_ridge_test = compute_test_mse(X_val, y_val, lambda_ = lmbd, beta = beta) + MSE_ridge.append(mse_ridge_test) + + fig = plt.figure() + ax = fig.gca(projection='3d') ##get current axis + lambda_ = np.ravel(lambda_) + eta = np.ravel(eta) + ax.zaxis.set_major_locator(LinearLocator(5)) + ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f')) + ax.xaxis.set_major_formatter(FormatStrFormatter('%.02f')) + ax.yaxis.set_major_formatter(FormatStrFormatter('%.03f')) + ax.plot_trisurf(lambda_, eta, MSE_ridge, cmap='viridis', edgecolor='none') + ax.set_xlabel(r'$\lambda$') + ax.set_ylabel(r'$\eta$') + ax.set_title(r'MSE Ridge') + ax.view_init(30, 60) + plt.show() + + if scheme == 'separate': + + if method == 'ols': + + eta = np.logspace(-5, 0, 10) + + for i in eta: + + beta = SGD(X_train, y_train, learning_rate=i, lambda_ = 0.01, method = method) + mse_ols_test, mse_ridge_test = compute_test_mse(X_val, y_val, beta = beta) + MSE_ols.append(mse_ols_test) + + print('The learning rate {} performs best for the OLS' .format(eta[MSE_ols.index(min(MSE_ols))])) + print('Corresponding minimum MSE for OLS: {}'.format(min(MSE_ols))) + plt.semilogx(eta, MSE_ols) + plt.xlabel(r'Learning rate, $\eta$') + plt.ylabel('MSE OLS') + plt.title('Stochastic Gradient Descent') + plt.show() + if scheme == 'separate': + + if method == 'ridge': + + eta = np.logspace(-5, 0, 10) + + for i in eta: + + beta = SGD(X_train, y_train, learning_rate=i, lambda_ = 0.01, method = method) + mse_ols_test, mse_ridge_test = compute_test_mse(X_val, y_val, beta = beta) + MSE_ols.append(mse_ridge_test) + + print('The learning rate {} performs best for Ridge' .format(eta[MSE_ols.index(min(MSE_ols))])) + print('Corresponding minimum MSE for Ridge: {}'.format(min(MSE_ols))) + + plt.plot(eta, MSE_ols) + plt.xlabel(r'Learning rate, $\eta$') + plt.ylabel('MSE Ridge') + plt.title('Stochastic Gradient Descent') + plt.show() + + + +# plot_MSE(method='ridge', scheme = 'joint') + +# plot_MSE(method='ols', scheme = 'separate') + +# plot_MSE(method='ridge', scheme = 'separate') + + +####Predict OLS, Ridge on test data after tuning learning rate and lambda on validation data + +def plot_scatter(y_true, method = 'ols'): + if method == 'ols': + beta = SGD(X_train, y_train, learning_rate=0.07, lambda_ = 0, method = method, n_epochs=300) + if method == 'ridge': + beta = SGD(X_train, y_train, learning_rate=0.0001, lambda_ = 0, method = method, n_epochs=300) + y_pred = np.dot(X_test, beta) + mse_ols_test, mse_ridge_test = compute_test_mse(X_test, y_true, beta = beta) + print('Test MSE OLS: {}' .format(mse_ols_test)) + print('Test MSE Ridge: {}' .format(mse_ridge_test)) + a = plt.axes(aspect='equal') + plt.scatter(y_pred, y_pred, color= 'blue', label = "True values") + plt.scatter(y_pred, y_true, color = 'red', label = "Predicted values") + plt.xlabel('True y values') + plt.ylabel('Predicted y') + plt.title(f"Prediction - {method}") + plt.legend() + # if method == 'ols': + # plt.savefig(os.path.join(os.path.dirname(__file__), 'Plots', 'ols_reg_pred.png'), transparent=True, bbox_inches='tight') + # if method == 'ridge': + # plt.savefig(os.path.join(os.path.dirname(__file__), 'Plots', 'ridge_reg_pred.png'), transparent=True, bbox_inches='tight') + + plt.show() + +plot_scatter(y_test, method='ols') + +plot_scatter(y_test, method='ridge') diff --git a/doc/src/week39/codes/test12.py b/doc/src/week39/codes/test12.py new file mode 100644 index 000000000..da046faa1 --- /dev/null +++ b/doc/src/week39/codes/test12.py @@ -0,0 +1,741 @@ +Aimport os +import sys +import pytest +import numba +from matplotlib import cm +from matplotlib.ticker import LinearLocator, FormatStrFormatter +import matplotlib.pyplot as plt +from mpl_toolkits.mplot3d import Axes3D +import numpy as np +import functools +import time +from numba import jit, njit +from PIL import Image +import pandas as pd +import seaborn as sns +sns.set() +import math + +from sklearn.model_selection import train_test_split, cross_val_score, KFold +from sklearn.preprocessing import StandardScaler +from sklearn.linear_model import SGDRegressor, LinearRegression, LogisticRegression, Lasso, Ridge +from sklearn import datasets +from sklearn.metrics import confusion_matrix, mean_squared_error, r2_score +from sklearn.utils import resample + + +# Bootstrap + +def Bootstrap(x1,x2, y, N_boot=500, method = 'ols', degrees = 5, random_state = 42): + """ + Computes bias^2, variance and the mean squared error using bootstrap resampling method + for the provided data and the method. + + Arguments: + x1: 1D numpy array, covariate + x2: 1D numpy array, covariate + N_boot: integer type, the number of bootstrap samples + method: string type, accepts 'ols', 'ridge' or 'lasso' as arguments + degree: integer type, polynomial degree for generating the design matrix + random_state: integer, ensures the same split when using the train_test_split functionality + + Returns: Bias_vec, Var_vec, MSE_vec, betaVariance_vec + numpy arrays. Bias, Variance, MSE and the variance of beta for the predicted model + """ + ##split x1, x2 and y arrays as a train and test data and generate design matrix + x1_train, x1_test,x2_train, x2_test, y_train, y_test = train_test_split(x1,x2, y, test_size=0.2, random_state = random_state) + y_pred_test = np.zeros((y_test.shape[0], N_boot)) + X_test = designMatrix(x1_test, x2_test, degrees) + + betaMatrix = np.zeros((X_test.shape[1], N_boot)) + + ##resample and fit the corresponding method on the train data + for i in range(N_boot): + x1_,x2_, y_ = resample(x1_train, x2_train, y_train) + X_train = designMatrix(x1_, x2_, degrees) + scaler = StandardScaler() + scaler.fit(X_train) + X_train = scaler.transform(X_train) + X_train[:, 0] = 1 + X_test = designMatrix(x1_test, x2_test, degrees) + X_test = scaler.transform(X_test) + X_test[:, 0] = 1 + + if method == 'ols': + manual_regression = linregOwn(method = 'ols') + beta = manual_regression.fit(X_train, y_) + if method == 'ridge': + manual_regression = linregOwn(method = 'ridge') + beta = manual_regression.fit(X_train, y_, lambda_ = 0.05) + if method == 'lasso': + manual_regression = linregOwn(method = 'lasso') + beta = manual_regression.fit(X_train, y_, lambda_ = 0.05) + + ##predict on the same test data + y_pred_test[:, i] = np.dot(X_test, beta) + betaMatrix[:, i] = beta + y_test = y_test.reshape(len(y_test),1) + + Bias_vec = [] + Var_vec = [] + MSE_vec = [] + betaVariance_vec = [] + R2_score = [] + y_test = y_test.reshape(len(y_test),1) + MSE = np.mean( np.mean((y_test - y_pred_test)**2, axis=1, keepdims=True) ) + bias = np.mean( (y_test - np.mean(y_pred_test, axis=1, keepdims=True))**2 ) + variance = np.mean( np.var(y_pred_test, axis=1, keepdims=True) ) + betaVariance = np.var(betaMatrix, axis=1) + print("-------------------------------------------------------------") + print("Degree: %d" % degrees) + print('MSE:', np.round(MSE, 3)) + print('Bias^2:', np.round(bias, 3)) + print('Var:', np.round(variance,3)) + print('{} >= {} + {} = {}'.format(MSE, bias, variance, bias+variance)) + print("-------------------------------------------------------------") + + Bias_vec.append(bias) + Var_vec.append(variance) + MSE_vec.append(MSE) + betaVariance_vec.append(betaVariance) + return Bias_vec, Var_vec, MSE_vec, betaVariance_vec + + + +class CrossValidation: + """ + A class of cross-validation technique. Performs cross-validation with shuffling. + """ + def __init__(self, LinearRegression, DesignMatrix): + """ + Initialization + + Arguments: + LinearRegression: Instance from the class created by either linregOwn or linregSKl + DesignMatrix: Function that generates design matrix + """ + self.LinearRegression = LinearRegression + self.DesignMatrix = DesignMatrix + + def kFoldCV(self, x1, x2, y, k = 10, lambda_ = 0, degree = 5): + """ + Performs shuffling of the data, holds a split of the data as a test set at each split and evaluates the model + on the rest of the data. + Calculates the MSE , R2_score, variance, bias on the test data and MSE on the train data. + + Arguments: + x1: 1D numpy array + x2: 1D numpy array + y: 1D numpy array + k: integer, the number of splits + lambda_: float type, shrinkage parameter for ridge and lasso methods. + degree: integer type, the number of polynomials, complexity parameter + + """ + self.lambda_ = lambda_ + M = x1.shape[0]//k ## Split input data x in k folds of size M + + + ##save the statistic in the list + MSE_train = [] + MSE_k = [] + R2_k = [] + var_k = [] + bias_k = [] + + ##shuffle the data randomly + shf = np.random.permutation(x1.size) + x1_shuff = x1[shf] + x2_shuff = x2[shf] + y_shuff = y[shf] + + for i in range(k): + # x_k and y_k are the hold out data for fold k + x1_k = x1_shuff[i*M:(i+1)*M] + x2_k = x2_shuff[i*M:(i+1)*M] + y_k = y_shuff[i*M:(i+1)*M] + + ## Generate train data and then scale both train and test + index_true = np.array([True for i in range(x1.shape[0])]) + index_true[i*M:(i+1)*M] = False + X_train = self.DesignMatrix(x1_shuff[index_true], x2_shuff[index_true], degree) + y_train = y_shuff[index_true] + scaler = StandardScaler() + scaler.fit(X_train) + X_train = scaler.transform(X_train) + X_train[:, 0] = 1 + + ### Fit the regression on the train data + beta = self.LinearRegression.fit(X_train, y_train, lambda_) + y_predict_train = np.dot(X_train, beta) + MSE_train.append(np.sum( (y_train-y_predict_train)**2)/len(y_train)) + + ## Predict on the hold out data and calculate statistic of interest + X_k = self.DesignMatrix(x1_k, x2_k, degree) + X_k = scaler.transform(X_k) + X_k[:, 0] = 1 + y_predict = np.dot(X_k,beta) + MSE_k.append(np.sum((y_k-y_predict)**2, axis=0, keepdims=True)/len(y_predict)) + R2_k.append(1.0 - np.sum((y_k - y_predict)**2, axis=0, keepdims=True) / np.sum((y_k - np.mean(y_k))**2, axis=0, keepdims=True) ) + var_k.append(np.var(y_predict,axis=0, keepdims=True)) + bias_k.append((y_k - np.mean(y_predict, axis=0, keepdims=True))**2 ) + + means = [np.mean(MSE_k), np.mean(R2_k), np.mean(var_k), + np.mean(bias_k),np.mean(MSE_train)] + #print('MSE_test: {}' .format(np.round(np.mean(MSE_k),3))) + #print('R2: {}' .format(np.round(np.mean(R2_k),3))) + #print('Variance of the predicted outcome: {}' .format(np.round(np.mean(var_k),3))) + #print('Bias: {}' .format(np.round(np.mean(bias_k),3))) + #print('MSE_train {}' .format(np.round(np.mean(MSE_train),3))) + return means + + +# Franke Function + +def franke(x, y): + """ + Computes Franke function. + Franke's function has two Gaussian peaks of different heights, and a smaller dip. + It is used as a test function in interpolation problems. + + Franke's function is normally defined on the grid [0, 1] for each x, y. + + Arguments of the function: + x : numpy array + y : numpy array + + Output of the function: + f : Franke function values at specific coordinate points of x and y + """ + f = (0.75 * np.exp(-((9*x - 2)**2)/4 - ((9*y - 2)**2)/4 ) + + 0.75 * np.exp(-((9*x + 1)**2)/49 - (9*y + 1) /10) + + 0.5 * np.exp(-((9*x - 7)**2)/4 - ((9*y - 3)**2)/4 ) + - 0.2 * np.exp(-((9*x - 4)**2) - ((9*y - 7)**2) )) + return f + + + + +class linregOwn: + """ + A class of linear regressions. Perform ordinarly least squares (OLS) and Ridge regression manually. Lasso + is performed using scikit-learn functionality. + """ + def __init__(self, method = 'ols'): + """ + Constructor + + Determines the method used in the fitting + + Arguments: + method: string type. Accepts either 'ols', 'ridge' or 'lasso'. + + """ + self.method = method + self.yHat = None + self.X = None + self.y = None + self.beta = None + + self._MSE = None + self._R2 = None + self._betaVariance = None + self.lambda_ = None + + def fit(self, X_train, y_train, lambda_ = 0): + """ + Performs the fit of OLS, Ridge or Lasso, depending on the argument provided initially. + + Arguments: + X_train: Covariate matrix of the train data set, i.e. design matrix of + the shape m x p where m is the number of rows and p is the number of columns + (i.e. p is the complexity parameter). + y_train: Outcome variable, 1D numpy array + lambda_: float type. Shrinkage parameter for ridge and lasso methods. The higher value, higher shrinkage. + lambda_ is set to 0 for the OLS regression + + """ + self.X_train = X_train + self.y_train = y_train + self.lambda_ = lambda_ + if self.method == 'ols': + self._olsFit(X_train, y_train) + if self.method == 'ridge': + self._ridgeFit(X_train, y_train, lambda_) + if self.method == 'lasso': + self._lassoFitSKL(X_train, y_train, lambda_) + return self.beta + + def _olsFit(self, X_train, y_train): + """ + Performs the ordinary least squares (OLS) fit on the provided data using singular value decomposition(SVD). + + + Arguments: + + X_train: Covariate matrix of the train data set, i.e. design matrix of + the shape m x p where m is the number of rows and p is the number of columns + (i.e. p is the complexity parameter). + y_train: Outcome variable, 1D numpy array + + Returns: + beta : numpy.array + The beta parameters from the performed fit + """ + self.X_train = X_train + self.y_test = y_train + U, S, VT = np.linalg.svd(self.X_train, full_matrices=True) + S_inverse = np.zeros(shape=self.X_train.shape) + ##S is a vector, with shape of the number of columns + S_inverse[:S.shape[0], :S.shape[0]] = np.diag(1/S) + self.beta = np.dot(VT.T, np.dot(S_inverse.T, np.dot(U.T, self.y_train))) + #self.beta = np.linalg.inv(np.dot(X.T,X)).dot(X.T, y) + + def _ridgeFit(self, X_train, y_train, lambda_): + """ + Performs the ridge regression fit + + Arguments: + X_train: Covariate matrix of the train data set, design matrix of + the shape m x p (m_train_rows, p_columns). + y_train: Outcome variable, 1D numpy array, dimension m x 1 + lambda_: Integer type. The shrinkage parameter + + Returns: + beta : numpy.array + The beta parameters from the performed fit + """ + self.X_train = X_train + self.y_train = y_train + self.lambda_ = lambda_ + self.beta = np.dot(np.linalg.inv(np.dot(X_train.T,X_train) + self.lambda_ * np.eye(X_train.shape[1])), np.dot(X_train.T,y_train)) + + def _lassoFitSKL(self, X_train, y_train, lambda_): + """ + Performs lasso fit using scikit-learn functionality. + + Arguments: + X_train: Covariate matrix of the train data set, design matrix of + the shape m x p (m_train_datapoints, p_parameters). + y_train: Outcome variable, 1D numpy array, dimension m x 1 + lambda_: Integer type. The shrinkage parameter + + Returns: + self.beta : numpy.array + The beta parameters from the performed fit + """ + self.regression = Lasso(fit_intercept=True, max_iter=1000000, alpha=self.lambda_) + self.regression.fit(X_train,y_train) + self.beta = self.regression.coef_ + self.beta[0] = self.regression.intercept_ + + def predict(self, X_test): + """ + Performs prediction of the fitted model on the provided test data set. + + Arguments: + X_test: Design matrix, covariate matrix, dimension k x p (k_test_rows, p_columns) + + Returns: self.yHat + numpy 1D array, prediction values of dimension k x p + """ + self.X_test = X_test + self._predictOwntest(X_test) + return self.yHat + + def _predictOwntest(self, X_test): + """ + Performs manual prediction of the given model on the train data. + """ + self.X_test = X_test + self.yHat = np.dot(self.X_test, self.beta) + + def MSE(self, y_test): + """ + Calculates the mean squared error (MSE) manually after the fit and prediction have been implemented. + + Arguments: + y_test: Outcome variable, 1D numpy array, dimension k x 1 (k_test_rows, 1_column) + + Returns: self._MSE + The mean squared error of the predicted model + """ + self.y_test = y_test + if self.yHat is None : + self._predictOwntest(X_test) + N = self.yHat.size + self._MSE = (np.sum((self.y_test - self.yHat)**2))/N + return self._MSE + + def R2(self, y_test): + """ + Calculates R2 score manually after the fit and prediction have been implemented. + + Arguments: + y_test: Outcome variable, 1D numpy array, dimension k x 1 (k_test_rows, 1_column) + + Returns: self._R2 + The R2 score of the predicted model + """ + self.y_test = y_test + if self.yHat is None: + self._predictOwntest(X_test) + yMean = (1.0 / self.y_test.size) * np.sum(self.y_test) + self._R2 = 1.0 - np.sum((self.y_test - self.yHat)**2) / np.sum((self.y_test - yMean)**2) + return self._R2 + + def CI(self, y_test): + """ + Calculates confidence intervals manually after the fit and prediction have been implemented. + + Arguments: + y_test: Outcome variable, 1D numpy array, dimension k x 1 (k_test_rows, 1_column) + + Returns: var, Lower, Upper + Variance, Lower and Upper bounds of the confidence intervals for the parameter self.beta + """ + self.y_test = y_test + if self.yHat is None: + self._predictOwntest(X_test) + sigma2 = np.sum(((self.y_test - self.yHat)**2))/(self.y_test.size - self.beta.size) + var = np.diag(np.linalg.inv(np.dot(self.X_test.T, self.X_test))) * sigma2 + Lower = self.beta - 1.96*np.sqrt(var) + Upper = self.beta + 1.96*np.sqrt(var) + return var, Lower, Upper + + + ###Implementation through scikitlearn +class linregSKL: + def __init__(self, method = 'ols'): + """ + A class of linear regressions. Perform ordinarly least squares (OLS) and Ridge and Lasso + using scikit-learn functionality. + + """ + self.method = method + self.yHat = None + self.X = None + self.y = None + self.beta = None + + self._MSE = None + self._R2 = None + self._betaVariance = None + + + def fit(self, X_train, y_train, lambda_ = 0): + self.X_train = X_train + self.y_train = y_train + if self.method == 'ols': + self._olsSKLfit(X_train, y_train) + if self.method == 'ridge': + self._sklRidgeFit(X_train, y_train, lambda_) + if self.method == 'lasso': + self._SKLlassoFit(X_train, y_train, lambda_) + return self.beta + + def _olsSKLfit(self, X_train, y_train): + self.X_train = X_train + self.y_train = y_train + ##We already have standardized data from design matrix + self.ols = LinearRegression().fit(self.X_train, self.y_train) + self.beta = self.ols.coef_ + self.beta[0] = self.ols.intercept_ + + def _SKLlassoFit(self, X_train, y_train, lambda_): + self.regression = Lasso(fit_intercept=True, max_iter=100000, alpha=self.lambda_) + self.regression.fit(X_train,y_train) + self.beta = self.regression.coef_ + self.beta[0] = self.regression.intercept_ + + def _sklRidgeFit(self, X_train, y_train, lambda_): + self.regression = Ridge(fit_intercept=True, alpha=self.lambda_) + self.regression.fit(X,y) + self.beta = self.regression.coef_ + self.beta[0] = self.regression.intercept_ + + def predict(self, X_test): + self.X_test = X_test + if self.method == 'ols': + self._sklPredict(X_test) + return self.yHat + + def _sklPredict(self, X_test): + self.X_test = X_test + ## Since our data contains 1-s, we should subtract intercept, since scikit learn additionally + ##generates the 1-s + self.yHat = self.ols.predict(self.X_test) - self.beta[0] + + def MSE(self, y_test): + self.y_test = y_test + if self.yHat is None : + self._sklPredict(X_test) + self._MSE = mean_squared_error(self.y_test, self.yHat) + return self._MSE + + def R2(self, y_test): + self.y_test = y_test + if self.yHat is None : + self._sklPredict() + self._R2 = r2_score(self.y_test, self.yHat) + return self._R2 + + + +def designMatrix(x, y, k=5): + """ + Generates the design matrix (covariates of polynomial degree k). + Intercept is included in the design matrix. + Scaling does not apply to the intercept term. + if k = 2, generated column vectors: 1, x, y, x^2, xy, y^2 + if k = 3, generated column vectors: 1, x, y, x^2, xy, y^2, x^3, x^2y, xy^2, y^3 + ... + + Arguments: + x: 1D numpy array + y: 1D numpy array + k: integer type. complexity parameter (i.e polynomial degree) + """ + + xb = np.ones((x.size, 1)) + + for i in range(1, k+1): + for j in range(i+1): + xb = np.c_[xb, (x**(i-j))*(y**j)] + + xb[:, 0] = 1 + return xb + + + + +# Stochastic Gradient Descent + +from matplotlib.ticker import LinearLocator, FormatStrFormatter + +def compute_square_loss(X, y, theta): + loss = 0 #Initialize the average square loss + + m = len(y) + loss = (1.0/m)*(np.linalg.norm((X.dot(theta) - y)) ** 2) + return loss + + +def gradient_ridge(X, y, beta, lambda_): + return 2*(np.dot(X.T, (X.dot(beta) - y))) + 2*lambda_*beta + +def gradient_ols(X, y, beta): + m = X.shape[0] + + grad = 2/m * X.T.dot(X.dot(beta) - y) + + return grad + +def learning_schedule(t): + t0, t1 = 5, 50 + return t0/(t+t1) + + +def iterate_minibatches(inputs, targets, batchsize, shuffle=True): + assert inputs.shape[0] == targets.shape[0] + if shuffle: + indices = np.random.permutation(inputs.shape[0]) + for start_idx in range(0, inputs.shape[0], batchsize): + end_idx = min(start_idx + batchsize, inputs.shape[0]) + if shuffle: + excerpt = indices[start_idx:end_idx] + else: + excerpt = slice(start_idx, end_idx) + yield inputs[excerpt], targets[excerpt] + + +###sgd +def SGD(X, y, learning_rate = 0.02, n_epochs = 100, lambda_ = 0.01, batch_size = 20, method = 'ols'): + num_instances, num_features = X.shape[0], X.shape[1] + beta = np.random.randn(num_features) ##initialize beta + + for epoch in range(n_epochs+1): + + for batch in iterate_minibatches(X, y, batch_size, shuffle=True): + + X_batch, y_batch = batch + + # for i in range(batch_size): + # learning_rate = learning_schedule(n_epochs*epoch + i) + + if method == 'ols': + gradient = gradient_ols(X_batch, y_batch, beta) + beta = beta - learning_rate*gradient + if method == 'ridge': + gradient = gradient_ridge(X_batch, y_batch, beta, lambda_ = lambda_) + beta = beta - learning_rate*gradient + + mse_ols_train = compute_square_loss(X, y, beta) + mse_ridge_train = compute_square_loss(X, y, beta) + lambda_*np.dot(beta.T, beta) + + return beta + +def compute_test_mse(X_test, y_test, beta, lambda_ = 0.01): + mse_ols_test = compute_square_loss(X_test, y_test, beta) + mse_ridge_test = compute_square_loss(X_test, y_test, beta) + lambda_*np.dot(beta.T, beta) + return mse_ols_test, mse_ridge_test + + +# # Part A + +# In[10]: + + +# a + +##Make synthetic data +n = 1000 +np.random.seed(20) +x1 = np.random.rand(n) +x2 = np.random.rand(n) +X = designMatrix(x1, x2, 4) +y = franke(x1, x2) + +##Train-validation-test samples. +# We choose / play with hyper-parameters on the validation data and then test predictions on the test data + +X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=1) +X_train, X_val, y_train, y_val = train_test_split(X_train, y_train, test_size=0.25, random_state=1) # 0.25 x 0.8 = 0.2 + +scaler = StandardScaler() +scaler.fit(X_train) +X_train = scaler.transform(X_train) +X_test = scaler.transform(X_test) +X_val = scaler.transform(X_val) + +X_train[:, 0] = 1 +X_test[:, 0] = 1 +X_val[:, 0] = 1 + + + +linreg = linregOwn(method='ols') +#print('Invert OLS:', linreg.fit(X_train, y_train)) +beta = SGD(X_train, y_train, learning_rate=0.07) +#print('SGD OLS:', beta) + + +linreg = linregOwn(method='ridge') +#print('Invert Ridge:', linreg.fit(X_train, y_train, lambda_= 0.01)) +beta = SGD(X_train, y_train, learning_rate=0.0004, method='ridge') +#print('SGD Ridge:', beta) + + +sgdreg = SGDRegressor(max_iter = 100, penalty=None, eta0=0.1) +sgdreg.fit(X_train[:, 1:],y_train.ravel()) +#print('sklearn:', sgdreg.coef_) +#print('sklearn intercept:', sgdreg.intercept_) + + +def plot_MSE(method = 'ridge', scheme = None): + eta = np.logspace(-5, -3, 10) + lambda_ = np.logspace(-5, -1, 10) + MSE_ols = [] + MSE_ridge = [] + + if scheme == 'joint': + + if method == 'ridge': + + for lmbd in lambda_: + + for i in eta: + + beta = SGD(X_train, y_train, learning_rate=i, lambda_ = lmbd, method = method) + mse_ols_test, mse_ridge_test = compute_test_mse(X_val, y_val, lambda_ = lmbd, beta = beta) + MSE_ridge.append(mse_ridge_test) + + fig = plt.figure() + ax = fig.gca(projection='3d') ##get current axis + lambda_ = np.ravel(lambda_) + eta = np.ravel(eta) + ax.zaxis.set_major_locator(LinearLocator(5)) + ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f')) + ax.xaxis.set_major_formatter(FormatStrFormatter('%.02f')) + ax.yaxis.set_major_formatter(FormatStrFormatter('%.03f')) + ax.plot_trisurf(lambda_, eta, MSE_ridge, cmap='viridis', edgecolor='none') + ax.set_xlabel(r'$\lambda$') + ax.set_ylabel(r'$\eta$') + ax.set_title(r'MSE Ridge') + ax.view_init(30, 60) + plt.show() + + if scheme == 'separate': + + if method == 'ols': + + eta = np.logspace(-5, 0, 10) + + for i in eta: + + beta = SGD(X_train, y_train, learning_rate=i, lambda_ = 0.01, method = method) + mse_ols_test, mse_ridge_test = compute_test_mse(X_val, y_val, beta = beta) + MSE_ols.append(mse_ols_test) + + print('The learning rate {} performs best for the OLS' .format(eta[MSE_ols.index(min(MSE_ols))])) + print('Corresponding minimum MSE for OLS: {}'.format(min(MSE_ols))) + plt.semilogx(eta, MSE_ols) + plt.xlabel(r'Learning rate, $\eta$') + plt.ylabel('MSE OLS') + plt.title('Stochastic Gradient Descent') + plt.show() + if scheme == 'separate': + + if method == 'ridge': + + eta = np.logspace(-5, 0, 10) + + for i in eta: + + beta = SGD(X_train, y_train, learning_rate=i, lambda_ = 0.01, method = method) + mse_ols_test, mse_ridge_test = compute_test_mse(X_val, y_val, beta = beta) + MSE_ols.append(mse_ridge_test) + + print('The learning rate {} performs best for Ridge' .format(eta[MSE_ols.index(min(MSE_ols))])) + print('Corresponding minimum MSE for Ridge: {}'.format(min(MSE_ols))) + + plt.plot(eta, MSE_ols) + plt.xlabel(r'Learning rate, $\eta$') + plt.ylabel('MSE Ridge') + plt.title('Stochastic Gradient Descent') + plt.show() + + + + +####Predict OLS, Ridge on test data after tuning learning rate and lambda on validation data + +def plot_scatter(y_true, method = 'ols'): + if method == 'ols': + beta = SGD(X_train, y_train, learning_rate=0.07, lambda_ = 0, method = method, n_epochs=300) + if method == 'ridge': + beta = SGD(X_train, y_train, learning_rate=0.0001, lambda_ = 0, method = method, n_epochs=300) + y_pred = np.dot(X_test, beta) + mse_ols_test, mse_ridge_test = compute_test_mse(X_test, y_true, beta = beta) + print('Test MSE OLS: {}' .format(mse_ols_test)) + print('Test MSE Ridge: {}' .format(mse_ridge_test)) + a = plt.axes(aspect='equal') + plt.scatter(y_pred, y_pred, color= 'blue', label = "True values") + plt.scatter(y_pred, y_true, color = 'red', label = "Predicted values") + plt.xlabel('True y values') + plt.ylabel('Predicted y') + plt.title(f"Prediction - {method}") + plt.legend() + # if method == 'ols': + # plt.savefig(os.path.join(os.path.dirname(__file__), 'Plots', 'ols_reg_pred.png'), transparent=True, bbox_inches='tight') + # if method == 'ridge': + # plt.savefig(os.path.join(os.path.dirname(__file__), 'Plots', 'ridge_reg_pred.png'), transparent=True, bbox_inches='tight') + + plt.show() + +plot_scatter(y_test, method='ols') + +plot_scatter(y_test, method='ridge') + + + diff --git a/doc/src/week43/LatexFigures/Graph b/doc/src/week43/LatexFigures/Graph new file mode 100644 index 000000000..06703d636 --- /dev/null +++ b/doc/src/week43/LatexFigures/Graph @@ -0,0 +1,12 @@ +// A Round Graph +digraph { + A [label=Alex] + B [label=Rishu] + C [label=Mohe] + D [label=Satyam] + A -> B + A -> 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100644 index 000000000..9bf344653 --- /dev/null +++ b/doc/src/week43/LatexFigures/fig2.tex @@ -0,0 +1,64 @@ +\documentclass[border=0.125cm]{standalone} +\usepackage{tikz} +\usetikzlibrary{positioning} +\begin{document} + +\tikzset{% + every neuron/.style={ + circle, + draw, + minimum size=1cm + }, + neuron missing/.style={ + draw=none, + scale=4, + text height=0.333cm, + execute at begin node=\color{black}$\vdots$ + }, +} + +\begin{tikzpicture}[x=1.5cm, y=1.5cm, >=stealth] + +\foreach \m/\l [count=\y] in {1,2,3,missing,4} + \node [every neuron/.try, neuron \m/.try] (input-\m) at (0,2.5-\y) {}; + +\foreach \m [count=\y] in {1,missing,2} + \node [every neuron/.try, neuron \m/.try ] (hidden-\m) at (2,2-\y*1.25) {}; + +\foreach \m [count=\y] in {1,missing,2} + \node [every neuron/.try, neuron \m/.try ] (output-\m) at (4,1.5-\y) {}; + +\foreach \l [count=\i] in {1,2,3,n} + \draw [<-] (input-\i) -- ++(-1,0) + node [above, midway] {$x_\l$}; + +\foreach \l [count=\i] in {1,m} + \node [above] at (hidden-\i.north) {$h^{1}_\l$}; + +\foreach \l [count=\i] in {1,k} + \node [above] at (hidden-\i.north) {$h^{2}_\l$}; + + +\foreach \l [count=\i] in {1,n} + \draw [->] (output-\i) -- ++(1,0) + node [above, midway] {$\tilde{y}_\l$}; + +\foreach \i in {1,...,4} + \foreach \j in {1,...,2} + \draw [->] (input-\i) -- (hidden-\j); + +\foreach \i in {1,...,2} + \foreach \j in {1,...,2} + \draw [->] (hidden-\i) -- (hidden-\j); + + +\foreach \i in {1,...,2} + \foreach \j in {1,...,2} + \draw [->] (hidden-\i) -- (output-\j); + +\foreach \l [count=\x from 0] in {Input, First hidden, Second hidden, Ouput} + \node [align=center, above] at (\x*2,2) {\l \\ layer}; + +\end{tikzpicture} + +\end{document} diff --git a/doc/src/week43/LatexFigures/fig3.tex b/doc/src/week43/LatexFigures/fig3.tex new file mode 100644 index 000000000..22c777fcd --- /dev/null +++ b/doc/src/week43/LatexFigures/fig3.tex @@ -0,0 +1,55 @@ +\documentclass[border=0.125cm]{standalone} +\usepackage{tikz} +\usetikzlibrary{positioning} +\begin{document} + +\tikzset{% + every neuron/.style={ + circle, + draw, + minimum size=1cm + }, + neuron missing/.style={ + draw=none, + scale=4, + text height=0.333cm, + execute at begin node=\color{black}$\vdots$ + }, +} + +\begin{tikzpicture}[x=1.5cm, y=1.5cm, >=stealth] + +\foreach \m/\l [count=\y] in {1,2,3,missing,4} + \node [every neuron/.try, neuron \m/.try] (input-\m) at (0,2.5-\y) {}; + +\foreach \m [count=\y] in {1,missing,2} + \node [every neuron/.try, neuron \m/.try ] (hidden-\m) at (2,2-\y*1.25) {}; + +\foreach \m [count=\y] in {1,missing,2} + \node [every neuron/.try, neuron \m/.try ] (output-\m) at (4,1.5-\y) {}; + +\foreach \l [count=\i] in {1,2,3,n} + \draw [<-] (input-\i) -- ++(-1,0) + node [above, midway] {$I_\l$}; + +\foreach \l [count=\i] in {1,n} + \node [above] at (hidden-\i.north) {$H_\l$}; + +\foreach \l [count=\i] in {1,n} + \draw [->] (output-\i) -- ++(1,0) + node [above, midway] {$O_\l$}; + +\foreach \i in {1,...,4} + 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zhf?n>9arC9rHs0VxBfG7dkw3(Mm#^8AD63mQTJk7B2WxyruZlF?YO$F&iC^9+Z^Z5 z?GT)Q{z6`gTfP}B)FzUvGo#U<;cRFAhH0|K(V#gI@~Uc+9hnXIRUH4jiJ&AUq;Z;r z0NPnuv5@t+^bWd*$90jKU4jhqkSLhjYCUK5yP!)J1Y4euEm{Dsk`gDEyeq!}=a0Ms+4lzR+gPxu1-oYXNH))mp~1{(OOv=EAJ^bsb@BB(z4 zlGPCIf*`Vfkb?1Y+m6ny?~~N?;?inrAHeeC!g}dXoJg!dVJ2At355on%eBqAbabun zLd?|nx5mNfvjYrWp`F^#6TX&eSrCw-Pnyv_7)(u-W6+$1tHX1iyLv0kizAaA{}^$} zC~=HHKT~hbB@uH&^#$OEf9t?!Q!g9em4+HgJA1xx`arb)TF(FO(ENx|325t3Nxt zoSpqsh2s+|*!FDT18QQ1p#|fZLjO-l0C+7z2QX#^4qc7C4I9|1;XP%x`Wm}_+<7*Q zujx_7&hy>VLzQmmYtt|O`*S*Z`ZC(##&@BHUGnY9G{bH2-H}pbWqjBDaoz7usfoV2 z4W;{3bbA!XS?G43-LXPN5?|eM)=1~|D!NZaSEs`6I$`sb=eK>> fig = plt.figure(figsize=(12, 12)) + >>> draw_neural_net(fig.gca(), .1, .9, .1, .9, [4, 7, 2]) + + :parameters: + - ax : matplotlib.axes.AxesSubplot + The axes on which to plot the cartoon (get e.g. by plt.gca()) + - left : float + The center of the leftmost node(s) will be placed here + - right : float + The center of the rightmost node(s) will be placed here + - bottom : float + The center of the bottommost node(s) will be placed here + - top : float + The center of the topmost node(s) will be placed here + - layer_sizes : list of int + List of layer sizes, including input and output dimensionality + ''' + n_layers = len(layer_sizes) + v_spacing = (top - bottom)/float(max(layer_sizes)) + h_spacing = (right - left)/float(len(layer_sizes) - 1) + # Nodes + for n, layer_size in enumerate(layer_sizes): + layer_top = v_spacing*(layer_size - 1)/2. + (top + bottom)/2. + for m in range(layer_size): + circle = plt.Circle((n*h_spacing + left, layer_top - m*v_spacing), v_spacing/4., + color='w', ec='k', zorder=4) + ax.add_artist(circle) + # Edges + for n, (layer_size_a, layer_size_b) in enumerate(zip(layer_sizes[:-1], layer_sizes[1:])): + layer_top_a = v_spacing*(layer_size_a - 1)/2. + (top + bottom)/2. + layer_top_b = v_spacing*(layer_size_b - 1)/2. + (top + bottom)/2. + for m in range(layer_size_a): + for o in range(layer_size_b): + line = plt.Line2D([n*h_spacing + left, (n + 1)*h_spacing + left], + [layer_top_a - m*v_spacing, layer_top_b - o*v_spacing], c='k') + ax.add_artist(line) + + +fig = plt.figure(figsize=(12, 12)) +ax = fig.gca() +ax.axis('off') +draw_neural_net(ax, .1, .9, .1, .9, [4, 7, 2]) +fig.savefig('nn.png') +plt.show() diff --git a/doc/src/week43/chapter10.do.txt b/doc/src/week43/chapter10.do.txt deleted file mode 100644 index b10741ef0..000000000 --- a/doc/src/week43/chapter10.do.txt +++ /dev/null @@ -1,3065 +0,0 @@ -======= Recurrent Neural Networks ======= - -"Overview video":"https://www.youtube.com/watch?v=SEnXr6v2ifU&ab_channel=AlexanderAmini". -See also lecture on Thursday October 22 and examples from "week 42":"https://compphysics.github.io/MachineLearning/doc/pub/week42/html/week42.html". - -"IN5400 at UiO Lecture":"https://www.uio.no/studier/emner/matnat/ifi/IN5400/v20/material/week10/in5400_2020_week10_recurrent_neural_network.pdf" - -"CS231 at Stanford Lecture":"https://www.youtube.com/watch?v=6niqTuYFZLQ&list=PLzUTmXVwsnXod6WNdg57Yc3zFx_f-RYsq&index=10&ab_channel=StanfordUniversitySchoolofEngineering" - -===== Recurrent neural networks: Overarching view ===== - -Till now our focus has been, including convolutional neural networks -as well, on feedforward neural networks. The output or the activations -flow only in one direction, from the input layer to the output layer. - -A recurrent neural network (RNN) looks very much like a feedforward -neural network, except that it also has connections pointing -backward. - -RNNs are used to analyze time series data such as stock prices, and -tell you when to buy or sell. In autonomous driving systems, they can -anticipate car trajectories and help avoid accidents. More generally, -they can work on sequences of arbitrary lengths, rather than on -fixed-sized inputs like all the nets we have discussed so far. For -example, they can take sentences, documents, or audio samples as -input, making them extremely useful for natural language processing -systems such as automatic translation and speech-to-text. - - - - -!split -===== Set up of an RNN ===== - - -Text to come. - - -!split -===== A simple example ===== - -!bc pycod -# Start importing packages -import pandas as pd -import numpy as np -import matplotlib.pyplot as plt -import tensorflow as tf -from tensorflow.keras import datasets, layers, models -from tensorflow.keras.layers import Input -from tensorflow.keras.models import Model, Sequential -from tensorflow.keras.layers import Dense, SimpleRNN, LSTM, GRU -from tensorflow.keras import optimizers -from tensorflow.keras import regularizers -from tensorflow.keras.utils import to_categorical - - - -# convert into dataset matrix -def convertToMatrix(data, step): - X, Y =[], [] - for i in range(len(data)-step): - d=i+step - X.append(data[i:d,]) - Y.append(data[d,]) - return np.array(X), np.array(Y) - -step = 4 -N = 1000 -Tp = 800 - -t=np.arange(0,N) -x=np.sin(0.02*t)+2*np.random.rand(N) -df = pd.DataFrame(x) -df.head() - -plt.plot(df) -plt.show() - -values=df.values -train,test = values[0:Tp,:], values[Tp:N,:] - -# add step elements into train and test -test = np.append(test,np.repeat(test[-1,],step)) -train = np.append(train,np.repeat(train[-1,],step)) - -trainX,trainY =convertToMatrix(train,step) -testX,testY =convertToMatrix(test,step) -trainX = np.reshape(trainX, (trainX.shape[0], 1, trainX.shape[1])) -testX = np.reshape(testX, (testX.shape[0], 1, testX.shape[1])) - -model = Sequential() -model.add(SimpleRNN(units=32, input_shape=(1,step), activation="relu")) -model.add(Dense(8, activation="relu")) -model.add(Dense(1)) -model.compile(loss='mean_squared_error', optimizer='rmsprop') -model.summary() - -model.fit(trainX,trainY, epochs=100, batch_size=16, verbose=2) -trainPredict = model.predict(trainX) -testPredict= model.predict(testX) -predicted=np.concatenate((trainPredict,testPredict),axis=0) - -trainScore = model.evaluate(trainX, trainY, verbose=0) -print(trainScore) - -index = df.index.values -plt.plot(index,df) -plt.plot(index,predicted) -plt.axvline(df.index[Tp], c="r") -plt.show() -!ec - - -!split -===== An extrapolation example ===== - -The following code provides an example of how recurrent neural -networks can be used to extrapolate to unknown values of physics data -sets. Specifically, the data sets used in this program come from -a quantum mechanical many-body calculation of energies as functions of the number of particles. - - -!bc pycod - -# For matrices and calculations -import numpy as np -# For machine learning (backend for keras) -import tensorflow as tf -# User-friendly machine learning library -# Front end for TensorFlow -import tensorflow.keras -# Different methods from Keras needed to create an RNN -# This is not necessary but it shortened function calls -# that need to be used in the code. -from tensorflow.keras import datasets, layers, models -from tensorflow.keras.layers import Input -from tensorflow.keras import regularizers -from tensorflow.keras.models import Model, Sequential -from tensorflow.keras.layers import Dense, SimpleRNN, LSTM, GRU -# For timing the code -from timeit import default_timer as timer -# For plotting -import matplotlib.pyplot as plt - - -# The data set -datatype='VaryDimension' -X_tot = np.arange(2, 42, 2) -y_tot = np.array([-0.03077640549, -0.08336233266, -0.1446729567, -0.2116753732, -0.2830637392, -0.3581341341, -0.436462435, -0.5177783846, - -0.6019067271, -0.6887363571, -0.7782028952, -0.8702784034, -0.9649652536, -1.062292565, -1.16231451, - -1.265109911, -1.370782966, -1.479465113, -1.591317992, -1.70653767]) - -!ec - -!split -===== Formatting the Data ===== - -The way the recurrent neural networks are trained in this program -differs from how machine learning algorithms are usually trained. -Typically a machine learning algorithm is trained by learning the -relationship between the x data and the y data. In this program, the -recurrent neural network will be trained to recognize the relationship -in a sequence of y values. This is type of data formatting is -typically used time series forcasting, but it can also be used in any -extrapolation (time series forecasting is just a specific type of -extrapolation along the time axis). This method of data formatting -does not use the x data and assumes that the y data are evenly spaced. - -For a standard machine learning algorithm, the training data has the -form of (x,y) so the machine learning algorithm learns to assiciate a -y value with a given x value. This is useful when the test data has x -values within the same range as the training data. However, for this -application, the x values of the test data are outside of the x values -of the training data and the traditional method of training a machine -learning algorithm does not work as well. For this reason, the -recurrent neural network is trained on sequences of y values of the -form ((y1, y2), y3), so that the network is concerned with learning -the pattern of the y data and not the relation between the x and y -data. As long as the pattern of y data outside of the training region -stays relatively stable compared to what was inside the training -region, this method of training can produce accurate extrapolations to -y values far removed from the training data set. - - -# -# The idea behind formatting the data in this way comes from [this resource](https://machinelearningmastery.com/time-series-prediction-lstm-recurrent-neural-networks-python-keras/) and [this one](https://fairyonice.github.io/Understand-Keras%27s-RNN-behind-the-scenes-with-a-sin-wave-example.html). -# -# The following method takes in a y data set and formats it so the "x data" are of the form (y1, y2) and the "y data" are of the form y3, with extra brackets added in to make the resulting arrays compatable with both Keras and Tensorflow. -# -# Note: Using a sequence length of two is not required for time series forecasting so any lenght of sequence could be used (for example instead of ((y1, y2) y3) you could change the length of sequence to be 4 and the resulting data points would have the form ((y1, y2, y3, y4), y5)). While the following method can be used to create a data set of any sequence length, the remainder of the code expects the length of sequence to be 2. This is because the data sets are very small and the higher the lenght of the sequence the less resulting data points. - -!bc pycod -# FORMAT_DATA -def format_data(data, length_of_sequence = 2): - """ - Inputs: - data(a numpy array): the data that will be the inputs to the recurrent neural - network - length_of_sequence (an int): the number of elements in one iteration of the - sequence patter. For a function approximator use length_of_sequence = 2. - Returns: - rnn_input (a 3D numpy array): the input data for the recurrent neural network. Its - dimensions are length of data - length of sequence, length of sequence, - dimnsion of data - rnn_output (a numpy array): the training data for the neural network - Formats data to be used in a recurrent neural network. - """ - - X, Y = [], [] - for i in range(len(data)-length_of_sequence): - # Get the next length_of_sequence elements - a = data[i:i+length_of_sequence] - # Get the element that immediately follows that - b = data[i+length_of_sequence] - # Reshape so that each data point is contained in its own array - a = np.reshape (a, (len(a), 1)) - X.append(a) - Y.append(b) - rnn_input = np.array(X) - rnn_output = np.array(Y) - - return rnn_input, rnn_output - - -# ## Defining the Recurrent Neural Network Using Keras -# -# The following method defines a simple recurrent neural network in keras consisting of one input layer, one hidden layer, and one output layer. - -def rnn(length_of_sequences, batch_size = None, stateful = False): - """ - Inputs: - length_of_sequences (an int): the number of y values in "x data". This is determined - when the data is formatted - batch_size (an int): Default value is None. See Keras documentation of SimpleRNN. - stateful (a boolean): Default value is False. See Keras documentation of SimpleRNN. - Returns: - model (a Keras model): The recurrent neural network that is built and compiled by this - method - Builds and compiles a recurrent neural network with one hidden layer and returns the model. - """ - # Number of neurons in the input and output layers - in_out_neurons = 1 - # Number of neurons in the hidden layer - hidden_neurons = 200 - # Define the input layer - inp = Input(batch_shape=(batch_size, - length_of_sequences, - in_out_neurons)) - # Define the hidden layer as a simple RNN layer with a set number of neurons and add it to - # the network immediately after the input layer - rnn = SimpleRNN(hidden_neurons, - return_sequences=False, - stateful = stateful, - name="RNN")(inp) - # Define the output layer as a dense neural network layer (standard neural network layer) - #and add it to the network immediately after the hidden layer. - dens = Dense(in_out_neurons,name="dense")(rnn) - # Create the machine learning model starting with the input layer and ending with the - # output layer - model = Model(inputs=[inp],outputs=[dens]) - # Compile the machine learning model using the mean squared error function as the loss - # function and an Adams optimizer. - model.compile(loss="mean_squared_error", optimizer="adam") - return model - -!ec - -!split -===== Predicting New Points With A Trained Recurrent Neural Network ===== - -!bc pycod -def test_rnn (x1, y_test, plot_min, plot_max): - """ - Inputs: - x1 (a list or numpy array): The complete x component of the data set - y_test (a list or numpy array): The complete y component of the data set - plot_min (an int or float): the smallest x value used in the training data - plot_max (an int or float): the largest x valye used in the training data - Returns: - None. - Uses a trained recurrent neural network model to predict future points in the - series. Computes the MSE of the predicted data set from the true data set, saves - the predicted data set to a csv file, and plots the predicted and true data sets w - while also displaying the data range used for training. - """ - # Add the training data as the first dim points in the predicted data array as these - # are known values. - y_pred = y_test[:dim].tolist() - # Generate the first input to the trained recurrent neural network using the last two - # points of the training data. Based on how the network was trained this means that it - # will predict the first point in the data set after the training data. All of the - # brackets are necessary for Tensorflow. - next_input = np.array([[[y_test[dim-2]], [y_test[dim-1]]]]) - # Save the very last point in the training data set. This will be used later. - last = [y_test[dim-1]] - - # Iterate until the complete data set is created. - for i in range (dim, len(y_test)): - # Predict the next point in the data set using the previous two points. - next = model.predict(next_input) - # Append just the number of the predicted data set - y_pred.append(next[0][0]) - # Create the input that will be used to predict the next data point in the data set. - next_input = np.array([[last, next[0]]], dtype=np.float64) - last = next - - # Print the mean squared error between the known data set and the predicted data set. - print('MSE: ', np.square(np.subtract(y_test, y_pred)).mean()) - # Save the predicted data set as a csv file for later use - name = datatype + 'Predicted'+str(dim)+'.csv' - np.savetxt(name, y_pred, delimiter=',') - # Plot the known data set and the predicted data set. The red box represents the region that was used - # for the training data. - fig, ax = plt.subplots() - ax.plot(x1, y_test, label="true", linewidth=3) - ax.plot(x1, y_pred, 'g-.',label="predicted", linewidth=4) - ax.legend() - # Created a red region to represent the points used in the training data. - ax.axvspan(plot_min, plot_max, alpha=0.25, color='red') - plt.show() - -# Check to make sure the data set is complete -assert len(X_tot) == len(y_tot) - -# This is the number of points that will be used in as the training data -dim=12 - -# Separate the training data from the whole data set -X_train = X_tot[:dim] -y_train = y_tot[:dim] - - -# Generate the training data for the RNN, using a sequence of 2 -rnn_input, rnn_training = format_data(y_train, 2) - - -# Create a recurrent neural network in Keras and produce a summary of the -# machine learning model -model = rnn(length_of_sequences = rnn_input.shape[1]) -model.summary() - -# Start the timer. Want to time training+testing -start = timer() -# Fit the model using the training data genenerated above using 150 training iterations and a 5% -# validation split. Setting verbose to True prints information about each training iteration. -hist = model.fit(rnn_input, rnn_training, batch_size=None, epochs=150, - verbose=True,validation_split=0.05) - -for label in ["loss","val_loss"]: - plt.plot(hist.history[label],label=label) - -plt.ylabel("loss") -plt.xlabel("epoch") -plt.title("The final validation loss: {}".format(hist.history["val_loss"][-1])) -plt.legend() -plt.show() - -# Use the trained neural network to predict more points of the data set -test_rnn(X_tot, y_tot, X_tot[0], X_tot[dim-1]) -# Stop the timer and calculate the total time needed. -end = timer() -print('Time: ', end-start) -!ec - -!split -===== Other Things to Try ===== - - -Changing the size of the recurrent neural network and its parameters -can drastically change the results you get from the model. The below -code takes the simple recurrent neural network from above and adds a -second hidden layer, changes the number of neurons in the hidden -layer, and explicitly declares the activation function of the hidden -layers to be a sigmoid function. The loss function and optimizer can -also be changed but are kept the same as the above network. These -parameters can be tuned to provide the optimal result from the -network. For some ideas on how to improve the performance of a -"recurrent neural network":"https://danijar.com/tips-for-training-recurrent-neural-networks". - -!bc pycod -def rnn_2layers(length_of_sequences, batch_size = None, stateful = False): - """ - Inputs: - length_of_sequences (an int): the number of y values in "x data". This is determined - when the data is formatted - batch_size (an int): Default value is None. See Keras documentation of SimpleRNN. - stateful (a boolean): Default value is False. See Keras documentation of SimpleRNN. - Returns: - model (a Keras model): The recurrent neural network that is built and compiled by this - method - Builds and compiles a recurrent neural network with two hidden layers and returns the model. - """ - # Number of neurons in the input and output layers - in_out_neurons = 1 - # Number of neurons in the hidden layer, increased from the first network - hidden_neurons = 500 - # Define the input layer - inp = Input(batch_shape=(batch_size, - length_of_sequences, - in_out_neurons)) - # Create two hidden layers instead of one hidden layer. Explicitly set the activation - # function to be the sigmoid function (the default value is hyperbolic tangent) - rnn1 = SimpleRNN(hidden_neurons, - return_sequences=True, # This needs to be True if another hidden layer is to follow - stateful = stateful, activation = 'sigmoid', - name="RNN1")(inp) - rnn2 = SimpleRNN(hidden_neurons, - return_sequences=False, activation = 'sigmoid', - stateful = stateful, - name="RNN2")(rnn1) - # Define the output layer as a dense neural network layer (standard neural network layer) - #and add it to the network immediately after the hidden layer. - dens = Dense(in_out_neurons,name="dense")(rnn2) - # Create the machine learning model starting with the input layer and ending with the - # output layer - model = Model(inputs=[inp],outputs=[dens]) - # Compile the machine learning model using the mean squared error function as the loss - # function and an Adams optimizer. - model.compile(loss="mean_squared_error", optimizer="adam") - return model - -# Check to make sure the data set is complete -assert len(X_tot) == len(y_tot) - -# This is the number of points that will be used in as the training data -dim=12 - -# Separate the training data from the whole data set -X_train = X_tot[:dim] -y_train = y_tot[:dim] - - -# Generate the training data for the RNN, using a sequence of 2 -rnn_input, rnn_training = format_data(y_train, 2) - - -# Create a recurrent neural network in Keras and produce a summary of the -# machine learning model -model = rnn_2layers(length_of_sequences = 2) -model.summary() - -# Start the timer. Want to time training+testing -start = timer() -# Fit the model using the training data genenerated above using 150 training iterations and a 5% -# validation split. Setting verbose to True prints information about each training iteration. -hist = model.fit(rnn_input, rnn_training, batch_size=None, epochs=150, - verbose=True,validation_split=0.05) - - -# This section plots the training loss and the validation loss as a function of training iteration. -# This is not required for analyzing the couple cluster data but can help determine if the network is -# being overtrained. -for label in ["loss","val_loss"]: - plt.plot(hist.history[label],label=label) - -plt.ylabel("loss") -plt.xlabel("epoch") -plt.title("The final validation loss: {}".format(hist.history["val_loss"][-1])) -plt.legend() -plt.show() - -# Use the trained neural network to predict more points of the data set -test_rnn(X_tot, y_tot, X_tot[0], X_tot[dim-1]) -# Stop the timer and calculate the total time needed. -end = timer() -print('Time: ', end-start) -!ec - -!split -===== Other Types of Recurrent Neural Networks ===== - -Besides a simple recurrent neural network layer, there are two other -commonly used types of recurrent neural network layers: Long Short -Term Memory (LSTM) and Gated Recurrent Unit (GRU). For a short -introduction to these layers see URL:"https://medium.com/mindboard/lstm-vs-gru-experimental-comparison-955820c21e8b" -and URL:"https://medium.com/mindboard/lstm-vs-gru-experimental-comparison-955820c21e8b". - -The first network created below is similar to the previous network, -but it replaces the SimpleRNN layers with LSTM layers. The second -network below has two hidden layers made up of GRUs, which are -preceeded by two dense (feeddorward) neural network layers. These -dense layers "preprocess" the data before it reaches the recurrent -layers. This architecture has been shown to improve the performance -of recurrent neural networks (see the link above and also -URL:"https://arxiv.org/pdf/1807.02857.pdf". - -!bc pycod -def lstm_2layers(length_of_sequences, batch_size = None, stateful = False): - """ - Inputs: - length_of_sequences (an int): the number of y values in "x data". This is determined - when the data is formatted - batch_size (an int): Default value is None. See Keras documentation of SimpleRNN. - stateful (a boolean): Default value is False. See Keras documentation of SimpleRNN. - Returns: - model (a Keras model): The recurrent neural network that is built and compiled by this - method - Builds and compiles a recurrent neural network with two LSTM hidden layers and returns the model. - """ - # Number of neurons on the input/output layer and the number of neurons in the hidden layer - in_out_neurons = 1 - hidden_neurons = 250 - # Input Layer - inp = Input(batch_shape=(batch_size, - length_of_sequences, - in_out_neurons)) - # Hidden layers (in this case they are LSTM layers instead if SimpleRNN layers) - rnn= LSTM(hidden_neurons, - return_sequences=True, - stateful = stateful, - name="RNN", use_bias=True, activation='tanh')(inp) - rnn1 = LSTM(hidden_neurons, - return_sequences=False, - stateful = stateful, - name="RNN1", use_bias=True, activation='tanh')(rnn) - # Output layer - dens = Dense(in_out_neurons,name="dense")(rnn1) - # Define the midel - model = Model(inputs=[inp],outputs=[dens]) - # Compile the model - model.compile(loss='mean_squared_error', optimizer='adam') - # Return the model - return model - -def dnn2_gru2(length_of_sequences, batch_size = None, stateful = False): - """ - Inputs: - length_of_sequences (an int): the number of y values in "x data". This is determined - when the data is formatted - batch_size (an int): Default value is None. See Keras documentation of SimpleRNN. - stateful (a boolean): Default value is False. See Keras documentation of SimpleRNN. - Returns: - model (a Keras model): The recurrent neural network that is built and compiled by this - method - Builds and compiles a recurrent neural network with four hidden layers (two dense followed by - two GRU layers) and returns the model. - """ - # Number of neurons on the input/output layers and hidden layers - in_out_neurons = 1 - hidden_neurons = 250 - # Input layer - inp = Input(batch_shape=(batch_size, - length_of_sequences, - in_out_neurons)) - # Hidden Dense (feedforward) layers - dnn = Dense(hidden_neurons/2, activation='relu', name='dnn')(inp) - dnn1 = Dense(hidden_neurons/2, activation='relu', name='dnn1')(dnn) - # Hidden GRU layers - rnn1 = GRU(hidden_neurons, - return_sequences=True, - stateful = stateful, - name="RNN1", use_bias=True)(dnn1) - rnn = GRU(hidden_neurons, - return_sequences=False, - stateful = stateful, - name="RNN", use_bias=True)(rnn1) - # Output layer - dens = Dense(in_out_neurons,name="dense")(rnn) - # Define the model - model = Model(inputs=[inp],outputs=[dens]) - # Compile the mdoel - model.compile(loss='mean_squared_error', optimizer='adam') - # Return the model - return model - -# Check to make sure the data set is complete -assert len(X_tot) == len(y_tot) - -# This is the number of points that will be used in as the training data -dim=12 - -# Separate the training data from the whole data set -X_train = X_tot[:dim] -y_train = y_tot[:dim] - - -# Generate the training data for the RNN, using a sequence of 2 -rnn_input, rnn_training = format_data(y_train, 2) - - -# Create a recurrent neural network in Keras and produce a summary of the -# machine learning model -# Change the method name to reflect which network you want to use -model = dnn2_gru2(length_of_sequences = 2) -model.summary() - -# Start the timer. Want to time training+testing -start = timer() -# Fit the model using the training data genenerated above using 150 training iterations and a 5% -# validation split. Setting verbose to True prints information about each training iteration. -hist = model.fit(rnn_input, rnn_training, batch_size=None, epochs=150, - verbose=True,validation_split=0.05) - - -# This section plots the training loss and the validation loss as a function of training iteration. -# This is not required for analyzing the couple cluster data but can help determine if the network is -# being overtrained. -for label in ["loss","val_loss"]: - plt.plot(hist.history[label],label=label) - -plt.ylabel("loss") -plt.xlabel("epoch") -plt.title("The final validation loss: {}".format(hist.history["val_loss"][-1])) -plt.legend() -plt.show() - -# Use the trained neural network to predict more points of the data set -test_rnn(X_tot, y_tot, X_tot[0], X_tot[dim-1]) -# Stop the timer and calculate the total time needed. -end = timer() -print('Time: ', end-start) - - -# ### Training Recurrent Neural Networks in the Standard Way (i.e. learning the relationship between the X and Y data) -# -# Finally, comparing the performace of a recurrent neural network using the standard data formatting to the performance of the network with time sequence data formatting shows the benefit of this type of data formatting with extrapolation. - -# Check to make sure the data set is complete -assert len(X_tot) == len(y_tot) - -# This is the number of points that will be used in as the training data -dim=12 - -# Separate the training data from the whole data set -X_train = X_tot[:dim] -y_train = y_tot[:dim] - -# Reshape the data for Keras specifications -X_train = X_train.reshape((dim, 1)) -y_train = y_train.reshape((dim, 1)) - - -# Create a recurrent neural network in Keras and produce a summary of the -# machine learning model -# Set the sequence length to 1 for regular data formatting -model = rnn(length_of_sequences = 1) -model.summary() - -# Start the timer. Want to time training+testing -start = timer() -# Fit the model using the training data genenerated above using 150 training iterations and a 5% -# validation split. Setting verbose to True prints information about each training iteration. -hist = model.fit(X_train, y_train, batch_size=None, epochs=150, - verbose=True,validation_split=0.05) - - -# This section plots the training loss and the validation loss as a function of training iteration. -# This is not required for analyzing the couple cluster data but can help determine if the network is -# being overtrained. -for label in ["loss","val_loss"]: - plt.plot(hist.history[label],label=label) - -plt.ylabel("loss") -plt.xlabel("epoch") -plt.title("The final validation loss: {}".format(hist.history["val_loss"][-1])) -plt.legend() -plt.show() - -# Use the trained neural network to predict the remaining data points -X_pred = X_tot[dim:] -X_pred = X_pred.reshape((len(X_pred), 1)) -y_model = model.predict(X_pred) -y_pred = np.concatenate((y_tot[:dim], y_model.flatten())) - -# Plot the known data set and the predicted data set. The red box represents the region that was used -# for the training data. -fig, ax = plt.subplots() -ax.plot(X_tot, y_tot, label="true", linewidth=3) -ax.plot(X_tot, y_pred, 'g-.',label="predicted", linewidth=4) -ax.legend() -# Created a red region to represent the points used in the training data. -ax.axvspan(X_tot[0], X_tot[dim], alpha=0.25, color='red') -plt.show() - -# Stop the timer and calculate the total time needed. -end = timer() -print('Time: ', end-start) - -!ec - - - - - -======= Solving ODEs with Deep Learning ======= - -The Universal Approximation Theorem states that a neural network can -approximate any function at a single hidden layer along with one input -and output layer to any given precision. - - - -===== Ordinary Differential Equations ===== - -An ordinary differential equation (ODE) is an equation involving functions having one variable. - -In general, an ordinary differential equation looks like - -!bt -\begin{equation} \label{ode} -f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 -\end{equation} -!et - -where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$. - -The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in (ref{ode}). -The highest order of derivative, that is the value of $n$, determines to the order of the equation. -The equation is referred to as a $n$-th order ODE. -Along with (ref{ode}), some additional conditions of the function $g(x)$ are typically given -for the solution to be unique. - - -===== The trial solution ===== - -Let the trial solution $g_t(x)$ be - -!bt -\begin{equation} - g_t(x) = h_1(x) + h_2(x,N(x,P)) -\end{equation} -!et - - -where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set -of conditions, $N(x,P)$ a neural network with weights and biases -described by $P$ and $h_2(x, N(x,P))$ some expression involving the -neural network. The role of the function $h_2(x, N(x,P))$, is to -ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is -evaluated at the values of $x$ where the given conditions must be -satisfied. The function $h_1(x)$ should alone make $g_t(x)$ satisfy -the conditions. - -But what about the network $N(x,P)$? - - -As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation. - - - -===== Minimization process ===== - -For the minimization to be defined, we need to have a cost function at hand to minimize. - -It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in (ref{ode}). -We can choose to consider the mean squared error as the cost function for an input $x$. -Since we are looking at one input, the cost function is just $f$ squared. -The cost function $c\left(x, P \right)$ can therefore be expressed as - -!bt -C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 -!et - -If $N$ inputs are given as a vector $\bm{x}$ with elements $x_i$ for $i = 1,\dots,N$, -the cost function becomes - -!bt -\begin{equation} \label{cost} - C\left(\bm{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 -\end{equation} -!et - -The neural net should then find the parameters $P$ that minimizes the cost function in -(ref{cost}) for a set of $N$ training samples $x_i$. - - -===== Minimizing the cost function using gradient descent and automatic differentiation ===== - -To perform the minimization using gradient descent, the gradient of $C\left(\bm{x}, P\right)$ is needed. -It might happen so that finding an analytical expression of the gradient of $C(\bm{x}, P)$ from (ref{cost}) gets too messy, depending on which cost function one desires to use. - -Luckily, there exists libraries that makes the job for us through automatic differentiation. -Automatic differentiation is a method of finding the derivatives numerically with very high precision. - - - -===== Example: Exponential decay ===== - -An exponential decay of a quantity $g(x)$ is described by the equation - -!bt -\begin{equation} \label{solve_expdec} - g'(x) = -\gamma g(x) -\end{equation} -!et - -with $g(0) = g_0$ for some chosen initial value $g_0$. - -The analytical solution of (ref{solve_expdec}) is - -!bt -\begin{equation} - g(x) = g_0 \exp\left(-\gamma x\right) -\end{equation} -!et - -Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (ref{solve_expdec}). - - - -===== The function to solve for ===== - -The program will use a neural network to solve - -!bt -\begin{equation} \label{solveode} -g'(x) = -\gamma g(x) -\end{equation} -!et - -where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values. - -In this example, $\gamma = 2$ and $g_0 = 10$. - - -===== The trial solution ===== -To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be - -!bt -g_t(x, P) = h_1(x) + h_2(x, N(x, P)) -!et - -with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer. - - -===== Setup of Network ===== - -In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}}, P_{\text{output}} \}$. -If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer. - -The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. -If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix. - -Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron. - -It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of (ref{solveode}). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution: - -!bt -\begin{equation} \label{trial} -g_t(x, P) = g_0 + x \cdot N(x, P) -\end{equation} -!et - - -===== Reformulating the problem ===== - -We wish that our neural network manages to minimize a given cost function. - -A reformulation of out equation, (ref{solveode}), must therefore be done, -such that it describes the problem a neural network can solve for. - -The neural network must find the set of weights and biases $P$ such that the trial solution in (ref{trial}) satisfies (ref{solveode}). - -The trial solution - -!bt -g_t(x, P) = g_0 + x \cdot N(x, P) -!et - -has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that - -!bt -\begin{equation} \label{nnmin} -g_t'(x, P) = - \gamma g_t(x, P) -\end{equation} -!et - -is fulfilled as *best as possible*. - - -===== More technicalities ===== - -The left hand side and right hand side of (ref{nnmin}) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible. -This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. -In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network. - -This gives the following cost function our neural network must solve for: - -!bt -\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} -!et - -(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$) - -or, in terms of weights and biases for the hidden and output layer in our network: - -!bt -\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} -!et - -for an input value $x$. - - -===== More details ===== - -If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes - -!bt -\begin{equation} \label{min} -\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} -\end{equation} -!et - -Letting $\bm{x}$ be a vector with elements $x_i$ and $C(\bm{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes - -!bt -\min_{P} C(\bm{x}, P) -!et - -In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as - -$$ -\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\bm{x}, \{P_{\text{hidden} }, P_{\text{output} }\}) -$$ - - -===== A possible implementation of a neural network ===== - -For simplicity, it is assumed that the input is an array $\bm{x} = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills (ref{min}). - -First, the neural network must feed forward the inputs. -This means that $\bm{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. -The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\text{hidden} }$. - - -===== Technicalities ===== - -For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is: - -!bt -\begin{aligned} -z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ -&= -\begin{pmatrix} -b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -x_j -\end{pmatrix} -\end{aligned} -!et - - -===== Final technicalities I ===== - -The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector: - -!bt -\begin{aligned} -\bm{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ -&= -\begin{pmatrix} - b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -x_1 & x_2 & \dots & x_N -\end{pmatrix} \\ -&= \bm{p}_{i, \text{hidden}}^T X -\end{aligned} -!et - - -===== Final technicalities II ===== - -The vector $\bm{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to (ref{min}). - -After having found $\bm{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\bm{z})$. - -In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: - -!bt -f(z) = \frac{1}{1 + \exp{(-z)}} -!et - -It is possible to use other activations functions for the hidden layer also. - -The output $\bm{x}_i^{\text{hidden}}$ from each $i$-th hidden neuron is: - -$$ -\bm{x}_i^{\text{hidden} } = f\big( \bm{z}_{i}^{\text{hidden}} \big) -$$ - -The outputs $\bm{x}_i^{\text{hidden} } $ are then sent to the output layer. - -The output layer consists of one neuron in this case, and combines the -output from each of the neurons in the hidden layers. The output layer -combines the results from the hidden layer using some weights $w_i^{\text{output}}$ -and biases $b_i^{\text{output}}$. In this case, -it is assumes that the number of neurons in the output layer is one. - - -===== Final technicalities III ===== - - -The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously. - -!bt -\begin{aligned} -z_{1,j}^{\text{output}} & = -\begin{pmatrix} -b_1^{\text{output}} & \bm{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -\bm{x}_j^{\text{hidden}} -\end{pmatrix} -\end{aligned} -!et - - -===== Final technicalities IV ===== - -Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer: - -!bt -\bm{z}_{1}^{\text{output}} = -\begin{pmatrix} -b_1^{\text{output}} & \bm{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -\bm{x}_1^{\text{hidden}} & \bm{x}_2^{\text{hidden}} & \dots & \bm{x}_N^{\text{hidden}} -\end{pmatrix} -!et - -In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\bm{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\bm{z}_{1}^{\text{output}}$ is the final output of the network. - - -===== Back propagation ===== - -The next step is to decide how the parameters should be changed such that they minimize the cost function. - -The chosen cost function for this problem is - -!bt -C(\bm{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 -!et - -In order to minimize the cost function, an optimization method must be chosen. - -Here, gradient descent with a constant step size has been chosen. - - -===== Gradient descent ===== - -The idea of the gradient descent algorithm is to update parameters in -a direction where the cost function decreases goes to a minimum. - -In general, the update of some parameters $\bm{\omega}$ given a cost -function defined by some weights $\bm{\omega}$, $C(\bm{x}, -\bm{\omega})$, goes as follows: - -!bt -\bm{\omega}_{\text{new} } = \bm{\omega} - \lambda \nabla_{\bm{\omega}} C(\bm{x}, \bm{\omega}) -!et - -for a number of iterations or until $ \big|\big| \bm{\omega}_{\text{new} } - \bm{\omega} \big|\big|$ becomes smaller than some given tolerance. - -The value of $\lambda$ decides how large steps the algorithm must take -in the direction of $ \nabla_{\bm{\omega}} C(\bm{x}, \bm{\omega})$. -The notation $\nabla_{\bm{\omega}}$ express the gradient with respect -to the elements in $\bm{\omega}$. - -In our case, we have to minimize the cost function $C(\bm{x}, P)$ with -respect to the two sets of weights and biases, that is for the hidden -layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output} -}$ . - -This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by - -!bt -\begin{aligned} -P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\bm{x}, P) \\ -P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\bm{x}, P) -\end{aligned} -!et - - -===== The code for solving the ODE ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Assuming one input, hidden, and output layer -def neural_network(params, x): - - # Find the weights (including and biases) for the hidden and output layer. - # Assume that params is a list of parameters for each layer. - # The biases are the first element for each array in params, - # and the weights are the remaning elements in each array in params. - - w_hidden = params[0] - w_output = params[1] - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - ## Hidden layer: - - # Add a row of ones to include bias - x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_input) - x_hidden = sigmoid(z_hidden) - - ## Output layer: - - # Include bias: - x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) - - z_output = np.matmul(w_output, x_hidden) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial(x,params, g0 = 10): - return g0 + x*neural_network(params,x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The cost function: -def cost_function(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input, hidden, and output layer -def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): - ## Set up initial weights and biases - - # For the hidden layer - p0 = npr.randn(num_neurons_hidden, 2 ) - - # For the output layer - p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included - - P = [p0, p1] - - print('Initial cost: %g'%cost_function(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of two arrays; - # one for the gradient w.r.t P_hidden and - # one for the gradient w.r.t P_output - cost_grad = cost_function_grad(P, x) - - P[0] = P[0] - lmb * cost_grad[0] - P[1] = P[1] - lmb * cost_grad[1] - - print('Final cost: %g'%cost_function(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - # Set seed such that the weight are initialized - # with same weights and biases for every run. - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = 10 - num_iter = 10000 - lmb = 0.001 - - # Use the network - P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) - - # Print the deviation from the trial solution and true solution - res = g_trial(x,P) - res_analytical = g_analytic(x) - - print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - - - -===== The network with one input layer, specified number of hidden layers, and one output layer ===== - -It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers. - -The number of neurons within each hidden layer are given as a list of integers in the program below. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# The neural network with one input layer and one output layer, -# but with number of hidden layers specified by the user. -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - - N_hidden = np.size(deep_params) - 1 # -1 since params consists of - # parameters to all the hidden - # layers AND the output layer. - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial_deep(x,params, g0 = 10): - return g0 + x*deep_neural_network(params, x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The same cost function as before, but calls deep_neural_network instead. -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(deep_neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input and one output layer, -# but with specified number of hidden layers from the user. -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # The number of elements in the list num_hidden_neurons thus represents - # the number of hidden layers. - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weights and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = np.array([10,10]) - num_iter = 10000 - lmb = 0.001 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - res = g_trial_deep(x,P) - res_analytical = g_analytic(x) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','dnn']) - plt.ylabel('g(x)') - plt.show() -!ec - - - -===== Example: Population growth ===== - -A logistic model of population growth assumes that a population converges toward an equilibrium. -The population growth can be modeled by - -!bt -\begin{equation} \label{log} - g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment. -Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant. - -In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability -and high execution time (this might be more apparent in the examples solving PDEs), -using a library like TensorFlow is recommended. -Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method. - - -===== Setting up the problem ===== - -Here, we will model a population $g(t)$ in an environment having carrying capacity $A$. -The population follows the model - -!bt -\begin{equation} \label{solveode_population} -g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(0) = g_0$. - -In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$. - - -===== The trial solution ===== - -We will get a slightly different trial solution, as the boundary conditions are different -compared to the case for exponential decay. - -A possible trial solution satisfying the condition $g(0) = g_0$ could be - -$$ -h_1(t) = g_0 + t \cdot N(t,P) -$$ - -with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$. - -The analytical solution is - -$$ -g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} -$$ - - -===== The program using Autograd ===== - -The network will be the similar as for the exponential decay example, but with some small modifications for our problem. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Function to get the parameters. -# Done such that one can easily change the paramaters after one's liking. -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def deep_neural_network(P, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = P[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = P[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = f(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# The right side of the ODE: -def f(x, g_trial): - alpha,A, g0 = get_parameters() - return alpha*g_trial*(A - g_trial) - -# The trial solution using the deep neural network: -def g_trial_deep(x, params): - alpha,A, g0 = get_parameters() - return g0 + x*deep_neural_network(params,x) - -# The analytical solution: -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100, 50, 25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - plt.show() -!ec - - -===== Using forward Euler to solve the ODE ===== - -A straightforward way of solving an ODE numerically, is to use Euler's method. - -Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$: - -$$ -f(x + \Delta x) \approx f(x) + \Delta x f'(x) -$$ - -In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields - -!bt -\begin{aligned} - g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ - &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) -\end{aligned} -!et -along with the condition that $g(0) = g_0$. - -Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$. - -For $i \geq 1$, we have that -!bt -\begin{aligned} -t_i &= i\Delta t \\ -&= (i - 1)\Delta t + \Delta t \\ -&= t_{i-1} + \Delta t -\end{aligned} -!et - -Now, if $g_i = g(t_i)$ then - -!bt -\begin{equation} - \begin{aligned} - g_i &= g(t_i) \\ - &= g(t_{i-1} + \Delta t) \\ - &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ - &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) - \end{aligned} -\end{equation} \label{odenum} -!et -for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$. - -Equation (ref{odenum}) could be implemented in the following way, -extending the program that uses the network using Autograd: - -!bc pycod -# Assume that all function definitions from the example program using Autograd -# are located here. - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100,50,25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - ## Find an approximation to the funtion using forward Euler - - alpha, A, g0 = get_parameters() - dt = T/(Nt - 1) - - # Perform forward Euler to solve the ODE - g_euler = np.zeros(Nt) - g_euler[0] = g0 - - for i in range(1,Nt): - g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1])) - - # Print the errors done by each method - diff1 = np.max(np.abs(g_euler - g_analytical)) - diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical)) - - print('Max absolute difference between Euler method and analytical: %g'%diff1) - print('Max absolute difference between deep neural network and analytical: %g'%diff2) - - # Plot results - plt.figure(figsize=(10,10)) - - plt.plot(t,g_euler) - plt.plot(t,g_analytical) - plt.plot(t,g_dnn_ag[0,:]) - - plt.legend(['euler','analytical','dnn']) - plt.xlabel('Time t') - plt.ylabel('g(t)') - - plt.show() -!ec - - - - -===== Example: Solving the one dimensional Poisson equation ===== - -The Poisson equation for $g(x)$ in one dimension is - -!bt -\begin{equation} \label{poisson} - -g''(x) = f(x) -\end{equation} -!et - -where $f(x)$ is a given function for $x \in (0,1)$. - -The conditions that $g(x)$ is chosen to fulfill, are -!bt -\begin{align*} - g(0) &= 0 \\ - g(1) &= 0 -\end{align*} -!et - -This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. -The results from the networks can then be compared to the analytical solution. -In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks. - - -===== The specific equation to solve for ===== - -Here, the function $g(x)$ to solve for follows the equation - -!bt --g''(x) = f(x),\qquad x \in (0,1) -!et - -where $f(x)$ is a given function, along with the chosen conditions - -!bt -\begin{aligned} -g(0) = g(1) = 0 -\end{aligned}\label{cond} -!et - -In this example, we consider the case when $f(x) = (3x + x^2)\exp(x)$. - -For this case, a possible trial solution satisfying the conditions could be - -!bt -g_t(x) = x \cdot (1-x) \cdot N(P,x) -!et - -The analytical solution for this problem is - -!bt -g(x) = x(1 - x)\exp(x) -!et - - -===== Solving the equation using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - max_diff = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%max_diff) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - - -===== Comparing with a numerical scheme ===== - -The Poisson equation is possible to solve using Taylor series to approximate the second derivative. - -Using Taylor series, the second derivative can be expressed as - -$$ -g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) -$$ - -where $\Delta x$ is a small step size and $E_{\Delta x}(x)$ being the error term. - -Looking away from the error terms gives an approximation to the second derivative: - -!bt -\begin{equation} \label{approx} -g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} -\end{equation} -!et - -If $x_i = i \Delta x = x_{i-1} + \Delta x$ and $g_i = g(x_i)$ for $i = 1,\dots N_x - 2$ with $N_x$ being the number of values for $x$, (ref{approx}) becomes - -!bt -\begin{aligned} -g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ -&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} -\end{aligned} -!et - -Since we know from our problem that - -!bt -\begin{aligned} --g''(x) &= f(x) \\ -&= (3x + x^2)\exp(x) -\end{aligned} -!et - -along with the conditions $g(0) = g(1) = 0$, -the following scheme can be used to find an approximate solution for $g(x)$ numerically: - -!bt -\begin{equation} - \begin{aligned} - -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ - -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) - \end{aligned} -\end{equation} \label{odesys} -!et - -for $i = 1, \dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\exp(x_i)$, which is given for our specific problem. - -The equation can be rewritten into a matrix equation: - -!bt -\begin{aligned} -\begin{pmatrix} -2 & -1 & 0 & \dots & 0 \\ --1 & 2 & -1 & \dots & 0 \\ -\vdots & & \ddots & & \vdots \\ -0 & \dots & -1 & 2 & -1 \\ -0 & \dots & 0 & -1 & 2\\ -\end{pmatrix} -\begin{pmatrix} -g_1 \\ -g_2 \\ -\vdots \\ -g_{N_x - 3} \\ -g_{N_x - 2} -\end{pmatrix} -&= -\Delta x^2 -\begin{pmatrix} -f(x_1) \\ -f(x_2) \\ -\vdots \\ -f(x_{N_x - 3}) \\ -f(x_{N_x - 2}) -\end{pmatrix} \\ -\bm{A}\bm{g} &= \bm{f}, -\end{aligned} -!et - -which makes it possible to solve for the vector $\bm{g}$. - - -===== Setting up the code ===== - -We can then compare the result from this numerical scheme with the output from our network using Autograd: - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - - ## Perform the computation using the numerical scheme - - dx = 1/(Nx - 1) - - # Set up the matrix A - A = np.zeros((Nx-2,Nx-2)) - - A[0,0] = 2 - A[0,1] = -1 - - for i in range(1,Nx-3): - A[i,i-1] = -1 - A[i,i] = 2 - A[i,i+1] = -1 - - A[Nx - 3, Nx - 4] = -1 - A[Nx - 3, Nx - 3] = 2 - - # Set up the vector f - f_vec = dx**2 * f(x[1:-1]) - - # Solve the equation - g_res = np.linalg.solve(A,f_vec) - - g_vec = np.zeros(Nx) - g_vec[1:-1] = g_res - - # Print the differences between each method - max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical)) - max_diff2 = np.max(np.abs(g_vec - g_analytical)) - print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1) - print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.plot(x,g_vec) - plt.plot(x,g_analytical) - plt.plot(x,g_dnn_ag[0,:]) - - plt.legend(['numerical scheme','analytical','dnn']) - plt.show() - -!ec - - - - -===== Partial Differential Equations ===== - -A partial differential equation (PDE) has a solution here the function -is defined by multiple variables. The equation may involve all kinds -of combinations of which variables the function is differentiated with -respect to. - -In general, a partial differential equation for a function $g(x_1,\dots,x_N)$ with $N$ variables may be expressed as - -!bt -\begin{equation} \label{PDE} - f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 -\end{equation} -!et - -where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given. - - -===== Type of problem ===== - -The problem our network must solve for, is similar to the ODE case. -We must have a trial solution $g_t$ at hand. - -For instance, the trial solution could be expressed as -!bt -\begin{align*} - g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) -\end{align*} -!et -where $h_1(x_1,\dots,x_N)$ is a function that ensures $g_t(x_1,\dots,x_N)$ satisfies some given conditions. -The neural network $N(x_1,\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$ is an expression using the output from the neural network in some way. - -The role of the function $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$, is to ensure that the output of $N(x_1,\dots,x_N,P)$ is zero when $g_t(x_1,\dots,x_N)$ is evaluated at the values of $x_1,\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\dots,x_N)$ should alone make $g_t(x_1,\dots,x_N)$ satisfy the conditions. - - - -===== Network requirements ===== - -The network tries then the minimize the cost function following the -same ideas as described for the ODE case, but now with more than one -variables to consider. The concept still remains the same; find a set -of parameters $P$ such that the expression $f$ in (ref{PDE}) is as -close to zero as possible. - -As for the ODE case, the cost function is the mean squared error that -the network must try to minimize. The cost function for the network to -minimize is - -!bt -\begin{equation*} -C\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - - -===== More details ===== - -If we let $\bm{x} = \big( x_1, \dots, x_N \big)$ be an array containing the values for $x_1, \dots, x_N$ respectively, the cost function can be reformulated into the following: -!bt -\[ - C\left(\bm{x}, P\right) = f\left( \left( \bm{x}, \frac{\partial g(\bm{x}) }{\partial x_1}, \dots , \frac{\partial g(\bm{x}) }{\partial x_N}, \frac{\partial g(\bm{x}) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\bm{x}) }{\partial x_N^n} \right) \right)^2 -\] -!et - -If we also have $M$ different sets of values for $x_1, \dots, x_N$, that is $\bm{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)$ for $i = 1,\dots,M$ being the rows in matrix $X$, the cost function can be generalized into -!bt -\begin{equation*} -C\left(X, P \right) = \sum_{i=1}^M f\left( \left( \bm{x}_i, \frac{\partial g(\bm{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\bm{x}_i) }{\partial x_N}, \frac{\partial g(\bm{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\bm{x}_i) }{\partial x_N^n} \right) \right)^2. -\end{equation*} -!et - - -===== Example: The diffusion equation ===== - -In one spatial dimension, the equation reads -!bt -\begin{equation*} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -where a possible choice of conditions are -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x)$ being some given function. - - -===== Defining the problem ===== - -For this case, we want to find $g(x,t)$ such that - -!bt -\begin{equation} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} \label{diffonedim} -!et - -and - -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x) = \sin(\pi x)$. - -First, let us set up the deep neural network. -The deep neural network will follow the same structure as discussed in the examples solving the ODEs. -First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions. - - - - -===== Setting up the network using Autograd ===== - -The only change to do here, is to extend our network such that -functions of multiple parameters are correctly handled. In this case -we have two variables in our function to solve for, that is time $t$ -and position $x$. The variables will be represented by a -one-dimensional array in the program. The program will evaluate the -network at each possible pair $(x,t)$, given an array for the desired -$x$-values and $t$-values to approximate the solution at. - -!bc pycod -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] -!ec - - -===== Setting up the network using Autograd; The trial solution ===== - -The cost function must then iterate through the given arrays -containing values for $x$ and $t$, defines a point $(x,t)$ the deep -neural network and the trial solution is evaluated at, and then finds -the Jacobian of the trial solution. - -A possible trial solution for this PDE is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) -$$ - -with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$. - -To fulfill the conditions, $A(x,t)$ could be: - -$$ -h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) -$$ -since $(0) = u(1) = 0$ and $u(x) = \sin(\pi x)$. - - -===== Why the jacobian? ===== - -The Jacobian is used because the program must find the derivative of -the trial solution with respect to $x$ and $t$. - -This gives the necessity of computing the Jacobian matrix, as we want -to evaluate the gradient with respect to $x$ and $t$ (note that the -Jacobian of a scalar-valued multivariate function is simply its -gradient). - -In Autograd, the differentiation is by default done with respect to -the first input argument of your Python function. Since the points is -an array representing $x$ and $t$, the Jacobian is calculated using -the values of $x$ and $t$. - -To find the second derivative with respect to $x$ and $t$, the -Jacobian can be found for the second time. The result is a Hessian -matrix, which is the matrix containing all the possible second order -mixed derivatives of $g(x,t)$. - -!bc pycod -# Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum -!ec - - -===== Setting up the network using Autograd; The full program ===== - -Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution. - -The analytical solution of our problem is - -$$ -g(x,t) = \exp(-\pi^2 t)\sin(\pi x) -$$ - -A possible way to implement a neural network solving the PDE, is given below. -Be aware, though, that it is fairly slow for the parameters used. -A better result is possible, but requires more iterations, and thus longer time to complete. - - -Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. -Using TensorFlow results in a much better execution time. Try it! - -!bc pycod -import autograd.numpy as np -from autograd import jacobian,hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the network - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## Define the trial solution and cost function -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum /( np.size(x)*np.size(t) ) - -## For comparison, define the analytical solution -def g_analytic(point): - x,t = point - return np.exp(-np.pi**2*t)*np.sin(np.pi*x) - -## Set up a function for training the network to solve for the equation -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [100, 25] - num_iter = 250 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - g_dnn_ag = np.zeros((Nx, Nt)) - G_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - g_dnn_ag[i,j] = g_trial(point,P) - - G_analytical[i,j] = g_analytic(point) - - # Find the map difference between the analytical and the computed solution - diff_ag = np.abs(g_dnn_ag - G_analytical) - print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = g_dnn_ag[:,indx1] - res2 = g_dnn_ag[:,indx2] - res3 = g_dnn_ag[:,indx3] - - # Slice the analytical results - res_analytical1 = G_analytical[:,indx1] - res_analytical2 = G_analytical[:,indx2] - res_analytical3 = G_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - - -===== Example: Solving the wave equation with Neural Networks ===== - -The wave equation is -!bt -\begin{equation*} - \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -with $c$ being the specified wave speed. - -Here, the chosen conditions are -!bt -\begin{align*} - g(0,t) &= 0 \\ - g(1,t) &= 0 \\ - g(x,0) &= u(x) \\ - \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) -\end{align*} -!et -where $\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions. - - -===== The problem to solve for ===== - -The wave equation to solve for, is - -!bt -\begin{equation} \label{wave} -\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} -!et - -where $c$ is the given wave speed. -The chosen conditions for this equation are - -!bt -\begin{aligned} -g(0,t) &= 0, &t \geq 0 \\ -g(1,t) &= 0, &t \geq 0 \\ -g(x,0) &= u(x), &x\in[0,1] \\ -\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] -\end{aligned} \label{condwave} -!et - -In this example, let $c = 1$ and $u(x) = \sin(\pi x)$ and $v(x) = -\pi\sin(\pi x)$. - - - -===== The trial solution ===== -Setting up the network is done in similar matter as for the example of solving the diffusion equation. -The only things we have to change, is the trial solution such that it satisfies the conditions from (ref{condwave}) and the cost function. - -The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) -$$ - -where - -$$ -h_1(x,t) = (1-t^2)u(x) + tv(x) -$$ - -Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example. - - -===== The analytical solution ===== - -The analytical solution for our specific problem, is - -$$ -g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) -$$ - - -===== Solving the wave equation - the full program using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def v(x): - return -np.pi*np.sin(np.pi*x) - -def h1(point): - x,t = point - return (1 - t**2)*u(x) + t*v(x) - -def g_trial(point,P): - x,t = point - return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point) - -## Define the cost function -def cost_function(P, x, t): - cost_sum = 0 - - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t_hessian = g_t_hessian_func(point,P) - - g_t_d2x = g_t_hessian[0][0] - g_t_d2t = g_t_hessian[1][1] - - err_sqr = ( (g_t_d2t - g_t_d2x) )**2 - cost_sum += err_sqr - - return cost_sum / (np.size(t) * np.size(x)) - -## The neural network -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## The analytical solution -def g_analytic(point): - x,t = point - return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t) - -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [50,20] - num_iter = 1000 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - res = np.zeros((Nx, Nt)) - res_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - res[i,j] = g_trial(point,P) - - res_analytical[i,j] = g_analytic(point) - - diff = np.abs(res - res_analytical) - print("Max difference between analytical and solution from nn: %g"%np.max(diff)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = res[:,indx1] - res2 = res[:,indx2] - res3 = res[:,indx3] - - # Slice the analytical results - res_analytical1 = res_analytical[:,indx1] - res_analytical2 = res_analytical[:,indx2] - res_analytical3 = res_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - - -===== Resources on differential equations and deep learning ===== - -o "Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al":"https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf" -o "Neural networks for solving differential equations by A. Honchar":"https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c" -o "Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener":"http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf" -o "Introduction to Partial Differential Equations by A. Tveito, R. Winther":"https://www.springer.com/us/book/9783540225515" - - - - - diff --git a/doc/src/week43/odenn.do.txt b/doc/src/week43/odenn.do.txt deleted file mode 100644 index 9d3d9268e..000000000 --- a/doc/src/week43/odenn.do.txt +++ /dev/null @@ -1,1229 +0,0 @@ -TITLE: Solving Differential Equations with Deep Learning -AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and Facility for Rare ion Beams, Michigan State University -DATE: today - -!split -===== Ordinary Differential Equations ===== - -An ordinary differential equation (ODE) is an equation involving functions having one variable. - -In general, an ordinary differential equation looks like - -!bt -\begin{equation} \label{ode} -f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 -\end{equation} -!et - -where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$. - -The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in (ref{ode}). -The highest order of derivative, that is the value of $n$, determines to the order of the equation. -The equation is referred to as a $n$-th order ODE. -Along with (ref{ode}), some additional conditions of the function $g(x)$ are typically given -for the solution to be unique. - -!split -===== The trial solution ===== - -Let the trial solution $g_t(x)$ be - -!bt -\begin{equation} - g_t(x) = h_1(x) + h_2(x,N(x,P)) -\end{equation} -!et - - -where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set -of conditions, $N(x,P)$ a neural network with weights and biases -described by $P$ and $h_2(x, N(x,P))$ some expression involving the -neural network. The role of the function $h_2(x, N(x,P))$, is to -ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is -evaluated at the values of $x$ where the given conditions must be -satisfied. The function $h_1(x)$ should alone make $g_t(x)$ satisfy -the conditions. - -But what about the network $N(x,P)$? - - -As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation. - - -!split -===== Minimization process ===== - -For the minimization to be defined, we need to have a cost function at hand to minimize. - -It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in (ref{ode}). -We can choose to consider the mean squared error as the cost function for an input $x$. -Since we are looking at one input, the cost function is just $f$ squared. -The cost function $c\left(x, P \right)$ can therefore be expressed as - -!bt -C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 -!et - -If $N$ inputs are given as a vector $\bm{x}$ with elements $x_i$ for $i = 1,\dots,N$, -the cost function becomes - -!bt -\begin{equation} \label{cost} - C\left(\bm{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 -\end{equation} -!et - -The neural net should then find the parameters $P$ that minimizes the cost function in -(ref{cost}) for a set of $N$ training samples $x_i$. - -!split -===== Minimizing the cost function using gradient descent and automatic differentiation ===== - -To perform the minimization using gradient descent, the gradient of $C\left(\bm{x}, P\right)$ is needed. -It might happen so that finding an analytical expression of the gradient of $C(\bm{x}, P)$ from (ref{cost}) gets too messy, depending on which cost function one desires to use. - -Luckily, there exists libraries that makes the job for us through automatic differentiation. -Automatic differentiation is a method of finding the derivatives numerically with very high precision. - - -!split -===== Example: Exponential decay ===== - -An exponential decay of a quantity $g(x)$ is described by the equation - -!bt -\begin{equation} \label{solve_expdec} - g'(x) = -\gamma g(x) -\end{equation} -!et - -with $g(0) = g_0$ for some chosen initial value $g_0$. - -The analytical solution of (ref{solve_expdec}) is - -!bt -\begin{equation} - g(x) = g_0 \exp\left(-\gamma x\right) -\end{equation} -!et - -Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (ref{solve_expdec}). - - -!split -===== The function to solve for ===== - -The program will use a neural network to solve - -!bt -\begin{equation} \label{solveode} -g'(x) = -\gamma g(x) -\end{equation} -!et - -where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values. - -In this example, $\gamma = 2$ and $g_0 = 10$. - -!split -===== The trial solution ===== -To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be - -!bt -g_t(x, P) = h_1(x) + h_2(x, N(x, P)) -!et - -with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer. - -!split -===== Setup of Network ===== - -In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}}, P_{\text{output}} \}$. -If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer. - -The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. -If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix. - -Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron. - -It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of (ref{solveode}). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution: - -!bt -\begin{equation} \label{trial} -g_t(x, P) = g_0 + x \cdot N(x, P) -\end{equation} -!et - -!split -===== Reformulating the problem ===== - -We wish that our neural network manages to minimize a given cost function. - -A reformulation of out equation, (ref{solveode}), must therefore be done, -such that it describes the problem a neural network can solve for. - -The neural network must find the set of weights and biases $P$ such that the trial solution in (ref{trial}) satisfies (ref{solveode}). - -The trial solution - -!bt -g_t(x, P) = g_0 + x \cdot N(x, P) -!et - -has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that - -!bt -\begin{equation} \label{nnmin} -g_t'(x, P) = - \gamma g_t(x, P) -\end{equation} -!et - -is fulfilled as *best as possible*. - -!split -===== More technicalities ===== - -The left hand side and right hand side of (ref{nnmin}) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible. -This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. -In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network. - -This gives the following cost function our neural network must solve for: - -!bt -\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} -!et - -(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$) - -or, in terms of weights and biases for the hidden and output layer in our network: - -!bt -\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} -!et - -for an input value $x$. - -!split -===== More details ===== - -If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes - -!bt -\begin{equation} \label{min} -\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} -\end{equation} -!et - -Letting $\bm{x}$ be a vector with elements $x_i$ and $C(\bm{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes - -!bt -\min_{P} C(\bm{x}, P) -!et - -In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as - -$$ -\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\bm{x}, \{P_{\text{hidden} }, P_{\text{output} }\}) -$$ - -!split -===== A possible implementation of a neural network ===== - -For simplicity, it is assumed that the input is an array $\bm{x} = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills (ref{min}). - -First, the neural network must feed forward the inputs. -This means that $\bm{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. -The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\text{hidden} }$. - -!split -===== Technicalities ===== - -For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is: - -!bt -\begin{aligned} -z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ -&= -\begin{pmatrix} -b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -x_j -\end{pmatrix} -\end{aligned} -!et - -!split -===== Final technicalities I ===== - -The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector: - -!bt -\begin{aligned} -\bm{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ -&= -\begin{pmatrix} - b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -x_1 & x_2 & \dots & x_N -\end{pmatrix} \\ -&= \bm{p}_{i, \text{hidden}}^T X -\end{aligned} -!et - -!split -===== Final technicalities II ===== - -The vector $\bm{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to (ref{min}). - -After having found $\bm{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\bm{z})$. - -In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: - -!bt -f(z) = \frac{1}{1 + \exp{(-z)}} -!et - -It is possible to use other activations functions for the hidden layer also. - -The output $\bm{x}_i^{\text{hidden}}$ from each $i$-th hidden neuron is: - -$$ -\bm{x}_i^{\text{hidden} } = f\big( \bm{z}_{i}^{\text{hidden}} \big) -$$ - -The outputs $\bm{x}_i^{\text{hidden} } $ are then sent to the output layer. - -The output layer consists of one neuron in this case, and combines the -output from each of the neurons in the hidden layers. The output layer -combines the results from the hidden layer using some weights $w_i^{\text{output}}$ -and biases $b_i^{\text{output}}$. In this case, -it is assumes that the number of neurons in the output layer is one. - -!split -===== Final technicalities III ===== - - -The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously. - -!bt -\begin{aligned} -z_{1,j}^{\text{output}} & = -\begin{pmatrix} -b_1^{\text{output}} & \bm{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -\bm{x}_j^{\text{hidden}} -\end{pmatrix} -\end{aligned} -!et - -!split -===== Final technicalities IV ===== - -Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer: - -!bt -\bm{z}_{1}^{\text{output}} = -\begin{pmatrix} -b_1^{\text{output}} & \bm{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -\bm{x}_1^{\text{hidden}} & \bm{x}_2^{\text{hidden}} & \dots & \bm{x}_N^{\text{hidden}} -\end{pmatrix} -!et - -In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\bm{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\bm{z}_{1}^{\text{output}}$ is the final output of the network. - -!split -===== Back propagation ===== - -The next step is to decide how the parameters should be changed such that they minimize the cost function. - -The chosen cost function for this problem is - -!bt -C(\bm{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 -!et - -In order to minimize the cost function, an optimization method must be chosen. - -Here, gradient descent with a constant step size has been chosen. - -!split -===== Gradient descent ===== - -The idea of the gradient descent algorithm is to update parameters in -a direction where the cost function decreases goes to a minimum. - -In general, the update of some parameters $\bm{\omega}$ given a cost -function defined by some weights $\bm{\omega}$, $C(\bm{x}, -\bm{\omega})$, goes as follows: - -!bt -\bm{\omega}_{\text{new} } = \bm{\omega} - \lambda \nabla_{\bm{\omega}} C(\bm{x}, \bm{\omega}) -!et - -for a number of iterations or until $ \big|\big| \bm{\omega}_{\text{new} } - \bm{\omega} \big|\big|$ becomes smaller than some given tolerance. - -The value of $\lambda$ decides how large steps the algorithm must take -in the direction of $ \nabla_{\bm{\omega}} C(\bm{x}, \bm{\omega})$. -The notation $\nabla_{\bm{\omega}}$ express the gradient with respect -to the elements in $\bm{\omega}$. - -In our case, we have to minimize the cost function $C(\bm{x}, P)$ with -respect to the two sets of weights and biases, that is for the hidden -layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output} -}$ . - -This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by - -!bt -\begin{aligned} -P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\bm{x}, P) \\ -P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\bm{x}, P) -\end{aligned} -!et - -!split -===== The code for solving the ODE ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Assuming one input, hidden, and output layer -def neural_network(params, x): - - # Find the weights (including and biases) for the hidden and output layer. - # Assume that params is a list of parameters for each layer. - # The biases are the first element for each array in params, - # and the weights are the remaning elements in each array in params. - - w_hidden = params[0] - w_output = params[1] - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - ## Hidden layer: - - # Add a row of ones to include bias - x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_input) - x_hidden = sigmoid(z_hidden) - - ## Output layer: - - # Include bias: - x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) - - z_output = np.matmul(w_output, x_hidden) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial(x,params, g0 = 10): - return g0 + x*neural_network(params,x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The cost function: -def cost_function(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input, hidden, and output layer -def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): - ## Set up initial weights and biases - - # For the hidden layer - p0 = npr.randn(num_neurons_hidden, 2 ) - - # For the output layer - p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included - - P = [p0, p1] - - print('Initial cost: %g'%cost_function(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of two arrays; - # one for the gradient w.r.t P_hidden and - # one for the gradient w.r.t P_output - cost_grad = cost_function_grad(P, x) - - P[0] = P[0] - lmb * cost_grad[0] - P[1] = P[1] - lmb * cost_grad[1] - - print('Final cost: %g'%cost_function(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - # Set seed such that the weight are initialized - # with same weights and biases for every run. - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = 10 - num_iter = 10000 - lmb = 0.001 - - # Use the network - P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) - - # Print the deviation from the trial solution and true solution - res = g_trial(x,P) - res_analytical = g_analytic(x) - - print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - - -!split -===== The network with one input layer, specified number of hidden layers, and one output layer ===== - -It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers. - -The number of neurons within each hidden layer are given as a list of integers in the program below. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# The neural network with one input layer and one output layer, -# but with number of hidden layers specified by the user. -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - - N_hidden = np.size(deep_params) - 1 # -1 since params consists of - # parameters to all the hidden - # layers AND the output layer. - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial_deep(x,params, g0 = 10): - return g0 + x*deep_neural_network(params, x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The same cost function as before, but calls deep_neural_network instead. -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(deep_neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input and one output layer, -# but with specified number of hidden layers from the user. -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # The number of elements in the list num_hidden_neurons thus represents - # the number of hidden layers. - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weights and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = np.array([10,10]) - num_iter = 10000 - lmb = 0.001 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - res = g_trial_deep(x,P) - res_analytical = g_analytic(x) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','dnn']) - plt.ylabel('g(x)') - plt.show() -!ec - - - -!split -===== Example: Population growth, comparing Autograd, and Euler's scheme ===== - -A logistic model of population growth assumes that a population converges toward an equilibrium. -The population growth can be modeled by - -!bt -\begin{equation} \label{log} - g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment. -Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant. - -In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability -and high execution time (this might be more apparent in the examples solving PDEs), -a network has been constructed using TensorFlow also. -For comparison, the forward Euler method has been implemented in order to see how the networks performs compared to a numerical scheme. - -!split -===== Setting up the problem ===== - -Here, we will model a population $g(t)$ in an environment having carrying capacity $A$. -The population follows the model - -!bt -\begin{equation} \label{solveode_population} -g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(0) = g_0$. - -In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$. - -!split -===== The trial solution ===== - -We will get a slightly different trial solution, as the boundary conditions are different -compared to the case for exponential decay. - -A possible trial solution satisfying the condition $g(0) = g_0$ could be - -$$ -h_1(t) = g_0 + t \cdot N(t,P) -$$ - -with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$. - -The analytical solution is - -$$ -g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} -$$ - -!split -===== The program using Autograd ===== - -The network will be the similar as for the exponential decay example, but with some small modifications for our problem. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Function to get the parameters. -# Done such that one can easily change the paramaters after one's liking. -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def deep_neural_network(P, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = P[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = P[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = f(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# The right side of the ODE: -def f(x, g_trial): - alpha,A, g0 = get_parameters() - return alpha*g_trial*(A - g_trial) - -# The trial solution using the deep neural network: -def g_trial_deep(x, params): - alpha,A, g0 = get_parameters() - return g0 + x*deep_neural_network(params,x) - -# The analytical solution: -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100, 50, 25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - plt.show() -!ec - -!split -===== Using forward Euler to solve the ODE ===== - -A straight-forward way of solving an ODE numerically, is to use Euler's method. - -Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$: - -$$ -f(x + \Delta x) \approx f(x) + \Delta x f'(x) -$$ - -In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields - -!bt -\begin{aligned} - g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ - &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) -\end{aligned} -!et -along with the condition that $g(0) = g_0$. - -Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$. - -For $i \geq 1$, we have that -!bt -\begin{aligned} -t_i &= i\Delta t \\ -&= (i - 1)\Delta t + \Delta t \\ -&= t_{i-1} + \Delta t -\end{aligned} -!et - -Now, if $g_i = g(t_i)$ then - -!bt -\begin{equation} - \begin{aligned} - g_i &= g(t_i) \\ - &= g(t_{i-1} + \Delta t) \\ - &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ - &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) - \end{aligned} -\end{equation} \label{odenum} -!et -for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$. - -Equation (ref{odenum}) could be implemented in the following way, -extending the program that uses the network using Autograd: - -!bc pycod -# Assume that all function definitions from the example program using Autograd -# are located here. - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100,50,25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - ## Find an approximation to the funtion using forward Euler - - alpha, A, g0 = get_parameters() - dt = T/(Nt - 1) - - # Perform forward Euler to solve the ODE - g_euler = np.zeros(Nt) - g_euler[0] = g0 - - for i in range(1,Nt): - g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1])) - - # Print the errors done by each method - diff1 = np.max(np.abs(g_euler - g_analytical)) - diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical)) - - print('Max absolute difference between Euler method and analytical: %g'%diff1) - print('Max absolute difference between deep neural network and analytical: %g'%diff2) - - # Plot results - plt.figure(figsize=(10,10)) - - plt.plot(t,g_euler) - plt.plot(t,g_analytical) - plt.plot(t,g_dnn_ag[0,:]) - - plt.legend(['euler','analytical','dnn']) - plt.xlabel('Time t') - plt.ylabel('g(t)') - - plt.show() -!ec - - - -!split -===== Using TensorFlow ===== - -TensorFlow is a library widely used in the machine learning community. - -Tensorflow is an open source library machine learning library -developed by the Google Brain team for internal use. It was released -under the Apache 2.0 open source license in November 9, 2015. - -Tensorflow is a computational framework that allows you to construct -machine learning models at different levels of abstraction, from -high-level, object-oriented APIs like Keras, down to the C++ kernels -that Tensorflow is built upon. The higher levels of abstraction are -simpler to use, but less flexible, and our choice of implementation -should reflect the problems we are trying to solve. - - - -To install tensorflow on Unix/Linux systems, use pip as -!bc pycod -pip3 install tensorflow -!ec -and/or if you use _anaconda_, just write (or install from the graphical user interface) -(current release of CPU-only TensorFlow) -!bc pycod -conda create -n tf tensorflow -conda activate tf -!ec -To install the current release of GPU TensorFlow -!bc pycod -conda create -n tf-gpu tensorflow-gpu -conda activate tf-gpu -!ec - -!split -===== Using Keras ===== - -Keras is a high level "neural network":"https://en.wikipedia.org/wiki/Application_programming_interface" -that supports Tensorflow, CTNK and Theano as backends. -If you have Anaconda installed you may run the following command -!bc pycod -conda install keras -!ec -You can look up the "instructions here":"https://keras.io/" for more information. - - - - - -!bc pycod - - -import numpy as np -import matplotlib.pyplot as plt -import tensorflow as tf -from math import * -import time -from keras.optimizers import Adam -from keras.models import Model -from keras.layers import Dense, Input, LeakyReLU -from keras import optimizers -%matplotlib inline - -# Analytical Solution to position -def ana_cos(r0,t,k=1,m=1): - w0 = sqrt(k/m) - return r0*np.cos(w0*t) - -# Trial Function for Neural Net -def trial_func(x,y,y0=1): - return x*y + y0 - -# Loss Function for Position and Velocity -def combined_right(trialv,trialy,k=1,m=1): - return -k/m*trialy, trialv - -# Loss Wrapper in order to pass the Loss Function to Neural Net -def con_loss_wrapper(input_tensor): - def con_loss_function(y,y_pred): - trialy = trial_func(input_tensor,y_pred) - trialv = trial_func(input_tensor,y_pred) - righty, rightv = combined_right(trialv,trialy) - - leftv = tf.gradients(trialv,input_tensor)[0] - lefty = tf.gradients(leftv,input_tensor)[0] - - - loss = tf.reduce_mean((tf.math.squared_difference(lefty,righty))) - return loss - return con_loss_function - -# Creates the input data for the Neural Net -def create_input_data(x0=0,xmax=1,num_batch=5,len_batch=15): - input_data = np.linspace(x0,xmax,num_batch*len_batch) - input_data = input_data.reshape(num_batch,len_batch) - - return input_data - -# Creates the Neural Net -def create_net(data,len_batch,lr,epochs,right_side,loss,n_hidden_layer=50): - - input_tensor = Input(shape=(len_batch,)) - - hidden1 = Dense(n_hidden_layer,activation='tanh', - kernel_initializer='random_uniform',bias_initializer='random_uniform')(input_tensor) - hidden2 = Dense(n_hidden_layer,activation='tanh', - kernel_initializer='random_uniform',bias_initializer='random_uniform')(hidden1) - hidden3 = Dense(n_hidden_layer,activation='tanh', - kernel_initializer='random_uniform',bias_initializer='random_uniform')(hidden2) - hidden4 = Dense(n_hidden_layer,activation='tanh', - kernel_initializer='random_uniform',bias_initializer='random_uniform')(hidden3) - output = Dense(len_batch)(hidden4) - - model = Model(input_tensor,output) - - gd = optimizers.adam(lr=lr) # May need to change first 'lr' to 'learning_rate' depending on TF/Keras version - model.compile(loss=loss(input_tensor),optimizer=gd) - model.fit(data,np.zeros((data.shape[0])),epochs=epochs) - - res = model.predict(data) - - del model - - return res - -# Define Function for Mean Squared Error -def mean_squared_error(analytical, results): - mse = 0 - for i in range(len(analytical)): - mse += (analytical[i] - results[i])**2 - - mse = mse/len(analytical) - return mse - - -# Define Constants -num_batch = 1000 -len_batch = 1 - -# Create input data -data = create_input_data(x0=0,xmax=10,num_batch=num_batch,len_batch=len_batch) - -# Create and Run the Neural Net -nn_start_time = time.time() - -velocity = create_net(data,len_batch=len_batch,lr=0.001,n_hidden_layer=50, - epochs=1000,right_side=combined_right,loss=con_loss_wrapper) - -nn_end_time = time.time() - - -# Reshape Neural Net Output for easy graphing and analysis -n = num_batch*len_batch -velocity = velocity.reshape(1,n) -t = data.reshape(1,n)[0] -results_v = trial_func(t,velocity)[0] - -# Create Comparison data from Analytical Solution -analyt = ana_cos(1,t,m=1,k=1) - - -# Plot -plt.plot(t,analyt,label='analytical') -plt.plot(t,results_v,label='net') -plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left') - -# Find Mean Squared Error -print("The Mean Squared Error of the Neural Net Solution is", mean_squared_error(analyt, results_v), "with a runtime of", nn_end_time-nn_start_time,"seconds.") - -!ec