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Homework #3 : Raw files

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BIN  ex3.pdf
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59 ex3/displayData.m
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+function [h, display_array] = displayData(X, example_width)
+%DISPLAYDATA Display 2D data in a nice grid
+% [h, display_array] = DISPLAYDATA(X, example_width) displays 2D data
+% stored in X in a nice grid. It returns the figure handle h and the
+% displayed array if requested.
+
+% Set example_width automatically if not passed in
+if ~exist('example_width', 'var') || isempty(example_width)
+ example_width = round(sqrt(size(X, 2)));
+end
+
+% Gray Image
+colormap(gray);
+
+% Compute rows, cols
+[m n] = size(X);
+example_height = (n / example_width);
+
+% Compute number of items to display
+display_rows = floor(sqrt(m));
+display_cols = ceil(m / display_rows);
+
+% Between images padding
+pad = 1;
+
+% Setup blank display
+display_array = - ones(pad + display_rows * (example_height + pad), ...
+ pad + display_cols * (example_width + pad));
+
+% Copy each example into a patch on the display array
+curr_ex = 1;
+for j = 1:display_rows
+ for i = 1:display_cols
+ if curr_ex > m,
+ break;
+ end
+ % Copy the patch
+
+ % Get the max value of the patch
+ max_val = max(abs(X(curr_ex, :)));
+ display_array(pad + (j - 1) * (example_height + pad) + (1:example_height), ...
+ pad + (i - 1) * (example_width + pad) + (1:example_width)) = ...
+ reshape(X(curr_ex, :), example_height, example_width) / max_val;
+ curr_ex = curr_ex + 1;
+ end
+ if curr_ex > m,
+ break;
+ end
+end
+
+% Display Image
+h = imagesc(display_array, [-1 1]);
+
+% Do not show axis
+axis image off
+
+drawnow;
+
+end
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69 ex3/ex3.m
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+%% Machine Learning Online Class - Exercise 3 | Part 1: One-vs-all
+
+% Instructions
+% ------------
+%
+% This file contains code that helps you get started on the
+% linear exercise. You will need to complete the following functions
+% in this exericse:
+%
+% lrCostFunction.m (logistic regression cost function)
+% oneVsAll.m
+% predictOneVsAll.m
+% predict.m
+%
+% For this exercise, you will not need to change any code in this file,
+% or any other files other than those mentioned above.
+%
+
+%% Initialization
+clear ; close all; clc
+
+%% Setup the parameters you will use for this part of the exercise
+input_layer_size = 400; % 20x20 Input Images of Digits
+num_labels = 10; % 10 labels, from 1 to 10
+ % (note that we have mapped "0" to label 10)
+
+%% =========== Part 1: Loading and Visualizing Data =============
+% We start the exercise by first loading and visualizing the dataset.
+% You will be working with a dataset that contains handwritten digits.
+%
+
+% Load Training Data
+fprintf('Loading and Visualizing Data ...\n')
+
+load('ex3data1.mat'); % training data stored in arrays X, y
+m = size(X, 1);
+
+% Randomly select 100 data points to display
+rand_indices = randperm(m);
+sel = X(rand_indices(1:100), :);
+
+displayData(sel);
+
+fprintf('Program paused. Press enter to continue.\n');
+pause;
+
+%% ============ Part 2: Vectorize Logistic Regression ============
+% In this part of the exercise, you will reuse your logistic regression
+% code from the last exercise. You task here is to make sure that your
+% regularized logistic regression implementation is vectorized. After
+% that, you will implement one-vs-all classification for the handwritten
+% digit dataset.
+%
+
+fprintf('\nTraining One-vs-All Logistic Regression...\n')
+
+lambda = 0.1;
+[all_theta] = oneVsAll(X, y, num_labels, lambda);
+
+fprintf('Program paused. Press enter to continue.\n');
+pause;
+
+
+%% ================ Part 3: Predict for One-Vs-All ================
+% After ...
+pred = predictOneVsAll(all_theta, X);
+
+fprintf('\nTraining Set Accuracy: %f\n', mean(double(pred == y)) * 100);
+
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88 ex3/ex3_nn.m
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+%% Machine Learning Online Class - Exercise 3 | Part 2: Neural Networks
+
+% Instructions
+% ------------
+%
+% This file contains code that helps you get started on the
+% linear exercise. You will need to complete the following functions
+% in this exericse:
+%
+% lrCostFunction.m (logistic regression cost function)
+% oneVsAll.m
+% predictOneVsAll.m
+% predict.m
+%
+% For this exercise, you will not need to change any code in this file,
+% or any other files other than those mentioned above.
+%
+
+%% Initialization
+clear ; close all; clc
+
+%% Setup the parameters you will use for this exercise
+input_layer_size = 400; % 20x20 Input Images of Digits
+hidden_layer_size = 25; % 25 hidden units
+num_labels = 10; % 10 labels, from 1 to 10
+ % (note that we have mapped "0" to label 10)
+
+%% =========== Part 1: Loading and Visualizing Data =============
+% We start the exercise by first loading and visualizing the dataset.
+% You will be working with a dataset that contains handwritten digits.
+%
+
+% Load Training Data
+fprintf('Loading and Visualizing Data ...\n')
+
+load('ex3data1.mat');
+m = size(X, 1);
+
+% Randomly select 100 data points to display
+sel = randperm(size(X, 1));
+sel = sel(1:100);
+
+displayData(X(sel, :));
+
+fprintf('Program paused. Press enter to continue.\n');
+pause;
+
+%% ================ Part 2: Loading Pameters ================
+% In this part of the exercise, we load some pre-initialized
+% neural network parameters.
+
+fprintf('\nLoading Saved Neural Network Parameters ...\n')
+
+% Load the weights into variables Theta1 and Theta2
+load('ex3weights.mat');
+
+%% ================= Part 3: Implement Predict =================
+% After training the neural network, we would like to use it to predict
+% the labels. You will now implement the "predict" function to use the
+% neural network to predict the labels of the training set. This lets
+% you compute the training set accuracy.
+
+pred = predict(Theta1, Theta2, X);
+
+fprintf('\nTraining Set Accuracy: %f\n', mean(double(pred == y)) * 100);
+
+fprintf('Program paused. Press enter to continue.\n');
+pause;
+
+% To give you an idea of the network's output, you can also run
+% through the examples one at the a time to see what it is predicting.
+
+% Randomly permute examples
+rp = randperm(m);
+
+for i = 1:m
+ % Display
+ fprintf('\nDisplaying Example Image\n');
+ displayData(X(rp(i), :));
+
+ pred = predict(Theta1, Theta2, X(rp(i),:));
+ fprintf('\nNeural Network Prediction: %d (digit %d)\n', pred, mod(pred, 10));
+
+ % Pause
+ fprintf('Program paused. Press enter to continue.\n');
+ pause;
+end
+
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BIN  ex3/ex3data1.mat
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BIN  ex3/ex3weights.mat
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175 ex3/fmincg.m
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+function [X, fX, i] = fmincg(f, X, options, P1, P2, P3, P4, P5)
+% Minimize a continuous differentialble multivariate function. Starting point
+% is given by "X" (D by 1), and the function named in the string "f", must
+% return a function value and a vector of partial derivatives. The Polack-
+% Ribiere flavour of conjugate gradients is used to compute search directions,
+% and a line search using quadratic and cubic polynomial approximations and the
+% Wolfe-Powell stopping criteria is used together with the slope ratio method
+% for guessing initial step sizes. Additionally a bunch of checks are made to
+% make sure that exploration is taking place and that extrapolation will not
+% be unboundedly large. The "length" gives the length of the run: if it is
+% positive, it gives the maximum number of line searches, if negative its
+% absolute gives the maximum allowed number of function evaluations. You can
+% (optionally) give "length" a second component, which will indicate the
+% reduction in function value to be expected in the first line-search (defaults
+% to 1.0). The function returns when either its length is up, or if no further
+% progress can be made (ie, we are at a minimum, or so close that due to
+% numerical problems, we cannot get any closer). If the function terminates
+% within a few iterations, it could be an indication that the function value
+% and derivatives are not consistent (ie, there may be a bug in the
+% implementation of your "f" function). The function returns the found
+% solution "X", a vector of function values "fX" indicating the progress made
+% and "i" the number of iterations (line searches or function evaluations,
+% depending on the sign of "length") used.
+%
+% Usage: [X, fX, i] = fmincg(f, X, options, P1, P2, P3, P4, P5)
+%
+% See also: checkgrad
+%
+% Copyright (C) 2001 and 2002 by Carl Edward Rasmussen. Date 2002-02-13
+%
+%
+% (C) Copyright 1999, 2000 & 2001, Carl Edward Rasmussen
+%
+% Permission is granted for anyone to copy, use, or modify these
+% programs and accompanying documents for purposes of research or
+% education, provided this copyright notice is retained, and note is
+% made of any changes that have been made.
+%
+% These programs and documents are distributed without any warranty,
+% express or implied. As the programs were written for research
+% purposes only, they have not been tested to the degree that would be
+% advisable in any important application. All use of these programs is
+% entirely at the user's own risk.
+%
+% [ml-class] Changes Made:
+% 1) Function name and argument specifications
+% 2) Output display
+%
+
+% Read options
+if exist('options', 'var') && ~isempty(options) && isfield(options, 'MaxIter')
+ length = options.MaxIter;
+else
+ length = 100;
+end
+
+
+RHO = 0.01; % a bunch of constants for line searches
+SIG = 0.5; % RHO and SIG are the constants in the Wolfe-Powell conditions
+INT = 0.1; % don't reevaluate within 0.1 of the limit of the current bracket
+EXT = 3.0; % extrapolate maximum 3 times the current bracket
+MAX = 20; % max 20 function evaluations per line search
+RATIO = 100; % maximum allowed slope ratio
+
+argstr = ['feval(f, X']; % compose string used to call function
+for i = 1:(nargin - 3)
+ argstr = [argstr, ',P', int2str(i)];
+end
+argstr = [argstr, ')'];
+
+if max(size(length)) == 2, red=length(2); length=length(1); else red=1; end
+S=['Iteration '];
+
+i = 0; % zero the run length counter
+ls_failed = 0; % no previous line search has failed
+fX = [];
+[f1 df1] = eval(argstr); % get function value and gradient
+i = i + (length<0); % count epochs?!
+s = -df1; % search direction is steepest
+d1 = -s'*s; % this is the slope
+z1 = red/(1-d1); % initial step is red/(|s|+1)
+
+while i < abs(length) % while not finished
+ i = i + (length>0); % count iterations?!
+
+ X0 = X; f0 = f1; df0 = df1; % make a copy of current values
+ X = X + z1*s; % begin line search
+ [f2 df2] = eval(argstr);
+ i = i + (length<0); % count epochs?!
+ d2 = df2'*s;
+ f3 = f1; d3 = d1; z3 = -z1; % initialize point 3 equal to point 1
+ if length>0, M = MAX; else M = min(MAX, -length-i); end
+ success = 0; limit = -1; % initialize quanteties
+ while 1
+ while ((f2 > f1+z1*RHO*d1) | (d2 > -SIG*d1)) & (M > 0)
+ limit = z1; % tighten the bracket
+ if f2 > f1
+ z2 = z3 - (0.5*d3*z3*z3)/(d3*z3+f2-f3); % quadratic fit
+ else
+ A = 6*(f2-f3)/z3+3*(d2+d3); % cubic fit
+ B = 3*(f3-f2)-z3*(d3+2*d2);
+ z2 = (sqrt(B*B-A*d2*z3*z3)-B)/A; % numerical error possible - ok!
+ end
+ if isnan(z2) | isinf(z2)
+ z2 = z3/2; % if we had a numerical problem then bisect
+ end
+ z2 = max(min(z2, INT*z3),(1-INT)*z3); % don't accept too close to limits
+ z1 = z1 + z2; % update the step
+ X = X + z2*s;
+ [f2 df2] = eval(argstr);
+ M = M - 1; i = i + (length<0); % count epochs?!
+ d2 = df2'*s;
+ z3 = z3-z2; % z3 is now relative to the location of z2
+ end
+ if f2 > f1+z1*RHO*d1 | d2 > -SIG*d1
+ break; % this is a failure
+ elseif d2 > SIG*d1
+ success = 1; break; % success
+ elseif M == 0
+ break; % failure
+ end
+ A = 6*(f2-f3)/z3+3*(d2+d3); % make cubic extrapolation
+ B = 3*(f3-f2)-z3*(d3+2*d2);
+ z2 = -d2*z3*z3/(B+sqrt(B*B-A*d2*z3*z3)); % num. error possible - ok!
+ if ~isreal(z2) | isnan(z2) | isinf(z2) | z2 < 0 % num prob or wrong sign?
+ if limit < -0.5 % if we have no upper limit
+ z2 = z1 * (EXT-1); % the extrapolate the maximum amount
+ else
+ z2 = (limit-z1)/2; % otherwise bisect
+ end
+ elseif (limit > -0.5) & (z2+z1 > limit) % extraplation beyond max?
+ z2 = (limit-z1)/2; % bisect
+ elseif (limit < -0.5) & (z2+z1 > z1*EXT) % extrapolation beyond limit
+ z2 = z1*(EXT-1.0); % set to extrapolation limit
+ elseif z2 < -z3*INT
+ z2 = -z3*INT;
+ elseif (limit > -0.5) & (z2 < (limit-z1)*(1.0-INT)) % too close to limit?
+ z2 = (limit-z1)*(1.0-INT);
+ end
+ f3 = f2; d3 = d2; z3 = -z2; % set point 3 equal to point 2
+ z1 = z1 + z2; X = X + z2*s; % update current estimates
+ [f2 df2] = eval(argstr);
+ M = M - 1; i = i + (length<0); % count epochs?!
+ d2 = df2'*s;
+ end % end of line search
+
+ if success % if line search succeeded
+ f1 = f2; fX = [fX' f1]';
+ fprintf('%s %4i | Cost: %4.6e\r', S, i, f1);
+ s = (df2'*df2-df1'*df2)/(df1'*df1)*s - df2; % Polack-Ribiere direction
+ tmp = df1; df1 = df2; df2 = tmp; % swap derivatives
+ d2 = df1'*s;
+ if d2 > 0 % new slope must be negative
+ s = -df1; % otherwise use steepest direction
+ d2 = -s'*s;
+ end
+ z1 = z1 * min(RATIO, d1/(d2-realmin)); % slope ratio but max RATIO
+ d1 = d2;
+ ls_failed = 0; % this line search did not fail
+ else
+ X = X0; f1 = f0; df1 = df0; % restore point from before failed line search
+ if ls_failed | i > abs(length) % line search failed twice in a row
+ break; % or we ran out of time, so we give up
+ end
+ tmp = df1; df1 = df2; df2 = tmp; % swap derivatives
+ s = -df1; % try steepest
+ d1 = -s'*s;
+ z1 = 1/(1-d1);
+ ls_failed = 1; % this line search failed
+ end
+ if exist('OCTAVE_VERSION')
+ fflush(stdout);
+ end
+end
+fprintf('\n');
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52 ex3/lrCostFunction.m
@@ -0,0 +1,52 @@
+function [J, grad] = lrCostFunction(theta, X, y, lambda)
+%LRCOSTFUNCTION Compute cost and gradient for logistic regression with
+%regularization
+% J = LRCOSTFUNCTION(theta, X, y, lambda) computes the cost of using
+% theta as the parameter for regularized logistic regression and the
+% gradient of the cost w.r.t. to the parameters.
+
+% Initialize some useful values
+m = length(y); % number of training examples
+
+% You need to return the following variables correctly
+J = 0;
+grad = zeros(size(theta));
+
+% ====================== YOUR CODE HERE ======================
+% Instructions: Compute the cost of a particular choice of theta.
+% You should set J to the cost.
+% Compute the partial derivatives and set grad to the partial
+% derivatives of the cost w.r.t. each parameter in theta
+%
+% Hint: The computation of the cost function and gradients can be
+% efficiently vectorized. For example, consider the computation
+%
+% sigmoid(X * theta)
+%
+% Each row of the resulting matrix will contain the value of the
+% prediction for that example. You can make use of this to vectorize
+% the cost function and gradient computations.
+%
+% Hint: When computing the gradient of the regularized cost function,
+% there're many possible vectorized solutions, but one solution
+% looks like:
+% grad = (unregularized gradient for logistic regression)
+% temp = theta;
+% temp(1) = 0; % because we don't add anything for j = 0
+% grad = grad + YOUR_CODE_HERE (using the temp variable)
+%
+
+
+
+
+
+
+
+
+
+
+% =============================================================
+
+grad = grad(:);
+
+end
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66 ex3/oneVsAll.m
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+function [all_theta] = oneVsAll(X, y, num_labels, lambda)
+%ONEVSALL trains multiple logistic regression classifiers and returns all
+%the classifiers in a matrix all_theta, where the i-th row of all_theta
+%corresponds to the classifier for label i
+% [all_theta] = ONEVSALL(X, y, num_labels, lambda) trains num_labels
+% logisitc regression classifiers and returns each of these classifiers
+% in a matrix all_theta, where the i-th row of all_theta corresponds
+% to the classifier for label i
+
+% Some useful variables
+m = size(X, 1);
+n = size(X, 2);
+
+% You need to return the following variables correctly
+all_theta = zeros(num_labels, n + 1);
+
+% Add ones to the X data matrix
+X = [ones(m, 1) X];
+
+% ====================== YOUR CODE HERE ======================
+% Instructions: You should complete the following code to train num_labels
+% logistic regression classifiers with regularization
+% parameter lambda.
+%
+% Hint: theta(:) will return a column vector.
+%
+% Hint: You can use y == c to obtain a vector of 1's and 0's that tell use
+% whether the ground truth is true/false for this class.
+%
+% Note: For this assignment, we recommend using fmincg to optimize the cost
+% function. It is okay to use a for-loop (for c = 1:num_labels) to
+% loop over the different classes.
+%
+% fmincg works similarly to fminunc, but is more efficient when we
+% are dealing with large number of parameters.
+%
+% Example Code for fmincg:
+%
+% % Set Initial theta
+% initial_theta = zeros(n + 1, 1);
+%
+% % Set options for fminunc
+% options = optimset('GradObj', 'on', 'MaxIter', 50);
+%
+% % Run fmincg to obtain the optimal theta
+% % This function will return theta and the cost
+% [theta] = ...
+% fmincg (@(t)(lrCostFunction(t, X, (y == c), lambda)), ...
+% initial_theta, options);
+%
+
+
+
+
+
+
+
+
+
+
+
+
+% =========================================================================
+
+
+end
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35 ex3/predict.m
@@ -0,0 +1,35 @@
+function p = predict(Theta1, Theta2, X)
+%PREDICT Predict the label of an input given a trained neural network
+% p = PREDICT(Theta1, Theta2, X) outputs the predicted label of X given the
+% trained weights of a neural network (Theta1, Theta2)
+
+% Useful values
+m = size(X, 1);
+num_labels = size(Theta2, 1);
+
+% You need to return the following variables correctly
+p = zeros(size(X, 1), 1);
+
+% ====================== YOUR CODE HERE ======================
+% Instructions: Complete the following code to make predictions using
+% your learned neural network. You should set p to a
+% vector containing labels between 1 to num_labels.
+%
+% Hint: The max function might come in useful. In particular, the max
+% function can also return the index of the max element, for more
+% information see 'help max'. If your examples are in rows, then, you
+% can use max(A, [], 2) to obtain the max for each row.
+%
+
+
+
+
+
+
+
+
+
+% =========================================================================
+
+
+end
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42 ex3/predictOneVsAll.m
@@ -0,0 +1,42 @@
+function p = predictOneVsAll(all_theta, X)
+%PREDICT Predict the label for a trained one-vs-all classifier. The labels
+%are in the range 1..K, where K = size(all_theta, 1).
+% p = PREDICTONEVSALL(all_theta, X) will return a vector of predictions
+% for each example in the matrix X. Note that X contains the examples in
+% rows. all_theta is a matrix where the i-th row is a trained logistic
+% regression theta vector for the i-th class. You should set p to a vector
+% of values from 1..K (e.g., p = [1; 3; 1; 2] predicts classes 1, 3, 1, 2
+% for 4 examples)
+
+m = size(X, 1);
+num_labels = size(all_theta, 1);
+
+% You need to return the following variables correctly
+p = zeros(size(X, 1), 1);
+
+% Add ones to the X data matrix
+X = [ones(m, 1) X];
+
+% ====================== YOUR CODE HERE ======================
+% Instructions: Complete the following code to make predictions using
+% your learned logistic regression parameters (one-vs-all).
+% You should set p to a vector of predictions (from 1 to
+% num_labels).
+%
+% Hint: This code can be done all vectorized using the max function.
+% In particular, the max function can also return the index of the
+% max element, for more information see 'help max'. If your examples
+% are in rows, then, you can use max(A, [], 2) to obtain the max
+% for each row.
+%
+
+
+
+
+
+
+
+% =========================================================================
+
+
+end
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6 ex3/sigmoid.m
@@ -0,0 +1,6 @@
+function g = sigmoid(z)
+%SIGMOID Compute sigmoid functoon
+% J = SIGMOID(z) computes the sigmoid of z.
+
+g = 1.0 ./ (1.0 + exp(-z));
+end
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531 ex3/submit.m
@@ -0,0 +1,531 @@
+function submit(partId)
+%SUBMIT Submit your code and output to the ml-class servers
+% SUBMIT() will connect to the ml-class server and submit your solution
+
+ fprintf('==\n== [ml-class] Submitting Solutions | Programming Exercise %s\n==\n', ...
+ homework_id());
+ if ~exist('partId', 'var') || isempty(partId)
+ partId = promptPart();
+ end
+
+ % Check valid partId
+ partNames = validParts();
+ if ~isValidPartId(partId)
+ fprintf('!! Invalid homework part selected.\n');
+ fprintf('!! Expected an integer from 1 to %d.\n', numel(partNames) + 1);
+ fprintf('!! Submission Cancelled\n');
+ return
+ end
+
+ if ~exist('ml_login_data.mat','file')
+ [login password] = loginPrompt();
+ save('ml_login_data.mat','login','password');
+ else
+ load('ml_login_data.mat');
+ [login password] = quickLogin(login, password);
+ save('ml_login_data.mat','login','password');
+ end
+
+ if isempty(login)
+ fprintf('!! Submission Cancelled\n');
+ return
+ end
+
+ fprintf('\n== Connecting to ml-class ... ');
+ if exist('OCTAVE_VERSION')
+ fflush(stdout);
+ end
+
+ % Setup submit list
+ if partId == numel(partNames) + 1
+ submitParts = 1:numel(partNames);
+ else
+ submitParts = [partId];
+ end
+
+ for s = 1:numel(submitParts)
+ thisPartId = submitParts(s);
+ [login, ch, signature, auxstring] = getChallenge(login, thisPartId);
+ if isempty(login) || isempty(ch) || isempty(signature)
+ % Some error occured, error string in first return element.
+ fprintf('\n!! Error: %s\n\n', login);
+ return
+ end
+
+ % Attempt Submission with Challenge
+ ch_resp = challengeResponse(login, password, ch);
+
+ [result, str] = submitSolution(login, ch_resp, thisPartId, ...
+ output(thisPartId, auxstring), source(thisPartId), signature);
+
+ partName = partNames{thisPartId};
+
+ fprintf('\n== [ml-class] Submitted Assignment %s - Part %d - %s\n', ...
+ homework_id(), thisPartId, partName);
+ fprintf('== %s\n', strtrim(str));
+
+ if exist('OCTAVE_VERSION')
+ fflush(stdout);
+ end
+ end
+
+end
+
+% ================== CONFIGURABLES FOR EACH HOMEWORK ==================
+
+function id = homework_id()
+ id = '3';
+end
+
+function [partNames] = validParts()
+ partNames = { 'Vectorized Logistic Regression ', ...
+ 'One-vs-all classifier training', ...
+ 'One-vs-all classifier prediction', ...
+ 'Neural network prediction function' ...
+ };
+end
+
+function srcs = sources()
+ % Separated by part
+ srcs = { { 'lrCostFunction.m' }, ...
+ { 'oneVsAll.m' }, ...
+ { 'predictOneVsAll.m' }, ...
+ { 'predict.m' } };
+end
+
+function out = output(partId, auxdata)
+ % Random Test Cases
+ X = [ones(20,1) (exp(1) * sin(1:1:20))' (exp(0.5) * cos(1:1:20))'];
+ y = sin(X(:,1) + X(:,2)) > 0;
+ Xm = [ -1 -1 ; -1 -2 ; -2 -1 ; -2 -2 ; ...
+ 1 1 ; 1 2 ; 2 1 ; 2 2 ; ...
+ -1 1 ; -1 2 ; -2 1 ; -2 2 ; ...
+ 1 -1 ; 1 -2 ; -2 -1 ; -2 -2 ];
+ ym = [ 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 ]';
+ t1 = sin(reshape(1:2:24, 4, 3));
+ t2 = cos(reshape(1:2:40, 4, 5));
+
+ if partId == 1
+ [J, grad] = lrCostFunction([0.25 0.5 -0.5]', X, y, 0.1);
+ out = sprintf('%0.5f ', J);
+ out = [out sprintf('%0.5f ', grad)];
+ elseif partId == 2
+ out = sprintf('%0.5f ', oneVsAll(Xm, ym, 4, 0.1));
+ elseif partId == 3
+ out = sprintf('%0.5f ', predictOneVsAll(t1, Xm));
+ elseif partId == 4
+ out = sprintf('%0.5f ', predict(t1, t2, Xm));
+ end
+end
+
+% ====================== SERVER CONFIGURATION ===========================
+
+% ***************** REMOVE -staging WHEN YOU DEPLOY *********************
+function url = site_url()
+ url = 'http://www.coursera.org/ml';
+end
+
+function url = challenge_url()
+ url = [site_url() '/assignment/challenge'];
+end
+
+function url = submit_url()
+ url = [site_url() '/assignment/submit'];
+end
+
+% ========================= CHALLENGE HELPERS =========================
+
+function src = source(partId)
+ src = '';
+ src_files = sources();
+ if partId <= numel(src_files)
+ flist = src_files{partId};
+ for i = 1:numel(flist)
+ fid = fopen(flist{i});
+ if (fid == -1)
+ error('Error opening %s (is it missing?)', flist{i});
+ end
+ line = fgets(fid);
+ while ischar(line)
+ src = [src line];
+ line = fgets(fid);
+ end
+ fclose(fid);
+ src = [src '||||||||'];
+ end
+ end
+end
+
+function ret = isValidPartId(partId)
+ partNames = validParts();
+ ret = (~isempty(partId)) && (partId >= 1) && (partId <= numel(partNames) + 1);
+end
+
+function partId = promptPart()
+ fprintf('== Select which part(s) to submit:\n', ...
+ homework_id());
+ partNames = validParts();
+ srcFiles = sources();
+ for i = 1:numel(partNames)
+ fprintf('== %d) %s [', i, partNames{i});
+ fprintf(' %s ', srcFiles{i}{:});
+ fprintf(']\n');
+ end
+ fprintf('== %d) All of the above \n==\nEnter your choice [1-%d]: ', ...
+ numel(partNames) + 1, numel(partNames) + 1);
+ selPart = input('', 's');
+ partId = str2num(selPart);
+ if ~isValidPartId(partId)
+ partId = -1;
+ end
+end
+
+function [email,ch,signature,auxstring] = getChallenge(email, part)
+ str = urlread(challenge_url(), 'post', {'email_address', email, 'assignment_part_sid', [homework_id() '-' num2str(part)], 'response_encoding', 'delim'});
+
+ str = strtrim(str);
+ r = struct;
+ while(numel(str) > 0)
+ [f, str] = strtok (str, '|');
+ [v, str] = strtok (str, '|');
+ r = setfield(r, f, v);
+ end
+
+ email = getfield(r, 'email_address');
+ ch = getfield(r, 'challenge_key');
+ signature = getfield(r, 'state');
+ auxstring = getfield(r, 'challenge_aux_data');
+end
+
+function [result, str] = submitSolution(email, ch_resp, part, output, ...
+ source, signature)
+
+ params = {'assignment_part_sid', [homework_id() '-' num2str(part)], ...
+ 'email_address', email, ...
+ 'submission', base64encode(output, ''), ...
+ 'submission_aux', base64encode(source, ''), ...
+ 'challenge_response', ch_resp, ...
+ 'state', signature};
+
+ str = urlread(submit_url(), 'post', params);
+
+ % Parse str to read for success / failure
+ result = 0;
+
+end
+
+% =========================== LOGIN HELPERS ===========================
+
+function [login password] = loginPrompt()
+ % Prompt for password
+ [login password] = basicPrompt();
+
+ if isempty(login) || isempty(password)
+ login = []; password = [];
+ end
+end
+
+
+function [login password] = basicPrompt()
+ login = input('Login (Email address): ', 's');
+ password = input('Password: ', 's');
+end
+
+function [login password] = quickLogin(login,password)
+ disp(['You are currently logged in as ' login '.']);
+ cont_token = input('Is this you? (y/n - type n to reenter password)','s');
+ if(isempty(cont_token) || cont_token(1)=='Y'||cont_token(1)=='y')
+ return;
+ else
+ [login password] = loginPrompt();
+ end
+end
+
+function [str] = challengeResponse(email, passwd, challenge)
+ str = sha1([challenge passwd]);
+end
+
+% =============================== SHA-1 ================================
+
+function hash = sha1(str)
+
+ % Initialize variables
+ h0 = uint32(1732584193);
+ h1 = uint32(4023233417);
+ h2 = uint32(2562383102);
+ h3 = uint32(271733878);
+ h4 = uint32(3285377520);
+
+ % Convert to word array
+ strlen = numel(str);
+
+ % Break string into chars and append the bit 1 to the message
+ mC = [double(str) 128];
+ mC = [mC zeros(1, 4-mod(numel(mC), 4), 'uint8')];
+
+ numB = strlen * 8;
+ if exist('idivide')
+ numC = idivide(uint32(numB + 65), 512, 'ceil');
+ else
+ numC = ceil(double(numB + 65)/512);
+ end
+ numW = numC * 16;
+ mW = zeros(numW, 1, 'uint32');
+
+ idx = 1;
+ for i = 1:4:strlen + 1
+ mW(idx) = bitor(bitor(bitor( ...
+ bitshift(uint32(mC(i)), 24), ...
+ bitshift(uint32(mC(i+1)), 16)), ...
+ bitshift(uint32(mC(i+2)), 8)), ...
+ uint32(mC(i+3)));
+ idx = idx + 1;
+ end
+
+ % Append length of message
+ mW(numW - 1) = uint32(bitshift(uint64(numB), -32));
+ mW(numW) = uint32(bitshift(bitshift(uint64(numB), 32), -32));
+
+ % Process the message in successive 512-bit chs
+ for cId = 1 : double(numC)
+ cSt = (cId - 1) * 16 + 1;
+ cEnd = cId * 16;
+ ch = mW(cSt : cEnd);
+
+ % Extend the sixteen 32-bit words into eighty 32-bit words
+ for j = 17 : 80
+ ch(j) = ch(j - 3);
+ ch(j) = bitxor(ch(j), ch(j - 8));
+ ch(j) = bitxor(ch(j), ch(j - 14));
+ ch(j) = bitxor(ch(j), ch(j - 16));
+ ch(j) = bitrotate(ch(j), 1);
+ end
+
+ % Initialize hash value for this ch
+ a = h0;
+ b = h1;
+ c = h2;
+ d = h3;
+ e = h4;
+
+ % Main loop
+ for i = 1 : 80
+ if(i >= 1 && i <= 20)
+ f = bitor(bitand(b, c), bitand(bitcmp(b), d));
+ k = uint32(1518500249);
+ elseif(i >= 21 && i <= 40)
+ f = bitxor(bitxor(b, c), d);
+ k = uint32(1859775393);
+ elseif(i >= 41 && i <= 60)
+ f = bitor(bitor(bitand(b, c), bitand(b, d)), bitand(c, d));
+ k = uint32(2400959708);
+ elseif(i >= 61 && i <= 80)
+ f = bitxor(bitxor(b, c), d);
+ k = uint32(3395469782);
+ end
+
+ t = bitrotate(a, 5);
+ t = bitadd(t, f);
+ t = bitadd(t, e);
+ t = bitadd(t, k);
+ t = bitadd(t, ch(i));
+ e = d;
+ d = c;
+ c = bitrotate(b, 30);
+ b = a;
+ a = t;
+
+ end
+ h0 = bitadd(h0, a);
+ h1 = bitadd(h1, b);
+ h2 = bitadd(h2, c);
+ h3 = bitadd(h3, d);
+ h4 = bitadd(h4, e);
+
+ end
+
+ hash = reshape(dec2hex(double([h0 h1 h2 h3 h4]), 8)', [1 40]);
+
+ hash = lower(hash);
+
+end
+
+function ret = bitadd(iA, iB)
+ ret = double(iA) + double(iB);
+ ret = bitset(ret, 33, 0);
+ ret = uint32(ret);
+end
+
+function ret = bitrotate(iA, places)
+ t = bitshift(iA, places - 32);
+ ret = bitshift(iA, places);
+ ret = bitor(ret, t);
+end
+
+% =========================== Base64 Encoder ============================
+% Thanks to Peter John Acklam
+%
+
+function y = base64encode(x, eol)
+%BASE64ENCODE Perform base64 encoding on a string.
+%
+% BASE64ENCODE(STR, EOL) encode the given string STR. EOL is the line ending
+% sequence to use; it is optional and defaults to '\n' (ASCII decimal 10).
+% The returned encoded string is broken into lines of no more than 76
+% characters each, and each line will end with EOL unless it is empty. Let
+% EOL be empty if you do not want the encoded string broken into lines.
+%
+% STR and EOL don't have to be strings (i.e., char arrays). The only
+% requirement is that they are vectors containing values in the range 0-255.
+%
+% This function may be used to encode strings into the Base64 encoding
+% specified in RFC 2045 - MIME (Multipurpose Internet Mail Extensions). The
+% Base64 encoding is designed to represent arbitrary sequences of octets in a
+% form that need not be humanly readable. A 65-character subset
+% ([A-Za-z0-9+/=]) of US-ASCII is used, enabling 6 bits to be represented per
+% printable character.
+%
+% Examples
+% --------
+%
+% If you want to encode a large file, you should encode it in chunks that are
+% a multiple of 57 bytes. This ensures that the base64 lines line up and
+% that you do not end up with padding in the middle. 57 bytes of data fills
+% one complete base64 line (76 == 57*4/3):
+%
+% If ifid and ofid are two file identifiers opened for reading and writing,
+% respectively, then you can base64 encode the data with
+%
+% while ~feof(ifid)
+% fwrite(ofid, base64encode(fread(ifid, 60*57)));
+% end
+%
+% or, if you have enough memory,
+%
+% fwrite(ofid, base64encode(fread(ifid)));
+%
+% See also BASE64DECODE.
+
+% Author: Peter John Acklam
+% Time-stamp: 2004-02-03 21:36:56 +0100
+% E-mail: pjacklam@online.no
+% URL: http://home.online.no/~pjacklam
+
+ if isnumeric(x)
+ x = num2str(x);
+ end
+
+ % make sure we have the EOL value
+ if nargin < 2
+ eol = sprintf('\n');
+ else
+ if sum(size(eol) > 1) > 1
+ error('EOL must be a vector.');
+ end
+ if any(eol(:) > 255)
+ error('EOL can not contain values larger than 255.');
+ end
+ end
+
+ if sum(size(x) > 1) > 1
+ error('STR must be a vector.');
+ end
+
+ x = uint8(x);
+ eol = uint8(eol);
+
+ ndbytes = length(x); % number of decoded bytes
+ nchunks = ceil(ndbytes / 3); % number of chunks/groups
+ nebytes = 4 * nchunks; % number of encoded bytes
+
+ % add padding if necessary, to make the length of x a multiple of 3
+ if rem(ndbytes, 3)
+ x(end+1 : 3*nchunks) = 0;
+ end
+
+ x = reshape(x, [3, nchunks]); % reshape the data
+ y = repmat(uint8(0), 4, nchunks); % for the encoded data
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % Split up every 3 bytes into 4 pieces
+ %
+ % aaaaaabb bbbbcccc ccdddddd
+ %
+ % to form
+ %
+ % 00aaaaaa 00bbbbbb 00cccccc 00dddddd
+ %
+ y(1,:) = bitshift(x(1,:), -2); % 6 highest bits of x(1,:)
+
+ y(2,:) = bitshift(bitand(x(1,:), 3), 4); % 2 lowest bits of x(1,:)
+ y(2,:) = bitor(y(2,:), bitshift(x(2,:), -4)); % 4 highest bits of x(2,:)
+
+ y(3,:) = bitshift(bitand(x(2,:), 15), 2); % 4 lowest bits of x(2,:)
+ y(3,:) = bitor(y(3,:), bitshift(x(3,:), -6)); % 2 highest bits of x(3,:)
+
+ y(4,:) = bitand(x(3,:), 63); % 6 lowest bits of x(3,:)
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % Now perform the following mapping
+ %
+ % 0 - 25 -> A-Z
+ % 26 - 51 -> a-z
+ % 52 - 61 -> 0-9
+ % 62 -> +
+ % 63 -> /
+ %
+ % We could use a mapping vector like
+ %
+ % ['A':'Z', 'a':'z', '0':'9', '+/']
+ %
+ % but that would require an index vector of class double.
+ %
+ z = repmat(uint8(0), size(y));
+ i = y <= 25; z(i) = 'A' + double(y(i));
+ i = 26 <= y & y <= 51; z(i) = 'a' - 26 + double(y(i));
+ i = 52 <= y & y <= 61; z(i) = '0' - 52 + double(y(i));
+ i = y == 62; z(i) = '+';
+ i = y == 63; z(i) = '/';
+ y = z;
+
+ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+ % Add padding if necessary.
+ %
+ npbytes = 3 * nchunks - ndbytes; % number of padding bytes
+ if npbytes
+ y(end-npbytes+1 : end) = '='; % '=' is used for padding
+ end
+
+ if isempty(eol)
+
+ % reshape to a row vector
+ y = reshape(y, [1, nebytes]);
+
+ else
+
+ nlines = ceil(nebytes / 76); % number of lines
+ neolbytes = length(eol); % number of bytes in eol string
+
+ % pad data so it becomes a multiple of 76 elements
+ y = [y(:) ; zeros(76 * nlines - numel(y), 1)];
+ y(nebytes + 1 : 76 * nlines) = 0;
+ y = reshape(y, 76, nlines);
+
+ % insert eol strings
+ eol = eol(:);
+ y(end + 1 : end + neolbytes, :) = eol(:, ones(1, nlines));
+
+ % remove padding, but keep the last eol string
+ m = nebytes + neolbytes * (nlines - 1);
+ n = (76+neolbytes)*nlines - neolbytes;
+ y(m+1 : n) = '';
+
+ % extract and reshape to row vector
+ y = reshape(y, 1, m+neolbytes);
+
+ end
+
+ % output is a character array
+ y = char(y);
+
+end
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