Complete and hand in this completed worksheet (including its outputs and any supporting code outside of the worksheet) with your assignment submission. For more details see the assignments page on the course website.
In this exercise you will:
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# Run some setup code for this notebook.
import random
import numpy as np
from cs231n.data_utils import load_CIFAR10
import matplotlib.pyplot as plt
# This is a bit of magic to make matplotlib figures appear inline in the
# notebook rather than in a new window.
%matplotlib inline
plt.rcParams['figure.figsize'] = (10.0, 8.0) # set default size of plots
plt.rcParams['image.interpolation'] = 'nearest'
plt.rcParams['image.cmap'] = 'gray'
# Some more magic so that the notebook will reload external python modules;
# see http://stackoverflow.com/questions/1907993/autoreload-of-modules-in-ipython
%load_ext autoreload
%autoreload 2
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# Load the raw CIFAR-10 data.
cifar10_dir = 'cs231n/datasets/cifar-10-batches-py'
X_train, y_train, X_test, y_test = load_CIFAR10(cifar10_dir)
# As a sanity check, we print out the size of the training and test data.
print 'Training data shape: ', X_train.shape
print 'Training labels shape: ', y_train.shape
print 'Test data shape: ', X_test.shape
print 'Test labels shape: ', y_test.shape
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# Visualize some examples from the dataset.
# We show a few examples of training images from each class.
classes = ['plane', 'car', 'bird', 'cat', 'deer', 'dog', 'frog', 'horse', 'ship', 'truck']
num_classes = len(classes)
samples_per_class = 7
for y, cls in enumerate(classes):
idxs = np.flatnonzero(y_train == y)
idxs = np.random.choice(idxs, samples_per_class, replace=False)
for i, idx in enumerate(idxs):
plt_idx = i * num_classes + y + 1
plt.subplot(samples_per_class, num_classes, plt_idx)
plt.imshow(X_train[idx].astype('uint8'))
plt.axis('off')
if i == 0:
plt.title(cls)
plt.show()
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# Split the data into train, val, and test sets. In addition we will
# create a small development set as a subset of the training data;
# we can use this for development so our code runs faster.
num_training = 49000
num_validation = 1000
num_test = 1000
num_dev = 500
# Our validation set will be num_validation points from the original
# training set.
mask = range(num_training, num_training + num_validation)
X_val = X_train[mask]
y_val = y_train[mask]
# Our training set will be the first num_train points from the original
# training set.
mask = range(num_training)
X_train = X_train[mask]
y_train = y_train[mask]
# We will also make a development set, which is a small subset of
# the training set.
mask = np.random.choice(num_training, num_dev, replace=False)
X_dev = X_train[mask]
y_dev = y_train[mask]
# We use the first num_test points of the original test set as our
# test set.
mask = range(num_test)
X_test = X_test[mask]
y_test = y_test[mask]
print 'Train data shape: ', X_train.shape
print 'Train labels shape: ', y_train.shape
print 'Validation data shape: ', X_val.shape
print 'Validation labels shape: ', y_val.shape
print 'Test data shape: ', X_test.shape
print 'Test labels shape: ', y_test.shape
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# Preprocessing: reshape the image data into rows
X_train = np.reshape(X_train, (X_train.shape[0], -1))
X_val = np.reshape(X_val, (X_val.shape[0], -1))
X_test = np.reshape(X_test, (X_test.shape[0], -1))
X_dev = np.reshape(X_dev, (X_dev.shape[0], -1))
# As a sanity check, print out the shapes of the data
print 'Training data shape: ', X_train.shape
print 'Validation data shape: ', X_val.shape
print 'Test data shape: ', X_test.shape
print 'dev data shape: ', X_dev.shape
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# Preprocessing: subtract the mean image
# first: compute the image mean based on the training data
mean_image = np.mean(X_train, axis=0)
print mean_image[:10] # print a few of the elements
plt.figure(figsize=(4,4))
plt.imshow(mean_image.reshape((32,32,3)).astype('uint8')) # visualize the mean image
plt.show()
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# second: subtract the mean image from train and test data
X_train -= mean_image
X_val -= mean_image
X_test -= mean_image
X_dev -= mean_image
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# third: append the bias dimension of ones (i.e. bias trick) so that our SVM
# only has to worry about optimizing a single weight matrix W.
X_train = np.hstack([X_train, np.ones((X_train.shape[0], 1))])
X_val = np.hstack([X_val, np.ones((X_val.shape[0], 1))])
X_test = np.hstack([X_test, np.ones((X_test.shape[0], 1))])
X_dev = np.hstack([X_dev, np.ones((X_dev.shape[0], 1))])
print X_train.shape, X_val.shape, X_test.shape, X_dev.shape
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# Evaluate the naive implementation of the loss we provided for you:
from cs231n.classifiers.linear_svm import svm_loss_naive
import time
# generate a random SVM weight matrix of small numbers
W = np.random.randn(3073, 10) * 0.0001
loss, grad = svm_loss_naive(W, X_dev, y_dev, 0.00001)
print 'loss: %f' % (loss, )
print(grad)
The grad returned from the function above is right now all zero. Derive and implement the gradient for the SVM cost function and implement it inline inside the function svm_loss_naive. You will find it helpful to interleave your new code inside the existing function.
To check that you have correctly implemented the gradient correctly, you can numerically estimate the gradient of the loss function and compare the numeric estimate to the gradient that you computed. We have provided code that does this for you:
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# Once you've implemented the gradient, recompute it with the code below
# and gradient check it with the function we provided for you
# Compute the loss and its gradient at W.
loss, grad = svm_loss_naive(W, X_dev, y_dev, 0.0)
# Numerically compute the gradient along several randomly chosen dimensions, and
# compare them with your analytically computed gradient. The numbers should match
# almost exactly along all dimensions.
from cs231n.gradient_check import grad_check_sparse
f = lambda w: svm_loss_naive(w, X_dev, y_dev, 0.0)[0]
grad_numerical = grad_check_sparse(f, W, grad)
# do the gradient check once again with regularization turned on
# you didn't forget the regularization gradient did you?
loss, grad = svm_loss_naive(W, X_dev, y_dev, 1e2)
f = lambda w: svm_loss_naive(w, X_dev, y_dev, 1e2)[0]
grad_numerical = grad_check_sparse(f, W, grad)
It is possible that once in a while a dimension in the gradcheck will not match exactly. What could such a discrepancy be caused by? Is it a reason for concern? What is a simple example in one dimension where a gradient check could fail? Hint: the SVM loss function is not strictly speaking differentiable
Your Answer: This is in the case of when, analytically speaking, there is a difference in slopes because the function is not differentiable at possibly n(feature) points. At when y = 0 line intersects with any of the features, and when any of the features(in graphical sense these are the slopes of the lines) and weights create lines that intersect with each other above the y = 0 line. Since we are taking a piecewise function the slope may not match in those cases if one numerically computed slope(would likely lead to a subgradient) is compared to a calculus computed slope(would lead to the slope of one of the piecewise functions).
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# Next implement the function svm_loss_vectorized; for now only compute the loss;
# we will implement the gradient in a moment.
tic = time.time()
loss_naive, grad_naive = svm_loss_naive(W, X_dev, y_dev, 0.00001)
toc = time.time()
print 'Naive loss: %e computed in %fs' % (loss_naive, toc - tic)
from cs231n.classifiers.linear_svm import svm_loss_vectorized
tic = time.time()
loss_vectorized, _ = svm_loss_vectorized(W, X_dev, y_dev, 0.00001)
toc = time.time()
print 'Vectorized loss: %e computed in %fs' % (loss_vectorized, toc - tic)
# The losses should match but your vectorized implementation should be much faster.
print 'difference: %f' % (loss_naive - loss_vectorized)
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# Complete the implementation of svm_loss_vectorized, and compute the gradient
# of the loss function in a vectorized way.
# The naive implementation and the vectorized implementation should match, but
# the vectorized version should still be much faster.
tic = time.time()
_, grad_naive = svm_loss_naive(W, X_dev, y_dev, 0.00001)
toc = time.time()
print 'Naive loss and gradient: computed in %fs' % (toc - tic)
tic = time.time()
_, grad_vectorized = svm_loss_vectorized(W, X_dev, y_dev, 0.00001)
toc = time.time()
print 'Vectorized loss and gradient: computed in %fs' % (toc - tic)
# The loss is a single number, so it is easy to compare the values computed
# by the two implementations. The gradient on the other hand is a matrix, so
# we use the Frobenius norm to compare them.
difference = np.linalg.norm(grad_naive - grad_vectorized, ord='fro')
print 'difference: %f' % difference
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# In the file linear_classifier.py, implement SGD in the function
# LinearClassifier.train() and then run it with the code below.
from cs231n.classifiers import LinearSVM
svm = LinearSVM()
tic = time.time()
loss_hist = svm.train(X_train, y_train, learning_rate=1e-7, reg=5e4,
num_iters=1500, verbose=True)
toc = time.time()
print 'That took %fs' % (toc - tic)
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# A useful debugging strategy is to plot the loss as a function of
# iteration number:
plt.plot(loss_hist)
plt.xlabel('Iteration number')
plt.ylabel('Loss value')
plt.show()
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# Write the LinearSVM.predict function and evaluate the performance on both the
# training and validation set
y_train_pred = svm.predict(X_train)
print 'training accuracy: %f' % (np.mean(y_train == y_train_pred), )
y_val_pred = svm.predict(X_val)
print 'validation accuracy: %f' % (np.mean(y_val == y_val_pred), )
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# Use the validation set to tune hyperparameters (regularization strength and
# learning rate). You should experiment with different ranges for the learning
# rates and regularization strengths; if you are careful you should be able to
# get a classification accuracy of about 0.4 on the validation set.
learning_rates = [1e-7, 5e-5]
regularization_strengths = [5e4, 1e5]
# results is dictionary mapping tuples of the form
# (learning_rate, regularization_strength) to tuples of the form
# (training_accuracy, validation_accuracy). The accuracy is simply the fraction
# of data points that are correctly classified.
results = {}
best_val = -1 # The highest validation accuracy that we have seen so far.
best_svm = None # The LinearSVM object that achieved the highest validation rate.
################################################################################
# TODO: #
# Write code that chooses the best hyperparameters by tuning on the validation #
# set. For each combination of hyperparameters, train a linear SVM on the #
# training set, compute its accuracy on the training and validation sets, and #
# store these numbers in the results dictionary. In addition, store the best #
# validation accuracy in best_val and the LinearSVM object that achieves this #
# accuracy in best_svm. #
# #
# Hint: You should use a small value for num_iters as you develop your #
# validation code so that the SVMs don't take much time to train; once you are #
# confident that your validation code works, you should rerun the validation #
# code with a larger value for num_iters. #
################################################################################
for lr in learning_rates:
for rs in regularization_strengths:
svm = LinearSVM()
loss_hist = svm.train(X_train, y_train, learning_rate=lr, reg=rs,
num_iters=1500, verbose=True)
training_accuracy = np.mean(svm.predict(X_train) == y_train)
validation_accuracy = np.mean(svm.predict(X_val) == y_val)
if best_val < validation_accuracy:
best_val = validation_accuracy
best_svm = svm
results[(lr, rs)] = (training_accuracy, validation_accuracy)
################################################################################
# END OF YOUR CODE #
################################################################################
# Print out results.
for lr, reg in sorted(results):
train_accuracy, val_accuracy = results[(lr, reg)]
print 'lr %e reg %e train accuracy: %f val accuracy: %f' % (
lr, reg, train_accuracy, val_accuracy)
print 'best validation accuracy achieved during cross-validation: %f' % best_val
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# Visualize the cross-validation results
import math
x_scatter = [math.log10(x[0]) for x in results]
y_scatter = [math.log10(x[1]) for x in results]
# plot training accuracy
marker_size = 100
colors = [results[x][0] for x in results]
plt.subplot(2, 1, 1)
plt.scatter(x_scatter, y_scatter, marker_size, c=colors)
plt.colorbar()
plt.xlabel('log learning rate')
plt.ylabel('log regularization strength')
plt.title('CIFAR-10 training accuracy')
# plot validation accuracy
colors = [results[x][1] for x in results] # default size of markers is 20
plt.subplot(2, 1, 2)
plt.scatter(x_scatter, y_scatter, marker_size, c=colors)
plt.colorbar()
plt.xlabel('log learning rate')
plt.ylabel('log regularization strength')
plt.title('CIFAR-10 validation accuracy')
plt.show()
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# Evaluate the best svm on test set
y_test_pred = best_svm.predict(X_test)
test_accuracy = np.mean(y_test == y_test_pred)
print 'linear SVM on raw pixels final test set accuracy: %f' % test_accuracy
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# Visualize the learned weights for each class.
# Depending on your choice of learning rate and regularization strength, these may
# or may not be nice to look at.
w = best_svm.W[:-1,:] # strip out the bias
w = w.reshape(32, 32, 3, 10)
w_min, w_max = np.min(w), np.max(w)
classes = ['plane', 'car', 'bird', 'cat', 'deer', 'dog', 'frog', 'horse', 'ship', 'truck']
for i in xrange(10):
plt.subplot(2, 5, i + 1)
# Rescale the weights to be between 0 and 255
wimg = 255.0 * (w[:, :, :, i].squeeze() - w_min) / (w_max - w_min)
plt.imshow(wimg.astype('uint8'))
plt.axis('off')
plt.title(classes[i])
Describe what your visualized SVM weights look like, and offer a brief explanation for why they look they way that they do.
Your answer: The visualized SVM weights look like a merged version of multiple images of the same class. If you look at the horse picture, it's clear there are 2 horse heads situated in the same image. This is to accomodate for all orientations of the horse. However, because linear SVM can only accomodate 1 picture formation for each class, the kNN results of these generated images would still not yield a high % of accuracy.
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