(Your student-id is the number shown on your student card.)
E.g. if you work with 3 people, the notebook should be named: 12301230_3434343_1238938934_lab1.ipynb.
This will be parsed by a regexp, so please double check your filename.
Before you turn this problem in, please make sure everything runs correctly. First, restart the kernel (in the menubar, select Kernel$\rightarrow$Restart) and then run all cells (in the menubar, select Cell$\rightarrow$Run All).
Make sure you fill in any place that says YOUR CODE HERE or "YOUR ANSWER HERE", as well as your names and email adresses below.
In [ ]:
NAME = ""
NAME2 = ""
NAME3 = ""
EMAIL = ""
EMAIL2 = ""
EMAIL3 = ""
Notes on implementation:
$\newcommand{\bx}{\mathbf{x}}$ $\newcommand{\bw}{\mathbf{w}}$ $\newcommand{\bt}{\mathbf{t}}$ $\newcommand{\by}{\mathbf{y}}$ $\newcommand{\bm}{\mathbf{m}}$ $\newcommand{\bb}{\mathbf{b}}$ $\newcommand{\bS}{\mathbf{S}}$ $\newcommand{\ba}{\mathbf{a}}$ $\newcommand{\bz}{\mathbf{z}}$ $\newcommand{\bv}{\mathbf{v}}$ $\newcommand{\bq}{\mathbf{q}}$ $\newcommand{\bp}{\mathbf{p}}$ $\newcommand{\bh}{\mathbf{h}}$ $\newcommand{\bI}{\mathbf{I}}$ $\newcommand{\bX}{\mathbf{X}}$ $\newcommand{\bT}{\mathbf{T}}$ $\newcommand{\bPhi}{\mathbf{\Phi}}$ $\newcommand{\bW}{\mathbf{W}}$ $\newcommand{\bV}{\mathbf{V}}$
In [3]:
%pylab inline
plt.rcParams["figure.figsize"] = [9,5]
Scenario: you have a friend with one big problem: she's completely blind. You decided to help her: she has a special smartphone for blind people, and you are going to develop a mobile phone app that can do machine vision using the mobile camera: converting a picture (from the camera) to the meaning of the image. You decide to start with an app that can read handwritten digits, i.e. convert an image of handwritten digits to text (e.g. it would enable her to read precious handwritten phone numbers).
A key building block for such an app would be a function predict_digit(x) that returns the digit class of an image patch $\bx$. Since hand-coding this function is highly non-trivial, you decide to solve this problem using machine learning, such that the internal parameters of this function are automatically learned using machine learning techniques.
The dataset you're going to use for this is the MNIST handwritten digits dataset (http://yann.lecun.com/exdb/mnist/). You can download the data with scikit learn, and load it as follows:
In [5]:
from sklearn.datasets import fetch_mldata
# Fetch the data
mnist = fetch_mldata('MNIST original')
data, target = mnist.data, mnist.target.astype('int')
# Shuffle
indices = np.arange(len(data))
np.random.seed(123)
np.random.shuffle(indices)
data, target = data[indices].astype('float32'), target[indices]
# Normalize the data between 0.0 and 1.0:
data /= 255.
# Split
x_train, x_valid, x_test = data[:50000], data[50000:60000], data[60000: 70000]
t_train, t_valid, t_test = target[:50000], target[50000:60000], target[60000: 70000]
MNIST consists of small 28 by 28 pixel images of written digits (0-9). We split the dataset into a training, validation and testing arrays. The variables x_train, x_valid and x_test are $N \times M$ matrices, where $N$ is the number of datapoints in the respective set, and $M = 28^2 = 784$ is the dimensionality of the data. The second set of variables t_train, t_valid and t_test contain the corresponding $N$-dimensional vector of integers, containing the true class labels.
Here's a visualisation of the first 8 digits of the trainingset:
In [6]:
def plot_digits(data, num_cols, targets=None, shape=(28,28)):
num_digits = data.shape[0]
num_rows = int(num_digits/num_cols)
for i in range(num_digits):
plt.subplot(num_rows, num_cols, i+1)
plt.imshow(data[i].reshape(shape), interpolation='none', cmap='Greys')
if targets is not None:
plt.title(int(targets[i]))
plt.colorbar()
plt.axis('off')
plt.tight_layout()
plt.show()
plot_digits(x_train[0:40000:5000], num_cols=4, targets=t_train[0:40000:5000])
In multiclass logistic regression, the conditional probability of class label $j$ given the image $\bx$ for some datapoint is given by:
$ \log p(t = j \;|\; \bx, \bb, \bW) = \log q_j - \log Z$
where $\log q_j = \bw_j^T \bx + b_j$ (the log of the unnormalized probability of the class $j$), and $Z = \sum_k q_k$ is the normalizing factor. $\bw_j$ is the $j$-th column of $\bW$ (a matrix of size $784 \times 10$) corresponding to the class label, $b_j$ is the $j$-th element of $\bb$.
Given an input image, the multiclass logistic regression model first computes the intermediate vector $\log \bq$ (of size $10 \times 1$), using $\log q_j = \bw_j^T \bx + b_j$, containing the unnormalized log-probabilities per class.
The unnormalized probabilities are then normalized by $Z$ such that $\sum_j p_j = \sum_j \exp(\log p_j) = 1$. This is done by $\log p_j = \log q_j - \log Z$ where $Z = \sum_i \exp(\log q_i)$. This is known as the softmax transformation, and is also used as a last layer of many classifcation neural network models, to ensure that the output of the network is a normalized distribution, regardless of the values of second-to-last layer ($\log \bq$)
Warning: when computing $\log Z$, you are likely to encounter numerical problems. Save yourself countless hours of debugging and learn the log-sum-exp trick.
The network's output $\log \bp$ of size $10 \times 1$ then contains the conditional log-probabilities $\log p(t = j \;|\; \bx, \bb, \bW)$ for each digit class $j$. In summary, the computations are done in this order:
$\bx \rightarrow \log \bq \rightarrow Z \rightarrow \log \bp$
Given some dataset with $N$ independent, identically distributed datapoints, the log-likelihood is given by:
$ \mathcal{L}(\bb, \bW) = \sum_{n=1}^N \mathcal{L}^{(n)}$
where we use $\mathcal{L}^{(n)}$ to denote the partial log-likelihood evaluated over a single datapoint. It is important to see that the log-probability of the class label $t^{(n)}$ given the image, is given by the $t^{(n)}$-th element of the network's output $\log \bp$, denoted by $\log p_{t^{(n)}}$:
$\mathcal{L}^{(n)} = \log p(t = t^{(n)} \;|\; \bx = \bx^{(n)}, \bb, \bW) = \log p_{t^{(n)}} = \log q_{t^{(n)}} - \log Z^{(n)}$
where $\bx^{(n)}$ and $t^{(n)}$ are the input (image) and class label (integer) of the $n$-th datapoint, and $Z^{(n)}$ is the normalizing constant for the distribution over $t^{(n)}$.
Derive the equations for computing the (first) partial derivatives of the log-likelihood w.r.t. all the parameters, evaluated at a single datapoint $n$.
You should start deriving the equations for $\frac{\partial \mathcal{L}^{(n)}}{\partial \log q_j}$ for each $j$. For clarity, we'll use the shorthand $\delta^q_j = \frac{\partial \mathcal{L}^{(n)}}{\partial \log q_j}$.
For $j = t^{(n)}$: $ \delta^q_j = \frac{\partial \mathcal{L}^{(n)}}{\partial \log p_j} \frac{\partial \log p_j}{\partial \log q_j}
For $j \neq t^{(n)}$: $ \delta^q_j = \frac{\partial \mathcal{L}^{(n)}}{\partial \log Z} \frac{\partial \log Z}{\partial Z} \frac{\partial Z}{\partial \log q_j} = - \frac{\partial \log Z}{\partial Z} \frac{\partial Z}{\partial \log q_j} $
Complete the above derivations for $\delta^q_j$ by furtherly developing $\frac{\partial \log Z}{\partial Z}$ and $\frac{\partial Z}{\partial \log q_j}$. Both are quite simple. For these it doesn't matter whether $j = t^{(n)}$ or not.
For $j = t^{(n)}$: \begin{align} \delta^q_j &= \end{align} For $j \neq t^{(n)}$: \begin{align} \delta^q_j &= \end{align} YOUR ANSWER HERE
Given your equations for computing the gradients $\delta^q_j$ it should be quite straightforward to derive the equations for the gradients of the parameters of the model, $\frac{\partial \mathcal{L}^{(n)}}{\partial W_{ij}}$ and $\frac{\partial \mathcal{L}^{(n)}}{\partial b_j}$. The gradients for the biases $\bb$ are given by:
$ \frac{\partial \mathcal{L}^{(n)}}{\partial b_j} = \frac{\partial \mathcal{L}^{(n)}}{\partial \log q_j} \frac{\partial \log q_j}{\partial b_j} = \delta^q_j \cdot 1 = \delta^q_j $
The equation above gives the derivative of $\mathcal{L}^{(n)}$ w.r.t. a single element of $\bb$, so the vector $\nabla_\bb \mathcal{L}^{(n)}$ with all derivatives of $\mathcal{L}^{(n)}$ w.r.t. the bias parameters $\bb$ is:
$ \nabla_\bb \mathcal{L}^{(n)} = \mathbf{\delta}^q $
where $\mathbf{\delta}^q$ denotes the vector of size $10 \times 1$ with elements $\mathbf{\delta}_j^q$.
The (not fully developed) equation for computing the derivative of $\mathcal{L}^{(n)}$ w.r.t. a single element $W_{ij}$ of $\bW$ is:
$ \frac{\partial \mathcal{L}^{(n)}}{\partial W_{ij}} = \frac{\partial \mathcal{L}^{(n)}}{\partial \log q_j} \frac{\partial \log q_j}{\partial W_{ij}} = \mathbf{\delta}_j^q \frac{\partial \log q_j}{\partial W_{ij}} $
What is $\frac{\partial \log q_j}{\partial W_{ij}}$? Complete the equation above.
If you want, you can give the resulting equation in vector format ($\nabla_{\bw_j} \mathcal{L}^{(n)} = ...$), like we did for $\nabla_\bb \mathcal{L}^{(n)}$.
YOUR ANSWER HERE
Implement the gradient calculations you derived in the previous question. Write a function logreg_gradient(x, t, w, b) that returns the gradients $\nabla_{\bw_j} \mathcal{L}^{(n)}$ (for each $j$) and $\nabla_{\bb} \mathcal{L}^{(n)}$, i.e. the first partial derivatives of the log-likelihood w.r.t. the parameters $\bW$ and $\bb$, evaluated at a single datapoint (x, t).
The computation will contain roughly the following intermediate variables:
$ \log \bq \rightarrow Z \rightarrow \log \bp\,,\, \mathbf{\delta}^q $
followed by computation of the gradient vectors $\nabla_{\bw_j} \mathcal{L}^{(n)}$ (contained in a $784 \times 10$ matrix) and $\nabla_{\bb} \mathcal{L}^{(n)}$ (a $10 \times 1$ vector).
For maximum points, ensure the function is numerically stable.
In [ ]:
# 1.1.2 Compute gradient of log p(t|x;w,b) wrt w and b
def logreg_gradient(x, t, w, b):
# YOUR CODE HERE
raise NotImplementedError()
return logp[:,t].squeeze(), dL_dw, dL_db.squeeze()
In [ ]:
np.random.seed(123)
# scalar, 10 X 768 matrix, 10 X 1 vector
w = np.random.normal(size=(28*28,10), scale=0.001)
# w = np.zeros((784,10))
b = np.zeros((10,))
# test gradients, train on 1 sample
logpt, grad_w, grad_b = logreg_gradient(x_train[0:1,:], t_train[0:1], w, b)
print("Test gradient on one point")
print("Likelihood:\t", logpt)
print("\nGrad_W_ij\t",grad_w.shape,"matrix")
print("Grad_W_ij[0,152:158]=\t", grad_w[152:158,0])
print("\nGrad_B_i shape\t",grad_b.shape,"vector")
print("Grad_B_i=\t", grad_b.T)
print("i in {0,...,9}; j in M")
assert logpt.shape == (), logpt.shape
assert grad_w.shape == (784, 10), grad_w.shape
assert grad_b.shape == (10,), grad_b.shape
In [ ]:
# It's always good to check your gradient implementations with finite difference checking:
# Scipy provides the check_grad function, which requires flat input variables.
# So we write two helper functions that provide can compute the gradient and output with 'flat' weights:
from scipy.optimize import check_grad
np.random.seed(123)
# scalar, 10 X 768 matrix, 10 X 1 vector
w = np.random.normal(size=(28*28,10), scale=0.001)
# w = np.zeros((784,10))
b = np.zeros((10,))
def func(w):
logpt, grad_w, grad_b = logreg_gradient(x_train[0:1,:], t_train[0:1], w.reshape(784,10), b)
return logpt
def grad(w):
logpt, grad_w, grad_b = logreg_gradient(x_train[0:1,:], t_train[0:1], w.reshape(784,10), b)
return grad_w.flatten()
finite_diff_error = check_grad(func, grad, w.flatten())
print('Finite difference error grad_w:', finite_diff_error)
assert finite_diff_error < 1e-3, 'Your gradient computation for w seems off'
def func(b):
logpt, grad_w, grad_b = logreg_gradient(x_train[0:1,:], t_train[0:1], w, b)
return logpt
def grad(b):
logpt, grad_w, grad_b = logreg_gradient(x_train[0:1,:], t_train[0:1], w, b)
return grad_b.flatten()
finite_diff_error = check_grad(func, grad, b)
print('Finite difference error grad_b:', finite_diff_error)
assert finite_diff_error < 1e-3, 'Your gradient computation for b seems off'
In [ ]:
Write a function sgd_iter(x_train, t_train, w, b) that performs one iteration of stochastic gradient descent (SGD), and returns the new weights. It should go through the trainingset once in randomized order, call logreg_gradient(x, t, w, b) for each datapoint to get the gradients, and update the parameters using a small learning rate of 1E-6. Note that in this case we're maximizing the likelihood function, so we should actually performing gradient ascent... For more information about SGD, see Bishop 5.2.4 or an online source (i.e. https://en.wikipedia.org/wiki/Stochastic_gradient_descent)
In [ ]:
def sgd_iter(x_train, t_train, W, b):
# YOUR CODE HERE
raise NotImplementedError()
return logp_train, W, b
In [ ]:
# Sanity check:
np.random.seed(1243)
w = np.zeros((28*28, 10))
b = np.zeros(10)
logp_train, W, b = sgd_iter(x_train[:5], t_train[:5], w, b)
In [ ]:
def test_sgd(x_train, t_train, w, b):
# YOUR CODE HERE
raise NotImplementedError()
np.random.seed(1243)
w = np.zeros((28*28, 10))
b = np.zeros(10)
w,b = test_sgd(x_train, t_train, w, b)
In [ ]:
# YOUR CODE HERE
raise NotImplementedError()
Describe in less than 100 words why these weights minimize the loss YOUR ANSWER HERE
In [ ]:
# YOUR CODE HERE
raise NotImplementedError()
You discover that the predictions by the logistic regression classifier are not good enough for your application: the model is too simple. You want to increase the accuracy of your predictions by using a better model. For this purpose, you're going to use a multilayer perceptron (MLP), a simple kind of neural network. The perceptron wil have a single hidden layer $\bh$ with $L$ elements. The parameters of the model are $\bV$ (connections between input $\bx$ and hidden layer $\bh$), $\ba$ (the biases/intercepts of $\bh$), $\bW$ (connections between $\bh$ and $\log q$) and $\bb$ (the biases/intercepts of $\log q$.
The conditional probability of the class label $j$ is given by:
$\log p(t = j \;|\; \bx, \bb, \bW) = \log q_j - \log Z$
where $q_j$ are again the unnormalized probabilities per class, and $Z = \sum_j q_j$ is again the probability normalizing factor. Each $q_j$ is computed using:
$\log q_j = \bw_j^T \bh + b_j$
where $\bh$ is a $L \times 1$ vector with the hidden layer activations (of a hidden layer with size $L$), and $\bw_j$ is the $j$-th column of $\bW$ (a $L \times 10$ matrix). Each element of the hidden layer is computed from the input vector $\bx$ using:
$h_j = \sigma(\bv_j^T \bx + a_j)$
where $\bv_j$ is the $j$-th column of $\bV$ (a $784 \times L$ matrix), $a_j$ is the $j$-th element of $\ba$, and $\sigma(.)$ is the so-called sigmoid activation function, defined by:
$\sigma(x) = \frac{1}{1 + \exp(-x)}$
Note that this model is almost equal to the multiclass logistic regression model, but with an extra 'hidden layer' $\bh$. The activations of this hidden layer can be viewed as features computed from the input, where the feature transformation ($\bV$ and $\ba$) is learned.
State (shortly) why $\nabla_{\bb} \mathcal{L}^{(n)}$ is equal to the earlier (multiclass logistic regression) case, and why $\nabla_{\bw_j} \mathcal{L}^{(n)}$ is almost equal to the earlier case.
Like in multiclass logistic regression, you should use intermediate variables $\mathbf{\delta}_j^q$. In addition, you should use intermediate variables $\mathbf{\delta}_j^h = \frac{\partial \mathcal{L}^{(n)}}{\partial h_j}$.
Given an input image, roughly the following intermediate variables should be computed:
$ \log \bq \rightarrow Z \rightarrow \log \bp \rightarrow \mathbf{\delta}^q \rightarrow \mathbf{\delta}^h $
where $\mathbf{\delta}_j^h = \frac{\partial \mathcal{L}^{(n)}}{\partial \bh_j}$.
Give the equations for computing $\mathbf{\delta}^h$, and for computing the derivatives of $\mathcal{L}^{(n)}$ w.r.t. $\bW$, $\bb$, $\bV$ and $\ba$.
You can use the convenient fact that $\frac{\partial}{\partial x} \sigma(x) = \sigma(x) (1 - \sigma(x))$.
YOUR ANSWER HERE
You derived equations for finding the maximum likelihood solution of the parameters. Explain, in a few sentences, how you could extend this approach so that it optimizes towards a maximum a posteriori (MAP) solution of the parameters, with a Gaussian prior on the parameters.
YOUR ANSWER HERE
Implement a MLP model with a single hidden layer of 20 neurons. Train the model for 10 epochs. Plot (in one graph) the conditional log-probability of the trainingset and validation set after each two iterations, as well as the weights.
In [41]:
# Write all helper functions here
def qj(w, h, b, j):
return np.exp((w.T)[j]*h + b[j])
def Z(q_j):
np.sum(q_j)
# 2
def delta_qj(w, h, b, j, t_n):
if j == t_n:
I = 1
else:
I = 0
q_j = qj(w, h, b, j)
Z = Z(q_j)
return I - q_j / Z
# 3
def partial_L_wij(w, h, b, j, t_n, i):
return h[i] * delta_qj(w, h, b, j, t_n)
# 1
def delta_h(W, delta_q):
return W * delta_q
#4
def partial_L_Vhi(h, x, W, delta_q, i, k):
return h[i] * (1 - h[i]) * x[k] * delta_h(W, delta_q)[i]
#5
def partial_L_ai(h, W, delta_q, i):
return h[i] * (1 - h[i]) * delta_h(W, delta_q)[i]
def gradient_L(W, w, h, x, t_n, j, i, k, delta_q):
return [partial_L_wij(w, h, b, j, t_n, i), partial_L_Vhi(h, x, W, delta_q, i, k), partial_L_ai(h, W, delta_q, i)]
def w_t1(w_t, eta, gradient_L):
return w_t - eta*gradient_L
def sigmoid(x):
return 1 / (1 + np.exp(-x))
def dsig(s):
return s * (1 - s)
def calc_h(x, v, a):
return sigmoid(np.matmul(x, v) + a)
def logq_j(w_j, h, Z):
return w_j.T*h + b_j - log(Z)
def logp_j(w_j, h, Z, q_j):
return logq_j(w_j, h, Z) - Z(q_j)
In [43]:
# Write training code here:
# Plot the conditional loglikelihoods for the train and validation dataset after every iteration.
# Plot the weights of the first layer.
x_test = x_train[0:1,:]
t_test = t_train[0:1]
K = x_train.shape[1] # No. of input units
L = 20 # No. of hidden units
M = 10 # No. of output units
V = np.zeros((K, L))
W = np.zeros((L, M))
a = np.zeros((L,))
b = np.zeros((M,))
h_test = calc_h(x_test, V, a)
logreg_gradient(h_test, t_test, W, b)
YOUR ANSWER HERE
You receive an additional 10 bonus points if you manage to train a model with very high accuracy: at most 2.5% misclasified digits on the test set. Note that the test set contains 10000 digits, so you model should misclassify at most 250 digits. This should be achievable with a MLP model with one hidden layer. See results of various models at : http://yann.lecun.com/exdb/mnist/index.html. To reach such a low accuracy, you probably need to have a very high $L$ (many hidden units), probably $L > 200$, and apply a strong Gaussian prior on the weights. In this case you are allowed to use the validation set for training.
You are allowed to add additional layers, and use convolutional networks, although that is probably not required to reach 2.5% misclassifications.
In [ ]:
predict_test = np.zeros(len(t_test))
# Fill predict_test with the predicted targets from your model, don't cheat :-).
# YOUR CODE HERE
raise NotImplementedError()
In [ ]:
assert predict_test.shape == t_test.shape
n_errors = np.sum(predict_test != t_test)
print('Test errors: %d' % n_errors)