Due Date: March 10, 8:59pm
Please see the guidelines at https://www.cs.toronto.edu/~lczhang/338/homework.html
Please hand in 2 files:
csc338_a3.py. If you are using Jupyter Notebook to complete the work, your notebook can be exported as a .py file (File -> Download As -> Python). Your code will be auto-graded using Python 3.6, so please make sure that your code runs. There will be a 20% penalty if you need a remark due to small issues that renders your code untestable.csc338_a3_written.pdf containing your solutions to the written parts of the assignment. Your solution can be hand-written, but must be legible. Graders may deduct marks for illegible or poorly presented solutions.Submit the assignment on MarkUs by 9pm on the due date. See the syllabus for the course policy regarding late assignments. All assignments must be done individually.
import math
import numpy as np
M1 = np.eye(5)
M2 = np.array([[-3., 1.],
[ 1., 3.]])
M3 = np.array([[3., 1.],
[1., 3.]])
M4 = np.array([[3., 1.],
[1., 3.],
[0., 0.]])
M1_pd = False
M2_pd = False
M3_pd = False
M4_pd = False
Suppose that $A$ is a symmetric positive definite matrix. Show that the function
$$ ||x||_A = (x^T A x)^{\frac{1}{2}} $$
on a vector $x$ satisfies the three properties of a vector norm.
Include your solution in your pdf writeup.
Write a function cholesky_factorize that returns the Cholesky factorization of a matrix, according to its docstring
def cholesky_factorize(A):
"""Return the Cholesky Factorization L of A, where
* A is an nxn symmetric, positive definite matrix
* L is lower triangular, with positive diagonal entries
* $A = LL^T$
>>> M = np.array([[8., 3., 2.],
[3., 5., 1.],
[2., 1., 3.]])
>>> L = cholesky_factorize(M.copy())
>>> np.matmul(L, L.T)
array([[8., 3., 2.],
[3., 5., 1.],
[2., 1., 3.]])
"""
Recall that we proved the Sherman-Morrison formula in class:
$$(A - uv^T)^{-1} = A^{-1} + A^{-1}u(1-v^TA^{-1}u)^{-1}v^T A^{-1}$$
A related formula is the Woodbury formula for a rank-k update of the matrix $A$. Specifically, if $U$ and $V$ are $n \times k$ matrices, then
$$(A - UV^T)^{-1} = A^{-1} + A^{-1}U(I-V^TA^{-1}U)^{-1}V^T A^{-1}$$
Prove that the Woodbury formula is correct. Include your solution in your pdf writeup.
A rank 1 matrix has the form $xy^T$ where $x$ and $y$ are column vectors. Suppose $A$ and $B$ are non-sigular matrices. Show that $A - B$ is rank 1 if and only if $A^{-1} - B^{-1}$ is also rank 1.
Include your solution in your pdf writeup.
Hint: Use the Sherman-Morrison formula. This question is not supposed to be easy, so leave aside time to think!
For this question, we will use the code from assignment 2.
def backward_substitution(A, b):
"""Return a vector x with np.matmul(A, x) == b, where
* A is an nxn numpy matrix that is upper-triangular and nonsingular
* b is an nx1 numpy vector
"""
n = A.shape[0]
x = np.zeros_like(b, dtype=np.float)
for i in range(n-1, -1, -1):
s = 0
for j in range(n-1, i, -1):
s += A[i,j] * x[j]
x[i] = (b[i] - s) / A[i,i]
return x
def eliminate(A, b, k):
"""Eliminate the k-th row of A, in the system np.matmul(A, x) == b,
so that A[i, k] = 0 for i < k. The elimination is done in place."""
n = A.shape[0]
for i in range(k + 1, n):
m = A[i, k] / A[k, k]
for j in range(k, n):
A[i, j] = A[i, j] - m * A[k, j]
b[i] = b[i] - m * b[k]
def gauss_elimination(A, b):
"""Return a vector x with np.matmul(A, x) == b using
the Gauss Elimination algorithm, without partial pivoting."""
for k in range(A.shape[0] - 1):
eliminate(A, b, k)
x = backward_substitution(A, b)
return x
Recall the following system for question 2(d) of assignment 2.
When we tried to solve this system using gauss elimiation without pivoting, we obtained an answer q4_x0 below that is incorrect. Compute the residual of q4_x0. Save the residual in the variable q4_r0
e = pow(2, -100)
A = np.array([[e, 1],
[1, 1]])
b = np.array([1+e, 2])
q4_x0 = gauss_elimination(A.copy(), b.copy())
q4_r0 = np.zeros(2)
We will use iterative refinement to produce a better solution q4_x1 to the system $Ax = b$.
First, what is the new system that we need to solve? Save the $A$ and $b$ of the new system
in the variable q4_A1 and q4_b1.
Compute the residual q4_r1 of your new solution.
q4_A1 = np.zeros([2,2]) # todo
q4_b1 = np.zeros(2) # todo
# gauss_elimination(q4_A1, q4_b1)
q4_x1 = np.zeros(2)
q4_r1 = np.zeros(2)
How does your new solution q4_x1 compare to your old solution q4_x0?
Store either "better" or "worse" in the variable new_solution_is, depending on whether you think q4_x1 is a better or worse solution than q4_x0 to the initial system $Ax = b$.
new_solution_is = None # either "better" or "worse"
Write a function solve_normal_equation that takes an $m \times n$ matrix $A$ and a vector $b$,
and solves the linear least squares problem $A x \approx b$ using the Normal Equation method.
You should use the cholesky_factorize method that you wrote in question 3. You may need
other functions that you wrote in assignment 2.
def solve_normal_equation(A, b):
"""
Return the solution x to the linear least squares problem
$$Ax \approx b$$ using normal equations,
where A is an (m x n) matrix, with m > n, rank(A) = n, and
b is a vector of size (m)
"""
Consider the linear least square system below, where $A$ is a $50 \times 3 $ matrix, and $b$ is a vector of size $50$. The system is derived from the "iris" dataset.
Use the solve_normal_equation function you wrote in part (a) to find the vector x
that minimizes the norm of matmul(A, x) - b.
Store the solution in the variable q5b_x and the residual in q5b_r. You may use
the function np.linalg.norm.
irisA = np.array([[4.8, 6.3, 5.7, 5.1, 7.7, 5.6, 4.9, 4.4, 6.4, 6.2, 6.7, 4.5, 6.3,
4.8, 5.8, 4.7, 5.4, 7.4, 6.4, 6.3, 5.1, 5.7, 6.5, 5.5, 7.2, 6.9,
6.8, 6. , 5. , 5.4, 5.6, 6.1, 5.9, 5.6, 6. , 6. , 4.4, 6.9, 6.7,
5.1, 6. , 6.3, 4.6, 6.7, 5. , 6.7, 5.8, 5.1, 5.2, 6.1],
[3.1, 2.5, 3. , 3.7, 3.8, 3. , 3.1, 2.9, 2.9, 2.9, 3.1, 2.3, 2.5,
3.4, 2.7, 3.2, 3.7, 2.8, 2.8, 3.3, 2.5, 4.4, 3. , 2.6, 3.2, 3.2,
3. , 2.9, 3.6, 3.9, 3. , 2.6, 3.2, 2.8, 2.7, 2.2, 3.2, 3.1, 3.1,
3.8, 2.2, 2.7, 3.1, 3. , 3.5, 2.5, 4. , 3.5, 3.4, 3. ],
[1.6, 4.9, 4.2, 1.5, 6.7, 4.5, 1.5, 1.4, 4.3, 4.3, 5.6, 1.3, 5. ,
1.6, 5.1, 1.3, 1.5, 6.1, 5.6, 6. , 3. , 1.5, 5.8, 4.4, 6. , 5.7,
5.5, 4.5, 1.4, 1.3, 4.1, 5.6, 4.8, 4.9, 5.1, 5. , 1.3, 4.9, 4.4,
1.5, 4. , 4.9, 1.5, 5.2, 1.3, 5.8, 1.2, 1.4, 1.4, 4.9]])
irisb = np.array([0.2, 1.5, 1.2, 0.4, 2.2, 1.5, 0.1, 0.2, 1.3, 1.3, 2.4, 0.3, 1.9,
0.2, 1.9, 0.2, 0.2, 1.9, 2.1, 2.5, 1.1, 0.4, 2.2, 1.2, 1.8, 2.3,
2.1, 1.5, 0.2, 0.4, 1.3, 1.4, 1.8, 2. , 1.6, 1.5, 0.2, 1.5, 1.4,
0.3, 1. , 1.8, 0.2, 2.3, 0.3, 1.8, 0.2, 0.2, 0.2, 1.8])
A = irisA.T # transpose
b = irisb
q5b_x = None
q5b_r = None
Consider an alternative value of $x$ for the system $Ax \approx b$
defined in part (b), with x = np.array([-0.2, 0.2, 0.5]).
Compute the norm of the residual $Ax - b$ for this alternate value of $x$,
and save the result in the variable q5c_r.
You should see that q5b_r < q5c_r.
x = np.array([-0.2, 0.2, 0.5])
q5c_r = None
Show that if an $m \times n$ matrix $A$ with $m > n$ has $rank(n)$, then $A^T A$ is positive definite.
Consider the following linear least squares problem. What is the minimum
value of $||Ax - b||_2$? Save the minimum norm of the
residual in the variable q5a_r.
What is the solution to the problem? Save the value in the variable q5a_x.
Do this entire problem by hand. You should verify that the residual matches
the value you stored in q5_r.
A = np.array([[1., 2., 3.],
[0., 3., 4.],
[0., 0., 6.],
[0., 0., 0.],
[0., 0., 0.]])
b = np.array([[1., 2., 3., 4., 5.]])
q5a_r = 0
q5a_x = np.array([0., 0., 0.])
Write a function householder_v that returns the vector $v$ that defines
the Householder transform
$$H = I - 2 \frac{v v^T} {v^T v}$$
that eliminates all but the first element of a vector $a$.
def householder_v(a):
"""Return the vector $v$ that defines the Householder Transform
H = I - 2 np.matmul(v, v.T) / np.matmul(v.T, v)
that will eliminate all but the first element of the
input vector a.
Example:
>>> householder_v(np.array([2., 1., 2.]))
array([5., 1., 2.])
"""
Write a function apply_householder that applies the Householder
transform defined by a vector $v$ to a vector $u$. You should not
compute the Householder transform matrix $H$. You should only
need to compute vector-vector dot products and vector-scalar multiplications.
def apply_householder(v, u):
"""Return the result of the Householder transformation defined
by the vector $v$ applied to the vector $u$. You should not
compute the Householder matrix H directly.
Example:
>>> apply_householder(np.array([5., 1., 2.]), np.array([2., 1., 2.]))
array([-3., 0., 0.])
>>> apply_householder(np.array([5., 1., 2.]), np.array([2., 3., 4.]))
array([-5. , 1.6, 1.2])
"""
Write a function apply_householder_matrix that applies the Householder
transform defined by a vector $v$ to all the columns of a matrix $U$.
You should not compute the Householder transform matrix $H$.
Do not use for loops. Instead, you may find the numpy function np.outer useful.
def apply_householder_matrix(v, U):
"""Return the result of the Householder transformation defined
by the vector $v$ applied to all the vectors in the matrix U.
You should not compute the Householder matrix H directly.
Example:
>>> v = np.array([5., 1., 2.])
>>> U = np.array([[2., 2.],
[1., 3.],
[2., 4.]])
>>> apply_householder_matrix(v, U)
array([[-3. , -5. ],
[ 0. , 1.6],
[ 0. , 1.2]])
"""
Write a function solve_qr_householder that takes an $m \times n$ matrix $A$ and a vector $b$,
and solves the linear least squares problem $A x \approx b$ using Householder QR Factorization.
You may use np.linalg.solve to solve any square system of the form $Ax = b$ that you produce.
You should use the helper function qr_householder that takes a matrix A and a vector b and
performs the Householder QR Factorization using the functions you wrote in parts (b-d).
def qr_householder(A, b):
"""Return the matrix [R O]^T, and vector [c1 c2]^T equivalent
to the system $Ax \approx b$. This algorithm is similar to
Algorithm 3.1 in the textbook.
"""
for k in range(A.shape[1]):
v = householder_v(A[k:, k])
if np.linalg.norm(v) != 0:
A[k:, k:] = apply_householder_matrix(v, A[k:, k:])
b[k:] = apply_householder(v, b[k:])
# now, A is upper-triangular
return A, b # A is uppe
def solve_qr_householder(A, b):
"""
Return the solution x to the linear least squares problem
$$Ax \approx b$$ using Householder QR decomposition.
Where A is an (m x n) matrix, with m > n, rank(A) = n, and
b is a vector of size (m)
"""
Use the function solve_qr_householder to solve the Iris system from Q5(b).
Save your result for $x$ in the variable q6f_x.
Do you get the same result or a different result from Q5(b)? Change the value
of the variable q6f_compare to either the string "same" or "different" depending on
whether your results were the same or different.
A = irisA.T # transpose
b = irisb
q6f_x = None
q6f_compare = None
Show that for $H = I - 2 \frac{v v^T}{v^T v}$, we have $H = H^T = H^{-1}$.
Include your solution in your PDF writeup.
For the next few questions, we will use the MNIST dataset for digit recognition
Download the files mnist_images.npy and mnist_labels.npy from the course
website, and place them into the same folder as your ipynb file.
The code below loads the data, splits it into "train" and "test" sets,
and plots the a subset of the data. We will use train_images and train_labels
to set up Linear Least Squares problems. We will use test_images and test_labels
to test the models that we build.
mnist_images = np.load("mnist_images.npy")
test_images = mnist_images[4500:] # 500 images
train_images = mnist_images[:4500] # 4500 images
mnist_labels = np.load("mnist_labels.npy")
test_labels = mnist_labels[4500:]
train_labels = mnist_labels[:4500]
def plot_mnist(remove_border=False):
""" Plot the first 40 data points in the MNIST train_images. """
import matplotlib.pyplot as plt
plt.figure(figsize=(10, 5))
for i in range(4 * 10):
plt.subplot(4, 10, i+1)
if remove_border:
plt.imshow(train_images[i,4:24,4:24])
else:
plt.imshow(train_images[i])
# plot_mnist() # please comment out this line before submission
How many examples of each digit are in train_images? You should use the
information in train_labels.
If you are not well versed in numpy, some of the code in the later part of this
question might be helpful. You might also find the sum function helpful.
Save your result in the array mnist_digits, where
mnist_digits[0] should contain the number of digit 0 in train_images,
mnist_digits[1] should contain the number of digit 1 in train_images, etc.
mnist_digits = []
We will build a rudimentary model to predict whether a digit is the
digit 0. Our features will be the intensity at each pixel.
There are 28 * 28 = 784 pixels in each image. However, in order
to obtain a matrix $A$ that is of full rank, we will ignore the
pixels along the border. That is, we will only use 400 pixels
in the center of the image.
Look at a few of the MNIST images using the plot_mnist function
written for you. Why would our matrix $A$ not be full rank if we
use all 784 pixels in our model?
If this question doesn't make sense yet, you might want to come back to it after completing the rest of this question.
Include your solution in your PDF writeup.
# Plot the MNIST images, with only the center pixels that we will use
# in our digit classification model.
#plot_mnist(True) # please comment out this line before submission
We will now build a rudimentary model to predict whether a digit is the
digit 0. To obtain a matrix of full rank, we will use the 400 pixels
at the center of each image as features. Our target will
be whether our digit is 0.
In short, the model we will build looks like this:
$$x_1 p_1 + x_2 p_2 + ... + x_{400} p_{400} = y$$
Where $p_i$ is the pixel intensity at pixel $i$ (the ordering of the pixel's
doesn't actually matter), and the value of $y$ determines whether our digit is a 0 or not.
We will solve for the coefficients $x_i$ by solving the linear
least squares problem $Ax \approx b$, where $A$ is constructed using
the pixel intensities of the images in train_images, and $y$ is constructed
using the labels for those images. For convenience, we will set $y = 1$ for
images of the digit 0, and $y=0$ for other digits.
I should stress that in CSC411, you will learn that this is not the proper way to build a digit detector. However, digit detection is quite fun, so we might as well use to tools that we have to try and solve the problem.
The code below obtains the matrices $A$ and the vector $b$ of our least squares problem, where $A$ is a $m \times n$ matrix and $b$ is a vector of length $m$.
What is the value of $m$ and $n$? Save the values in the variables mnist_m and mnist_n.
A = train_images[:, 4:24, 4:24].reshape([-1, 20*20])
b = (train_labels == 0).astype(np.float32)
mnist_m = None
mnist_n = None
Use the Normal Equations method to solve the system $Ax=b$. Save the result in the variable mnist_x1. Save the norm of the residuals of this solution in the variable mnist_r1.
Then, use the Householder QR decomposition method to solve the same system. Save the result in the variable mnist_x2. Save the norm of the residuals of this solution in the variable mnist_r2.
mnist_x1 = None
mnist_r1 = None
mnist_x2 = None
mnist_r2 = None
Consider test_images[0]. Is this image of the digit 0? Set the value of test_image_0 to either True or False
depending on your result.
Let $p$ be the vector containing the values of the 400 center pixels in test_image[0]. The features are extracted
for you in the variabel p. Use the solution mnist_x1 to estimate the target $y$ for the vector $p$.
Save the (float) value of the predicted value of $y$ in the variable test_image_0_y.
# import matplotlib.pyplot as plt # Please comment before submitting
# plt.imshow(test_images[0]) # Please comment before submitting
p = test_images[0, 4:24, 4:24].reshape([-1, 20*20])
test_image_0 = None
test_image_0_y = None
Write code to predict the value of $y$ for every image in test_images. Save
your result in test_image_y.
Do not use a loop.
test_image_y = None
We will want to turn the continuous estimates of $y$ into discrete predictions about whether an image is of the digit 0.
We will do this by selecting a cutoff. That is, we will predict that a test image is of the digit 0 if the prediction $y$ for that digit is at least $0.5$.
Create a numpy array test_image_pred with test_image_pred[i] == 1 if test_image[i]
is of the digit 0, and test_image_pred[i] == 0 otherwise. Then, run
the code to compute the test_accuracy, or the portion of the times that
our prediction matches the actual label.
HINT: You might find the code in Part(c) helpful.
(This is somewhat of an arbitrary cutoff. You will learn the proper way to do this prediction problem in CSC411.)
test_image_pred = None
test_accuracy = sum(test_image_pred == (test_labels == 0).astype(float)) / len(test_labels)
test_accuracy
So far, we built a linear least squares model that determines whether an image is of the digit 0. Let's go a step further, and build such a model for every digit!
Complete the function mnist_classifiers that uses train_images and train_labels,
and uses the function solve_normal_equation to build a linear least squares model for
every digit. The function should return a matrix $xs$ of shape $10 \times 400$,
with xs[0] == mnist_x1 from earlier.
Please comment out any code you use to test mnist_classifier. It will help make testing
your code faster, so I can get your assignments back to you more quickly.
def mnist_classifiers():
"""Return the coefficients for linear least squares models for every digit.
Example:
>>> xs = mnist_classifiers()
>>> np.all(xs[0] == mnist_x1)
True
"""
# you can use train_images and train_labels here, and make
# whatever edits to this function as you wish.
A = train_images[:, 4:24, 4:24].reshape([-1, 20*20])
xs = []
# ...
return np.stack(xs)
The code below makes predictions based on the result of your mnist_classifier.
That is, for every test image, the code runs all 10 models to see whether the
test image contains each of the 10 digits. We make a discrete prediction about
which digit the image contains by looking at which model yields the largest
value of $y$ for the image.
The code then compares the result against the actual labels, computes the accuracy measure: the fraction of predictions that is correct. Just for fun, look at the prediction accuracy of our model, but please comment any code you write before submitting your assignment.
Again, in CSC411 you will learn much better ways of classifying digits. You will achieve a much better test accuracy than what we have here.
def prediction_accuracy(xs):
"""Return the prediction
"""
testA = test_images[:, 4:24, 4:24].reshape([-1, 20*20])
ys = np.matmul(testA, xs.T)
pred = np.argmax(ys, axis=1)
return sum(pred == test_labels) / len(test_labels)