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# Numerical methods Assignment 2

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CSC338. Assignment 2

What to Hand In
• Python File containing all your code, named a2.py.
• PDF file named a2_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.
If you are using Jupyter Notebook to complete the work, your notebook can be exported as a .py file (File ->
There will be a 15% penalty if you need a remark due to small issues that renders your code untestable. (Please note
the penalty is higher than in A1!)
Make sure to remove or comment out all matplotlib or other expensive code before submitting your
code! Expensive code can render your code untestable, and you will incur the 15% penalty for remark.
Submit the assignment on MarkUs by 10pm 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
Question 1.
For this question, we will work in the floating-point system F(β = 10, p = 3, L = −10, U = 10).
Consider the system Ax = b where
A =

0.03 21.2
2.18 −0.30
, b =

21.5
21.5

Part (a) – 2pt
Solve the system by hand, exactly, using infinite precision arithmetic. That is, the addition, multiplication, division,
and other operations that you do will be exact (rather than the floating-point version). Do not use pivoting.
Include your step-by-step solution in your PDF writeup. Your final result can be rounded to 5 decimal places, but do
not round any intermediate computations since we are using infinite precision arithmetic.
Part (b) – 2pt
Solve the system using floating-point operations and the rounding rule is chopping. In other words, replace
every operation you used in part (a) with the floating-point operation in the system F(β = 10, p = 3, L = −10, U = 10).
Do not use pivoting.
Part (c) – 3pt
Solve this similar system using floating-point operations, but with partial pivoting.
Which result is closer to the true result that you obtained in Part (a)?
1
Question 2.
Part (a) [2 pt]
Prove that ||A||∞ can be computed by computing the maximum row sum:
||A||∞ = max
i
Xn
j=1
|aij |
Part (b) [2 pt]
Suppose that A and B are two n × n matrices, and both are well-conditioned. Is A(B−1
) also well-conditioned? Why
Part (c) [4 pt]
Describe an efficient algorithm to compute d
T BT A−1Bd
Where:
• A is an invertible n × n matrix,
• B is an n × n matrix, and
• d is an n × 1 vectors
Question 3
Part (a) – 2 pt
Suppose that A is a symmetric positive definite matrix. Show that the function
||x||A = (x
T Ax)
1
2
on a vector x satisfies the three properties of a vector norm.
Part (b) – 6 pt
Complete the 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 = LLˆT\$
>>> M = np.array([[8., 3., 2.],
[3., 5., 1.],
[2., 1., 3.]])
>>> L = cholesky_factorize(M)
>>> np.matmul(L, L.T)
array([[8., 3., 2.],
2
[3., 5., 1.],
[2., 1., 3.]])
“””
Question 4
Part (a) – 4 pts
Complete the function solve_rank_one_update that solves the system (A − uvT
)x = b, assuming that the factorization A = LU has already been done for you. You are welcome to add any helper functions that you wish, including
functions that you wrote in homeworks 3 and 4. Just make sure that you include the helper functions in your python
script submission.
def solve_rank_one_update(L, U, b, u, v):
“””Return the solution x to the system (A – u vˆT)x = b, where
A = LU, using the approach we derived in class using
the Sherman Morrison formula. You may assume that
the LU factorization of A has already been computed for you, and
that the parameters of the function have:
* L is an invertible nxn lower triangular matrix
* U is an invertible nxn upper triangular matrix
* b is a vector of size n
* u and b are also vectors of size n
>>> A = np.array([[2., 0., 1.],
[1., 1., 0.],
[2., 1., 2.]])
>>> L, U = lu_factorize(A) # from homework 3
>>> L
array([[1. , 0. , 0. ],
[0.5, 1. , 0. ],
[1. , 1. , 1. ]])
>>> U
array([[ 2. , 0. , 1. ],
[ 0. , 1. , -0.5],
[ 0. , 0. , 1.5]])
>>> b = np.array([1., 1., 0.])
>>> u = np.array([1., 0., 0.])
>>> v = np.array([0., 2., 0.])
>>> x = solve_rank_one_update(L, U, b, u, v)
>>> x
array([1. , 0. , -1.])
>>> np.matmul((A – np.outer(u, v)), x)
array([1. , 1. , 0.])
“””
Part (b) – 2 pt
Explain why using solve_rank_one_update does not give us accurate results in the below example:
def run_example():
A = np.array([[2., 0., 1.],
[1., 1., 0.],
[1., 1., 1.]])
L = np.array([[1., 0., 0.],
[0.5, 1., 0.],
3
[0.5, 1., 1.]])
U = np.array([[2., 0., 1.],
[0., 1., -0.5],
[0., 0., 1.]])
b = np.array([1, 1, -1])
u = np.array([0, 0, 0.9999999999999999])
v = np.array([0, 0, 0.9999999999999999])
x = solve_rank_one_update(L, U, b, u, v)
print(np.matmul((A – np.outer(u, v)), x) – b)
Part (c) – 4 pt
A rank 1 matrix has the form xyT 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.
Hint: Use the Sherman-Morrison formula. This question is not supposed to be easy, so leave aside time to think!
Question 5.
Part (a) – 3 pt
Write a function householder_v that returns the vector v that defines the Householder transform
H = I − 2
vvT
v
T v
that eliminates all but the first element of a vector a. You may use the function np.linalg.norm.
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. Choose the \$v\$ that does not result in
cancellation.
Do not modify the vector `a`.
Example:
>>> a = np.array([2., 1., 2.])
>>> householder_v(a)
array([5., 1., 2.])
>>> a
array([2., 1., 2.])
“””
Part (b) – 2 pt
Show that a Householder Transformation H is orthogonal. Include your solution in your PDF writeup.
4
Part (c) – 2 pt
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])
“””
Part (d) – 3 pt
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]])
“””
Part (e) – 3 pt
Write a function solve_qr_householder that takes an m × n matrix A and a vector b, and solves the linear least
squares problem Ax ≈ 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]ˆT, and vector [c1 c2]ˆT 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]):
5
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
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)
“””
Question 6
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.
test_images = mnist_images[4500:] # 500 images
train_images = mnist_images[:4500] # 4500 images
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
Remember to comment out any plotting related code before you submit your assignment!
# plot_mnist()
Part (a) – 1 pt
How many examples of each digit are in train_images? You should use the information in train_labels.
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 = []
6
Part (b) – 2 pt
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.
# 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
Part (c) – 1 pt
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:
x1p1 + x2p2 + … + x400p400 = y
Where pi
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 xi by solving the linear least squares problem Ax ≈ 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.
We should stress that in real machine learning courses, 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 × 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
Part (d) – 1 pt
Use the Householder QR decomposition method to solve the system. Save the result in the variable mnist_x. Save
the norm of the residuals of this solution in the variable mnist_r.
mnist_x = None
mnist_r = None
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Part (e) – 1 pt
Consider test_images[0]. Is this image of the digit 0? Set the value of test_image_0 to either True or False
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_x 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
Part (f) – 2 pt
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
Part (g) – 1 pt
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 a machine
learning course like CSC311.)
test_image_pred = None
# test_accuracy = sum(test_image_pred == (test_labels == 0).astype(float)) / len(test_labels)
# print(test_accuracy)
Part (h) – 4 pt
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 the functions you wrote
in Q3 to build a linear least squares model for every digit. The function should return a matrix xs of shape 10 × 400,
with xs[0] == mnist_x1 from earlier.
Make sure to comment out any code you use to test mnist_classifier, or your code might not be
testable.
def mnist_classifiers():
“””Return the coefficients for linear least squares models for every digit.
Example:
8
>>> 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)
Just for fun. . .
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
Again, in machine learning and statistics courses you will learn ways to classifying digits that are better and more
principled. You’ll 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)
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Numerical methods Assignment 2
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