CS4803/7643 Problem Set 2 solved

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This assignment introduces you to some theoretical results that address which neural networks can
represent what. It walks you through proving some simplified versions of more general results. In
particular, this assignment focuses on piecewise linear neural networks, which are the most common
type at the moment. The general strategy will be to construct a neural network that has the desired
properties by choosing appropriate sets of weights.
1 Convolution Basics
The convolution layer inside of a CNN is intrinsically an affine transformation: A vector is received
as input and is multiplied with a matrix to produce an output (to which a bias vector is usually
added before passing the result through a nonlinearity). This operation can be represented as
y = Ax, in which A describes the affine transformation.
1. [2 points] We will first revisit the convolution layer as discussed in the class. Consider a
convolution layer with a 3×3 kernel W, operated on a single input channel X, represented as:
W =

w(0,0) w(0,1) w(0,2)
w(1,0) w(1,1) w(1,2)
w(2,0) w(2,1) w(2,2)

 , X =

x(0,0) x(0,1) x(0,2)
x(1,0) x(1,1) x(1,2)
x(2,0) x(2,1) x(2,2)

 (1)
Now let us work out a stride-2 convolution layer, with zero-padding size of 1. Consider
‘flattening’ the input tensor X in row-major order as:
X =

x(0,0), x(0,1), x(0,2), …, x(2,0), x(2,1), x(2,2)>
Write down the convolution as a matrix operation A that: Y = AX. Output Y is also
flattened in row-major order.
2. [2 points] Recall that transposed convolution can help us upsample the feature size spatially.
Consider a transposed convolution layer with a 2×2 kernel W operated on a single input
channel X, represented as:
W =

w(0,0), w(0,1)
w(1,0), w(1,1)
, X =

x(0,0), x(0,1)
x(1,0), x(1,1)
We ‘flattened’ the input tensor in row-major order as X = [x(0,0), x(0,1), x(1,0), x(1,1)].
Write down the affine transformation A corresponding to a transposed convolution layer with
kernel W, stride 2, no padding. Output Y is also flattened in row-major order.
2 Logic
1. [4 points] Implement AND and OR for pairs of binary inputs using a single linear threshold
neuron with weights w ∈ R
, bias b ∈ R, and x ∈ {0, 1}
f(x) = (
1 if wT x + b ≥ 0
0 if wT x + b < 0
That is, find wAND and bAND such that
x1 x2 fAND(x)
0 0 0
0 1 0
1 0 0
1 1 1
Also find wOR and bOR such that
x1 x2 fOR(x)
0 0 0
0 1 1
1 0 1
1 1 1
2. [4 points] Consider the bidirectional implication “⇔” (i.e., “if and only if”, often abbreviated
as “iff”), which has the following truth table
x1 x2 f⇔(x)
0 0 1
0 1 0
1 0 0
1 1 1
Show that “⇔” can NOT be represented using a linear model with the same form as (4).
[Hint: To see why, plot the examples from above in a plane and think about drawing a linear
boundary that separates them.]
3 Depth – Composing Linear Pieces [Extra Credit for 4803; Regular
Credit for 7643]
An interesting property of networks with piecewise linear activations like the ReLU is that on the
whole they compute piecewise linear functions. For instance, each layer h
(i) below computes a linear
function of the previous layer, h
(i−1), and the resulting function is piecewise-linear in x:
(1)(x) = W(1)x + b
(1) (5)
(2)(x) = W(2) max{h
(1)(x), 0} + b
(2) (6)
(3)(x) = W(3) max{h
(2)(x), 0} + b
(3) (7)
Now we’ll turn to a more recent result that highlights the Deep in Deep Learning. Depth (composing more functions) results in a favorable combinatorial explosion in the “number of things that a
neural net can represent”. For example, to classify a cat it seems useful to first find parts of a cat:
eyes, ears, tail, fur, etc. The function which computes a probability of cat presence should be a
function of these components because this allows everything you learn about eyes to generalize to
all instances of eyes instead of just a single instance. Below you will detail one formalizable sense
of this combinatorial explosion for a particular class of piecewise linear networks.
Figure 1
Consider y = σ(x) = |x| for scalar x ∈ X ⊆ R and y ∈ Y ⊆ R (Fig. 1). It has one linear region on x < 0 and another on x > 0 and the activation identifies these regions, mapping both of
them to y > 0. More precisely, for each linear region of the input, σ(·) is a bijection. There is
a mapping to and from the output space and the corresponding input space. However, given an
output y, it’s impossible to tell which linear region of the input it came from, thus σ(·) identifies
(maps on top of each other) the two linear regions of its input. This is the crucial definition because
when a function identifies multiple regions of its domain that means any subsequent computation
applies to all of those regions. When these regions come from an input space like the space of images, functions which identify many regions where different images might fall (e.g., slightly different
images of a cat) automatically transfer what they learn about a particular cat to cats in the other
More formally, we will say that σ(·) identifies a set of M disjoint input regions R = {R1, . . . , RM}
(e.g., R = {(−1, 0),(0, 1)}) with Ri ⊆ X onto one output region O ⊆ Y (e.g., (0, 1)) if for all
Ri ∈ R there is a bijection from Ri to O.
1. [4 points] Start by applying the above notion of identified regions to linear regions of one
layer of a particular neural net that uses absolute value functions as activations. Let x ∈ R
y ∈ R
d 2
, and pick weights W(1) ∈ R
d×d and bias b
(1) ∈ R
d as follows:
ij =
2 if i = j
0 if i 6= j
i = −1 (9)
Then one layer of a neural net with absolute value activation functions is given by
f1(x) = |W(1)x + b| (10)
Note that this is an absolute value function applied piecewise and not a norm.
How many regions of the input are identified onto O = (0, 1)d by f1(·)? Prove it. 3
1Recall that a bijection from X to Y is a function µ : X → Y such that for all y ∈ Y there exists a unique x ∈ X
with µ(x) = y.
2Outputs are in some feature space, not a label space. Normally a linear classifier would be placed on top of what
we are here calling y.
3Absolute value activations are chosen to make the problem simpler, but a similar result holds for ReLU units.
Also, O could be the positive orthant (unbounded above).
2. [4 points] Next consider what happens when two of these functions are composed. Suppose
g identifies ng regions of (0, 1)d onto (0, 1)d and f identifies nf regions of (0, 1)d onto (0, 1)d
How many regions of its input does f ◦ g(·) identify onto (0, 1)d
3. [6 points] Finally consider a series of L layers identical to the one in question 4.1.
h1 = |W1x + b1| (11)
h2 = |W2h1 + b2| (12)
. (13)
hL = |WLhL−1 + bL| (14)
Let x ∈ (0, 1)d and f(x) = hL. Note that each hi
is implicitly a function of x. Show that
f(x) identifies 2
Ld regions of its input.
4 Conclusion to Theory Part
Now compare the number of identified regions for an L layer net to that of an L − 1 layer net. The
L layer net can separate its input space into 2
d more linear regions than the L − 1 layer net. On
the other hand, the number of parameters and the amount of computation time grows linearly in
the number of layers. In this very particular sense (which doesn’t always align well with practice)
deeper is better.
To summarize this problem set, you’ve shown a number of results about the representation power
of different neural net architectures. First, neural nets (even single neurons) can represent logical
operations. Second, neural nets we use today compute piecewise linear functions of their input.
Third, the representation power of neural nets increases exponentially with the number of layers.
The point of the exercise was to convey intuition that removes some of the magic from neural nets
representations. Specifically, neural nets can decompose problems logically, and “simple” piecewise
linear functions can be surprisingly powerful.
5 Paper Review
The second of our paper reviews for this course comes from some interesting work at Facebook AI
Research (FAIR) last year exploring the limits of weakly supervised learning and improving upon
their results on the ImageNet dataset by appending data from Instagram using hashtags as ground
truth labels. The paper can be viewed here.
Guidelines: Please restrict your answers to no more than 350 words combined. The evaluation
rubric for this section is as follows :
1. [2 points] The paper shows that training with very weak (noisy) labels still leads to robust
learning of features that generalize well. Why is deep learning so robust to noise? Provide
some conjectures, relating it to what you know about machine/deep learning and optimization.
2. [2 points] Another finding is that learning features using similar label spaces (i.e. on categories that overlap with what you are hoping to generalize to) is more successful than learning
features on dissimilar label spaces. Why might that be the case?
6 Coding: Gradients With Respect to Input [65 points]
The coding part of this assignment will have you implement different ideas which all rely on
gradients of some neural net output with respect to the input image. To get started go to