Mentally estimating square roots

What's $\sqrt{10}$? I'd reckon it's about $3 + \frac{10-9}{2 \times 3} = 3.166$. The actual answer is 3.162. What about $\sqrt{18}$? Should be about $4 + \frac{18-16}{2 \times 4} = 4.250$. The actual answer is 4.242. I found out about this calculation from this video but there was no explanation for why it works. Here's an explanation.

So the method here is as follows.

1. Let the number you want to find the square root of be $a^2$.
2. Let the largest square number which is less than $a^2$ be $b^2$ such that $b^2 <= a^2 < (b+1)^2$. For example if $a^2$ is 10 then $b^2$ is 9, if $a^2$ is 18 then $b^2$ is 16.
3. The square root of $a^2$ is approximately $b + \frac{a^2 - b^2}{2 b}$.

This method is easy to carry out mentally but why does it work? The trick here is that the graph of the square root function grows so slowly that we can approximate the curve between two adjacent square numbers as a line.



We can use the line to approximate the square root of any number between two square numbers. The first thing we need to know is the gradient of the line. The vertical distance between two adjacent square numbers on the square root curve is 1, since the two square numbers are the squares of two consecutive numbers. The horizontal distance changes and becomes larger as the adjacent square numbers become larger but we can calculate it as follows:

$$(b+1)^2 - b^2 = b^2 + 2b + 1 - b^2 = 2b + 1$$

So the horizontal distance is twice the square root of the smaller square number plus one. Therefore the gradient of the line is $\frac{1}{2b+1}$. Once we know by how much the line grows vertically for every horizontal unit, we can then determine how much higher than $b$ the point on the line will be at $a$ by multiplying the gradient by $a^2-b^2$, as shown below:



Since the difference in height is less than 1, it is going to be the part of the square root that comes after the decimal point, with the whole number part being $b$.

It might be hard to mentally divide by an odd number in $\frac{a^2-b^2}{2b+1}$ so we further approximate it as $\frac{a^2-b^2}{2b}$ instead. And that's why this method works.

Comparing numpy scalars directly is time consuming, use .tolist() before a comparison

This is something that I found out about recently when going through the elements of a numpy array in order to do some checks on each numbers. Turns out you shouldn't just do this

for x in nparr:
    if x == 0:
        something something

as that uses a lot more time than doing this

for x in nparr.tolist():
    if x == 0:
        something something

This is because a for loop iterating over a numpy array does not result in a sequence of Python constants but in a sequence of numpy scalars which would result in comparing a numpy array to a constant. Converting the array into a list first before the for loop will then result in a sequence of constants.

Here is some profiling I've done using cProfile to check different ways to do an 'if' on a numpy array element:

import cProfile
import numpy as np

runs = 1000000

print('Comparing numpy to numpy')
x = np.array(1.0, np.float32)
y = np.array(1.0, np.float32)
cProfile.run('''
for _ in range(runs):
    if x == y:
        pass
''')
print()

print('Comparing numpy to constant')
x = np.array(1.0, np.float32)
cProfile.run('''
for _ in range(runs):
    if x == 1.0:
        pass
''')
print()

print('Comparing constant to constant')
x = 1.0
cProfile.run('''
for _ in range(runs):
    if x == 1.0:
        pass
''')
print()

print('Comparing numpy.tolist() to constant')
x = np.array(1.0, np.float32)
cProfile.run('''
for _ in range(runs):
    if x.tolist() == 1.0:
        pass
''')
print()

print('Comparing numpy to numpy.array(constant)')
x = np.array(1.0, np.float32)
cProfile.run('''
for _ in range(runs):
    if x == np.array(1.0, np.float32):
        pass
''')
print()

print('Comparing numpy.tolist() to numpy.tolist()')
x = np.array(1.0, np.float32)
y = np.array(1.0, np.float32)
cProfile.run('''
for _ in range(runs):
    if x.tolist() == y.tolist():
        pass
''')
print()

Here are the results in order of speed:

Comparing constant to constant:	0.088 seconds
Comparing numpy.tolist() to constant:	0.288 seconds
Comparing numpy.tolist() to numpy.tolist():	0.508 seconds
Comparing numpy to numpy:	0.684 seconds
Comparing numpy to constant:	1.192 seconds
Comparing numpy to numpy.array(constant):	1.203 seconds

It turns out that it is always faster to first convert your numpy scalars into constants via .tolist() than to do anything with them as numpy scalars.

The McLauren series and Taylor series (approximating complicated functions with simple polynomials)

The McLauren series

Imagine you had a function $f(x)$ that you knew was a polynomial, but whose details were unknown and you could only apply operations to the function without being able to read it. How could you find the coefficients of this polynomial? We know that for coefficients $a_i$:

$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + ...$

If we find $f(0)$ then we can find $a_0$.

$f(0) = a_0 + a_1 0 + a_2 0^2 + a_3 0^3 + a_4 0^4 + ... = a_0 + 0 + 0 + ... = a_0$

That was easy, but how can we find $a_1$? We need an operation that gets rid of $a_0$ and also the $x$ in the term $a_1 x$. That operation turns out to be differentiation with respect to $x$:

$f'(x) = a_1 + 2 a_2 x + 3 a_3 x^2 + 4 a_4 x^3 + ...$

Great! Now we can find $a_1$ by replacing $x$ with 0:

$f'(0) = a_1 + 2 a_2 0 + 3 a_3 0^2 + 4 a_4 0^3 + ... = a_1$

We can find $a_2$ by repeating these two steps:

$f''(x) = 2 a_2 + 2 \cdot 3 a_3 x + 3 \cdot 4 a_4 x^2 + ...$
$f''(0) = 2 a_2 + 2 \cdot 3 a_3 0 + 3 \cdot 4 a_4 0^2 + ... = 2 a_2$

What we found is twice of $a_2$ which means that we need to divide by 2 to get $a_2$. The differentiation operation is multiplying constants by each coefficient and the constants get bigger and bigger the more times we apply differentiation. You might notice that what's happening is that $a_0$ was multiplied by 1, $a_1$ was also multiplied by 1, $a_2$ has been multiplied by 2, $a_3$ will be multiplied by 6, $a_4$ by 24, and so on. These are factorials, sequences of whole numbers multiplied together ($3! = 1 \times 2 \times 3 = 6$). Which means that we need to divide by the next factorial after each round of differentiation and substitution by zero.

In general we can find the $i$th coefficient in an unknown polynomial function by doing the following:

$a_i = \frac{f^i(0)}{i!}$

That's great. Now to test it. Let's see if a complex function is actually a polynomial in disguise. Take something simple such as $f(x) = e^x$. This doesn't look like a polynomial, but it may be represented by a polynomial with an infinite number of terms. Let's find what are the coefficients of the hidden polynomial in $e^x$.

$f(x) = e^x = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + ...$
$\frac{f(0)}{0!} = \frac{e^0}{1} = a_0$
$\frac{f(0)}{0!} = 1$

OK, so $a_0$ is 1. Let's find the rest of the coefficients.

$a_1 = \frac{f'(0)}{1!} = \frac{e^0}{1} = 1$
$a_2 = \frac{f''(0)}{2!} = \frac{e^0}{2} = \frac{1}{2}$
$a_3 = \frac{f'''(0)}{3!} = \frac{e^0}{6} = \frac{1}{6}$
$...$

So the first few terms of the polynomial hidden within $e^x$ are:
$f(x) = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + ...$

Does this partial polynomial look anything like $e^x$ when plotted as a graph?



Pretty good within a boundary! Note how the curve has a perfect fit at $x = 0$ and that it gets worse as we move away from there. Adding more terms to the polynomial will enlarge the area around $x = 0$ that is close to the curve but it will always be radiating out from there.

Let's try for $f(x) = cos(x)$ now.

$a_0 = \frac{f(0)}{0!} = \frac{cos(0)}{1} = 1$
$a_1 = \frac{f'(0)}{1!} = \frac{-sin(0)}{1} = 0$
$a_2 = \frac{f''(0)}{2!} = \frac{-cos(0)}{2} = -\frac{1}{2}$
$a_3 = \frac{f'''(0)}{3!} = \frac{sin(0)}{6} = 0$
$...$

So the first few terms of the polynomial hidden within $cos(x)$ is
$f(x) = 1 - \frac{1}{2} x^2 + ...$

The Taylor series

As you can see, this "polynomialisation" of functions is a neat way to approximate a function we might not know how to implement exactly but know how to differentiate and how to find its value when $x$ is 0. But what if we don't know what $f(0)$ is such as in $ln(x)$ or $\frac{1}{x}$? A slight modification to our method allows us to use any value of $x$ and not just 0. Let's call this value $b$. By slightly modifying the polynomial we expect to be hiding inside the function, we can make the polynomial act the same way when $f(b)$ is used instead of $f(0)$:

$f(x) = a_0 + a_1 (x - b) + a_2 (x - b)^2 + a_3 (x - b)^3 + a_4 (x - b)^4 + ...$
$f(b) = a_0 + a_1 (b - b) + a_2 (b - b)^2 + a_3 (b - b)^3 + a_4 (b - b)^4 + ...$
$f(b) = a_0$

$f'(x) = a_1 + 2 a_2 (x - b) + 3 a_3 (x - b)^2 + 4 a_4 (x - b)^3 + ...$
$f'(b) = a_1$

$a_i = \frac{f^i(b)}{i!}$

The catch here is that we are now finding coefficients to the polynomial $a_0 + a_1 (x - b) + a_2 (x - b)^2 + ...$ and not of $a_0 + a_1 x + a_2 x^2 + ...$, but that's OK. Let's try this on $ln(x)$ with $b = 1$.

$a_0 = \frac{f(1)}{0!} = \frac{ln(1)}{1} = 0$
$a_1 = \frac{f'(1)}{1!} = \frac{\frac{1}{1}}{1} = 1$
$a_2 = \frac{f''(1)}{2!} = \frac{-\frac{1}{1^2}}{1} = -1$
$a_3 = \frac{f'''(1)}{3!} = \frac{\frac{2}{1^3}}{1} = 2$
$...$

So the first few terms of the polynomial hidden within $ln(x)$ is
$f(x) = (x - 1) - (x - 1)^2 + 2 (x - 1)^3 + ...$



Adding more terms will approximate the original function better and better but what if we didn't have to? Remember how I said in the previous section that the polynomial approximates the original function best close to $x = 0$. Well now we can approximate it best around any point $b$ and not just around 0. This means that if our function has multiple known values, such as $cos(x)$ which is known to be 1 at $x = 0$, 0 at $x = \frac{\pi}{2}$, -1 at $x = \pi$, etc., then we can use several short polynomials centered at different points in the function instead of one large polynomial that approximates it well over a large interval. Let's try approximating $cos(x)$ around $x = \pi$, which means that we'll set $b$ to $\pi$.

$a_0 = \frac{f(\pi)}{0!} = \frac{cos(\pi)}{1} = -1$
$a_1 = \frac{f'(\pi)}{1!} = \frac{-sin(\pi)}{1} = 0$
$a_2 = \frac{f''(\pi)}{2!} = \frac{-cos(\pi)}{2} = \frac{1}{2}$
$a_3 = \frac{f'''(\pi)}{3!} = \frac{sin(\pi)}{6} = 0$
$...$

So the first few terms of the polynomial hidden within $cos(x)$ which best approximates the area around $x = \pi$ is
$f(x) = -1 + \frac{1}{2} (x - \pi)^2 + ...$



This is useful when implementing mathematical functions on a computer. You keep several simple polynomials defined at different points in the domain of the function and then pick the closest one to the $x$ you need to evaluate. You can then compute an approximation that isn't too bad without requiring a lot of computational time.

Hyperparameter tuning using Scikit-Optimize

One of my favourite academic discoveries this year was Scikit-Optimize, a nifty little Python automatic hyperparameter tuning library that comes with a lot of features I found missing in other similar libraries.

So as explained in an earlier blog post, automatic hyperparameter tuning is about finding the right hyperparameters for a machine learning algorithm automatically. Usually this is done manually using human experience but even simple MonteCarlo search random guesses can result in better performance than human tweaking (see here). So automatic methods were developed that try to explore the space of hyperparameters and their resulting performance after training the machine learning model and then try to home in on the best performing hyperparameters. Of course each time you want to evaluate a new hyperparameter combination you'd need to retrain and evaluate your machine learning model, which might take a very long time to finish, so it's important that a good hyperparameter combination is found with as little evaluations as possible. To do this we'll use Bayesian Optimization, a process where a separate simpler model is trained to predict the resulting performance of the whole hyperparameter space. We check this trained model to predict which hyperparameters will give the best resulting performance and actually evaluate them by training our machine learning model with them. The actual resulting performance is then used to update the hyperparameter space model so that it makes better predictions and then we'll get a new promising hyperparameter combination from it. This is repeated for a set number of times. Now the most common hyperparameter space model to use is a Gaussian Process which maps continuous numerical hyperparameters to a single number which is the predicted performance. This is not very good when your hyperparameters contain categorical data such as a choice of activation function. There is a paper that suggests that random forests are much better at mapping general hyperparameter combinations to predicted performance.

Now that we got the theory out of the way, let's see how to use the library. We'll apply it on a gradient descent algorithm that needs to minimize the squared function. For this we'll need 3 hyperparameters: the range of the initial value to be selected randomly, the learning rate, and the number of epochs to run. So we have two continuous values and one discrete integer value.

import random

def cost(x):
    return x**2

def cost_grad(x):
    return 2*x

def graddesc(learning_rate, max_init_x, num_epochs):
    x = random.uniform(-max_init_x, max_init_x)
    for _ in range(num_epochs):
        x = x - learning_rate*cost_grad(x)
    return x

Now we need to define the skopt optimizer:

import skopt

opt = skopt.Optimizer(
            [
                skopt.space.Real(0.0, 10.0, name='max_init_x'),
                skopt.space.Real(1.0, 1e-10, 'log-uniform', name='learning_rate'),
                skopt.space.Integer(1, 20, name='num_epochs'),
            ],
            n_initial_points=5,
            base_estimator='RF',
            acq_func='EI',
            acq_optimizer='auto',
            random_state=0,
        )

The above code is specifying 3 hyperparameters:

• the maximum initial value that is a real number (continuous) and that can be between 10 and 0
• the learning rate that is also a real number but that is also on a logarithmic scale (so that you are equally likely to try very large values and very small values) and can be between 1 and 1e-10
• the number of epochs that is an integer (whole number) and that can be between 1 and 20

It is also saying that the hyperparameter space model should be initialized based on 5 random hyperparameter combinations (you train the hyperparameter space model on 5 randomly chosen hyperparameters in order to be able to get the first suggested hyperparameter), that this model should be a random forest (RF), that the acquisition function (the function to decide which hyperparameter combination is the most promising to try next according to the hyperparameter space model) is the expected improvement (EI) of the hyperparameter combination, that the acquisition optimizer (the optimizer to find the next promising hyperparameter combination) is automatically chosen, and that the random state is set to a fixed number (zero) so that it always gives the same random values each time you run it.

Next we will use the optimizer to find good hyperparameter combinations.

best_cost = 1e100
best_hyperparams = None
for trial in range(5 + 20):
    [max_init_x, learning_rate, num_epochs] = opt.ask()
    [max_init_x, learning_rate, num_epochs] = [max_init_x.tolist(), learning_rate.tolist(), num_epochs.tolist()]
    next_hyperparams = [max_init_x, learning_rate, num_epochs]
    next_cost = cost(graddesc(max_init_x, learning_rate, num_epochs))
    if next_cost < best_cost:
        best_cost = next_cost
        best_hyperparams = next_hyperparams
    print(trial+1, next_cost, next_hyperparams)
    opt.tell(next_hyperparams, next_cost)
print(best_hyperparams)

The nice thing about this library is that you can use an 'ask/tell' system where you ask the library to give you the next hyperparameters to try and then you do something with them in order to get the actual performance value and finally you tell the library what this performance value is. This lets you do some nifty things such as ask for another value if the hyperparameters resulted in an invalid state in the machine learning model or even to save your progress and continue later.

In the above code we're running a for loop to run the number of times we want to evaluate different hyperparameters. We need to run it for the 5 random values we specified before to initialize the hyperparameter space model plus another 20 evaluations to actually optimize the hyperameters. Now skopt does something funny which is that it returns not plain Python values for hyperparameters but rather each number is represented as a numpy scalar. Because of this we convert each numpy scalar back into a plain Python float or int using ".tolist()". We ask for the next hyperparamters to try, convert them to plain Python values, get their resulting cost after running gradient descent, store the best hyperparameters found up to now, and tell the library what the given hyperparameters' resulting performance was. At the end we print the best hyperparamter combination found.

Some extra stuff:

• You can ask for categorical hyperparameters using "skopt.space.Categorical(['option1', 'option2'], name='options')" which will return one of the values in the list when calling "ask".
• You can ask for a different hyperparameter combination in case of an invalid one by changing "ask" to give you several hyperparameter suggestions rather than just one and then trying each one of them until one works by using "opt.ask(num_hyperpars)" (you can also incrementally ask for more values and always take the last one).
• You can save and continue by saving all the hyperparameters evaluated and their corresponding performance value in a text file. You then later resupply the saved hyperparameters and their performance using "tell" for each one. This is much faster than actually evaluating them on the machine learning model so straight supplying known values will be ready quickly. Just be careful that you also call "ask" before each "tell" in order to get the same exact behaviour from the optimizer or else the next "tell" will give different values from what it would have given had you not loaded the previous ones manually.

Saving/Loading a Tensorflow model using HDF5 (h5py)

The normal way to save the parameters of a neural network in Tensorflow is to create a tf.train.Saver() object and then calling the object's "save" and "restore" methods. But this can be a bit cumbersome and you might prefer to have more control on what and how things get saved and loaded. The standard file format for saving large tensors (such as the parameters of a neural network) is to use HDF5.

Saving is easy. Here is the code you'll need:

with h5py.File('model.hdf5', 'w') as f:
    for var in tf.trainable_variables():
        key = var.name.replace('/', ' ')
        value = session.run(var)
        f.create_dataset(key, data=value)

Notice that we need to use the session in order to get a parameter value.

If you're using variable scopes then your variable names will have slashes in them and here we're replacing slashes with spaces. The reason is because the HDF5 format treats key values as directories where folder names are separated by slashes. This means that you need to traverse the keys recursively in order to arrive at the data (one folder name at a time) if you do not know the full name at the start. This replacement of slashes simplifies the code for loading a model later.

Notice also that you can filter the variables to save as you like as well as save extra stuff. I like to save the Tensorflow version in the file in order to be able to check for incompatible variable names in contrib modules (RNNs had some variable names changed in version 1.2).

Now comes the loading part. Loading is a tiny bit more involved because it requires that you make you neural network code include stuff for assigning values to the variables. All you need to do whilst constructing your Tensorflow graph is to include the following code:

param_setters = dict()
for var in tf.trainable_variables():
    placeholder = tf.placeholder(var.dtype, var.shape, var.name.split(':')[0]+'_setter')
    param_setters[var.name] = (tf.assign(var, placeholder), placeholder)

What this code does is it creates separate placeholder and assign nodes for each variable in your graph. In order to modify a variable you need to run the corresponding assign node in a session and pass the value through the corresponding placeholder. All the corresponding assign nodes and placeholders are kept in a dictionary called param_setters. We're also naming the placeholder the same as the variable but with '_setter' at the end.

Notice that param_setters is a dictionary mapping variable names to a tuple consisting of the assign node and the placeholder.

Now we can load the HDF5 file as follows:

with h5py.File('model.hdf5', 'r') as f:
    for (name, val) in f.items()
        name = name.replace(' ', '/')
        val = np.array(val)
        session.run(param_setters[name][0], { param_setters[name][1]: val })

What's happening here is that we're loading each parameter from the file and replacing the spaces in names back into slashes. We then run the corresponding assign node for the given parameter name in param_setters and set it to the loaded value.

Fancy indexing in Tensorflow: Getting a different element from every row in a matrix

Let's say you have the following 4 by 3 matrix:

M = np.array([[  1,  2,  3 ],
              [  4,  5,  6 ],
              [  7,  8,  9 ],
              [ 10, 11, 12 ]])

When in Tensorflow we'd also have this line:

M = tf.constant(M)

Let's say that you want to get the first element of every row. In both Numpy and Tensorflow you can just do this:

x = M[:, 0]

which means 'get every row and from every row get the element at index 0'. "x" is now equal to:

np.array([1, 4, 7, 10])

Now let's say that instead of the first element of every row, you wanted the third element of the first row, the second element of the second row, the first element of the third row, and the second element of the fourth row. In Numpy, this is how you do that:

idx = [2,1,0,1]
x = M[[0,1,2,3], idx]

or equivalently:

x = M[np.arange(M.shape[0]), idx]

This is just breaking up the 'coordinates' of the desired elements into separate lists for each axis. "x" is now equal to:

np.array([3, 5, 7, 11])

Unfortunately this sort of fancy indexing isn't available in Tensorflow. Instead we have the function "tf.gather_nd" which lets us provide a list of 'coordinates'. Unlike Numpy, tf.gather_nd does not take separate lists per axis but expects a single list with one coordinate per item like this:

idx = tf.constant([[0,2],[1,1],[2,0],[3,1]], tf.int32)
x = tf.gather_nd(M, idx)

This is typically inconvenient as we usually have a single vector of indexes rather then a list of coordinates. It would be better to be able to just put a "range" like we did with Numpy. We can use the range and then join it to the vector of indexes sideways using "tf.stack", like this:

idx = tf.constant([2,1,0,1], tf.int32)
x = tf.gather_nd(M, tf.stack([tf.range(M.shape[0]), idx], axis=1))

Kind of bulky but at least it's possible. I miss Theano's Numpy-like interface.

Why Bahdanau's neural attention requires two layers

The neural attention model was first described for the task of machine translation in Bahdanau's Neural machine translation by jointly learning to align and translate. The source sentence is translated into the target sentence by attending to different words in the source sentence as the target sentence is generated one word at a time. The attended source sentence is defined as follows:

$$
\begin{align}
c_i &= \sum_j \alpha_{ij} s_j \\
\alpha_{ij} &= \frac{e^{z_{ij}}}{\sum_k e^{z_{ik}}} \\
z_{ij} &= W \tanh(V (s_j ++ p_{i-1}))
\end{align}
$$
where $c_i$ is the attended source sentence taken from the weighted sum of the source sentence vectors $s_j$ and the weights $\alpha_{ij}$, $p_{i-1}$ is the prefix vector produced by the 'decoder' RNN that remembers what has been generated thus far, $++$ means concatenation, and $W$ and $V$ are two weight matrices.

So the question we're asking is, why do we need to use two layers to produce $z_{ij}$? Can we do with just one layer?

In reality what happens when you use a single layer is that the attention weights will remain the same across time steps such that, although the attention will be different on different words in the source sentence, as the target sentence gets generated these words will keep receiving the same attention they did throughout the whole generation process. The reason for this is that softmax, which is the function the produces $\alpha_{ij}$, is shift invariant, that is, does not change if you add the same number to each of its inputs.

Let's say $z$ is defined as [ 1, 2, 3 ]. Then $\alpha$ will be
$$
\begin{matrix}
(\frac{e^1}{e^1 + e^2 + e^3}) & (\frac{e^2}{e^1 + e^2 + e^3}) & (\frac{e^3}{e^1 + e^2 + e^3})
\end{matrix}
$$
but if we add the constant k to each of the three numbers then the result will still be the same
$$
\begin{matrix}
& (\frac{e^{1+k}}{e^{1+k} + e^{2+k} + e^{3+k}}) & (\frac{e^{2+k}}{e^{1+k} + e^{2+k} + e^{3+k}}) & (\frac{e^{3+k}}{e^{1+k} + e^{2+k} + e^{3+k}}) \\
=& (\frac{e^1e^k}{e^1e^k + e^2e^k + e^3e^k}) & (\frac{e^2e^k}{e^1e^k + e^2e^k + e^3e^k}) & (\frac{e^3e^k}{e^1e^k + e^2e^k + e^3e^k}) \\
=& (\frac{e^1e^k}{e^k(e^1 + e^2 + e^3)}) & (\frac{e^2e^k}{e^k(e^1 + e^2 + e^3)}) & (\frac{e^3e^k}{e^k(e^1 + e^2 + e^3)}) \\
=& (\frac{e^1}{e^1 + e^2 + e^3}) & (\frac{e^2}{e^1 + e^2 + e^3}) & (\frac{e^3}{e^1 + e^2 + e^3})
\end{matrix}
$$

This proves that adding the same constant to every number in $z$ will leave the softmax unaltered. Softmax is shift invariant.

Now let's say that $z$ is determined by a single neural layer such that $z_{ij} = W (s_j ++ p_{i-1}) = W_0 s_j + W_1 p_{i-1}$. We can draw a matrix of all possible $z$ where the columns are the source vectors $s$ and the rows are the decoder prefix states $p$.
$$
\begin{matrix}
(W_0 s_0 + W_1 p_0) & (W_0 s_1 + W_1 p_0) & (W_0 s_2 + W_1 p_0) & \dots \\
(W_0 s_0 + W_1 p_1) & (W_0 s_1 + W_1 p_1) & (W_0 s_2 + W_1 p_1) & \dots \\
(W_0 s_0 + W_1 p_2) & (W_0 s_1 + W_1 p_2) & (W_0 s_2 + W_1 p_2) & \dots \\
\dots & \dots & \dots & \dots
\end{matrix}
$$

Given a single row, the $p$s are always the same, which means that the only source of variation between $z$s of the same prefix $p$ is from the source vectors $s$. This makes sense.

Now take the first two rows. What is the result of subtracting the second from the first?

$$
\begin{matrix}
(W_0 s_0 + W_1 p_0 - W_0 s_0 - W_1 p_1) & (W_0 s_1 + W_1 p_0 - W_0 s_1 - W_1 p_1) & (W_0 s_2 + W_1 p_0 - W_0 s_2 - W_1 p_1) & \dots \\
(W_1(p_0-p_1)) & (W_1(p_0-p_1)) & (W_1(p_0-p_1)) & \dots
\end{matrix}
$$

Notice how all the columns have the same difference, which means that the second column can be rewritten as:

$$
\begin{matrix}
(W_0 s_0 + W_1 p_0 + W_1(p_0-p_1)) & (W_0 s_1 + W_1 p_0 + W_1(p_0-p_1)) & (W_0 s_2 + W_1 p_0 + W_1(p_0-p_1)) & \dots
\end{matrix}
$$

We know that adding the same constant to every $z$ will leave the softmax unaltered, which means that every time step in the decoder RNN will lead to the same attention vector. The individual attention values will be different, but they will not change throughout the whole generation process. Using two layers with a non-linear activation function in the middle will disrupt this as the difference between two consecutive $z$ will now be different at each time step.