# The Inequality

Math and computer science are full of inequalities, but there is one that shows up more often in my work than any other. Of course, I’m talking about

$\displaystyle 1+x \leq e^{x}$

This is The Inequality. I’ve been told on many occasions that the entire field of machine learning reduces to The Inequality combined with the Chernoff bound (which is proved using The Inequality).

Why does it show up so often in machine learning? Mostly because in analyzing an algorithm you want to bound the probability that some bad event happens. Bad events are usually phrased similarly to

$\displaystyle \prod_{i=1}^m (1-p_i)$

And applying The Inequality we can bound this from above by

$\displaystyle\prod_{i=1}^m (1-p_i) \leq \prod_{i=1}^m e^{-p_i} = e^{-\sum_{i=1}^m p_i}$

The point is that usually $m$ is the size of your dataset, which you get to choose, and by picking larger $m$ you make the probability of the bad event vanish exponentially quickly in $m$. (Here $p_i$ is unrelated to how I am about to use $p_i$ as weights).

Of course, The Inequality has much deeper implications than bounds for the efficiency and correctness of machine learning algorithms. To convince you of the depth of this simple statement, let’s see its use in an elegant proof of the arithmetic geometric inequality.

Theorem: (The arithmetic-mean geometric-mean inequality, general version): For all non-negative real numbers $a_1, \dots, a_n$ and all positive $p_1, \dots, p_n$ such that $p_1 + \dots + p_n = 1$, the following inequality holds:

$\displaystyle a_1^{p_1} \cdots a_n^{p_n} \leq p_1 a_1 + \dots + p_n a_n$

Note that when all the $p_i = 1/n$ this is the standard AM-GM inequality.

Proof. This proof is due to George Polya (in Hungarian, Pólya György).

We start by modifying The Inequality $1+x \leq e^x$ by a shift of variables $x \mapsto x-1$, so that the inequality now reads $x \leq e^{x-1}$. We can apply this to each $a_i$ giving $a_i \leq e^{a_i - 1}$, and in fact,

$\displaystyle a_1^{p_1} \cdots a_n^{p_n} \leq e^{\sum_{i=1}^n p_ia_i - p_i} = e^{\left ( \sum_{i=1}^n p_ia_i \right ) - 1}$

Now we have something quite curious: if we call $A$ the sum $p_1a_1 + \dots + p_na_n$, the above shows that $a_1^{p_1} \cdots a_n^{p_n} \leq e^{A-1}$. Moreover, again because $A \leq e^{A-1}$ that shows that the right hand side of the inequality we’re trying to prove is also bounded by $e^{A-1}$. So we know that both sides of our desired inequality (and in particular, the max) is bounded from above by $e^{A-1}$. This seems like a conundrum until we introduce the following beautiful idea: normalize by the thing you think should be the larger of the two sides of the inequality.

Define new variables $b_i = a_i / A$ and notice that $\sum_i p_i b_i = 1$ just by unraveling the definition. Call this sum $B = \sum_i p_i b_i$. Now we know that

$b_1^{p_1} \cdots b_n^{p_n} = \left ( \frac{a_1}{A} \right )^{p_1} \cdots \left ( \frac{a_n}{A} \right )^{p_n} \leq e^{B - 1} = e^0 = 1$

Now we unpack the pieces, and multiply through by $A^{p_1}A^{p_2} \cdots A^{p_n} = A$, the result is exactly the AM-GM inequality.

$\square$

Even deeper, there is only one case when The Inequality is tight, i.e. when $1+x = e^x$, and that is $x=0$. This allows us to use the proof above to come to a full characterization of the case of equality in the proof above. Indeed, the crucial step was that $(a_i / A) = e^{A-1}$, which is only true when $(a_i / A) = 1$, i.e. when $a_i = A$. Spending a few seconds thinking about this gives the characterization of equality if and only if $a_1 = a_2 = \dots = a_n = A$.

So this is excellent: the arithmetic-geometric inequality is a deep theorem with applications all over mathematics and statistics. Adding another layer of indirection for impressiveness, one can use the AM-GM inequality to prove the Cauchy-Schwarz inequality rather directly. Sadly, the Wikipedia page for the Cauchy-Schwarz inequality hardly does it justice as far as the massive number of applications. For example, many novel techniques in geometry and number theory are proved directly from C-S. More, in fact, than I can hope to learn.

Of course, no article about The Inequality could be complete without a proof of The Inequality.

Theorem: For all $x \in \mathbb{R}$, $1+x \leq e^x$.

Proof. The proof starts by proving a simpler theorem, named after Bernoulli, that $1+nx \leq (1+x)^n$ for every $x [-1, \infty)$ and every $n \in \mathbb{N}$. This is relatively straightforward by induction. The base case is trivial, and

$\displaystyle (1+x)^{n+1} = (1+x)(1+x)^n \geq (1+x)(1+nx) = 1 + (n+1)x + nx^2$

And because $nx^2 \geq 0$, we get Bernoulli’s inequality.

Now for any $z \geq 0$ we can set $x = z/n$, and get $(1+z) = (1+nx) \leq (1+\frac{z}{n})^n$ for every $n$. Note that Bernoulli’s inequality is preserved for larger and larger $n$ because $x \geq 0$. So taking limits of both sides as $n \to \infty$ we get the definition of $e^z$ on the right hand side of the inequality. We can prove a symmetrical inequality for $-x$ when $x < 0$, and this proves the theorem.

$\square$

What other insights can we glean about The Inequality? For one, it’s a truncated version of the Taylor series approximation

$\displaystyle e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

Indeed, the Taylor remainder theorem tells us that the first two terms approximate $e^x$ around zero with error depending on some constant times $e^x x^2 \geq 0$. In other words, $1+x$ is a lower bound on $e^x$ around zero. It is perhaps miraculous that this extends to a lower bound everywhere, until you realize that exponentials grow extremely quickly and lines do not.

One might wonder whether we can improve our approximation with higher order approximations. Indeed we can, but we have to be a bit careful. In particular, $1+x+x^2/2 \leq e^x$ is only true for nonnegative $x$ (because the remainder theorem now applies to $x^3$, but if we restrict to odd terms we win: $1+x+x^2/2 + x^3/6 \leq e^x$ is true for all $x$.

What is really surprising about The Inequality is that, at least in the applications I work with, we rarely see higher order approximations used. For most applications, The difference between an error term which is quadratic and one which is cubic or quartic is often not worth the extra work in analyzing the result. You get the same theorem: that something vanishes exponentially quickly.

If you’re interested in learning more about the theory of inequalities, I wholeheartedly recommend The Cauchy-Schwarz Master Class. This book is wonderfully written, and chock full of fun exercises. I know because I do exercises from books like this one on planes and trains. It’s my kind of sudoku 🙂

# Simulating a Biased Coin with a Fair Coin

This is a guest post by my friend and colleague Adam Lelkes. Adam’s interests are in algebra and theoretical computer science. This gem came up because Adam gave a talk on probabilistic computation in which he discussed this technique.

Problem: simulate a biased coin using a fair coin.

Solution: (in Python)

def biasedCoin(binaryDigitStream, fairCoin):
for d in binaryDigitStream:
if fairCoin() != d:
return d


Discussion: This function takes two arguments, an iterator representing the binary expansion of the intended probability of getting 1 (let us denote it as $p$) and another function that returns 1 or 0 with equal probability. At first glance this might seem like an overcomplicated way of solving this problem: why can’t the probability be a floating point number?

The point is that $p$ can have infinite precision! Assuming that fairCoin() gives us a perfectly random stream of 1’s and 0’s (independently and with probability 1/2) and we can read each bit of the binary expansion of $p$, this function returns 1 with probability exactly $p$ even if $p$ is irrational or a fraction with infinite decimal expansion. If we used floating point arithmetic there would be a small chance we get unlucky and exhaust the precision available. We would only get an approximation of the true bias at best.

Now let us explain why this algorithm works. We keep tossing our fair coins to get a sequence of random bits, until one of our random bits is different from the corresponding bit in the binary expansion of $p$. If we stop after $i$ steps, that means that the first $i-1$ bits in the two binary sequences were the same, which happens with probability $\frac{1}{2^{i-1}}$. Given that this happens, in the $i$th step we will return the $i$th bit of $p$; let us denote this bit by $p_i$. Then the probability of returning 1 is $\sum_{i=1}^\infty \frac{p_i}{2^{i-1}}$, which is the binary expansion of $p$.

This algorithm is also efficient. By efficient here we mean that the expected running time is constant. Of course, to show this we need to make some assumption about the computational complexity of calculating the bits of $p$. If we assume that the bits of $p$ are efficiently computable in the sense that the time required to compute $p_i$ is bounded by a polynomial in $i$, then this algorithm does run in constant expected time.

Indeed, the expected running time is $\sum_{i=0}^\infty \frac{i^n}{2^i}$. Showing that this sum is a constant is an easy calculus exercise: using the ratio test we get that

$\displaystyle \textup{limsup}_{i \to \infty} \left | \frac{\frac{(i+1)^n}{2^{i+1}}}{\frac{i^n}{2^i}} \right | = \limsup_{i\to\infty} \frac{\left(\frac{i+1}{i}\right)^n}{2} = \frac{1}{2} < 1$,

thus the series is convergent.

Now that we proved that our algorithm works, it’s time to try it! Let’s say that we want to simulate a coin which gives “heads” with probability 1/3.
We need to construct our binary digit stream. Since 1/3 is 0.010101… in binary, we could use the following simple generator:

def oneThird():
while True:
yield 0
yield 1


However, we might want to have a more general generator that gives us the binary representation of any number. The following function, which takes a number between 0 and 1 as its argument, does the job:

def binaryDigits(fraction):
while True:
fraction *= 2
yield int(fraction)
fraction = fraction % 1


We also need a fair coin simulator. For this simulation, let’s just use Python’s built-in pseudo-random number generator:

def fairCoin():
return random.choice([0,1])


Let us toss our biased coin 10000 times and take the sum. We expect the sum to be around 3333. Indeed, when I tried

>>> sum(biasedCoin(oneThird(), fairCoin) for i in range(10000))
3330


It might be worth noting oneThird() is approximately ten times faster than binaryDigits(fractions.Fraction(1,3)), so when a large number of biased coins is needed, you can hardwire the binary representation of $p$ into the program.

# The Two-Dimensional Fourier Transform and Digital Watermarking

We’ve studied the Fourier transform quite a bit on this blog: with four primers and the Fast Fourier Transform algorithm under our belt, it’s about time we opened up our eyes to higher dimensions.

Indeed, in the decades since Cooley & Tukey’s landmark paper, the most interesting applications of the discrete Fourier transform have occurred in dimensions greater than 1. But for all our work we haven’t yet discussed what it means to take an “n-dimensional” Fourier transform. Our past toiling and troubling will pay off, though, because the higher Fourier transform and its 1-dimensional cousin are quite similar. Indeed, the shortest way to describe the $n$-dimensional transform is as the 1-dimensional transform with inner products of vector variables replacing regular products of variables.

In this post we’ll flush out these details. We’ll define the multivariable Fourier transform and it’s discrete partner, implement an algorithm to compute it (FFT-style), and then apply the transform to the problem of digitally watermarking images.

As usual, all the code, images, and examples used in this post are available on this blog’s Github page.

## Sweeping Some Details Under the Rug

We spent our first and second primers on Fourier analysis describing the Fourier series in one variable, and taking a limit of the period to get the Fourier transform in one variable. By all accounts, it was a downright mess of notation and symbol manipulation that culminated in the realization that the Fourier series looks a lot like a Riemann sum. So it was in one dimension, it is in arbitrary dimension, but to save our stamina for the applications we’re going to treat the $n$-dimensional transform differently. We’ll use the 1-dimensional transform as a model, and magically generalize it to operate on a vector-valued variable. Then the reader will take it on faith that we could achieve the same end as a limit of some kind of multidimensional Fourier series (and all that nonsense with Schwarz functions and tempered distributions is left to the analysts), or if not we’ll provide external notes with the full details.

So we start with a real-valued (or complex-valued) function $f : \mathbb{R}^n \to \mathbb{R}$, and we write the variable as $x = (x_1, \dots, x_n)$, so that we can stick to using the notation $f(x)$. Rather than think of the components of $x$ as “time variables” as we did in the one-dimensional case, we’ll usually think of $x$ as representing physical space. And so the periodic behavior of the function $f$ represents periodicity in space. On the other hand our transformed variables will be “frequency” in space, and this will correspond to a vector variable $\xi = (\xi_1, \dots, \xi_n)$. We’ll come back to what the heck “periodicity in space” means momentarily.

Remember that in one dimension the Fourier transform was defined by

$\displaystyle \mathscr{F}f(s) = \int_{-\infty}^\infty e^{-2\pi ist}f(t) dt$.

And it’s inverse transform was

$\displaystyle \mathscr{F}^{-1}g(t) = \int_{-\infty}^\infty e^{2\pi ist}f(s) ds$.

Indeed, with the vector $x$ replacing $t$ and $\xi$ replacing $s$, we have to figure out how to make an analogous definition. The obvious thing to do is to take the place where $st$ is multiplied and replace it with the inner product of $x$ and $\xi$, which for this post I’ll write $x \cdot \xi$ (usually I write $\left \langle x, \xi \right \rangle$). This gives us the $n$-dimensional transform

$\displaystyle \mathscr{F}f(\xi) = \int_{\mathbb{R}^n} e^{-2\pi i x \cdot \xi}f(x) dx$,

and its inverse

$\displaystyle \mathscr{F}^{-1}g(t) = \int_{\mathbb{R}^n} e^{2\pi i x \cdot \xi}f( \xi ) d \xi$

Note that the integral is over all of $\mathbb{R}^n$. To give a clarifying example, if we are in two dimensions we can write everything out in coordinates: $x = (x_1, x_2), \xi = (\xi_1, \xi_2)$, and the formula for the transform becomes

$\displaystyle \mathscr{F}f(\xi_1, \xi_2) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-2 \pi i (x_1 \xi_1 + x_2 \xi_2)} f(\xi_1, \xi_2) dx_1 dx_2$.

Now that’s a nasty integral if I’ve ever seen one. But for our purposes in this post, this will be as nasty as it gets, for we’re primarily concerned with image analysis. So representing things as vectors of arbitrary dimension is more compact, and we don’t lose anything for it.

## Periodicity in Space? It’s All Mostly the Same

Because arithmetic with vectors and arithmetic with numbers is so similar, it turns out that most of the properties of the 1-dimensional Fourier transform hold in arbitrary dimension. For example, the duality of the Fourier transform and its inverse holds, because for vectors $e^{-2 \pi i x \cdot (-\xi)} = e^{2 \pi i x \cdot \xi}$. So just like in on dimension, we have

$\mathscr{F}f(-\xi) = \mathscr{F}^{-1}f(\xi)$

And again we have correspondences between algebraic operations: convolution in the spatial domain corresponds to convolution in the frequency domain, the spectrum is symmetric about the origin, etc.

At a more geometric level, though, the Fourier transform does the same sort of thing as it did in the one-dimensional case. Again the complex exponentials form the building blocks of any function we want, and performing a Fourier transform on an $n$-dimensional function decomposes that function into its frequency components. So a function that is perfectly periodic corresponds to a Fourier spectrum that’s perfectly concentrated at a point.

But what the hell, the reader might ask, is ‘periodicity in space’? Since we’re talking about images anyway, the variables we care about (the coordinates of a pixel) are spatial variables. You could, if you were so inclined, have a function of multiple time variables, and to mathematicians a physical interpretation of dimension is just that, an interpretation. But as confusing as it might sound, it’s actually not so hard to understand the Fourier transform when it’s specialized to image analysis. The idea is that complex exponentials $e^{\pm 2 \pi i s \cdot \xi}$ oscillate in the $x$ variable for a fixed $\xi$ (and since $\mathscr{F}$ has $\xi$ as its input, we do want to fix $\xi$). The brief mathematical analysis goes like this: if we fix $\xi$ then the complex exponential is periodic with magnitudinal peaks along parallel lines spaced out at a distance of $1/ \left \| \xi \right \|$ apart. In particular any image is a sum of a bunch of these “complex exponential with a fixed $\xi$” images that look like stripes with varying widths and orientations (what you see here is just the real part of a particular complex exponential).

Any image can be made from a sum of a whole lot of images like the ones on top. They correspond to single points in the Fourier spectrum (and their symmetries), as on bottom.

What you see on top is an image, and on bottom its Fourier spectrum. That is, each brightly colored pixel corresponds to a point $[x_1, x_2]$ with a large magnitude for that frequency component $|\mathscr{F}f[x_1, x_2]|$.

It might be a bit surprising that every image can be constructed as a sum of stripey things, but so was it that any sound can be constructed as a sum of sines and cosines. It’s really just a statement about a basis of some vector space of functions. The long version of this story is laid out beautifully in pages 4 – 7 of these notes. The whole set of notes is wonderful, but this section is mathematically tidy and needs no background; the remainder of the notes outline the details about multidimensional Fourier series mentioned earlier, as well as a lot of other things. In higher dimensions the “parallel lines” idea is much the same, but with lines replaced by hyperplanes normal to the given vector.

## Discretizing the Transform

Recall that for a continuous function $f$ of one variable, we spent a bit of time figuring out how to find a good discrete approximation of $f$, how to find a good discrete approximation of the Fourier transform $\mathscr{F}f$, and how to find a quick way to transition between the two. In brief: $f$ was approximated by a vector of samples $(f[0], f[1], \dots, f[N])$, reconstructed the original function (which was only correct at the sampled points) and computed the Fourier transform of that, calling it the discrete Fourier transform, or DFT. We got to this definition, using square brackets to denote list indexing (or vector indexing, whatever):

Definition: Let $f = (f[1], \dots f[N])$ be a vector in $\mathbb{R}^N$. Then the discrete Fourier transform of $f$ is defined by the vector $(\mathscr{F}f[1], \dots, \mathscr{F}f[N])$, where

$\displaystyle \mathscr{F}f[j] = \sum_{k=0}^{N-1} f[k]e^{-2 \pi i jk/N}$

Just as with the one-dimensional case, we can do the same analysis and arrive at a discrete approximation of an $n$-dimensional function. Instead of a vector it would be an $N \times N \times \dots \times N$ matrix, where there are $n$ terms in the matrix, one for each variable. In two dimensions, this means the discrete approximation of a function is a matrix of samples taken at evenly-spaced intervals in both directions.

Sticking with two dimensions, the Fourier transform is then a linear operator taking matrices to matrices (which is called a tensor if you want to scare people). It has its own representation like the one above, where each term is a double sum. In terms of image analysis, we can imagine that each term in the sum requires us to look at every pixel of the original image

Definition: Let $f = (f[s,t])$ be a vector in $\mathbb{R}^N \times \mathbb{R}^M$, where $s$ ranges from $0, \dots, N-1$ and $t$ from $0, \dots, M-1$. Then the discrete Fourier transform of $f$ is defined by the vector $(\mathscr{F}f[s,t])$, where each entry is given by

$\displaystyle \mathscr{F}f[x_1, x_2] = \sum_{s=0}^{N-1} \sum_{t=0}^{M-1} f[s, t] e^{-2 \pi i (s x_1 / N + t x_2 / M)}$

In the one-dimensional case the inverse transform had a sign change in the exponent and an extra $1/N$ normalization factor. Similarly, in two dimensions the inverse transform has a normalization factor of $1/NM$ (1 over the total number of samples). Again we use a capital $F$ to denote the transformed version of $f$. The higher dimensional transforms are analogous: you get $n$ sums, one for each component, and the normalization factor is the inverse of the total number of samples.

$\displaystyle \mathscr{F}^{-1}F[x_1, x_2] = \frac{1}{NM} \sum_{s=0}^{N-1} \sum_{t=0}^{M-1} f[s,t] e^{2 \pi i (sx_1 / N + tx_2 / M)}$

Unfortunately, the world of the DFT disagrees a lot on the choice of normalization factor. It turns out that all that really matters is that the exponent is negated in the inverse, and that the product of the constant terms on both the transform and its inverse is $1/NM$. So some people will normalize both the Fourier transform and its inverse by $1/ \sqrt{NM}$. The reason for this is that it makes the transform and its inverse more similar-looking (it’s just that, cosmetic). The choice of normalization isn’t particularly important for us, but beware: non-canonical choices are out there, and they do affect formulas by adding multiplicative constants.

## The Fast Fourier Transform, Revisited

Now one might expect that there is another clever algorithm to drastically reduce the runtime of the 2-dimensional DFT, akin to the fast Fourier transform algorithm (FFT). But actually there is almost no additional insight required to understand the “fast” higher dimensional Fourier transform algorithm, because all the work was done for us in the one dimensional case.

All that we do is realize that each of the inner summations is a 1-dimensional DFT. That is, if we write the inner-most sum as a function of two parameters

$\displaystyle g(s, x_2) = \sum_{t=0}^{M-1} f(s,t) e^{-2 \pi i (tx_2 / M)}$

then the 2-dimensional FFT is simply

$\displaystyle \mathscr{F}f[x_1, x_2] = \sum_{s=0}^{N-1} g(s, x_2) e^{-2 \pi i (sx_1/N)}$

But now notice, that we can forget that $g(s,x_2)$ was ever a separate, two-dimensional function. Indeed, since it only depends on the $x_2$ parameter from out of the sum this is precisely the formula for a 1-dimensional DFT! And so if we want to compute the 2-dimensional DFT using the 1-dimensional FFT algorithm, we can compute the matrix of 1-dimensional DFT entries for all choices of $s, x_2$ by fixing each value of $s$ in turn and running FFT on the resulting “column” of values. If you followed the program from our last FFT post, then the only difficulty is in understanding how the data is shuffled around and which variables are fixed during the computation of the sub-DFT’s.

To remedy the confusion, we give an example. Say we have the following 3×3 matrix whose DFT we want to compute. Remember, these values are the sampled values of a 2-variable function.

$\displaystyle \begin{pmatrix} f[0,0] & f[0,1] & f[0,2] \\ f[1,0] & f[1,1] & f[1,2] \\ f[2,0] & f[2,1] & f[2,2] \end{pmatrix}$

The first step in the algorithm is to fix a choice of row, $s$, and compute the DFT of the resulting row. So let’s fix $s = 0$, and then we have the resulting row

$\displaystyle f_0 = (f[0,0], f[0,1], f[0,2])$

It’s DFT is computed (intentionally using the same notation as the inner summation above), as

$\displaystyle g[0,x_2] = (\mathscr{F}f_0)[x_2] = \sum_{t=0}^{M-1} f_0[t] e^{- 2 \pi i (t x_2 / M)}$

Note that $f_0[t] = f[s,t]$ for our fixed choice of $s=0$. And so if we do this for all $N$ rows (all 3 rows, in this example), we’ll have performed $N$ FFT’s of size $M$ to get a matrix of values

$\displaystyle \begin{pmatrix} g[0,0] & g[0,1] & g[0,2] \\ g[1,0] & g[1,1] & g[1,2] \\ g[2,0] & g[2,1] & g[2,2] \end{pmatrix}$

Now we want to compute the rest of the 2-dimensional DFT to the end, and it’s easy: now each column consists of the terms in the outermost sum above (since $s$ is the iterating variable). So if we fix a value of $x_2$, say $x_2 = 1$, we get the resulting column

$\displaystyle g_1 = (g[0, 1], g[1,1], g[2,1])$

and computing a DFT on this row gives

$\displaystyle \mathscr{F}f[x_1, 1] = \sum_{s=0}^{N-1} g_1[s] e^{-2 \pi i sx_1 / N}$.

Expanding the definition of $g$ as a DFT gets us back to the original formula for the 2-dimensional DFT, so we know we did it right. In the end we get a matrix of the computed DFT values for all $x_1, x_2$.

Let’s analyze the runtime of this algorithm: in the first round of DFT’s we computed $N$ DFT’s of size $M$, requiring a total of $O(N M \log M)$, since we know FFT takes time $O(M \log M)$ for a list of length $M$. In the second round we did it the other way around, computing $M$ DFT’s of size $N$ each, giving a total of

$O(NM \log M + NM \log N) = O(NM (\log N + \log M)) = O(NM \log (NM))$

In other words, if the size of the image is $n = NM$, then we are achieving an $O(n \log n)$-time algorithm, which was precisely the speedup that the FFT algorithm gave us for one-dimension. We also know a lower bound on this problem: we can’t do better than $NM$ since we have to look at every pixel at least once. So we know that we’re only a logarithmic factor away from a trivial lower bound. And indeed, all other known DFT algorithms have the same runtime. Without any assumptions on the input data (or any parallelization), nobody knows of a faster algorithm.

Now let’s turn to the code. If we use our FFT algorithm from last time, the pure Python one (read: very slow), then we can implement the 2D Fourier transform in just two lines of Python code. Full disclosure: we left out some numpy stuff in this code for readability. You can view the entire source file on this blog’s Github page.

def fft2d(matrix):
fftRows = [fft(row) for row in matrix]
return transpose([fft(row) for row in transpose(fftRows)])


And we can test it on a simple matrix with one nonzero value in it:

A = [[0,0,0,0], [0,1,0,0], [0,0,0,0], [0,0,0,0]]
for row in fft2d(A):
print(', '.join(['%.3f + %.3fi' % (x.real, x.imag) for x in row]))


The output is (reformatted in LaTeX, obviously):

$\displaystyle \begin{pmatrix} 1 & -i & -1 & i \\ -i & -1 & i & 1 \\ -1 & i & 1 & -i \\ i & 1 & -i & -1 \end{pmatrix}$

The reader can verify by hand that this is correct (there’s only one nonzero term in the double sum, so it just boils down to figuring out the complex exponential $e^{2 \pi i (x_1 + x_2 / 4)}$). We leave it as an additional exercise to the reader to implement the inverse transform, as well as to generalize this algorithm to higher dimensional DFTs.

## Some Experiments and Animations

As we did with the 1-dimensional FFT, we’re now going to switch to using an industry-strength FFT algorithm for the applications. We’ll be using the numpy library and its “fft2” function, along with scipy’s ndimage module for image manipulation. Getting all of this set up was a nightmare (thank goodness for people who guide users like me through this stuff, but even then the headache seemed unending!). As usual, all of the code and images used in the making of this post is available on this blog’s Github page.

And so we can start playing with a sample image, a still from one of my favorite television shows:

The Fourier transform of this image (after we convert it to grayscale) can be computed in python:

def fourierSpectrumExample(filename):
unshiftedfft = numpy.fft.fft2(A)
spectrum = numpy.log10(numpy.absolute(unshiftedfft) + numpy.ones(A.shape))
misc.imsave("%s-spectrum-unshifted.png" % (filename.split('.')[0]), spectrum)


With the result:

The Fourier spectrum of Sherlock and Watson (and London).

A few notes: we use the ndimage library to load the image and flatten the colors to grayscale. Then, after we compute the spectrum, we shift and take a logarithm. This is because the raw spectrum values are too massive; plotting them without modification makes the image contrast too high.

Something is odd, though, because the brightest regions are on the edges of the image, where we might expect the highest-frequency elements to be. Actually, it turns out that a raw DFT (as computed by numpy, anyhow) is “shifted.” That is, the indices are much like they were in our original FFT post, so that the “center” of the spectrum (the lowest frequency component) is actually in the corner of the image array.

The numpy folks have a special function designed to alleviate this called fftshift. Applying it before we plot the image gives the following spectrum:

Now that’s more like it. For more details on what’s going on with shifting and how to use the shifting functions, see this matlab thread. (As a side note, the “smudges” in this image are interesting. We wonder what property of the original image contributes to the smudges)

Shifted or unshifted, this image represents the frequency spectrum of the image. In other words, we could take the inverse DFT of each pixel (and its symmetric partner) of this image separately, add them all together, and get back to our original image! We did just that using a different image (one of size 266 x 189, requiring a mere 25137 frequency components), to produce this video of the process:

Many thanks to James Hance for his relentlessly cheerful art (I have a reddish version of this particular masterpiece on my bedroom wall).

For the interested reader, I followed this youtube video’s recommended workflow to make the time-lapsed movie, along with some additional steps to make the videos play side by side. It took quite a while to generate and process the images, and the frames take up a lot of space. So instead of storing all the frames, the interested reader may find the script used to generate the frames on this blog’s Github page (along with all of the rest of the code used in this blog post).

## Digital Watermarking

Now we turn to the main application of Fourier transforms to this post, the task of adding an invisible digital watermark to an image. Just in case the reader lives in a cave, a watermark is a security device used to protect the ownership or authenticity of a particular good. Usually they’re used on money to prevent counterfeits, but they’re often applied to high-resolution images on the web to protect copyrights. But perhaps more than just protect existing copyrights, watermarks as they’re used today are ugly, and mostly prevent people from taking the image (paid for or not) in the first place. Here’s an example from a big proponent of ugly watermarks, Shutterstock.com.

Now if you were the business of copyright litigation, you’d make a lot of money by suing people who took your clients’ images without permission. So rather than prevent people from stealing in the first place, you could put in an invisible watermark into all of your images and then crawl the web looking for stolen images with your watermark. It would be easy enough to automate (Google already did most of the work for you, if you just want to use Google’s search by image feature).

Now I’m more on the side of Fair Use For All, so I wouldn’t hope for a company to actually implement this and make using the internet that much scarier of a place. But the idea makes for an interesting thought experiment and blog post. The idea is simply to modify the spectrum of an image by adding in small, artificial frequency components. That is, the watermarked image will look identical to the original image to a human, but the Fourier spectrum will contain suspicious entries that we can extract if we know where to look.

Implementing the watermarking feature is quite easy, so let’s do that first. Let’s work again with James Hance’s fine artwork.

Let’s call our image’s pixel matrix $A$ and say we’re working with grayscale images for simplicity (for color, we just do the same thing to all three color channels). Then we can define a watermark matrix $W$ by the following procedure:

1. Pick a radius $r$, a length $L$, a watermark strength $\alpha$, and a secret key $k$.
2. Using $k$ as a seed to a random number generator, define a random binary vector $v$ of length $L$.
3. Pick a subset $S$ of the circle of coordinates centered at the image’s center of radius $r$, chosen or rejected based on the entries of $v$.
4. Let $W$ be the matrix of all zeros (of the same dimension as $A$ with 1’s in the entries of $S$.
5. Compute the watermarked image as $\mathscr{F}^{-1}(\mathscr{F}(A) + \alpha W)$. That is, compute the DFT of $A$, add $\alpha W$ to it, and then compute the inverse Fourier transform of the result.

The code for this is simple enough. To create a random vector:

import random
def randomVector(seed, length):
random.seed(secretKey)
return [random.choice([0,1]) for _ in range(length)]


To make the watermark (and flush out all of the technical details of how it’s done:

def makeWatermark(imageShape, radius, secretKey, vectorLength=50):
watermark = numpy.zeros(imageShape)
center = (int(imageShape[0] / 2) + 1, int(imageShape[1] / 2) + 1)

vector = randomVector(secretKey, vectorLength)

x = lambda t: center[0] + int(radius * math.cos(t * 2 * math.pi / vectorLength))
y = lambda t: center[1] + int(radius * math.sin(t * 2 * math.pi / vectorLength))
indices = [(x(t), y(t)) for t in range(vectorLength)]

for i,location in enumerate(indices):
watermark[location] = vector[i]

return watermark


We use the usual parameterization of the circle as $t \mapsto (\cos(2 \pi t / n), \sin(2 \pi t / n)$ scaled to the appropriate radius. Here’s what the watermark looks like as a spectrum:

It’s hard to see the individual pixels, so click it to enlarge.

And then applying a given watermark to an image is super simple.

def applyWatermark(imageMatrix, watermarkMatrix, alpha):
shiftedDFT = fftshift(fft2(imageMatrix))
watermarkedDFT = shiftedDFT + alpha * watermarkMatrix
watermarkedImage = ifft2(ifftshift(watermarkedDFT))

return watermarkedImage


And that’s all there is to it! One might wonder how the choice of $\alpha$ affects the intensity of the watermark, and indeed here we show a few example values of this method applied to Hance’s piece:

Click to enlarge. The effects are most visible in the rightmost image where alpha = 1,000,000

It appears that it’s not until $\alpha$ becomes egregiously large (over 10,000) that we visibly notice the effects. This could be in part due to the fact that this is an image of a canvas (which has lots of small textures in the background). But it’s good to keep in mind the range of acceptable values when designing a decoding mechanism.

Indeed, a decoding mechanism is conceptually much messier; it’s the art to the encoding mechanism’s science. This paper details one possible way to do it, which is essentially to scale everything up or down to 512×512 pixels and try circles of every possible radius until you find one (or don’t) which is statistically similar to the your random vector. And note that since we have the secret key we can generate the exact same random vector. So what the author of that paper suggests is to extract each circle of pixels from the Fourier spectrum, treating it as a single vector with first entry at angle 0. Then you do some statistical magic (compute cross-correlation or some other similarity measure) between the extracted pixels and your secret-key-generated random vector. If they’re sufficiently similar, then you’ve found your watermark, and otherwise there’s no watermark present.

The code required to do this only requires a few extra lines that aren’t present in the code we’re already presented in this article (numpy does cross-correlation for you), so we leave it as an exercise to the reader: write a program that determines if an image contains our watermark, and test the algorithm on various $\alpha$ and with modifications of the image like rotation, scaling, cropping, and jpeg compression. Part of the benefit of Fourier-based techniques is the resilience of the spectrum to mild applications of these transformations.

Next time we’ll use the Fourier transform to do other cool things to images, like designing filters and combining images in interesting ways.

Until then!

# Optimism in the Face of Uncertainty: the UCB1 Algorithm

The software world is always atwitter with predictions on the next big piece of technology. And a lot of chatter focuses on what venture capitalists express interest in. As an investor, how do you pick a good company to invest in? Do you notice quirky names like “Kaggle” and “Meebo,” require deep technical abilities, or value a charismatic sales pitch?

This author personally believes we’re not thinking as big as we should be when it comes to innovation in software engineering and computer science, and that as a society we should value big pushes forward much more than we do. But making safe investments is almost always at odds with innovation. And so every venture capitalist faces the following question. When do you focus investment in those companies that have proven to succeed, and when do you explore new options for growth? A successful venture capitalist must strike a fine balance between this kind of exploration and exploitation. Explore too much and you won’t make enough profit to sustain yourself. Narrow your view too much and you will miss out on opportunities whose return surpasses any of your current prospects.

In life and in business there is no correct answer on what to do, partly because we just don’t have a good understanding of how the world works (or markets, or people, or the weather). In mathematics, however, we can meticulously craft settings that have solid answers. In this post we’ll describe one such scenario, the so-called multi-armed bandit problem, and a simple algorithm called UCB1 which performs close to optimally. Then, in a future post, we’ll analyze the algorithm on some real world data.

As usual, all of the code used in the making of this post are available for download on this blog’s Github page.

## Multi-Armed Bandits

The multi-armed bandit scenario is simple to describe, and it boils the exploration-exploitation tradeoff down to its purest form.

Suppose you have a set of $K$ actions labeled by the integers $\left \{ 1, 2, \dots, K \right \}$. We call these actions in the abstract, but in our minds they’re slot machines. We can then play a game where, in each round, we choose an action (a slot machine to play), and we observe the resulting payout. Over many rounds, we might explore the machines by trying some at random. Assuming the machines are not identical, we naturally play machines that seem to pay off well more frequently to try to maximize our total winnings.

This is the most general description of the game we could possibly give, and every bandit learning problem has these two components: actions and rewards. But in order to get to a concrete problem that we can reason about, we need to specify more details. Bandit learning is a large tree of variations and this is the point at which the field ramifies. We presently care about two of the main branches.

How are the rewards produced? There are many ways that the rewards could work. One nice option is to have the rewards for action $i$ be drawn from a fixed distribution $D_i$ (a different reward distribution for each action), and have the draws be independent across rounds and across actions. This is called the stochastic setting and it’s what we’ll use in this post. Just to pique the reader’s interest, here’s the alternative: instead of having the rewards be chosen randomly, have them be adversarial. That is, imagine a casino owner knows your algorithm and your internal beliefs about which machines are best at any given time. He then fixes the payoffs of the slot machines in advance of each round to screw you up! This sounds dismal, because the casino owner could just make all the machines pay nothing every round. But actually we can design good algorithms for this case, but “good” will mean something different than absolute winnings. And so we must ask:

How do we measure success? In both the stochastic and the adversarial setting, we’re going to have a hard time coming up with any theorems about the performance of an algorithm if we care about how much absolute reward is produced. There’s nothing to stop the distributions from having terrible expected payouts, and nothing to stop the casino owner from intentionally giving us no payout. Indeed, the problem lies in our measurement of success. A better measurement, which we can apply to both the stochastic and adversarial settings, is the notion of regret. We’ll give the definition for the stochastic case, and investigate the adversarial case in a future post.

Definition: Given a player algorithm $A$ and a set of actions $\left \{1, 2, \dots, K \right \}$, the cumulative regret of $A$ in rounds $1, \dots, T$ is the difference between the expected reward of the best action (the action with the highest expected payout) and the expected reward of $A$ for the first $T$ rounds.

We’ll add some more notation shortly to rephrase this definition in symbols, but the idea is clear: we’re competing against the best action. Had we known it ahead of time, we would have just played it every single round. Our notion of success is not in how well we do absolutely, but in how well we do relative to what is feasible.

## Notation

Let’s go ahead and draw up some notation. As before the actions are labeled by integers $\left \{ 1, \dots, K \right \}$. The reward of action $i$ is a $[0,1]$-valued random variable $X_i$ distributed according to an unknown distribution and possessing an unknown expected value $\mu_i$. The game progresses in rounds $t = 1, 2, \dots$ so that in each round we have different random variables $X_{i,t}$ for the reward of action $i$ in round $t$ (in particular, $X_{i,t}$ and $X_{i,s}$ are identically distributed). The $X_{i,t}$ are independent as both $t$ and $i$ vary, although when $i$ varies the distribution changes.

So if we were to play action 2 over and over for $T$ rounds, then the total payoff would be the random variable $G_2(T) = \sum_{t=1}^T X_{2,t}$. But by independence across rounds and the linearity of expectation, the expected payoff is just $\mu_2 T$. So we can describe the best action as the action with the highest expected payoff. Define

$\displaystyle \mu^* = \max_{1 \leq i \leq K} \mu_i$

We call the action which achieves the maximum $i^*$.

A policy is a randomized algorithm $A$ which picks an action in each round based on the history of chosen actions and observed rewards so far. Define $I_t$ to be the action played by $A$ in round $t$ and $P_i(n)$ to be the number of times we’ve played action $i$ in rounds $1 \leq t \leq n$. These are both random variables. Then the cumulative payoff for the algorithm $A$ over the first $T$ rounds, denoted $G_A(T)$, is just

$\displaystyle G_A(T) = \sum_{t=1}^T X_{I_t, t}$

and its expected value is simply

$\displaystyle \mathbb{E}(G_A(T)) = \mu_1 \mathbb{E}(P_1(T)) + \dots + \mu_K \mathbb{E}(P_K(T))$.

Here the expectation is taken over all random choices made by the policy and over the distributions of rewards, and indeed both of these can affect how many times a machine is played.

Now the cumulative regret of a policy $A$ after the first $T$ steps, denoted $R_A(T)$ can be written as

$\displaystyle R_A(T) = G_{i^*}(T) - G_A(T)$

And the goal of the policy designer for this bandit problem is to minimize the expected cumulative regret, which by linearity of expectation is

$\mathbb{E}(R_A(T)) = \mu^*T - \mathbb{E}(G_A(T))$.

Before we continue, we should note that there are theorems concerning lower bounds for expected cumulative regret. Specifically, for this problem it is known that no algorithm can guarantee an expected cumulative regret better than $\Omega(\sqrt{KT})$. It is also known that there are algorithms that guarantee no worse than $O(\sqrt{KT})$ expected regret. The algorithm we’ll see in the next section, however, only guarantees $O(\sqrt{KT \log T})$. We present it on this blog because of its simplicity and ubiquity in the field.

## The UCB1 Algorithm

The policy we examine is called UCB1, and it can be summed up by the principle of optimism in the face of uncertainty. That is, despite our lack of knowledge in what actions are best we will construct an optimistic guess as to how good the expected payoff of each action is, and pick the action with the highest guess. If our guess is wrong, then our optimistic guess will quickly decrease and we’ll be compelled to switch to a different action. But if we pick well, we’ll be able to exploit that action and incur little regret. In this way we balance exploration and exploitation.

The formalism is a bit more detailed than this, because we’ll need to ensure that we don’t rule out good actions that fare poorly early on. Our “optimism” comes in the form of an upper confidence bound (hence the acronym UCB). Specifically, we want to know with high probability that the true expected payoff of an action $\mu_i$ is less than our prescribed upper bound. One general (distribution independent) way to do that is to use the Chernoff-Hoeffding inequality.

As a reminder, suppose $Y_1, \dots, Y_n$ are independent random variables whose values lie in $[0,1]$ and whose expected values are $\mu_i$. Call $Y = \frac{1}{n}\sum_{i}Y_i$ and $\mu = \mathbb{E}(Y) = \frac{1}{n} \sum_{i} \mu_i$. Then the Chernoff-Hoeffding inequality gives an exponential upper bound on the probability that the value of $Y$ deviates from its mean. Specifically,

$\displaystyle \textup{P}(Y + a < \mu) \leq e^{-2na^2}$

For us, the $Y_i$ will be the payoff variables for a single action $j$ in the rounds for which we choose action $j$. Then the variable $Y$ is just the empirical average payoff for action $j$ over all the times we’ve tried it. Moreover, $a$ is our one-sided upper bound (and as a lower bound, sometimes). We can then solve this equation for $a$ to find an upper bound big enough to be confident that we’re within $a$ of the true mean.

Indeed, if we call $n_j$ the number of times we played action $j$ thus far, then $n = n_j$ in the equation above, and using $a = a(j,T) = \sqrt{2 \log(T) / n_j}$ we get that $\textup{P}(Y > \mu + a) \leq T^{-4}$, which converges to zero very quickly as the number of rounds played grows. We’ll see this pop up again in the algorithm’s analysis below. But before that note two things. First, assuming we don’t play an action $j$, its upper bound $a$ grows in the number of rounds. This means that we never permanently rule out an action no matter how poorly it performs. If we get extremely unlucky with the optimal action, we will eventually be convinced to try it again. Second, the probability that our upper bound is wrong decreases in the number of rounds independently of how many times we’ve played the action. That is because our upper bound $a(j, T)$ is getting bigger for actions we haven’t played; any round in which we play an action $j$, it must be that $a(j, T+1) = a(j,T)$, although the empirical mean will likely change.

With these two facts in mind, we can formally state the algorithm and intuitively understand why it should work.

UCB1:
Play each of the $K$ actions once, giving initial values for empirical mean payoffs $\overline{x}_i$ of each action $i$.
For each round $t = K, K+1, \dots$:

Let $n_j$ represent the number of times action $j$ was played so far.
Play the action $j$ maximizing $\overline{x}_j + \sqrt{2 \log t / n_j}$.
Observe the reward $X_{j,t}$ and update the empirical mean for the chosen action.

And that’s it. Note that we’re being super stateful here: the empirical means $x_j$ change over time, and we’ll leave this update implicit throughout the rest of our discussion (sorry, functional programmers, but the notation is horrendous otherwise).

Before we implement and test this algorithm, let’s go ahead and prove that it achieves nearly optimal regret. The reader uninterested in mathematical details should skip the proof, but the discussion of the theorem itself is important. If one wants to use this algorithm in real life, one needs to understand the guarantees it provides in order to adequately quantify the risk involved in using it.

Theorem: Suppose that UCB1 is run on the bandit game with $K$ actions, each of whose reward distribution $X_{i,t}$ has values in [0,1]. Then its expected cumulative regret after $T$ rounds is at most $O(\sqrt{KT \log T})$.

Actually, we’ll prove a more specific theorem. Let $\Delta_i$ be the difference $\mu^* - \mu_i$, where $\mu^*$ is the expected payoff of the best action, and let $\Delta$ be the minimal nonzero $\Delta_i$. That is, $\Delta_i$ represents how suboptimal an action is and $\Delta$ is the suboptimality of the second best action. These constants are called problem-dependent constants. The theorem we’ll actually prove is:

Theorem: Suppose UCB1 is run as above. Then its expected cumulative regret $\mathbb{E}(R_{\textup{UCB1}}(T))$ is at most

$\displaystyle 8 \sum_{i : \mu_i < \mu^*} \frac{\log T}{\Delta_i} + \left ( 1 + \frac{\pi^2}{3} \right ) \left ( \sum_{j=1}^K \Delta_j \right )$

Okay, this looks like one nasty puppy, but it’s actually not that bad. The first term of the sum signifies that we expect to play any suboptimal machine about a logarithmic number of times, roughly scaled by how hard it is to distinguish from the optimal machine. That is, if $\Delta_i$ is small we will require more tries to know that action $i$ is suboptimal, and hence we will incur more regret. The second term represents a small constant number (the $1 + \pi^2 / 3$ part) that caps the number of times we’ll play suboptimal machines in excess of the first term due to unlikely events occurring. So the first term is like our expected losses, and the second is our risk.

But note that this is a worst-case bound on the regret. We’re not saying we will achieve this much regret, or anywhere near it, but that UCB1 simply cannot do worse than this. Our hope is that in practice UCB1 performs much better.

Before we prove the theorem, let’s see how derive the $O(\sqrt{KT \log T})$ bound mentioned above. This will require familiarity with multivariable calculus, but such things must be endured like ripping off a band-aid. First consider the regret as a function $R(\Delta_1, \dots, \Delta_K)$ (excluding of course $\Delta^*$), and let’s look at the worst case bound by maximizing it. In particular, we’re just finding the problem with the parameters which screw our bound as badly as possible, The gradient of the regret function is given by

$\displaystyle \frac{\partial R}{\partial \Delta_i} = - \frac{8 \log T}{\Delta_i^2} + 1 + \frac{\pi^2}{3}$

and it’s zero if and only if for each $i$, $\Delta_i = \sqrt{\frac{8 \log T}{1 + \pi^2/3}} = O(\sqrt{\log T})$. However this is a minimum of the regret bound (the Hessian is diagonal and all its eigenvalues are positive). Plugging in the $\Delta_i = O(\sqrt{\log T})$ (which are all the same) gives a total bound of $O(K \sqrt{\log T})$. If we look at the only possible endpoint (the $\Delta_i = 1$), then we get a local maximum of $O(K \sqrt{\log T})$. But this isn’t the $O(\sqrt{KT \log T})$ we promised, what gives? Well, this upper bound grows arbitrarily large as the $\Delta_i$ go to zero. But at the same time, if all the $\Delta_i$ are small, then we shouldn’t be incurring much regret because we’ll be picking actions that are close to optimal!

Indeed, if we assume for simplicity that all the $\Delta_i = \Delta$ are the same, then another trivial regret bound is $\Delta T$ (why?). The true regret is hence the minimum of this regret bound and the UCB1 regret bound: as the UCB1 bound degrades we will eventually switch to the simpler bound. That will be a non-differentiable switch (and hence a critical point) and it occurs at $\Delta = O(\sqrt{(K \log T) / T})$. Hence the regret bound at the switch is $\Delta T = O(\sqrt{KT \log T})$, as desired.

## Proving the Worst-Case Regret Bound

Proof. The proof works by finding a bound on $P_i(T)$, the expected number of times UCB chooses an action up to round $T$. Using the $\Delta$ notation, the regret is then just $\sum_i \Delta_i \mathbb{E}(P_i(T))$, and bounding the $P_i$‘s will bound the regret.

Recall the notation for our upper bound $a(j, T) = \sqrt{2 \log T / P_j(T)}$ and let’s loosen it a bit to $a(y, T) = \sqrt{2 \log T / y}$ so that we’re allowed to “pretend” a action has been played $y$ times. Recall further that the random variable $I_t$ has as its value the index of the machine chosen. We denote by $\chi(E)$ the indicator random variable for the event $E$. And remember that we use an asterisk to denote a quantity associated with the optimal action (e.g., $\overline{x}^*$ is the empirical mean of the optimal action).

Indeed for any action $i$, the only way we know how to write down $P_i(T)$ is as

$\displaystyle P_i(T) = 1 + \sum_{t=K}^T \chi(I_t = i)$

The 1 is from the initialization where we play each action once, and the sum is the trivial thing where just count the number of rounds in which we pick action $i$. Now we’re just going to pull some number $m-1$ of plays out of that summation, keep it variable, and try to optimize over it. Since we might play the action fewer than $m$ times overall, this requires an inequality.

$P_i(T) \leq m + \sum_{t=K}^T \chi(I_t = i \textup{ and } P_i(t-1) \geq m)$

These indicator functions should be read as sentences: we’re just saying that we’re picking action $i$ in round $t$ and we’ve already played $i$ at least $m$ times. Now we’re going to focus on the inside of the summation, and come up with an event that happens at least as frequently as this one to get an upper bound. Specifically, saying that we’ve picked action $i$ in round $t$ means that the upper bound for action $i$ exceeds the upper bound for every other action. In particular, this means its upper bound exceeds the upper bound of the best action (and $i$ might coincide with the best action, but that’s fine). In notation this event is

$\displaystyle \overline{x}_i + a(P_i(t), t-1) \geq \overline{x}^* + a(P^*(T), t-1)$

Denote the upper bound $\overline{x}_i + a(i,t)$ for action $i$ in round $t$ by $U_i(t)$. Since this event must occur every time we pick action $i$ (though not necessarily vice versa), we have

$\displaystyle P_i(T) \leq m + \sum_{t=K}^T \chi(U_i(t-1) \geq U^*(t-1) \textup{ and } P_i(t-1) \geq m)$

We’ll do this process again but with a slightly more complicated event. If the upper bound of action $i$ exceeds that of the optimal machine, it is also the case that the maximum upper bound for action $i$ we’ve seen after the first $m$ trials exceeds the minimum upper bound we’ve seen on the optimal machine (ever). But on round $t$ we don’t know how many times we’ve played the optimal machine, nor do we even know how many times we’ve played machine $i$ (except that it’s more than $m$). So we try all possibilities and look at minima and maxima. This is a pretty crude approximation, but it will allow us to write things in a nicer form.

Denote by $\overline{x}_{i,s}$ the random variable for the empirical mean after playing action $i$ a total of $s$ times, and $\overline{x}^*_s$ the corresponding quantity for the optimal machine. Realizing everything in notation, the above argument proves that

$\displaystyle P_i(T) \leq m + \sum_{t=K}^T \chi \left ( \max_{m \leq s < t} \overline{x}_{i,s} + a(s, t-1) \geq \min_{0 < s' < t} \overline{x}^*_{s'} + a(s', t-1) \right )$

Indeed, at each $t$ for which the max is greater than the min, there will be at least one pair $s,s'$ for which the values of the quantities inside the max/min will satisfy the inequality. And so, even worse, we can just count the number of pairs $s, s'$ for which it happens. That is, we can expand the event above into the double sum which is at least as large:

$\displaystyle P_i(T) \leq m + \sum_{t=K}^T \sum_{s = m}^{t-1} \sum_{s' = 1}^{t-1} \chi \left ( \overline{x}_{i,s} + a(s, t-1) \geq \overline{x}^*_{s'} + a(s', t-1) \right )$

We can make one other odd inequality by increasing the sum to go from $t=1$ to $\infty$. This will become clear later, but it means we can replace $t-1$ with $t$ and thus have

$\displaystyle P_i(T) \leq m + \sum_{t=1}^\infty \sum_{s = m}^{t-1} \sum_{s' = 1}^{t-1} \chi \left ( \overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t) \right )$

Now that we’ve slogged through this mess of inequalities, we can actually get to the heart of the argument. Suppose that this event actually happens, that $\overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t)$. Then what can we say? Well, consider the following three events:

(1) $\displaystyle \overline{x}^*_{s'} \leq \mu^* - a(s', t)$
(2) $\displaystyle \overline{x}_{i,s} \geq \mu_i + a(s, t)$
(3) $\displaystyle \mu^* < \mu_i + 2a(s, t)$

In words, (1) is the event that the empirical mean of the optimal action is less than the lower confidence bound. By our Chernoff bound argument earlier, this happens with probability $t^{-4}$. Likewise, (2) is the event that the empirical mean payoff of action $i$ is larger than the upper confidence bound, which also occurs with probability $t^{-4}$. We will see momentarily that (3) is impossible for a well-chosen $m$ (which is why we left it variable), but in any case the claim is that one of these three events must occur. For if they are all false, we have

$\displaystyle \begin{matrix} \overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t) & > & \mu^* - a(s',t) + a(s',t) = \mu^* \\ \textup{assumed} & (1) \textup{ is false} & \\ \end{matrix}$

and

$\begin{matrix} \mu_i + 2a(s,t) & > & \overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t) \\ & (2) \textup{ is false} & \textup{assumed} \\ \end{matrix}$

But putting these two inequalities together gives us precisely that (3) is true:

$\mu^* < \mu_i + 2a(s,t)$

This proves the claim.

By the union bound, the probability that at least one of these events happens is $2t^{-4}$ plus whatever the probability of (3) being true is. But as we said, we’ll pick $m$ to make (3) always false. Indeed $m$ depends on which action $i$ is being played, and if $s \geq m > 8 \log T / \Delta_i^2$ then $2a(s,t) \leq \Delta_i$, and by the definition of $\Delta_i$ we have

$\mu^* - \mu_i - 2a(s,t) \geq \mu^* - \mu_i - \Delta_i = 0$.

Now we can finally piece everything together. The expected value of an event is just its probability of occurring, and so

\displaystyle \begin{aligned} \mathbb{E}(P_i(T)) & \leq m + \sum_{t=1}^\infty \sum_{s=m}^t \sum_{s' = 1}^t \textup{P}(\overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t)) \\ & \leq \left \lceil \frac{8 \log T}{\Delta_i^2} \right \rceil + \sum_{t=1}^\infty \sum_{s=m}^t \sum_{s' = 1}^t 2t^{-4} \\ & \leq \frac{8 \log T}{\Delta_i^2} + 1 + \sum_{t=1}^\infty \sum_{s=1}^t \sum_{s' = 1}^t 2t^{-4} \\ & = \frac{8 \log T}{\Delta_i^2} + 1 + 2 \sum_{t=1}^\infty t^{-2} \\ & = \frac{8 \log T}{\Delta_i^2} + 1 + \frac{\pi^2}{3} \\ \end{aligned}

The second line is the Chernoff bound we argued above, the third and fourth lines are relatively obvious algebraic manipulations, and the last equality uses the classic solution to Basel’s problem. Plugging this upper bound in to the regret formula we gave in the first paragraph of the proof establishes the bound and proves the theorem.

$\square$

## Implementation and an Experiment

The algorithm is about as simple to write in code as it is in pseudocode. The confidence bound is trivial to implement (though note we index from zero):

def upperBound(step, numPlays):
return math.sqrt(2 * math.log(step + 1) / numPlays)


And the full algorithm is quite short as well. We define a function ub1, which accepts as input the number of actions and a function reward which accepts as input the index of the action and the time step, and draws from the appropriate reward distribution. Then implementing ub1 is simply a matter of keeping track of empirical averages and an argmax. We implement the function as a Python generator, so one can observe the steps of the algorithm and keep track of the confidence bounds and the cumulative regret.

def ucb1(numActions, reward):
payoffSums = [0] * numActions
numPlays = [1] * numActions
ucbs = [0] * numActions

# initialize empirical sums
for t in range(numActions):
payoffSums[t] = reward(t,t)
yield t, payoffSums[t], ucbs

t = numActions

while True:
ucbs = [payoffSums[i] / numPlays[i] + upperBound(t, numPlays[i]) for i in range(numActions)]
action = max(range(numActions), key=lambda i: ucbs[i])
theReward = reward(action, t)
numPlays[action] += 1
payoffSums[action] += theReward

yield action, theReward, ucbs
t = t + 1


The heart of the algorithm is the second part, where we compute the upper confidence bounds and pick the action maximizing its bound.

We tested this algorithm on synthetic data. There were ten actions and a million rounds, and the reward distributions for each action were uniform from $[0,1]$, biased by $1/k$ for some $5 \leq k \leq 15$. The regret and theoretical regret bound are given in the graph below.

The regret of ucb1 run on a simple example. The blue curve is the cumulative regret of the algorithm after a given number of steps. The green curve is the theoretical upper bound on the regret.

Note that both curves are logarithmic, and that the actual regret is quite a lot smaller than the theoretical regret. The code used to produce the example and image are available on this blog’s Github page.

## Next Time

One interesting assumption that UCB1 makes in order to do its magic is that the payoffs are stochastic and independent across rounds. Next time we’ll look at an algorithm that assumes the payoffs are instead adversarial, as we described earlier. Surprisingly, in the adversarial case we can do about as well as the stochastic case. Then, we’ll experiment with the two algorithms on a real-world application.

Until then!