# Learning a single-variable polynomial, or the power of adaptive queries

Problem: Alice chooses a secret polynomial $p(x)$ with nonnegative integer coefficients. Bob wants to discover this polynomial by querying Alice for the value of $p(x)$ for some integer $x$ of Bob’s choice. What is the minimal number of queries Bob needs to determine $p(x)$ exactly?

Solution: Two queries. The first is $p(1)$, and if we call $N = p(1) + 1$, then the second query is $p(N)$.

To someone who is familiar with polynomials, this may seem shocking, and I’ll explain why it works in a second. After all, it’s very easy to prove that if Bob gives Alice all of his queries at the same time (if the queries are not adaptive), then it’s impossible to discover what $p(x)$ is using fewer than $\textup{deg}(p) + 1$ queries. This is due to a fact called polynomial interpolation, which we’ve seen on this blog before in the context of secret sharing. Specifically, there is a unique single-variable degree $d$ polynomial passing through $d+1$ points (with distinct $x$-values). So if you knew the degree of $p$, you could determine it easily. But Bob doesn’t know the degree of the polynomial, and there’s no way he can figure it out without adaptive queries! Indeed, if Bob tries and gives a set of $d$ queries, Alice could have easily picked a polynomial of degree $d+1$. So it’s literally impossible to solve this problem without adaptive queries.

The lovely fact is that once you allow adaptiveness, the number of queries you need doesn’t even depend on the degree of the secret polynomial!

Okay let’s get to the solution. It was crucial that our polynomial had nonnegative integer coefficients, because we’re going to do a tiny bit of number theory. Let $p(x) = a_0 + a_1 x + \dots + a_d x^d$. First, note that $p(1)$ is exactly the sum of the coefficients $\sum_i a_i$, and in particular $p(1) + 1$ is larger than any single coefficient. So call this $N$, and query $p(N)$. This gives us a number $y_0$ of the form

$\displaystyle y_0 = a_0 + a_1N + a_2N^2 + \dots + a_dN^d$

And because $N$ is so big, we can compute $a_0$ easily by computing $y_0 \mod N$. Now set $y_1 = (y_0 - a_0) / N$, and this has the form $a_1 + a_2N + \dots + a_dN^{d-1}$. We can compute modulus again to get $a_1$, and repeat until we have all the coefficients. We’ll stop once we get a $y_i$ that is zero.

As a small technical note, this is a polynomial-time algorithm in the number of bits needed to write down $p(x)$. So this demonstrates the power of adaptive queries: we get from something which is uncomputable with any number of queries to something which is efficiently computable with a constant number of queries.

The obvious follow-up question is: can you come up with an efficient algorithm if we allow the coefficients to be negative integers?

# The Complexity of Communication

One of the most interesting questions posed in the last thirty years of computer science is to ask how much “information” must be communicated between two parties in order for them to jointly compute something. One can imagine these two parties living on distant planets, so that the cost of communicating any amount of information is very expensive, but each person has an integral component of the answer that the other does not.

Since this question was originally posed by Andrew Yao in 1979, it has led to a flurry of applications in many areas of mathematics and computer science. In particular it has become a standard tool for proving lower bounds in many settings such as circuit design and streaming algorithms. And if there’s anything theory folks love more than a problem that can be solved by an efficient algorithm, it’s a proof that a problem cannot be solved by any efficient algorithm (that’s what I mean by “lower bound”).

Despite its huge applicability, the basic results in this area are elementary. In this post we’ll cover those basics, but once you get past these basic ideas and their natural extensions you quickly approach the state of the art and open research problems. Attempts to tackle these problems in recent years have used sophisticated techniques in Fourier analysis, Ramsey theory, and geometry. This makes it a very fun and exciting field.

As a quick side note before we start, the question we’re asking is different from the one of determining the information content of a specific message. That is the domain of information theory, which was posed (and answered) decades earlier. Here we’re trying to determine the complexity of a problem, where more complex messages require more information about their inputs.

## The Basic Two-Player Model

The most basic protocol is simple enough to describe over a dinner table. Alice and Bob each have one piece of information $x,y$, respectively, say they each have a number. And together they want to compute some operation that depends on both their inputs, for example whether $x > y$. But in the beginning Alice has access only to her number $x$, and knows nothing about $y$. So Alice sends Bob a few bits. Depending on the message Bob computes something and replies, and this repeats until they have computed an answer. The question is: what is the minimum number of bits they need to exchange in order for both of them to be able to compute the right answer?

There are a few things to clarify here: we’re assuming that Alice and Bob have agreed on a protocol for sending information before they ever saw their individual numbers. So imagine ten years earlier Alice and Bob were on the same planet, and they agreed on the rules they’d follow for sending/replying information once they got their numbers. In other words, we’re making a worst-case assumption on Alice and Bob’s inputs, and as usual it will be measured as a function of $n$, the lengths of their inputs. Then we take a minimum (asymptotically) over all possible protocols they could follow, and this value is the “communication complexity” of the problem. Computing the exact communication complexity of a given problem is no simple task, since there’s always the nagging question of whether there’s some cleverer protocol than the one you came up with. So most of the results are bounds on the communication complexity of a problem.

Indeed, we can give our first simple bound for the “$x$ greater than $y$” problem we posed above. Say the strings $x,y$ both have $n$ bits. What Alice does is send her entire string $x$ to Bob, and Bob then computes the answer and sends the answer bit back to Alice. This requires $n + 1$ bits of communication. This proves that the communication complexity of “$x > y$” is bounded from above by $n+1$. A much harder question is, can we do any better?

To make any progress on upper or lower bounds we need to be a bit more formal about the communication model. Basically, the useful analysis happens when the players alternate sending single bits, and this is only off by small constant factors from a more general model. This is the asymptotic analysis, that we only distinguish between things like linear complexity $O(n)$ versus sublinear options like $\log(n)$ or $\sqrt{n}$ or even constant complexity $O(1)$. Indeed, the protocol we described for $x > y$ is the stupidest possible protocol for the problem, and it’s actually valid for any problem. For this problem it happens to be optimal, but we’re just trying to emphasize that nontrivial bounds are all sub-linear in the size of the inputs.

On to the formal model.

Definition: player is a computationally unbounded Turing machine.

And we really mean unbounded. Our players have no time or space constraints, and if they want they can solve undecidable problems like the halting problem or computing Kolmogorov complexity. This is to emphasize that the critical resource is the amount of communication between players. Moreover, it gives us a hint that lower bounds in this model won’t come form computational intractability, but instead will be “information-theoretic.”

Definition: Let $\Sigma^*$ be the set of all binary strings. A communication protocol is a pair of functions $A,B: \Sigma^* \times \Sigma^* \to \{ 0,1 \}$.

The input to these functions $A(x, h)$ should be thought of as follows: $x$ is the player’s secret input and $h$ is the communication history so far. The output is the single bit that they will send in that round (which can be determined by the length of $h$ since only one bit is sent in each round). The protocol then runs by having Alice send $b_1 = A(x, \{ \})$ to Bob, then Bob replies with $b_2 = B(y, b_1)$, Alice continues with $b_3 = A(x, b_1b_2)$, and so on. We implicitly understand that the content of a communication protocol includes a termination condition, but we’ll omit this from the notation. We call the length of the protocol the number of rounds.

Definition: A communication protocol $A,B$ is said to be valid for a boolean function $f(x,y)$ if for all strings $x, y$, the protocol for $A, B$ terminates on some round $t$ with $b_t = 1$ if and only if $f(x,y) = 1$.

So to define the communication complexity, we let the function $L_{A,B}(n)$ be the maximum length of the protocol $A, B$ when run on strings of length $n$ (the worst-case for a given input size). Then the communication complexity of a function $f$ is the minimum of $L_{A,B}$ over all valid protocols $A, B$. In symbols,

$\displaystyle CC_f(n) = \min_{A,B \textup{ is valid for } f} L_{A,B}(n)$

We will often abuse the notation by writing the communication complexity of a function as $CC(f)$, understanding that it’s measured asymptotically as a function of $n$.

## Matrices and Lower Bounds

Let’s prove a lower bound, that to compute the equality function you need to send a linear number of bits in the worst case. In doing this we’ll develop a general algebraic tool.

So let’s write out the function $f$ as a binary matrix $M(f)$ in the following way. Write all $2^n$ inputs of length $n$ in some fixed order along the rows and columns of the matrix, and let entry $i,j$ be the value of $f(i,j)$. For example, the 6-bit function $f$ which computes whether the majority of the two player’s bits are ones looks like this:

The key insight to remember is that if the matrix of a function has a nice structure, then one needs very little communication to compute it. Let’s see why.

Say in the first round the row player sends a bit $b$. This splits the matrix into two submatrices $A_0, A_1$ by picking the rows of $A_0$ to be those inputs for which the row player sends a $b=0$, and likewise for $A_1$ with $b=1$. If you’re willing to rearrange the rows of the matrix so that $A_0$ and $A_1$ stack on top of each other, then this splits the matrix into two rectangles. Now we can switch to the column player and see which bit he sends in reply to each of the possible choices for $b$ (say he sends back $b'$). This separately splits each of $A_0, A_1$ into two subrectangles corresponding to which inputs for the column player make him send the specific value of $b'$. Continuing in this fashion we recurse until we find a submatrix consisting entirely of ones or entirely of zeros, and then we can say with certainty what the value of the function $f$ is.

It’s difficult to visualize because every time we subdivide we move around the rows and columns within the submatrix corresponding to the inputs for each player. So the following would be a possible subdivision of an 8×8 matrix (with the values in the rectangles denoting which communicated bits got you there), but it’s sort of a strange one because we didn’t move the inputs around at all. It’s just a visual aid.

If we do this for $t$ steps we get $2^t$ subrectangles. A crucial fact is that any valid communication protocol for a function has to give a subdivision of the matrix where all the rectangles are constant. or else there would be two pairs of inputs $(x,y), (x', y')$, which are labeled identically by the communication protocol, but which have different values under $f$.

So naturally one expects the communication complexity of $f$ would require as many steps as there are steps in the best decomposition, that is, the decomposition with the fewest levels of subdivision. Indeed, we’ll prove this and introduce some notation to make the discourse less clumsy.

Definition: For an $m \times n$ matrix $M$, a rectangle is a submatrix $A \times B$ where $A \subset \{ 1, \dots m \}, B \subset \{ 1, \dots, n \}$. A rectangle is called monochromatic if all entires in the corresponding submatrix $\left.M\right|_{A \times B}$ are the same. A monochromatic tiling of $M$ is a partition of $M$ into disjoint monochromatic rectangles. Define $\chi(f)$ to be the minimum number of rectangles in any monochromatic tiling of $M(f)$.

As we said, if there are $t$ steps in a valid communication protocol for $f$, then there are $2^t$ rectangles in the corresponding monochromatic tiling of $M(f)$. Here is an easy consequence of this.

Proposition: If $f$ has communication complexity $CC(f)$, then there is a monochromatic tiling of $M(f)$ with at most $2^{CC(f)}$ rectangles. In particular, $\log(\chi(f)) \leq CC(f)$.

Proof. Pick any protocol that achieves the communication complexity of $f$, and apply the process we described above to subdivide $M(f)$. This will take exactly $CC(f)$, and produce no more than $2^{CC(f)}$ rectangles.

$\square$

This already gives us a bunch of theorems. Take the EQ function, for example. Its matrix is the identity matrix, and it’s not hard to see that every monochromatic tiling requires $2^n$ rectangles, one for each entry of the diagonal. I.e., $CC(EQ) \geq n$. But we already know that one player can just send all his bits, so actually $CC(EQ) = \Theta(n)$. Now it’s not always so easy to compute $\chi(f)$. The impressive thing to do is to use efficiently computable information about $M(f)$ to give bounds on $\chi(f)$ and hence on $CC(f)$. So can we come up with a better lower bound that depends on something we can compute? The answer is yes.

Theorem: For every function $f$, $\chi(f) \geq \textup{rank }M(f)$.

Proof. This just takes some basic linear algebra. One way to think of the rank of a matrix $A$ is as the smallest way to write $A$ as a linear combination of rank 1 matrices (smallest as in, the smallest number of terms needed to do this). The theorem is true no matter which field you use to compute the rank, although in this proof and in the rest of this post we’ll use the real numbers.

If you give me a monochromatic tiling by rectangles, I can view each rectangle as a matrix whose rank is at most one. If the entries are all zeros then the rank is zero, and if the entries are all ones then (using zero elsewhere) this is by itself a rank 1 matrix. So adding up these rectangles as separate components gives me an upper bound on the rank of $A$. So the minimum way to do this is also an upper bound on the rank of $A$.

$\square$

Now computing something like $CC(EQ)$ is even easier, because the rank of $M(EQ) = M(I_{2^n})$ is just $2^n$.

## Upper Bounds

There are other techniques to show lower bounds that are stronger than the rank and tiling method (because they imply the rank and tiling method). See this survey for a ton of details. But I want to discuss upper bounds a bit, because the central open conjecture in communication complexity is an upper bound.

The Log-Rank Conjecture: There is a universal constant $c$, such that for all $f$, the communication complexity $CC(f) = O((\log \textup{rank }M(f))^c)$.

All known examples satisfy the conjecture, but unfortunately the farthest progress toward the conjecture is still exponentially worse than the conjecture’s statement. In 1997 the record was due to Andrei Kotlov who proved that $CC(f) \leq \log(4/3) \textup{rank }M(f)$. It was not until 2013 that any (unconditional) improvements were made to this, when Shachar Lovett proved that $CC(f) = O(\sqrt{\textup{rank }M(f)} \cdot \log \textup{rank }M(f))$.

The interested reader can check out this survey of Shachar Lovett from earlier this year (2014) for detailed proofs of these theorems and a discussion of the methods. I will just discuss one idea from this area that ties in nicely with our discussion: which is that finding an efficient communication protocol for a low-rank function $f$ reduces to finding a large monochromatic rectangle in $M(f)$.

Theorem [Nisan-Wigderson 94]: Let $c(r)$ be a function. Suppose that for any function $f: X \times Y \to \{ 0,1 \}$, we can find a monochromatic rectangle of size $R \geq 2^{-c(r)} \cdot | X \times Y |$ where $r = \textup{rank }M(f)$. Then any such $f$ is computable by a deterministic protocol with communication complexity.

$\displaystyle O \left ( \log^2(r) + \sum_{i=0}^{\log r} c(r/2^i) \right )$

Just to be concrete, this says that if $c(r)$ is polylogarithmic, then finding these big rectangles implies a protocol also with polylogarithmic complexity. Since the complexity of the protocol is a function of $r$ alone, the log-rank conjecture follows as a consequence. The best known results use the theorem for larger $c(r) = r^b$ for some $b < 1$, which gives communication complexity also $O(r^b)$.

The proof of the theorem is detailed, but mostly what you’d expect. You take your function, split it up into the big monochromatic rectangle and the other three parts. Then you argue that when you recurse to one of the other three parts, either the rank is cut in half, or the size of the matrix is much smaller. In either case, you can apply the theorem once again. Then you bound the number of leaves in the resulting protocol tree by looking at each level $i$ where the rank has dropped to $r/2^i$. For the full details, see page 4 of the Shachar survey.

## Multiple Players and More

In the future we’ll cover some applications of communication complexity, many of which are related to computing in restricted models such as parallel computation and streaming computation. For example, in parallel computing you often have processors which get arbitrary chunks of data as input and need to jointly compute something. Lower bounds on the communication complexity can help you prove they require a certain amount of communication in order to do that.

But in these models there are many players. And the type of communication matters: it can be point-to-point or broadcast, or something more exotic like MapReduce. So before we can get to these applications we need to define and study the appropriate generalizations of communication complexity to multiple interacting parties.

Until then!

# Making Hybrid Images

The Mona Lisa

Leonardo da Vinci’s Mona Lisa is one of the most famous paintings of all time. And there has always been a discussion around her enigmatic smile. He used a trademark Renaissance technique called sfumato, which involves many thin layers of glaze mixed with subtle pigments. The striking result is that when you look directly at Mona Lisa’s smile, it seems to disappear. But when you look at the background your peripherals see a smiling face.

One could spend decades studying the works of these masters from various perspectives, but if we want to hone in on the disappearing nature of that smile, mathematics can provide valuable insights. Indeed, though he may not have known the relationship between his work and da Vinci’s, hundreds of years later Salvador Dali did the artist’s equivalent of mathematically isolating the problem with his painting, “Gala Contemplating the Mediterranean Sea.”

Gala Contemplating the Mediterranean Sea (Salvador Dali, 1976)

Here you see a woman in the foreground, but step back quite far from the picture and there is a (more or less) clear image of Abraham Lincoln. Here the question of gaze is the blaring focus of the work. Now of course Dali and da Vinci weren’t scribbling down equations and computing integrals; their artistic expression was much less well-defined. But we the artistically challenged have tools of our own: mathematics, science, and programming.

In 2006 Aude Oliva, Antonio Torralba, and Philippe. G. Schyns used those tools to merge the distance of Dali and the faded smiles of da Vinci into one cohesive idea. In their 2006 paper they presented the notion of a “hybrid image,” presented below.

The Mona Lisas of Science

If you look closely, you’ll see three women, each of which looks the teensiest bit strange, like they might be trying to suppress a smile, but none of them are smiling. Blur your eyes or step back a few meters, and they clearly look happy. The effect is quite dramatic. At the risk of being overly dramatic, these three women are literally modern day versions of Mona Lisa, the “Mona Lisas of Science,” if you will.

Another, perhaps more famous version of their technique, since it was more widely publicized, is their “Marilyn Einstein,” which up close is Albert Einstein and from far away is Marilyn Monroe.

Marilyn Einstein

This one gets to the heart of the question of what the eye sees at close range versus long range. And it turns out that you can address this question (and create brilliant works of art like the ones above) with some basic Fourier analysis.

## Intuitive Fourier analysis (and references)

The basic idea of Fourier analysis is the idea that smooth functions are hard to understand, and realization of how great it would be if we could decompose them into simpler pieces. Decomposing complex things into simpler parts is one of the main tools in all of mathematics, and Fourier analysis is one of the clearest examples of its application.

In particular, the things we care about are functions $f(x)$ with specific properties I won’t detail here like “smoothness” and “finiteness.” And the building blocks are the complex exponential functions

$\displaystyle e^{2 \pi i kx}$

where $k$ can be any integer. If you have done some linear algebra (and ignore this if you haven’t), then I can summarize the idea succinctly by saying the complex exponentials form an orthonormal basis for the vector space of square-integrable functions.

Back in colloquial language, what the Fourier theorem says is that any function of the kind we care about can be broken down into (perhaps infinitely many) pieces of this form called Fourier coefficients (I’m abusing the word “coefficient” here). The way it’s breaking down is also pleasingly simple: it’s a linear combination. Informally that means you’re just adding up all the complex exponentials with specific weights for each one. Mathematically, the conversion from the function to its Fourier coefficients is called the Fourier transform, and the set of all Fourier coefficients together is called the Fourier spectrum. So if you want to learn about your function $f$, or more importantly modify it in some way, you can inspect and modify its spectrum instead. The reason this is useful is that Fourier coefficients have very natural interpretations in sound and images, as we’ll see for the latter.

We wrote $f(x)$ and the complex exponential as a function of one real variable, but you can do the same thing for two variables (or a hundred!). And, if you’re willing to do some abusing and ignore the complexness of complex numbers, then you can visualize “complex exponentials in two variables” as images of stripes whose orientation and thickness correspond to two parameters (i.e., the $k$ in the offset equation becomes two coefficients). The video below shows how such complex exponentials can be used to build up an image of striking detail. The left frame shows which complex exponential is currently being added, and the right frame shows the layers all put together. I think the result is quite beautiful.

This just goes to show how powerful da Vinci’s idea of fine layering is: it’s as powerful as possible because it can create any image!

Now for digital images like the one above, everything is finite. So rather than have an infinitely precise function and a corresponding infinite set of Fourier coefficients, you get a finite list of sampled values (pixels) and a corresponding grid of Fourier coefficients. But the important and beautiful theorem is, and I want to emphasize how groundbreakingly important this is:

If you give me an image (or any function!) I can compute the decomposition very efficiently.

And the same theorem lets you go the other way: if you give me the decomposition, I can compute the original function’s samples quite easily. The algorithm to do this is called the Fast Fourier transform, and if any piece of mathematics or computer science has a legitimate claim to changing the world, it’s the Fast Fourier transform. It’s hard to pinpoint specific applications, because the transform is so ubiquitous across science and engineering, but we definitely would not have cell phones, satellites, internet, or electronics anywhere near as small as we do without the Fourier transform and the ability to compute it quickly.

Constructing hybrid images is one particularly nice example of manipulating the Fourier spectrum of two images, and then combining them back into a single image. That’s what we’ll do now.

As a side note, by the nature of brevity, the discussion above is a big disservice to the mathematics involved. I summarized and abused in ways that mathematicians would object to. If you want to see a much better treatment of the material, this blog has a long series of posts developing Fourier transforms and their discrete analogues from scratch. See our four primers, which lead into the main content posts where we implement the Fast Fourier transform in Python and use it to apply digital watermarks to an image. Note that in those posts, as in this one, all of the materials and code used are posted on this blog’s Github page.

## High and low frequencies

For images, interpreting ranges of Fourier coefficients is easy to do. You can imagine the coefficients lying on a grid in the plane like so:

Each dot in this grid corresponds to how “intense” the Fourier coefficient is. That is, it’s the magnitude of the (complex) coefficient of the corresponding complex exponential. Now the points that are closer to the origin correspond informally to the broad, smooth changes in the image. These are called “low frequency” coefficients. And points that are further away correspond to sharp changes and edges, and are likewise called “high frequency” components. So the if you wanted to “hybridize” two images, you’d pick ones with complementary intensities in these regions. That’s why Einstein (with all his wiry hair and wrinkles) and Monroe (with smooth features) are such good candidates. That’s also why, when we layered the Fourier components one by one in the video from earlier, we see the fuzzy shapes emerge before the fine details.

Moreover, we can “extract” the high frequency Fourier components by simply removing the low frequency ones. It’s a bit more complicated than that, since you want the transition from “something” to “nothing” to be smooth in sone sense. A proper discussion of this would go into sampling and the Nyquist frequency, but that’s beyond the scope of this post. Rather, we’ll just define a family of “filtering functions” without motivation and observe that they work well.

Definition: The Gaussian filter function with variance $\sigma$ and center $(a, b)$ is the function

$\displaystyle g(x,y) = e^{-\frac{(x - a)^2 + (y - b)^2}{2 \sigma^2}}$

It looks like this

image credit Wikipedia

In particular, at zero the function is 1 and it gradually drops to zero as you get farther away. The parameter $\sigma$ controls the rate at which it vanishes, and in the picture above the center is set to $(0,0)$.

Now what we’ll do is take our image, compute its spectrum, and multiply coordinatewise with a certain Gaussian function. If we’re trying to get rid of high-frequency components (called a “low-pass filter” because it lets the low frequencies through), we can just multiply the Fourier coefficients directly by the filter values $g(x,y)$, and if we’re doing a “high-pass filter” we multiply by $1 - g(x,y)$.

Before we get to the code, here’s an example of a low-pass filter. First, take this image of Marilyn Monroe

Now compute its Fourier transform

Apply the low-pass filter

And reverse the Fourier transform to get an image

In fact, this is a common operation in programs like photoshop for blurring an image (it’s called a Gaussian blur for obvious reasons). Here’s the python code to do this. You can download it along with all of the other resources used in making this post on this blog’s Github page.

import numpy
from numpy.fft import fft2, ifft2, fftshift, ifftshift
from scipy import misc
from scipy import ndimage
import math

def makeGaussianFilter(numRows, numCols, sigma, highPass=True):
centerI = int(numRows/2) + 1 if numRows % 2 == 1 else int(numRows/2)
centerJ = int(numCols/2) + 1 if numCols % 2 == 1 else int(numCols/2)

def gaussian(i,j):
coefficient = math.exp(-1.0 * ((i - centerI)**2 + (j - centerJ)**2) / (2 * sigma**2))
return 1 - coefficient if highPass else coefficient

return numpy.array([[gaussian(i,j) for j in range(numCols)] for i in range(numRows)])

def filterDFT(imageMatrix, filterMatrix):
shiftedDFT = fftshift(fft2(imageMatrix))
filteredDFT = shiftedDFT * filterMatrix
return ifft2(ifftshift(filteredDFT))

def lowPass(imageMatrix, sigma):
n,m = imageMatrix.shape
return filterDFT(imageMatrix, makeGaussianFilter(n, m, sigma, highPass=False))

def highPass(imageMatrix, sigma):
n,m = imageMatrix.shape
return filterDFT(imageMatrix, makeGaussianFilter(n, m, sigma, highPass=True))

if __name__ == "__main__":
lowPassedMarilyn = lowPass(marilyn, 20)
misc.imsave("low-passed-marilyn.png", numpy.real(lowPassedMarilyn))


The first function samples the values from a Gaussian function with the specified parameters, discretizing the function and storing the values in a matrix. Then the filterDFT function applies the filter by doing coordinatewise multiplication (note these are all numpy arrays). We can do the same thing with a high-pass filter, producing the edgy image below

And if we compute the average of these two images, we basically get back to the original.

So the only difference between this and a hybrid image is that you take the low-passed part of one image and the high-passed part of another. Then the art is in balancing the parameters so as to make the averaged image look right. Indeed, with the following picture of Einstein and the above shot of Monroe, we can get a pretty good recreation of the Oliva-Torralba-Schyns piece. I think with more tinkering it could be even better (I did barely any centering/aligning/resizing to the original images).

Albert Einstein, Marilyn Monroe, and their hybridization.

And here’s the code for it

def hybridImage(highFreqImg, lowFreqImg, sigmaHigh, sigmaLow):
highPassed = highPass(highFreqImg, sigmaHigh)
lowPassed = lowPass(lowFreqImg, sigmaLow)

return highPassed + lowPassed


Interestingly enough, doing it in reverse doesn’t give quite as pleasing results, but it still technically works. So there’s something particularly important that the high-passed image does have a lot of high-frequency components, and vice versa for the low pass.

You can see some of the other hybrid images Oliva et al constructed over at their web gallery.

## Next Steps

How can we take this idea further? There are a few avenues I can think of. The most obvious one would be to see how this extends to video. Could one come up with generic parameters so that when two videos are hybridized (frame by frame, using this technique) it is only easy to see one at close distance? Or else, could we apply a three-dimensional transform to a video and modify that in some principled way? I think one would not likely find anything astounding, but who knows?

Second would be to look at the many other transforms we have at our disposal. How does manipulating the spectra of these transforms affect the original image, and can you make images that are hybridized in senses other than this one?

And finally, can we bring this idea down in dimension to work with one-dimensional signals? In particular, can we hybridize music? It could usher in a new generation of mashup songs that sound different depending on whether you wear earmuffs :)

Until next time!

# When Greedy Algorithms are Perfect: the Matroid

Greedy algorithms are by far one of the easiest and most well-understood algorithmic techniques. There is a wealth of variations, but at its core the greedy algorithm optimizes something using the natural rule, “pick what looks best” at any step. So a greedy routing algorithm would say to a routing problem: “You want to visit all these locations with minimum travel time? Let’s start by going to the closest one. And from there to the next closest one. And so on.”

Because greedy algorithms are so simple, researchers have naturally made a big effort to understand their performance. Under what conditions will they actually solve the problem we’re trying to solve, or at least get close? In a previous post we gave some easy-to-state conditions under which greedy gives a good approximation, but the obvious question remains: can we characterize when greedy algorithms give an optimal solution to a problem?

The answer is yes, and the framework that enables us to do this is called a matroid. That is, if we can phrase the problem we’re trying to solve as a matroid, then the greedy algorithm is guaranteed to be optimal. Let’s start with an example when greedy is provably optimal: the minimum spanning tree problem. Throughout the article we’ll assume the reader is familiar with the very basics of linear algebra and graph theory (though we’ll remind ourselves what a minimum spanning tree is shortly). For a refresher, this blog has primers on both subjects. But first, some history.

## History

Matroids were first introduced by Hassler Whitney in 1935, and independently discovered a little later by B.L. van der Waerden (a big name in combinatorics). They were both interested in devising a general description of “independence,” the properties of which are strikingly similar when specified in linear algebra and graph theory. Since then the study of matroids has blossomed into a large and beautiful theory, one part of which is the characterization of the greedy algorithm: greedy is optimal on a problem if and only if the problem can be represented as a matroid. Mathematicians have also characterized which matroids can be modeled as spanning trees of graphs (we will see this momentarily). As such, matroids have become a standard topic in the theory and practice of algorithms.

## Minimum Spanning Trees

It is often natural in an undirected graph $G = (V,E)$ to find a connected subset of edges that touch every vertex. As an example, if you’re working on a power network you might want to identify a “backbone” of the network so that you can use the backbone to cheaply travel from any node to any other node. Similarly, in a routing network (like the internet) it costs a lot of money to lay down cable, it’s in the interest of the internet service providers to design analogous backbones into their infrastructure.

A minimal subset of edges in a backbone like this is guaranteed to form a tree. This is simply because if you have a cycle in your subgraph then removing any edge on that cycle doesn’t break connectivity or the fact that you can get from any vertex to any other (and trees are the maximal subgraphs without cycles). As such, these “backbones” are called spanning trees. “Span” here means that you can get from any vertex to any other vertex, and it suggests the connection to linear algebra that we’ll describe later, and it’s a simple property of a tree that there is a unique path between any two vertices in the tree.

An example of a spanning tree

When your edges $e \in E$ have nonnegative weights $w_e \in \mathbb{R}^{\geq 0}$, we can further ask to find a minimum cost spanning tree. The cost of a spanning tree $T$ is just the sum of its edges, and it’s important enough of a definition to offset.

Definition: A minimum spanning tree $T$ of a weighted graph $G$ (with weights $w_e \geq 0$ for $e \in E$) is a spanning tree which minimizes the quantity

$w(T) = \sum_{e \in T} w_e$

There are a lot of algorithms to find minimal spanning trees, but one that will lead us to matroids is Kruskal’s algorithm. It’s quite simple. We’ll maintain a forest $F$ in $G$, which is just a subgraph consisting of a bunch of trees that may or may not be connected. At the beginning $F$ is just all the vertices with no edges. And then at each step we add to $F$ the edge $e$ whose weight is smallest and also does not introduce any cycles into $F$. If the input graph $G$ is connected then this will always produce a minimal spanning tree.

Theorem: Kruskal’s algorithm produces a minimal spanning tree of a connected graph.

Proof. Call $F_t$ the forest produced at step $t$ of the algorithm. Then $F_0$ is the set of all vertices of $G$ and $F_{n-1}$ is the final forest output by Kruskal’s (as a quick exercise, prove all spanning trees on $n$ vertices have $n-1$ edges, so we will stop after $n-1$ rounds). It’s clear that $F_{n-1}$ is a tree because the algorithm guarantees no $F_i$ will have a cycle. And any tree with $n-1$ edges is necessarily a spanning tree, because if some vertex were left out then there would be $n-1$ edges on a subgraph of $n-1$ vertices, necessarily causing a cycle somewhere in that subgraph.

Now we’ll prove that $F_{n-1}$ has minimal cost. We’ll prove this in a similar manner to the general proof for matroids. Indeed, say you had a tree $T$ whose cost is strictly less than that of $F_{n-1}$ (we can also suppose that $T$ is minimal, but this is not necessary). Pick the minimal weight edge $e \in T$ that is not in $F_{n-1}$. Adding $e$ to $F_{n-1}$ introduces a unique cycle $C$ in $F_{n-1}$. This cycle has some strange properties. First, $e$ has the highest cost of any edge on $C$. For otherwise, Kruskal’s algorithm would have chosen it before the heavier weight edges. Second, there is another edge in $C$ that’s not in $T$ (because $T$ was a tree it can’t have the entire cycle). Call such an edge $e'$. Now we can remove $e'$ from $F_{n-1}$ and add $e$. This can only increase the total cost of $F_{n-1}$, but this transformation produces a tree with one more edge in common with $T$ than before. This contradicts that $T$ had strictly lower weight than $F_{n-1}$, because repeating the process we described would eventually transform $F_{n-1}$ into $T$ exactly, while only increasing the total cost.

$\square$

Just to recap, we defined sets of edges to be “good” if they did not contain a cycle, and a spanning tree is a maximal set of edges with this property. In this scenario, the greedy algorithm performed optimally at finding a spanning tree with minimal total cost.

## Columns of Matrices

Now let’s consider a different kind of problem. Say I give you a matrix like this one:

$\displaystyle A = \begin{pmatrix} 2 & 0 & 1 & -1 & 0 \\ 0 & -4 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 7 \end{pmatrix}$

In the standard interpretation of linear algebra, this matrix represents a linear function $f$ from one vector space $V$ to another $W$, with the basis $(v_1, \dots, v_5)$ of $V$ being represented by columns and the basis $(w_1, w_2, w_3)$ of $W$ being represented by the rows. Column $j$ tells you how to write $f(v_j)$ as a linear combination of the $w_i$, and in so doing uniquely defines $f$.

Now one thing we want to calculate is the rank of this matrix. That is, what is the dimension of the image of $V$ under $f$? By linear algebraic arguments we know that this is equivalent to asking “how many linearly independent columns of $A$ can we find”? An interesting consequence is that if you have two sets of columns that are both linearly independent and maximally so (adding any other column to either set would necessarily introduce a dependence in that set), then these two sets have the same size. This is part of why the rank of a matrix is well-defined.

If we were to give the columns of $A$ costs, then we could ask about finding the minimal-cost maximally-independent column set. It sounds like a mouthful, but it’s exactly the same idea as with spanning trees: we want a set of vectors that spans the whole column space of $A$, but contains no “cycles” (linearly dependent combinations), and we want the cheapest such set.

So we have two kinds of “independence systems” that seem to be related. One interesting question we can ask is whether these kinds of independence systems are “the same” in a reasonable way. Hardcore readers of this blog may see the connection quite quickly. For any graph $G = (V,E)$, there is a natural linear map from $E$ to $V$, so that a linear dependence among the columns (edges) corresponds to a cycle in $G$. This map is called the incidence matrix by combinatorialists and the first boundary map by topologists.

The map is easy to construct: for each edge $e = (v_i,v_j)$ you add a column with a 1 in the $j$-th row and a $-1$ in the $i$-th row. Then taking a sum of edges gives you zero if and only if the edges form a cycle. So we can think of a set of edges as “independent” if they don’t contain a cycle. It’s a little bit less general than independence over $\mathbb{R}$, but you can make it exactly the same kind of independence if you change your field from real numbers to $\mathbb{Z}/2\mathbb{Z}$. We won’t do this because it will detract from our end goal (to analyze greedy algorithms in realistic settings), but for further reading this survey of Oxley assumes that perspective.

So with the recognition of how similar these notions of independence are, we are ready to define matroids.

## The Matroid

So far we’ve seen two kinds of independence: “sets of edges with no cycles” (also called forests) and “sets of linearly independent vectors.” Both of these share two trivial properties: there are always nonempty independent sets, and every subset of an independent set is independent. We will call any family of subsets with this property an independence system.

Definition: Let $X$ be a finite set. An independence system over $X$ is a family $\mathscr{I}$ of subsets of $X$ with the following two properties.

1. $\mathscr{I}$ is nonempty.
2. If $I \in \mathscr{I}$, then so is every subset of $I$.

This is too general to characterize greedy algorithms, so we need one more property shared by our examples. There are a few things we do, but here’s one nice property that turns out to be enough.

Definition: A matroid $M = (X, \mathscr{I})$ is a set $X$ and an independence system $\mathscr{I}$ over $X$ with the following property:

If $A, B$ are in $\mathscr{I}$ with $|A| = |B| + 1$, then there is an element $x \in A \setminus B$ such that $B \cup \{ a \} \in \mathscr{I}$.

In other words, this property says if I have an independent set that is not maximally independent, I can grow the set by adding some suitably-chosen element from a larger independent set. We’ll call this the extension property. For a warmup exercise, let’s prove that the extension property is equivalent to the following (assuming the other properties of a matroid):

For every subset $Y \subset X$, all maximal independent sets contained in $Y$ have equal size.

Proof. For one direction, if you have two maximal sets $A, B \subset Y \subset X$ that are not the same size (say $A$ is bigger), then you can take any subset of $A$ whose size is exactly $|B| + 1$, and use the extension property to make $B$ larger, a contradiction. For the other direction, say that I know all maximal independent sets of any $Y \subset X$ have the same size, and you give me $A, B \subset X$. I need to find an $a \in A \setminus B$ that I can add to $B$ and keep it independent. What I do is take the subset $Y = A \cup B$. Now the sizes of $A, B$ don’t change, but $B$ can’t be maximal inside $Y$ because it’s smaller than $A$ ($A$ might not be maximal either, but it’s still independent). And the only way to extend $B$ is by adding something from $A$, as desired.

$\square$

So we can use the extension property and the cardinality property interchangeably when talking about matroids. Continuing to connect matroid language to linear algebra and graph theory, the maximal independent sets of a matroid are called bases, the size of any basis is the rank of the matroid, and the minimal dependent sets are called circuits. In fact, you can characterize matroids in terms of the properties of their circuits, which are dual to the properties of bases (and hence all independent sets) in a very concrete sense.

But while you could spend all day characterizing the many kinds of matroids and comatroids out there, we are still faced with the task of seeing how the greedy algorithm performs on a matroid. That is, suppose that your matroid $M = (X, \mathscr{I})$ has a nonnegative real number $w(x)$ associated with each $x \in X$. And suppose we had a black-box function to determine if a given set $S \subset X$ is independent. Then the greedy algorithm maintains a set $B$, and at every step adds a minimum weight element that maintains the independence of $B$. If we measure the cost of a subset by the sum of the weights of its elements, then the question is whether the greedy algorithm finds a minimum weight basis of the matroid.

The answer is even better than yes. In fact, the answer is that the greedy algorithm performs perfectly if and only if the problem is a matroid! More rigorously,

Theorem: Suppose that $M = (X, \mathscr{I})$ is an independence system, and that we have a black-box algorithm to determine whether a given set is independent. Define the greedy algorithm to iteratively adds the cheapest element of $X$ that maintains independence. Then the greedy algorithm produces a maximally independent set $S$ of minimal cost for every nonnegative cost function on $X$, if and only if $M$ is a matroid.

It’s clear that the algorithm will produce a set that is maximally independent. The only question is whether what it produces has minimum weight among all maximally independent sets. We’ll break the theorem into the two directions of the “if and only if”:

Part 1: If $M$ is a matroid, then greedy works perfectly no matter the cost function.
Part 2: If greedy works perfectly for every cost function, then $M$ is a matroid.

Proof of Part 1.

Call the cost function $w : X \to \mathbb{R}^{\geq 0}$, and suppose that the greedy algorithm picks elements $B = \{ x_1, x_2, \dots, x_r \}$ (in that order). It’s easy to see that $w(x_1) \leq w(x_2) \leq \dots \leq w(x_r)$. Now if you give me any list of $r$ independent elements $y_1, y_2, \dots, y_r \in X$ that has $w(y_1) \leq \dots \leq w(y_r)$, I claim that $w(x_i) \leq w(y_i)$ for all $i$. This proves what we want, because if there were a basis of size $r$ with smaller weight, sorting its elements by weight would give a list contradicting this claim.

To prove the claim, suppose to the contrary that it were false, and for some $k$ we have $w(x_k) > w(y_k)$. Moreover, pick the smallest $k$ for which this is true. Note $k > 1$, and so we can look at the special sets $S = \{ x_1, \dots, x_{k-1} \}$ and $T = \{ y_1, \dots, y_k \}$. Now $|T| = |S|+1$, so by the matroid property there is some $j$ between $1$ and $r$ so that $S \cup \{ y_j \}$ is an independent set (and $y_j$ is not in $S$). But then $w(y_j) \leq w(y_k) < w(x_k)$, and so the greedy algorithm would have picked $y_j$ before it picks $x_k$ (and the strict inequality means they’re different elements). This contradicts how the greedy algorithm runs, and hence proves the claim.

Proof of Part 2.

We’ll prove this contrapositively as follows. Suppose we have our independence system and it doesn’t satisfy the last matroid condition. Then we’ll construct a special weight function that causes the greedy algorithm to fail. So let $A,B$ be independent sets with $|A| = |B| + 1$, but for every $a \in A \setminus B$ adding $a$ to $B$ never gives you an independent set.

Now what we’ll do is define our weight function so that the greedy algorithm picks the elements we want in the order we want (roughly). In particular, we’ll assign all elements of $A \cap B$ a tiny weight we’ll call $w_1$. For elements of $B - A$ we’ll use $w_2$, and for $A - B$ we’ll use $w_3$, with $w_4$ for everything else. In a more compact notation:

We need two things for this weight function to screw up the greedy algorithm. The first is that $w_1 < w_2 < w_3 < w_4$, so that greedy picks the elements in the order we want. Note that this means it’ll first pick all of $A \cap B$, and then all of $B - A$, and by assumption it won’t be able to pick anything from $A - B$, but since $B$ is assumed to be non-maximal, we have to pick at least one element from $X - (A \cup B)$ and pay $w_4$ for it.

So the second thing we want is that the cost of doing greedy is worse than picking any maximally independent set that contains $A$ (and we know that there has to be some maximal independent set containing $A$). In other words, if we call $m$ the size of a maximally independent set, we want

$\displaystyle |A \cap B| w_1 + |B-A|w_2 + (m - |B|)w_4 > |A \cap B|w_1 + |A-B|w_3 + (m-|A|)w_4$

This can be rearranged (using the fact that $|A| = |B|+1$) to

$\displaystyle w_4 > |A-B|w_3 - |B-A|w_2$

The point here is that the greedy picks too many elements of weight $w_4$, since if we were to start by taking all of $A$ (instead of all of $B$), then we could get by with one fewer. That might not be optimal, but it’s better than greedy and that’s enough for the proof.

So we just need to make $w_4$ large enough to make this inequality hold, while still maintaining $w_2 < w_3$. There are probably many ways to do this, and here’s one. Pick some $0 < \varepsilon < 1$, and set

It’s trivial that $w_1 < w_2$ and $w_3 < w_4$. For the rest we need some observations. First, the fact that $|A-B| = |B-A| + 1$ implies that $w_2 < w_3$. Second, both $|A-B|$ and $|B-A|$ are nonempty, since otherwise the second property of independence systems would contradict our assumption that augmenting $B$ with elements of $A$ breaks independence. Using this, we can divide by these quantities to get

$\displaystyle w_4 = 2 > 1 = \frac{|A-B|(1 + \varepsilon)}{|A-B|} - \frac{|B-A|\varepsilon}{|B-A|}$

This proves the claim and finishes the proof.

$\square$

As a side note, we proved everything here with respect to minimizing the sum of the weights, but one can prove an identical theorem for maximization. The only part that’s really different is picking the clever weight function in part 2. In fact, you can convert between the two by defining a new weight function that subtracts the old weights from some fixed number $N$ that is larger than any of the original weights. So these two problems really are the same thing.

This is pretty amazing! So if you can prove your problem is a matroid then you have an awesome algorithm automatically. And if you run the greedy algorithm for fun and it seems like it works all the time, then that may be hinting that your problem is a matroid. This is one of the best situations one could possibly hope for.

But as usual, there are a few caveats to consider. They are both related to efficiency. The first is the black box algorithm for determining if a set is independent. In a problem like minimum spanning tree or finding independent columns of a matrix, there are polynomial time algorithms for determining independence. These two can both be done, for example, with Gaussian elimination. But there’s nothing to stop our favorite matroid from requiring an exponential amount of time to check if a set is independent. This makes greedy all but useless, since we need to check for independence many times in every round.

Another, perhaps subtler, issue is that the size of the ground set $X$ might be exponentially larger than the rank of the matroid. In other words, at every step our greedy algorithm needs to find a new element to add to the set it’s building up. But there could be such a huge ocean of candidates, all but a few of which break independence. In practice an algorithm might be working with $X$ implicitly, so we could still hope to solve the problem if we had enough knowledge to speed up the search for a new element.

There are still other concerns. For example, a naive approach to implementing greedy takes quadratic time, since you may have to look through every element of $X$ to find the minimum-cost guy to add. What if you just have to have faster runtime than $O(n^2)$? You can still be interested in finding more efficient algorithms that still perform perfectly, and to the best of my knowledge there’s nothing that says that greedy is the only exact algorithm for your favorite matroid. And then there are models where you don’t have direct/random access to the input, and lots of other ways that you can improve on greedy. But those stories are for another time.

Until then!