The title of this blog and enumerative combinatorics

by t0rajir0u, Aug 13, 2008, 5:28 AM

The title of this blog is inspired by a quote from Professor Stevens at PROMYS, who likes to say that "mathematicians are annoyingly precise." I really liked this sentiment when I first heard it, especially since PROMYS is where I first learned to really be precise (that is, rigorous) about mathematics.

Professor Stevens also told us to "think deeply of simple things." At PROMYS (as at Ross), that meant thinking very hard about the integers - about addition, about multiplication, about what it meant for an integer to be greater than another integer - but of course this principle applies everywhere. So I'd like to think deeply about this simple thing:

Why does $1 + p + p^2 + ... + p^n$ equal $\frac {p^{n + 1} - 1}{p - 1}$ ?

This is such a basic fact (it's certainly an indispensable bit of competition math) that it doesn't really appear to be worth paying attention to. The usual proofs boil down to manipulation or induction. At a later stage, one can also view this identity as the statement of the factorization

$p^{n + 1} - 1 = (p - 1)(p^n + p^{n - 1} + ... + 1)$

which is also basic enough. So why even bother thinking about this identity further?

Maybe an "algebraic" proof just isn't satisfying. Is there a combinatorial way to interpret the above identity? Let me give an example. In proving that

$\sum_{k = 0}^{n} {n \choose k} = 2^n$

we could just cite the binomial theorem on $(1 + 1)^n$ and be done with it, or we could recall that the LHS counts the number of subsets of size $k$ of a set of size $n$ for every $k$ , and therefore counts the total number of subsets. The RHS also counts the total number of subsets: for each of the $n$ elements of a set, either that element is or is not in that set. This is perhaps the prototypical example of counting the same thing in two ways, or the method of bijection.

Bijection is the preferred proof technique of enumerative combinatorialists. "Manipulative" techniques - series, in other words - don't really tell you what's going on from this point of view. Let me illustrate this point by pushing the above identity further. We started with the generating function $(x + 1)^n$ and substituted $x = 1$ . Differentiating before substituting we obtain the identity

$\sum_{k = 1}^{n} k {n \choose k} = n \cdot 2^{n - 1}$ .

(Sums like this are also discussed in Engel.) So what's the combinatorial interpretation here? Well, what we're doing on the LHS is essentially picking a subset and then distinguishing an element of a subset. We can think of this in terms of picking a committee out of $n$ people and then picking a President of that committee. We can also do this by picking the President first, and then picking the committee out of the $n - 1$ people left; that's the RHS. Would you rather elect a President or differentiate?

Differentiating again or picking a committee and electing a President and a Secretary immediately leads to the identity

$\sum_{k = 2}^{n} k(k - 1) {n \choose k} = n(n - 1) 2^{n - 2}$

and $r$ differentiations or picking a committee and electing a board of directors of size $r$ gives

$\sum_{k = r}^{n} {n \choose k} {k \choose r} = {n \choose r} 2^{n - r}$

which is a nice lemma in its own right, and summing over all possible values of $r$ we have

$\sum_{r = 0}^{n} \sum_{k = r}^{n} {n \choose k} {k \choose r} = (1 + 2)^n = 3^n$

because we are splitting up our set into $3$ categories: non-committee, committee but non-director, and director. In terms of generating functions, what we did boils down to computing a Taylor series (or using the binomial theorem twice), but isn't bureaucracy so much simpler than calculus? Or algebra? (See also the similar argument here.)

Now that we've discussed at length the kind of argument we're looking for, what could be counted by $1 + p + p^2 + ... + p^n$ ? Well, since we're dividing by $p - 1$ on the RHS, perhaps it's better to multiply out, since multiplying is more natural when counting than division: hence we want to combinatorially interpret

$(p - 1) + (p - 1)p + ... + (p - 1)p^n = p^{n + 1} - 1$ .

The RHS might now remind us of base- $p$ representation, which is good inspiration, but we don't have to refer to it explicitly: let's continue the bureaucracy interpretation. Suppose that we are trying to divide some subset of $n + 1$ people among $p - 1$ committees (so each person has $p$ states; a committee or no committee) such that at least one person is in a committee. On the one hand, there are $p^{n + 1} - 1$ ways to do this. On the other hand, given any such division let's order the people and ignore the leftmost people who aren't in a committee (that is, zero digits); say there are $k$ of them. There are $p - 1$ ways to distribute the first person who is in a committee and $p^{n - k}$ ways to distribute the rest. There it is!

Perhaps that was too simple of an answer. Here's another based on the observation that when $p$ is (a power of a) prime, $p^k$ counts the number of points in the affine space $\mathbb{F}_p A^k$ of dimension $k$ over the finite field $\mathbb{F}_p$ (which means we're implicitly taking $p$ a power of a prime; this turns out not to matter). In other words, $1 + p + p^2 + ... + p^n$ counts the number of points in the affine spaces over $\mathbb{F}_p$ with dimension at most $n$ .

It turns out that we've already encountered a structure that behaves like this: it is the projective space $\mathbb{F}_p P^n$ . Recall that this is defined as the equivalence classes of points in $\mathbb{F}_p A^{n + 1}$ up to multiplication by a nonzero constant, excluding the origin. In other words, this is the set of lines through the origin. Since we have the advantage of working over finite fields, this set of lines is finite, and so we can count it. But first: over any field, recall that when $n = 1$ we have a projective line, which is an affine line plus a point at infinity. This occured because the equivalence classes of ordered pairs $(x : y)$ can be "normalized" to the ordered pairs $(t : 1)$ when $y \neq 0$ by appropriate multiplication, but when $y = 0$ we have an extra point $(1 : 0)$ . Generally, a projective space of dimension $n$ contains a copy of an affine space of dimension $n$ (when the last coordinate is not zero) and a projective $n - 1$ -space at infinity (when it is). Symbolically,

$KP^n \simeq KA^n + KP^{n - 1}$

in the appropriate sense, for any field $K$ and for all $n$ . Expanding the RHS,

$KP^n \simeq KA^n + KA^{n - 1} + KA^{n - 2} + ...$ ,

so in fact the projective space $KP^n$ contains a copy of each affine space with dimension less than $n$ ! (This essentially occurs because we're monitoring how many of the starting components of the vector are zero, which is the same observation we made about people not in committees.) It therefore follows that when $K$ is the finite field $\mathbb{F}_p$ we know that

$|\mathbb{F}_p P^n| = p^n + p^{n - 1} + ... + 1$ .

On the other hand, there are $p^{n + 1} - 1$ points in $\mathbb{F}_p A^{n + 1}$ (the affine space we start our construction in) besides the origin, and $p - 1$ nonzero elements of $\mathbb{F}_p$ we can divide by. In other words,

$|\mathbb{F}_p P^n| = \frac {p^{n + 1} - 1}{p - 1}$ .

Voila!

Of course, we haven't pushed far enough. The combinatorial interpretation in terms of projective space, while pretty, does not provide an obvious generalization to the identity

$1 + 2p + 3p^2 + 4p^3 + ... + np^{n - 1} = \frac { (n + 1)p^n (p - 1) - (p^{n + 1} - 1) }{ (p - 1)^2 }$

which we obtain by differentiating with respect to $p$ . Does bureaucracy have more hope? Well, first we have to wonder if we're even asking the right question. Perhaps a different formulation of this question (in which we want to ask how to evaluate sums such as $\sum n \cdot 2^n$ ) would be more susceptible to a counting argument; indeed, the complexity of the RHS suggests that perhaps we want to do some simplification.

Instead of starting with algebra, though, let's just push our combinatorial argument. Suppose we now want to exclude configurations where only one person is in a committee; that is, at least two people must be in a committee. This can clearly be done in $p^{n + 1} - {n \choose 0} - {n + 1 \choose 1} (p - 1)$ ways (where we are using binomial coefficients to suggest the eventual form of the generalization). On the other hand, in any committee assignment we can consider the positions (from left to right) of the first two people in a committee. For fixed $k, 1 \le k \le n$ the second person can be in position $k + 1$ whereas the first person can be in any of the first $k$ positions, leaving $n - k$ other people. Since the first two people must be in committees, this can be done in $k(p - 1)^2 p^{n - k - 1}$ ways. Thus we conclude that

$\sum_{k = 0}^{n} k(p - 1)^2 p^{n - k} = p^{n + 1} - 1 - (n + 1)(p - 1) \implies$
$\sum_{k = 0}^{n} k p^{n - k} = \frac {p^{n + 1} - 1 - (n + 1)(p - 1)}{(p - 1)^2}$ .

This is, in fact, the identity we get if we take $x^n + x^{n - 1} p + ... + xp^{n - 1} + p^n = \frac {x^{n + 1} - p^{n + 1}}{x - p}$ and differentiate with respect to $x$ instead of $p$ , and then plug in $x = 1$ . It generalizes immediately: if we require that at least $r$ people be in a committee (or differentiate $r$ times), we conclude that

$\sum_{k = 0}^{n} {k \choose r} p^{n - k} = \frac {p^{n + 1} - 1 - (n + 1)(p - 1) - {n + 1 \choose 2} (p - 1)^2 - ...}{(p - 1)^{r + 1}}$ .

The polynomial that begins to appear on the RHS is precisely the expansion of $p^{n + 1} - 1$ in powers of $p - 1$ . Note that unlike in the other example, where the generating functions approach requires less work, repeated use of the quotient rule gets very tedious on the problem as is. Of course, the answer suggests that the correct approach is to expand the RHS in powers of $x - p$ , and indeed we have already seen an example of how the binomial theorem and differentiation are closely related.

Back to the binomial theorem: here's another example of a simple problem, but this time I'd like to use it to introduce a more sophisticated technique in the guise of a simple one:

Why does $\sum_{k = 0}^{n} {n \choose k} ( - 1)^k$ equal $0$ ?

Remember, an answer like $(1 - 1)^k = 0$ is no longer satisfactory. What this statement means is that there are exactly as many subsets of even size as subsets of odd size, so ideally, what we'd like is a bijection between the two. This is very easy when $n$ is odd: the complement of an even set is odd, and the complement of an odd set is even. But what happens when $n$ is even?

Ideally we'd still like to produce a bijection. Here's an idea: if we add an element to a set, we change its parity! Well, unless that element's already there. But if we remove an element from a set, we change its parity! That's a remarkably simple idea, and it works beautifully. But what may not be obvious is that this strategy and the complement are really the same idea.

The operation we performed above was almost a union - except when our chosen element was in a set, in which case it became the union minus the intersection. The union minus the intersection of two sets is called their symmetric difference, a slightly less intuitive notion than union and intersection, but a very useful one. The complement of a subset is just its symmetric difference with the whole set, so our above two strategies are not really different.

But the story doesn't end here. Let me introduce a convenient notation for subsets of a set with $n$ elements. We will put them in some order and associate with each subset an $n$ -digit binary string, with a $1$ if the $k^{th}$ element is contained in that subset and a $0$ otherwise. For example, the subset $\{ 1, 3, 4 \} \in \{ 1, 2, 3, 4, 5 \}$ is the binary string $10110$ . This notation has two huge conveniences, easily verified by comparing truth tables:

- The intersection of two subsets is the componentwise multiplication of their binary strings $\bmod 2$ .
- The symmetric difference of two subsets is the componentwise addition of their binary strings $\bmod 2$ .

Several properties I was too lazy to prove now follow immediately: intersection distributes over symmetric difference, symmetric difference is associative, etc. Intersection and symmetric difference together turn the power set of a set into a structure called a Boolean ring. This turns out to characterize all finite Boolean rings.

This is a lovely reformulation of the problem, but we still don't quite have a nice way of expressing "the parity of a subset." The next step is to think of an $n$ -digit binary string as a vector in $\mathbb{F}_2^n$ (a vector space, to be distinguished from an affine space!), in which case the parity of a subset can be equivalently phrased as the dot product of a vector with $\left < 1, 1, 1... \right >$ (the vector that represents the entire set) or as the dot product of a vector with itself. It's the first formulation that leads us to the (in hindsight, obvious) conclusion that parity is additive, and therefore symmetric difference is now a completely obvious choice of technique: adding an odd vector (taking the symmetric difference with an odd subset) changes parity.

This has been an extremely "simple" use of linear algebra in combinatorics, but my hope is that it has been simple enough that the ideas are well-motivated. Another simple example can be found here, and an alternate approach to a generalization of the above question can be found here.

The natural extension of the above question is to ask about the number of subsets with size congruent to $0, 1, 2 \bmod 3$ . Unfortunately, as we have seen, the power set of a finite set is naturally equipped with a notion of parity but doesn't appear to handle any larger primes. In particular, the vector space $\mathbb{F}_3^n$ no longer has a simple interpretation in terms of subsets: we knew that $0$ meant "not in this set" and $1$ meant "in this set," but what exactly does $2$ mean?

The standard ("manipulative") technique to answer this question is to apply the roots of unity filter to the coefficients of the polynomial $P(x) = (1 + x)^n$ . For example, the number of subsets with size congruent to $0 \bmod 3$ (which we will call $0$ -sets) is the sum of the binomial coefficients ${n \choose 0}, {n \choose 3}, ...$ , which is

$S_0 = \frac {P(1) + P(\omega) + P(\omega^2)}{3} = \frac {2^n + ( - \omega^2)^n + ( - \omega)^n}{3}$

where $\omega$ is a primitive third root of unity. (What the above manipulation does is force monomials $x^k$ to evaluate to zero if $k \equiv 1, 2 \bmod 3$ and evaluate to $1$ otherwise, therefore "filtering" out precisely the sum of the coefficients of the terms $x^{3r}$ .) If $n \equiv 1, 5 \bmod 6$ , the above evaluates to $\frac {2^n + 1}{3}$ ; if $n \equiv 2, 4 \bmod 6$ , it's $\frac {2^n - 1}{3}$ ; if $n \equiv 3 \bmod 6$ , it's $\frac {2^n - 2}{3}$ ; finally, if $n \equiv 0 \bmod 6$ it's $\frac {2^n + 2}{3}$ . Can we capture this behavior by a purely combinatorial argument, without dealing with roots of unity at all?

First we might as well give the expressions for the number of subsets with size congruent to $1, 2 \bmod 3$ (which we will call $1$ - and $2$ -sets respectively). While the above technique is well-known, I believe less people are aware of the following two sums, which can be determined by guesswork, by looking at the coefficients in a DFT matrix, or by using Dirichlet characters:

$S_1 = \frac {P(1) + \omega^2 P(\omega) + \omega P(\omega^2)}{3} = \frac {2^n + \omega^2 ( - \omega^2)^n + \omega ( - \omega)^n}{3}$
$S_2 = \frac {P(1) + \omega P(\omega) + \omega^2 P(\omega^2)}{3} = \frac {2^n + \omega ( - \omega^2)^n + \omega^2 ( - \omega)^n}{3}$ .

There is another way to derive the above sums which is perhaps more natural in that we are naturally led to the use of third roots of unity, and it generalizes readily to higher moduli. Rename the above sums $S_{0,n}, S_{1,n}, S_{2,n}$ . Call a subset of a set with $n$ elements with cardinality congruent to $0 \bmod 3$ a $0,n$ -subset and similarly for $1$ and $2$ . We can determine these sums by working inductively: starting with a set with $n$ elements and adding a new element, a $0,n + 1$ -subset is either a $0,n$ -subset without the new element or a $2,n$ -subset with a new element, and similarly; in other words, we have the system

$S_{0,n + 1} = S_{0,n} + S_{2,n}$
$S_{1,n + 1} = S_{0,n} + S_{1,n}$
$S_{2,n + 1} = S_{2,n} + S_{1,n}$

which, letting $\mathbf{S}_n = \left[ \begin{array}{c} S_{0,n} \\ S_{1,n} \\ S_{2,n} \end{array} \right]$ , we can write as the matrix recurrence

$\mathbf{S}_{n + 1} = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] \mathbf{S}_{n}$ .

We dealt with companion matrices earlier in dealing with linear recurrences; this is another special type of matrix known as a circulant matrix, whose eigenvalues turn out to be determined by a discrete Fourier transform. Its eigenvalues are $2, 1 + \omega, 1 + \omega^2$ , which explains the patterns we observed directly (and this observation generalizes), but eigendecomposition is not in the purely combinatorial spirit of this post.

The casework is too tedious to go through a second or third time. What we're basically seeing here is that these three sums are nearly equal except for a tiny error term that behaves in a very annoying way. We should therefore expect this to manifest itself in "near-bijections" between the corresponding three sets.

Actually, half of the time there are just bijections. When $n \equiv 0 \bmod 3$ the complement of a $1$ -set is a $2$ -set, when $n \equiv 1 \bmod 3$ the complement of a $1$ -set is a $0$ -set, and when $n \equiv 2 \bmod 3$ the complement of a $0$ -set is a $2$ -set, so in any case at least two of these sums are equal at a time. Perhaps we should just stick to one case; let's try $n \equiv 0 \bmod 6$ . Our algebraic arguments tell us we should have $S_0 = S_1 + 1 = S_2 + 1$ , which means we should try to find an injection from either $S_1$ or $S_2$ to $S_0$ that misses exactly one set.

... and it is here that I am stuck! I have not been able to construct such an injection, and again I must end a post in disappointment. I will keep thinking about this problem, but in the meantime I suppose this is food for thought.

Practice Problem 1: Recall the combinatorial definition of $F_{n + 1}$ as the number of ways to tile a $1 \times n$ rectangle using $1 \times 1$ and $1 \times 2$ tiles. Give combinatorial arguments for the following identities (not arranged in order of difficulty):

- $F_n^2 + F_{n + 1}^2 = F_{2n + 1}$
- $F_{n + 1}^2 + 2F_{n + 1} F_n = F_{2n + 2}$
- $1 + F_0 + F_1 + ... + F_n = F_{n + 2}$
- $F_n | F_{mn}$
- $F_{n + 1} F_{n - 1} = ( - 1)^n + F_n^2$

Practice Problem 2: Let $p$ be a prime and let $a$ be an integer not divisible by $p$ . How many ways are there to arrange $p$ beads of $a$ different colors on a necklace? Conclude that $p | a^p - a$ .

Practice Problem 3: Give a combinatorial interpretation for the equality of exponential generating functions

$\left( x - \frac {x^3}{3!} + \frac {x^5}{5!} \mp ... \right)^2 + \left( 1 - \frac {x^2}{2!} + \frac {x^4}{4!} \mp ... \right)^2 = 1$ ;

in other words, a combinatorial proof of the Pythagorean theorem.

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This is most likely not pretty enough for what you're looking for, but here goes:

Given some universe $U$ (here the set $\{1, \cdots, n\}$ ) and some subset $S \subset U$ , I'll define the map "reverse with respect to $S$ " as $A \mapsto (A \cap S) \cup (U \setminus (A \cup S))$ (blah I just don't like the notation for complements so I write $U \setminus$ instead). In other words, the set $A$ gets broken up into the part in $S$ and the part not in $S$ ; we keep the part in $S$ and complement the part not in $S$ (the complement being with respect to $U \setminus S$ ). Now, is this the sort of an injection that might be useful?

If the set has a $1$ , reverse wrt $\{1\}$ .
If the set has no $1$ , no $2$ , reverse wrt $\{1, 2\}$ .
If the set has no $1$ , a $2$ , a $3$ , reverse wrt $\{1, 2, 3\}$ .
If the set has no $1$ , a $2$ , no $3$ , no $4$ , reverse wrt $\{1, 2, 3, 4\}$ .

$\cdots$

If the set has no $1$ , a $2$ , no $3$ , a $4$ , $\cdots$ , no $2k - 1$ , a $2k$ , a $2k + 1$ , reverse wrt $\{1, \cdots, 2k + 1\}$ .
If the set has no $1$ , a $2$ , $\cdots$ , a $2k$ , no $2k + 1$ , no $2k + 2$ , reverse wrt $\{1, \cdots, 2k + 2\}$ .

$\cdots$

Of course this can be restated in terms of the boolean ring structure, though I don't think it makes this operation seem immediately more natural.

by MysticTerminator, Aug 15, 2008, 3:50 AM

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I can never read your posts too well...

Is there a typo in the second practice problem? I don't think $a$ can't not be divisible by a prime...

by Brut3Forc3, Aug 17, 2008, 4:49 AM

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Oops - fixed.

by t0rajir0u, Aug 17, 2008, 7:03 AM

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Wow this is pretty cool. I've thought of some similar things myself.

In light of proving algebraic identities using combinatorial reasoning, here is a nice proof of the fact that
$\sum_{d|n} \phi (d) = n$
Let's look at the cyclic group with $n$ elements. How many elements have order $n$ ? Well, if $x$ is a generator, we want $x^k$ , where $\gcd(k,n) = 1$ , and so there are $\phi(n)$ elements of order $n$ .

In general, if $d | n$ , then we want elements of the form $x^{(k)(\frac {n}{d})}$ , where $\gcd(k,d) = 1$ . The number of such elements is $\phi(d)$ .

Now we add up the number of elements of order $d$ for each $d|n$ to get:
$\sum_{d|n} \phi (d)$
as the total number of elements in the group. Of course, $n$ is the total number of elements in the group, and this proves our claim

You could of course do this in a more algebraic way, but in the spirit of this post, this way is nice.

by calc rulz, Nov 9, 2008, 7:55 PM

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Here's another way of thinking about the "subsets modulo 3" problem. When you're trying to prove three sets nearly equal instead of two, what you should look for is a map $\phi$ which has period $3$ instead of $2$ , and has very few fixed points.

If $n$ is small enough (say $n=2$ ), it's easy to write down the bijection (or shall I say trijection) explicitly: $\emptyset \rightarrow \{2\} \rightarrow \{1, 2\} \rightarrow \emptyset$ .

If $n$ is even, you can extend this to a trijection on almost the whole set by thinking of $\{1, \dots 2m\}$ as made up of $m$ blocks of length $2$ . Letting $k$ be the smallest positive integer so that $\{2k-1, 2k\} \cap S \neq \{2k-1\}$ , we perform the $n=2$ trijection on that block of length $2$ .

This gives a map whose only fixed point is the set $\{1,3, \dots 2m-1\}$ , so the three sets have equal cardinality except for that one fixed point (which has $m$ elements).

If $n$ is odd, you do the same thing, only now there's an element left over not in any block of length $2$ , so you actually have two fixed points $\{1, 3, \dots 2m-1\}$ and $\{1, 3, \dots, 2m+1\}$ .

by kevinatcausa, Feb 26, 2009, 7:12 PM

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Annoying precision

The title of this blog and enumerative combinatorics

by t0rajir0u, Aug 13, 2008, 5:28 AM

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