Counting Permutations by Putting Balls into Boxes

Transcription

1 Counting Permutations by Putting Balls into Boxes Ira M. Gessel Brandeis University Conference June 19, 2007

2 I will tell you shamelessly what my bottom line is: It is placing balls into boxes. Gian-Carlo Rota, Indiscrete Thoughts

3 Eulerian polynomials If π is a permutation of [n] = {1, 2,..., n}, a descent of π is an i, with 1 i n 1, such that π(i) > π(i + 1). Example: has two descents. Let S n be the group of permutations of [n]. How many permutations in S n have i descents? Let us define the Eulerian polynomials E n (t) = t des(π). π S n Then E 1 (t) = 1, E 2 (t) = 1 + t, E 3 (t) = 1 + 4t + t 2.

4 Theorem. k n t k 1 = k=1 π S n t des(π) (1 t) n+1 Proof. k n is the number of placements of n balls, labeled 1, 2,..., n, into k boxes. We will associate to each placement a permutation π S n so that the total contribution from π is t des(π) /(1 t) n+1. We represent a placement more compactly as The balls in each box are in increasing order.

5 We remove the bars from the placement to get the permutation So we can think of a placement as a permutation with bars in it: a barred permutation. Which barred permutations correspond to ?

6 We remove the bars from the placement to get the permutation So we can think of a placement as a permutation with bars in it: a barred permutation. Which barred permutations correspond to ? We need at least one bar in every descent, so we start with

7 We remove the bars from the placement to get the permutation So we can think of a placement as a permutation with bars in it: a barred permutation. Which barred permutations correspond to ? We need at least one bar in every descent, so we start with Then we put any number of additional bars in each of the 6 spaces to get We assign the weight t to each bar. Then the contribution from this permutation is t(1 + t + t 2 + ) 6 = t (1 t) 6

8 In general, the contribution from a permutation π of [n] is t des(π) (1 t) n+1 so k n t k 1 = k=1 π S n t des(π) (1 t) n+1. Note that if there are k boxes then there are k 1 bars.

9 The Method of Barred Permutations A barred permutation is a permutation of balls numbered 1 to n with bars in it. Between (and before and after) the bars are boxes and between (and before and after) the balls are spaces. spaces boxes We can count barred permutations in two ways: 1) Start with bars, and put balls into boxes. 2) Start with a permutation, and put bars into spaces. Note: The method of barred permutations is closely related to the method of P-partitions (MacMahon, Knuth, Stanley).

10 2-descents We consider barred permutations in which consecutive balls cannot be in the same box How many placements of n balls in k boxes are there?

11 2-descents We consider barred permutations in which consecutive balls cannot be in the same box How many placements of n balls in k boxes are there? Ball 1: Ball 2: Ball 3: Ball n: k boxes k 1 boxes k 1 boxes... k 1 boxes So there are k(k 1) n 1 placements.

12 If we start with a permutation (for example, ) we must put a bar in each descent, but also in any space where m is followed by m + 1:

13 If we start with a permutation (for example, ) we must put a bar in each descent, but also in any space where m is followed by m + 1:

14 If we start with a permutation (for example, ) we must put a bar in each descent, but also in any space where m is followed by m + 1: Then we put an arbitrary number of additional bars in each space:

15 If we start with a permutation (for example, ) we must put a bar in each descent, but also in any space where m is followed by m + 1: Then we put an arbitrary number of additional bars in each space: We call i a 2-descent of π if π(i) + 2 > π(i + 1). Then by the same reasoning as before, k(k 1) n 1 t k 1 = k=1 π S n t 2-des(π) (1 t) n+1.

16 We could define r-descents similarly: π(i) + r > π(i + 1). The same reasoning would give k(k 1)(k 2) (k r+2)(k r+1) n r+1 t k 1 = k=1 π S n t r-des(π) (1 t) n+1. (Foata-Schützenberger)

17 We could define r-descents similarly: π(i) + r > π(i + 1). The same reasoning would give k(k 1)(k 2) (k r+2)(k r+1) n r+1 t k 1 = k=1 π S n t r-des(π) (1 t) n+1. (Foata-Schützenberger) Note that k(k 1)(k 2) (k r + 2)(k r + 1) n r+1 is the chromatic polynomial of the graph with vertex set [n] in which two vertices are adjacent if and only if they differ by at most r 1; i.e., they are not allowed in the same box. A similar result holds for the chromatic polynomial of any chordal graph.

18 Signed Permutations A signed permutation of [n] is permutation of [n] in which the entries may have minus signs: It s convenient to write ī for i so we ll write this signed permutation as Sometimes it s useful to think of a signed permutation as a permutation of the set { n, n + 1,..., n 1, n} (with or without 0) with the property that π( i) = π(i). We denote by B n the set (or group) of signed permutations of [n]. (This is the hyperoctahedral group, the Coxeter group of type B n.) Descents of signed permutations are defined as usual except that if π(1) < 0 then 0 is a descent of π. (Think of π(0) = 0.)

19 Theorem. (Steingrímsson) (2k + 1) n t k = k=0 π B n t des(π) (1 t) n+1

20 Theorem. (Steingrímsson) (2k + 1) n t k = k=0 π B n t des(π) (1 t) n+1 To prove this formula, we count barred signed permutations In the first box (box 0) only positive numbers can appear but in the other boxes, positive and negative numbers can appear. How many barred permutations have k bars? For each of the n balls, we can make it positive and put it in any of k + 1 boxes or make it negative and put it in any of k boxes. So there are (k + 1) + k = 2k + 1 possibilities for each ball, so (2k + 1) n in all.

21 Flag descents Adin, Brenti, and Roichman (2001) defined the flag-descent number of a signed permutation π by fdes(π) = 2 des π χ(π(1) < 0). In other words all descents are counted twice, except that a descent in position 0 is counted only once. They proved π B k n t k 1 = n t fdes(π) (1 t)(1 t 2 ) n. k=1

22 Flag descents Adin, Brenti, and Roichman (2001) defined the flag-descent number of a signed permutation π by fdes(π) = 2 des π χ(π(1) < 0). In other words all descents are counted twice, except that a descent in position 0 is counted only once. They proved π B k n t k 1 = n t fdes(π) (1 t)(1 t 2 ) n. k=1 Comparing this with our formula for Eulerian polynomials, π S k n t k 1 = n t des(π) (1 t) n+1, k=1 we see that their result is equivalent to t fdes(π) = (1 + t) n t des(π). π B n π S n

23 Their proof was by induction, but we can give a direct proof using barred permutations. Let s first go back to the enumeration of signed permutations by descents. Instead of looking at the sequence π(1) π(2) π(n), let s look at π( n) π( 1) π(0) π(1) π(n), where π( i) = π(i) and in particular, π(0) = 0. We consider only symmetric" barred permutations, for example or These symmetric barred permutations have 2(n + 1) spaces and 2k bars for some k. (So if we want to weight all the bars equally, we should weight each one by t rather than t). If we know the right half of a such a barred permutation (the part to the right of the 0) then the left half is determined. So what have we gained?

24 We can give a slightly different argument from before that the number of barred permutations with 2k bars (corresponding to k bars before) is (2k + 1) n. With 2k bars there are now 2k + 1 spaces, so we put the numbers 1, 2,..., n arbitrarily into these 2k + 1 spaces, and then the locations of 1, 2,..., n (and 0) are determined

25 We can give a slightly different argument from before that the number of barred permutations with 2k bars (corresponding to k bars before) is (2k + 1) n. With 2k bars there are now 2k + 1 spaces, so we put the numbers 1, 2,..., n arbitrarily into these 2k + 1 spaces, and then the locations of 1, 2,..., n (and 0) are determined 3 2 1

26 We can give a slightly different argument from before that the number of barred permutations with 2k bars (corresponding to k bars before) is (2k + 1) n. With 2k bars there are now 2k + 1 spaces, so we put the numbers 1, 2,..., n arbitrarily into these 2k + 1 spaces, and then the locations of 1, 2,..., n (and 0) are determined

27 For flag descents, we do the same thing, but without 0. For example, if we start with the permutation 3 2 1, adding in its negative half gives There are now 7 spaces, rather than 8, and they are paired, except for the central space. Then if we count the barred permutations corresponding to this permutation (weighting each bar with t), the bars in the n noncentral spaces come in pairs, but the bars in the center space do not

28 For flag descents, we do the same thing, but without 0. For example, if we start with the permutation 3 2 1, adding in its negative half gives There are now 7 spaces, rather than 8, and they are paired, except for the central space. Then if we count the barred permutations corresponding to this permutation (weighting each bar with t), the bars in the n noncentral spaces come in pairs, but the bars in the center space do not So the sum of the weights of the barred permutation corresponding to a given permutation π is t fdes(π) (1 t)(1 t 2 ) n

29 To get the other side of the equation, we note that the number of bars need not be even; it can be any number. If there are k 1 bars, there are k boxes, and thus k n ways to put 1, 2,..., n into the boxes, and as before the locations of 1, 2,..., n are determined.

30 To get the other side of the equation, we note that the number of bars need not be even; it can be any number. If there are k 1 bars, there are k boxes, and thus k n ways to put 1, 2,..., n into the boxes, and as before the locations of 1, 2,..., n are determined. So we have Adin, Brenti, and Roichman s identity k n t k 1 = k=1 π B n t fdes(π) (1 t)(1 t 2 ) n.

31 We do not have a bijective proof of π B n t fdes(π) = (1 + t) n σ S n t des(σ). However, our approach gives a refinement of this formula, by telling us which permutations in B n correspond to each permutation in S n. For σ S n, let B(σ) be the set of 2 n signed permutations in B n obtained from σ by the following procedure: First we cut σ into two parts:

32 We do not have a bijective proof of π B n t fdes(π) = (1 + t) n σ S n t des(σ). However, our approach gives a refinement of this formula, by telling us which permutations in B n correspond to each permutation in S n. For σ S n, let B(σ) be the set of 2 n signed permutations in B n obtained from σ by the following procedure: First we cut σ into two parts: Next we reverse and negate the first part:

33 We do not have a bijective proof of π B n t fdes(π) = (1 + t) n σ S n t des(σ). However, our approach gives a refinement of this formula, by telling us which permutations in B n correspond to each permutation in S n. For σ S n, let B(σ) be the set of 2 n signed permutations in B n obtained from σ by the following procedure: First we cut σ into two parts: Next we reverse and negate the first part: Finally we shuffle the two parts:

34 Then π B(σ) t fdes(π) = (1 + t) n t des(σ)

35 Then π B(σ) t fdes(π) = (1 + t) n t des(σ) Proof sketch. The set of barred permutations of σ (in S n ) is the same as the set of barred permutations of elements of B(σ) (in B n ). Therefore t des(σ) π B(σ) (1 t) n+1 = tfdes(π) (1 t)(1 t 2 ) n. Example