Counting Permutations by Putting Balls into...
Transcript of Counting Permutations by Putting Balls into...
Counting Permutationsby Putting Balls into Boxes
Ira M. Gessel
Brandeis University
C&O@40 ConferenceJune 19, 2007
I will tell you shamelessly what my bottom line is: It is placingballs into boxes.
—Gian-Carlo Rota, Indiscrete Thoughts
Eulerian polynomials
If π is a permutation of [n] = {1, 2, . . . , n}, a descent of π is ani , with 1 ≤ i ≤ n − 1, such that π(i) > π(i + 1).
Example: 1 3 • 2 6 • 4 5 has two descents.Let Sn be the group of permutations of [n]. How manypermutations in Sn have i descents? Let us define the Eulerianpolynomials
En(t) =∑
π∈Sn
tdes(π).
Then E1(t) = 1, E2(t) = 1 + t , E3(t) = 1 + 4t + t2.
Theorem.∞∑
k=1
kntk−1 =
∑π∈Sn
tdes(π)
(1− t)n+1
Proof. kn is the number of placements of n balls, labeled 1, 2,. . . , n, into k boxes. We will associate to each placement apermutation π ∈ Sn so that the total contribution from π istdes(π)/(1− t)n+1. We represent a placement
2 5 1 3 4
more compactly as2 5 | 1 | | 3 | 4 |
The balls in each box are in increasing order.
We remove the bars from the placement 2 5 | 1 | | 3 | 4 | to getthe permutation 2 5 1 3 4.
So we can think of a placement as a permutation with bars in it:a barred permutation.
Which barred permutations correspond to 2 5 1 3 4?
We needat least one bar in every descent, so we start with
2 5 | 1 3 4
Then we put any number of additional bars in each of the 6spaces to get 2 5 | 1 | | 3 | 4 | . We assign the weight t to eachbar. Then the contribution from this permutation is
t(1 + t + t2 + · · · )6 =t
(1− t)6
We remove the bars from the placement 2 5 | 1 | | 3 | 4 | to getthe permutation 2 5 1 3 4.
So we can think of a placement as a permutation with bars in it:a barred permutation.
Which barred permutations correspond to 2 5 1 3 4? We needat least one bar in every descent, so we start with
2 5 | 1 3 4
Then we put any number of additional bars in each of the 6spaces to get 2 5 | 1 | | 3 | 4 | . We assign the weight t to eachbar. Then the contribution from this permutation is
t(1 + t + t2 + · · · )6 =t
(1− t)6
We remove the bars from the placement 2 5 | 1 | | 3 | 4 | to getthe permutation 2 5 1 3 4.
So we can think of a placement as a permutation with bars in it:a barred permutation.
Which barred permutations correspond to 2 5 1 3 4? We needat least one bar in every descent, so we start with
2 5 | 1 3 4
Then we put any number of additional bars in each of the 6spaces to get 2 5 | 1 | | 3 | 4 | . We assign the weight t to eachbar. Then the contribution from this permutation is
t(1 + t + t2 + · · · )6 =t
(1− t)6
In general, the contribution from a permutation π of [n] is
tdes(π)
(1− t)n+1
so∞∑
k=1
kntk−1 =
∑π∈Sn
tdes(π)
(1− t)n+1 .
Note that if there are k boxes then there are k − 1 bars.
The Method of Barred Permutations
A barred permutation is a permutation of balls numbered 1 to nwith bars in it. Between (and before and after) the bars areboxes and between (and before and after) the balls are spaces.
spaces
boxes
0
21
05
2
| 11|2
3
| 43|4
34|5
5|6
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 tothe method of P-partitions (MacMahon, Knuth, Stanley).
2-descents
We consider barred permutations in which consecutive ballscannot be in the same box.
3 | 5 7 || 1 | 2 6 | 4
How many placements of n balls in k boxes are there?
Ball 1: k boxesBall 2: k − 1 boxesBall 3: k − 1 boxes
. . .Ball n: k − 1 boxes
So there are k(k − 1)n−1 placements.
2-descents
We consider barred permutations in which consecutive ballscannot be in the same box.
3 | 5 7 || 1 | 2 6 | 4
How many placements of n balls in k boxes are there?
Ball 1: k boxesBall 2: k − 1 boxesBall 3: k − 1 boxes
. . .Ball n: k − 1 boxes
So there are k(k − 1)n−1 placements.
If we start with a permutation (for example, 4 1 2 5 3 6) we mustput a bar in each descent, but also in any space where m isfollowed by m + 1:
4 | 1 | 2 5 | 3 6
Then we put an arbitrary number of additional bars in eachspace:
|| 4 | 1 || 2 5 | 3 || 6 |
We call i a 2-descent of π if π(i) + 2 > π(i + 1). Then by thesame reasoning as before,
∞∑k=1
k(k − 1)n−1tk−1 =
∑π∈Sn
t2-des(π)
(1− t)n+1 .
If we start with a permutation (for example, 4 1 2 5 3 6) we mustput a bar in each descent, but also in any space where m isfollowed by m + 1:
4 | 1 | 2 5 | 3 6
Then we put an arbitrary number of additional bars in eachspace:
|| 4 | 1 || 2 5 | 3 || 6 |
We call i a 2-descent of π if π(i) + 2 > π(i + 1). Then by thesame reasoning as before,
∞∑k=1
k(k − 1)n−1tk−1 =
∑π∈Sn
t2-des(π)
(1− t)n+1 .
If we start with a permutation (for example, 4 1 2 5 3 6) we mustput a bar in each descent, but also in any space where m isfollowed by m + 1:
4 | 1 | 2 5 | 3 6
Then we put an arbitrary number of additional bars in eachspace:
|| 4 | 1 || 2 5 | 3 || 6 |
We call i a 2-descent of π if π(i) + 2 > π(i + 1). Then by thesame reasoning as before,
∞∑k=1
k(k − 1)n−1tk−1 =
∑π∈Sn
t2-des(π)
(1− t)n+1 .
If we start with a permutation (for example, 4 1 2 5 3 6) we mustput a bar in each descent, but also in any space where m isfollowed by m + 1:
4 | 1 | 2 5 | 3 6
Then we put an arbitrary number of additional bars in eachspace:
|| 4 | 1 || 2 5 | 3 || 6 |
We call i a 2-descent of π if π(i) + 2 > π(i + 1). Then by thesame reasoning as before,
∞∑k=1
k(k − 1)n−1tk−1 =
∑π∈Sn
t2-des(π)
(1− t)n+1 .
We could define r -descents similarly: π(i) + r > π(i + 1). Thesame reasoning would give
∞∑k=1
k(k−1)(k−2) · · · (k−r+2)(k−r+1)n−r+1tk−1 =
∑π∈Sn
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 thechromatic polynomial of the graph with vertex set [n] in whichtwo vertices are adjacent if and only if they differ by at mostr − 1; i.e., they are not allowed in the same box. A similar resultholds for the chromatic polynomial of any chordal graph.
We could define r -descents similarly: π(i) + r > π(i + 1). Thesame reasoning would give
∞∑k=1
k(k−1)(k−2) · · · (k−r+2)(k−r+1)n−r+1tk−1 =
∑π∈Sn
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 thechromatic polynomial of the graph with vertex set [n] in whichtwo vertices are adjacent if and only if they differ by at mostr − 1; i.e., they are not allowed in the same box. A similar resultholds for the chromatic polynomial of any chordal graph.
Signed Permutations
A signed permutation of [n] is permutation of [n] in which theentries may have minus signs:
−4 2 − 1 − 5 3
It’s convenient to write i for −i so we’ll write this signedpermutation as
4 2 1 5 3
Sometimes it’s useful to think of a signed permutation as apermutation of the set {−n,−n + 1, . . . , n − 1, n} (with orwithout 0) with the property that π(−i) = −π(i). We denote byBn the set (or group) of signed permutations of [n]. (This is thehyperoctahedral group, the Coxeter group of type Bn.)
Descents of signed permutations are defined as usual exceptthat if π(1) < 0 then 0 is a descent of π. (Think of π(0) = 0.)
Theorem. (Steingrímsson)
∞∑k=0
(2k + 1)ntk =
∑π∈Bn
tdes(π)
(1− t)n+1
To prove this formula, we count barred signed permutations.
2 4 | 3 1 5 | | 6
In the first box (box 0) only positive numbers can appear but inthe other boxes, positive and negative numbers can appear.
How many barred permutations have k bars? For each of the nballs, we can make it positive and put it in any of k + 1 boxes ormake 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 inall.
Theorem. (Steingrímsson)
∞∑k=0
(2k + 1)ntk =
∑π∈Bn
tdes(π)
(1− t)n+1
To prove this formula, we count barred signed permutations.
2 4 | 3 1 5 | | 6
In the first box (box 0) only positive numbers can appear but inthe other boxes, positive and negative numbers can appear.
How many barred permutations have k bars? For each of the nballs, we can make it positive and put it in any of k + 1 boxes ormake 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 inall.
Flag descentsAdin, Brenti, and Roichman (2001) defined the flag-descentnumber 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 adescent in position 0 is counted only once. They proved
∞∑k=1
kntk−1 =
∑π∈Bn
t fdes(π)
(1− t)(1− t2)n .
Comparing this with our formula for Eulerian polynomials,∞∑
k=1
kntk−1 =
∑π∈Sn
tdes(π)
(1− t)n+1 ,
we see that their result is equivalent to∑π∈Bn
t fdes(π) = (1 + t)n∑
π∈Sn
tdes(π).
Flag descentsAdin, Brenti, and Roichman (2001) defined the flag-descentnumber 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 adescent in position 0 is counted only once. They proved
∞∑k=1
kntk−1 =
∑π∈Bn
t fdes(π)
(1− t)(1− t2)n .
Comparing this with our formula for Eulerian polynomials,∞∑
k=1
kntk−1 =
∑π∈Sn
tdes(π)
(1− t)n+1 ,
we see that their result is equivalent to∑π∈Bn
t fdes(π) = (1 + t)n∑
π∈Sn
tdes(π).
Their proof was by induction, but we can give a direct proofusing barred permutations. Let’s first go back to theenumeration of signed permutations by descents. Instead oflooking 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 consideronly “symmetric" barred permutations, for example
| 1 || 3 2 | 0 | 2 3 ||1|
or| 1 || 3 | 2 0 2 | 3 ||1|
These symmetric barred permutations have 2(n + 1) spacesand 2k bars for some k . (So if we want to weight all the barsequally, we should weight each one by
√t rather than t).
If we know the right half of a such a barred permutation (thepart to the right of the 0) then the left half is determined. Sowhat have we gained?
We can give a slightly different argument from before that thenumber of barred permutations with 2k bars (corresponding tok bars before) is (2k + 1)n. With 2k bars there are now 2k + 1spaces, so we put the numbers 1, 2,. . . , n arbitrarily into these2k + 1 spaces, and then the locations of −1, −2, . . . , −n (and0) are determined
| | | | | | | |
| | | 3 | 2 | | |1|
| 1 | | 3 | 2 0 2 | 3 | |1|
We can give a slightly different argument from before that thenumber of barred permutations with 2k bars (corresponding tok bars before) is (2k + 1)n. With 2k bars there are now 2k + 1spaces, so we put the numbers 1, 2,. . . , n arbitrarily into these2k + 1 spaces, and then the locations of −1, −2, . . . , −n (and0) are determined
| | | | | | | |
| | | 3 | 2 | | |1|
| 1 | | 3 | 2 0 2 | 3 | |1|
We can give a slightly different argument from before that thenumber of barred permutations with 2k bars (corresponding tok bars before) is (2k + 1)n. With 2k bars there are now 2k + 1spaces, so we put the numbers 1, 2,. . . , n arbitrarily into these2k + 1 spaces, and then the locations of −1, −2, . . . , −n (and0) are determined
| | | | | | | |
| | | 3 | 2 | | |1|
| 1 | | 3 | 2 0 2 | 3 | |1|
For flag descents, we do the same thing, but without 0. Forexample, if we start with the permutation 3 2 1, adding in itsnegative half gives
1 2 3 3 2 1
There are now 7 spaces, rather than 8, and they are paired,except for the central space. Then if we count the barredpermutations corresponding to this permutation (weightingeach bar with t), the bars in the n noncentral spaces come inpairs, but the bars in the center space do not.
|| 2 | 1 3 ||| 3 1 | 2 ||
So the sum of the weights of the barred permutationcorresponding to a given permutation π is
t fdes(π)
(1− t)(1− t2)n
For flag descents, we do the same thing, but without 0. Forexample, if we start with the permutation 3 2 1, adding in itsnegative half gives
1 2 3 3 2 1
There are now 7 spaces, rather than 8, and they are paired,except for the central space. Then if we count the barredpermutations corresponding to this permutation (weightingeach bar with t), the bars in the n noncentral spaces come inpairs, but the bars in the center space do not.
|| 2 | 1 3 ||| 3 1 | 2 ||
So the sum of the weights of the barred permutationcorresponding to a given permutation π is
t fdes(π)
(1− t)(1− t2)n
To get the other side of the equation, we note that the numberof bars need not be even; it can be any number. If there arek − 1 bars, there are k boxes, and thus kn ways to put 1, 2, . . . ,n into the boxes, and as before the locations of 1, 2, . . . , n aredetermined.
So we have Adin, Brenti, and Roichman’s identity
∞∑k=1
kntk−1 =
∑π∈Bn
t fdes(π)
(1− t)(1− t2)n .
To get the other side of the equation, we note that the numberof bars need not be even; it can be any number. If there arek − 1 bars, there are k boxes, and thus kn ways to put 1, 2, . . . ,n into the boxes, and as before the locations of 1, 2, . . . , n aredetermined.
So we have Adin, Brenti, and Roichman’s identity
∞∑k=1
kntk−1 =
∑π∈Bn
t fdes(π)
(1− t)(1− t2)n .
We do not have a bijective proof of∑π∈Bn
t fdes(π) = (1 + t)n∑
σ∈Sn
tdes(σ).
However, our approach gives a refinement of this formula, bytelling us which permutations in Bn correspond to eachpermutation in Sn.
For σ ∈ Sn, let B(σ) be the set of 2n signed permutations in Bnobtained from σ by the following procedure:
• First we cut σ into two parts:
1425637 → 142 5637
• Next we reverse and negate the first part:
241 5637
• Finally we shuffle the two parts:
2564137
We do not have a bijective proof of∑π∈Bn
t fdes(π) = (1 + t)n∑
σ∈Sn
tdes(σ).
However, our approach gives a refinement of this formula, bytelling us which permutations in Bn correspond to eachpermutation in Sn.
For σ ∈ Sn, let B(σ) be the set of 2n signed permutations in Bnobtained from σ by the following procedure:
• First we cut σ into two parts:
1425637 → 142 5637
• Next we reverse and negate the first part:
241 5637
• Finally we shuffle the two parts:
2564137
We do not have a bijective proof of∑π∈Bn
t fdes(π) = (1 + t)n∑
σ∈Sn
tdes(σ).
However, our approach gives a refinement of this formula, bytelling us which permutations in Bn correspond to eachpermutation in Sn.
For σ ∈ Sn, let B(σ) be the set of 2n signed permutations in Bnobtained from σ by the following procedure:
• First we cut σ into two parts:
1425637 → 142 5637
• Next we reverse and negate the first part:
241 5637
• Finally we shuffle the two parts:
2564137
Then ∑π∈B(σ)
t fdes(π) = (1 + t)ntdes(σ)
Proof sketch. The set of barred permutations of σ (in Sn) is the
same as the set of barred permutations of elements of B(σ) (inBn). Therefore
tdes(σ)
(1− t)n+1 =
∑π∈B(σ) t fdes(π)
(1− t)(1− t2)n .
Example.
3 | 1 ||| 2 → 2 3 | 1 || 1 | 3 2
3 1 2 → 1 3 2
Then ∑π∈B(σ)
t fdes(π) = (1 + t)ntdes(σ)
Proof sketch. The set of barred permutations of σ (in Sn) is the
same as the set of barred permutations of elements of B(σ) (inBn). Therefore
tdes(σ)
(1− t)n+1 =
∑π∈B(σ) t fdes(π)
(1− t)(1− t2)n .
Example.
3 | 1 ||| 2 → 2 3 | 1 || 1 | 3 2
3 1 2 → 1 3 2