Eric Fusy (LIX, Baxter permutations and meandersfusy/Talks/baxter_meanders.pdf · Baxter...

Baxter permutations and meanders

Eric Fusy (LIX, Ecole Polytechnique)

Journees Viennot, 28-29 juin 2012, Labri

Meanders on two lines• A 2-line meander

Meanders on two lines• A 2-line meander encoded by a permutation

1 234 567 8 9


• Monotone 2-line meander:can be obtained from two monotone lines (one in x, the other in y)

1 234 567 8 9



associatedpermutation

1 234 567 8 9




Which permutations can be obtained this way ?

1 234 567 8 9




Which permutations can be obtained this way ?Maps odd numbers to odd numbers, even numbers to even numbers

1 234 567 8 9

Permutations for monotone 2-line meanders[Baxter’64, Boyce’67&’81]

Permutations for monotone 2-line meanders

left orright?

[Baxter’64, Boyce’67&’81]


left



then has togo left

left`



``



``

left orright?



``

left orright?

r

then has togo right



``

left orright?

r r



``

left orright?

r r

`



``

left orright?

r r

`

then has togo left



``

left orright?

r r

``



``

left orright?

r r

``

white points are either:

or


rising descending


``

r r

``


or


rising descending


``

r r

``


rising descending

or


``

r r

``


Permutations mapping even to even, odd to odd, and satisfying condition

rising descending

or

shown on the right are called complete Baxter permutations


``

r r

``


Permutations mapping even to even, odd to odd, and satisfying condition

Theorem ([Boyce’81] reformulated bijectively):Monotone 2-line meanders with 2n− 1 crossings are in bijection withcomplete Baxter permutations on 2n− 1 elements

rising descending

or

shown on the right are called complete Baxter permutations

Inverse constructionFrom a complete Baxter permutation to a monotone 2-line meander


or



1) Draw the blue curve

or




2) Draw the red curve

or




2) Draw the red curve

The two curves meet only at the permutation points(because of the empty area-property at white points)

or

Complete and reduced Baxter permutations

complete reduced


complete reduced

• complete one can be recovered from reduced one


complete reduced


case of a descent


complete reduced


case of a rise


complete reduced

• complete one can be recovered from reduced one• reduced one is characterized by forbidden patterns

2− 41− 3 and 3− 14− 2


complete reduced

• complete one can be recovered from reduced one• reduced one is characterized by forbidden patterns

2− 41− 3 and 3− 14− 2• permutation on white points (called anti-Baxter)is characterized by forbidden patterns

2− 14− 3 and 3− 41− 2

Counting results• Baxter permutations

[Chung et al’78] [Mallows’79]

- Number of reduced Baxter permutations with n elements

bn =

n−1∑r=0

2

n(n+ 1)2

(n+ 1

r

)(n+ 1

r + 1

)(n+ 1

r + 2

)




bn =

n−1∑r=0

2

n(n+ 1)2

(n+ 1

r

)(n+ 1

r + 1

)(n+ 1

r + 2

)

- Bijective proof: [Viennot’81], [Dulucq-Guibert’98]

5 7 6 2 1 4 3 ⇔




bn =

n−1∑r=0

2

n(n+ 1)2

(n+ 1

r

)(n+ 1

r + 1

)(n+ 1

r + 2

)


• Subfamilies- alternating [Cori-Dulucq-Viennot’86], [Dulucq-Guibert’98]

- doubly alternating [Guibert-Linusson’00]CatkCatk if n = 2k CatkCatk+1 if n = 2k + 1

Catk where k = bn/2c

5 7 6 2 1 4 3 ⇔




bn =

n−1∑r=0

2

n(n+ 1)2

(n+ 1

r

)(n+ 1

r + 1

)(n+ 1

r + 2

)


• Subfamilies- alternating [Cori-Dulucq-Viennot’86], [Dulucq-Guibert’98]

- doubly alternating [Guibert-Linusson’00]

• anti-Baxter permutations

CatkCatk if n = 2k CatkCatk+1 if n = 2k + 1

Catk where k = bn/2c

an =

b(n+1)/2c∑i=0

(−1)i(n+ 1− i

i

)bn+1−i[Asinowski et al’10]

5 7 6 2 1 4 3 ⇔

Local conditions for monotone 2-line meanders


Conditions- two (bipartite) matchings missing a (black) point

(one matching above, one below the blue line)

⇔

- white points are either or

rising descending




⇔


Proof of ⇐Assume there is a red loop (say, clockwise):

rising descending




⇔



then the leftmost and therighmost point on the loopare of different colors

rising descending




⇔




impossible

rising descending




⇔




impossible

⇒ we have a 2-line meander

rising descending


Conditions


⇔

- white nodes are either or

Proof of ⇐: construct permutation step by step

- two (bipartite) matchings missing a (black) point


Conditions


⇔



1



Conditions


⇔



12



Conditions


⇔



123



Conditions


⇔



123 4



Conditions


⇔



123 4

important observation:

i i+1︸︷︷︸already labelled

By similar argument as toshow there is no red loop



Conditions


⇔



123 4



ii+1 ︸︷︷︸already labelled

By similar argument as toshow there is no red loop



Conditions


⇔



123 4



5




Conditions


⇔



123 4



56




Conditions


⇔



123 4



56


7



Conditions


⇔



123 4



56


78



Conditions


⇔



123 4



56


789



Conditions


⇔



123 4



56


789 10



Conditions


⇔



123 4



56


789 10 11



Conditions


⇔



123 4



56


789 10 1112



Conditions


⇔



123 4



56


789 10 111213



Conditions


⇔



123 4



56


789 10 111213

Similarly:

or


Encoding a monotone 2-line meander


⇓


⇓

⇓


⇓

⇓

0 1 1 0 1 1

1 0 1 1 1 0

0 1 1 1 0 1


⇓

⇓

0 1 1 0 1 1

1 0 1 1 1 0

0 1 1 1 0 1 ⇒0

1


⇓

⇓

0 1 1 0 1 1

1 0 1 1 1 0

0 1 1 1 0 1 ⇒0

1

# rises

each path haslength n− 1


⇓

⇓

0 1 1 0 1 1

1 0 1 1 1 0

0 1 1 1 0 1 ⇒0

1

# rises


close to encoding in [Viennot’81,Dulucq-Guibert’98]


⇓

⇓

0 1 1 0 1 1

1 0 1 1 1 0

0 1 1 1 0 1 ⇒0

1

# rises


close to encoding in [Viennot’81,Dulucq-Guibert’98]exactly coincides with encoding in [Felsner-F-Noy-Orden’11](uses “equatorial line” in separating decompositions of quadrangulations)

Enumeration using the LGV lemmar


A1

A2

A3

B1

B2

B3



A1

A2

A3

B1

B2

B3

Let ai,j = # (upright lattice paths from Ai to Bj) =(

n−1x(Bj)−x(Ai)

)By the Lindstroem-Gessel-Viennot Lemma (used in [Viennot’81])

the number bn,r of such nonintersecting triples of paths is

bn,r = Det(ai,j) =

∣∣∣∣∣∣∣∣(n−1r

) (n−1r+1

) (n−1r+2

)(n−1r−1) (

n−1r

) (n−1r+1

)(n−1r−2) (

n−1r−1) (

n−1r

)∣∣∣∣∣∣∣∣=

2n(n+1)2

(n+1r

)(n+1r+1

)(n+1r+2

)



A1

A2

A3

B1

B2

B3

Let ai,j = # (upright lattice paths from Ai to Bj) =(

n−1x(Bj)−x(Ai)

)By the Lindstroem-Gessel-Viennot Lemma (used in [Viennot’81])

the number bn,r of such nonintersecting triples of paths is

bn,r = Det(ai,j) =

∣∣∣∣∣∣∣∣(n−1r

) (n−1r+1

) (n−1r+2

)(n−1r−1) (

n−1r

) (n−1r+1

)(n−1r−2) (

n−1r−1) (

n−1r

)∣∣∣∣∣∣∣∣=

2n(n+1)2

(n+1r

)(n+1r+1

)(n+1r+2

)

bn,r is also the number of reduced Baxter permutations of sizen with r rises

Alternating (reduced) Baxter permutations

1) Case n even, n = 2k[Cori-Dulucq-Viennot’86], [Dulucq-Guibert’98]

π= 1 23498 67 510


alternation ⇔ middle word is 010101 . . . 0

1) Case n even, n = 2k[Cori-Dulucq-Viennot’86], [Dulucq-Guibert’98]

π= 1 23498 67 510



1) Case n even, n = 2k

middle path is

[Cori-Dulucq-Viennot’86], [Dulucq-Guibert’98]

π= 1 23498 67 510



1) Case n even, n = 2k

middle path is

⇒


There are CatkCatk alternating(reduced) Baxter permutations of size n

k

π= 1 23498 67 510



2) Case n odd, n = 2k + 1

middle path is

⇒


There are CatkCatk+1 alternating(reduced) Baxter permutations of size n

k

π= 12 3 49 8 6 7 5

k

Doubly alternating (reduced) Baxter permutations[Guibert-Linusson’00]1) Case n even, n = 2k

1 2438675π =

alternation of π: middle word is 010101 . . . 0

alternation of π−1: black points are or

middle path is

Doubly alternating (reduced) Baxter permutations[Guibert-Linusson’00]1) Case n even, n = 2k

1 2438675π =

alternation of π: middle word is 010101 . . . 0

alternation of π−1: black points are or

⇒ There are Catk doubly alternating(reduced) Baxter permutations of size nmirror

of eachother

middle path is

k

Doubly alternating (reduced) Baxter permutations[Guibert-Linusson’00]2) Case n odd, n = 2k + 1

1 2 43 8 675π=

alternation of π: middle word is 010101 . . . 1alternation of π−1: black points are or

⇒ There are Catk doubly alternating(reduced) Baxter permutations of size nmirror

of eachother

9

not intop-word

change of convention

not inbottom-word

k

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