Towards the distribution of the size of the largest non-crossing - PowerPoint PPT Presentation

Towards the distribution of the size of the largest non-crossing matchings in random bipartite graphs Marcos KIWI, U. Chile joint work with Martin LOEBL, Charles U.

Problem w 1 w 2 w 3 w n G ∼ G m 1 m 2 m 3 m n L ( G ) = 5

Instance 1: Longest Increasing Sequence (LIS) Problem � 1 � 2 3 4 5 6 7 8 π = 3 6 1 2 8 7 4 5 1 2 3 4 5 6 7 8 π (3) = 1 π (4) = 2 π (1) = 3 π (7) = 4 π (8) = 5 π (2) = 6 π (6) = 7 π (5) = 8 L ( G π ) = 4

Instance 2: Longest Common Subsequence (LCS) Problem α = a b c b d a d c β = a a c b d d c c L ( G α,β ) = 6

Known Results LIS model Baik-Deift-Johanson - J. AMS’99 L − 2 √ n asymptotically, is Tracy-Widom. n 1 / 6 LCS model Loebl-K.-Matousek - AIM’05 √ 1 ∃ γ k = lim n E ( L ) and γ k k → 2 as k → ∞ . n → + ∞ r -regular graph model Loebl-K. - RS&A’02 1 √ rn E ( L ) → 2 as n → ∞ when r = o ( n 1 / 4 ). Erd¨ os-Renyi model Sepp¨ al¨ ainen - Ann. App. Prob.’97 1 2 n L → 1+1 / √ p a.s. as n → ∞ when 0 < p < 1.

Focus of this talk We consider the uniform distribution over r -regular bipartite multigraphs with n nodes per color class. We try to derive/characterize the distribution of L , ... not only its expectation.

Gessel’s Identity If g 1 ( n ; d ) denotes the number of permutations of [ n ] with LIS at most d , and 1 � x � 2 m + ν � I ν ( x ) = , m !( m + ν )! 2 m ∈ N it holds that g 1 ( n ; d ) x 2 n = det � � � I | r − s | (2 x ) 1 ≤ r , s ≤ d . n ! 2 n ∈ N Equivalently, for N ( t ) Poisson of rate t and π uniform over S N ( t ) , e − t t n √ · g 1 ( n ; d ) = e − t det � � � Pr [ L ( G π ) ≤ d ] = I | r − s | (2 t ) 1 ≤ r , s ≤ d . n ! n ! n ≥ 0

Goal Derive a similar relation but for r -regular bipartite multigraphs

Step 1: Starting point Consider a r -regular n -node per color class bipartite multigraph G such that L ( G ) ≤ d u 1 u 2 u 3 v 1 v 2 v 3

Step 2: Obtain an associated permutation u 1 u 2 u 3 v 1 v 2 v 3 ... consider the following permutation π of [ rn ] u 2 u 1 u 2 u 1 u 2 u 1 3 1 1 2 2 3 v 2 v 1 v 1 v 2 v 1 v 2 1 1 2 2 3 3 Note that: π belongs to a restricted class of permutations, and LIS ( π ) ≤ d .

Step 3: Obtain an associated pair of Young Tableaux u 1 u 2 u 1 u 2 u 1 u 2 1 1 2 2 3 3 v 1 v 2 v 1 v 2 v 1 v 2 1 1 2 2 3 3 ... apply the RSK algorithm to π P 1 3 Q 1 5 2 5 2 6 4 3 6 4 Note that ( P , Q ) belongs to a restricted class of equal λ -shape Young tableaux, where λ is a partition of rn .

Step 4: Obtain an associated lattice walk 1 3 1 5 P Q 2 5 2 6 4 3 6 4 ... consider the following walk ω in Z d � 0 Note that ω belongs to a restricted class of 2 rn step closed walks.

Summarizing We want to determine exactly the number of walks in Z d that: (i) Start at � 0. (ii) Steps in positive direction come before steps in negative direction. (iii) The i -th block of r steps (i.e. steps ir +1 , . . . , ( i +1) r ) are in non-increasing order of dimension. (iv) End at � 0. (v) Stay in the region x 1 ≥ x 2 ≥ . . . ≥ x d .

Step 5: Define a parity reversing involution + Walks satisfying (i)-(v) − Walks satisfying (i)-(iii) ending at Toeplitz points

Conclusion Let W ( d , r , 2 rn ; T ( π )) denote the 2 rn -step walks in Z d satisfying ( i )-( iii ) and ending at Toeplitz point T ( π ) g r ( n ; d ) denote the number of r -regular n node per color class bipartite multi-graphs G such that L ( G ) ≤ d Then, � g r ( n ; d ) = sign ( π ) | W ( d , r , 2 rn ; T ( π )) | . π ∈ S d

Consequences Applying the Inclusion-Exclusion principle: Walks with 1 st block step bad Walks with block steps unconstrained Walks with 2 nd block step bad Walks with (2 r ) th block step bad ... and obtain g r ( n ; d ) for some small values of r and n .

Symmetrization technique (e.g. r = d = 2 case) L ′ L w w ′

Induced mapping From w ∈ W ( d = 2 , r , 2 rn ; T ( π )) to walks With steps: r even r odd r / 2 up r + 1 steps r / 2 down Ending at: (2 rn , 0) if π = id , and (2 rn , 2) if π = (12),

Kernel Method (case r odd) Consider the Laurent polynomial P r ( u ) = 1 u r − 1 + . . . + u r − 1 + u r . 1 u r + Note that 1 ( zP r ( u )) n = [ z 2 n ] | W (2 , r , 2 rn ; (0 , 0)) | = [ z 2 n ] � 1 − zP r ( u ) , n ∈ N 1 ( zP r ( u )) n = [ z 2 n u 2 ] � | W (2 , r , 2 rn ; ( − 1 , 1)) | = [ z 2 n u 2 ] 1 − zP r ( u ) . n ∈ N Banderier & Flajolet - TCS’02 ): Thus (see r � � 1 [ z 2 n − 1 ] G r , 2 ( z ) = [ z 2 n − 1 ] � g r ( n ; 2) = u ′ j ( z ) u j ( z ) − u j ( z ) , j =1 where u 1 , . . . , u r are the “small branches” of the characteristic equation u r − zu r P r ( u ) = 0.

Consequences ( d = 2 case) √ 1 1 − 4 z 2 ) of the characteristic r = 1: One small branch u 1 ( z ) = 2 z (1 − equation with P 1 ( u ) = u − 1 + u . 1 − √ 1 − 4 z = 1 + z 2 + 2 z 4 + 5 z 6 + 14 z 8 + 42 z 10 + 132 z 12 + 429 z 14 + 1430 z 16 + 4862 z 18 . . . G 1 , 2 ( z ) = 2 z 2 √ 1 1 − 2 z − 3 z 2 ) of the r = 2: One small branch u 1 ( z ) = 2 z (1 − z − characteristic equation with P 2 ( u ) = u − 1 + 1 + u . � 1 − 2 z − 3 z 2 1 + z − = 1 + z 2 + z 3 + 3 z 4 + 6 z 5 + 15 z 6 + 36 z 7 + 91 z 8 + 232 z 9 + 603 z 10 . . . G 2 , 2 ( z ) = 2 z (1 + z ) r = 3: G 3 , 2 ( z ) = 1 + z 2 + 4 z 4 + 34 z 6 + 364 z 8 + 4269 z 10 + 52844 z 12 + 679172 z 14 + 8976188 z 16 . . . r = 4: G 4 , 2 ( z ) = 1 + z 2 + z 3 + 5 z 4 + 16 z 5 + 65 z 6 + 260 z 7 + 1085 z 8 + 4600 z 9 + 19845 z 10 . . . Related to EIS A000108 , A005043 , and A007043 .

To conclude ... Is there a generating function approach for the d > 2 case? Steps are now Z d vectors satisfying x 1 + . . . + x d = ± r . (0 , 0 , 2) d = 3, r = 2 (0 , 1 , 1) (1 , 0 , 1) (0 , 2 , 0) (1 , 1 , 0) (2 , 0 , 0) If only positive steps are taken, how many n step walks starting at � 0 end in nr � I ? What about the number of signed sums of walks ending in Toeplitz points?

THE END!