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

m1 m2 m3 mn

G ∼ G

w1 w2 w3 wn

L(G) = 5

Instance 1: Longest Increasing Sequence (LIS) Problem

π = 1 2 3 4 5 6 7 8 3 6 1 2 8 7 4 5

• 1

3 4 6 7 2 5 8 π(1) = 3 π(3) = 1 π(4) = 2 π(7) = 4 π(8) = 5 π(2) = 6 π(6) = 7 π(5) = 8

L(Gπ) = 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(Gα,β) = 6

Known Results

LIS model

Baik-Deift-Johanson - J. AMS’99

L−2√n n1/6

asymptotically, is Tracy-Widom. LCS model

Loebl-K.-Matousek - AIM’05

∃γk = lim

n→+∞

1 nE (L) and γk √ k → 2 as k → ∞. r-regular graph model

Loebl-K. - RS&A’02

1 √rnE (L) → 2 as n → ∞ when r = o(n1/4).

Erd¨

• s-Renyi model

Sepp¨ al¨ ainen - Ann. App. Prob.’97

1 nL → 2 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

• nly its expectation.

Gessel’s Identity

If g1(n; d) denotes the number of permutations of [n] with LIS at most d, and Iν(x) =

• m∈N

1 m!(m + ν)! x 2 2m+ν , it holds that

• n∈N

g1(n; d) n!2 x2n = det

• I|r−s|(2x)
• 1≤r,s≤d .

Equivalently, for N(t) Poisson of rate t and π uniform over SN(t), Pr [L(Gπ) ≤ d] =

• n≥0

e−ttn n! ·g1(n; d) n! = e−t det

• I|r−s|(2

√ t)

• 1≤r,s≤d .

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

v1 v2 v3 u1 u3 u2

Step 2: Obtain an associated permutation

v1 v2 v3 u1 u3 u2

... consider the following permutation π of [rn]

v 1

1

v2

1

u1

1

u1

2

u2

1

u2

2

u1

3

v 2

2

v 1

3

v 2

3

u2

3

v 1

2

Note that: π belongs to a restricted class of permutations, and LIS(π) ≤ d.

Step 3: Obtain an associated pair of Young Tableaux

v1 1 v2 1 u1 1 u1 2 u2 1 u2 2 u1 3 v2 2 v1 3 v2 3 u2 3 v1 2

... apply the RSK algorithm to π

1 2 3 4 5 6 2 4 6 3 5 1 Q P

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 2 3 4 5 6 2 4 6 3 5 1 Q P

... consider the following walk ω in Zd

• Note that ω belongs to a restricted class of 2rn step closed walks.

Summarizing

We want to determine exactly the number of walks in Zd 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 x1 ≥ x2 ≥ . . . ≥ xd.

Step 5: Define a parity reversing involution

− + Walks satisfying (i)-(iii) ending at Toeplitz points Walks satisfying (i)-(v)

Conclusion

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

• π∈Sd

sign(π) |W (d, r, 2rn; T(π))| .

Consequences

Applying the Inclusion-Exclusion principle:

Walks with block steps unconstrained Walks with (2r)th block step bad Walks with 2nd block step bad Walks with 1st block step bad

... and obtain gr(n; d) for some small values of r and n.

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

w L′ L w′

Induced mapping

From w ∈ W (d = 2, r, 2rn; T(π)) to walks With steps:

r/2 up r even r odd r + 1 steps r/2 down

Ending at: (2rn, 0) if π = id, and (2rn, 2) if π = (12),

Kernel Method (case r odd)

Consider the Laurent polynomial Pr(u) = 1

ur + 1 ur−1 + . . . +ur−1+ur.

Note that

|W (2, r, 2rn; (0, 0))| = [z2n]

• n∈N

(zPr(u))n = [z2n] 1 1 − zPr(u) , |W (2, r, 2rn; (−1, 1))| = [z2nu2]

• n∈N

(zPr(u))n = [z2nu2] 1 1 − zPr(u) .

Thus (see

Banderier & Flajolet - TCS’02 ):

gr(n; 2) = [z2n−1]Gr,2(z) = [z2n−1]

r

• j=1

u′

j(z)

• 1

uj(z) − uj(z)

• ,

where u1, . . . , ur are the “small branches” of the characteristic equation ur − zurPr(u) = 0.

Consequences (d = 2 case)

r = 1: One small branch u1(z) =

1 2z (1−

√ 1−4z2) of the characteristic equation with P1(u) = u−1 + u.

G1,2(z) = 1 − √1 − 4z 2z2 = 1 + z2 + 2z4 + 5z6 + 14z8 + 42z10 + 132z12 + 429z14 + 1430z16 + 4862z18 . . .

r = 2: One small branch u1(z) =

1 2z (1−z−

√ 1 − 2z−3z2) of the characteristic equation with P2(u) = u−1 + 1 + u.

G2,2(z) = 1 + z −

• 1 − 2z − 3z2

2z(1 + z) = 1 + z2 + z3 + 3z4 + 6z5 + 15z6 + 36z7 + 91z8 + 232z9 + 603z10 . . .

r = 3:

G3,2(z) = 1 + z2 + 4z4 + 34z6 + 364z8 + 4269z10 + 52844z12 + 679172z14 + 8976188z16 . . .

r = 4:

G4,2(z) = 1 + z2 + z3 + 5z4 + 16z5 + 65z6 + 260z7 + 1085z8 + 4600z9 + 19845z10 . . .

Related to EIS A000108, A005043, and A007043.

To conclude ...

Is there a generating function approach for the d > 2 case? Steps are now Zd vectors satisfying x1 + . . . + xd = ±r.

d = 3, r = 2 (1, 1, 0) (0, 1, 1) (0, 2, 0) (2, 0, 0) (0, 0, 2) (1, 0, 1)

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!