Applications of Tuttes tree decomposition in the enumeration of - - PowerPoint PPT Presentation

Applications

• f Tutte’s tree decomposition

in the enumeration

• f bipartite graph families

Juanjo Ru´ e, Kerstin Weller

Nice Random Graphs Workshop

Objects: graphs

Labelled Graph = vertices+edges

1 3 2 4 5

Simple Graph = NO multiples edges, NO loops A family G under study must be: STABLE (closed under permutation of the labels) CLOSED under Tutte’s decomposition ( G ∈ G if and only if its 3-connected components are in G) PROTOTYPE: planar graphs ← SP-graphs

Contents

◮ Connectivity and enumeration ◮ A combinatorial trick:

The Dissymmetry Theorem for trees

◮ The Ising model and the equations

for bipartite Series-Parallel graphs

Connectivity and enumeration

The Symbolic Method ` a la Flajolet

COMBINATORIAL RELATIONS between CLASSES ↕⇕↕ EQUATIONS between GENERATING FUNCTIONS Construction EGF Union A ∪ B A(x) + B(x) Labelled Product A × B A(x) · B(x) Sequence Seq (A)

1 1−A(x)

Set Set (A) exp (A(x)) Pointing A• \ \ A◦ x ∂

∂xA(x) \ \ ∂ ∂xA(x)

Substitution A ◦ B A (B(x)) GF will be exponential in x (vertices), and ordinary in y (edges) ∑

n,m≥0

an,m xn n! ym

The Strategy

General graph Connected graph 2-connected graph 3-connected graph Networks Map World Steinitz's Theorem

For SP - graphs, there are NOT 3-connected components

General graphs from connected graphs

Let C be a family of connected graphs. G : graphs such that their connected components are in C. G = Set(C) = ⇒ G(x, y) = exp(C(x, y))

Connected graphs from 2-connected graphs

A vertex-rooted connected graph is a tree of blocks. C• = v × Set(B◦(v ← C•)) = ⇒ xC′(x, y) = x exp B′(xC′(x, y), y)

2-connected graphs from 3-connected graphs

We study networks: graphs with two privileged virtual vertices (poles) such that adding and edge between them makes it 2-connected 2(1 + y)∂B ∂y (x, y) = x2 (1 + D(x, y)) 3 types: SERIES, PARALLEL and T-composition D(x, y) = y + S(x, y) + P(x, y) + H(x, y) S(x, y) = (D(x, y) − S(x, y))xD(x, y) P(x, y) = yexp≥1 (S(x, y) + H(x, y)) + exp≥2 (S(x, y) + H(x, y)) H(x, y) = 2 x2 ∂ ∂z T(x, D(x, y))

Simplifying a little...

Eliminating variables: D = (1 + y) exp ( xD2 1 + xD + 1 2x2 ∂ ∂z T(x, D) ) − 1 ∂B ∂y = x2 2 1 + D 1 + y

A set of equations

INPUT: T(x, z)        1 2x2 ∂ ∂z T(x, D) − log (1 + D 1 + y ) + xD2 1 + xD = 0 ∂B ∂y (x, y) = x2 2 (1 + D 1 + y ) → INTEGRATION    xC′(x, y) = x exp B′(xC′(x, y), y) G(x, y) = exp(C(x, y)) To have in mind: this integration step could become VERY difficult if we consider enriched families of graphs.

The integration step

We have this integral: ∂B ∂y (x, y) = x2 2 (1 + D 1 + y ) → B(x, y) = x2 2 ∫ y 1 + D(x, s) 1 + s ds Surprisingly, we get an EXACT expression: T(x, D(x, y)) − 1 2xD + 1 2 log(1 + xD) + x2 2 ( D + 1 2D2 + (1 + D) log ( 1 + y 1 + D )) . Are there combinatorial reasons to get such an easy formula?

A combinatorial trick: The Dissymmetry Theorem for trees

A toy example: trees

We apply the previous grammar to count ROOTED trees ⇒ T = • × Set(T ) → T(x) = xeT(x) To forget the root, we just integrate: (xU′(x) = T(x)) ∫ x T(s) s ds = { T(s) = u T ′(s) ds = du } = ∫ T(x)

T(0)

1−u du = T(x) − 1 2T(x)2

The Dissymmetry Theorem for trees

That can be explained only by combinatorial means: T ∪ T•→• ≃ T•−• ∪ T• where T∗ is the family of trees T with an extra rooted structure. U(x) = U•→•(x) + U•−•(x) − U•→•(x) = 1 2T(x)2 + T(x) − T(x)2 For tree-like families we have an extension of this result: A ∪ A•→• ≃ A•−• ∪ A•

Tutte’s tree-like decomposition

blocks admit a tree-like decomposition ` a la Tutte: This is a 2-connected SP-graph.

Tutte’s tree-like decomposition

blocks admit a tree-like decomposition ` a la Tutte: We split it using 2-cuts

Tutte’s tree-like decomposition

blocks admit a tree-like decomposition ` a la Tutte: R R R R R R R M M M M M R The tree has two type of vertices: M, R

Encoding using GFs

Developed by Chapuy, Fusy, Kang, Shoilekova. B(x, y) = BR(x, y) + BM(x, y) − BR−M(x, y) B R

M R-M

B B

The Ising model and the equations for bipartite Series-Parallel graphs

Bipartite graphs

Is it possible to apply the same arguments for bipartite graphs? No problem from connected to 2-connected From 2-connected to 3-connected we have problems... We need to study something more complicated: Ising model on graphs.

An statistical model

Consider graphs with two types of edges: • − • and ◦ − •, and write the GF ∑

n,r,s≥0

dn,r,s xn n! yr

• −•ys
• −•.

Observe that permuting colors on vertices do NOT change the type of the edges: vertex coloring is an auxiliar tool Let us try to get B(x, y•−•, y◦−•) in terms of certain networks: 2(1 + y•−•) ∂B ∂y•−• (x, y•−•, y◦−•) + 2(1 + y◦−•) ∂B ∂y◦−• (x, y•−•, y◦−•) = x2(1 + D•−•(x, y•−•, y◦−•) + D◦−•(x, y•−•, y◦−•)) ...It is better to integrate this COMBINATORIALLY...

The networks in the model

We distinguish depending on the type of the root: D◦−• = y◦−• + S◦−• + P◦−• D•−• = y•−• + S•−• + P•−• S•−• = (D•−• − S•−•)xS•−• + (D◦−• − S◦−•)xS◦−• S◦−• = (D•−• − S•−•)xS◦−• + (D◦−• − S◦−•)xS•−• P◦−• = y◦−•Set≥1(S◦−•) + Set≥2(S◦−•) P•−• = y•−•Set≥1(S•−•) + Set≥2(S•−•)

From networks to 2-connected graphs

Now we can write BR, BM and BRM, but we have more cases: B R

M R-M

B B Then we get a long expression in terms of the GF for networks.

Going back to the bipartite model

Working a little, we can get two equations relating the bipartite families D◦−•(x, y, 0) = D2 and D•−•(x, y, 0) = D1 D1 = −1 + exp ( x ( D13x − D1 D22x + D12 + D22) D12x2 − D22x2 + 2xD1 + 1 ) D2 = y + y ( exp ( xD2 ( xD12 − xD22 + 2D1 ) D12x2 − D22x2 + 2xD1 + 1 ) − 1 ) + exp ( xD2 ( xD12 − xD22 + 2D1 ) D12x2 − D22x2 + 2xD1 + 1 ) − 1. And then, we can substitute in the previous expressions for B

Asymptotic enumeration

Applying the machinery of analytic combinatorics we have the following theorem [R., Weller, 2014] The number of connected and general bipartite (labelled) series-parallel graphs with n vertices is asymptotically equal to cn ∼ c · n−5/2 · γn · n! gn ∼ g · n−5/2 · γn · n! where γ ≈ 4,22044, c ≈ 0,021446 and g ≈ 0,026499 are computable constants. The constant growth for SP graphs is γ′ ≈ 9,07359

...And random graphs

We have then a precise probability of a random SP-graph being bipartite [R., Weller, 2014] The probability that a uniformly connected random SP-graph with n vertices is bipartite is 0,3167 · (2,1499)−n(1 + o(1)) We can also obtain limiting distributions for the number of edges, cutvertices, blocks (Normal distribution, linear expectation).

Moltes gr` acies!

Applications

• f Tutte’s tree decomposition

in the enumeration

• f bipartite graph families

Juanjo Ru´ e, Kerstin Weller

Nice Random Graphs Workshop