Concatenating bipartite graphs Paul Seymour (Princeton) joint with - - PowerPoint PPT Presentation

Concatenating bipartite graphs

Paul Seymour (Princeton) joint with Maria Chudnovsky, Patrick Hompe, Alex Scott and Sophie Spirkl

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Caccetta-Häggkvist conjecture, 1978

For every integer k ≥ 1, if G is an n-vertex digraph and every vertex has out-degree at least n/k, then G has girth at most k.

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Caccetta-Häggkvist conjecture, 1978

For every integer k ≥ 1, if G is an n-vertex digraph and every vertex has out-degree at least n/k, then G has girth at most k. True for k ≤ 2; open for k ≥ 3.

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Caccetta-Häggkvist conjecture, 1978

For every integer k ≥ 1, if G is an n-vertex digraph and every vertex has out-degree at least n/k, then G has girth at most k. True for k ≤ 2; open for k ≥ 3.

Conjecture:

For every integer k ≥ 1, if G is a bipartite digraph with n vertices in each part, and every vertex has out-degree more than n/(k + 1), then G has girth at most 2k.

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Caccetta-Häggkvist conjecture, 1978

For every integer k ≥ 1, if G is an n-vertex digraph and every vertex has out-degree at least n/k, then G has girth at most k. True for k ≤ 2; open for k ≥ 3.

Conjecture:

For every integer k ≥ 1, if G is a bipartite digraph with n vertices in each part, and every vertex has out-degree more than n/(k + 1), then G has girth at most 2k.

Theorem (S., Spirkl, 2018)

This is true for k = 1, 2, 3, 4, 6, and all k ≥ 224,539.

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Caccetta-Häggkvist conjecture, 1978

For every integer k ≥ 1, if G is an n-vertex digraph and every vertex has out-degree at least n/k, then G has girth at most k. True for k ≤ 2; open for k ≥ 3.

Conjecture:

For every integer k ≥ 1, if G is a bipartite digraph with n vertices in each part, and every vertex has out-degree more than n/(k + 1), then G has girth at most 2k.

Theorem (S., Spirkl, 2018)

This is true for k = 1, 2, 3, 4, 6, and all k ≥ 224,539. True for regular digraphs.

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Caccetta-Häggkvist conjecture, 1978

For every integer k ≥ 1, if G is an n-vertex digraph and every vertex has out-degree at least n/k, then G has girth at most k. True for k ≤ 2; open for k ≥ 3.

Conjecture:

For every integer k ≥ 1, if G is a bipartite digraph with n vertices in each part, and every vertex has out-degree more than n/(k + 1), then G has girth at most 2k.

Theorem (S., Spirkl, 2018)

This is true for k = 1, 2, 3, 4, 6, and all k ≥ 224,539. True for regular digraphs. Implied by the Caccetta-Häggkvist conjecture.

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Let G be a digraph with bipartition (A, B), where every vertex in A has

• ut-degree at least x|B|, and every vertex in B has out-degree at least

y|A|. Which values of x, y guarantee girth at most four?

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Let G be a digraph with bipartition (A, B), where every vertex in A has

• ut-degree at least x|B|, and every vertex in B has out-degree at least

y|A|. Which values of x, y guarantee girth at most four? Not (1/3, 1/3).

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Let G be a digraph with bipartition (A, B), where every vertex in A has

• ut-degree at least x|B|, and every vertex in B has out-degree at least

y|A|. Which values of x, y guarantee girth at most four? Not (1/3, 1/3). Not (

1 2t+1, t 2t+1) if t ≥ 1 is an integer.

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Let G be a digraph with bipartition (A, B), where every vertex in A has

• ut-degree at least x|B|, and every vertex in B has out-degree at least

y|A|. Which values of x, y guarantee girth at most four? Not (1/3, 1/3). Not (

1 2t+1, t 2t+1) if t ≥ 1 is an integer.

x y

(1/3, 1/3) (1/5, 2/5) (1/7, 3/7) (2/5, 1/5) (3/7, 1/7)

BAD GOOD

? ?

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Theorem

If x, y > 0 and 2x + y > 1 then (x, y) is good.

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Theorem

If x, y > 0 and 2x + y > 1 then (x, y) is good.

Conjecture

If x, y > 0 and kx + y > 1 then G has girth at most 2k.

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Theorem

If x, y > 0 and 2x + y > 1 then (x, y) is good.

Conjecture

If x, y > 0 and kx + y > 1 then G has girth at most 2k. (This implies the Caccetta-Häggkvist conjecture for the same value of k.)

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Different question: Now G is a graph with three disjoint sets of vertices A, B, C. Every vertex in A has at least x|B| neighbours in B; every vertex in B has at least y|C| neighbours in C. What is the maximum z such that we can guarantee some vertex in C can reach z|A| vertices in A by two-edge paths? Call this φ(x, y).

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Different question: Now G is a graph with three disjoint sets of vertices A, B, C. Every vertex in A has at least x|B| neighbours in B; every vertex in B has at least y|C| neighbours in C. What is the maximum z such that we can guarantee some vertex in C can reach z|A| vertices in A by two-edge paths? Call this φ(x, y).

Theorem

max(x, y) ≤ φ(x, y) .

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Different question: Now G is a graph with three disjoint sets of vertices A, B, C. Every vertex in A has at least x|B| neighbours in B; every vertex in B has at least y|C| neighbours in C. What is the maximum z such that we can guarantee some vertex in C can reach z|A| vertices in A by two-edge paths? Call this φ(x, y).

Theorem

max(x, y) ≤ φ(x, y) ≤ ⌈kx⌉ + ⌈ky⌉ − 1 k for every integer k ≥ 1.

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Different question: Now G is a graph with three disjoint sets of vertices A, B, C. Every vertex in A has at least x|B| neighbours in B; every vertex in B has at least y|C| neighbours in C. What is the maximum z such that we can guarantee some vertex in C can reach z|A| vertices in A by two-edge paths? Call this φ(x, y).

Theorem

max(x, y) ≤ φ(x, y) ≤ ⌈kx⌉ + ⌈ky⌉ − 1 k for every integer k ≥ 1.

Theorem

φ(x, y) ≤ z iff φ(y, 1 − z) ≤ 1 − x.

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Different question: Now G is a graph with three disjoint sets of vertices A, B, C. Every vertex in A has at least x|B| neighbours in B; every vertex in B has at least y|C| neighbours in C. What is the maximum z such that we can guarantee some vertex in C can reach z|A| vertices in A by two-edge paths? Call this φ(x, y).

Theorem

max(x, y) ≤ φ(x, y) ≤ ⌈kx⌉ + ⌈ky⌉ − 1 k for every integer k ≥ 1.

Theorem

φ(x, y) ≤ z iff φ(y, 1 − z) ≤ 1 − x.

Theorem

φ(x, y) = φ(y, x).

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What if also, every vertex in B has at least x|A| neighbours in A, and every vertex in C has at least y|B| neighbours in B? Let ψ(x, y) be the best z.

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What if also, every vertex in B has at least x|A| neighbours in A, and every vertex in C has at least y|B| neighbours in B? Let ψ(x, y) be the best z.

Theorem

For every integer k > 0, if

1 k+1 < x ≤ 1 k then ψ(x, x) = 1 k .

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What if also, every vertex in B has at least x|A| neighbours in A, and every vertex in C has at least y|B| neighbours in B? Let ψ(x, y) be the best z.

Theorem

For every integer k > 0, if

1 k+1 < x ≤ 1 k then ψ(x, x) = 1 k .

x ψ(x, x)

(1, 1) (1/2, 1/2) (1/3, 1/3) (1/4, 1/4)

x φ(x, x)

(1, 1) (1/2, 1/2) (1/3, 1/3) (1/4, 1/4)

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Does x > 1/3 imply φ(x, x) ≥ 1/2?

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Does x > 1/3 imply φ(x, x) ≥ 1/2?

Conjecture:

For all integers k ≥ 1, if x, y ∈ (0, 1] with x + ky > 1 and kx + y ≥ 1, then φ(x, y) ≥ 1/k.

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Does x > 1/3 imply φ(x, x) ≥ 1/2?

Conjecture:

For all integers k ≥ 1, if x, y ∈ (0, 1] with x + ky > 1 and kx + y ≥ 1, then φ(x, y) ≥ 1/k.

Theorem

If x > 1/3 then φ(x, x) > 3/7.

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Theorem

If x > .3672 then φ(x, x) ≥ 1/2. More generally, if y(1 + √ 2x)2/2 + x > 1 then φ(x, y) ≥ 1/2.

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Theorem

If x > .3672 then φ(x, x) ≥ 1/2. More generally, if y(1 + √ 2x)2/2 + x > 1 then φ(x, y) ≥ 1/2.

Theorem

If x ≤ 1/3 and y < 1/2 and y < (1 − x)2/(2 − 4x + 6x2), then φ(x, y) < 1/2.

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Theorem

If x > .3672 then φ(x, x) ≥ 1/2. More generally, if y(1 + √ 2x)2/2 + x > 1 then φ(x, y) ≥ 1/2.

Theorem

If x ≤ 1/3 and y < 1/2 and y < (1 − x)2/(2 − 4x + 6x2), then φ(x, y) < 1/2.

Theorem

For all integers k ≥ 1, if x + ky > 1 and kx + y ≥ 1, then ψ(x, y) ≥ 1/k.

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Theorem

If x > .3672 then φ(x, x) ≥ 1/2. More generally, if y(1 + √ 2x)2/2 + x > 1 then φ(x, y) ≥ 1/2.

Theorem

If x ≤ 1/3 and y < 1/2 and y < (1 − x)2/(2 − 4x + 6x2), then φ(x, y) < 1/2.

Theorem

For all integers k ≥ 1, if x + ky > 1 and kx + y ≥ 1, then ψ(x, y) ≥ 1/k.

Theorem

If x + ky < 1 and kx + y < 1, then ψ(x, y) < 1

k .

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When ψ, φ ≥ 1/2

x y (1/3, 1/3) (3/7, 1/7) ψ < 1/2 ψ ≥ 1/2

? ?

(1/8, 11/24) x y (8/17, 1/6) φ ≥ 1/2 φ < 1/2

?

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When ψ, φ ≥ 2/3

x y (1/2, 1/2) (3/5, 1/5)

? ? ?

ψ ≥ 2/3 ψ < 2/3 (1/2, 1/3) (1/4, 1/2) (1/5, 3/5) x y (1/2, 1/2) (1/3, 1/2) y = 2/3 φ < 2/3 φ ≥ 2/3

? ?

(5/9, 4/13)

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When ψ, φ ≥ 1/3

x y (1/4, 1/4) (2/19, 6/19) ψ < 1/3 ψ ≥ 1/3 ? ? (2/19, 6/19) x y y = 1/3 φ ≥ 1/3 φ < 1/3 (2/13, 6/19) ?

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When ψ, φ ≥ 3/4

x y (1/2, 1/2) ψ ≥ 3/4 ψ ≤ 3/4 (1/3, 1/2) (1/2, 1/3) (2/3, 1/3) (1/3, 2/3) (2/3, 1/6) x y (1/2, 1/2) φ ≥ 3/4 φ ≤ 3/4 (2/3, 1/3) (1/3, 2/3) (1/4, 2/3) (2/3, 1/4) y = 3/4

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