Covering random graphs by monochromatic cycles Rajko Nenadov (joint - - PowerPoint PPT Presentation

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Covering random graphs by monochromatic cycles Rajko Nenadov (joint - - PowerPoint PPT Presentation

Covering random graphs by monochromatic cycles Rajko Nenadov (joint with D. Korndi, F. Mousset, N. kori , and B. Sudakov) Warmup Theorem (Gerencsr, Gyrfs 1967) The vertex set of any 2 -edge-coloured complete graph K n can be


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Covering random graphs by monochromatic cycles

Rajko Nenadov (joint with D. Korándi, F. Mousset, N. Škorić, and B. Sudakov)

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Warmup

Theorem (Gerencsér, Gyárfás 1967)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue path.

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Warmup

Theorem (Gerencsér, Gyárfás 1967)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue path. Take a maximal red-blue-path:

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Warmup

Theorem (Gerencsér, Gyárfás 1967)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue path. Take a maximal red-blue-path:

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Warmup

Theorem (Gerencsér, Gyárfás 1967)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue path. Take a maximal red-blue-path:

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Warmup

Theorem (Gerencsér, Gyárfás 1967)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue path. Take a maximal red-blue-path:

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Warmup

Theorem (Gerencsér, Gyárfás 1967)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue path. Take a maximal red-blue-path:

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Warmup

Theorem (Gerencsér, Gyárfás 1967)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue path. Take a maximal red-blue-path:

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Warmup

Theorem (Gerencsér, Gyárfás 1967)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue path. Take a maximal red-blue-path:

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Covering and partitioning by monochromatic cycles

For an edge-coloured graph G, let cp(G) = minimum no. of vertex-disjoint monochromatic cycles covering V (G) cc(G) = minimum no. of monochromatic cycles covering V (G) cc(G)  cp(G)

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Covering and partitioning by monochromatic cycles

For an edge-coloured graph G, let cp(G) = minimum no. of vertex-disjoint monochromatic cycles covering V (G) cc(G) = minimum no. of monochromatic cycles covering V (G) cc(G)  cp(G) For a graph G, let cpr(G) = maximum of cp(G) over all r-colourings of G ccr(G) = maximum of cc(G) over all r-colourings of G

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Conjecture (Lehel 1979)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue cycle, cp2(Kn) = 2.

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Conjecture (Lehel 1979)

The vertex set of any 2-edge-coloured complete graph Kn can be partitioned into a red and a blue cycle, cp2(Kn) = 2.

I Gyárfás (1983) ! cover by two cycles intersecting in at most

  • ne vertex;

I Łuczak, Rödl, Szemerédi (1998) ! proof for large n; I Allen (2008) ! proof for smaller n; I Bessy, Thomassé (2010) ! proof for all n.

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More colours

Conjecture (Erdős, Gyárfás, Pyber 1991)

For every r 2 cpr(Kn)  r.

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More colours

Conjecture (Erdős, Gyárfás, Pyber 1991)

For every r 2 cpr(Kn)  r.

I Erdős, Gyárfás, Pyber (1991) ! cpr(Kn) = O(r2 log r) I Gyárfás, Ruszinkó, Sárközy, Szemerédi (2006) ! O(r log r).

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More colours

Conjecture (Erdős, Gyárfás, Pyber 1991)

For every r 2 cpr(Kn)  r.

I Erdős, Gyárfás, Pyber (1991) ! cpr(Kn) = O(r2 log r) I Gyárfás, Ruszinkó, Sárközy, Szemerédi (2006) ! O(r log r). I Pokrovskiy (2012) ! the conjecture is wrong

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What about non-complete graphs?

Similar results hold in

I complete bipartite graphs I graphs with sufficiently large minimum degree I graphs with bounded independence number

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What about non-complete graphs?

Similar results hold in

I complete bipartite graphs I graphs with sufficiently large minimum degree I graphs with bounded independence number

These graphs are all very dense.

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Tree partitioning of random graphs

Theorem (Kohayakawa, Mota, Schacht, 2017+)

If p (log n/n)1/2 then whp every 2-colouring of Gn,p contains a partition into two monochromatic trees, tp2(Gn,p)  2.

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Tree partitioning of random graphs

Theorem (Kohayakawa, Mota, Schacht, 2017+)

If p (log n/n)1/2 then whp every 2-colouring of Gn,p contains a partition into two monochromatic trees, tp2(Gn,p)  2.

I Haxell, Kohayakawa (1996) ! tpr(Kn)  r

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Tree partitioning of random graphs

Theorem (Kohayakawa, Mota, Schacht, 2017+)

If p (log n/n)1/2 then whp every 2-colouring of Gn,p contains a partition into two monochromatic trees, tp2(Gn,p)  2.

I Haxell, Kohayakawa (1996) ! tpr(Kn)  r I The statement is false if p ⌧ (log n/n)1/2.

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Tree partitioning of random graphs

Theorem (Kohayakawa, Mota, Schacht, 2017+)

If p (log n/n)1/2 then whp every 2-colouring of Gn,p contains a partition into two monochromatic trees, tp2(Gn,p)  2.

I Haxell, Kohayakawa (1996) ! tpr(Kn)  r I The statement is false if p ⌧ (log n/n)1/2. I Proved by Bal and DeBiasio (2016) for p (log n/n)1/3.

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Cycle covering of random graphs

Theorem (Korándi, Mousset, N., Škorić, Sudakov)

Given r 2 and ✏ > 0, if p n−1/r+✏ then whp ccr(Gn,p)  Cr6 log r.

I Note: this is covering, not partitioning.

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This is almost tight: if p ⌧ n−1/r then ccr(Gn,p) = !(1).

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This is almost tight: if p ⌧ n−1/r then ccr(Gn,p) = !(1). Construction for r = 2: 1 2 3 k . . .

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This is almost tight: if p ⌧ n−1/r then ccr(Gn,p) = !(1). Construction for r = 2: 1 2 3 k . . . Pr[v has at least two neighbours in {1, . . . , k}]  k

2

  • p2

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This is almost tight: if p ⌧ n−1/r then ccr(Gn,p) = !(1). Construction for r = 2: 1 2 3 k . . . Pr[v has at least two neighbours in {1, . . . , k}]  k

2

  • p2

For any constant k: Pr[such v exists]  n ✓k 2 ◆ p2 ! 0

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This is almost tight: if p ⌧ n−1/r then ccr(Gn,p) = !(1). Construction for r = 2: 1 2 3 k . . . Pr[v has at least two neighbours in {1, . . . , k}]  k

2

  • p2

For any constant k: Pr[such v exists]  n ✓k 2 ◆ p2 ! 0 A similar construction works for r > 2.

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Theorem

If p n−1/r+✏ then ccr(Gn,p)  f (r).

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Theorem

If p n−1/r+✏ then ccr(Gn,p)  f (r). Proof idea. Show that:

  • 1. constantly many monochromatic cycles can cover all but

O(1/p) vertices;

  • 2. every set of O(1/p) can be covered by constantly many

monochromatic cycles.

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Covering all but O(1/p) vertices

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Covering all but O(1/p) vertices

Split the vertices randomly into constantly many small parts.

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Covering all but O(1/p) vertices

Goal: cover each part using vertices from other parts (except for O(1/p) vertices).

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Covering all but O(1/p) vertices

Each vertex has a majortity colour to the top (at least np/r neighbours in that colour).

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Covering all but O(1/p) vertices

red majority blue majority green majority Classify the vertices according to the majority colour.

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Covering all but O(1/p) vertices

red majority blue majority green majority We handle each colour independently.

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Covering all but O(1/p) vertices

red majority Each vertex has at least np/r red edges going to the right.

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Covering all but O(1/p) vertices

red majority ↵np2 If two vertices have ↵np2 red common neighbours, place an auxiliary edge between them (here ↵ > 0 is a small constant).

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In this way, we obtain an auxiliary graph on the red-majority vertices.

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In this way, we obtain an auxiliary graph on the red-majority vertices. Using Hall’s condition a cycle in the auxiliary graph can be transformed into a red cycle in the real graph, covering at least the same vertices.

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In this way, we obtain an auxiliary graph on the red-majority vertices. Using Hall’s condition a cycle in the auxiliary graph can be transformed into a red cycle in the real graph, covering at least the same vertices.

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In this way, we obtain an auxiliary graph on the red-majority vertices. Using Hall’s condition a cycle in the auxiliary graph can be transformed into a red cycle in the real graph, covering at least the same vertices.

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Goal: show that the auxiliary graph contains cycles covering all but O(1/p) vertices.

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Goal: show that the auxiliary graph contains cycles covering all but O(1/p) vertices.

Lemma (Structural lemma)

Let C be large enough and let X1, . . . , Xr+1 be disjoint subsets of C/p vertices in the auxiliary graph. Then there are i 6= j such that the auxiliary graph has an edge going from Xi to Xj.

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Goal: show that the auxiliary graph contains cycles covering all but O(1/p) vertices.

Lemma (Structural lemma)

Let C be large enough and let X1, . . . , Xr+1 be disjoint subsets of C/p vertices in the auxiliary graph. Then there are i 6= j such that the auxiliary graph has an edge going from Xi to Xj. In other words: the complement of the auxiliary graph does not contain a complete (r + 1)-partite graph with parts of size C/p.

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Covering all but O(1/p) vertices

The proof of the first step is thus completed by showing:

Lemma

Let G be a graph whose complement does not contain a complete k-partite graph with parts of size m. Then G contains k2 vertex disjoint cycles covering all but k2m vertices.

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Covering C/p vertices

Next step: show that every subset of C/p vertices can be covered by a constant number of cycles.

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Covering C/p vertices

Suppose |X|  C/p. Here’s the strategy:

I We again define an auxiliary graph on X, but this time, an

edge-coloured one.

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Covering C/p vertices

Suppose |X|  C/p. Here’s the strategy:

I We again define an auxiliary graph on X, but this time, an

edge-coloured one.

I Place a red auxiliary edge between u and v if Gn,p contains

“many” short red paths from u to v. (Same for other colours.)

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Covering C/p vertices

Suppose |X|  C/p. Here’s the strategy:

I We again define an auxiliary graph on X, but this time, an

edge-coloured one.

I Place a red auxiliary edge between u and v if Gn,p contains

“many” short red paths from u to v. (Same for other colours.)

I A monochromatic cycle in the auxiliary graph will correspond

to a monochromatic cycle in Gn,p.

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Covering C/p vertices

Suppose |X|  C/p. Here’s the strategy:

I We again define an auxiliary graph on X, but this time, an

edge-coloured one.

I Place a red auxiliary edge between u and v if Gn,p contains

“many” short red paths from u to v. (Same for other colours.)

I A monochromatic cycle in the auxiliary graph will correspond

to a monochromatic cycle in Gn,p.

I Moreover, the auxiliary graph will have bounded independence

number.

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Covering C/p vertices

Suppose |X|  C/p. Here’s the strategy:

I We again define an auxiliary graph on X, but this time, an

edge-coloured one.

I Place a red auxiliary edge between u and v if Gn,p contains

“many” short red paths from u to v. (Same for other colours.)

I A monochromatic cycle in the auxiliary graph will correspond

to a monochromatic cycle in Gn,p.

I Moreover, the auxiliary graph will have bounded independence

number.

I Thus it can be partitioned into constantly many

monochromatic cycles (Sárközy 2010).

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There is no independent set of size 6

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There is no independent set of size 6

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There is no independent set of size 6

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There is no independent set of size 6

Ω(np2)

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There is no independent set of size 6

Ω(np2)

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There is no independent set of size 6

Ω(np2) Ω(n2p4)

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There is no independent set of size 6

Ω(np2) Ω(n2p4)

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There is no independent set of size 6

Ω(np2) Ω(n2p4)

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There is no independent set of size 6

Ω(np2) Ω(n2p4) Ω(n3p6)

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There is no independent set of size 6

Ω( log n

p )

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There is no independent set of size 6

Ω( log n

p )

Ω( log n

p )

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There is no independent set of size 6

Ω( log n

p )

Ω( log n

p )

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Open problems

Cycles:

I Partitioning instead of covering

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Open problems

Cycles:

I Partitioning instead of covering I Better p (get rid of the ✏)

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Open problems

Cycles:

I Partitioning instead of covering I Better p (get rid of the ✏)

Trees:

I if p (log n/n)1/r then whp tpr(Gn,p)  r

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