[PPT] - Acyclic Edge Coloring Using Entropy Compression Louis Esperet PowerPoint Presentation

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Acyclic Edge Coloring Using Entropy Compression

Louis Esperet (G-SCOP, Grenoble, France) Aline Parreau (LIFL, Lille, France)

Bordeaux Graph Workshop, November 2012

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Acyclic Edge Colorings of graphs

An acyclic edge coloring of a graph is a coloring of the edges such that:

two edges sharing a vertex have different color,
there are no bicolored cycles.

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Acyclic Edge Colorings of graphs

An acyclic edge coloring of a graph is a coloring of the edges such that:

two edges sharing a vertex have different color,
there are no bicolored cycles.
a′(G): minimum number of colors in an acyclic edge coloring of G.
If G has maximum degree ∆:

a′(G) ≥ ∆.

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Result

If G has maximum degree ∆, a′(G) ≤ ∆ + 2. Conjecture Alon, Sudakov and Zaks, 2001

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Result

If G has maximum degree ∆, a′(G) ≤ ∆ + 2. Conjecture Alon, Sudakov and Zaks, 2001 Using the Lov´ asz Local Lemma and variations:

a′(G) ≤ 64∆ (Alon, McDiarmid and Reed, 1991)
a′(G) ≤ 16∆ (Molloy and Reed, 1998)
a′(G) ≤ 9.62∆ (Ndreca, Procacci and Scoppola, 2012)

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Result

If G has maximum degree ∆, a′(G) ≤ ∆ + 2. Conjecture Alon, Sudakov and Zaks, 2001 Using the Lov´ asz Local Lemma and variations:

a′(G) ≤ 64∆ (Alon, McDiarmid and Reed, 1991)
a′(G) ≤ 16∆ (Molloy and Reed, 1998)
a′(G) ≤ 9.62∆ (Ndreca, Procacci and Scoppola, 2012)

If G has maximum degree ∆, a′(G) ≤ 4∆. Theorem Esperet and P., 2012 Method of ”entropy compression” based on the proof by Moser and Tardos of LLL and extended by Grytczuk, Kozik and Micek.

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

e

C

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

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Algorithm

Order the edge set. While there is an uncolored edge:

Select the smallest uncolored edge e
Give a random color in {1, ..., 4∆} to e (not appearing in N[e])
If e lies in a bicolored cycle C, uncolor e and all the other edges of

C, except two edges.

G

We prove that this algorithm ends with non zero probability. ⇒ Any graph has an acyclic edge coloring with 4∆ colors.

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:-

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:-

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ...

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ... 17:-

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

e

C

Record

1:- 2:- ... 17:- 18:Cycle C is uncolored

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ... 17:- 18:Cycle C is uncolored

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ... 17:- 18:Cycle C is uncolored 19:-

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ... 17:- 18:Cycle C is uncolored 19:- ...

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ... 17:- 18:Cycle C is uncolored 19:- ... 276:-

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ... 17:- 18:Cycle C is uncolored 19:- ... 276:- 277:Cycle C′ is uncolored

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ... 17:- 18:Cycle C is uncolored 19:- ... 276:- 277:Cycle C′ is uncolored

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Record

1:- 2:- ... 17:- 18:Cycle C is uncolored 19:- ... 276:- 277:Cycle C′ is uncolored 278:- ...

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Final partial coloring Φt Record

1:- 2:- ... 17:- 18:Cycle C is uncolored 19:- ... 276:- 277:Cycle C′ is uncolored 278:- ... t:-

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Recording

We assume the algorithm is still running after t steps.

→ bad scenario

We record in a compact way what happens during the algorithm.

G

Final partial coloring Φt Record

1:- 2:- ... 17:- 18:Cycle C is uncolored 19:- ... 276:- 277:Cycle C′ is uncolored 278:- ... t:-

1 record + 1 final partial coloring = 1 bad scenario

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Rewrite the history

1. Top-down reading → set of colored edges at each step.

1:- 2:- ... 17:- 18:C is uncolored 19:- ... 276:- 277:C′ is uncolored 278:- ... t:-

1

Sets of colored edges

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Rewrite the history

1. Top-down reading → set of colored edges at each step.
2. Down-top reading→ partial coloring at each step and scenario.

1:- 2:- ... 17:- 18:C is uncolored 19:- ... 276:- 277:C′ is uncolored 278:- ... t:-

1

Sets of colored edges

2

partial colorings and scenario

ei

C

Step i

2

→

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Rewrite the history

1. Top-down reading → set of colored edges at each step.
2. Down-top reading→ partial coloring at each step and scenario.

1:- 2:- ... 17:- 18:C is uncolored 19:- ... 276:- 277:C′ is uncolored 278:- ... t:-

1

Sets of colored edges

2

partial colorings and scenario

ei

C

Step i

2

→ ei

C

ei gets Step i − 1

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Rewrite the history

1. Top-down reading → set of colored edges at each step.
2. Down-top reading→ partial coloring at each step and scenario.

1:- 2:- ... 17:- 18:C is uncolored 19:- ... 276:- 277:C′ is uncolored 278:- ... t:-

1

Sets of colored edges

2

partial colorings and scenario

⇒ 1 record + 1 final partial coloring = 1 bad scenario

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Summary

1 record +1 partial coloring = 1 bad scenario

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Summary

1 record +1 partial coloring = 1 bad scenario

≤ (4∆ + 1)m

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Summary

1 record +1 partial coloring = 1 bad scenario

≤ (4∆ + 1)m ? ?

How many possible records ?

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Compact records of cycles

We know one edge e of C.

C

e

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Compact records of cycles

We know one edge e of C.

C

e

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Compact records of cycles

We know one edge e of C.

C

e 2

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Compact records of cycles

We know one edge e of C.

C

e 2 3

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Compact records of cycles

We know one edge e of C.
No choice for the last edge

C

e 2 3 1 3 5 4

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Compact records of cycles

We know one edge e of C.
No choice for the last edge

C

e 2 3 1 3 5 4

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Compact records of cycles

We know one edge e of C.
No choice for the last edge

C

e 2 3 1 3 5 4

i:C is uncolored i:231354

⇔

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Compact records of cycles

We know one edge e of C.
No choice for the last edge

C

e 2 3 1 3 5 4

≤ ∆ ≤ ∆ i:C is uncolored i:231354

⇔

Cycle coded by a word on {1, ..., ∆}2k−2 where 2k is the length of C.

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Number of records

( − , − , ..., − , 231354 , − , ..., − , 4213 , − , ..., − ) Record

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Number of records

( − , − , ..., − , 231354 , − , ..., − , 4213 , − , ..., − ) Record 0111111 01111

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Number of records

( − , − , ..., − , 231354 , − , ..., − , 4213 , − , ..., − ) Record 0111111 01111 0 ↔ :

an edge is colored

1 ↔ :

an edge is uncolored

t

Number of colored edges

Partial Dyck word of length ≤ 2t and blocks of ones of even size.

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Number of records

( − , − , ..., − , 231354 , − , ..., − , 4213 , − , ..., − ) Record 0111111 01111 0 ↔ :

an edge is colored

1 ↔ :

an edge is uncolored

t

Number of colored edges

Partial Dyck word of length ≤ 2t and blocks of ones of even size. → Number of such words : 2t/t3/2

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Number of records

( − , − , ..., − , 231354 , − , ..., − , 4213 , − , ..., − ) Record 0111111 01111 0 ↔ :

an edge is colored

1 ↔ :

an edge is uncolored

t

Number of colored edges

Partial Dyck word of length ≤ 2t and blocks of ones of even size. → Number of such words : 2t/t3/2 → Number of records : (2∆)t/t3/2

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End of the proof

1 record +1 partial coloring = 1 bad scenario

(4∆ + 1)m

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End of the proof

1 record +1 partial coloring = 1 bad scenario

(4∆ + 1)m (2∆)t/t3/2

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End of the proof

1 record +1 partial coloring = 1 bad scenario

(4∆ + 1)m (2∆)t/t3/2

(4∆+1)m(2∆)t t3/2

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End of the proof

1 record +1 partial coloring = 1 bad scenario

(4∆ + 1)m (2∆)t/t3/2

(4∆+1)m(2∆)t t3/2

Number of scenarios: (2∆)t
Number of bad scenarios:

(4∆+1)m(2∆)t t3/2

= o((2∆)t)

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End of the proof

1 record +1 partial coloring = 1 bad scenario

(4∆ + 1)m (2∆)t/t3/2

(4∆+1)m(2∆)t t3/2

Number of scenarios: (2∆)t
Number of bad scenarios:

(4∆+1)m(2∆)t t3/2

= o((2∆)t)

⇒ For t large enough, there are good scenarios. ⇔ The algorithm stops with nonzero probability !

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Conclusion

If G has maximum degree ∆ and girth g:

a′(G) ≤ 4∆;
if g ≥ 7, a′(G) ≤ 3.74∆;
if g ≥ 53, a′(G) ≤ 3.14∆;
if g ≥ 220, a′(G) ≤ 3.05∆.

Theorem Esperet and P., 2012

Procedure in expected polynomial time using (4 + ǫ)∆ colors.
Holds also for list coloring.
Can be applied for any coloring with ”forbidden” configurations.

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Conclusion

If G has maximum degree ∆ and girth g:

a′(G) ≤ 4∆;
if g ≥ 7, a′(G) ≤ 3.74∆;
if g ≥ 53, a′(G) ≤ 3.14∆;
if g ≥ 220, a′(G) ≤ 3.05∆.

Theorem Esperet and P., 2012

Procedure in expected polynomial time using (4 + ǫ)∆ colors.
Holds also for list coloring.
Can be applied for any coloring with ”forbidden” configurations.

Thanks !

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