Coloration acyclique des arêtes d un graphe en utilisant la compression d entropie

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1 1/20 Coloration acyclique des arêtes d un graphe en utilisant la compression d entropie Louis Esperet (G-SCOP, Grenoble, France) Aline Parreau (LIFL, Lille, France) Séminaire de l équipe GrAMA LIRIS, Vendredi 25 janvier 2013

2 Proper Edge Colorings of graphs A proper edge coloring of a graph is a coloring of the edges such that two edges sharing a vertex have different colors. 2/20

3 2/20 Proper Edge Colorings of graphs A proper edge coloring of a graph is a coloring of the edges such that two edges sharing a vertex have different colors. χ (G): minimum number of colors in a proper edge coloring of G. If G has maximum degree : χ (G).

4 2/20 Proper Edge Colorings of graphs A proper edge coloring of a graph is a coloring of the edges such that two edges sharing a vertex have different colors. χ (G): minimum number of colors in a proper edge coloring of G. If G has maximum degree : χ (G). Theorem Vizing, 1964 If G has maximum degree, χ (G) + 1.

5 3/20 Acyclic edge coloring of graphs An acyclic edge coloring of a graph is: a proper edge coloring, without bicolored cycles.

6 3/20 Acyclic edge coloring of graphs An acyclic edge coloring of a graph is: a proper edge coloring, without bicolored cycles. a (G): minimum number of colors in an acyclic edge coloring of G. If G has maximum degree, a (G).

7 3/20 Acyclic edge coloring of graphs An acyclic edge coloring of a graph is: a proper edge coloring, without bicolored cycles. a (G): minimum number of colors in an acyclic edge coloring of G. If G has maximum degree, a (G). Conjecture Alon, Sudakov and Zaks, 2001 If G has maximum degree, a (G) + 2.

8 4/20 Lovász Local Lemma Theorem Lovász Local Lemma A 1,...,A k bad events, each occurs with small probability, each event is independent of almost all the others, with nonzero probability, no bad event occurs.

9 4/20 Lovász Local Lemma Theorem Lovász Local Lemma A 1,...,A k bad events, each occurs with small probability, each event is independent of almost all the others, with nonzero probability, no bad event occurs. Acyclic edge coloring: Take a uniform random coloring with K colors. Bad event: a cycle is bicolored or two adjacent edges have the same color. Dependancy: one edge is not in too many cycles.

10 5/20 Results Using the Lovász 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)

11 5/20 Results Using the Lovász 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) Using entropy compression : Theorem Esperet and P., 2012 If G has maximum degree, a (G) 4.

12 6/20 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

13 6/20 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

14 6/20 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

15 6/20 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

16 6/20 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

17 6/20 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

18 6/20 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

19 6/20 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

20 6/20 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

21 6/20 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

22 6/20 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

23 6/20 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

24 6/20 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.

25 7/20 Recording Execution determined by set of drawn colors : scenario Assume the algorithm is still running after t steps. bad scenario

26 7/20 Recording Execution determined by set of drawn colors : scenario Assume the algorithm is still running after t steps. bad scenario We record in a compact way what happens during the algorithm. Record G

27 7/20 Recording Execution determined by set of drawn colors : scenario Assume the algorithm is still running after t steps. bad scenario We record in a compact way what happens during the algorithm. G 1:- Record

28 7/20 Recording Execution determined by set of drawn colors : scenario Assume the algorithm is still running after t steps. bad scenario We record in a compact way what happens during the algorithm. G 1:- 2:- Record

29 7/20 Recording Execution determined by set of drawn colors : scenario Assume the algorithm is still running after t steps. bad scenario We record in a compact way what happens during the algorithm. G 1:- 2:-... Record

30 7/20 Recording Execution determined by set of drawn colors : scenario Assume the algorithm is still running after t steps. bad scenario We record in a compact way what happens during the algorithm. G 1:- 2: :- Record

31 7/20 Recording Execution determined by set of drawn colors : scenario 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: :- 18:Cycle C is uncolored

32 7/20 Recording Execution determined by set of drawn colors : scenario 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: :- 18:Cycle C is uncolored

33 7/20 Recording Execution determined by set of drawn colors : scenario 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: :- 18:Cycle C is uncolored 19:-

34 7/20 Recording Execution determined by set of drawn colors : scenario 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: :- 18:Cycle C is uncolored 19:-...

35 7/20 Recording Execution determined by set of drawn colors : scenario 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: :- 18:Cycle C is uncolored 19: :-

36 7/20 Recording Execution determined by set of drawn colors : scenario 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: :- 18:Cycle C is uncolored 19: :- 277:Cycle C is uncolored

37 7/20 Recording Execution determined by set of drawn colors : scenario 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: :- 18:Cycle C is uncolored 19: :- 277:Cycle C is uncolored

38 7/20 Recording Execution determined by set of drawn colors : scenario 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: :- 18:Cycle C is uncolored 19: :- 277:Cycle C is uncolored 278:-...

39 7/20 Recording Execution determined by set of drawn colors : scenario 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 Φ t Record 1:- 2: :- 18:Cycle C is uncolored 19: :- 277:Cycle C is uncolored 278:-... t:-

40 7/20 Recording Execution determined by set of drawn colors : scenario 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 Φ t Record 1:- 2: :- 18:Cycle C is uncolored 19: :- 277:Cycle C is uncolored 278:-... t:- 1 record + 1 final partial coloring = 1 bad scenario

41 Rewrite the history I X i : set of uncolored edges after step i reading of the record to get X i : X0 = E 8/20

42 8/20 Rewrite the history I X i : set of uncolored edges after step i reading of the record to get X i : X0 = E Step i: 2 cases i:- i:c is uncolored X i+1 = X i {smallest edge of X i } X i+1 = X i + {C except two edges}

43 8/20 Rewrite the history I X i : set of uncolored edges after step i reading of the record to get X i : X0 = E Step i: 2 cases i:- i:c is uncolored X i+1 = X i {smallest edge of X i } X i+1 = X i + {C except two edges} With the record, we can find the edge e i which is colored at step i.

44 9/20 Rewrite the history II: partial colorings Φ i : partial coloring after step i Inverse reading of the record to get Φ i : Φt is known

45 9/20 Rewrite the history II: partial colorings Φ i : partial coloring after step i Inverse reading of the record to get Φ i : Φt is known Step i: 2 cases i:- i:c is uncolored Φ i 1 = Φ i with e i uncolored Φ i 1 = Φ i with C recolored and e i uncolored e i C Step i

46 9/20 Rewrite the history II: partial colorings Φ i : partial coloring after step i Inverse reading of the record to get Φ i : Φt is known Step i: 2 cases i:- i:c is uncolored Φ i 1 = Φ i with e i uncolored Φ i 1 = Φ i with C recolored and e i uncolored e i C e i C e i gets Step i Step i 1

47 9/20 Rewrite the history II: partial colorings Φ i : partial coloring after step i Inverse reading of the record to get Φ i : Φt is known Step i: 2 cases i:- i:c is uncolored Φ i 1 = Φ i with e i uncolored Φ i 1 = Φ i with C recolored and e i uncolored e i C e i C e i gets Step i Step i 1 With Φ t and the record, we can find the partial colorings and the scenario.

48 10/20 Rewrite the history - Summary 1. Top-down reading set of colored edges at each step. 1:- 2: :- 18:C is uncolored 19: :- 277:C is uncolored 278:-... t:- 1 Sets of colored edges

49 10/20 Rewrite the history - Summary 1. Top-down reading set of colored edges at each step. 2. Buttum-up reading partial coloring at each step and scenario. 1:- 2: :- 18:C is uncolored 19: :- 277:C is uncolored 278:-... t:- 1 Sets of colored edges 2 partial colorings and scenario

50 10/20 Rewrite the history - Summary 1. Top-down reading set of colored edges at each step. 2. Buttum-up reading partial coloring at each step and scenario. 1:- 2: :- 18:C is uncolored 19: :- 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

51 11/20 Summary 1 record +1 partial coloring = 1 bad scenario

52 11/20 Summary 1 record +1 partial coloring = 1 bad scenario (4 + 1) m

53 11/20 Summary 1 record +1 partial coloring = 1 bad scenario (4 + 1) m?? How many possible records?

54 12/20 Compact records of cycles We know one edge e of C. C e

55 12/20 Compact records of cycles We know one edge e of C. C e

56 12/20 Compact records of cycles We know one edge e of C. C 2 e

57 12/20 Compact records of cycles We know one edge e of C. C 3 2 e

58 12/20 Compact records of cycles We know one edge e of C. No choice for the last edge 4 C 3 2 e

59 12/20 Compact records of cycles We know one edge e of C. No choice for the last edge 4 C 3 2 e

60 12/20 Compact records of cycles We know one edge e of C. No choice for the last edge 4 C 3 2 e i:c is uncolored i:231354

61 12/20 Compact records of cycles We know one edge e of C. No choice for the last edge 4 C 2 3 e i:c is uncolored i: Cycle coded by a word on {1,..., } 2k 2 where 2k is the length of C.

62 13/20 Number of records Record (,,...,, ,,...,, 4213,,..., )

63 13/20 Number of records Record (,,...,, ,,...,, 4213,,..., )

64 13/20 Number of records Record (,,...,, ,,...,, 4213,,..., ) 0 : an edge is colored Number of colored edges : an edge is uncolored t

65 13/20 Number of records Record (,,...,, ,,...,, 4213,,..., ) 0 : an edge is colored Number of colored edges : an edge is uncolored Partial Dyck word of length 2t and descents of even size. t

66 13/20 Number of records Record (,,...,, ,,...,, 4213,,..., ) 0 : an edge is colored Number of colored edges : an edge is uncolored Partial Dyck word of length 2t and descents of even size > 2. t

67 13/20 Number of records Record (,,...,, ,,...,, 4213,,..., ) 0 : an edge is colored 1 : an edge is uncolored Number of colored edges Partial Dyck word of length 2t and descents of even size > 2. Number of such words : 2 t /t 3/2 t

68 13/20 Number of records Record (,,...,, ,,...,, 4213,,..., ) 0 : an edge is colored 1 : an edge is uncolored Number of colored edges Partial Dyck word of length 2t and descents of even size > 2. Number of such words : 2 t /t 3/2 Number of records : (2 ) t /t 3/2 t

69 14/20 End of the proof 1 record +1 partial coloring = 1 bad scenario (4 + 1) m

70 14/20 End of the proof 1 record +1 partial coloring = 1 bad scenario (2 ) t /t 3/2 (4 + 1) m

71 14/20 End of the proof 1 record +1 partial coloring = 1 bad scenario (2 ) t /t 3/2 (4 + 1) m (4 +1)m (2 ) t t 3/2

72 14/20 End of the proof 1 record +1 partial coloring = 1 bad scenario (2 ) t /t 3/2 (4 + 1) m (4 +1)m (2 ) t t 3/2 Number of scenarios: (2 ) t Number of bad scenarios: (4 +1) m (2 ) t t 3/2 = o((2 ) t )

73 14/20 End of the proof 1 record +1 partial coloring = 1 bad scenario (2 ) t /t 3/2 (4 + 1) m (4 +1)m (2 ) t t 3/2 Number of scenarios: (2 ) t Number of bad scenarios: (4 +1) m (2 ) t t 3/2 = o((2 ) t ) For t large enough, there are good scenarios. The algorithm stops with nonzero probability! There is a coloring in 4 colors.

74 15/20 Algorithmic aspect To have a small propability of a bad event in t steps, we should have: Equivalently: t can be exponential in m. bad scenarios all scenarios = (4 + 1)m (2 ) t /t 3/2 (2 ) t < δ t 3/2 > (4 + 1)m δ

75 15/20 Algorithmic aspect To have a small propability of a bad event in t steps, we should have: Equivalently: t can be exponential in m. bad scenarios all scenarios = (4 + 1)m (2 ) t /t 3/2 (2 ) t < δ t 3/2 > But if we have (4 + ɛ) colors : (4 + 1)m δ bad scenarios all scenarios = ((4 + ɛ) + 1)m (2 ) t /t 3/2 ((2 + ɛ) ) t < δ

76 15/20 Algorithmic aspect To have a small propability of a bad event in t steps, we should have: Equivalently: t can be exponential in m. bad scenarios all scenarios = (4 + 1)m (2 ) t /t 3/2 (2 ) t < δ t 3/2 > But if we have (4 + ɛ) colors : t is polynomial in m. (4 + 1)m δ bad scenarios all scenarios = ((4 + ɛ) + 1)m (2 ) t /t 3/2 ((2 + ɛ) ) t < δ

77 16/20 With larger girth Girth: size of the smallest cycle in G. Girth l All the uncolored cycles have size at least l All the descents in the Dyck word have size 2k for some k l/2

78 16/20 With larger girth Girth: size of the smallest cycle in G. Girth l All the uncolored cycles have size at least l All the descents in the Dyck word have size 2k for some k l/2 There are fewer Dyck words!

79 17/20 Counting Dyck Words E: set of integers D t,e : Dyck words of length 2t with descents in E Couting D t,e counting plane rooted trees on t + 1 vertices where each vertex as a number of children in E. Generating function f (x) = t D t,e x t : f (x) = x + x i E f (x) i Using analytic combinatorics (Flageolet and Sedgwick, 2009), the asymptotic behaviour of D t,e is γ t E t 3/2.

80 18/20 Some results Theorem Esperet and P., 2012 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.

81 19/20 History of entropy compression Moser in 2009 and Moser,Tardos in 2010: constructive proof of LLL. Example of SAT with small intersections between clauses. Same ideas applied to nonrepetitive sequences (Grytczuk, Kozik, Micek, 2012) nonrepetitive coloring (Dujmović, Joret, Kozik, Wood, 2012)

82 19/20 History of entropy compression Moser in 2009 and Moser,Tardos in 2010: constructive proof of LLL. Example of SAT with small intersections between clauses. Same ideas applied to nonrepetitive sequences (Grytczuk, Kozik, Micek, 2012) nonrepetitive coloring (Dujmović, Joret, Kozik, Wood, 2012) Entropy compression? Input: large random vector Output: smaller record

83 19/20 History of entropy compression Moser in 2009 and Moser,Tardos in 2010: constructive proof of LLL. Example of SAT with small intersections between clauses. Same ideas applied to nonrepetitive sequences (Grytczuk, Kozik, Micek, 2012) nonrepetitive coloring (Dujmović, Joret, Kozik, Wood, 2012) Entropy compression? Input: large random vector Output: smaller record Works well since : we can remove a lot of colors ( add entropy); while being able to recover the coloring (give compact record).

84 20/20 Generalization Coloring with forbidden configuration H i : H i graph with a specific coloring c i ; For any vertex v of H i, there are k i fixed vertices that determines c i, l i = V (H i ) k i (number of vertices that will be uncolored), E = {l i } d l : max number of configurations H i, l i = l, containing a vertex ; Bound : γ E sup d 1/l l Example: star coloring, bound in 3 2 3/2.

85 20/20 Generalization Coloring with forbidden configuration H i : H i graph with a specific coloring c i ; For any vertex v of H i, there are k i fixed vertices that determines c i, l i = V (H i ) k i (number of vertices that will be uncolored), E = {l i } d l : max number of configurations H i, l i = l, containing a vertex ; Bound : γ E sup d 1/l l Example: star coloring, bound in 3 2 3/2. Thanks!