3-Coloring of Random Graphs

Transcription

1 3-Coloring of Random Graphs Josep Díaz 1 UPC 1 Research partially supported by the Future and Emerging Technologies programme of the EU under contract (DELIS)

2 Coloring problems. Given a graph G = (V, E) a legal coloring is an assignment of colors to V in such a way that no adjacent vertices get the same color. The k-colorability problem of G Given a graph G = (V, E) the k-col problem is to decide if there is a legal coloring of G with at most k colors. The Chromatic number of G Given a graph G = (V, E) the chromatic number problem is to decide which is the minimal number of colors χ(g) required to color G legally.

3 The Erdős Rényi model It is known that G(n, p) and G(n, m) are interchangeable for not extremely dense or extremely sparse graphs. i.e. If n = V then for p = d n and m = d 2 n then G n,p and G n,m are interchangeable. Average degree in G n,d/n = p ( n 2) dn/2 = d. Edge-density in G n, dn = d 2 2

4 The beginning of colorability in random graphs. Erdős Rényi: On the evolution of random graphs (1960) 1. If d = 1 ɛ the a.a.s. all connected components of G(n, d/n) have at most one cycle and O(log n) vertices. For G G(n, 1 ɛ n ), χ(g) 3 aas 2. If d = 1 + ɛ then a.a.s. there exist a unique connected components of G(n, d/n) with Ω(n) cycles and in particular one odd cycle For G G(n, 1+ɛ n ), χ(g) 3 aas Question: If G G(n, 1 n ), χ(g) = 3 aas? Answer: 29 years later Luczak, Wierman: If G G(n, 1 n ), χ(g) = 3 aas.

5 Some of the work done on colorability in random graphs. Grimmett, McDiarmid (75); Matula (79); McDiarmid (79); Bollobás,Thomason (83); Bollobás (83); Fernandez de la Vega (83); Gazmuri (84); Shamir, Upfal (84); McDiarmid (84); Shamir, Spencer (87), Bollobás (88); Luczak, Wierman (89); Kučera, Matula (90); McDiarmid (90); Cheeseman, Kanefsky, Taylor (91); Culberson, Gent (91); Kučera (91); Luczak(91), Luczak (91); Molloy (92); Frieze, Luczak (92); Molloy, Reed (95), Molloy (96); Pittel, Spencer, Wormald (96); Alon, Krivelevich (97); Achlioptas, Molloy (97); Dunne, Zito (98); Achlioptas, Molloy (99); Achlioptas, Friedgut (99); Achlioptas (99); Kaporis, Kirousis, Stamatiou (00); Krivelevich, Sudakov, Vu, Wormald (01); Mulet, Pagnani, Weight, Zechina (02); van Mourik, Saad (02); Achlioptas, Moore (02), Cooper, Frieze, Reed, Riordan (02); Dubois, Mandler (03); Achlioptas, Moore (03), Achlioptas, Naor (04), Achlioptas, Moore (04); Krzaka la, Pagnani, Weight (04); Monasson (04); Shi, Wormald (04); Krzaka la (05); Díaz, Kaporis, Kirousis, Pérez, Stamatiou (05), Kremkes, Wormald (05).

6 Phase transition: Experimental results Cheeseman, Kanefefky, Taylor (91) and Culberson, Gent (91): If G G(n, d/n) (G has average degree d): Empirical observation: for each fixed k, there is a threshold average degree d k such that If d < d k, then G is a.a.s. k-colorable, while if d > d k then G is a.a.s. non-k-colorable. For k = 3 experimental evidence of d k = 4.7

7 Phase transition, the statistical mechanics point of view Braunstein, Mulet, Pagnani, Weight, Zecchina (03): Analytic (non-rigorous) verification of the previous experimental results by the cavity and replica symmetry breaking methods of Statistical Physics suggest that d Pr[3-Col] d

8 The Achlioptas Friedgut theorem Theorem k 3 there is a sequence d k (n) such that ɛ : A random graph with average degree d k (n) ɛ is a.a.s. 3-colorable. A random graph with average degree d k (n) + ɛ is a.a.s. non-3-colorable. In other words, 3-col has a sharp treshold. Open question: Does d k (n) converge? If yes, to what value?

9 Corollary to the Achlioptas Friedgut theorem Recall a sequence of events E holds: a.a.s. if lim n Pr [E] = 1, w.u.p.p. if lim inf n Pr [E] > 0 Given d, for G G(n, d/n), if for all n, Pr [G is k-col.] > ɛ, then Pr [G is k-col.] = 1 aas. For k-col. wupp aas 1 ε Pr[k-Col] 0 dk d

10 3-Col threshold: The basic upper bound technique. Let G be a random graph and C(G) the random class of its legal 3-colorings. Then Pr [G is 3-col.] = Pr [ C(G) 1] E [ C(G) ]. Compute E [ C(G) ] (easy) and find the values of d for which E [ C(G) ] = 0. We get a trivial upper bound of 5.42 Why experiments suggest d ? Because of the Lottery paradox: A class of graphs with small probability but with many legal 3-colorings contributes too much to E [ C(G) ].

11 Upper bound: Improvements to the first moment Kaporis, Kirousis, Stamatiou (00): thin C(G) so that the unrealistic expectation of the cardinality of C(G), due to the Lottery paradox, gets smaller. They consider rigid 3-colorings, (colorings where one change of color to a higher one in the 012 ordering destroys the legality) NO RIGID RIGID d

12 Best upper bound to 3-Col threshold. Dubois, Mandler (03) (unpublished): refine the space of graphs to the Typical graphs (graphs with a Poisson degree sequence). It is know that a.a.s. any G G(n, p) has degree sequence within an ɛ of a Poisson. Therefore the few graphs in G(n, p) which have not Poisson degree sequence have many 3-colorings and force the 1st moment to give bad estimates. They use the decimation technique: repetitively depleted G of vertices of degree 2 or less, as these vertices do not interfere with the colorability. d

13 3-Col threshold: Lower bounds. Observation: G is k-colorable if G does not contain a subgraph with min. degree k Pittel, Spencer, Wormald (96): Study the size of a gigant k-core in G(n, d/n) and proved d Achlioptas, Moore (03): analyzing the Brélaz s heuristic Using the DEM) got d which is the best known bound for 3-Col on G(n, d/n) d

14 The Brélaz heuristic. D. Brélaz (79). Input: A graph G together with a list of possible k colors for each of its vertices. At every step, choose a vertex with the least number of colors in its lists, select a color from the list and assign it to the vertex. At ties, choose a vertex with the largest number of uncolored neighbors. Delete the vertex and also delete the selected color from neighboring vertices. If the graph becomes empty, return yes ; if a vertex with an empty list appears, return no.

15 Technique for lower bounds for 3-Col threshold. Let d k denote the lower bound for d k that we try to compute. Consider Brélatz or a variation. Notice the previous version of Brélatz does not have backtracking, so failure is reported after the choice of a color, this failure is considered permanent. Prove that for the max. value of d < d k, the coloring algorithm a.a.s. succeeds. The more sophisticated the heuristic is, the more difficult its probabilistic analysis is.

16 Chromatic number of G(n, p) with p fixed. Grimmett, McDiarmid (75) gave the first lower bound of χ(g) by determining d k s.t. G has not and independent set of size n/k, so χ(g) > k. Many papers latter, Bollobás (88) and Kučera, Matula (88): Theorem For fixed p, for any G G(n, p). χ(g) n 2 ln n.

17 Chromatic number of G(n, p/n). Let G G(n, d/n): Luczak (91): Theorem d R +, k d N such that aas χ(g) is either k d or k d + 1. Luczak gives no information on these two values, neither do the above asymptotics of the threshold d k. Main task: Make the asymptotics of d k finer so that the two possible values of the chromatic number are found.

18 Chromatic number of G(n, p/n). Achlioptas, Naor (91): Determine k d as the smallest k such that d < 2k lg k. Then aas χ(g) {k d, k d + 1}. Moreover, if d [(2k 1) lg k, 2k lg k] then aas χ(g) = k d + 1 They determine the exact value for about half of the d s.

19 The second moment method Let X be a non-negative counting random variable that depends on n. As n grows large E [X ] may also grow large, but yet Pr [X > 0] may approach zero. However, if E [ X 2] does not approach infinity too fast compared to E [X ], then it may turn that Pr [X > 0] stays away from zero: Pr[X > 0] (E [X ])2 E [X 2 ] > 0. If E [ X 2] = Θ(E [X ]) then Pr[X > 0] > 0 wupp.

20 The solution by Achlioptas-Naor Achlioptas and Naor (2004). Let k d be the smallest integer k such that d < 2k ln k. Almost all G(n, d/n) have chromatic number either k d or k d + 1. Method of Proof: Second moment where X counts the number of balanced k-colorings of G(n, d/n). E [ X 2] turns out to be a sum of exponential terms. Locating the term with the largest base, which essentially gives the value of the sum, is tricky. They use a martingale-based concentration result.

21 The case of general k-col By the first moment method: d k < 2k ln k. By Luczak: ɛ > 0, d k > 2(k ɛ) ln k (asymptotically in k) Also, by Luczak: the chromatic number of graphs in G(n, d/n) ranges a.a.s. within a window of only two possible integer values. So d k 2k ln k (the asymptotic is w.r.t k).

22 The geometry of k-colorings. Krz aka la, Pagnani, Weight (04): For G G(n, d/n), consider the space of all k-color assignments (legal or not) to the vertices of G. A cluster is a set of colorings s.t. it is possible to go from one coloring to other by changing the colors of at most 2 vertices. For d < (1 ɛ)k ln k, all legal k-colorings form a unique cluster in this space, sooner or latter a greedy change of colors in the vertices is going to hit the cluster.

23 As d increases, the clusters break down into exponentially many legal clusters and exponentially many almost legal clusters. Therefore, any change in the colors of a few vertices gives rise to more illegally colored edges. As a consequence, local search algorithms are not expected to produce results for average degrees beyond the breaking down point of the unique cluster. There seem to be scant hopes to sufficiently improve the lower bound of d k by algorithmic techniques. This is denoted the hard-cololorability region.

24 Random regular graphs. Different model from the Erdős Rényi. G n,d class of d-regular graphs on n labeled vertices.

25 Pairings and multigraphs. Let C n,d be the set of random d-pairings on n vertices. Theorem If a property holds aas for C n,d, then it also holds aas for G n,d.

26 3-colorability of G n,d Frieze, Luczak-92: If G G n,d then χ(g) It says nothing about fixed values of d d 2 ln d aas (in d) Molloy-92: If k(1 1 k )d/2 < 1 then G G n,d is not k-colorable aas d 6, aas G G n,d is not 3-colorable. Uses the first moment method in a clever way to prove that the expected number of r-colorings of a d-regular configuration is r(1 1/r) d/2.

27 3-colorability of G n,d Achlioptas, Moore-04: If G G n,4 then wupp χ(g) = 3. The proof is analysis of Brélaz s heuristics + the fact that aas G G n,4 is not bipartite. Recall Achlioptas and Moore-04: The chromatic number of random regular graphs of degree d a.a.s. ranges in {k d, k d + 1, k d + 2}, where k d is the smallest integer k such that d < 2k ln k.

28 3-colorability of G n,d Shi, Wormald (04): If G G n,d, aas: χ(g) = 3 if d = 4 χ(g) = 3 or 4 if d = 5 χ(g) = 4 if d = 6 χ(g) = 4 or 5 if 7 d 9 χ(g) = 5 or 6 if d = 10 The proof is algorithmic: Properly color all short cycles of G up to some length l Greedily color the remaining vertices following some priority given by an ordering of the possible labels. At any moment, each vertex is labeled (i, j), where i is the number of non-colored neighbors and j is the number of available colors.

29 3-colorability of random 5-regular. Non-rigorous results from physics by Krz aka la, Pagnani, Weigt (04): almost all 5-regular graphs have chromatic number 3. Moreover, the solution space of 3-colorings of 5-regular graphs has many clusters. Therefore, algorithmic techniques are expected to be hard. Until recently, the only rigorous result for 5-regular graphs is that almost all of them have chromatic number 3 or 4. The second moment fails when X counts 3-colorings, even if they are balanced.

30 Why the 2nd moment fails for 4-reg. graphs? By linearity of expectation and by summing over pairs of 3-colorings we have: E [ X 2] = i E i P i where E i is the number of pairs of 3-colorings with a given pattern of color assignments (characterized by a parameter i) and P i is the probability that a fixed pair of color assignments with pattern E i is legal.

31 The term E i P i that is equal (ignoring sub-exponential factors) to (E [X ]) 2 is the barycentric term that corresponds to a completely symmetric pattern. But unfortunately unlike the case of G(n, p) graphs, the barycentric term is not the prevalent one in the sum. Because there is a slight bias towards pairs of colorings that give the same color to a vertex.

32 Recent results on 3-colorability of G n,5. Consider rainbow colorings, where each vertex has neighbors with both the other two legal colors. D., Kaporis, Kirousis, Pérez and Sotiropoulos-05: If G G n,5 then χ(g) = 5 wupp Kemkes and Wormald-05 : Padded up this probability to 1 (asymptotically).

33 The proof Used balanced rainbow colorings. First moment: E [X ] = { (G, C) G C 5,n, C = G } C 5,n Second moment: E [ X 2] { (G, C 1, C 2 ) G C 5,n, C 1, C 2 = G } = C 5,n

34 First moment: Exact expression For any v its spectra s = (s 1, s +1 ) (the distribution of the semi-edges of v according to the distribution of the semi-edges matched to. s=(2,3) Notice thar for each v and each C, there is only one s. For i Z 3, let d(i, s) = {v with color i and spectra s} n.

35 First moment: Exact expression Notice that d(i, s) i,s is the degree sequence of nodes in G, colored by C, normalized by n. E [X ] = d(i,s) i,s ( n (d(i, s)n) i,s ) i,s ( ) 5 d(i,s)n ( 5n s 6!)3 E [X ] = i I T i

36 First moment: Asymptotics Asymptotic expression: E [X ] poly i (n)(f i ) n i I Exponential behaviour: If F i0 F i ( i I ) then E [X ] (F i0 ) n, (where X Y means log X log Y ) Polynomial( factors: E [X ] = Θ n r/2) poly i0 (n)(f i0 ) n (Laplace integration technique)

37 Result on C n,5 Doing similarly with E [ X 2], we get (E [X ]) 2 E [X 2 ] which gives Random 5-regular configurations can be 3-coloured wupp.

38 Result on G n,5 Let Pr and E be the probability and expectation conditional to the event G C n,5 is simple, then after a messy computation E [X ] = Θ(E [X ]) and E [ X 2 ] = Θ(E [ X 2] ) Pr [X > 0] (E [X ])2 E [X 2 ] Theorem If G G n,5 then χ(g) = 3 wupp = Θ(1). Therefore

39 From wupp to aas In the previous result X is such that var [X ] is too large, so we could not get the concentration result Pr [X > 0] 1. Kemkes and Wormald, used the small subgraph conditioning (Robinson, Wormald (83)) The idea is the following: Let Y 1, Y 2,..., Y k de r.v. such that Y i = number of cycles with length i. In C n,5 Y i has a Poisson distribution independent of n. Let Z a r.v. counting the number of rainbow, balanced 3-colorings conditioned to Y 1,..., Y k, for arbitrary fixed k. Then var [Z] = o(e [Z] 2 ).

40 Theorem If G G n,5 then χ(g) = 3 aas

41 Conclusions. An exciting field. Many people work on the topic. Sinergies between mathematics, physics and computer science. It developed in parallel to similar work for 3-SAT, k-sat and MaxSAT. Plenty of open problems to work on. Thanks for your attention.