ADFOCS 2006 Saarbrücken. Phase Transition. Joel Spencer

Transcription

1 ADFOCS 2006 Saarbrücken The Erdős-Rényi Phase Transition Joel Spencer 1

2 TP! trivial being! I have received your letter, you should have written already a week ago. The spirit of Cantor was with me for some length of time during the last few days, the results of our encounters are the following... letter, Paul Erdős to Paul Turán November 11,

3 Paul Erdős and Alfred Rényi On the Evolution of Random Graphs Magyar Tud. Akad. Mat. KutatóInt. Közl volume 8, 17-61, 1960 Γ n,n(n) : n vertices, random N(n) edges [...] the largest component of Γ n,n(n) is of order log n for N(n) N(n) n 1 2 n c< 1 2, of order n2/3 for and of order n for N(n) n c> 1 2. This double jump when c passes the value 1 2 is one of the most striking facts concerning random graphs. 3

4 The Double Jump G(n, p), p = n c (or 2 c n edges) (Average Degree c, natural model) c<1 Biggest Component O(ln n) C 1 C 2... All Components simple (= tree/unicyclic) c =1 Biggest Component Θ(n 2/3 ) C 1 n 2/3 nontrivial distribution C 2 / C 1 nontrivial distribution Complexity of C 1 nontrivial distribution c>1 Giant Component C 1 yn, y = y(c) > 0 All other C i = O(ln n) andsimple 4

5 Galton-Watson Birth Process Root node Eve Each node has Po(c) children (Poisson: Pr[Po(c) =k] =e c c k /k!) T = T c is total size c<1 T finite c =1 T finite E[T ] infinite (heavy tail) c>1 Pr[T = ] =y = y(c) > 0 5

6 Galton-Watson Exact Pr[T c = u] = e uc (uc) u 1 u! Pr[T 1 = u] = e u u u 1 =Θ(u 3/2 ) u! For c>1, Pr[T = ] =y = y(c) > 0 where 1 y = e cy For c<1, α := ce c < 1 Pr[T c >u]=o(α u ) Exponential Tail Duality: d<1 <cwith de d = c c Conditioning on Po(c) process being finite gives the Po(d) process 6

7 Math Physics Bond Percolation Z d. Bond open with probability p There exists a critical probability p c Subcritical, p<p c. All C finite, E[ C( 0) ] finite Pr[ C( 0) u] exponential tail Supercritical, p>p c. Unique Infinite Component E[ C( 0) ] infinite Pr[ C( 0) u finite] exponential tail Critical, p = p c. All C finite, E[ C( 0) ] infinite, heavy tail 7

8 Random 3-SAT n Boolean x 1,...,x n L = {x 1, x 1,...,x n, x n } literals Random Clauses C i = y i1 y i2 y i3, y ij L f(m) :=Pr[C 1 C m satisfiable] Conjecture: There exists critical c 0 Subcritical, c<c 0, f(cn) 1 Supercritical, c>c 0, f(cn) 0 Friedgut: Criticality, but possibly nonuniform 8

9 Evolution of n-cube Ajtai, Komlos, Szemeredi Bollobas, Luczak, Kohayakawa Borgs, Chayes, Slade, JS, van der Hofstad p = c/n c<1 subcritical c>1giantω(2 n ) component Much more! 9

10 The Critical Window p = 1 n + λn 4/3 λ(n) Subcritical Biggest Component o(n 2/3 ) C 1 C 2... All Components simple λ constant. The Critical Window Biggest Component Θ(n 2/3 ) C 1 n 2/3 nontrivial distribution C 2 / C 1 nontrivial distribution Complexity of C 1 nontrivial distribution λ(n) + Supercritical Dominant Component C 1 n 2/3 High Complexity All other C i = o(n 2/3 )andsimple 10

11 What is the Critical Window? Difficult in General When Dominant Component is Emerging Subcritical: Biggest about same size Supercritical: Biggest second Susceptibility χ(g) =E[ C(0) ] = 1 n Ci 2 Largest Component starts to dominate Subcritical: n 1 C 1 2 χ(g) Critical: n 1 C 1 2 = O(χ(G)) Supercritical: n 1 C 1 2 χ(g) 11

12 Computer Experiment (Try It!) n = vertices. Start: Empty Add random edges Parametrize e/ ( ) n 2 =(1+λn 1/3 )/n Merge-Find for Component Size/Complexity 4 λ +4, C i = c i n 2/3 See biggest merge into dominant 12

13 A Strange Physics Components c i n 2/3, c j n 2/3 λ λ + dλ, p p + n 4/3 dλ Merge with probability c i c j dλ Increment Complexity 1 2 c2 i dλ 13

14 An Open Question What is the critical window for random k-sat That is, can you find a parametrization so that m = f 0 (n)+λf 1 (n) Subcritical λ,pr[sat] 1 Supercritical λ +, Pr[SAT] 0 Critical Window λ fixed, Pr[SAT] g(λ) where lim g(λ) = 1 and lim λ (Maybe m = c 0 n + λn 1 β exponent β.) g(λ) =0 λ + for some critical 14

15 Galton-Watson Near Criticality Pr[T 1 u] cu 1/2 Pr[T 1+ɛ = ] 2ɛ Conditioning on finite, T 1+ɛ becomes T 1 ɛ+o(ɛ) Pr[T 1 ɛ u] Pr[ >T 1+ɛ u] If u = o(ɛ 2 ) (can t see ɛ): Pr[ >T 1+ɛ u] Pr[T 1 u] cu 1/2 If u =Θ(ɛ 2 ) (somewhat see ɛ): Pr[ >T 1+ɛ u] = Θ(Pr[T 1 u]) = Θ(u 1/2 ) If u ɛ 2 (strong ɛ effect): Pr[ >T 1+ɛ u] Pr[T 1 u]e uɛ2 /2 Pr[T 1±ɛ = u] Pr[T 1 = u] = [e ɛ ] u (1 ± ɛ) u 1 [(1 ± ɛ)e ɛ ] u = e (1+o(1))uɛ2 /2 15

16 Galton-Watson as Walk Z i Po(c), i =1, 2,... Y 0 = 1 (Eve) Y i = Y i 1 + Z i 1(Z i children and dies) Fictitious Continuation T =mintwith Y t =0 (If no such t, T = ) c<1 negative drift, T finite c>1 positive drift, maybe T infinite c = 1 zero drift, delicate 16

17 C(v) ing(n, p) as BFS Walk Y 0 =1(Root v) Y i = Y i 1 + Z i 1 (pop queue/add Z i new) where Z i BIN[n (i 1) Y i 1,p] The Link: When p c n and i 1+Y i 1 = o(n) Z i is roughly Po(c) C(v), T c similar while small Ecological Limitation: Success in BFS in G(n, p) is selflimiting. Eating your seed corn 17

18 Rough (but Accurate) Link p = c n, c>1 C(v) like T c if finite With probability y, T c infinite Corresponding C(v) become large All merge to form giant yn component p = 1+ɛ n, o(1) = ɛ n 1/3 With probability 2ɛ, T c infinite Corresponding C(v) become large All merge to form dominant 2ɛn component Finite T c have small C(v) <ɛ 2+ ɛ n 1/3 small/dominant dichotomy 18

19 The (easy!) subcritical case G G(n, p), p = n c, c<1 C(v) dominated by Galton Watson T c Pr[T c >u] <α u = o(n 1 )foru = K ln n Therefore: NO C(v) >Kln n 19

20 Why Θ(n 2/3 )atp = 1 n Ignore Ecological Limitation (so rough!) Pr[ C(v) u] Pr[T 1 u] =Θ(u 1/2 ) X u := number v with C(v) u E[X u ]=Θ(nu 1/2 ) X u 0 X u u Pr[X u 0] = O(nu 3/2 ) = O(1) when u = Θ(n 2/3 ) 20

21 BFS on G(n, p) Root 0, Nonroots 1,...,n 1 T j = i: Vertex j joins BFS tree at i-th opportunity (fictitious continuation!) T j geometric If X = k then precisely k 1ofT j k A 1 := Pr[BIN[n 1, 1 (1 p) k ]=k 1 Take p c n with c>1 A 1 very small unless k = O(ln n) or k yn with 1 e cy = y Thus: All components small or giant 21

22 The Giant Exists and is Unique! t = O(ln n) same as Galton-Watson Pr[ C(v) = O(ln n)] Pr[T c < ] =1 y Karp Approach: Keep generating components After O(1) tries (still n reservoir) get giant Now n = n(1 y), p = d/n, d<1 <c Now subcritical, nomoregiants! Duality: G(n, c/n) minus giant component is like G(n,d/n )(c, d conjugate) 22

23 BFS on G(n, p) conditioned Condition: Precisely k 1ofT j k WLOG T j k for 1 j k 1 T j T j, truncated geometric Pr[T j = u] = p(1 p)u 1 1 (1 p) k Z t := number of T j = t (join queue at time t) Y 0 =1, Y t = Y t 1 1+Z t (queue size) TREE: Y t > 0for1 t<k. Pr[ C(0) = k] =A 1 Pr[TREE] 23

24 Example: Connected on 0, 1, 2, 3, 4, N N Y Y N N N - - N Y N - - N - Y - - Y T 3 = T 4 =1, T 1 =3,T 2 = T 5 =4 A 1 : All T j 6 Z =(2, 0, 1, 2, 0, 0) Walk Y =(1, 2, 1, 1, 2, 1, 0) TREE: BFS doesn t terminate early Tree Edges 03, 04, 41, 12, 15 24

25 Pr[TREE] with p c k Left Z i Poisson c 1 e c Galton-Watson Pr[ESC] 1 e c Right Z i = Z k i; Y i = Y k i Y 0 =0, Y i = Y i 1 +1 Z i Z i Poisson ce c 1 e c Pr[ESC ] 1 ce c 1 e c Chernoff: Y i > 0inmiddle Pr[TREE] Pr[ESC] Pr[ESC ] 1 (c+1)e c 25

26 It is six in the morning. The house is asleep. Nice music is playing. I prove and conjecture. PaulErdős, in letter to Vera Sós 26