Tianjin. June Phase Transition. Joel Spencer

Transcription

1 Tianjin June 2007 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 (Traditional) 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 The Five Phases Subcritical: p = c n and c<1 Barely subcritical: p 1 n and p = 1 n λ(n)n 4/3 with λ(n) The Critical Window p = 1 n + λn 4/3 λ arbitrary real, but constant. Barely supercritical: p n 1 and p = n 1 +λ(n)n 4/3 with λ(n) Supercritical: p = c n and c>1 5

6 Barely Subcritical p n 1 and p = n 1 λ(n)n 4/3 with λ(n) All components simple. Top k components about same size C 1 = o(n 2/3 ) Barely Supercritical p n 1 and p = n 1 + λ(n)n 4/3 with λ(n) Dominant Component C 1 n 2/3, High Complexity All other C n 2/3, Simple Duality: Remove Dominant Component and get Subcritical Picture. 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 Key topic: p = p c ± ɛ as ɛ 0. 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 Critical Window???: m 0 (n) withf(m 0 )= 1 2. Is there scaling m = m 0 + λn α to see f(m) go 1to 0. 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 Critical p 0 n 1 At p 0 (1 ɛ) all small At p 0 (1 + ɛ). For ɛ =Ω(n 100 ) and more: Giant 2ɛn. Second open Critical Window (dominant emerges): open 9

10 Poisson Birth Process Root node Eve Parameter c Each node has Po(c) children (Poisson: Pr[Po(c) =k] =e c c k /k!) Z t Po(c), iid t-th node has Z t children Queue Size Y t. Y 0 = 1 (Eve) Y t = Y t 1 + Z t 1 (Has children and dies) Fictional Continuation: Y t defined though process stops when some Y s =0. Size T = T po c is minimal t with Y t =0. T = : All Y t > 0. T = T c is total size 10

11 Binomial Birth Process Parameters m, p Z t B[m, p], iid T = T bin m,p total size. For m large, p small, mp moderate: Binomial is very close to Poisson c = mp. Binomial Birth Process very close to Poisson Birth Process 11

12 Graph Birth Process Parameters n, p Generate C(v) ing(n, p). BFS Queue: Y 0 =1, Y t = Y t 1 + Z t 1 Points Born: Z t B[N t 1,p] Dead Points (popped): t Live Points (in Queue): Y t Neutral Points(in Reservoir): N t t + Y t + N t = n N 0 = n 1, N t = N t 1 Z t, N t B[n 1, (1 p) t ] T = T gr n,p: minimal t with Y t =0 T = t implies N t = n t 12

13 Poisson Birth Trichotomy c<1 T finite c =1 T finite E[T ] infinite (heavy tail) c>1 Pr[T = ] =y = y(c) > 0 13

14 Poisson Birth 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 1 c < 1 Pr[T c >u]=o(α u ) Exponential Tail 14

15 Poisson Birth Near Criticality c =1+ɛ, T = T po c Pr[T = ] 2ɛ Pr[T = u] (2π) 1/2 u 3/2 (ce 1 c ) k ln[ce 1 c ] ɛ 2 /2 u small: u = o(ɛ 2 ) Pr[T c = u] Pr[T 1 = u] =Θ(u 3/2 ) Scaling: u = Aɛ 2 Pr[ >T 1+ɛ >Aɛ 2 ]=ɛe (1+o(1))A/2 Pr[T 1 ɛ >Aɛ 2 ]=ɛe (1+o(1))A/2 15

16 Poisson Birth Graph Birth Z 1 B[n 1,p] roughly Po(c), c = pn. Ecological Limitation: Z t B[N t 1,p]. Process succeeds, N t 1 gets smaller Fewer new vertices Death is inevitable Upper: Pr[T gr n,p u] Pr[T bin n 1,p u] Proof: Replenish reservoir Lower: Pr[T gr n,p u] Pr[T bin n u,p u] Proof: Hold reservoir to n u. 16

17 Why n 4/3 for Critical Window p =(1+ɛ)/n, ɛ>0, ɛ = o(1). Pr[T po 1+ɛ = ] 2ɛ. The 2ɛn points going to infinity merge to form dominant component. T po finite is O(ɛ 2 ), corresponds to component sizes O(ɛ 2 ). Finite/Infinite Poisson Dichotomy becomes Small/Dominant Graph Dichotomy if ɛ 2 2nɛ, orɛ n 1/3. 17

18 The Barely Subcritical Region p =(1 ɛ)/n, ɛ = λn 1/3, Pr[ C(v) u] Pr[T 1 ɛ u] u = Kɛ 2 ln n Pr = o(n 1 ) No Such component. More delicately: Parametrize u = Kɛ 2 ln λ = Kn 2/3 λ 2 ln λ K big: Pr[ C(v) u] =O(ɛλ 10 ) Expected nɛλ 10 = n 2/3 λ 9 vertices in components of size Kn 2/3 λ 2 ln λ No such component! 18

19 Barely Supercritical p =(1+ɛ)/n, ɛ = λn 1/3, λ + Trichotomy on Component Size Small: C <Kɛ 2 ln n [can be impoved!] Large: (1 δ)2ɛn < C < (1 + δ)2ɛn Awkward: All else No Middle Ground No Awkward Components Suffices: Pr[C(v) awkward]=o(n 1 ) 19

20 No Middle Ground Y t = n t N t = B[n 1, 1 (1 p) t ] (t 1) At start E[Y t ] ɛt [Negligible EcoLim] When t ɛ 2 ln n, E[Y t ] Var[Y t ] 1/2 t 1/2, Pr[Y t =0]=o(n 10 ) Later E[Y t ]=(n 1)[1 (1 p) t ] (t 1) ɛt t2 2n For t 2ɛn, E[Y t ] 0, dominant component. C(v) = t implies Y t =0. For t yɛn, y 2: Pr[ C(v) = t] Pr[Y t =0]=o(n 10 ). 20

21 Escape Probability S := Kɛ 2 ln n, α := Pr[ C(v) S] Pr[ C(v) S] Pr[T bin n 1,p S] np =1+ɛ, S ɛ 2 so 2ɛ Pr[T bin n S,p S] Pr[ C(v) S] (Here ɛ n 1/3 ln 1/3 n but with care...) As Sp = o(ɛ) EcoLim negligible! p(n S) =1+ɛ + o(ɛ) sopr 2ɛ Sandwich: Escape Prob 2ɛ 21

22 Almost Done Not Small implies Large 2ɛn Expected 2ɛn in Large components BUT Can we have two of size 2ɛn half the time? 22

23 Sprinkling Add sprinkle of n 4/3, p p + If G(n, p) had two Large they would merge That would give 4ɛn in G(n, p + ) But p + =(1+ɛ + o(ɛ))/n has nothing 4ɛn Conclusion: G(n, p) has precisely one Large component It has size 2ɛn As no middle ground: All other component sizes Kɛ 2 ln n. So Large Component is Dominant Component 23

24 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 24

25 It is six in the morning. The house is asleep. Nice music is playing. I prove and conjecture. PaulErdős, in letter to Vera Sós 25