By Yun Long, Asaf Nachmias, Weiyang Ning, Yuval Peres

The Swendsen-Wang dynamics is a Markov chain time-honored through physicists to pattern from the Boltzmann-Gibbs distribution of the Ising version. Cooper, Dyer, Frieze and Rue proved that at the whole graph Kn the blending time of the chain is at such a lot O( O n) for all non-critical temperatures. during this paper the authors exhibit that the blending time is Q (1) in excessive temperatures, Q (log n) in low temperatures and Q (n 1/4) at criticality. in addition they supply an higher certain of O(log n) for Swendsen-Wang dynamics for the q-state ferromagnetic Potts version on any tree of n vertices

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**Additional info for A Power Law of Order 1/4 for Critical Mean Field Swendsen-wang Dynamics**

**Example text**

Assume that p = 1+ m . Write τ = min{t : Yt = 0}, then for any small α > 0 −1 2 −2 . E eαY −2 | τ ≥ −2 ≤ Ce2α +α Proof. We have that P(τ ≥ −2 ) ≥ c . 16. This has been done in this section several times so we omit the details. Thus, it suﬃces to bound from above EeαY −2 1{τ ≥ −2 } . d. Bin(m, p) − 1 increments, it suﬃces to bound the same expectation for Wt . Write γ = min{t : Wt = 0 or Wt ≥ −1 }. 32) EeαW −2 1{τ ≥ −2 ,γ< −2 } ≤ P(Wγ ≥ −1 )E[eαWγ eα(W −2 −1 −2 ,γ< −2 } −2 . we have that . } (which implies Wγ ≥ −2 −Wγ ) | Wγ ≥ −1 ,γ < −1 −2 ].

5 m this random graph is 3 32 YUN LONG, ASAF NACHMIAS, WEIYANG NING, and YUVAL PERES in the subcritical regime, and the probability of having such a component is smaller than 1/2 (in fact, it is exponentially small). 12. This ﬁnishes the proof of (ii). The proof of (iii) goes in similar lines of (ii). 5 m} . 5 m) and bound the rest by E( j≥1 |Cj |2 )2 (instead of j ≥ 2). 12 shows that E( j≥1 |Cj |2 )2 is controlled by (E j≥1 |Cj |2 )2 . 5 E( m} ≤ Ce−c 3 m m4 4 = O(m2 −2 ). 12 to estimate the remaining subcritical graph.

J≥1 7. SUBCRITICAL CASE (ii) 2 c 47 − 1 + ≤ x0 ≤ 1(in case c > 1). In this case, the random graph is supercritical with θ ≥ 1 + 2c . 6, we have c 0 G( 1+x 2 n, n ) E |Cj+ |2 2 ≤ E|C1+ | = φ(x0 )n + O(n) j≥1 2 + E|C1+ | − φ(x0 )n E|C1+ | + φ(x0 )n + O(n). 5, we have E|C1+ | − φ(x0 )n = O( n). 1, we have φ(x0 )n ≤ δx0 n. So we have E ≤ δ 2 x20 n2 + O(n3/2 ) ≤ (δ 2 + o(1))x20 n2 . |Cj+ |2 j≥1 (iii) that E 2 c −1− ≤ x0 ≤ + 2 j≥1 |Cj | = 2 c −1+ (or 1 − 1+x0 2 nE|Cv |. So E ≤ x0 ≤ 1 in case c = 1). Recall + 2 j≥1 |Cj | reaches its maximum at x0 = − 1 + for 1 < c < 2 or x0 = 1 for c = 1.