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EECS 495: Randomized Algorithms Lecture 7 Using Chernoﬀ Reading: Text: Chapter 4; Balanced Allo- Oblivious Routing cations by Azar, Broder, Karlin, and Upfal Given: • hypercube network Balls and Bins • permutation destinations Problem: n balls, place obliviously into n Output: bins • set of oblivious routes with min # steps Algorithm: Random. Claim: W/prob. (1 − n−c ), fullest bin has Algorithm: (deterministic): Bit ﬁxing (left ln n to right) (1 + o(1)) ln ln n balls. Proof: Let Ejk be event that bin j has more Question: Randomized alg.? than k balls Recall load-balancing: m jobs, n = m machines; to distribute load oblivioiusly, n i n−i n 1 1 we randomly routed jobs to machines. Pr[Ejk ] = 1− i=k i n n Idea: Load-balance paths! n i ne i 1 ≤ i=k i n First try: random destination, n e i bit-ﬁxing = i=k i ∞ • T (el ) = # paths using el e k e i ≤ k k • By symmetry, all T (el ) equal i=0 e k 1 • Expected path length n/2 = k 1 − e/k e k • LOE, total expected path length N n/2 ≤ 2 k • N n edges in hypercube 1 ≤ O n2 So E[T (el )] = 1/2, so delay at most n/2. for k = O(log n/ log log n). Result follows by Claim: Delay ≤ 6n with high prob. union bound. Proof: Chernoﬀ: Delay X ≤ l T (el ), so 1 • Pr[X > (1 + δ)µ] ≤ exp(−(1 + δ)µ) Intuition: • Pr[X > 6n] ≤ exp(−6n) • At most n/4 bins have 4 balls → prob. WRONG! get bin with 5 balls at most (1/4)(1/4) = 1/16 Proof: Chernoﬀ: Fix packet i, let Hij indi- cate if routes for i and j share an edge. Inde- • Expect n/16 bins have 5 balls → prob. pendent and j Hij ≤ i T (el ). get bin with 6 balls at most 1/162 k−3 • delay of ﬁxed packet i ≥ 6n with prob. • Expect n/22 bins have k balls ≤ exp(−6n) ≤ 2−6n • Whp, no bin has more than log log n • prob. any N = 2n packets gets delay balls more than 2−6n at most 2−5n (union bound) Problem: Assume system behaves as ex- • time to route any packet at most length pects to analyze next layer, must cope with plus delay, at most 7n conditioning. Claim: Bound binomial: Pr[B(n, p) > → w/prob. ≥ 1 − 2−5n , every packet reaches 2np] < 2−np/3 random dest. in 7n or fewer steps. Proof: B(n, p) = n Xi where i=1 But wanted to reach d(i)! Idea: Route to random intermediate desti- 1 : w/prob. p Xi = nation. 0 : otherwise • doubles path length • Chernoﬀ: µ • destroys bad perms. eδ Pr[X > (1 + δ)µ] < (1 + δ)(1+δ) Run backwards, same time bound, so packets fail to reach ﬁnal dest. in at most 14n steps with prob. at most 2−5n+1 ≤ 1/N . • µ = np, δ = 1, implies claim Claim: With prob. ≥ 1 − 1/N , every packet Note: Let Yi depend on X1 , . . . , Xi−1 . If reaches dest. in at most 14n steps. Pr[Yi |X1 , . . . , Xi−1 ] < p for all i, then X sto- Note: Didn’t allow phase 2 to delay phase 1; castichally dominates Y , so can use above must have packets wait at intermediate dest. bound. for 7n steps. Analysis: • vi (t) = # bins of height ≥ i at time t Power of Two Choices • ht = height of t’th ball Algorithm: For each ball, pick two bins ran- i domly, place in less-loaded bin. • bounds β4 = 1/4, βi = 2βi−1 = 2−2 2 2 Claim: Let Qi be event that vi ≤ βi n. Then • but then bins no longer grow because Qi happens whp. – prob. ball is tall for any i ≥ i∗ at Proof: By induction. most ((log n)/n)2 • Let Yt = 1 if ht ≥ i+1 and vi (t−1) ≤ βi n – prob. two particular balls both tall at most ((log n)/n)4 (events nega- Yt indicates t’th ball placed in a bad bin tively correlated) even though there were enough good bins. – union bound, prob. two distinct tall • Then Pr[Yi = 1] ≤ βi2 balls o(1) • Then t Yt stochastically dominated by and bins don’t grown beyond i∗ + 1 if X with p = βi2 only see one tall ball i∗ • By Chernoﬀ, • βi∗ n = 2−2 ≥ log n → i∗ = log log n − log log log n Pr[ Yt ≥ 2nβi2 = βi+1 n] < 2−βi+1 n/6 t • so max load is O(log log n) 2 which is O(1/n ) so long as βi+1 n ≥ Relationship to random graphs: bins are c log n nodes, two choices are edges. Bound comes from low expected degree and When Qi holds, then t Yt is # tall balls: “small” giant component. • # tall bins ≤ # tall balls Note: Extensions: • so • Pick more bins: O(logd log n) Pr[¬Qi+1 |Qi ] ≤ Pr[ Yt > βi+1 n|Qi ] • Pick more bins and consistent tie- t breaking: O(log log n/d) ≤ Pr[ Yt > βi+1 n]/ Pr[Qi ] t ≤ 1/(n2 Pr[Qi ]) Deal with conditioning: Pr[¬Qi+1 ] = Pr[¬Qi+1 |Qi ] Pr[Qi ] + Pr[¬Qi+1 |¬Qi ] Pr[¬Qi ] 1 ≤ + Pr[¬Qi ] n2 1 ≤ n by induction. Deal with large i: • ok until that i∗ s.t. bound on # tall bins dips below log n 3

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