1. 1/36 Randomized algorithms Instructor: YE, Deshi yedeshi@zju.edu.cn

2. 2/36 Probability We define probability in terms of a sample space S, which is a set whose elements are called elementary events. Each elementary event can be viewed as a possible outcome of an experiment. An event is a subset of the sample space S. Example: flipping two distinguishable coins Sample space: S = {HH, HT, TH, TT}. Event: the event of obtaining one head and one tail is {HT, TH}. Null event: ø. Two events A and B are mutually exclusive if A ∩ B = ⊘. A probability distribution Pr{} on a sample space S is a mapping from events of S to real numbers such that Pr{A} ≥ 0 for any event A. Pr{S} = 1. Pr{A ∪ B} = Pr{A} + Pr{B} for any two mutually exclusive events A and B.

3. 3/36 Axioms of probability Using Ā to denote the event S - A (the complement of A), we have Pr{Ā} = 1 - Pr{A}. For any two events A and B Discrete probability distributions: A probability distribution is discrete if it is defined over a finite or countably infinite sample space. Let S be the sample space. Then for any event A, Uniform probability distribution on S: Pr{s} = 1/ |S| Continuous uniform probability distribution: For any closed interval [c, d], where a ≤ c ≤ d ≤ b,

4. 4/36 Probability Conditional probability of an event A given that another event B occurs is defined to be Two events are independent if Bayes's theorem, Pr{B}=Pr{B ∩ A} + Pr{B ∩ Ā} =Pr{A} Pr {B | A} + Pr{Ā}Pr{B | Ā}.

5. 5/36 Discrete random variables For a random variable X and a real number x, we define the event X = x to be {s ∈ S : X(s) = x}; Thus Probability density function of random variable X: f (x) = Pr{X = x}. Pr{X = x} ≥ 0 and Σ x Pr{X = x} = 1. If X and Y are random variables, the function f (x, y) = Pr{X = x and Y = y} For a fixed value y,

6. 6/36 Expected value of a random variable Expected value (or, synonymously, expectation or mean) of a discrete random variable X is Example: Consider a game in which you flip two fair coins. You earn $3 for each head but lose $2 for each tail. The expected value of the random variable X representing your earnings is E[X] = 6 · Pr{2 H's} + 1 · Pr{1 H, 1 T} - 4 · Pr{2 T's} =6(1/4) + 1(1/2) - 4(1/4) =1. Linearity of expectation: when n random variables X 1, X 2,..., X n are mutually independent,

7. 7/36 First success Waiting for a first success. Coin is heads with probability p and tails with probability 1-p. How many independent flips X until first heads? Useful property. If X is a 0/1 random variable, E[X] = Pr[X = 1]. Pf. j-1 tails1 head

8. 8/36 Variance and standard deviation The variance of a random variable X with mean E[X] is: If n random variables X 1, X 2,..., X n are pairwise independent, then The standard deviation of a random variable X is the positive square root of the variance of X.

9. 9/36 Randomization Randomization. Allow fair coin flip in unit time. Why randomize? Can lead to simplest, fastest, or only known algorithm for a particular problem. Ex. Symmetry breaking protocols, graph algorithms, quicksort, hashing, load balancing, Monte Carlo integration, cryptography.

10. 10/36 Maximum 3-Satisfiability MAX-3SAT. Given 3-SAT formula, find a truth assignment that satisfies as many clauses as possible. Remark. NP-hard problem. Simple idea. Flip a coin, and set each variable true with probability ½, independently for each variable.

11. 11/36 Claim. Given a 3-SAT formula with k clauses, the expected number of clauses satisfied by a random assignment is 7k/8. Pf. Consider random variable Let Z = weight of clauses satisfied by assignment Z j. Maximum 3-Satisfiability: Analysis linearity of expectation

12. 12/36 E[Z j ] E[Z j ] is equal to the probability that C j is satisfied. C j is not satisfied, each of its three variables must be assigned the value that fails to make it true, since the variables are set independently, the probability of this is (1/2)3=1/8. Thus C j is satisfied with probability 1 – 1/8 = 7/8. Thus E[Z j ] = 7/8.

13. 13/36 Maximum 3-SAT: Analysis Q. Can we turn this idea into a 7/8-approximation algorithm? In general, a random variable can almost always be below its mean. Lemma. The probability that a random assignment satisfies  7k/8 clauses is at least 1/(8k). Pf. Let p j be probability that exactly j clauses are satisfied; let p be probability that  7k/8 clauses are satisfied.

14. 14/36 Analysis con. Let k΄ denote the largest natural number that is strictly smaller than 7k/8. Then 7k/8 - k΄ ≥ 1/8, thus, k΄ ≤ 7k/8 – 1/8. Because k΄ is a natural number, and the remaining of 7k mod 8 is at least 1. Rearranging terms yields p  1 / (8k).

15. 15/36 Maximum 3-SAT: Analysis Johnson's algorithm. Repeatedly generate random truth assignments until one of them satisfies  7k/8 clauses. Theorem. Johnson's algorithm is a 7/8-approximation algorithm. Pf. By previous lemma, each iteration succeeds with probability at least 1/(8k). By the waiting-time bound, the expected number of trials to find the satisfying assignment is at most 8k. ▪

16. 16/36 Maximum Satisfiability Extensions. Allow one, two, or more literals per clause. Find max weighted set of satisfied clauses. Theorem. [Asano-Williamson 2000] There exists a 0.784-approximation algorithm for MAX-SAT. Theorem. [Karloff-Zwick 1997, Zwick+computer 2002] There exists a 7/8-approximation algorithm for version of MAX-3SAT where each clause has at most 3 literals. Theorem. [Håstad 1997] Unless P = NP, no  -approximation algorithm for MAX-3SAT (and hence MAX-SAT) for any  > 7/8. very unlikely to improve over simple randomized algorithm for MAX-3SAT

17. 17/36 Randomized Divide-and-Conquer

18. 18/36 Finding the Median We are given a set of n numbers S={a 1,a 2,...,a n }. The median is the number that would be in the middle position if we were to sort them. The median of S is equal to k th largest element in S, where k = (n+1)/2 if n is odd, and k=n/2 if n is even. Remark. O(n log n) time if we simply sort the number first. Question: Can we improve it?

19. 19/36 Selection problem Selection problem. Given a set of n numbers S and a number k between 1 and n. Return the k th largest element in S. Select (S, k) choose a splitter a i  S uniformly at random foreach (a  S) { if (a < a i ) put a in S - else if (a > a i ) put a in S + } If |S - |=k-1 then a i was the desired answer Else if |S - |≥k-1 then The kth largest element lies in S - Recursively call Select(S -,k) Else suppose |S - |= l < k-1 then The kth largest element lies in S + Recursively call Select(S +,k-1- l ) Endif

20. 20/36 Analysis Remark. Regardless how the splitter is chosen, the algorithm above returns the k th largest element of S. Choosing a good Splitter. A good choice: a splitter should produce sets S - and S + that are approximately equal in size. For example: we always choose the median as the splitter. Then each iteration, the size of problem shrink half. Let cn be the running time for selecting a uniformed number. Then the running time is T(n) ≤ T(n/2) + cn Hence T(n) = O(n).

21. 21/36 Analysis con. Funny!! The median is just what we want to find. However, if for any fixed constant b > 0, the size of sets in the recursive call would shrink by a factor of at least (1- b) each time. Thus the running time T(n) would be bounded by the recurrence T(n) ≤ T((1-b)n) + cn. We could also get T(n) = O(n). A bad choice. If we always chose the minimum element as the splitter, then T(n) ≤ T(n-1) + cn Which implies that T(n) = O(n 2 ).

22. 22/36 Random Splitters However, we choose the splitters randomly. How should we analysis the running time of this performance? Key idea. We expect the size of the set under consideration to go down by a fixed constant fraction every iteration, so we would get a convergent series and hence a linear bound running time.

23. 23/36 Analyzing the randomized algorithm We say that the algorithm is in phase j when the size of the set under consideration is at most n(3/4) j but greater than n(3/4) j+1. So, to reach phase j, we kept running the randomized algorithm after the phase j – 1 until it is phase j. How much calls (or iterations) in each phases? Central: if at least a quarter of the elements are smaller than it and at least a quarter of the elements are larger than it.

24. 24/36 Analyzing Observe. If a central element is chosen as a splitter, then at least a quarter of the set will be thrown away, the set shrink by ¾ or better. Moreover, at least half elements could be central, so the probability that our random choice of splitter produces a central element is ½. Q: when will the central will be found in each phase? A: by waiting-time bound, the expected number of iterations before a central element is found is 2. Remark. The running time in one iteration of the algorithm is at most cn.

25. 25/36 Analyzing Let X be a random variable equal to the number of steps taken by the algorithm. We can write it as the sum X=X 0 +X 1 +..., where X j is the expected number of steps spent by the algorithm in phase j. In Phase j, the set has size at most n(¾ ) j and the number of iterations is 2, thus, E[X j ] ≤ 2 cn(¾ ) j So, Theorem. The expected running time of Select(n,k) is O(n).

26. 26 Contention Resolution in a Distributed System Contention resolution. Given n processes P 1, …, P n, each competing for access to a shared database. If two or more processes access the database simultaneously, all processes are locked out. Devise protocol to ensure all processes get through on a regular basis. Restriction. Processes can't communicate. Challenge. Need symmetry-breaking paradigm. P1P1 P2P2 PnPn......

27. 27 Contention Resolution: Randomized Protocol Protocol. Each process requests access to the database at time t with probability p = 1/n. Claim. Let S[i, t] = event that process i succeeds in accessing the database at time t. Then 1/(e  n)  Pr[S(i, t)]  1/(2n). Pf. By independence, Pr[S(i, t)] = p (1-p) n-1. n Setting p = 1/n, we have Pr[S(i, t)] = 1/n (1 - 1/n) n-1. ▪ Useful facts from calculus. As n increases from 2, the function: n (1 - 1/n) n-1 converges monotonically from 1/4 up to 1/e n (1 - 1/n) n-1 converges monotonically from 1/2 down to 1/e. process i requests access none of remaining n-1 processes request access value that maximizes Pr[S(i, t)] between 1/e and 1/2

28. 28 Contention Resolution: Randomized Protocol Claim. The probability that process i fails to access the database in en rounds is at most 1/e. After e  n(c ln n) rounds, the probability is at most n -c. Pf. Let F[i, t] = event that process i fails to access database in rounds 1 through t. By independence and previous claim, we have Pr[F(i, t)]  (1 - 1/(en)) t. n Choose t =  e  n  : n Choose t =  e  n   c ln n  :

29. 29 Contention Resolution: Randomized Protocol Claim. The probability that all processes succeed within 2e  n ln n rounds is at least 1 - 1/n. Pf. Let F[t] = event that at least one of the n processes fails to access database in any of the rounds 1 through t. Choosing t = 2  en   c ln n  yields Pr[F[t]]  n · n -2 = 1/n. ▪ Union bound. Given events E 1, …, E n, union bound previous slide

30. Global Minimum Cut

31. 31 Global Minimum Cut Global min cut. Given a connected, undirected graph G = (V, E) find a cut (A, B) of minimum cardinality. Applications. Partitioning items in a database, identify clusters of related documents, network reliability, network design, circuit design, TSP solvers. Network flow solution. n Replace every edge (u, v) with two antiparallel edges (u, v) and (v, u). n Pick some vertex s and compute min s-v cut separating s from each other vertex v  V. False intuition. Global min-cut is harder than min s-t cut.

32. 32 Contraction Algorithm Contraction algorithm. [Karger 1995] n Pick an edge e = (u, v) uniformly at random. n Contract edge e. – replace u and v by single new super-node w – preserve edges, updating endpoints of u and v to w – keep parallel edges, but delete self-loops n Repeat until graph has just two nodes v 1 and v 2. n Return the cut (all nodes that were contracted to form v 1 ). uvw  contract u-v a bc e f c a b f d

33. 33 Contraction Algorithm Claim. The contraction algorithm returns a min cut with prob  2/n 2. Pf. Consider a global min-cut (A*, B*) of G. Let F* be edges with one endpoint in A* and the other in B*. Let k = |F*| = size of min cut. n In first step, algorithm contracts an edge in F* probability k / |E|. Every node has degree  k since otherwise (A*, B*) would not be min-cut.  |E|  ½ kn. n Thus, algorithm contracts an edge in F* with probability  2/n. A* B* F*

34. 34 Contraction Algorithm Claim. The contraction algorithm returns a min cut with prob  2/n 2. Pf. Consider a global min-cut (A*, B*) of G. Let F* be edges with one endpoint in A* and the other in B*. Let k = |F*| = size of min cut. n Let G' be graph after j iterations. There are n' = n-j supernodes. n Suppose no edge in F* has been contracted. The min-cut in G' is still k. Since value of min-cut is k, |E'|  ½ kn'. n Thus, algorithm contracts an edge in F* with probability  2/n'. n Let E j = event that an edge in F* is not contracted in iteration j.

35. 35 Contraction Algorithm Amplification. To amplify the probability of success, run the contraction algorithm many times. Claim. If we repeat the contraction algorithm n 2 ln n times with independent random choices, the probability of failing to find the global min-cut is at most 1/n 2. Pf. By independence, the probability of failure is at most (1 - 1/x) x  1/e

36. 36 Global Min Cut: Context Remark. Overall running time is slow since we perform  (n 2 log n) iterations and each takes  (m) time. Improvement. [Karger-Stein 1996] O(n 2 log 3 n). n Early iterations are less risky than later ones: probability of contracting an edge in min cut hits 50% when n / √2 nodes remain. n Run contraction algorithm until n / √2 nodes remain. n Run contraction algorithm twice on resulting graph, and return best of two cuts. Extensions. Naturally generalizes to handle positive weights. Best known. [Karger 2000] O(m log 3 n). faster than best known max flow algorithm or deterministic global min cut algorithm