PRAMs. M 1 M 2 M p. globaler Speicher

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1 PRAMs A PRAM (parallel random access machine) consists of p many identical processors M,..., M p (RAMs). Processors can read from/write to a shared (global) memory. Processors work synchronously. M M 2 M p globaler Speicher

2 PRAMs Different variants: CRCW (Concurrent Read Concurrent Write) CREW (Concurrent Read Exclusive Write) EREW (Exclusive Read Exclusive Write) ERCW (Exclusive Read Concurrent Write) randomized PRAMs: Processors may toss coins.

3 The Class NC Problems which are efficiently parallelizable. NC is called Nick s Class (after Nick Pippenger). Definition: A problem belongs to the class NC, if it can be solved on a PRAM such that for an input of length n, we: use only n d many processors (for a constant d), and spend time (log n) c (for a constant c). The class is robust to minor changes of the machine model: for instance, it doesn t matter, whether we use the CRCW/CREW/EREW/ERCW PRAM-model. The question NC? = P is still open.

4 Recall that an independent set of an undirected graph G = (V, E) is a subset I V such that (u, v) E for all u, v I. Goal: For a given undirected graph G = (V, E), find an independent set I V of G, which is maximal under inclusion, i.e., if I J for an independent set J then I = J. We want to do this in NC, i.e., in polylogarithmic time using polynomially many processors. Our first solution will be a randomized NC-algorithm. For a set U V of nodes let N(U) = {v V u U : (u, v) E} be the set of neighbors of U.

5 Luby s algorithm works in rounds. In every round we calculate an independent set I in the current graph G and remove I N(I ) (and all edges that are incident with a node from I N(I )) from G. We repeat this until the graph is empty. The calculated independent set is the union of the independent sets calculated in the rounds. A single round, where d(v) = N(v) for v V : In parallel: for every node v V, put v with probability 2d(v) into a set S (isolated nodes can be put into S into a preprocessing step), independently from the other nodes (i.e., Pr( k i= v i S) = k i= Pr(v i S)). In parallel: For every (u, v) E with u, v S, remove from S the node with the smaller degree (break ties arbitrarily). Call the remaining set I ; it is an independent set.

6 A single round can be done in constant time using O( V 2 ) processors. We will show that the expected value of the number of rounds is in O(log E ). First step: We show that the expected number of edges that are deleted in every round is at least 72 of the total number of edges.

7 Lemma For every node v: Pr(v I ) 4d(v) Proof: We will show that Then we obtain: Pr(v I v S) 2 Pr(v I ) = Pr(v I v S) Pr(v S) Pr(v S) = 2 4d(v).

8 We have Thus: Pr(v I v S) Pr( u L(v) : u S v S) where L(v) = {u N(v) d(u) d(v)}. Pr(v I v S) u L(v) = u L(v) = u L(v) u L(v) Pr(u S v S) Pr(u S) 2d(u) (independence) 2d(v), since L(v) N(v). 2

9 Definition: A node v V is good, if 2d(u) 6 u N(v) (intuition: many neighbors with small degree), otherwise v is bad. An edge (u, v) E is good, if u or v is good, otherwise it is bad. Lemma For a good node v V we have Pr(v N(I )) 36. Proof: Case : u N(v) : 2d(u) > 6. Then, by the previous lemma, Pr(v N(I )) Pr(u I ) 4d(u) > 2 > 36.

10 Case 2: u N(v) : 2d(u) 6. Then there exists M(v) N(v) with 6 u M(v) 2d(u) 3. Thus Pr(v N(I )) Pr( u M(v) : u I ) Pr(u I ) = u M(v) u M(v) 4d(u) (independence) 4d(u) u M(v) u M(v) 2d(u) u,w M(v),u w u,w M(v),u w u,w M(v) 2 w M(v) 2d(u) Pr(u I w I ) Pr(u S w S) 2d(w) 2d(w) 6 6 = 36

11 By the previous lemma, a good edge will be deleted with probability at least /36. Lemma At least half of all edges are good. Proof: Direct every edge towards its endpoint of higher degree, breaking ties arbitrarily. Claim: For every bad node v V, there are at least twice as many outgoing edges than incoming edges. Proof of the claim: Let N be the set of predecessors of v after directing the edges. If N d(v) 3, then u N(v) 2d(u) u N 2d(v) = 2 N d(v) 6, i.e., v would be good a contradiction.

12 Thus, N d(v) < 3, i.e., N < 2 (d(v) N ), which proves the claim. Hence, to every bad edge e (for which both endpoints are bad) we can assign a set P(e) = {e, e 2 } of two edges e e 2 such that e f P(e) P(f ) =. This proves the lemma.

13 Theorem Let X be the number of edges that are deleted (in a certain round). Then for the expected value E(X ) of X we have E(X ) E 72. Proof: Let X e =, if e is deleted, otherwise X e = 0. Then we have: E(X ) = E(X e ) E(X e ) e E e good e good 36 E 2 36

14 Let m be the total number of edges in our graph. For i 0 we define S i = number of edges that were removed in round i. X i = number of edges that are removed in round i. Thus, S 0 = 0, S i m, and S i+ = S i + X i+. The statement of the previous theorem can be restated as follows, where ε = 72 : E(X i+ S i = l) = k N k Pr(X i+ = k S i = l) ε(m l) Lemma E(X i+ ) ε m ε E(S i )

15 Proof: E(X i+ ) = k Pr(X i+ = k) k N = k Pr(X i+ = k S i = l) k,l N = k Pr(X i+ = k S i = l) Pr(S i = l) k,l N = Pr(S i = l) k Pr(X i+ = k S i = l) l N k N = Pr(S i = l) E(X i+ S i = l) l N Pr(S i = l) ε(m l) l N = ε m ε l Pr(S i = l) = ε m ε E(S i ) l N

16 Lemma E(S i ) m( ( ε) i ). Proof: Induction on i. The case i = 0 is clear. For i + we obtain: E(S i+ ) = E(S i ) + E(X i+ ) E(S i ) + εm ε E(S i ) = εm + ( ε)e(s i ) εm + m( ε)( ( ε) i ) = m( ( ε) i+ )

17 Lemma E(S i ) m + Pr(S i = m) Proof: E(S i ) = m j Pr(S i = j) j=0 m (m ) Pr(S i = j) + m Pr(S i = m) j=0 = m Pr(S i = m) + (m )( Pr(S i = m)) = m + Pr(S i = m)

18 By the two previous lemmas we have Pr(S i = m) m( ε) i. Thus, Pr(S i < m) m( ε) i. Choose k O(log m) such that m( ε) k. Then, for i k we have Pr(S i < m) m( ε) i ( ε) i k. Define f : N {0, } by f (x) = { if x < m 0 otherwise. Thus, E(f (S i )) = Pr(S i < m) ( ε) i k for i k.

19 The random variable R = f (S 0 ) + f (S ) + f (S 2 ) + counts the number of rounds in Luby s algorithm. We have E(R) = i 0 E(f (S i )) k + i k E(f (S i )) k + i k ( ε) i k = k + ε O(log m) We have shown Theorem The expected number of rounds in Luby s algorithm is in O(log m).

20 In the current version of Luby s algorithm we put a node v into S with probability 2d(v). For this we have to flip n = V many biased coins (with Pr(head) = 2d(v) and Pr(tail) = 2d(v) ) independently. It can be shown that n Ω() many truely random bits (fair coin flips) are necessary (and sufficient) in order to approximate these n independent biased coin flips sufficiently good. But: In the analysis of Luby s algorithm, we only used pairwise independence. We will show that Ω(log(n)) many random bits suffice in order to generate n = V biased coin flips (Pr(head) 2d(v) ) that are pairwise independent.

21 This leads to a derandomized version of Luby s algorithm: Assume that α log(n) random bits suffice in order to generate n biased coin flips, where α is a constant. Let R = {0, } α log(n), thus R = n α. A single round in Luby s algorithm is replaced by the following procedure: for all s = a a α log(n) R do in parallel simulate the next round of Luby s algorithm deterministically with a i = the i-th random bit endfor choose that simulation that removes the largest number of edges and go with the resulting graph to the next round

22 For every v V let δ(v) R such that 7 9 2d(v) = 2d(v) 9d(v) 2δ(v) 2d(v) First, we check that the analysis of Luby s algorithm also works when we replace d(v) by δ(v) everywhere (in particular, Pr(v S) := 2δ(v) ). Lemma ( v V : Pr(v I ) 4δ(v) ): Lemma 2 (v good Pr(v N(I )) /36): Recall that v is good if u N(v) good if one of its endpoints is good. 2δ(u) 6 and that an edge is

23 Instead of showing that at least half of the edges are good (Lemma 3), we will prove that at least /4 of all edges are good. Again, we direct every edge towards its endpoint with larger δ-value. Lemma Let n = N be the number of incoming edges of a node v. If then v is good. 7n 8d(v) 6 Proof u N(v) 2δ(u) u N 2δ(v) = n 2δ(v) 7 9 n 2d(v) 6

24 Thus, if v is bad then 7 8 n d(v) 6. We obtain n 3 7 d(v). Thus, d(v) n 7 3 n n = 4 3 n. Therefore, n 3 4 (d(v) n ), i.e., at least /4 of all edges is good. If X is the number of edges that are removed (in a certain round), then we obtain E(X ) 36 E 4 = 44 E

25 We have shown that Luby s algorithm works with δ(v) instead of d(v) as well. [ ] 7 Recall δ only has to satisfy 2δ(v) 9 2d(v),. 2d(v) Now choose a prime number p with 9n p 8n such a prime exists by Betrand s postulat. We may identify V with a subset of F p = {0,..., p }. [ ] The interval 7 9 2d(v), 2d(v) has size 2d(v) 7 9 2d(v) = 9d(v) 9n p, thus there exists a number of the form av 9n in this interval for some a v N. We can set 2δ(v) = av p. Determine a subset A v F p with A v = a v, where 7 9 2d(v) av p = 2δ(v) 2d(v)

26 Now choose (x, y) F p F p randomly (using O(log(n)) many random bits) and put v into S if and only if x + vy A v. Since for every y, z F p there is exactly one x F p with x + vy = z, namely x = z vy, we have Pr(v S) = p 2 {(x, y) x + vy A v} = p 2 z A v {(x, y) x + vy = z} = p 2 z A v p = a v p

27 Finally, we have to show pairwise independence: Let u v be two different nodes. Then Pr(u S v S) = p 2 {(x, y) x + uy A u x + vy A v } = { ( ) ( ) ( )} u x zu p 2 (x, y) = v y (z u,z v ) A u A v z v

28 The matrix has an inverse (the determinant is v u 0), thus for every (z u, z v ) there is exactly one (x, y) F p F p for We obtain ( u v ) ( x y ) = ( zu z v ) Pr(u S v S) = p 2 a ua v = a u p a v p = Pr(u S) Pr(v S). We have shown pairwise independence.