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NP-complete problems can be found in Coffman [37], Cormen et al [42], and Garey and Johnson [73] Theorem 41 (NP-Completeness) Let G = (V, E, w, c) be a task graph and P a parallel system The decision problem SCHED(G, P) associated with the scheduling problem is as follows Is there a schedule S for G on P with length sl(S) T , T Q+ SCHED(G, P) is NP-complete in the strong sense Proof First, it is argued that SCHED belongs to NP, then it is shown that SCHED is NP-hard by reducing the well-known NP-complete problem 3-PARTITION (Garey and Johnson [73]) in polynomial time to SCHED 3-PARTITION is NP-complete in the strong sense The reduction is inspired by a proof presented in Sarkar [167] The 3-PARTITION problem is stated as follows Given a set A of 3m positive integer numbers ai and a positive integer bound B such that 3m ai = mB with B/4 < i=1 ai < B/2 for i = 1, , 3m, can A be partitioned into m disjoint sets A1 , , Am (triplets) such that each Ai , i = 1, , m, contains exactly 3 elements of A, whose sum is B Clearly, for any given solution S of SCHED(G, P) it can be veri ed in polynomial time that S is feasible (Algorithm 5) and sl(S) T ; hence, SCHED(G, P) NP From an arbitrary instance of 3-PARTITION A = {a1 , a2 , , a3m }, an instance of SCHED(G, P) is constructed in the following way A so-called fork, or send, graph G = (V, E, w, c) is constructed as shown in Figure 42 It consists of |V| = 3m + 1 nodes (n0 , n1 , , n3m ), where node n0 is the entry node with 3m successors Hence, one edge goes from n0 to each of the other nodes of G and there are no other edges apart from these The edge weight of the 1 3m edges is c(e) = 2 e E, node n0 has weight w(n0 ) = 1, and the 3m nodes have weights that correspond to the integer numbers of A, w(ni ) = ai , i = 1, , 3m The number of processors of the target system is |P| = m and the time bound is set to T = B + 15 Clearly, the construction of the instance of SCHED is polynomial in the size of the instance of 3-PARTITION It is now shown how a schedule S is derived for SCHED(G, P) from an arbitrary instance of 3-PARTITION A = {a1 , a2 , , a3m }, that admits a solution to 3-PARTITION: let A1 , , Am (triplets) be m disjoint sets such that each Ai , i = 1, , m, contains exactly 3 elements of A, whose sum is B
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Figure 42 The constructed task graph
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TASK SCHEDULING
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Figure 43 Extract of the constructed schedule for P1 and Pi
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Node n0 is allocated to a processor, which shall be called P1 The remaining nodes n1 , , n3m are allocated in triplets to the processors P1 , , Pm Let ai1 , ai2 , ai3 be the elements of Ai The nodes ni1 , ni2 , ni3 , corresponding to the elements of triplet Ai , are allocated to processor Pi The resulting schedule is illustrated for P1 and Pi in Figure 43 What is the length of this schedule The time to execute ni1 , ni2 , ni3 on each processor is B, n0 is executed in 1 time unit, and the communication from n0 to 1 its successor nodes takes 2 time unit Hence, the nish time of all processors Pi , i = 2, , m, is tf (Pi ) = B + 15 Since the communication on P1 is local, the nish time on P1 is tf (P1 ) = B + 1 The resulting length of the constructed schedule S is sl(S) = B + 15 T and therefore the constructed schedule S is a solution to SCHED(G, P) Conversely, assume that an instance of SCHED admits a solution, given by the feasible schedule S with sl(S) T It will now be shown that S is necessarily of the same kind as the schedule constructed above In this schedule S, each processor can at most spend B time units in executing nodes from {n1 , , n3m } Otherwise the nish time of a processor, say, Pi , spending more that B time units is tf (Pi ) > w(n0 ) + B = 1 + B Due to the fact that all ai s (ie, node weights) are positive integers, this means that tf (Pi ) 2 + B However, this is larger than the bound T Furthermore, with 3m w(ni ) = mB and |P| = m, it follows that each processor i=1 spends exactly B time units in executing nodes from {n1 , , n3m } Finally, due to w(ni ) = ai , B/4 < ai < B/2, i = 1, , 3m, only three nodes can have the exact execution time of B Hence, the distribution of the nodes on the m processors corresponds to a solution of 3-PARTITION This proof demonstrated the NP-completeness of the scheduling problem SCHED in the strong sense (Garey and Johnson [73]) Alternatively, the NP-completeness can also be shown with a proof based on a reduction from PARTITION (Garey and Johnson [73]) Such a proof (Chr tienne [32]) is slightly less involved, but only demonstrates NP-completeness in the weak sense, since PARTITION is NP-complete in the weak sense In Exercise 43 you are asked to devise a proof based on PARTITION
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