## Presentation on theme: "1 EE5900 Advanced Embedded System For Smart Infrastructure Advanced Theory."— Presentation transcript:

3DM to Numerical 4DM First show that numerical 4DM is NP- complete. Reduce from 3DM. 4DM problem says that given four sets S1,S2,S3,S4, each of which consists of q distinct elements, and a collection C=S1S2S3S4, one asks whether there exists a subcollection C’ to partition the union of four sets and the sum of values of each set in C’ is B. 2

Reduce from 3DM to numerical 4DM Create four elements for each candidate set (x a,y b,z c ) in M. e 1 in S 1, e 2 in S 2, e 3 in S 3 and e 4 in S 4. If x a is in the candidate set, create an element e 1 with value either 2q 3 +aq 2 (core) or aq 2 (dummy). If y b is in the candidate set, create an element e 2 with value either bq (core) or q 3 +bq (dummy). If z c is in the candidate set, create an element e 3 with value either c (core) or q 3 +c (dummy). create an element e 4 with value 2q 3 -aq 2 -bq-c. If there is only one occurrence of a variable (e.g., x1) in M, then there is only one core element generated. If there are k occurrences (e.g., z7) in M, then there are k elements generated where contains one core element and k-1 dummy elements. Note that different elements can have the same value. Candidate sets in 4DM is created such that it contains either all core elements or all dummy elements. Enumerate all possible candidate sets. Set B=4q 3. 3

Reduction example Suppose that the candidate sets M in 3DM is as follows. (x 1,y 5,z 7 ), (x 2,y 2,z 7 ), (x 2,y 5,z 5 ) … (x 1,y 5,z 7 ) produces e 11 with value 2q 3 +q 2, e 21 with value 5q, e 31 with value q 3 +7, e 41 with value 2q 3 -q 2 -5q-7. (x 2,y 2,z 7 ) produces e 12 with value 2q 3 +2q 2, e 22 with value 2q, e 32 with value 7, e 42 with value 2q 3 -2q 2 -2q-7. (x 2,y 5,z 5 ) produces e 13 with value 2q 2, e 23 with value q 3 +5q, e 33 with value 5, e 43 with value 2q 3 -2q 2 -5q-5. If (x 1,y 5,z 7 ) is picked in M, we pick (e 11 e 21 e 32 e 41 ). Since (x 2,y 2,z 7 ), (x 2,y 5,z 5 ) are not picked, we pick (2q 2 q 3 +2q e 31 e 42 ) and (2q 2 e 23 e 33 e 43 ). The elements with values are those generated from other candidate sets in M. e 12 e 22 e 13 are not picked and they will be picked corresponding to some sets picked in M. 4

If direction When there is solution of 3DM problem, If a set is picked in 3DM, the corresponding core set is picked in numerical 4DM. Otherwise, the corresponding dummy set is picked. Each variable is picked exactly once in 3DM, so each core element is picked exactly once. Note that core elements generated from multiple sets in M could be combined together and picked (since we enumerate candidate sets in numerical 4DM). Given k occurrences of a variable in M, they are in k candidate sets in M. One of them is picked (so is the corresponding core element), and k-1 of them is not picked (so the corresponding k-1 dummy elements are picked). Thus, each generated element is picked exactly once. There is only one e4 for each set in M, which will be used to make the sum of values 4q 3. This is the subcollection of sets to partition the union of four sets and each set with the sum of values to be B. 5

Only if direction Given a solution to numerical 4DM, each core element is covered exactly once. There exists sets which contain only the core elements and one can pick the corresponding sets in M. 6

Numerical 4DM to Numerical 3DM Given a numerical 4DM instance, we have four sets S1, S2, S3, S4. We create three new sets T1, T2, T3. –T1 = S1 ⋃ P where P=S3 S4 are pairing elements –T2= S2 ⋃ S3 ⋃ F where F are fillers and |F|=q 2 -q –T3=S4 ⋃ P’ where P’ are copairing elements Form the collection as follows 7

Form Collection (t1,t2,t3) can be formed as the elements from the following sets through enumeration –(S1,S2,P’) –(P,S3,S4) –(P,F,P’) Weighting functions –w1’(p)=3B-w3(s3)-w4(s4) –w1’(s1)=w1(s1) –w2’(s2)=w2(s2)+2B –w2’(s3)=w3(s3)+4B –w2’(f)=0 –w3’(s4)=w4(s4) –w3’(p’)=4B+w3(s3)+w4(s4) 8

Only if direction When there is a solution for numerical 4DM, for each selected candidate set (s1,s2,s3,s4), we pick (t1,t2,t3) as (s1,s2,p’(s3,s4)) and (p(s3,s4),s3,s4). Since |S1|=|S2|=|S3|=|S4|=q, we have picked 2q sets for numerical 3DM. Since |T1|=|T2|=|T3|=q 2 +q, we need to pick q 2 -q sets. They are (p,f,p’). The sum of each set is 7B 9

If Direction When there is a solution for numerical 3DM, since there are only q 2 -q filler elements, we need to pick at least 2q sets in the form of (s1,s2,p’) or (p,s3,s4) to cover all elements. Note that |S1|=|S2|=|S3|=|S4|=q, so there are only q sets of (s1,s2,p’) and q sets of (p,s3,s4). Since |P|=q 2, there are only q sets of (p,s3,s4) which can be picked since other |P| are picked with fillers. Similarly, since |P’|=q 2, there are only q sets of (s1,s2,p’) which can be picked. One then correspondingly picks (s1,s2,s3,s4) Each set has sum 7B in numerical 3DM, and thus each set has sum B in numerical 4DM 10

Numerical 3DM to 3Partition Given a set S of 3m elements where each element a has a value v(s) and ∑ s ∈ S v(s)=mB, one asks whether S can be partitioned into m disjoint subsets S 1,S 2,…,S m such that for each subset ∑ s ∈ Si v(s)=B? Reduce from Numerical 3DM with sum target B’. Form the set S as T1 ⋃ T2 ⋃ T3 and set w(s)=w’(t)+3iB’ where i = 1,2,3. Set B=19B’. 11

Reduction When there is a solution for Numerical 3DM, one accordingly partition the sets and each set has sum 19B’ since the sum of elements in each set in the solution for numerical 3DM is B’. When there is a solution for 3Partition, one accordingly picks the sets in numerical 3DM. 12

3Partition to Nonpreemptive Scheduling Given an instance of 3partition, form an instance of nonpreemptive scheduling problem which contains 3m+1 tasks, T 1,T 2,…,T 3m+1 as follows. For each element s i, create a task T i with p=d=mB+m and c=v(s i ). Create a task T 3m+1 with p=B+1 and d=c=1. We claim that the task set is schedulable if and only if the 3partition instance is feasible. 13

Only if direction When the task set is schedulable –Task T 3m+1 is scheduled at time 0, B+1, 2(B+1), … –Consider the hyper period mB+m. All of the first 3m tasks need to be scheduled within it. –During this hyper period, T 3m+1 has run for m times with total time m. –Thus, mB time is for all other tasks. –The available time between the first and the second T 3m+1 is B. –The task set between them has total time bounded by B. Let S1 denote the corresponding set in S, so ∑ s ∈ S1 v(s) ≦ B –Similarly, ∑ s ∈ Si v(s) ≦ B for all 1 ≦ i ≦ m since T 3m+1 has run for m times –On the other hand, ∑ s ∈ S1 v(s) + ∑ s ∈ S2 v(s) +…+ ∑ s ∈ Sm v(s)=mB. One has that each ∑ s ∈ Si v(s)=B. 14

If direction When there is a feasible 3partition solution, –One can schedule T 3m+1 at time 0, B+1, 2(B+1),… –One then puts the other tasks according to the 3partition solution 15