W[2]-hardness of precedence constrained K-processor scheduling

W[2]-hardness of precedence constrained K-processor scheduling by H. L. Bodlaender and M. R. Fellows Presented by Hyun-Chul Chung

Overview • Problem Statements • Graph Transformation • Proof

Problem statements • Precedence Constrained K-processor scheduling • Instance: Set T of unit length tasks, partial order on T, a deadline D, number of processors K • Question: Does there exist a mapping such that for all , and for all i, ? • Parameter: K • Dominating Set • Since we know that Dominating Set is W[2]-Complete, we use this to show that the Precedence Constrained K-processor scheduling problem is W[2]-hard

Graph Transformation 1 • Let (G=(V,E), k) be an instance to Dominating Set. • . • Take • Define a directed acyclic graph H=(W,F) which consists of • The floor • The floor gadgets • The selector paths • The selector gadgets

Graph Transformation 2 • The floor • Take the path with length D : vertices and edges for all i , • The floor gadgets • “Parallel” to each floor vertex of the form take vertices and add edges and • The selector paths • For each i, , we take a path of length D-n+1. • This path will represent the ith vertex of the dominating set of G • Take vertices and edges for all i,

Graph Transformation 3 • The selector gadgets • If and , then take a vertex which is “parallel” to where and call these verices • Add edges and • Now the graph transformation is complete. The only thing left is to prove the following: • Task set W with partial order , deadline D, and number of processors K, is a yes-instance to Precedence Constrained K-Processor Scheduling iff G has a dominating set of size at least k.

Proof 1 • <=: Suppose is the dominating set of G. • Consider the following schedule f of W : • . • To an integer i not of the form , one floor vertex, no floor gadget vertex, at most k selector path vertices, and at most k selector gadget vertices are mapped, so for such i, • To an integer i of the form : • Case 1: is in the dominating set. • At most one floor vertex, one floor gadget vertex, k selector path vertices, and k-1 selector gadget vertices. • Case 2: is adjacent to vertex • Same as Case 1.

Proof 2 • =>: Suppose is a schedule, fulfilling the required proerties. • The only schedules that the floor and the floor gadgets can have is . • Call the interval the ith range where . • The ith range is “polluted” by the jth selector path, when there exists an integer in this range to which no vertex on this jth selector path is mapped. • Since each selector path has length of D-n+1, it can only polute n-1 ranges. In total, kn-k.

Proof 3 • So, there is at least one range that is not polluted. • Let’s call this the th range • Define numbers such that • It follows that for all selector path vertices: • So,

Proof 4 • Now, we show that for all q, belongs to the set or is adjacent to a vertex in this set. • Say, • The set contains one floor vertex, one floor gadget vertex, k selector path vertices and at most k-1 selector gadget vertices. What this means is that there is an l such that does not contain any vertex of the form • Claim that does not exist in the selector gadgets. • Notice that . • So, does not exist in the selector gadgets, otherwise it will be mapped to z.

Proof 5 • As does not exist in the selector gadgets, we have that or • This means that is a dominating set of G

Questions?

W[2]-hardness of precedence constrained K-processor scheduling