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r(U) = X w(e) e∈U\{e∈U|∃e′shortest path problem is NP-complete, as the next section will show.

2 Reduction from 3SAT

Originally, a reduction from PARTITION was given, which turned out to be incorrect and fundamentally flawed (see Appendix). Instead, we use the same reduction from 3SAT [2, LO2] as in the paper by Broersma et al. [1].

Proposition 1 The edge information reuse shortest path problem is NP-complete.

Proof: It is clear that the problem is in NP. Let U be a path from s to t. We can check whether the path taking edge information reuse into account has length at most some K by two traversals of U. We use f(e) as a pointer to mark for information reuse. In the first traversal, all f(e) for e ∈ U are marked as unused. In the second traversal the weights of unused f(e)’s are summed and f(e) marked as used. The 3SAT problem is, given a collection of m three-literals clauses Cj, 0 ≤ jvertex ui will assign true to xi, whereas a path ′ ′ ′ going through vertexu ¯i will assign true to the negationx ¯i. The edges (ui, ui) and (ui, u¯i) will share information with edges in clause subgraphs. 0 1 2 The subgraph for clause Cj consists of five vertices vj, vj , vj , vj , vj+1 for 0 ≤ j

2 The graph thus constructed has O(n + m) vertices and edges, and can clearly be computed in a polynomial number of steps, including the function f assigning (shared) information with the edges. Now, the claim is that the given 3SAT instance is satisfiable, if an only if the shortest st path with edge information reuse has length exactly n. Any path from s to t must have length at least n, since it has to pass through all vertices ui, 1 ≤ iset to true) in the variable subgraph. ✷

The graph constructed in the reduction is acyclic, which shows that the shortest st path problem with edge information reuse on acyclic graphs is NP-complete.

3 Remark

Alex Khodaverdian (personal communication, 24.5.2016) pointed out some problems with the original reduction from PARTITION (still retained in the appendix), which eventually showed this reduction to be fundamentally flawed. Fortunately, the 3SAT reduction from [1] shows the problem NP-complete, so that the original claim still holds. It might still be enlightening to find a different reduction. For the first version of this note, Gerhard Woeginger (personal communication, 15.9.2015) pointed out that NP-completeness follows straightforwardly from the NP-completeness of the problem of finding a simple st path with the smallest number of colors in a graph with colored edges, and supplied the reference [1]. Simply let the function f denote the edge colors and take all edge weights to be 1. A shortest st path with edge information reuse corresponds to a simple st path with the smallest number of colors.

References

[1] Hajo Broersma, Xueliang Li, Gerhard Woeginger, and Shengui Zhang. Paths and cycles in colored graphs. Australasian Journal on Combinatorics, 31:299–311, 2005.

[2] M. R. Garey and D. S. Johnson. Computers and Intractability: A Guide to the Theory of NP-Completeness. Freeman, 1979. With an addendum, 1991.

A The flawed reduction from PARTITION

We tried to show the edge information reuse shortest path problem NP-complete by a reduction from the NP-complete PARTITION problem [2, SP12].

Proposition 2 The edge information reuse shortest path problem is NP-complete.

3 v1 vn−1

0 0 0 0

w w wn w w wn w w wn 0 1 . . . −1 ′ 0 ′ 1 ′ . . . ′ −1 ′ ′′ 0 ′′ 1 ′′ . . . ′′ −1 ′′ s = u0 u1 u2 un−1 un = u0 u1 u2 un−1 un = u0 u1 u2 un−1 un = t

⌊B/2⌋ ⌊B/2⌋

Figure 1: The weighted, directed, acyclic graph constructed from the given PARTITION in- stance. The shortest path taking edge information reuse into account should have length B if and only if the PARTITION instance is feasible.

It is clear that the problem is in NP. Let U be a path from s to t. We can check whether the path taking edge information reuse into account has length at most some K by two traversals of U. We use f(e) as a pointer to mark for information reuse. In the first traversal, all f(e) for e ∈ U are marked as unused. In the second traversal the weights of unused f(e)’s are summed and f(e) marked as used. Now, let the item set A = {x0,...xn−1} with integer item weights w(xi) be an instance of PARTITION, and let B = Px∈A w(x). We construct a weighted, directed, acyclic graph G such that there is a shortest path with edge information reuse of length B if and only if the answer to PARTITION is “yes”, that is, if there is a subset P of A such that Px∈P w(x)= Px∈A−P w(x). The graph G consists if three parts, the first with vertices u0, u1, . . . , un and v1, . . . vn−1, ′ ′ ′ ′′ ′′ ′′ the second with vertices u0, u1, . . . , un, and the third part with vertices u0, u1, . . . , un, where ′ ′ ′′ ′′ un = u0 and un = u0. The source is s = u0 and the sink t = un. For each i = 0,n − 1 there ′ ′ ′′ ′′ are edges (ui, ui+1), (ui, ui+1), and (ui , ui+1) each with weight w(xi), and these three edges ′ ′ ′′ ′′ carry the same item information, that is f((ui, ui+1)) = f((ui, ui+1)) = f((ui , ui+1)). Different edges (ui, ui+1) and (uj, uj+1) carry different item information. The information associated ′ ′ with these edges designate the items of A. There are edges (u0, un) and (u0, un) both of weight ⌊B/2⌋ (which share no information). In the first part of G there is for each i = 1,...,n − 1 two edges (ui, vi) and (vi, ui+1) of weight 0. The construction is illustrated in Figure 1. It can clearly be carried out in O(n) operations. There is clearly a path from s to t in G. The purpose of first part of the graph is to select items for the set P . For any path through this part the information associated with non-zero weight edges designate items chosen for P . Each subset P is chosen by a unique path. Assume that a shortest st path with information reuse is longer than B. The path either passes through the vertices u1, u2, . . . , un−1, and possibly some of the vi vertices as well, or takes the shortcut edge (u0, un). In the first case, the length of the path to vertex un is at most ⌊B/2⌋, and there is no possibility for information reuse since all edge information occurs for the first time in this part of G. Let W be the sum of the weights of the (ui, ui+1) edges on the path. These weights ′ ′ ′′ ′′ will not be counted again since they share information with the edges (ui, ui+1) and (ui , ui+1). ′ ′ If W < ⌊B/2⌋ the shortest path must take the shortcut edge (u0, un) of weight ⌊B/2⌋ since ′′ B − W > ⌊B/2⌋. The path through the final part of G from u0 to t will have weight W − B for a total of B + ⌊B/2⌋. If the shortest path instead takes the shortcut edge (u0, un) of weight ⌊B/2⌋ (which means that there is no item xi ∈ A with w(xi) ≤ ⌊B/2⌋), then the path must also take the shortcut ′ ′ ′ edge (u0, un) since there is no edge information to reuse and the non-shortcut path from u0 ′ ′′ to un has length B. The part of the path from u0 to t in this case has cost B since no edge information can be reused, for a total of 2⌊B/2⌋ + B.

4 Thus, the length of a shortest path with information reuse is B if and only it is possible to find a path from u0 to un of length exactly ⌊B/2⌋ (that does not use the shortcut edge), and if the weight of the edges whose information has not been used in this part is likewise exactly ′′ ⌊B/2⌋. In this case, all edge information has been used, and the part of the path from u0 to t contributes weight 0. The solution P to the PARTITION is the information f(e) on the non-zero edges on the part of the path from s to un. The flaw in this argument is simply that subpaths of a shortest st-path do not have to be shortest paths themselves. Therefore, with information reuse, there is always a shortest path of length exactly B, and the shortcut edges of weight ⌊B/2⌋ do not have to be taken.

5