Hardness classification

Interestingly, the behavior of the DS2P varies drastically with the desired distance between the paths. Specifically, when ∆ = 1, the problem is very easy, while for ∆ ≥ 2 it becomes very hard. We proceed to illustrate this phenomenon in detail.

As mentioned in Subsection 2.1, ∆ = 1 leads to the vertex-disjoint shortest paths prob-lem. Polynomial algorithms are given for the vertex-disjoint shortest paths problem in undirected graphs with positive weights in [23]-[25]. Note that a digraph can be modified to fit the min-cost flow model [26]. We illustrate this modification procedure with the example illustrated in Figure 3.2 where the solid lines have cost of 1 and the dashed lines have cost of 10.

Suppose we want to find two vertex-disjoint shortest paths from Node 1 to Node 4 in Figure 3.2. If we simply implement the min-cost flow model, we will get {1,2,3,4} and

Figure 3.2: Digraph modification for ∆ = 1.

{1,3,5,4} whose total cost is zero. However, these two paths have a common vertex: Node 3. The modification guarantees that every node is visited only once, since the edge (3, 3⁰) can be only used once. As we have mentioned before, the min-cost flow model has property of having integer flow (more details in Section 3.2) and since the LP problem is in P, we conclude that the problem with ∆ = 1 is in P as well.

When considering DS2P and its possible relatives, it is interesting to mention that the following problem (stated in [23]) is N P-complete.

Problem 3.11. Given a graph G with positive edge weights and two pairs of vertices (s1, t1), (s2, t2), find whether there exist two disjoint (edge-disjoint or vertex-disjoint) paths P1 from s₁ to t₁ and P₂ from s₂ to t₂ such that P₁ is a shortest path.

Indeed, inspired by the reduction technique of Problem 3.11 in [23], we have the following theorem:

Theorem 3.12. Let N_s = {s} and N_t= {t}.

If ∆ = 1, then DS2P is in P

If ∆ ≥ 2, then DS2P is in N P-hard.

We claim that the DS2P problem with ∆ = 2 is also N P-hard by showing the 3SAT problem is reducible to it.

Proof. Denote the number of clauses and the number of literals in the expression by m and n repectively. For each clause (x_i∨ y_i∨ z_i) we construct a subgraph A_i (1 ≤ i ≤ m), and for each pair of literals v_j and ¯v_j we construct a subgraph B_j, 1 ≤ j ≤ n as shown in Figure 3.3.

A_i has nine edges and the three middle ones represent the corresponding literals x_i, y_i and z_i. B_j has two sides – the left side consists of the edges of ¯v_j and the right side consists of the edges of v_j. Denote the times that each literal v_j appears in all A_i’s by |v_j|, and then set the length of both sides of B_j be l_j = 2 max{|v_j|, |¯v_j|} + 1, so that both sides have l_j edges of v_j and ¯v_j repectively. Note that this construction process is a polynomial-time transformation.

Figure 3.3: 3SAT reduction subgraphs

Connect each A_i (1 ≤ i ≤ m) sequentially by coinciding one’s last vertex with the next one’s first vertex, and repeat the same process for Bj’s. Then connect each pair of vertices which are incident with the edge of the literal v_k (1 ≤ k ≤ n) in A_i’s to the ones in B_j’s incident with the edge of ¯v_k with dashed lines respectively and sequentially. We use solid lines to connect the source s to the initial points s₁ and s₂ of subgraphs A and B respectively and connect the sink t to the ending points t₁ and t₂ as well, and we use dashed lines to connect s₁ to t₁ and s₂ to t₂. Set the weight of all the solid lines to be 1 and the weight of all the dashed lines to be 3m +P

jl_j. For example, Figure 3.4 is contructed for (x1∨ ¯x2∨ x3) ∧ (¯x1∨ x1∨ x4) ∧ (¯x2∨ x3∨ ¯x4) and Figure 3.5 is contructed for (x1∨ x2∨ x2) ∧ (x1 ∨ ¯x2 ∨ ¯x2) ∧ (¯x1∨ x2∨ x2) ∧ (¯x1∨ ¯x2∨ ¯x2).

We claim that finding two disjoint shortest paths subject to the distance constraint ∆ = 2 from s to t in Figure 3.4 is equivalent to solving the 3SAT problem (x₁ ∨ ¯x₂∨ x₃) ∧ (¯x₁ ∨ x₁∨ x₄) ∧ (¯x₂∨ x₃∨ ¯x₄). Note that Figure 3.4 has the following properties.

Figure 3.4: The graph constructed for (x1∨ ¯x2∨ x3) ∧ (¯x1∨ x1∨ x4) ∧ (¯x2∨ x3∨ ¯x4) 1. The DS2P is always feasible: there are always two paths satisfying all the constraints,

that is {s, s₁, t₁, t} and {s, s₂, t₂, t} whose total cost is 2(3m +P

jl_j + 2);

2. Every variable admits a unqiue assignment: if the first path P1 goes through the edge x₁ in A₁, the second path P₂ cannot go through the edge ¯x₁ in B₁ due to the distance constraint ∆ = 2;

3. The DS2P optimum determines the satisfiability. To be convenient, we denote that one path goes through subgraphs A_i’s (or B_j’s) by A (or B) and otherwise by ¯A (or ¯B); and by A ∪ ¯B, we mean that, under the distance constraint, the two situations A and ¯B co-occur,

that is one of the two path goes through all A_i’s while the other does not go through all B_j’s. There are four scenarios:

(a) A ∪ B, and then the least cost is cost_a= 3m +P

and Scenario (a) implies the corresponding 3SAT problem has a true assignment.

In the example shown in Figure 3.4, there exist such two disjoint shortest path:

P1 : {s, s1, x1(A1), x4(A2), ¯x2(A3), t1, t}

and P₂ : {s, s₂, x₁(B₁), ¯x₂(B₂), x₃(B₃), x₄(B₄), t₂, t}.

If we set all the literals in P₂ to be true, that is

x₁ = 1, ¯x₂ = 1, x₃ = 1, and x₄ = 1,

then (x₁∨ ¯x₂∨ x₃) ∧ (¯x₁ ∨ x₁ ∨ x₄) ∧ (¯x₂∨ x₃∨ ¯x₄) = 1 and we solve the 3SAT problem.

In the example shown in Figure 3.5, there do not exist such two paths with the total cost 3m +P

jl_j+ 4 that satisfy the distance constraint. And it is not a coincidence that the corresponding 3SAT problem (x₁∨ x₂∨ x₂) ∧ (x₁∨ ¯x₂∨ ¯x₂) ∧ (¯x₁∨ x₂∨ x₂) ∧ (¯x₁∨ ¯x₂∨ ¯x₂) has no true assignment, meaning it is always false.

Considering these two example together, we can see that indeed the outcome of solving a DS2P problem implies the satisfiability of its corresponding 3SAT expression. If the outcome

Figure 3.5: The graph constructed for (x1∨x2∨x2)∧(x1∨ ¯x2∨ ¯x2)∧(¯x1∨x2∨x2)∧(¯x1∨ ¯x2∨ ¯x2) shows the cost equals 3m +P

jl_j + 4, then it has a true assignment which can be told by the path from s₂ to t₂; otherwise, the 3SAT expression is always false.

Graphs constructed for scenarios when ∆ ≥ 3 can be simply regard as subdivisions of the graph of scenario when ∆ = 2. Also note that this reduction applies regardless of the graph being directed or undirected, hence, for both types of graphs the DS2P problem with

∆ ≥ 2 is N P-hard.