A weighted graph $G= (V, E, W_1, W_2)$ has two cost functions $W_1, W_2 : E \rightarrow \mathbb{R}$. Each edge $e$ has two costs $W_1(e)$ and $W_2(e)$. We may think the costs as representing different resources, such as time and fuel. Similarly, a path $p =e_1, e_2, \dots e_n$ has two costs,

$$W_1(p) = \sum_i W_1(e_i) \qquad \text{and} \qquad W_2(p) = \sum_i W_2(e_i) \qquad$$

Consider the following decision problem: given two nodes $u, v \in V$ and two constants $c_1, c_2$, does there exist a path $p$ from $u$ to $v$ such that $W_1(p) \le c_1$ and $W_2(p) \le c_2$?

Is it NP complete or is a polynomial algorithm known to exist?

  • $\begingroup$ The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! $\endgroup$
    – Raphael
    Dec 19 '17 at 16:28
  • $\begingroup$ " we want to go from place A to place B quickly as possible" and " we need to find a route with at most t minutes" are not the same. Which is it? $\endgroup$
    – Raphael
    Dec 19 '17 at 16:29
  • $\begingroup$ If every edge has two costs, this is not simply Dijkstra, as there may be many "closest" nodes. $\endgroup$ Dec 20 '17 at 18:20

This is NP-hard by reduction from Partition.

Given an instance of the Partition problem consisting of $n$ nonnegative integer values $x_1, \dots, x_n$, we can generate an instance of your problem as follows:

  • Make $n+1$ vertices $s_0, \dots, s_n$. Let $u = s_0$ and $v = s_n$.
  • Make $n$ vertices $a_1, \dots, a_n$, and edges $s_{i-1}a_i$ for every $1 \le i \le n$ having weights $W_1(s_{i-1}a_i) = x_i$ and $W_2(s_{i-1}a_i) = 0$.
  • Make another $n$ vertices $b_1, \dots, b_n$, and edges $s_{i-1}b_i$ for every $1 \le i \le n$ having weights $W_1(s_{i-1}b_i) = 0$ and $W_2(s_{i-1}b_i) = x_i$.
  • Also add edges $a_is_i$ and $b_is_i$ having both weights 0 for every $1 \le i \le n$.
  • Set $c_1 = c_2 = \frac{1}{2}\sum_{i=1}^n x_i$.

The original Partition instance has a solution if and only if the above-constructed instance of your problem has a solution. Basically, for each number $x_i$, we must visit exactly one of the vertices $a_i$ or $b_i$, incurring a cost of $x_i$ for one of the two weights (and zero cost for the other weight).


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