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Consider the following problem:

Input: An undirected graph $G = (V, E)$, each edge has a non-negative weight $w_i$ and a non-negative value $v_i$. There are two vertices to represent start point $s$ and end point $t$. We are also given two values $W,V$.

Decide whether there is a simple path from $s$ to $t$ with total weight at most $W$ and total value at most $V$.

How do I show that the problem is NP-hard by reduction from PARTITION?

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  • $\begingroup$ What did you try? Consider encoding the decision to put an item in one or another part by different choices of arcs in an st-path. $\endgroup$ – Marcus Ritt May 28 at 1:59
  • $\begingroup$ I have tried to construct a complete graph, so that I can represent every combination of choices of integers, but since both constraints are at most, this graph will return a yes instance even if the partition instance is no $\endgroup$ – Deangelo Kingwell May 28 at 3:49
  • $\begingroup$ Make the graph more restricted. Try to do the reduction first for a single item, then for two. $\endgroup$ – Marcus Ritt May 28 at 11:53
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Here is the idea of the reduction; I'll let you fill in the details.

We are given an instance of the partition problem, consisting of numbers $x_1,\ldots,x_n$ summing to $S$. We construct a (non-simple) graph on the vertex set $s = v_0,\ldots,v_n = t$. For $i=1,\ldots,n$, there are two edges from $v_{i-1}$ to $v_i$: one with weight $x_i$ and value $0$, and one with weight $0$ and value $x_i$. Our target weight and value are both $S/2$.

I'll let you figure out why this works, and convert it to a simple graph.

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  • $\begingroup$ We can simply divide $v_i$ to $v_i$ and $v_i'$. Then, the edges from $v_{i-1}$ to $v_i$ and $v_i'$ should be $x_i\;/\;0$, and edges from $v_{i-1}'$ to $v_i$ and $v_i'$ should be $0\;/\;x_i$. $\endgroup$ – Deangelo Kingwell May 28 at 16:18
  • $\begingroup$ Right, that's the idea. $\endgroup$ – Yuval Filmus May 28 at 16:23

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