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I'm trying the following question from my homework:

We're given a graph $G$ and parameters $k,\hat{P}, \hat{W}\in \mathbb{N}$. Additionally, each $v \in V(G)$ has a profit and weight: $p_v, w_v\in \mathbb{N}$.

Suppose you're given an $f(k) \cdot n^{O(1)}$-time algorithm $\mathcal{A}$ which finds whether there is a vertex cover $X\subseteq V$ with: $$|X| \leq k,\quad \sum_{v\in X} P(v)\geq \hat{P},\quad \sum_{v\in X} W(v)\leq \hat{W}$$ where $$P(v) = \sum_{s\in \{v\}\cup N(v)} p_s $$ $$W(v) = \sum_{s\in \{v\}\cup N(v)} w_s $$

The task is to find a polynomial-time algorithm that solves Knapsack (the decision version) using $\mathcal{A}$.

My attempts - given n items with profits and weights, $p_i, w_i$:

  1. Define a graph $G$ with 0 edges, where each vertex corresponds to an item. Now run $\mathcal{A}$ on $G$ with $k=n$. Clearly, the problem here is that the running time would be $f(n) \cdot n^{O(1)}$ - depends of $f$ whether the time is polynomial.
  2. In this attempt we'll try to run $\mathcal{A}$ multiple times. Define a graph $G$, where each vertex corresponds to an item. Now we'll consider every possible option for edges between $v_1$ to the other vertices and for each option we run $\mathcal{A}$ with these graphs and $k=1$. Now do the same for all possible edges between $v_2$ and all other vertices, and so on. This will take $n\cdot 2^{n-1}\cdot f(1)\cdot n^{O(1)} = 2^{n-1}\cdot f(1)\cdot n^{O(1)}$ - again, not polynomial.

Is it even possible to come up with a polynomial-time algorithm using $\mathcal{A}$?

Any idea (even if not an exact solution) would be very appreciated!

Chi Pong
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1 Answers1

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Showing such an algorithm exists would prove $P = NP$.

We show that an algorithm $\mathcal{A}$ does indeed exist. An algorithm to decide vertex cover in $O(2^kkn)$ is known. We can modify it for the extended vertex cover from the question:

Ext-Vertex-Cover(G, k) {
    S = new k-sized stack
    return Ext-Vertex-Cover-Rec(G, k, S)
}

Ext-Vertex-Cover-Rec(G, k, S) {
   if (G contains no edges) return (check if S fulfills sum constraints)
   if (G contains ≥ kn edges) return false


let (u, v) be any edge of G
   push u onto S
   a = Ext-Vertex-Cover-Rec(G - {u}, k-1, S)
   pop S
   push v onto S
   b = Ext-Vertex-Cover-Rec(G - {v}, k-1, S)
   pop S
   return a or b
}

Push and pop is $O(1)$. Checking the sum constraints takes additional $O(kn)$ per invocation, keeping overall run-time at $O(2^kkn)$. Thus, for $f(k) = 2^kk$ this is a suitable algorithm $\mathcal{A}$.

Then, since knapsack is NP-complete, finding a polynomial-time algorithm for knapsack using $\mathcal{A}$ would imply $P = NP$.

idmean
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