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I'm trying to understand why exactly the 0/1 knapsack problem actually has the optimal substructure property.

Let $E$ be the set of items to consider and $v$ and $w$ the value and weight functions defined over $E$.

Now, suppose that, among all solutions weighing at most $W$, $S \subseteq E$ is the best solution. How can I prove that $S - \{x\}$ is the best solution weighing at most $W - w(x)$, considering the set of possible items as $E - \{x\}$? I tried to suppose that $S' \subseteq E - \{x\}$ is a better solution (that is, $v(S') > v(S - \{x\})$ and $w(S') \leq W - w(x)$), but I get stuck when I try to find a contradiction.

Any help?

Thanks in advance!

matheuscscp
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1 Answers1

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Suppose that some $S'\subseteq E-\{x\}$ is a better admissible solution than $S-\{x\}$. Then $S'\not= S$, the value is more, $v(S') > v(S-\{x\})$, and $x\not \in S'$, and $w(S')\leq W-w(x)$, as you indicated. Now add $x$ to $S'$ to obtain $S''=S'\cup \{x\}$. What will be the value and weight of this new subset? And how does it compare to the value and weight of $S$? Let's write it out:

$w(S'') = w(S')+w(x) \leq (W-w(x))+w(x) = W$ so $S''$ weighs less than $W$, so $S''$ is an admissible solution, so it makes sense to compare their values.

$v(S'') = v(S') + v(x)$. By assumption, $v(S')>v(S-\{x\})$, so, adding $v(x)$ to both sides of this last equation, $v(S')+v(x)>v(S-\{x\})+v(x) = v(S)$. Hence $v(S'')>v(S)$. But this is a contradiction, because $S''$ is supposedly a better admissible solution while you had presumed that $S$ was the best admissible solution. $\square$

By admissible I mean that the knapsack weighs less than the maximum weight.

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