The problem:

There are $N$ houses for sale. The $i$-th house costs $A_i$ dollars to buy. You have a budget of $B$ dollars to spend.

What is the maximum number of houses you can buy?

There is an intuitive way to solve this problem in a greedy way: we must at first sort the house costs and then buy houses until we are out of money and it works.

and the proof is:

Let's prove the correctness of this greedy algorithm. Let the solution produced by the greedy algorithm be $A = \{a_1, a_2, \ldots, a_k\}$ and an optimal solution be $O = \{o_1, o_2, \ldots, o_m\}$.

If $O$ and $A$ are the same, we are done with the proof. Let's assume that there is at least one element $o_j$ in $O$ that is not present in $A$. Because we always take the smallest element from the original set, we know that any element that is not in $A$ is greater than or equal to any $a_i$ in $A$. We could replace $o_j$ with the absent element in $A$ without worsening the solution, because there will always be element in $A$ that is not in $O$. We then increased number of elements in common between $A$ and $O$, hence we can repeat this operation only finite number of times. We could repeat this process until all the elements in $O$ are elements in $A$. Therefore, $A$ is as good as any optimal solution.

I don't understand the proof and I have couple of questions:

  1. What does it mean to replace $o_j$ with an absent element in $A$?
  2. Why do we need to do it?

Could you explain this proof in a simpler way with examples?

You can see this problem here.


1 Answer 1


I don't like this proof so much. Here is a different one.

We can assume without loss of generality that $A_1 \leq \cdots \leq A_N$. Consider an optimal solution $\{i_1,\ldots,i_\ell\}$, where $i_1 < \cdots < i_\ell$. In particular, $$ A_{i_1} + \cdots + A_{i_\ell} \leq B. $$ Intuitively, it is clear that $A_1 + \cdots + A_\ell \leq B$ (we will prove this formally in a moment), and so the greedy algorithm will also choose at least $\ell$ houses.

Now let us prove that $A_1 + \cdots + A_\ell \leq A_{i_1} + \cdots + A_{i_\ell}$.

I claim that $i_j \geq j$. Indeed, $$ i_j \geq 1 + i_{j-1} \geq 2 + i_{j-2} \geq \cdots \geq j-1 + i_1 \geq j. $$ This implies that $A_{i_j} \geq A_j$, and so $A_{i_1} + \cdots + A_{i_\ell} \geq A_1 + \cdots + A_\ell$.


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