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Problem

Given a range of integers $\{a,a+1,...,b-1,b\}$, find a subset of size $k$ such that the sum is equal to $s$.

Question

This problem came from evaluating some scheduling algorithms that I am interested in optimizing for some small home grown useless embedded system I am playing with. My problem is that I do not know if this problem is NP-Complete like the K-Sum problem. I am guessing it might be but it has been a while since I have dealt with proofs pertaining to NP problems. I remember something with SAT, but looking around did not jog any memories (at least any good ones).

How might I prove it is or is not NP-Complete?

tkellehe
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2 Answers2

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Nice question!

Given a range of integers $\{a,a+1,...,b−1,b\}$, find a subset of size $k$ such that the sum is equal to $s$.

Well, this problem can be solved in $O(n)$ time, where $n$ is the number of given integers.

(Continuity of $k$-sums). Given integers $\{a,a+1,...,a+n-1\}$, an integer $k$, $1\le k\le n$ and an integer $s$, there are $k$ integers whose sum is $s$ if and only if $ka+\dfrac{(k-1)k}2\le s\le k(a+n-1)-\dfrac{(k-1)k}2$

Here is the algorithm.

Check if $ka+\dfrac{(k-1)k}2\le s\le k(a+n-1)-\dfrac{(k-1)k}2$. If not, return none.

Let $sum = s$. Let $S$ be an empty set. For $i$ from 0 to $n-1$ do the following.

  • Let $temp = sum -(b - i)$.
  • If $temp\le (k-i-1)(a+n-1)-\dfrac{(k-i-2)(k-i-1)}2$, let $sum = temp$ and insert $b-i$ into $S$. Otherwise, do nothing.
  • If $sum = 0$, break the loop.

Return S.

Here are a few related exercises.

Exercise 1. Prove the proposition of continuity of $k$-sums.

Exercise 2. Show the algorithm is correct.

Exercise 3. Modify the algorithm so that it will try adding the smallest number first.

Exercise 3. Modify the algorithm so that it will run in $O(k)$ time.

John L.
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I suspect a polynomial-time algorithm exists. WLOG, you can assume $a=0$ (subtract $a$ from all numbers in the range, and subtract $ak$ from $s$). Then the problem becomes: given a range $\{0,\dots,b\}$, find a subset of size $k$ that sums to $s$. If you allow a multiset, it is easy to come up with a polynomial-time algorithm; take $k-1$ copies of $\lfloor s/(k-1) \rfloor$ and one copy of $s - \lfloor s/(k-1) \rfloor$. I suspect similar ideas can probably be applied to find a set as well (i.e., to avoid repeats). I suggest you spend some time working out the details to see.

D.W.
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