Find solution
How does this solution of the subset sum problem‍​​ work?

Solution to a number problem.?

  • Consider a set S of 15 integers with the following property: For every integer s in S, there exist a,b in S so that s = a + b. Prove that for every such S, there exists a subset T of S with <= 7 elements so that the sum of the elements of T equals zero. I don't have the full proof yet, but have done some research. I was hoping any of you people here can solve it. It would be greatly appreciated. Thanks in advance, Thom

  • Answer:

    If 0 is an element of the set, then for every a ∈ S, a+0 = a, and T = { 0 } has sum of zero, and the rest of the numbers can be anything at all. So we'll consider S sets which don't have 0 as an element. The numbers cannot all be positive, or there would not exist a,b such that a+b = Pmin, where Pmin is the smallest positive number in the set. By the same reasoning, they cannot all be negative. So some are positive and some are negative. Without 0, there must also be at least two of each. Otherwise, let P be the lone positive. All the others are negative, and no two of them would add up to P. Same thing for negatives. Can there be just 2 positives ? Let P1 and P2 be the only two positives. Either they are equal, or one is greater than the other. (Problem did not state if all elements are unique.) It is not possible that they are equal, because then there is no way to find a,b such that a+b = P1. (Cannot add 2 negatives, and don't have 0 to add to P2.) Let P2 > P1. There is no way to get a sum for P2, since only have smaller P1 and a bunch of negatives. So there must be at least 3 positives (and by symmetric reasoning, 3 negatives.) Well ... that's nowhere near a complete proof, but maybe it can help someone get started ... or I'll find something more.

Thom K at Yahoo! Answers Visit the source

Was this solution helpful to you?

Other answers

I found this solution in a book. let F <Z be a finite set of integers satisfying the following properties: 1. for any x in F, there exists y,z in F such that x=y+z 2. There exists n such that, for any natural number 1<=k<=n, and any choice x_1,...,x_k in F, their sum x_1+...+x_k is nonzero. Prove that card(F)>=2n+2 (American Math. Monthly) Solution: By hypothesis 0 is not in F. Let A=elements in F which are positive, B=elements in F which are negative, so that F=A union B. Consider the unoriented graph G whose vertices are in A and whose edges are defined as: x and y are adjacent if there exists z in F such that x=y+z. By 1., each vertex is adjacent to at least one edge. This implies that the graph G has a cycle [x_1,...x_k]. Assume that k is minimal with this property. This means that there exists z_i in F such that: x_1=x_2+z_1=x_3+z_2+z_1=... =x_1+z_k+...+z_1, which implies that z_1+....+z_k=0. The second hypothesis implies that k>n+1 and thus card(A)>=k>=n+1. A similar argument shows that card(B)>=n+1, and the claim follows Comments: 1) I think using graphs is a fancy way to solve it, while it can be viewed in a natural way why there is a cycle. 2) The problem that you ask is still different in a way. It requests to show that there are 7 DISTINCT numbers whose sum is non-zero, while from the problem that I gave, it would result that those numbers are not necessarily distinct. So either you have an error in the conclusion part of your problem or it is even a harder problem. 3) I don't see why he takes k minimal in the solution

Theta40

Related Q & A:

Just Added Q & A:

Find solution

For every problem there is a solution! Proved by Solucija.

  • Got an issue and looking for advice?

  • Ask Solucija to search every corner of the Web for help.

  • Get workable solutions and helpful tips in a moment.

Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.