The sharp-eyed among you may have noted that my denominator 240 solution for f(5) was spurious, since it has 10 parts and not 9. In the meantime I got my cplex program to use their populate feature (find all solutions by exhausting the tree), and it got 18. I noticed that they are the same as found by user "Grizzly" on dxdy. I'm having cplex find all solutions for f(6), but it's taking a while. The f(5) solutions are (after multiplying by 60): (1, 5, 6, 11, 13, 17, 19, 23, 24)/2 1, 2, 3, 3, 7, 8, 12, 12, 12 1, 2, 3, 4, 6, 9, 11, 12, 12 1, 2, 3, 4, 7, 8, 11, 12, 12 1, 2, 3, 5, 6, 9, 10, 12, 12 1, 2, 3, 5, 7, 8, 10, 12, 12 1, 2, 4, 5, 7, 8, 10, 11, 12 1, 3, 3, 4, 5, 9, 11, 12, 12 1, 3, 3, 4, 6, 8, 11, 12, 12 1, 3, 3, 5, 7, 8, 9, 12, 12 1, 3, 4, 5, 7, 8, 9, 11, 12 1, 3, 4, 6, 6, 8, 9, 11, 12 2, 3, 3, 4, 7, 8, 9, 12, 12 2, 3, 3, 5, 6, 7, 10, 12, 12 2, 3, 4, 5, 7, 8, 9, 10, 12 2, 3, 5, 6, 6, 7, 9, 10, 12 3, 3, 4, 5, 6, 7, 8, 12, 12 3, 4, 5, 6, 6, 7, 8, 9, 12 On Sat, Dec 23, 2017 at 1:07 PM, Veit Elser <ve10@cornell.edu> wrote:
On Dec 22, 2017, at 4:34 PM, Victor Miller <victorsmiller@gmail.com> wrote:
I just found a solution to the f(5) problem with denominator 240 (4 times the lcm). This seems to be new:
pieces = [2,9,14,17,21,27,31,34,39,46]/240
I ran a “divide and concur” solver on the f(5) problem. It manages to break symmetry just by choice of the random initial point (the iteration rule is deterministic). Here are the solutions (x 60), with counts, for 50 trials:
{1, {1, 2, 3, 3, 7, 8, 12, 12, 12}} {1, {1, 2, 3, 4, 6, 9, 11, 12, 12}} {1, {1, 2, 3, 4, 7, 8, 11, 12, 12}} {1, {1, 2, 3, 5, 6, 9, 10, 12, 12}} {1, {2, 3, 3, 5, 6, 7, 10, 12, 12}} {1, {2, 3, 4, 5, 7, 8, 9, 10, 12}} {1, {3, 4, 5, 6, 6, 7, 8, 9, 12}} {4, {1, 2, 4, 5, 7, 8, 10, 11, 12}} {4, {1, 3, 4, 6, 6, 8, 9, 11, 12}} {5, {1, 3, 3, 4, 5, 9, 11, 12, 12}} {5, {1, 3, 3, 5, 7, 8, 9, 12, 12}} {6, {1, 3, 4, 5, 7, 8, 9, 11, 12}} {9, {1, 2, 3, 5, 7, 8, 10, 12, 12}} {10, {1/2, 5/2, 7/2, 11/2, 13/2, 17/2, 19/2, 23/2, 12}}
Together with Victor’s denominator-240 solution that makes at least 15 solutions.
My method does not assume a given denominator, and is quite different from what I sketched in my previous email (which turned out to be flawed).
For n=5, f(5)=9, the variables form a real-valued tensor of shape 4x9x2x3x4x5 (so 4320 numbers). In each iteration, two constraint projections are applied to this tensor, designed so that a tensor is a solution iff it is fixed by both projections.
-Veit
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