On 4/16/2011 7:25 AM, Viktor Cerovski wrote: > Here is one partitioning: > > B0=:484 504 284 291 223 80 320 295 > B1=:696 107 371 175 144 650 337 > B2=:111 771 443 954 201 > B3=:937 777 767 > B4=:761 131 733 855 > B5=:772 903 804 > B6=:729 754 758 143 43 53 > B7=:729 322 832 154 443 > B8=:450 215 701 195 918 > B9=:788 698 858 137 > > with the bucket sums: > > >+/&.>B0;B1;B2;B3;B4;B5;B6;B7;B8;B9 > 2481 2480 2480 2481 2480 2479 2480 2480 2479 2481
Nice, Victor! Are you claiming that this is actually the smallest-difference solution to Marshall's 50-mass 10-bucket problem, or that it is just one of the solutions that is fairly close to the minimum solution? Did you examine all other solutions that come close, and selected this one as the minimal-difference solution? Is your approach all programmatic, or are there manual steps involved? How large of a problem will it scale to? Can your approach solve the 200-mass 10-bucket problem I posed earlier? I'm sill looking for a scalable approach to the general #m-mass nb-bucket problem. Do you think that there is a general solution that would scale just linearly with #m? Skip ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
