[Magictour] Re: Bounds

>Yes, we start with 64, and decrement at every jump.

that's uncommon. Can't you change it ?

>You can also view the problem as starting with 1, and getting the smallest
>possible sum:
>01 00 03 00 05 00 07 00 -> sum=1+3+5+7
>00 00 00 00 00 00 00 00
>00 02 00 04 00 06 00 00
>
>We can get:
>
>01 00 03 10 05 00 07 00 -> sum=1+3+5+7+10=26
>00 00 00 00 08 00 00 00
>00 02 09 04 00 06 00 00
>
>01 00 03 10 05 00 07 12 -> sum=1+3+5+7+10+12=38
>00 00 00 00 08 11 00 00
>00 02 09 04 01 06 00 00
>
>01 00 03 10 05 14 07 12 -> sum=1+3+5+7+10+12+14=52
>00 00 00 00 08 11 00 00
>00 02 09 04 01 06 13 00
>
>Are 26, 38 and 52 the lowest possible sums with 3,2 and 1 empty square ?

how will you ever complete these to get 260 ??

>If yes, these values will speedup the search !
>
>>Completable means to me, that the empty squares can be filled later
>with numbers bigger than the largest number in the row.
>>to solve(approximate) it numerically,
>>you could try to generate all SMKTs where you ignore the sums of the
>columns
>>and then find the maxima with 3,2,1 of all their rows.
>>It should be sufficient to generate a lot random such {rows=260}-tours
>
>Of course, finding the best sum with real tours is great, but can you prove
>that your MonteCarlo method is correct ?

a good evidence is not so much than a proof

>>do you have an estimate, how long it will take ?
>
>I have no idea for the moment !
>I think we need more computation results from Hugues, in order to define
the
>best strategy to find one.
>
>Perhaps can we start with a closed tour (which one ?), or is there a
special
>first move to be sure to get at least one solution in reasonable time ?

tours starting at b3 are usually faster to complete, since the first
3 moves are forced

>Perhaps the start/end approach is not the best way to work on this problem
?

let's fix the first k moves


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Received on Thu Dec 04 14:24:18 2001