[Magictour] Re: on rectangles

OK, I reformulate:

is a knight's tour possible on an 8*10 , where all row-sums
leave rest one when divided by 4 and all column sums are divisible
by 4 ?

a magic 8*10 rectangle must satisfy this (row-sums=405,column-sums=324),
but above condition is much weaker and it should be easy to construct
such a tour - if there is one.


OK, I looked at the first knight tours on a 8x10 . Tours satisfying the above
conditions
are very rare, but do exist.

here are the counts:



number of rows with row-sum mod 4 = 1
number of rows with row-sum mod 4 = 3
number of columns with column-sum mod 4 = 0
number of columns with column-sum mod 4 = 2
count of such tours



 0 8 10 0 ,1
 0 8 2 8 ,43
 0 8 4 6 ,327
 0 8 6 4 ,591
 0 8 8 2 ,101
 2 6 0 10 ,15
 2 6 10 0 ,51
 2 6 2 8 ,1196
 2 6 4 6 ,9488
 2 6 6 4 ,11982
 2 6 8 2 ,2982
 4 4 0 10 ,37
 4 4 10 0 ,176
 4 4 2 8 ,3485
 4 4 4 6 ,23145
 4 4 6 4 ,30327
 4 4 8 2 ,7517
 6 2 0 10 ,20
 6 2 10 0 ,90
 6 2 2 8 ,1294
 6 2 4 6 ,9908
 6 2 6 4 ,12813
 6 2 8 2 ,3510
 8 0 10 0 ,1
 8 0 2 8 ,30
 8 0 4 6 ,331
 8 0 6 4 ,673
 8 0 8 2 ,159


here is the tour 8 0 10 0,1 , the tour which answers my above question
to the
affirmative , the tour which proves that the (mod 4 ) arguments from
theorems 1..4
are not sufficiant to prove that no 8x10 magic tour exists.
The counts could also suggest, that 8*10 tours may exist, but are very rare.
I only searched closed tours starting in the corner, though.


 1 38 15 64 13 36 27 58 11 34
 16 73 80 37 28 65 12 35 26 57
 39 2 63 14 75 54 59 56 33 10
 72 17 74 79 66 29 76 53 46 25
 3 40 67 62 77 52 55 60 9 32
 18 71 78 51 68 61 30 45 24 47
 41 4 69 20 43 6 49 22 31 8
 70 19 42 5 50 21 44 7 48 23


GS



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Received on Thu Dec 04 14:24:18 2001