
Posted by Denis Borris on 16/9/2009, 4:45 pm, in reply to "impossible for m=2"
Yep; agree 100%. Only gets worse:
you can have say a,b,c,d,e,f,g,h,i = 1,2,7,8,9,10,11,12,13
2c = 14
b+d = 10
b+g = 13
So the trips start to end would be:
1,9,10
1,12,13
1,8,11
1,2,7
That's 41 + 3 (returns of a) = 44
I guess the only way for me to have properly
stated the puzzle is with something like:
"after arranging in ascending order, each trio
(positions 4-6, 7-9 ...) must cross together".


Message Thread:
![]()
« Back to thread