Talk:Bridge and torch problem

Semi Formal Analysis section
The section appears to be entirely WP:OR and also isn't particularly mathematically rigorous.

This is the vague discussion I remember from reading an analysis of it, I can't find the book it was in as yet.

To optimise we must find the sum. Labelling our 4 people(ABCD) in speed order, the total journeys where they are the slowest member is.

Because we know A=3, j>=1, k>=1 so we can take 5A+3l+m+n as a theoretical minimum and see how much over each of our steps is.

A B C D     SUM      Reduced Sum 0 0 0 5 5A+5l+5m+5n  2l+4m+4n 0 0 2 3 5A+5l+5m+3n  2l+4m+2n 0 0 4 1 5A+5l+5m+n   2l+4m 0 1 1 3 5A+5l+4m+3n  2l+3m+2n 0 1 3 1 5A+5l+4m+n   2l+3m 0 2 0 3 5A+5l+3m+3n  2l+2m+2n 0 2 2 1 5A+5l+3m+n   2l+2m 0 3 1 1 5A+5l+2m+n   2l+ m 1 0 1 3  5A+4l+4m+3n  2l+3m+2n 1 0 3 1 5A+4l+4m+n   l +3m 1 1 0 3 5A+4l+3m+3n  l +2m+2n 1 1 2 1 5A+4l+3m+n   l +2m 1 2 1 1 5A+4l+2m+n   l +m 1 3 0 1 5A+4l+1m+n   l        Possible Minimum 2 1 1 1 5A+3l+2m+n      m     Possible Minimum

So we can see that our minimum depends on whether the gap between 1 and 2 or 1 and 3 is the smallest.

{1,3,0,1} is created by the standard {12,1,34,2,12} and {12,2,34,1,12} solutions and {2,1,1,1} is formed from the various orders of 1 shuttling people across {12,1,13,1,14}, {12,1,14,1,11}, {13,1,12,1,14}, {13,1,14,1,12}, {14,1,12,1,13}, {14,1,13,1,12}. In this case m>l (5-2=3 > 2-1=1) so taking 34 across together in the middle is the swifter solution (15 as opposed to 17). if ABCD were {1,4,6,10) then Shuttling is 22minutes and the standard is 23minutes.SPACKlick (talk) 11:24, 29 July 2014 (UTC)

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