new puzzle challenge

04042015, 07:02 PM
Post: #33




RE: new puzzle challenge
(04022015 11:11 PM)Paul Dale Wrote: Early pruning of the search is probably also advisable, I suspect it doesn't save a lot of searching but every bit helps. If we are discarding, then the count would start from 3, as 1 and 2 are easy to discard as follows: If a vertex contains 1, the 2 adjacent variables can only be 18 and 19 (left and right, or viceversa doesn't matter), because n(i)+n(i1)>=19. The following corners would have to be: 38181 = 19 38191 = 18 But both numbers are already being used, so we can conclude the number 1 cannot appear on any vertex. Similar deal with number 2, the adjacent can only be 17, 18 and 19, and the following corners would be 17,18 or 19, it's easy to see that we have 3 valid numbers to fill in 4 spaces, so there's no solution, concluding that 2 cannot be on any vertex of the solution. There might be other logical rules like these ones to discard more options that could help a human solve the problem by hand. 

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