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Sujiko http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=13&t=1501 |
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Author: | azpaull [ Tue Aug 06, 2019 6:34 pm ] |
Post subject: | Re: Sujiko |
Thanks! Will give it a go this weekend! |
Author: | Andrew [ Sun Aug 11, 2019 7:53 pm ] |
Post subject: | Re: Sujiko |
Thanks HATMAN for this fun puzzle. Here is how I solved it: Solution: Note: |
Author: | HATMAN [ Mon Aug 12, 2019 6:31 pm ] |
Post subject: | Re: Sujiko |
Puzzle Question I've just published this question on the players forum (where they are into theoretical analysis) however, I'm not sure how many of you visit there, so: If the 25 sums give a feasible solution is it automatically unique in all cases? I.e. necessary is the same as sufficient. Whether it is solvable without heavy number crunching is a separate matter - although even then the analysis is only of spreadsheet level. I am attempting currently to create a counter example. |
Author: | azpaull [ Tue Aug 13, 2019 11:34 pm ] |
Post subject: | Re: Sujiko |
Thanks, HATMAN! That was a fun change - especially after your correction! |
Author: | HATMAN [ Wed Aug 14, 2019 7:22 pm ] |
Post subject: | Re: Sujiko |
I've been spending too much time on counter examples but it is definitely not unique when feasible in all cases, but I am sure that in the vast majority when clues are not symmetrical it is. Where the 25 sums are all 20 then then the four 20 in a nonet give seven solutions plus symmetries: C1 C2 C3 C4 C5 C6 C7 C8 C9 1 8 3 9 2 7 4 5 6 1 6 7 9 4 3 2 5 8 3 5 2 8 4 9 7 1 6 1 6 7 8 5 2 3 4 9 2 5 8 7 6 1 3 4 9 3 2 7 9 6 5 4 1 8 4 3 7 5 8 2 6 1 9 So centre values of 1 3 7 9 have no solutions 2 5 8 have a single solution and 4 and 6 have two slightly linked solutions. Using the 5 centre set in the 4 nonets we can form a solution with the four centre cells 1991 and another with four centre cells 3773, plus of course their symmetries. There may be other solutions using mix and match but as I am doing this semi-manually, life is too short. |
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