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 Author: pnm  [ Fri Jan 18, 2013 7:52 pm ] Post subject: harder 5x5 twin puzzles Another one that has been "on the list" for a long time: the "difficult" 5x5 twinsof Wednesday and Thursday were too easy (usually perfectly possible to solvethem by themselves).Starting this coming Wednesday, they should be quite a bit trickier (still doablein a few minutes, but at least not trivial anymore).Below is an example:Your opinions??

 Author: clm  [ Fri Jan 18, 2013 9:35 pm ] Post subject: Re: harder 5x5 twin puzzles pnm wrote:Another one that has been "on the list" for a long time: the "difficult" 5x5 twinsof Wednesday and Thursday were too easy (usually perfectly possible to solvethem by themselves).Starting this coming Wednesday, they should be quite a bit trickier (still doablein a few minutes, but at least not trivial anymore).Below is an example: ...Your opinions?? Interesting idea, Patrick, this is the way, it looks a little trickier, however in this particular case it is possible to solve the full twin puzzle using only the right puzzle, for instance (you can see it quickly), that is, the left puzzle is not providing information at all (in the past I talked in the Forum, IIRC in the thread "The essential info in a puzzle", about the difficulty of designning the twins so interleaved that it would be impossible to solve them without the cooperating info of both sides).

 Author: pnm  [ Fri Jan 18, 2013 9:59 pm ] Post subject: Re: harder 5x5 twin puzzles clm wrote:Interesting idea, Patrick, this is the way, it looks a little trickier, however in this particular case it is possible to solve the full twin puzzle using only the right puzzle, for instanceThanks for checking that, I'll raise the stakes a bit then A slightly harder example (the left puzzle should not be solvableon its own, for example):

 Author: clm  [ Fri Jan 18, 2013 11:36 pm ] Post subject: Re: harder 5x5 twin puzzles pnm wrote: ... I'll raise the stakes a bit then ...A slightly harder example (the left puzzle should not be solvableon its own, for example):... Yes, because the left puzzle by itself can have these two solutions (so not a unique solution):2143552314342514512313542and2415332415532411532441532but the right puzzle is solvable again alone (without the cooperation of the left side): c5 = 5 (addition rule, two leftmost columns) >>> c1 = 1, d1 = 5. Now, since: 9 + b2 + 6 + (15 - c3) = 30 (two upmost rows) >>> b2 = c3 and this quickly permits to eliminate a4 = 2, a5 = 1, b5 = 3 ... driving to the unique a4 = 1, a5 = 4, b5 = 1 >>> b1 = 4, a1 = 2 (e1 = 3), a2 = 3 >>> b2 = 2 = c3 ... finding inmediately the rest of the unique solution.

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