Calcudoku puzzle forumhttps://www.calcudoku.org/forum/ No-op 6 x 6, 10 July 2016https://www.calcudoku.org/forum/viewtopic.php?f=16&t=807 Page 1 of 1

 Author: paulv66  [ Sun Jul 10, 2016 2:31 pm ] Post subject: No-op 6 x 6, 10 July 2016 Having finished all my puzzles for the day, I decided to attempt the above puzzle manually (as I am not a subscriber). I managed to solve it, but it seems to me that there are two possible solutions, which I thought was not allowed.The two different solutions don't affect the numbers that go in each cell, but the operator in the 2 cell in the second row, which can be either minus or divide. Does this constitute a breach in the "unique solution" rule?

 Author: jaek  [ Sun Jul 10, 2016 4:35 pm ] Post subject: Re: No-op 6 x 6, 10 July 2016 paulv66 wrote:Having finished all my puzzles for the day, I decided to attempt the above puzzle manually (as I am not a subscriber). I managed to solve it, but it seems to me that there are two possible solutions, which I thought was not allowed.The two different solutions don't affect the numbers that go in each cell, but the operator in the 2 cell in the second row, which can be either minus or divide. Does this constitute a breach in the "unique solution" rule?The solution must have a unique latin square, but there are times when the operation isn't. I think it always ends up being subtraction vs division. 2,3,6 and 2,4 are two that come to mind.It's been argued that in light of this the operations shouldn't even be required. But it's a way to 'keep you honest', I suppose.

 Author: eclipsegirl  [ Sun Jul 10, 2016 10:52 pm ] Post subject: Re: No-op 6 x 6, 10 July 2016 This is also true of some two squares where multiplication or division would work.It is always when there is a 1 in the solution

 Author: jpoos  [ Mon Jul 11, 2016 12:56 pm ] Post subject: Re: No-op 6 x 6, 10 July 2016 The first time I came across a no-op puzzle, I actually tried to use the unique solution rule to rule out possibilities out, quickly leading me to a dead end. I figured then that that rule didn't apply to operators. This is probably for the best, since if it did apply, there would be quite some unusable combinations, like the ones eclipsegirl mentioned.

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