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No-op 6 x 6, 10 July 2016
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Posted on: Sun Jul 10, 2016 2:31 pm

Posts: 306
Joined: Tue Mar 01, 2016 10:03 pm
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?

Posted on: Sun Jul 10, 2016 4:35 pm

Posts: 227
Joined: Fri Jun 17, 2011 8:15 pm
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.

Posted on: Sun Jul 10, 2016 10:52 pm

Posts: 304
Joined: Wed Apr 16, 2014 9:20 pm
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

Posted on: Mon Jul 11, 2016 12:56 pm

Posts: 113
Joined: Sun Nov 03, 2013 10:28 pm
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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