Re: Some nice reasoning in a no-op puzzle

jpoos wrote:

On last sundays no-op puzzle (18 feb), I noticed some quite beautiful reasoning I had never used before, so I think it's worth sharing. ...

The rest of the puzzle will be easy from here on out.

There might have been easier ways to solve it, but I thought this was just to beautiful not to share.

Thank you. This is a perfect reasoning and I agree it's beautiful since it shows how many times we "overlook" the internal structure and the beautiful properties of the puzzle.

I often use this type of procedure (adding cages or combination of cages then eliminating possibilities) to arrive to the solution, specially in the no-op, the "fm 0", the bitwise OR and the exponentiation puzzles.

Anyway, for this particular puzzle, I would like to add a few comments. I departed from the same preconfiguration as yours but also

**adding the pair 4-5 for the cage "1" (1-) in column 6**, what will make very easy the solution: Here 1-2 is not possible because there is no place for a 5 in column 6, 2-3 is not possible for the same reason, 3-4 >>> 5-2 and 1-6 for the "3" cages, and 5-6 >>> 1-4 and then 2-3 for the "3" cages, incongruent combinations.

And once we have this, we quickly see that d5 is not a 1 because >>> c5 = 6, c6 =4, e5 = 2, e6 = 1 and f6 = 2 and that would force f5 = 3 making incongruent the cage "3" in f5-f6.

In these conditions d5 must be a 2 >>> d4 =1, c5 = 5, c6 =4, e5 = 1, e6 =2,

**f6 =1 and necessarily f5 = 3** (since a 6 is not possible). Thus cage f1-f2 = [2,6] (operator :). Also a5 = 4, b5 = 6, b4 = 4, f4 = 5, f3 = 4, d2 =4, d3 = 6, and a6 = 6, b6 = 5.

Next: cage "6" in row 1 is not a sum so

**it must be a product** (or a division with the same result of multiplying to 6), and using the multiplication rule in row 1, a1 x f1 = 6, but a1 = 3 (f1 = 2) is not possible so a1 =1 and f1 = 6 (f2 = 2). Now b3 cannot be a 2 (being cage "8" a sum and in order to have a3 + a4 = 6) so b1 =2, c1 = 3. Obvioulsy b2 cannot be a 3, so b3 = 3, b2 = 1, a4 = 3, a3 =2, a2 = 5, e2 = 3, c2 = 6, and the other 4 numbers are inmediately set.

In this particular puzzle, following this process makes all in some way "automatic" (once we have the additional information for the cage "1" in column 6).