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 Some nice reasoning in a no-op puzzle 
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Posted on: Thu Feb 22, 2018 7:42 pm




Posts: 119
Joined: Sun Nov 03, 2013 10:28 pm
Post Some nice reasoning in a no-op puzzle
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.
Here the puzzle, filled in with everything that should be 'obvious':
Image

Now I'm assuming we're all familiar with the trick of adding up all numbers in a certain row/column (as described here by clm: viewtopic.php?f=3&t=26), most often used in killer sudokus. Somewhat surprisingly though, we can used it in this no-op puzzle as well, something I had never done before.
We can use this trick on the left two columns, which have a total sum of 2*(1+2+3+4+5+6) = 42.
First off, we know that the cages with 11 and 14 must have addition as their operator.
Next, we see that the cage with an 8 must be either addition or multiplication. If we give it addition as its operator, its sum will be 8 (obviously), and if we give it multiplication as its operator, we must fill it with the numbers 1 2 and 4, giving it a sum of 7.
And finally, the cage with the 5 can be addition, multiplication or division.

Now, we look what happens if we fill that 5-cage with addition. The total sum of the left 2 columns without the b1-cell will then be, depending on the operator we use in the 8-cage, 11+14+7+5=37 or 11+14+8+5=38. This means the b1-cell will be filled with either a 42-37=5 or 42-38=4. But both the 4 and the 5 are already used in the top row!
This contradiction means that the 5-cage must be filled with multiplication or division. In both cases however, the only numbers that fit in the cage are two 1's and a 5, so a1=b2=1 and a2=5.
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 too beautiful not to share.


Last edited by jpoos on Fri Feb 23, 2018 11:43 pm, edited 1 time in total.



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Posted on: Thu Feb 22, 2018 8:24 pm




Posts: 317
Joined: Tue Mar 01, 2016 10:03 pm
Post Re: Some nice reasoning in a no-op puzzle
Thanks jpoos. I actually went through something similar in terms of thought process, although I probably didn't get there quite as quickly as you did! No-ops can be quite frustrating sometimes.


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Posted on: Fri Feb 23, 2018 10:53 pm




Posts: 713
Joined: Fri May 13, 2011 6:51 pm
Post 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).


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Posted on: Fri Feb 23, 2018 11:48 pm




Posts: 119
Joined: Sun Nov 03, 2013 10:28 pm
Post Re: Some nice reasoning in a no-op puzzle
Thank you for your solution clm, that indeed seems much easier.
I'm not sure if I figured out that 1-cage in column 6, though I do remember specifically looking at it. I do remember doing some trial and error on those 1's and 2's in the 4th and 5th column, and it was only near the end that I saw my particular solution.


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