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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: 135
Joined: Sun Nov 03, 2013 10:28 pm 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': 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. Posted on: Thu Feb 22, 2018 8:24 pm

Posts: 461
Location: Dublin, Ireland
Joined: Tue Mar 01, 2016 10:03 pm 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. Posted on: Fri Feb 23, 2018 10:53 pm

Posts: 735
Joined: Fri May 13, 2011 6:51 pm 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). Posted on: Fri Feb 23, 2018 11:48 pm

Posts: 135
Joined: Sun Nov 03, 2013 10:28 pm 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. Display posts from previous:  Sort by Page 1 of 1 [ 4 posts ]

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