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motris
motris
08 April 2009 @ 08:24 pm
Spot The Differences (Redux)  
So tonight I'm putting together a packet of "Not Sudoku" Puzzles for the WSC in two weeks (where it seems both Wayne Gould and Maki Kaji will be making guest appearances).

In putting together this hand-out, I realized it was time to fix the one puzzle that I had made overly difficult based on an error in my multi-cell subtraction considerations. I've made a much better subtraction-only puzzle that actually has a fair solution path.

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Stable Strangelet: believe[info]cuthalion on April 9th, 2009 04:42 am (UTC)
I got it, but I had to just guess and see where it took me (without realizing it! I thought it was the only possibility, but I missed an alternative - when the path I tried yielded a contradiction I realized the overlooked option must be right.
zundevil[info]zundevil on April 9th, 2009 06:18 pm (UTC)
I realize it should count as the default, but do you think putting something on the pdf about how digits can be repeated in cages would help? It goes without saying on the puzzle with the super huge multiplication cage (among others), but could alleviate possible confusion (on puzzles like this one, for example).
motris[info]motris on April 9th, 2009 06:24 pm (UTC)
Sure - that's good. I've fixed the rules and will have that for ~50% of the printouts since I've already assembled 50 or so.
Robert Hutchinson[info]ertchin on April 12th, 2009 01:10 am (UTC)
It took me a good ten minutes of staring, but I finally found (what I would consider) the big "in".
motris[info]motris on April 12th, 2009 01:36 am (UTC)
Great! Since no one has done it yet, here is the quick walkthrough from me:


The 3- in the rightmost column forces the bottom two cells in the three-cell region to be a 3- as well (14), (25), or (36), and with any of these for 2- overall, the extra cell in column 5 is a 1. (This part of the walkthrough should be familiar from before).

Now, the fifth column has another pair of 3-'s which makes the top value, R1C5, a 4. The remainder of the puzzle can be solved by recognizing that two 2- cells in a row means one must be "even with a 4" and one "odd with a 3". The 2- with the 4 in row 1 has a 2 or 6 left, but the other one must be 3 and 1 or 5. However, the second column must have the same "odd with a 3" and "even with a 4" in its 2- cages, so R1C3 must be 3 and R6C2 is 2 or 6.

With the 3 in R1C3, you can show column 3 has the 3- as {14}, the 1- as {56}, and the bottom cell is 2 (making R6C2=6). It trickles out from there, but the placement at R1C3 is the "improved" solution path I intend with an interesting but hard deduction.

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cyrebjr[info]cyrebjr on April 15th, 2009 01:03 am (UTC)
You switched the 2 and 6 in the last row. No biggie, though.