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19 March 2009 @ 12:02 am
03/19/09 - Divide and Conquer  
6x6 KenKen*

Yesterday's puzzle was a bit harder than typical for Wednesday, in part because I believed there was a simpler solution path than there actually was (each week I seem to prove at least once that I'm not mistake free - sorry). However, the goal yesterday was to make a puzzle that featured subtraction heavily and that certainly was true whether it was harder than intended or not.

This puzzle does the same with division with a pseudo-symmetric diagonal made up of several large divisions. Until I post a second variation that reasonably allows all cells to have division occurring, this is likely about how densely you can use division in a KenKen* without having a trivial puzzle or using lots of two-cell cages with division by 1.
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devjoedevjoe on March 19th, 2009 12:58 pm (UTC)
With my long solution yesterday vindicated, here's today's somewhat easier solution:

The 4-cell 3/ must be 6/2/1/1, so each row and column within it contains a 1 and either 6 or 2. The 13+ must contain a 6, so the first row of the 3/ contains 2 and 1, and by summing the whole row, we see that the top of the 3- is a 5, so the bottom is a 2. This also means the 13+ is 6+4+3.

Then the 1- in column 1 must be 4-3, and the 3-cell 3/ has 1,6 in the first column, so must have 2 in the other cell. (And the free 6 lets us place the 1 and 6 in the 3/.) The 3-cell 1- now can only be 5-3-1, so place the 4 in the last cell of the row.

The 2/ clues cannot contain a 5, so the only place for a 5 in column 2 is the bottom of the 2- (so the top is a 3). Then the only place for a 1 in the column is the top of the 2/, and the rest must be 4,2 and a 6 goes in the top cell of the column.

Now the only place for a 5 in row 4 is the last cell. This forces the rest of the column, the 4-cell 3/, and rows 5, 2, 1, and 6. The other 2/ can be filled, then the rest by elimination.

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