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#1: paul dahmer (paul) on Nov 16, 2007

solvable - no guessing required

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This is a very cool puzzle. I used edge logic on the 6 clue in column 1. It can be completely placed by looking at how it causes contradictions in column 4.

This will allow you to solve most of the puzzle, but you are left with 12 red cells and 12 dots that have yet to be placed. Looking at the top half of the puzle, you need a red in either R5C13 or R5C17, either R6C12 or R6C20, either R9C7 or R9C13, either R9C17 or R9C23, either R10C7 or R10C12, and either R11C20 or R11C23. That is 6 reds you have to place when accounting for the rows.

Now let's look at the columns. You will need a red in either R9C23 or R11C23, either R6C20 or R11C20, either R5C17 or R9C17, either R5C13 or R9C13, and either R6C12 or R10C12. That is 5 reds you have to place when accounting for all columns except column 7.

Given you have to place 5 reds in the columns and you have to place 6 reds for the rows, this only leaves 1 cell that can be filled in for column 7. Therefore, R9C7 would have to be a dot and R10C7 would have to be a red.

This is a great example of the summing technique.

The rest solves out normally.

Found to be logically solvable by Gator.

Definitely fun to solve.

Cute title, too

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