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#1: Joe (infrapinklizzard) on Oct 2, 2010 [HINT]

Purple logic goes pretty far.

There are a few ways to go forward using edge logic, but the key to this puzzle is two-way logic.


Sooner or later you will have to do two-way logic on the right red 1 in r9.

If it is in c13 then r10c13 is white. If it is in c15 instead, then r10c15 would be red which would make r10c13 white. Either way r10c13 is white.

The rest will then solve with line logic.

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This would have been a good entry in my Versus series.

got stuck but the hint from Joe helped and than it was real easy. thanks Joe.

i just thought it was fun to solve

Me too Tom.

Good solve Joe

That was a fun challenge. I was able to figure it out without looking at the hints. Yeah! :-)

I still struggle with the idea that two-way logic seems like trial and error. The fact that it's a 50/50 proposition rather than one greater than that doesn't lessen it for me. I'm not sure why it would be more acceptable than "pattern logic" - seeing that the solution follows a particular pattern and then solving according to that.

Not trying to start an argument, just expressing my thoughts.

Found to be logically solvable by Gator.

Nice puzzle Joe. I did the two-way logic with the 2 clue in row 4 first before I saw the one on the right side.

Martin, I know we've butted heads on this before, but your comment in this puzzle has made me think of a new way to explain.

Please forgive if I over-explain, but I want to make sure my assumptions are stated so we start on the same foot.

There is only a 50-50 chance if we were to mark one of those two pixels with no reason and then continue on. That is not what we are doing.

Two-way logic never actually gives you information about the two pixels about which you are hypothesizing. What it does do is give you information about zero or more *other* pixels. (Of course, zero is the worst-case scenario, when it fails.)

Two-way logic can only be used when you have two positions (let's call them A&B) and you know that one-and-only-one of them can be "black". (Or any color but white).

What is happening with two-way is that we look for the direct consequences of placing a pixel in position A. Then the same for position B. If any of the consequences are the same for *both* positions, we know that that pixel *must be* that consequence.

Note that by definition the original two pixels are excluded from having the same consequence.

Specifically, in this puzzle (after all line solving): A is c13r9 and B is C15r9. We know that one and only one can be red because of the red 1 clue to the right of the blue 2 in r9.

If A, then:
c13r10 will be white (since "A" is the red 1 under the blue 3 in c13), and
c15r9 ("B") will be white (since "A" is the rightmost red 1 in r9), which causes
c15r12 to be red (part of the red 3 in c15).
That's about as far as we can go in our head.
If B, then:
c13r9 ("A") will be white (since "B" is the rightmost red 1 in r9), and
c15r10 will be red (as part of the red 3 in c15), which causes
c13r10 to be white (since c15r10 is the rightmost red 1 in r10).

C13r10 will be white if either A or B is red. And since one of A or B MUST be red, c13r10 must be white.

This has been proven without making a mark on the page, so it counts as a logical move.

I hope this helps, and if not, I hope at least I didn't irritate you.
~Joe

I had no problem solving this. It was disappointing since I don't have a clue as to what it is.

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