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#1: Josh Greifer (joshgreifer) on Mar 19, 2010 [HINT]

The puzzle *can* be solved without guesswork:

The top left horizontal 4 can only go in one place.

After you've put that in, the "3" in the 2 3 2 row is constrained.

I solved this puzzle w/o guess work!

I'm not seeing how the top left 4 can only go in one place. Could you describe how you determine that? Thanks!

Hi Gator, there are five places the first 4 or the 4-4 can go, but all except placing it at the top left, lead to a "red dot" on row 2, or on another row.

If you lay it along columns 2,3,4,5, for example, it leads to this:

12345
1 XXXX
2.X.X. <-- row two (1,2,1,2) is illegal
3 . X
4 .


Similarly with the other positions -- all lead to an inconsistency.

Placing it on columns 5,6,7,8 seems okay at first, but eventually leads to an inconsistency with (if I remember correctly) the fourth row with a singe 2.

Josh

We do classify this type of reasoning as guessing. If you have to look more than 1 move ahead (rarely 2 if it is easy to see), these are not considered logically solvable without guessing.

Okay, I began to suspect that's what it means -- I'm new in these parts. This puzzle is definitely not soluble without thinking through more than one move ahead. Seems a little harsh to give a puzzle a question mark when you need to do that kind of analysis, though. I think an exclamation point would be better :)

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Found to require some guessing by jan.

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i would contend that i just solved it using logic alone without looking ahead to trigger the "requires some guessing". complicated edge logic? yes. guessing? no.

here are the main two spots that broke me away from stuck points:

- no matter where the first 4 in R1 goes, C4R2 will always be red (and must be part of the 2 in R2...that helps later)

- after you have three of the 4 in C13 and the 3 in C 12, look at how the 2 in C12 is going to come in...in R1-7 only one row (R5) ends in a 1, but there are two 1's in C13. this means that one of the ones will trigger the 2, one has to "hide" (meaning not trigger anything) on C13R5 AND the 4 can't trigger the 2 by extending into R7.

I don't know what that logic is called. but it is a form of edge logic. what do you think?

I would venture to change it to "solvable with deep lookahead"

Found to be solvable with deep lookahead by jan.

seems fair

David has some good points.
Here's my solve:

Line Logic gets symmetrical splashes on the bottom corners.

The two 1s in c13 cannot both extend into c12 without a conflict so one must be crossed by a 1. The only place that happens is in r5, so c13r5 must be red.

ll

Then two bits of deep lookahead:
the remaining 1 in c13 can go in three positions:
> r3 > it will start a zig-zag-- going down runs into an error in c11; going up also runs into an error in c11 -- so r3 is bad and must be white.
> r2 > it will start a zig-zag -- going down runs into an error in c11; going up also runs into an error in c11 -- so r2 is bad and must be white.

LL

Edge Logic 3 r9 = c4 w

LL

another deep-lookahead:
no matter where the 4 in r1 goes, r2c4 will be red.

LL to finish.

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