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#1: Kristen Vognild (kristen) on Dec 29, 2011 [HINT]

Again, edge logic in the first row to get started.
I enjoyed solving this one.

Absolutely no guessing.

Nice solve. There's lots of white to find and mark before the black can get anywhere.


Edge logic on r1 gets you c6-8 black and c1-3 white.

Edge logic on c10 gives you r4-5 black and r9-11 white.

Then look at r1 vs r2. The 5 in r1 must go fully to the left or to the right. If it were to straddle, it would create two blocks of black in r2 - impossible.
- if it were to go all the way left r2c1-3 & c9-10 would be white
- if it were to go all the way right, r2c1-5 would be white
Either way, r2c1-3 are white.

Edge logic on the 3 in r3 now makes c1-2 and c7 white.

Edge logic on the 5 in r1 again makes c4 white

Line logic to finish.

i did the same first 3 steps as lizzard

but then i did 2-way logic on the 5 clue in c10 (no matter if it goes all the way up or all the way down down [just like the 5 clue in r1, it also cant straddle the dots in c9] r9-11 are dots)

continuing the 2-way logic further on the 5 clue in c10, the 2 clue in c8 will now either go only in r3-4 or r7-8 (dot the rest of c8)

then i did edge logic on the 3 clue in r3 to get dots in c1-2 (i dont see how you got a dot in c7 which definitely would make this a lot easier to solve the rest)

2-way logic on the 5 clue in r1 makes r3c3-4 dots
then either of the following two-move look ahead steps:

1) extended edge logic on the 5 clue in r1 to make c4 (and therefore c5 as well) both dots (as mentioned by lizzard, but without the dot in r2c7 it takes an extra step)

or

2) if the 3 clue in r8 goes all the way to the right, two moves later the 4 clue in c3 cant be placed so the 5 clue from c10 must go up into r1

nice solve
no guessing

Found to be logically solvable by gator.

I did edge logic in row 1. Then two-way logic to make R2C1-R2C3 dots. Then edge logic on the 3 clue in row 3 to make R3C1,R3C2, and R3C7 dots. Then I saw that if the 5 clue in row 1 extended to the left it would cause row 3 to be invalid. So it had to extended to the right (as mentioned it has to go all the way left or right). Line logic to finish.

Nice puzzle. Looks like we tackled this one in similar ways.

Like Adam, I don't see how edge logic on the 3 in R3 gives a dot in C7. But something along the lines Adam described worked for me.

Technically, it is extended edge logic on the 3 in r3.

My layout of the full logic:
In order for c7r3 to be black, one of the three pixels in the 3 must be there. This is possible in three placements of the 3 in that line: c5-7, c6-8, & c7-9.

Regular edge logic rules out c5-7 and c6-8 as they create a block of three in r4.

Extended edge logic rules out c7-9. Even though it is ok as far as r4 is concerned, it forces r2c9 to be black. That places the 2 in r2 and forces r3c10 to be black, conflicting with r3.

(I guess that would be deep-lookahead as it goes one step past the normal extended edge logic. I take comfort in that Gator also used it.)

So, since all three possible positions using c7 are falsified, r3c7 must be white.

Weak.

You should lie down, then.

*snerk*

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