Puzzle Description:

homer

#1: Adam Nielson (monkeyboy) on Mar 12, 2009 [HINT]

This is a nice image overall, but VERY difficult. I used edge logic (trial and error) over and over practically the entire puzzle, starting with the 5 on top, then the 4 in the first column, and then over to the 3 on the right, and back to a 2 on the left, etc. I think this is definitely worth fixing up just a tad, to make it more logically solvable. Just a few black pixels here and there would do it wonders!

ditto #1. I used trial and error lots until things just started working.
tough but nice image.

Found to require much guessing by jan.

I agree with the comments above. Nice image, but it doesn't really work as a puzzle. It could use some more work to make it really solve well.

I think that the way the profile was done is actually a unique and fun way to do a portrait, however like it's already been said, a couple extra pixels could make this solveable.

Okay, so I used the "Peek at Solution" feature to do this one. Anyone who wants to do so without ruining the final product: the bottom row was the key to unlocking the rest of it logically. After I did all I could figure out, placing those two 2's made it solve easily.

The bottom row is 3 dots, 2 black, 3 dots, 2 black.

Wow. I can't believe I wrote such a nice comment in #1. :-)

ditto #6

as has been said this is a very nice image that should be fixed to make solvable

I do not agree with all of you-it is solvable by logic alone
Start with a 4 on the left,than move to 9 on the bottom,
than 3 on the right.After than line logic would do.
Good puzzle.

ditto #7 ;)

...i have to say that it was tough, but "logically" worked this one too...with a sort of internal edge logic. once line logic and a whole bunch of more traditional edge logic were exhausted I looked at the 5 in C4 and how the combination of first numbers in those bottom rows could stair-step off of it...basically, no matter where you slid that 5 and eye-balled where the numbers greater than 1 would sprout it will HAVE to trigger the 2 C5. Doesn't matter where that 2 ends up, just knowing that it is down there puts a bunch of dots in the upper half of C5.

I used edge logic on the left and right sides, and it solved easily. Now I want to sit back with a frosty Duff.

@ #10: I think you made an unwarranted assumption somewhere.
Edge logic on the 4 in c1 makes c2-3, c10-12 and c17-18 white, but makes nothing black.
The 9 in r16 cannot be narrowed by edge logic at this point.
The 3 in c15 can only make r17-18 white with edge logic.

To continue on from there,
EL 5 r2 = c2-5, c13-15 white
LL
EL 2 r3 = c14-15 w
LL
NOW edge logic on the 3 in c1 can rule out r13-16 since it would make three blocks in c2 where there's only two clues.
LL

Extended edge logic on the 2 in r18:
If it were in c12-13, then it would take up the 1s in those columns. So the black pixels in r16 would have to be the bottom pixels of the second-to-bottom clues in c12-13.
That, however would make a block of three in r15 where it could not be. Thus the 2 in r18 cannot be in c12-13.

LL

Extended edge logic 5 r2:
If it were in c6-10, then the 4 in c7 would trigger the 2 in r4. However, the forced white in r2c11 would make the 2 in that column go down into r4 making a block of two. That's a conflict since there's only one 2 clue left. Thus our assumption was wrong and so r2c6 must be white.

LL
(note that in r5, c9 must be white due to LL,
and in c5, r7 must be white due to LL)

That's all I see for moderate lookahead.

I also think it can be solved logically.

A case could be made for Deep Lookahead. There are a few options to place the initial edge logic, and a bit of mental lookahead will quickly eliminate the wrong choices.

I encourage you to go ahead and make the case. Anyone can post a detailed solve, and if it checks out then the rating will be changed.

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