Puzzle Description:

Nothin' gets it done quite like a paper clip.

#1: ant (agrest272) on Mar 1, 2010 [HINT]

So if you get stuck after placing 10 black pixels, look closely at column 5. Should solve nicely from there.

I used some edge logic on the top, left, and bottom.

I love it!

Found to be logically solvable by jan.

Delightful.

Ditto #2. Edge logic on the 4s on the top left corner and the 3 in the bottom row was all I needed. I don't see how your hint would help, ant.
Great 10x10 puzzle.

yea I'm not really sure where I was going with that hint.. My hint was assuming you already did the edge logic and the line logic on C9 and R5 and were stuck from there, but I guess if you do enough of these puzzles to recognize edge logic then you don't need a hint beyond that heh.

wonderful :) - nice and simple, yet still required some thinking. it's cool that you didn't go with the predictable vertical or horizontal angles.

so tiny, but yet so clever. very good one!

fun mini puzzle. :)

Nicely done, Ant.

I think the clue in #1 was reversed - it's the 10 *white* in column 10 and the clues in *row* 5 that led me to figure the clues in column 5 and the edge logic in row 10. After those, everything else fell into place.

Fun puzzle, btw, with a nice result.

HELP! I got C3&4R, C1R3&4, C8R7&8 and C7R7 as black. I don't know where to go from here. Please help.

I'm not sure what you're talking about, Ron. Line logic will give you r2c5, r5c2,5,&8 and r7c8 as black, and all of c10 as white.

Edge logic can be done on the left, bottom and top edges. The left and bottom are dead ends. The top will lead on to line logic and victory!

The advanced logic faq page is here: http://webpbn.com/index.cgi?page=solving.html

Specifically in this instance, look at the 4 in r1. Now look at the topmost vertical clues. (It is helpful to press the up arrow to top-justify them.) There are two 1s and the rest of the clues are >1. Therefore we know that it will cause some clues to run into r2.
Consider if we place the 4 all the way to the right. It would cross four 2s. So there would be a block of four pixels blacken right under them in r2.

However, now we must consider row 2. There are two 2 clues, not a 4. Therefore, the 4 in r1 cannot go all the way to right. In fact, it must go somewhere where it will generate no more than two black pixels in a row in r2.

If the 4 in r1 were placed any farther right than c6, it would generate three or more black pixels in r2, so r1c7-9 must be white.

The rest solves with line logic.

[Note: if a number on an edge is larger than any clue next to it, it is a red flag for edge logic.
Note also that 1s are a locus of interest in edge logic.]

Fun logic.

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