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#1: julius nadas (jnadas) on Nov 16, 2007 [HINT]

if you use two line edge logic it's not that hard.

it can not be done with single line logic.

you have to figure out where you can put the 5 on the left, knowing that there can only be one darkened square in the next column over. Tricky but not too hard.

A very nice little puzzle. Quite challenging.

Very nice! Loved it

challenging!! i'm afraid i used a bit of guess work with my "edge logic" to complete this one.

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I sure had to guess...how did the computer put this in as "logically solvable"? I did edge logic wherever I could and still got very little.
very tough!

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You have to do edge logic in several places. When you focus on the 5 on the left, pay attention to the 4 clues in the rows.

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I did edge logic on the left, dotting many squares in C1, which gave me most of the 5 (all but 1) in C2. I did edge logic on the far right side as well, getting the upper R corner. A few other scattered pixels placed, and then I was stuck.

Let me rephrase that. After I "got stuck," I did some interesting edge logic. I looked at the bottom 2 rows. 4 sets of 2's under 4 sets of 1's. The columns that end in (consecutively) 4224, 533, and 4554 are key (C6,7,8,9, and 13,14,15 and 17,18,19,20). You can't have any of the 2's from the bottom row be centered under these columns, thus being able to place dots in R 15 at C's 7, 8, 14, 18, and 19). Then you get a little more with edge logic, and finally it solved.

I think this is the best "hard" puzzle on this site, and compound that with the fact that it is a very nice image overall for such a small puzzle, plus it being only B/W, makes it for me, probably one of the best overall puzzles entirely.

I could do the whole thing without guessing but it was exceedingly tricky. Needed some C2 edge logic on the left, as well as some negative-space edge logic on the bottom and right (in other words, I couldn't figure out where, exactly, things went on the bottom and right at first, but I could figure out several places where they WEREN'T). Once I had the empties marked from the bottom and right I could use line logic for the rest, but generally only marking one or two squares at a time per line.

i agree with #12 this was definitely a tough puzzle for the size, probably the best hard one I've done as well

This one is getting filed under favorites for sure. Thanks for a great puzzle!

I just found this one looking for smaller puzzles with a difficulty of 4 and up. Fantastic puzzle. I used the "double-row edge logic" with the columns of 5 and 3 on the left and the 2s and 1s in the bottom two rows.

wow that was amazingly hard and very rewarding at the end
i wish there were more puzzles like this one

great little puzzle. had to start over twice even though i had figured out the double-row edge logic...just ran into the situation where red dots went streaming across the lines at me and the puzzle is small enough that starting over was easier than backtracking. thanks.

simply excelant. It can be solved by simple logic. The key is the third row.

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I just used some internal edge logic on the left to begin, plus some normal edge logic elsewhere.


After line logic gets you a total of four black pixels, internal edge logic on the 5 in c2:

>>Since we're looking at a clue not directly on the edge, we must take into account clues that will reach past the column we're looking at. Here, any clue over 3 or over will reach into c3 and possibly screw up the 1 clue in that column.

Thus the 5 in c2 cannot go above r4 without making a block of two in c3 (because of the 4s in rows 3 & 4). So c2r1-3 is white.

Also, c2r4 must be white for if it were black, we would end up with two blocks of one in c3 (from the 4 in r4 and the 4 in r7).

>>Internal edge logic can also take into account clues on both sides. Since we're looking at c2, we can look at the clues in both c1 and c3 to see if the horizontal clues can fit.

This one can be harder to notice, but the 5 cannot start in r5 either. If it did, then since the 4 in r7 will go into c3 and fulfill it, the 2s in r5&6 must go to the left. So far so good, as that will fulfill the 3 in r1. The 1 in r8 is fine. However, the 2 in r9 throws a wrench in the works. It cannot go either left or right as both those columns would have no more space. Therefore c2r5 is white.

Then c2r6&7 must be white as the 5 would cross two 4s, screwing up c3.



After a tiny bit of line logic,
edge logic on the left 2 in r15 makes c1 white.

Then a decent amount of LL

Edge logic on the 2s in r15 can make white c7-8, c14 & c18-19. Then LL.

Edge logic on the 3 in c22 makes r8-10 white.

Then LL to finish.

Fun.

A few spots of edge logic where you have to look at 2 rows at once. After you do the last one the solve is very fast and simple.

great images for such a small puzzle! Nice job!

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