Puzzle Description:

... Logic

#1: BlackCat (BlackCat) on Oct 12, 2010 [HINT]

It was doable with logic. Without the title though, I don't think I would have figured it out. Very clever.

Found to be logically solvable by Gator.

If someone really wants me to detail out all of the edge logic required to solve this one I will.

There is a lot of internal edge logic in this one. I sometimes had to do some affirmation logic, but I'm sure it was completely required or not. Great puzzle!

That was really hard. It took me forever to find that linchpin (as I think of it) that helped me "turn the corner" in the puzzle. I think it was figuring out that I needed to go in three columns in from the right when looking for the placement of the 4 in c20.

Glad you liked it!

I used some "guess" edge logic based on the patterns of numbers...solved quickly if you looked where there was a long string of ones followed by 2's or more. Well until you got to the bottom...but by then regular logic filled in the rest.

i loved it. absolutely fun to solve. cute image at the end.

tough. I had to look about three moves ahead or do extended edge logic in order to finish (actually it was near the beginning) this one....so what did I miss? Gator can you help?
It solved pretty easily after that though.

cool puz, Kai

This may not be the best or easiest way to get to the solution, but this is how I did it.

Details (you need to do line logic in between):
I first did the edge logic on the top row to find that R1C1 and R1C16-R1C20 were dots.

Next you can use edge logic on column 1 to make R2C1, R3C1, and R15C1-R20C1 dots.

I next tackled the 5 clue in row 20. With edge logic you can make R20C2-R20C4 and R20C16-R20C20 dots.

Next I looked at the 4 clue in column 20. With edge logic, you can make R16C20-R19C20 dots.

Next I looked at the 7 clue in row 19. With edge logic, you can make R19C16-R19C19 dots.

There are a few others places to do edge logic at this point, but I used affirmation logic to get me going at this point. Look at the 5 clue in row 20 again. Look at how the placement of the 5 clue affects the 7 clue in row 19. What we are looking for is things that are in common in row 19 no matter where I place the 5 clue in row 20. In every case, R19C2 and R19C3 will be dots. Also, R19C8-R19C10 are always filled in.

I then used internal edge logic with the 5 clue in column 3. Starting at the lowest place you can place it (R18C3), look at how rows 14-16 are affected. It forces R14C4 and R16C4 to be dots and R15C4 to be filled in, but this contradicts column 4. This lets us make R16C3-R18C3 dots.

I then used edge logic with the 3 clue in column 4 to make R18C4 and R19C4 dots.

I then used simple edge logic to make R18C19 and then R14C20 and R15C20 dots (I'll let you figure out what clues I used to do that).

I next looked at possible places I could put the 5 clue in row 18. Starting at the left, the first place would be R18C8. Notice how this affects the 5 clue in row 20 though. This will make R18C8 and R18C9 as places that the 5 clue cannot go (but you cannot make them dots yet as they could be part of the 2 clue). So now that 5 clue in row 18 can start in R18C10. This lets us fill in R18C14.

I then looked at the 2 clue in row 15. If it is to go all the way to the right, then it would have to go in R15C16 and R15C17. But looking at how this affects row 18, it cannot go here. So the 2 clue must be part of R15C12 which is already filled in. This also tells us that the 3 clue in the same row must go to the left. So R15C4 gets filled in and R15C10 is a dot.

At this point I was having trouble finding a next move until I took a closer look at column 1. Again we are going to use affirmation logic. Notice that no matter where we finish out the 6 clue, there are a lot of cells in column 2 that will always be dots. Either R9C2 or R13C2 has to be the 1 clue in column 2. Examining the affects on row 2 when we try to place the 6 clue, we will find R10C2-R12C2 and R14C2-R16C2 are always dots. Now a lot more line logic can be done.

Next simple edge logic in column 20 makes R6C20 and R7C20 dots. The rest will finish out with line logic.

I probably could spend more time and condense down the logic outlined above to fewer steps, but I think this puzzle is a good tool for practicing both edge and affirmation logic so I'll leave it.

Thank you very very much for the detailed description, Gator! I wanted to make this as a edge-logic-practice for everyone. Glad it worked this way.

What a cool puzzle, Kai. I knew it was going to be difficult but logical, so when I got stuck, I just kept looking for clues. I was able to solve it on my own, which surprised me, because I'm not very good at internal edge logic. Really cute image too, which made it very enjoyable. :-)

T... still waiting

So hard! I believed that it was logically solvable, so I persevered, and had fun. :) I feel like I've accomplished something now.

Excellent! Really enjoyed it!! Thanks!

Wisconsin has a face!

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