Puzzle Description:

An igloo: hemisphere of the North.

#1: Joe (infrapinklizzard) on Jun 14, 2010

Well, this is somewhat of a farewell puzzle as the future for me looks like work, work and more work. I'll still be solving small ones over breakfast, but actually making new ones will probably be few and far between. :(

This takes color logic edge logic, two-way logic and patience (but not as much as "Honeydew").



After line and basic color logic, the bottom blocks of color are filled in. There are only 1s unaccounted for in blue.

There are a few avenues of attack open now. Let's try the most obvious: color and edge logic on r1.

The blue 1 must be in c24 due to color logic. That limits the right of the black 8. Color logic rules out c9 and left. Edge logic rules out c11 and left. So c16-19 must be black.

Now color logic on the black 4 in r2 rules out c9 and left. Color logic on the blue 1 in r2 now places it in c16. Edge logic on the 8 in r1 again rules out c12. Edge logic on the 4 in r2 rules out c10-11 and thus it is in c12-15.

Line logic finishes the 8 in r1 and makes r3c11-12 black.

Color logic on r3 tells us c7-9 are white. That still means that r3c11-12 could be part of the 2 or the 3 (unless you've already ruled out r3c5 due to the 4 in c5). However, edge logic on the 2 shows c4-5 must be white, and therefore, so must be c10.

Color logic on r4 makes c11 blue, c10 black and c6 white. Color logic on c6 makes r5-6 and r11-12 white. Thus the blue 1 is in c6r10. Line logic to continue.

Color logic on c7 makes r9 white, r8 blue and r7 black. Line logic finishes c6&7 and all rows r7 and down. Color logic then forces c7r6 blue. Line logic finishes the left side.

Now we'll try edge logic on the black 3 in c34. Since there is only a 3 in c33, c34r2-6 must be white. There are only six white spaces left in that column now. The 3 must be either in the top three or bottom three in order for all the rows 9-12 to get their requisite black 1. If it goes in the top three (r7-9), it forces the 3 in c33 into r10-12, but then causes a conflict in r7c33. So c34r10-12 must be black.

Edge logic on the 3 in r33 makes r2-3 white.

Now some (extended) two-way logic. No matter where you place the 3 in r33, it will cause one black in r32 (ie. in r5 OR r7). Since there is only one black left, every other row must be white. Therefore, c32r2-4, r6 & r8-9 must be white. That forces the 3 in c33 to r7-9 to fulfill the right 1 in r9.
{Alternatively - you could use edge logic on the right 1 in r9. If it is in c32, it would force the 3 in c33 into r5-7. This would cause a conflict in c32r7. Thus r9c32 is white and r9c33 is black. Then line logic for the 3 in c33.}

Edge logic on the 2 in c31 makes r2 white. Extended two-way logic as before tells us that no matter where that 2 goes, either c30r3 or c30r5 MUST be black. Therefore c30r2,4,&6 must be white.

Now look at rows 2 & 3. Note that wherever you put those two 3s, one must be in c29. The farthest down the 2 in c29 can reach therefore is r4. Therefore c29r5-6 is white. Line logic then finishes the puzzle.

loved it
great combination of multiple logical techniques
excellent work
i hope you make more like this despite being busy with your work

Joe, this is a fantastic puzzle. Great fun to solve, and a beautiful image. The different kinds of logic made it fun. No guessing required.

I'm so sad that you won't have much time now for creating puzzles. One of these days, while you are busy at work, you will have an idea that will inspire a puzzle, and you won't be able to help yourself, and you'll just HAVE to create another wonderful puzzle for us. :-)

Thank you for all your awesome puzzles.

Found to be logically solvable by Gator.

Awesome logic puzzle Joe.

Going to miss your puzzles! Loved this one. I had to work at it and love that! Thanks!

Goto next topic

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