Puzzle Description:

#1: Kristen Vognild (kristen) on Dec 29, 2011

These are good exercises in edge logic. :)
I like the way this one loops around.

Look-ahead edge logic for the top row, then everything falls into place.

I would really like to know how you two did the edge logic, Kristen and Lollipop.

I have dots in the first row in columns 1, 11, 12, and 13. Did you place a dot in R1C5? It could be a dot, but doesn't have to be at this point. The 4 in R1 could start there or end there without conflict, looking ahead several moves. My instinct from past solving experience tells me to start the 4 at R1C6. But that's guessing, not logic. If I missed something obvious, I wanna know. :-)

Looking at the 4 and two 2's in the first two rows, I knew the 4 had to flank the 1's in the center of row 1. If you use the 1's at either edge, there isn't room for the 2's in the 2nd row.

The 4 can be placed in c2-5 and the 2s fit. I don't see the logic. I did just that while seeing if I could correctly guess at a solution, and received a contradiction after about 10 other moves in c5.

i might be missing something but i think i had to guess

even after reading what you wrote kristen i agree with sgusa that the information you gave didnt help solve this logically
(maybe there is something there but the way it was worded didnt help me see it)

Guessing is needed short of massive look-ahead.

Initial line logic gives r5c11 black (and r5c12 white)

Then edge logic on the 4 in r1 gives c1 and c10-13 white

Line logic gives a further black in r6c11 (and r6c12 white)

Edge logic on 3 in c1 makes r1-2, 6 white

r2c13 is white due to edge logic on either r2 or c13

And that's all I can find that's not an unbearably long look-ahead.



A hint to make this solvable: "r1c2 is white."

Then edge logic on the 4 in r1 and line logic.



Alternatively, adding a black pixel to the puzzle at r8c3 would make it solvable.

Thanks, Joe. I got stuck at the same place you did, so I don't feel too bad. I appreciate the help in getting this one solved.

ever have one of those puzzles where you FINALLY complete it logically, but aren't sure you'll be able to do it again to track the moves? holy cow that was difficult! i'll give it a shot and post my moves if i can do that again.

- LL black in C11R5 and white C12R5
- EL 4 in R1 dots in C1,11-13,R1
(joe, i didn't see C10 here)
- LL black in C11R6 and white C12R6
- EL 3 in C1 dots R2,6C1
- EL 2 in C18 dots C18R2

- no matter where the 4 in R11 goes, the 2 in R10 will always land in C9-10R10, they are black.
- LL dot C8,11R10 and black C9R9
- internal EL, 4 in C6: if it ends in R10 or R11, then it forms dots to it's left in R8 and 10, making it impossible to place the 2s in C5, so C6R10,11 are dots.
- LL black C5R6-7 then dot C2-4,8-10,13R6.
- LL dot C9R7.
- EL 2 in C13 to dot C13R7.
- LL black C3R11 and C10R11 then dot C3R10 and C10R9.
- LL black C4R4.

- whether C12R10 or C13R10 are black, C13R3 is always a dot.
- internal EL the 5 in C11: if it starts in R9 it will trigger the 2 in C10R7,8 or if it moves any higher the dot formed in C10R4 will do the same, so black C10R7,8
- LL black C11R7 and dot C12R7
- LL dot C11R2 then C12R2
- LL dot C7R7
- whether C12R10 or C13R10 are black, either resultant dot in R8 means the black C10R8 is part of the two, so dot C12,13R8
- whether C9R8 or C11R8 is black, C11R11 is black.
- LL dot C7R11
- whether C10R2 or C10R4 is black, C10R4 is a dot.
- LL black C9R2 and dot C9R5.
- whether C6R2 or C6R3 is black, C3R2 is a dot
- LL dot C2R2
- whether the 3 in C1 is high or low, C2R9 is a dot
- no matter where the 4 in R1 goes, C5R2 is black
- LL dot C7R2
- whether C9R1 or C9R3 is black, C7R3 is always a dot
- LL dot C6R3
- LL to end

thank you for posting your solve (excellent logical finds by the way)

just an fyi though... (for those who may have been confused)

step 5 should be C13 (not C18)
step 9 should be C6R6-7 (not C5R6-7)
step 13 should be C4R3 (not C4R4)
step 22 should be "...C9R4 is a dot" (not "C10R4 is a dot")

thanks bugaboo, i should be more careful...and maybe i shouldn't be doing these explanations at 2 in the morning. should i make those edits or let people use your translator?

just leave it... why do extra work? (as if spending that long giving a detailed step-by-step solving guide didnt waste more than enough time) haha

plus my comments would then not make sense

about the 2 am solve/explanation... been there done that
(smile)

all valid points...didn't help that the puzzle itself uses a non-5-increment grid...that always trips me up too.

Found to be logically solvable by gator.

Everything checks out (two move look-ahead or less). The hardest parts are the two-move look-ahead affirmation logic.

OK David - step 6 of your explanation doesn't make sense to me. Why couldn't the 2 in row 10 end up at C5/6 instead of C9/10 (considering what we know at that point, which is only 9 dots and 2 blacks)?

I agree with Debbie on this one. David says "no matter where the 4 in R11 goes, the 2 in R10 will always land in C9-10R10, they are black." But this isn't true. If the 4 in R11 is in C5-8, then the 2 in R10 is in C5-6.

Hmmm...that case can be excluded if you also consider the fact that if the 4 is there, then the 3 must be in C1-3. Then the we can't place both 1's to the left of the two as they must. I guess that doesn't actually increase the depth of the lookahead, but it increases the breadth. I guess we are still OK.

WOW. That one has way too much edge logic for me. But it's all edge logic. Impressive.

Goto next topic

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