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#1: Brian Bellis (mootpoint) on Mar 6, 2019 [HINT]

Joe. Is this solvable? I tried summing logic on both the reds and greens and just confused myself.

I just solved it, and didn't have to guess. :)

I couldn't, but i'm not marking it as MLA or Guessing until Joe answers.

Only R9 and R10 can begin with black, or end with black, which puts the black 2's in C1 and C20 in those rows, so both rows are completely black, and this also fills in black on R8 for C5 - C12, and C7 - C12 for R7.

Because R1 and R2 need to end in green, C17-18 is white for those rows, putting the red 6's on R3-8. The red 2's on R3 and R7 force the red 3's on C16 and C19 into R4-6, which finishes the red blots on those rows.

Line logic gives red for C3 on R4-7, C4 on R2-7, and C5 on R3-5. From there, C1 is white for R2-7 and C2 is white for R3 and R5. Since C2 needs a red 3, R4 is also white, as are R1-2, putting red 3 in R6-8.

Since red 3 on R7 is solved, R7C5 cannot be red, and because the black 3 on C5 is solved, it cannot be black, so it is white, and R2C5 is red. Also, due to white on C2, R1C1 is white, forcing R8C1 to be red, and R1C4-5 to be red in order to meet the red 3 on R1.

The white in C2 makes the 4 red on R4 start on C3, ending in C6.

C3 requires white in either R2 or R3, depending on which way the 5 red goes. Since white in R2 requires R2C6 to be red, R3C6 would have to be red to satisfy the C6 blot, which would force R3C3 to be white, and thus illegal, so the white has to be R3, and red on R8. Thus R8 starts black on C4, ending on C14. Since no green in C16, the green blot occupies just C15.

R1-7 each have 2 green, so total of 14 green squares remain. Since C8-13 each require 2 green, for total of 12, the remaining 2 green have to be in the C7 and C14 blots, meaning the C15 blot is only 1 green, and C15 solved.

The same logic used to determine white for R3C3 Also demands white for R2C3, otherwise C6 would be in an illegal situation, which finishes all the red, leaving only green and a bit of black remaining.

Green 2 on R7 is on C13-14, since C15 was previously solved with white on R7. Since the C14 green blot was discovered above to be only 1 square, C14 is solved, and C13 has green for R6-7. With the 2's on the rows and columns, the green keeps moving left and up in a step pattern to finish the green.

Since R6C7 and RCC10 are white, due to black 4's being R7-10 for those columns, the black 2 on R6 has to be C8-9, and the black 7 on R7 is discovered to start on C6 due to the other end stopping where it meets the green on C13, which completes the solve!

But I have NO IDEA what this is a picture of!!!

Well, it's not a plain plane by any means. It's quite colorful. I did a bit of guessing.

not quite done yet....but I'm gonna track it as I go in this window. Thus far, I've got all the red and pretty much all the black. There's one black from the 7 clue that can go either left to r7c6 or right to fall into r7c13. The only adv move I made was determining that the red "1" in c3 must be in r1, lest you get a contradiction in c6.....

Looks like the green might sum out though. You've got a total of 15 green spaces to find(by adding up the green clues in the rows and knowing the green blot clue in r8 is a "1" cuz you solved that row already), and 2 of the 15 possible are already placed. C's 8-13 contain 12 of the remaining 13 green....and at least one green must go in c7, so no more green can be in c's 14 or 15. Dot the rest of those 2 c's... And voila! LL does solve it out from there.

Soooooo.....no guessing! 1 tiny advanced move for the red 1 clue in c3, and summing to get the green. "mod look-ahead"

Fun solve, Brian! Thank you :)

effing puppy "snores"! that's his way of saying "Daddy, give me attention." He's not sleeping, really. He's on the couch going to town and very loud. Guess I need to leave the comp....

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Summing the green dots solves their placement after all color possible black and reds are in their places.

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Excellent Logic!!!!!! What an awesome solve!!

Found to be solvable with deep lookahead by valerie.

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After LL/CL finishes all but two of the black, the red on the right, and a whole one pixel of green,

Two-way logic 3 r1: whether it goes left or right, c6r2 must be red.
LL finishes the red and gets one more pixel each of the black and green.

Summing logic: We know from the horizontal clues that there must be exactly fifteen green pixels. (Fourteen are in the seven 2 clues, and the unknown clue is already completely placed as a block of one.) So:
> The known vertical clues add up to twelve (six 2 clues).
> That leaves three more green pixels. There are also three unknowns, so each of the blots must be one pixel.
= That makes c14r1-6 and c15r1-7 white.
LL to finish.

Found to be solvable with moderate lookahead by infrapinklizzard.

Have to count the greens, as mentioned. As by that time the only green block in the rows has been determined, we know exactly how many green square there are, and because that number is so low, we know what the green squares in the columns have to be.

Found to be solvable with moderate lookahead by infrapinklizzard.

For some reason, sometimes the rating either doesn't take, or it gets changed to votable again.

Joe, you managed to condense my 33 line explanation into just 12 lines! I've been solving these for a long time, but not so good at explaining my solve. :P

I've made a lot of tutorials. After a while you get more succinct.

Excellent puzzle, Brian. I loved the challenge.

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