Puzzle Description:

Hey! How can you be longer than I am?

#1: Spot (Pspaughtamus) on May 27, 2018 [SPOILER]

Shadow, rhyming with Colorado?

Taller shadow?

Juli. You and Susan are like a one two punch. Taller shadow Colorado is correct

I was up past my bedtime (2-3 am or later), I need to stop doing these new ones so late.

Love the puzzle.

ll/cl leaves only black

But then I see no way forward short of guessing. Am I missing something?

It doesn't rhyme. Colorado doesn't end with -ou sound but a clear cut -o sound.
Here are some of your other options:
mikado, credo, dodo, judo, pseudo, speedo, weirdo, commando, libido, tornado, torpedo, tuxedo

Wasn't aware shadow was pronounced shadoo. :)

You have to give me a bit of slack on these rhymes. Different regions have different pronunciations. Tallershadow and Colorado sounds like an almost perfect rhyme to me.

I'm from Colorado and some people pronounce it Call-o-rah-do. Some people say Kello-rad-do. I think Taller Shadow is an awesome rhyme for a puzzle, and is the coorect rhyme for Call-o-rad-do. Fun one, Brian. And I did get this one! Yay!

Very cool

I found no issue with Brian's rhyme. Although I pronounce the "a" in Colorado like the "o" in "clock", I've heard Colorado pronounced in a way that would make it rhyme perfectly well with the word, shadow, that is, with the short "a" like "bat". It was only natural to change how I pronounced it to hear the rhyme.

And lol, #8.

I had a very tough time with this. Here's what I did for the blacks:
the 6, 4, 5, 3 (C15-18) cannot all be going down to the bottom half together because of the 2s beginning R8. That means some rows above them cannot be a line of dots, which exluded some options for the 4 in R5. This placed some dots, and showed the 5 in C12 had to be in the bottom half.
I sort of continued to place the rest of it based on the fact you can't have 3s in any of those places where there are 2s.

Emily's hint doesn't quite track, but does at least provide a starting point.

At least one of C15-17 needs to go into R6 to avoid making R10 invalid. This means R6 C8-10 are white. C15 or C17 would need to go into R5 as well, but C16 wouldn't necessarily have to, so we can't start the 4 in R5 yet like Emily suggested. The 5s in R8-9 make coexisting on those rows possible.

C15: If the 6 goes down into R12, then R6C14 would need to be black, which would make R6 invalid, so R6C15 is black. LL.

[EDITED] I had a deep lookahead here that I thought solved the puzzle, but in re-reading it, I realized I missed a possibility, so I can no longer consider it valid. Un-setting solvability completely isn't currently working, but I have not actually determined deep lookahead will work yet.

Found to be solvable with deep lookahead by wombatilim.

Deep lookahead R2-3:

The 3 in R2 can either place a black in C13 or occupy C18-20. The 3 in R3 can either place a black in C13 or blacks in C18-19. If we white out R2 C18-19, we can't place the 3 in R3 on that side either, since it will cause a contradiction in R4. Blacks would now be placed in R2-3 C13. The 1s in C8-10 prevent us from putting the 2s in R2-3 into the same columns, so either R2 is in C8-9 and R3 in C10-11, or R2 is in C10-11 and R3 in C8-9. Either of these makes R2C12 white, which fully places both 3s into C13-15, and completes the 2s in C13-14 and the 6 in C15.

Next we see how this impacts R4-5. Since R3C18 would be white, R4C18 must also be white, and completing the 2 in C14 means R4C14 would be white. The black that would now be in R4C15 would have to be the 1 in R4. In R5, the black in R5C15 would be part of the 4, so R5 C8-11 would be white. If the R2 2 is in C8-9, this blocks R4 C8-9. If the R2 2 is in C10-11, it blocks R4C10. Neither of these leaves enough room for the 4 in R4, which means the original proposed placement cannot be true.

TL;DR: R2 C18-20 must be black to avoid invalidating R4 at a distant point in the future. LL solves the rest.

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