Puzzle Description:

1 word, 7 letters.

#1: Web Paint-By-Number Robot (webpbn) on Mar 29, 2023

Found to have a unique solution by blurglecruncheon.

Found to be solvable with moderate lookahead by blurglecruncheon.

Note that if we assume symmetry, the puzzle falls out readily--we place row 1-15, and then row 10 falls since we have a single black dot and two next to it, which must be the 3's.

But in the interest of rigor...

A lot of the solution below relies on how once we see something, it can be turned 90 degrees and we can add more.

We can get to 59% with edge logic on the 5s in the 5-2-6/6-2-5.

If R11C12 is a dot, then R11C7 and R15C15 and black, but then R15C7 has a contradiction.

Note some line logic places dots but with R7C14/15 being black and a dot, R7C13 must be a dot, or the 3-1-1-3 doesn't fit. The hinter doesn't help here. 67%.

If R11C12 is a dot, then more contradictions. R17C11 is black, but so are C15R16/R23. 70%.

If R12C6 is a dot, we get contradictions in C6-7 with the 3's in R6-7. 71%. Similarly for R6C12. 72%.

If R6C10 is black, C6/7 can't place the top 3. This places a lot: R6/7 C6/7, 9/10, 16/17, 19/20 and their reflections are all black dots.

But now R21C7 is a dot, or R18 has a contradiction. Similarly for R21C19. R18C10 is then a dot, as we know where all the 3's are now.

That causes the whole puzzle to fall with line logic.

(Again, this was actually tougher before added color. I'm disappointed my puzzles are tougher than I want them to be but I hope people who enjoy the logic enjoy this.)

Initial line & color logic solves 30%, with blue finished.

EL on the red and green 5s in R3, R23, C3, and C23 can completely place them. Resulting line & color logic completes red and green, and gets to 64%. Not sure where Andrew fell short in his 59%.

From here I'm not tracking the next step. If R11C12 is a dot, how can R11C7 be black?

I'm out of time for tonight, but I'll give this another look tomorrow.

rotator?

That would be my guess.

If R11C12 is a dot: The 7 in R11 is extended to C19, the 7 in C11 is extended to R18, and the 7 in C15 will not reach R18. This creates a contradiction in R18 where only 3s are available but the combination will join 4 together. You can repeat the logic with R15C14, R15C12, R14C11, and R12C15. LL.

R11C14, R15C12, R14C15, and R12C11 can use the same logic to be marked black. LL.

This puts it at 70% complete. Andrew's hints are working from here.

Found to be solvable with moderate lookahead by wombatilim.

Rotator indeed! Sorry I was away for a few days. I'll try to make something less difficult next time.

A tough, fun puzzle! I might argue that it is deep look ahead, some of the contradictions are very tricky to find. I enjoyed the challenge quite a lot!

For me, when I got stuck, looking for contradictions around the placement of the black 7s in rows 11 & 15 and columns 11 and 15 let me continue.

I admit I needed the help above to find the contradiction in R21C7.

Thanks for a wonderful puzzle!

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