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Comments on Puzzle #37812: Add a letter
By Brian Bellis (mootpoint)

peek at solution       solve puzzle
  version: 2    quality:   difficulty:   solvability: deep lookahead  

Puzzle Description:

Two words where one has a letter tacked on.

#1: Web Paint-By-Number Robot (webpbn) on Nov 18, 2023

New version published by mootpoint.
#2: Dan Tomlinson (goode2shoes) on Nov 18, 2023
I don't know but you could double "TROUBLE"!

#3: Bill Eisenmann (Bullet) on Nov 18, 2023 [SPOILER]
Or you can straddle a saddle or sketch a stretch, but these don't follow the rules ...
#4: Gary Webster (glwebste) on Nov 18, 2023 [SPOILER]
If he was spitting, you could also have a split, but he's not. And whatever he's spanning is bigger than a slit, so that's not it.

#5: Jota (jota) on Nov 18, 2023
#6: Brian Bellis (mootpoint) on Nov 19, 2023
Those are all great tries but not what I intended.
#7: gregg licht (Lgreg) on Nov 19, 2023 [SPOILER]
#8: Brian Bellis (mootpoint) on Nov 19, 2023
You got it Gregg!
#9: Joe (infrapinklizzard) on Nov 22, 2023 [HINT]
After LL/CL gets you to 49% with only red and black remaining

EL 6 c20 = r1-4 white
LL to 64%

deep lookahead:
IEL on the 3 in r13: if it were to go in c13-15, then
> it would force the 1,1 in r11 to be in c11 and c13 (and, importantly c12 to be white)
> which would force the 2 in c12 to make r13 to be black
= making a conflict (it would make a block of 4 in that row)
So The 3 can't be there and r13c15 must be white
LL to 66%

From here the black can be placed in at least 4 patterns without immediate conflict, so onto the red:

deep lookahead:
If the right 2 in r9 were in c13-14, then
> it would trigger the 1 and 3 in columns 13&14, and
> the 3 would trigger the 2 in r8, forcing it to the right
> which would trigger the 2 in c15, forcing it up
= which would then make a block of two in r7 where there cannot be one
so the assumption was wrong and r9c13 must be white
LL to 67%

IEL black 2 c14: if it were in r7-8, then it would force both row 7 and 8 in c16 to be red, causing a conflict. Therefore the black 2 in c14 must be in the range [12-14], making c14r13 black
LL to 68%

If the right 2 in r8 were to go right into c16, then it would force the 2 in r9 left into c14where it would make a block of one, causing a conflict. Therefore r8c16 must be white
LL to 69%

more deep lookahead:
if the 3 in r6 were in c15-17, then it would force:
> the 3 in c14 down into r9
> which would make r9c16 white
meanwhile, it would also
> make the rest of c15 to the top white
> Which makes c16r1-4 white
= a conflict as there's then no room for the second red 1 in that column
Therefore the assumption is wrong and r6c17 must be white
LL to finish
#10: Web Paint-By-Number Robot (webpbn) on Nov 22, 2023
Found to be solvable with deep lookahead by infrapinklizzard.
#11: Joe (infrapinklizzard) on Nov 22, 2023
As always, If you have a moderate lookahead solve, post it here
#12: David Bouldin (dbouldin) on May 6, 2024 [HINT]
Here is my solve minus the LL. I feel borderline moderate, but again will defer :)

- EL dots C4R1,2
- EL dots C3R3
- EL dots C8R9
- no matter where the 2 in R9 goes, C7,8R8 are dots
- IEL: if C14R12-14 are dots it violates C12, so C14R10 is a dot
- color EL: no matter where the 2 in R9 goes, C14R8 is always red
- whether C13R8 or C15R8 is red, C15R9 is always red
- EL in R13 dots C15R13 otherwise violates R11
- whether C17R4 or C16R6 is red, C15R1,2,3,5 are dots
- C16R4 and C16R6 can't both be red, so C16R9 is red.

Nice challenge!

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