Web Paint-by-Number Forum
Comments on Puzzle #31308: Spikeball
By James Male Hyman (jmhpro14)

peek at solution       solve puzzle
  version: 3    quality:   difficulty:   solvability: moderate lookahead  

Puzzle Description:

Spikeball Logo

#1: Web Paint-By-Number Robot (webpbn) on Jun 6, 2018

New version published by jmhpro14.
#2: Kristen Vognild (kristen) on Jun 6, 2018 [SPOILER]
That is, indeed, the logo! The picture still needs some tweaking. Maybe if you thickened the outline of the S?

https://www.coolstuff.com/Spikeball-Game
#3: James Male Hyman (jmhpro14) on Jun 6, 2018
Will do, thanks Kristen!
#4: Web Paint-By-Number Robot (webpbn) on Jun 6, 2018
New version published by jmhpro14.
#5: Joe (infrapinklizzard) on Jun 6, 2018 [HINT]
edge logic 17 r1 = c18-25 white
line logic
EL 9 c25 = r5-11, r22-28 w
LL
EL bottom 5 c25 = r23-28 w
LL
EL 7 r28 places it.
LL
EL 7 c1 = r6-7, r18-19 w
LL
EL 5 c22 = r18-20 w
LL
EL 5 r4 = c7-8, c18 w
LL
EL 5 c4 = it cannot go below r17 without causing a block of one in c5, so c4r18-20 must be white.
LL
two-way logic 5 c22: it can only be in r9-13 or r13-17. Either way, c21r18-20 must be black.
LL
EL 2 c20 = r6 w
(no LL)

internal EL left 3 r5: if it were in c10-12, the 5 in r4 would then leave no room for the right 3 in r5, so r5c12 must be white.
LL
EL left 2 r6 = c6 w
(no LL)
internal EL top 4 c5: if it were in r7-10, it would cause a block of one in c6 (in r8) so c5r7 must be white
LL
EL 5 c7: it's naive range is [r9-19] but it cannot go above r14 without creating a block of three in c8. So its real range is [r14-19] and so c7r15-18 must be black.
LL to finish.
#6: Web Paint-By-Number Robot (webpbn) on Jun 6, 2018
Found to be solvable with moderate lookahead by infrapinklizzard.
#7: Ki Bitzar (kibitzar) on Jun 7, 2018 [HINT]
Joe, you meant C21 R18-20 must be black. :)
#8: Joe (infrapinklizzard) on Jun 7, 2018
Ki, you are right. Changed. Usually two-way logic gives you white spaces and this time I spaced.

Goto next topic

You must register and log in to be able to participate in this discussion.