Puzzle Description:

#1: Joe (infrapinklizzard) on Feb 11, 2018 [HINT]

after ll/cl makes a three-toed footprint,
el 7 r47 places it
ll
color edge logic 18 r24 places it - since there are no blue clues above row 24, we can do edge logic on the (blue) 18 as if it were on the edge.
ll/cl
iel 17 r32 places it
ll/cl finishes the blue

Now there are three sections left, two red and one black, and they all require deep lookahead.

The 8 in c32 could fit in three places: r7-14, r8-15, or r16-23.
If it were in r16-23 then it would place the 2 in 31 and trigger the 2 in c30. The 2 in c30 would then cause a conflict in r18. So it can't be in this position, and c32r16-23 must be white.

ll finishes the black

The two red bits left have an obvious solution to veteran solvers, but that doesn't make that solution automatically logical. If we see the staircase and then fill it in without a conflict, it's still not *logically* solved because we started with an assumption. There might be another way to fit the clues that is also without conflict. The only way to go forward when starting with an assumption is to find a conflict which then tells us our assumption is wrong. That tells us that the clue we placed cannot be where we placed it and we may be able to mark one or more of those pixels as white.

deep lookahead for the red:
We need to start by pinning down the 2s in r31. We might intuit that they get split up one to each section, but we can't just assume that - they could both fit within either side.
So let's do some summing - the vertical clues for the left section add up to ten red pixels we need to place. The horizontal clues that we know must be in the left sections are in rows 28-30 and add up to eight. We need two more red pixels. So we now KNOW that the left 2 in r31 MUST be in the left section.

now let's continue on the left:
> if the left 3 in r28 were in c13-15, then it would force the 3 in r29 to be in c14-16.
> That would make the rest of columns 13-16 white.
> The 2s in r30 & r31 would then make c18 red in each row, conflicting with the 1 clue in that column.
= So the 3 cannot be in c13-15 and r28c13 must be white.

> if the left 3 in r28 were in c14-16, then it would force the 3 in r29 to be in c13-15.
> That would make the rest of columns 13-16 white.
> The 2s in r30 & r31 would then make c18 red in each row, conflicting with the 1 clue in that column.
= So the 3 cannot be in c14-16 and r28c14 must be white.

ll

el left 3 r28 = c15 w
ll finishes the left

And then we mirror the process:
> if the 3 in r 28 is in c37-39, then it would place the 2 in r29
> That would make the rest of c37-39 white
> The 2s in r30 & r31 would then make c35 red in each row, conflicting with the 1 clue in that column.
= So the 3 cannot be in c37-39 and c39 must be white.

ll to finish.

Found to be solvable with deep lookahead by infrapinklizzard.

So...was the game of darts invented 180 years ago? I imagine the title refers to the number 3x3x20 (a score being 20)?

Kristen - I think it is the maximum score you can get with three throws - Three darts, each getting into the triple 20 section of the board.

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