Puzzle Description:

he did

#1: Kristen Vognild (Kristen) on Dec 9, 2013 [SPOILER]

Is he really excited about his outfit, or are the colored ribbons there to make it more interesting to solve?

Tough puzzle but very colourful

I'm not sure it was color logic alone, I don't think so. Thanks!

agree Jota,, the top right was iffy

very cute snowman!

I got about 80% done with straight color logic. (At most some very easy edge logic - it's often easy to slip in unnoticed with the color logic).

That left all the green in the upper right, all the red in the lower left and all but one pixel of the red in the upper left.

Then edge logic on the red 4 in r2 makes r13w; then the upper left red can be done with line/color logic.

Edge logic on the 4 in r32 makes c8 white. Then line logic finishes the lower left.


Now some tough (and deep) logic on the upper right.

Some extended extended edge logic on the 4 in r4:
If it were to go in c17-20, then the 2s in the row above and below it will have to be placed within columns 26-30. That means that the 2|2 in r1-2 would have to fit into the rectangle [c23r1-c25r2]. They cannot, so r4c30 is white.

The same happens in c16-19, so r4c19 is also white.


Extended edge logic on the 4 in r4 (from the other side):
If it is in c21-24 then one of three things:
>> c25's 2 is in r3-4: leads to a conflict in r3 as c24 is forced up
>> c25's 2 is in r4-5 and c26's 3 is in r3-5: leads to a conflict in r5 as c24 is forced down
>> c25's 2 is in r4-5 and c26's 3 is in r4-6: leads to a conflict in c23 as r6 is forced right.
== So it cannot be in c21-24 and r4c21 must be white.

Then LL gives one green and seven white pixels.

Back to the right side and some more-involved logic.
If the 4 in r4 were to go in c15-18, then the 2|2 in c17-18 must both go in either r3 or 5.
>> In r3: that fulfills r3&4 and the 2|2 in r1-2 would have to fit within [rc23r2-c24r2] and [c29r1-c30r2]. The cross clues make this impossible.
>> In r5: that fulfills r4&5 and c25-28, so the 2|2|2 in r1-3 would have to fit within the rectangles [c22r2-c24r3] and [c29r1-c30r3]. This can be shown to be impossible with summing. C22 could not actually have a green because its clue is too long. That leaves four green pixels in [c23r2-c24r3]. So, one of the 2s must be in c29-30, but if it is, then there would have to be three pixels vertically. That conflicts with the horizontal clues.
==Therefore r4's 4 cannot be in c25-28 and r4c28 must be white.

Line logic places one green and five whites.

Edge logic on the 2 in r3: if it is in c22-23 then there is a conflict in r5 as c24 is forced down. So r3c22 is white. Then LL,

Edge logic on the 2 in c28: if it is in r1-2, there will be no room for the 1 in c30. So c28r1 must be white.

Then LL to finish.

Whew.

Found to be solvable with deep lookahead by infrapinklizzard.

Thanks Joe!

yep, thanks Joe!

Goto next topic

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