Puzzle Description:

My first attempt at sky diving. I was sure it would work. No broken bones so I guess it did "work"

#1: Kai-Uwe Zickerick (conzick) on Jul 12, 2013

FUNNY!!!

thx for entry Bran
a little tricky solve :)

Very cute! I used zigzag logic to start. :)

Hilarious Brian.:)

I agree with everyone above, except the zig zag logic, I did edge logic on R1.
Thanks!

That 2121... pattern with 3's for the row clues is a pattern we've seen before. Only question, slope up or down? If you are fortunate, you will choose WRONG and reach a conflict quickly and know which is correct. It is trial and error so deserves the moderate lookahead label.

no zig zag logic here either, but i did take me more than just edge logic in R1. i did some IEL also...watch what happens to C6 if the 7 in C12 goes too high.

I do not agree with the zig-zag logic. Just because it *can* fit does not give us enough confidence that it *must* go there.

F'rinstance, once we have eight of the 11 in c13 placed (after edge logic on the 5 in r2), we can see that there is a strong possibility of the 3s in rows 8-13 making a staircase, but where exactly? It could start in c1r8 and zig at r9, zag at c3, etc.

However, the 2|1 in columns 1-2 could be placed in rows 5-6. Then the zig-zag could start at c3r8, zig at r9, zag at c5, and etc. There is not enough info to start it.

Also, a zig-zag that's long enough to be a proper zig-zag has enough lookahead steps to be deep-lookahead.



However, this puzzle can be done with a bunch of little edge logic bits.

To get started, there is some internal edge logic on the 5 in r2 - it cannot be in any row from 1-13 or there would be more than one pixel in r3. Then line logic makes a cascade down to the left.

This is where the zig-zag might have been done, but it is not even necessary. Just a whole buch of edge logic bits one after another before we get to the next black pixel to be placed:

edge logic on the 2 in c1 makes r10-13 white (else there'd be two pixels in c2);
el on the 2 in c3 makes r11-13 white;
el on the 2 in c5 makes r12-13 white;
el on the 2 in c7 makes r13 white. (Line logic either here or as you go along fills in the spaces between these columns)

Then el on the 3 in r12 makes c6 white. That then places one black pixel with line logic. And then a bunch of white from that.

Then el on the 7 in c12 makes r8-12 white. (Because if the 7 were in r12, the row clues would not work.)

Line logic to finish.

Found to be solvable with moderate lookahead by infrapinklizzard.

While I have fallen from a few heights, I have never done it on purpose. It might be interesting to take an umbrella along on a cliff dive, though...

Should I try it from Mount Washington?

I don't suggest it, J....lol

Congrats on your placing, Brian!

ditto #12

Congrats on honourable mention Brian.:)

Thank you.

Nice challenging logic. I used some look-ahead to help place the 7, then the rest fell in easily. Fun story and image. :)

in omnia paratus

in omnia paratus

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