Puzzle Description:

Riding a bicycle

#1: Jota (jota) on Jan 9, 2019

Great solve!

This was inspired by SmarterEveryDay's video on the "Backwards Brain Bicycle". https://www.youtube.com/watch?v=MFzDaBzBlL0

After initial Line Logic,
Edge logic 5 c2 = r10-12, c19 White
LL
EL 4 c29 places it
LL to finish

Found to be solvable with moderate lookahead by infrapinklizzard.

A very interesting solve and a nice image! Alternate solution from Joe here.

I worried a bit about if the lookahead was moderate with the 4 in C29. I'm still foggy on the precise definition! So I looked for another way, which I hope people enjoy.

I eventually got a contradiction with a black dot at R10C29. My solution sort of echoes that, eventually.

At 91% with C22-29 R10-17 still to fill, I noted, first, edge logic made C22R10 and C29R17 dots.

Then I noticed any blackdot left of C25 in R12 would create a contradiction in R14 C28/29. Similarly for R14 and R12.

Similarly for C25/C27 and C27/25 -- the 4's and 2's horizontally put a 2 in C25 or27.

But now a black dot R16C23 means you can't place the black dot in R17.

Wow, I really have no idea what I was thinking. EL does NOT place the 4 in c29, it only rules out r17. Then LL
EL 4 r10 makes c22 white.
LL

Then, look at the 24 and 2,2 in r10 and 11. The crossing 1s in c25 and 26 mean that those must go in opposite directions. That takes up those 1s and so the rest of the unknowns in those columns (from r12-17) must be white

The same thing happens on the right side (rotated) so that the rest of r12 and r14 (from c22-24) must be white

Then, if r15c23 is black,
> the rest of that column must be white and
> r15c22 must be white,
> making r11c22 black
= which is a contradiction (a block of one where there are only 2 clues),
so our assumption is wrong and r15c23 must be white

LL to finish

Goto next topic

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