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Sep 13, 2020 at 9:08 comment added Max Alekseyev You said "first $W(k,r)/2$ numbers".
Sep 13, 2020 at 5:24 comment added domotorp We didn't use in the argument that it is even.
Sep 12, 2020 at 21:05 comment added Max Alekseyev Thanks, I see now. The only question remains - what if $W(k,r)$ is odd?
Sep 12, 2020 at 19:49 comment added domotorp Case 1: the circular AP's difference is a positive number $<W(k,r)/2$. Case 2: the circular AP's difference is a negative number $>W(k,r)/2$. In both cases the circular AP is just a regular (non-circular) AP, as otherwise it would change colors.
Sep 12, 2020 at 14:01 comment added Max Alekseyev I'm still not convinced. Can you add a proof for the general case?
Sep 12, 2020 at 13:41 comment added domotorp To give an example: If $k=3$ and $r=2$, then 12122121 becomes 34342121. A circular AP would need to make a small ($<W(k,r)$) jump and a big ($>W(k,r)$) jump.
Sep 12, 2020 at 12:40 comment added Max Alekseyev Why the constructed $2r$-coloring has no circular $k$-AP?
Sep 12, 2020 at 9:27 history answered domotorp CC BY-SA 4.0