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Inspired by this puzzle


Find the rule that each color chain on the left follows, but none on the right follow.

enter image description here

Raw text if you're colorblind or something (O = orange, Y = yellow, G = green, C = cyan):

Left:

COGGYCYOC
GYGYG
O
YYGGGGYYY
OC
GYO
GYYYGOG
OCGCOYO
CYOYOOY
GCGCGCGC

Right:

CC
GCYO
YYYYYOOOO
YCGOYCG
OYCGCY
GOC
CCCGYOO
OOY
CYOGYYCG
GOCOYOGGCO

If you want to see more cases, you may suggest up to 5 additional color chains, and I will sort them accordingly. As a rule of thumb, try to keep the length at $10$ blocks or less. You may suggest chains in this chat.

Hint 1:

If you reverse a chain, it will stay on the same side.

Hint 2:

If you recolor all green to be orange, everything will stay on the same side.

Hint 3:

The rule can be described as the equality of two quantities.

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  • 2
    $\begingroup$ Another Hint Please? $\endgroup$ – Holyprogrammer Aug 28 '18 at 6:12
  • $\begingroup$ @KhushrajRathod I will add a hint every 24 hours. $\endgroup$ – Riley Aug 28 '18 at 11:52
  • $\begingroup$ @Riley Maybe you can add a few 'caterpillar logic' style hints, i.e. we ask a combination and you put it on the left or on the right. $\endgroup$ – rhsquared Aug 28 '18 at 12:47
  • $\begingroup$ @rhsquared That could work. I'll do that in addition to hints I'm already giving. Not sure what the best way to do this is. Maybe each user is limited to 5 chains or something? $\endgroup$ – Riley Aug 28 '18 at 12:59
  • 1
    $\begingroup$ @rhsquared I've added rules for suggesting more chains to the question. $\endgroup$ – Riley Aug 28 '18 at 16:26
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+50
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The set on the left is defined by

having equal numbers of adjacent pairs of primary and secondary colored blocks. First, treating O/G and Y/C as equivalent gives a picture like this: simplification Counting a group of n as (n - 1) pairs, all the left sequences have the same number of primary and secondary pairs, while the right sequences differ.

Another way to phrase the answer:

Count the number of primary and secondary colored blocks, weighting each end block .5 less than normal. A solitary block's weighted zero, then. The totals should be the same.

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  • 1
    $\begingroup$ Yes, you got it! (And right after I added the bounty, too). I will accept this answer and award you the bounty. But not immediately so the question can receive more attention. $\endgroup$ – Riley Aug 30 '18 at 17:04
2
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This is just putting Mr. Fish's answer into mathematics:

A sequence is valid if, and only if, it consists of two partitions of $n$ (the total number of blocks) into sub-sequences $a$ and $b$, with $\mid a\mid+\mid b\mid=n$. Furthermore, $\mid\operatorname{np}(a)-\operatorname{np}(b)\mid\le1$, where $\operatorname{np}(k)$ gives the number of parts of $k$. Also, the total sums of each part minus $1$ in each sequence are equal.

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