Cooperative Zendo

Not the kind with cardstock and pawns. Mostly play by post Mafia so far.
kuuskytkolme
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Re: Cooperative Zendo

Post by kuuskytkolme »

Do the thes loop?

thethethethethethe
thethethethethethethe
thethethethethethethethe
thethethethethethethethethe
thethethethethethethethethethe

Should be no yes no yes, right?
I am number 63. I'm also ESL, please don't eat me.
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DanielH
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Re: Cooperative Zendo

Post by DanielH »

Love seemed to be a counterexamples to my idea of things appearing to alternate yes and no with repetition, but I don’t think it really is. My original idea was contradictory; lovelovelovelove would have needed to be both a yes and a no (depending whether you start with love or lovelove). It might alternate only if you start with something accepted; then any even repetition would be a no and any odd would be a yes. Therefore I expect that any repetition of 4n loves will be a no, and any repetition of 4n+2 will be a yes.
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Lambda
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Re: Cooperative Zendo

Post by Lambda »

thethethethethethe (6): no
thethethethethethethe (7): yes
thethethethethethethethe (8): yes
thethethethethethethethethe (9): yes
thethethethethethethethethethe (10): no
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DanielH
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Re: Cooperative Zendo

Post by DanielH »

*Stares at (8): yes*

Well that takes away the primary hypothesis and most of the variants I can think of.
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Lambda
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Re: Cooperative Zendo

Post by Lambda »

:3
kuuskytkolme
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Re: Cooperative Zendo

Post by kuuskytkolme »

let's try

0
00
000
0000
00000
000000
0000000
00000000
000000000
0000000000
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DanielH
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Re: Cooperative Zendo

Post by DanielH »

Wait, we don’t know what causes things to be accepted in the first place yet. Maybe the^8 is unrelated to the^1.

the^16
the^24
the^40
the^56

the^11
the^12
the^13
the^14

big^8
lovelove^8
Kappa
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Re: Cooperative Zendo

Post by Kappa »

I love how we're exponentiating words now. (But wait, do you mean to exponentiate them or just multiply?)
Throne3d
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Re: Cooperative Zendo

Post by Throne3d »

I assume it's meant algebraically, where if you exponentiate them you string them together (aa = a², aaa=a³), but if you multiply you just stick a number at the start (a × 2 = 2a).
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DanielH
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Re: Cooperative Zendo

Post by DanielH »

I did mean it algebraically, but not all algebra is commutative so perhaps a2 ≠ 2a. Just because it seems like this rule is anagram-invariant doesn’t mean it really is. Technically exponentiation usually binds more tightly than multiplication but I think the meaning is clear. I suppose I could have surrounded the words with parentheses, though.

However, we are supposed to be dealing with strings, which have different notational conventions.
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