Do the thes loop?
thethethethethethe
thethethethethethethe
thethethethethethethethe
thethethethethethethethethe
thethethethethethethethethethe
Should be no yes no yes, right?
Cooperative Zendo
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Re: Cooperative Zendo
I am number 63. I'm also ESL, please don't eat me.
Re: Cooperative Zendo
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.
Re: Cooperative Zendo
thethethethethethe (6): no
thethethethethethethe (7): yes
thethethethethethethethe (8): yes
thethethethethethethethethe (9): yes
thethethethethethethethethethe (10): no
thethethethethethethe (7): yes
thethethethethethethethe (8): yes
thethethethethethethethethe (9): yes
thethethethethethethethethethe (10): no
Re: Cooperative Zendo
*Stares at (8): yes*
Well that takes away the primary hypothesis and most of the variants I can think of.
Well that takes away the primary hypothesis and most of the variants I can think of.
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Re: Cooperative Zendo
let's try
0
00
000
0000
00000
000000
0000000
00000000
000000000
0000000000
0
00
000
0000
00000
000000
0000000
00000000
000000000
0000000000
I am number 63. I'm also ESL, please don't eat me.
Re: Cooperative Zendo
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
the^16
the^24
the^40
the^56
the^11
the^12
the^13
the^14
big^8
lovelove^8
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Re: Cooperative Zendo
I love how we're exponentiating words now. (But wait, do you mean to exponentiate them or just multiply?)
Re: Cooperative Zendo
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).
Re: Cooperative Zendo
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.
However, we are supposed to be dealing with strings, which have different notational conventions.