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"Darren New" <dne### [at] san rrcom> wrote in message
news:4b758c57$1@news.povray.org...
>
> You know, he says that, but I'm pretty sure a CA with an infinite amount
> of initialization counts as Turing-equivalent.
Wait... California is a Turing machine? Good Lord, that explains everything!
--
Jack
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Darren New wrote:
> as he described it, he needs to set
> up an infinite number of clock pulses before starting the emulation in
> order to clock the emulation. I'm not sure that's allowed
Yeah, that does sound a little suspect...
I think this is just a case of a flawed proof though. I'm pretty sure
the *result* is correct.
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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Orchid XP v8 wrote:
> I think this is just a case of a flawed proof though. I'm pretty sure
> the *result* is correct.
He's trying to prove that *his* CA is turing complete. I already know there
are lots of turing complete CAs. I'm not sure what "result" you think is
correct?
--
Darren New, San Diego CA, USA (PST)
Forget "focus follows mouse." When do
I get "focus follows gaze"?
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Captain Jack wrote:
> "Darren New" <dne### [at] sanrrcom> wrote in message
> news:4b758c57$1@news.povray.org...
>> You know, he says that, but I'm pretty sure a CA with an infinite amount
>> of initialization counts as Turing-equivalent.
>
> Wait... California is a Turing machine? Good Lord, that explains everything!
No. Canada.
--
Darren New, San Diego CA, USA (PST)
Forget "focus follows mouse." When do
I get "focus follows gaze"?
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"Darren New" <dne### [at] sanrrcom> wrote in message
news:4b75c435$1@news.povray.org...
> Captain Jack wrote:
>> Wait... California is a Turing machine? Good Lord, that explains
>> everything!
>
> No. Canada.
Ah, of course... all the pieces are falling into place. The entire country
is itself the tabulating device for Winter Olympics scoring then, I'm
assuming.
--
Jack
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> I was attempting to work out what the result of
>
> A sin t + B cos t
>
> is. I was convinced there was a standard identity for this, but I can't
> find it. In the end I came up with
>
> Sqrt(A^2 + B^2) sin (t + atan(B/A))
>
> but I don't even know if that's correct. And I still have to apply it to
> my original formula to figure out the result.
I think...
a*sin(t) + b*cos(t) = sqrt(a^2+b^2)*cos(t-atan(a/b))
notice that yours was reciprocal and positive.
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>> I think this is just a case of a flawed proof though. I'm pretty sure
>> the *result* is correct.
>
> He's trying to prove that *his* CA is turing complete. I already know
> there are lots of turing complete CAs. I'm not sure what "result" you
> think is correct?
That CA #30 (amoung others) is Turing-complete.
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Invisible wrote:
>>> I think this is just a case of a flawed proof though. I'm pretty sure
>>> the *result* is correct.
>>
>> He's trying to prove that *his* CA is turing complete. I already know
>> there are lots of turing complete CAs. I'm not sure what "result" you
>> think is correct?
>
> That CA #30 (amoung others) is Turing-complete.
Why would you think this if the proof is flawed?
--
Darren New, San Diego CA, USA (PST)
Forget "focus follows mouse." When do
I get "focus follows gaze"?
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>>>> I think this is just a case of a flawed proof though. I'm pretty
>>>> sure the *result* is correct.
>>>
>>> He's trying to prove that *his* CA is turing complete. I already know
>>> there are lots of turing complete CAs. I'm not sure what "result"
>>> you think is correct?
>>
>> That CA #30 (amoung others) is Turing-complete.
>
> Why would you think this if the proof is flawed?
1. It seems a logical and intuitive result.
2. Nobody has disproved it.
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Invisible wrote:
> 1. It seems a logical and intuitive result.
Uh, no?
> 2. Nobody has disproved it.
That's not how math works. "Hey, P=NP! Well, nobody disproved it."
--
Darren New, San Diego CA, USA (PST)
Forget "focus follows mouse." When do
I get "focus follows gaze"?
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