On Nov 8, 2010, at 6:29 PM, Ken Wesson wrote:
> On Mon, Nov 8, 2010 at 9:08 PM, Greg <g...@kinostudios.com> wrote:
>> Ah, I see that this wasn't immediately clear from my explanation: they
>> change by 1, but 3 wraps to 1. I.e. the chain is:
>>
>> 3 => 1 => 2 => 3 => 1 => etc...
>
> I assume only the first two and last two can be changed together.
Correct.
> But if only the first two and last two can be changed together, then
> anytime one of the end ones rotates by one, the middle one does as
> well. Note that the initial condition 3 3 3 satisfies (a + c) = b (mod
> 3) where the numbers are a b c. And if you raise the middle one and
> one of the other two mod 3, then you raise b mod 3 by 1 and (a + c)
> mod 3 by 1. So (a + c) = b (mod 3) is an invariant of the system. And
> 2 2 1 violates it since (a + c) = 3 != 2 (mod 3).
Woah, that seems pretty cool, but I'm not sure I understand it completely
because I'm unfamiliar with the notation... perhaps you could clear up my
understanding:
3 3 3 -> a b c
Then,
(a + c) = 3 + 3 = 6 = b (mod 3) = 3 (mod 3)
I'm not sure I understand what you mean by: 3 (mod 3)
I understand modulus in this way: 3 % 3 = 0
So I'm unclear on what 3 (mod 3) means...
- Greg
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