red and blue points on a circle

outermeasure · **Posted:** Mon, 21 May 2012 06:10:25 UTC

rdj5933mile5math64 wrote:

Could you tell me what your motivation was to add 1 for R and multiply by -1 for B?

That is from RRR equivalent to BB. Since RRR is not equivalent to any shorter R's, letting R be +1 seems reasonable, hence mod 3, ...

rdj5933mile5math64 wrote:

outermeasure wrote:

Spoiler:

This comment threw me off a little. I'm probably misinterpreting your hint, but how would a monovariant work here?

If you work at it further, you will find that unfortunately the two-bead RB is not the same as BR, even when they are the only points we have on the circle. Luckily we have a get-out-of-jail-free card: if all we care is ... then it turns out that ...

rdj5933mile5math64 · **Posted:** Mon, 21 May 2012 14:38:30 UTC

Let

Shadow wrote:

As an internet culture FYI, you don't need to give excuses as to why you're not responding immediately, it's not expected of you to do so, or even at all, especially on a problem you posed on a site dedicated mostly to homework things, hence there is no real major importance attached to the problem, it's just there for recreation, and so you're not shackled to it to such a degree.

Derp I misread the date on the post - I thought it was from last week Saturday. =_=

outermeasure wrote:

rdj5933mile5math64 wrote:

Could you tell me what your motivation was to add 1 for R and multiply by -1 for B?

That is from RRR equivalent to BB. Since RRR is not equivalent to any shorter R's, letting R be +1 seems reasonable, hence mod 3, ...

Hmmm I haven't seen a trick like that before. After you decided to do mod 3, then the multiplying by -1 for B is clear.

outermeasure wrote:

If you work at it further, you will find that unfortunately the two-bead RB is not the same as BR, even when they are the only points we have on the circle. Luckily we have a get-out-of-jail-free card: if all we care is ... then it turns out that ...

My b, I thought I had a proof that showed that rotations didn't matter as a result I didn't consider the BR and I only checked RB. :oops:

And we just need to show that we can't get from BB to RR, so we just need to show that everything in the BB equivalence class has value 1 when we apply the R and B functions, right?

outermeasure · **Posted:** Mon, 21 May 2012 14:59:20 UTC

rdj5933mile5math64 wrote:

outermeasure wrote:

rdj5933mile5math64 wrote:

Could you tell me what your motivation was to add 1 for R and multiply by -1 for B?

That is from RRR equivalent to BB. Since RRR is not equivalent to any shorter R's, letting R be +1 seems reasonable, hence mod 3, ...

Hmmm I haven't seen a trick like that before. After you decided to do mod 3, then the multiplying by -1 for B is clear.

Yes, it is a little trick that sometimes work --- letting the longer of the shortest op be +1.

rdj5933mile5math64 wrote:

outermeasure wrote:

If you work at it further, you will find that unfortunately the two-bead RB is not the same as BR, even when they are the only points we have on the circle. Luckily we have a get-out-of-jail-free card: if all we care is ... then it turns out that ...

My b, I thought I had a proof that showed that rotations didn't matter as a result I didn't consider the BR and I only checked RB. :oops:

And we just need to show that we can't get from BB to RR, so we just need to show that everything in the BB equivalence class has value 1 when we apply the R and B functions, right?

Indeed. Suffices to show BB and everything obtainable from it using the moves and rotation, all evaluates to the identity function on $\mathbb{Z}/3$ .

rdj5933mile5math64 · **Posted:** Mon, 21 May 2012 15:04:45 UTC

Awesome thanks again outermeasure!

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red and blue points on a circle

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