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PostPosted: Wed, 4 Apr 2012 15:27:24 UTC 
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I tried using a search function and I wasn't able to find anything.~

Question:
We have red and blue points on a circle. We can do any one of 2 things.

1. We can add a red point between any two existing points that are next to each other and change the color of the original points.

2. We can remove a red point and change the color of the 2 points next to it (there need to be two points next to the red point we are removing).


Prove that if there are initially 2 red points and no blue points on the circle, then we cannot get 2 blue points and no red points on the circle.

Mathematical Background:
Invariants, monovariants, injections bijections, generating functions.... (other olympiad level combinatorics stuff that I might have forgot :P )

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PostPosted: Wed, 4 Apr 2012 15:40:11 UTC 
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rdj5933mile5math64 wrote:
We have red and blue points on a circle.
.......
Prove that if there are initially 2 red points and no blue points on the circle...


You say there are blue points, then you say no blue points ?

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PostPosted: Wed, 4 Apr 2012 15:55:08 UTC 
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Denis wrote:
You say there are blue points, then you say no blue points ?


Sorry my wording was a little ambiguous. My first statement that you quoted was just to give a background of the problem. Like to establish that all points we put on a circle are either red or blue (for example we can't have green points) and that all of the points are on a circle.

To give a more direct answer to your question: we start with no blue points on the circle.

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PostPosted: Wed, 4 Apr 2012 17:14:19 UTC 
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rdj5933mile5math64 wrote:
I tried using a search function and I wasn't able to find anything.~

Question:
We have red and blue points on a circle. We can do any one of 2 things.

1. We can add a red point between any two existing points that are next to each other and change the color of the original points.

2. We can remove a red point and change the color of the 2 points next to it (there need to be two points next to the red point we are removing).


Prove that if there are initially 2 red points and no blue points on the circle, then we cannot get 2 blue points and no red points on the circle.

Mathematical Background:
Invariants, monovariants, injections bijections, generating functions.... (other olympiad level combinatorics stuff that I might have forgot :P )


Am I understanding the first move correctly? You probably mean only those two selected points have their colour changed, not (as written) every point on the circle prior to insertion of your red point? So adding a red point and removing it shouldn't change anything, rather than flipping the colour of all points except 2 consecutive ones.

Not really a hint, but a pre-hint:
Spoiler:
There is a (reasonably natural) way to choose a (pseudo-)representative of what you can get from any starting position.


Another hint, this time for real:
Spoiler:
the equivalence between RRR and BB suggests you play with mod 3, where R stands for +1 and B stands for ...

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Wed, 4 Apr 2012 18:46:53 UTC 
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outermeasure wrote:
I don't think so. (Am I understanding the first move correctly? You probably mean only those two selected points have their colour changed, not every point on the circle prior to insertion of your red point?)

That's correct :) . For example, we could have something like:
Spoiler:
rr
bbr
rrrr
rrbrb


outermeasure wrote:
Spoiler:
There is a (reasonably natural) way to choose a (pseudo-)representative of what you can get from any starting position.



Hmmm I'm sort of new to this website so to be safe I'll hide stuff.
Spoiler:
Looked up pseudo. hmmmmm ignoring what I said earlier about mathematical background sorry :( . We can make an equivalence relation of the set of all configurations, C. So if we let x,y be elements in C then x~y iff we can make "moves" on x to get to configuration y. So, at this point, I was thinking ok now what can I do. By taking a few examples, I thought C should be partitioned into different subsets then we can think about using the canonical form of each subset, which is just the configuration with the smallest points. But, that wasn't working as well as I'd hoped it might. Seeing as this was an old olympiad problem, the approach that I'm trying to use can't be right.


outermeasure wrote:
Spoiler:
Create "devices" that act on your representative.



Spoiler:
I'm assuming "devices"=invariants/monovariants.

So my thoughts
I was like hmmm what "matters"?
well
rrbb =/= rbrb - the relative position of the numbers matter.
bbr=brb=rbb - rotations shouldn't matter ( <--- led me to think: complex numbers)
but I couldn't get anywhere here :/
I'd really like to see a nice invariant/monovariant proof :mrgreen:
hmmm I've been trying a lot of somewhat contrived invariants but nothing has worked so far...

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PostPosted: Thu, 5 Apr 2012 01:38:28 UTC 
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Sorry for double posting :( I don't know why I can't edit :?

outermeasure wrote:
Another hint, this time for real:
Spoiler:
the equivalence between RRR and BB suggests you play with mod 3, where R stands for +1 and B stands for ...


Spoiler:
... you're hinting B=-1. Why? :confused: Neither addition nor multiplication work since under both circumstances they don't account for RRBB being distinct from RBRB. These two configurations are very different seeing as RR ->RBB -> RRBB and BB ->RRR -> RBRB.


EDIT: Grammar Error

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PostPosted: Thu, 5 Apr 2012 01:43:54 UTC 
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rdj5933mile5math64 wrote:
Sorry for double posting :( I don't know why I can't edit :?

outermeasure wrote:
Another hint, this time for real:
Spoiler:
the equivalence between RRR and BB suggests you play with mod 3, where R stands for +1 and B stands for ...


Spoiler:
... you're hinting B=-1. Why? :confused: Neither addition nor multiplication work since under both circumstances they don't account for RRBB being distinct from RBRB. These two configurations are very different seeing as RR ->RBB -> RRBB and BB ->RRR -> RBRB.


EDIT: Grammar Error


You don't have to hide everything. outermeasure used the spoiler tags because the hints make the problem markedly easier, and it's the only way to make it so you can get a little bit of help at a time rather than all of it all at once.

Also, you cannot edit after the first minute or so of posting, so if you have a grammar error which is nonessential, just let it be.

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PostPosted: Thu, 5 Apr 2012 02:00:41 UTC 
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Shadow wrote:
You don't have to hide everything. outermeasure used the spoiler tags because the hints make the problem markedly easier, and it's the only way to make it so you can get a little bit of help at a time rather than all of it all at once.

Also, you cannot edit after the first minute or so of posting, so if you have a grammar error which is nonessential, just let it be.


Oh lol the point of the post you quoted wasn't for a quick grammar fix. :P I was addressing a point that outermeasure had made (looking at the R's and B's using mod's) which I must have missed earlier. I then edited the post because there was a small typo.

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PostPosted: Thu, 5 Apr 2012 05:24:26 UTC 
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rdj5933mile5math64 wrote:
Sorry for double posting :( I don't know why I can't edit :?

outermeasure wrote:
Another hint, this time for real:
Spoiler:
the equivalence between RRR and BB suggests you play with mod 3, where R stands for +1 and B stands for ...


Spoiler:
... you're hinting B=-1. Why? :confused: Neither addition nor multiplication work since under both circumstances they don't account for RRBB being distinct from RBRB. These two configurations are very different seeing as RR ->RBB -> RRBB and BB ->RRR -> RBRB.


EDIT: Grammar Error


No, that's not what I am hinting at. Indeed you need something that distinguishes RBRB from RRBB, and that is easy to get hold of if you look at it the right way.

By the way, you can (mostly) forget about the representative part. I just put it there to motivate why my real hint of using mod 3 and assigning R to +1 is a good idea (and the "device" thing to simplify your representative, which you can actually ignore for this exercise but a good idea to keep in mind). By the way, that is a hint:
Spoiler:
Why did I use +1 instead of 1?

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Thu, 5 Apr 2012 18:24:24 UTC 
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outermeasure wrote:
I just put it there to motivate why my real hint of using mod 3 and assigning R to +1 is a good idea. By the way, that is a hint:
Spoiler:
Why did I use +1 instead of 1?


My (Wrong) Reasoning:
I thought that the +1 was used instead of the 1 to emphasize using B=-1 mod 3. My reasoning was that if B=R=1 mod 3, then BB is equivalent to AA so that wouldn't work. Since 1 is the multiplicative identity I thought that you were multiplying things together. As a result, B=0 mod 3 wouldn't work. Process of elimination left me with B=-1 mod 3.

I still don't know how to figure out exactly what to do with the mod 3.

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PostPosted: Thu, 12 Apr 2012 11:05:51 UTC 
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It has been a week (ok, minus 8 hours or so), and evidently my subtle hint didn't work terribly well..

It is very hard to give any more hint(s) without giving the entire solution away, so here is a big spoiler:
Spoiler:
Let R be the function "add 1". What is a good candidate for the function B?

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Mon, 16 Apr 2012 15:27:03 UTC 
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outermeasure wrote:
It has been a week (ok, minus 8 hours or so), and evidently my subtle hint didn't work terribly well..

It is very hard to give any more hint(s) without giving the entire solution away, so here is a big spoiler:
Spoiler:
Let R be the function "add 1". What is a good candidate for the function B?


Darn the hint still isn't helping. I know it's really tough to give a hint to a problem like this since it's either you get it or you don't, but nothing has "clicked" yet.

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PostPosted: Sat, 19 May 2012 10:20:32 UTC 
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Hmm... forgot about this thread, hence the late reply.

Spoiler:
Let B be the function "multiply by 2 (or equivalently, -1)". Now R and B are functions \mathbb{Z}/3\mathbb{Z}\to\mathbb{Z}/3\mathbb{Z} satisfying R^3=B^2=\mathop{\mathrm{id}}, and I'll leave you to check the details.


Spoiler:
You don't need the full invariance. You just need ...

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Mon, 21 May 2012 00:40:30 UTC 
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This solution is awesome! Thanks, outermeasure! :D

I've been really busy studying for finals, ACT classes, and ARML (I'm too fond of the the first 2~), so I'm sorry for the late reply.

outermeasure wrote:
Spoiler:
Let B be the function "multiply by 2 (or equivalently, -1)". Now R and B are functions \mathbb{Z}/3\mathbb{Z}\to\mathbb{Z}/3\mathbb{Z} satisfying R^3=B^2=\mathop{\mathrm{id}}, and I'll leave you to check the details.



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

outermeasure wrote:
Spoiler:
You don't need the full invariance. You just need ...


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

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PostPosted: Mon, 21 May 2012 00:50:21 UTC 
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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. :)

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