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rdj5933mile5math64
 Post subject: Intersecting Committees
PostPosted: Fri, 15 Jun 2012 17:44:51 UTC 
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6 students formed 30 committees. Every 2 of these 30 committees intersect, meaning that they each have at least one member in common (no 2 committees have the same members). Prove that they can form at least one more committee that intersects each of the existing 30 committees.

Math Background: Olympiad Combinatorics.

The difficult part of this problem is utilizing the fact that we have 30 committees each of which intersects.

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outermeasure
 Post subject: Re: Intersecting Committees
PostPosted: Fri, 15 Jun 2012 18:09:32 UTC 
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rdj5933mile5math64 wrote:
6 students formed 30 committees. Every 2 of these 30 committees intersect, meaning that they each have at least one member in common (no 2 committees have the same members). Prove that they can form at least one more committee that intersects each of the existing 30 committees.

Math Background: Olympiad Combinatorics.

The difficult part of this problem is utilizing the fact that we have 30 committees each of which intersects.


In fact, every maximal intersecting family on [n] has size 2^{n-1}, so you can form two more committees.

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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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rdj5933mile5math64
 Post subject: Re: Intersecting Committees
PostPosted: Fri, 15 Jun 2012 19:00:00 UTC 
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outermeasure wrote:
In fact, every maximal family on [n] has size 2^{n-1}


How do you prove that?

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outermeasure
 Post subject: Re: Intersecting Committees
PostPosted: Sat, 16 Jun 2012 02:49:57 UTC 
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rdj5933mile5math64 wrote:
How do you prove that?


Spoiler:
\forall A\in 2^{[n]}\quad A\cap A^c=\varnothing, so every maximal (1-)intersecting family on [n] has size at most 2^{n-1}.

On the other hand, if \mathcal{F}\subseteq 2^{[n]} is maximal intersecting and has size <2^{n-1}, then there exists A\subseteq [n] such that A,A^c\notin\mathcal{F}. Since \mathcal{F} is maximal, \exists B\in\mathcal{F} with A\cap B=\varnothing (so B\subseteq A^c) and \exists C\in\mathcal{F} with C\cap A^c=\varnothing, contradicting B\cap C\neq\varnothing.

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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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Soroban
 Post subject: Re: Intersecting Committees
PostPosted: Sat, 16 Jun 2012 02:59:48 UTC 
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Hello, rdj5933mile5math64!

I wondered exactly how we form 30 committees with only 6 students.
I began an exhaustive list so I could familiarize myself with on the problem.
And, as a result, I solved the problem by an inelegant method: Brute Force.

Quote:
6 students formed 30 committees. .Every 2 of these 30 committees intersect,
meaning that they each have at least one member in common.
No two committees have the same members.

Prove that they can form at least one more committee
that intersects each of the existing 30 committees.

Suppose the six students are: A,B,C,D,E,F.


There are {6\choose5} = 6 five-member committees.

. . \begin{array}{cccccc}ABCDE & ABCDF & ABCEF & ABD{E}F & ACD{E}F & BCD{E}F \end{array}


There are {6\choose4} = 15 four-member committees.

. . \begin{array}{ccc} ABCD & ABEF & BCDE \\ ABCE & ACDE & BCDE \\ ABCF & ACDF & BCDF \\ ABDE & ACEF & BCEF \\ ABDF & AD{E}F & CD{E}F \end{array}


There are {6\choose3} = 20 three-member committees.
But we can use only the "first ten".

. . \begin{array}{c|c} ABC & \cancel{D{E}F} \\ ABD & \cancel{CEF} \\ ABE & \cancel{CDF} \\ ABF & \cancel{CDE} \\ ACD & \cancel{BEF} \\ ACE & \cancel{BDF} \\ ACF & \cancel{BDE} \\ ADE & \cancel{BCF} \\ ADF & \cancel{BCE} \\ BDE & \cancel{ACF}\end{array}

Each of the"second ten" is the complement of a "first ten" set.
. . Hence, they are disjoint.


Take the 6 five-member committees, the 15 four-member committees,
. . and any 9 of the three-member committees.
We will have 30 committees with at least one member in common.

Then the tenth 3-member committee satisfies the problem.


Admittedly, an inelegant solution.
But one which can be used if totally desperate.
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outermeasure
 Post subject: Re: Intersecting Committees
PostPosted: Sat, 16 Jun 2012 03:10:10 UTC 
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Soroban wrote:
Take the 6 five-member committees, the 15 four-member committees,
. . and any 9 of the three-member committees.
We will have 30 committees with at least one member in common.

Then the tenth 3-member committee satisfies the problem.


Admittedly, an inelegant solution.
But one which can be used if totally desperate.


That is not a solution at all --- you have just listed some example of intersecting sets (incidentally you listed BDE|ACF twice when you want AEF|BCD) but not an exhaustive list. For example, it could be a "dictatorship" where 1\in[n] is in every committee, or generally a "junta" (where the membership of A in \mathcal{F} depends only on A\cap [k], some k<n).

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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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rdj5933mile5math64
 Post subject: Re: Intersecting Committees
PostPosted: Sat, 16 Jun 2012 17:10:01 UTC 
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outermeasure wrote:
rdj5933mile5math64 wrote:
How do you prove that?


Spoiler:
On the other hand, if \mathcal{F}\subseteq 2^{[n]} is maximal intersecting and has size <2^{n-1}, then there exists A\subseteq [n] such that A,A^c\notin\mathcal{F}. Since \mathcal{F} is maximal, \exists B\in\mathcal{F} with A\cap B=\varnothing (so B\subseteq A^c) and \exists C\in\mathcal{F} with C\cap A^c=\varnothing, contradicting B\cap C\neq\varnothing.


Very nice! Thanks for the solution! :D

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