Proof Problem Of The Day - The Chess Tournament!

In a chess tournament of nn players, everyone played against everyone. Luckily enough, there were no draws. Prove that it is possible to label the players as A1A_1, A2A_2, A3A_3, \cdots AnA_n such that AiA_i won against Ai+1A_{i+1} for i{1,2,3,n1}i \in \{1, 2, 3, \cdots n-1\}


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Note by Mursalin Habib
5 years ago

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BdMO 2014 - Sec 9

Consider the complete directed graph with nn vertices where ABA\longrightarrow B denotes AA won against BB. Then the problem can be rephrased in the language of Graph Theory as a well-known statement: Prove that every complete directed graph contains a Hamiltonian path.

We proceed by induction on nn. Verify the trivial cases n=1,2n=1,2. We assume that the statement is true for n=kn=k and consider the complete directed graph GG with n=k+1n=k+1 vertices. Take a vertex pp of GG. Now applying the inductive hypothesis, consider a Hamiltonian path {p1p2pk}\{p_1\to p_2\to\cdot\cdot\cdot\to p_k\} in G{p}G-\{p\} which clearly has kk vertices. Now let m{1,2,...,k}m\in\{1,2,...,k\} be the minimal element such that the edge ppmp \longrightarrow p_m exists. Then

{p1p2pm1ppmpk}\left\{p_1\to p_2\to \cdot\cdot\cdot\to p_{m-1}\to p\to p_m \to \cdot\cdot\cdot\to p_k\right\}

clearly is a Hamiltonian path. If such an mm doesn't exist, just append the edge pkpp_k\longrightarrow p. This completes the induction. \square

Jubayer Nirjhor - 5 years ago

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There's a shorter approach that uses strong induction.

Assume that the claim is true for a tournament of any size mkm\leq k. For a tournament of k+1k+1 players, consider any player (let's call him Nirjhor). Now divide the rest of the players into two sets; one that contains all the players Nirjhor has won against and another that contains all the players that contain all the players Nirjhor has lost against. Both of these sets have size k\leq k.

I hope everyone can take it from here.

Mursalin Habib - 5 years ago

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Great approach. But I don't see how this is shorter.

Jubayer Nirjhor - 5 years ago

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@Jubayer Nirjhor It isn't? It seemed shorter to me.

It certainly is much sleeker [to me] if not shorter.

Mursalin Habib - 5 years ago

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I also think this one sleeker as Nirjhor's proof went a bit over my head! (No Offense)

Sagnik Saha - 5 years ago

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Given 8 distinguishable rings ,find the number of possible S-ring arrangements on the four fingers(not the thumb) of one hand .(the order of the rings on each finger significant,but it is not required that each finger have a ring).

Souvik Das - 4 years, 1 month ago

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