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Proof Problem Of The Day - The Chess Tournament!

In a chess tournament of \(n\) players, everyone played against everyone. Luckily enough, there were no draws. Prove that it is possible to label the players as \(A_1\), \(A_2\), \(A_3\), \(\cdots\) \(A_n\) such that \(A_i\) won against \(A_{i+1}\) for \(i \in \{1, 2, 3, \cdots n-1\}\)


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Note by Mursalin Habib
2 years, 9 months 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 · 1 year, 10 months ago

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BdMO 2014 - Sec 9

Consider the complete directed graph with \(n\) vertices where \(A\longrightarrow B\) denotes \(A\) won against \(B\). 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 \(n\). Verify the trivial cases \(n=1,2\). We assume that the statement is true for \(n=k\) and consider the complete directed graph \(G\) with \(n=k+1\) vertices. Take a vertex \(p\) of \(G\). Now applying the inductive hypothesis, consider a Hamiltonian path \(\{p_1\to p_2\to\cdot\cdot\cdot\to p_k\}\) in \(G-\{p\}\) which clearly has \(k\) vertices. Now let \(m\in\{1,2,...,k\}\) be the minimal element such that the edge \(p \longrightarrow p_m\) exists. Then

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

clearly is a Hamiltonian path. If such an \(m\) doesn't exist, just append the edge \(p_k\longrightarrow p\). This completes the induction. \(\square\) Jubayer Nirjhor · 2 years, 9 months ago

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@Jubayer Nirjhor There's a shorter approach that uses strong induction.

Assume that the claim is true for a tournament of any size \(m\leq k\). For a tournament of \(k+1\) 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 \(\leq k\).

I hope everyone can take it from here. Mursalin Habib · 2 years, 9 months ago

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@Mursalin Habib Great approach. But I don't see how this is shorter. Jubayer Nirjhor · 2 years, 9 months 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 · 2 years, 9 months ago

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@Mursalin Habib I also think this one sleeker as Nirjhor's proof went a bit over my head! (No Offense) Sagnik Saha · 2 years, 9 months ago

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