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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TopNewestGiven 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 9Consider 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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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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– Jubayer Nirjhor · 2 years, 9 months ago

Great approach. But I don't see how this is shorter.Log in to reply

It certainly is much sleeker [to me] if not shorter. – Mursalin Habib · 2 years, 9 months ago

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– Sagnik Saha · 2 years, 9 months ago

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