How to resolve the algorithm Round-robin tournament schedule step by step in the Rust programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Round-robin tournament schedule step by step in the Rust programming language

Table of Contents

Problem Statement

A round-robin tournament is also known as an all-play-all-tournament; each participant plays every other participant once. For N participants the number of rounds is N-1 if N is an even number. When there are an odd number of participants then each round one contestor has no opponent (AKA as a "bye"). The number of rounds is N in that case. Write a program that prints out a tournament schedule for 12 participants (represented by numbers 1 to 12).

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Round-robin tournament schedule step by step in the Rust programming language

Source code in the rust programming language

fn round_robin(n: usize) {
    assert!(n >= 2);
    let mut n = n;
    let mut list1: Vec<usize> = (2..=n).collect();
    
    if n % 2 == 1 {
        list1.push(0); // 0 denotes a "bye".
        n += 1;
    }
    
    for r in 1..n {
        print!("Round {:2}:", r);
        let list2 = vec![1].into_iter().chain(list1.iter().cloned()).collect::<Vec<_>>();
        
        for i in 0..(n / 2) {
            print!(" ({:>2} vs {:<2})", list2[i], list2[n - i - 1]);
        }
        
        println!();
        list1.rotate_right(1);
    }
}

fn main() {
    println!("Round robin for 12 players:\n");
    round_robin(12);

    println!("\n\nRound robin for 5 players (0 denotes a bye):\n");
    round_robin(5);
}


  

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