The Josephus problem is a classic puzzle that involves a group of n people arranged in a circle. The problem is named after the Jewish historian Flavius Josephus, who according to legend, had to use this strategy to avoid being captured and killed by the Romans during the Jewish-Roman War.
The problem statement is as follows: There are n people standing in a circle, and you start eliminating every kth person, moving clockwise around the circle. The process continues until only one person remains. The task is to find the position of the last person standing.
For example, if there are 7 people in a circle and k = 3, the order in which they are eliminated is:
1 2 3 4 5 6 7 x (person 3 is eliminated) 1 2 4 5 6 7 x (person 6 is eliminated) 1 2 4 5 7 x (person 4 is eliminated) 1 2 5 7 x (person 2 is eliminated) 1 5 7 x (person 7 is eliminated) 1 5 (person 1 is the last one standing)
The Josephus problem has a closed-form solution, which is based on the binary representation of n and k. The last person standing can be found using the following recursive formula:
j(n, k) = (j(n – 1, k) + k – 1) % n + 1, with j(1, k) = 1
Here, j(n, k) represents the position of the last person standing for n people and k steps. The formula calculates the position of the last person standing by recursively calling the same function for (n – 1) people and adding k-1 to the result (to account for the people who have been eliminated so far). The modulus operator (%) is used to handle the case when the result of the addition exceeds the remaining number of people in the circle. Finally, the result is shifted by 1 to account for the 1-based indexing of the circle.
The Josephus problem has applications in computer science, such as in the design of data structures and algorithms that involve circular buffers and ring buffers. It is also a popular problem in recreational mathematics and has been studied by many mathematicians and computer scientists over the years.
fun josephus(n: Int, k: Int): Int {
if (n == 1) {
return 1
}
return (josephus(n - 1, k) + k - 1) % n + 1
}josephus function takes two arguments: n is the number of people in the circle and k is the number of steps to move before eliminating a person.
The base case of the recursion is when there is only one person left in the circle, in which case their position is 1. Otherwise, the function recursively calculates the position of the last person standing for (n - 1) people and adds (k - 1) to the result, as described in the formula. The modulus operator is used to ensure that the result is in the range 1..n, and the position is shifted by 1 to account for the 0-based indexing of the function.
fun josephus(n: Int, k: Int): Int {
if (n == 1) {
return 1
}
return (josephus(n - 1, k) + k - 1) % n + 1
}
fun main() {
val n = 7
val k = 3
val position = josephus(n, k)
println("The last person standing is at position $position.")
}This program calculates the position of the last person standing for a circle of 7 people and k = 3. The result is printed as “The last person standing is at position 4.”