A General Solution to Josephus Problem

Yoav Francis

July 2007

Topics in Number Theory, 2007/2COMAS – College of Management – Academic Studies, Rishon Lezion, Israel

Overview and Additional Problems

Motivation

• Given , find , where denotes the number of people in the initial circle.

• Let us first consider the case in which .

Recursive Solution to

Where:

A General Method for For a group of people, each time we iterate on a person, we can provide him with a new number. In such a scenario – the numbers 1,2 would become . The number 3 would then be killed. 4,5 would become . 6 would be killed, and so on.

Finally, would become , and would be killed and so on.

Eventually the person denoted as will be the sole survivor.

Illustration for :

A General Method for

• The person to be killed is denoted (Example: the 10th person is numbered “30”, the 9th is numbered “27” and so on)

• We can find out who is the survivor if we find the original number of the person numbered (let us mark )

• If then the person numbered necessarily had a previous number, which we can find in the following manner:

• or (Similarly to how we found the next number of and in the previous slide)

• Thus . The previous number of the person was or – meaning, it was (using )

A General Method for

We have received an algorithm for :

A General Method for

Let us note that this can be somewhat improved. The algorithm is not recursive, neither has a “closed” shape – but still allows us to compute the answer rather quickly for a large .

Luckily, we can improve it.

• Let us define and use it instead of .

• This change matches numbering from to 1 instead of from to .

• This yields the following:

A General Method for

We can now write the algorithm in the following manner:

This is a much simpler form as enters the computation in a simple manner.

The General Algorithm for Josephus Problem

In a matter of fact, we can use the same justifications we used to show that the survivor for (where every person is killed) is computed in the following way:

This is the general algorithm. Let us note that for we’d get in every phase , so for the case of we get:

The General Algorithm for Josephus Problem

We can attempt to bring the algorithm to a “nicer” form.

It cannot be brought to a closed form (but for the case of ), due to the “ceiling” function that we perform on every phase.

However – we can still tidy-up the algorithm. Let us recursively define:

These numbers do not have a “nice” form but for , but if we agree to accept the expression as a “known expression” – then the solution for the general case of the Josephus problem would be:

(Where is the smallest one for which the following holds: )

Additional Problems Derived From Josephus Problem

Finding the second-to-last person remaining in Josephus problem (For example, if we want to find the last 2 people standing).

Let us denote and get:

Additional Problems Derived From Josephus Problem

Another problem: Josephus finds himself in the position . Given that he can choose the “skip factor”(, which should he choose so he stays alive?

Let us define and assume that . By Bertrand’s Postulate there exists a prime number . We can also assume that .

We’d want to choose a q so the following holds:

• and:

Additional Problems Derived From Josephus Problem

Now, the people who would be removed (“killed”) are:

The position would not be killed (we may get but that is still valid for the problem)

Example:

• . Josephus got and will decide: (This holds for the 2 requirements of ).• LCM(1,….,10) = 2520

Thanks