Josephus problem


In computer science and mathematics, the Josephus problem is a theoretical problem related to a certain counting-out game. Such games are used to pick out a person from a group, e.g. eeny, meeny, miny, moe.
In the particular counting-out game that gives rise to the Josephus problem, a number of people are standing in a circle waiting to be executed. Counting begins at a specified point in the circle and proceeds around the circle in a specified direction. After a specified number of people are skipped, the next person is executed. The procedure is repeated with the remaining people, starting with the next person, going in the same direction and skipping the same number of people, until only one person remains, and is freed.
The problem—given the number of people, starting point, direction, and number to be skipped—is to choose the position in the initial circle to avoid execution.

History

The problem is named after Flavius Josephus, a Jewish historian and leader who lived in the 1st century. According to Josephus's firsthand account of the siege of Yodfat, he and his 40 soldiers were trapped in a cave by Roman soldiers. They chose suicide over capture, and settled on a serial method of committing suicide by drawing lots. Josephus states that by luck or possibly by the hand of God, he and another man remained until the end and surrendered to the Romans rather than killing themselves. This is the story given in Book 3, Chapter 8, part 7 of Josephus's The Jewish War :
The details of the mechanism used in this feat are rather vague. According to James Dowdy and Michael Mays, in 1612 Claude Gaspard Bachet de Méziriac suggested the specific mechanism of arranging the men in a circle and counting by threes to determine the order of elimination. This story has been often repeated and the specific details vary considerably from source to source. For instance, Israel Nathan Herstein and Irving Kaplansky have Josephus and 39 comrades stand in a circle with every seventh man eliminated. A history of the problem can be found in S. L. Zabell's Letter to the editor of the Fibonacci Quarterly.
As to intentionality, Josephus asked: “shall we put it down to divine providence or just to luck?” But the surviving Slavonic manuscript of Josephus tells a different story: that he “counted the numbers cunningly and so managed to deceive all the others”. Josephus had an accomplice; the problem was then to find the places of the two last remaining survivors. It is alleged that he placed himself and the other man in the 31st and 16th place respectively.

Variants and generalizations

A medieval version of the Josephus problem involves 15 Turks and 15 Christians aboard a ship in a storm which will sink unless half the passengers are thrown overboard. All 30 stand in a circle and every ninth person is to be tossed into the sea. The Christians need to determine where to stand to ensure that only the Turks are tossed. In other versions the roles of Turks and Christians are interchanged.
describe and study a "standard" variant: Determine where the last survivor stands if there are people to start and every second person is eliminated.
A generalization of this problem is as follows. It is supposed that every th person will be executed from a group of size, in which the th person is the survivor. If there is an addition of people to the circle, then the survivor is in the -th position if this is less than or equal to. If is the smallest value for which, then the survivor is in position.

Solution

In the following, denotes the number of people in the initial circle, and denotes the count for each step, that is, people are skipped and the -th is executed. The people in the circle are numbered from to, the starting position being and the counting being inclusive.

''k'' = 2

The problem is explicitly solved when every second person will be killed, i.e..
The solution is expressed recursively. Let denote the position of the survivor when there are initially people.
The first time around the circle, all of the even-numbered people die.
The second time around the circle, the new 2nd person dies, then the new 4th person, etc.; it is as though there were no first time around the circle.
If the initial number of people were even, then the person in position during the second time around the circle was originally in position . Let. The person at who will now survive was originally in position. This yields the recurrence
If the initial number of people were odd, then person 1 can be thought of as dying at the end of the first time around the circle. Again, during the second time around the circle, the new 2nd person dies, then the new 4th person, etc.
In this case, the person in position was originally in position. This yields the recurrence
When the values are tabulated of and a pattern emerges :
12345678910111213141516
1131357135791113151

This suggests that is an increasing odd sequence that restarts with whenever the index n is a power of 2.
Therefore, if m and are chosen so that and, then.
It is clear that values in the table satisfy this equation. Or it can be thought that after people are dead there are only people and it goes to the st person. This person must be the survivor. So. Below, a proof is given by induction.
Theorem: If and, then.
Proof: The strong induction is used on. The base case is true.
The cases are considered separately when is even and when is odd.
If is even, then choose and such that and. Note that.
is had where the second equality follows from the induction hypothesis.
If is odd, then choose and such that and. Note that.
is had where the second equality follows from the induction hypothesis. This completes the proof.
can be solved to get an explicit expression for :
The most elegant form of the answer involves the binary representation of size : can be obtained by a one-bit left cyclic shift of itself. If is represented in binary as, then the solution is given by. The proof of this follows from the representation of as or from the above expression for.
Implementation: If denotes the number of people, the safe position is given by the function, where and.
Now if the number is represented in binary format, the first bit denotes and remaining bits will denote. For example, when, its binary representation is:
n = 1 0 1 0 0 1
2m = 1 0 0 0 0 0
l = 0 1 0 0 1

/**
* @param n the number of people standing in the circle
* @return the safe position who will survive the execution
* f = 2L + 1 where N =2^M + L and 0 <= L < 2^M
*/
public int getSafePosition

Bitwise

The easiest way to find the safe position is by using bitwise operators. In this approach, shifting the most-significant set bit of to the least significant bit will return the safe position. Input must be a positive integer.
n = 1 0 1 0 0 1
f = 0 1 0 0 1 1

/**
* @param n the number of people standing in the circle
* @return the safe position who will survive the execution
*/
public int getSafePosition

''k'' = 3

In 1997, Lorenz Halbeisen and Norbert Hungerbühler discovered a closed-form for the case. They showed that there is a certain constant
that can be computed to arbitrary precision. Given this constant, choose to be the greatest integer such that . Then, the final survivor is
for all.
As an example computation, Halbeisen and Hungerbühler give . They compute:
This can be verified by looking at each successive pass on the numbers through :

The general case

is used to solve this problem in the general case by performing the first step and then using the solution of the remaining problem. When the index starts from one, then the person at shifts from the first person is in position, where is the total number of people. Let denote the position of the survivor. After the -th person is killed, a circle of remains, and the next count is started with the person whose number in the original problem was. The position of the survivor in the remaining circle would be if counting is started at ; shifting this to account for the fact that the starting point is yields the recurrence
which takes the simpler form
if the positions are numbered from to instead.
This approach has running time, but for small and large there is another approach. The second approach also uses dynamic programming but has running time. It is based on considering killing k-th, 2k-th,..., -th people as one step, then changing the numbering.
This improved approach takes the form