This is Part 3 of an Npart series.
 The Josephus Problem: Part 1: The Problem
 The Josephus Problem: Part 2: Two Examples
 The Josephus Problem: Part 3: Solving the Double Step Case
Table of Contents
 Solving the Double Step Case
 Algorithm D: Using Doubling Permutation
 Why Does Algorithm D Work?
 Algorithm D Summary
Solving the Double Step Case
The Josephus Problem for a step size of \(m = 2\) can be solved two ways:
 Algorithm D: Doubling Permutation Algorithm
 Algorithm S: Label Skipped Nodes Algorithm
In this blog post we cover Algorithm D, which makes use of a doubling permutation.
In the first section (Algorithm D: Using Doubling Permutation) we start by showing the recipe as applied to the case of \(n = 11, m = 2\).
The recipe is straightforward to apply, but it is not at all clear how it works. We start by introducing the algorithm, then carrying it out for the case of \(n = 11, m = 2\) (this is a familiar case that we have covered in prior posts).
We make observations about the procedure by comparing the case of \(n = 8, m = 2\) (a special case where the circle size is a power of two) before returning to the example of \(n = 11, m = 2\) to make generalizations and give a shortcut method for the general doublestep case.
Algorithm D: Using Doubling Permutation
In Exercise 29, Knuth asks a question that lays out a solution procedure for the special case of \(m = 2\), and asks the reader to prove it:
Prove: the cycle form of the Josephus permutation when \(m = 2\) can be obtained by expressing the "doubling permutation" of \(\{1, 2, \dots, 2n\}\), which takes \(j\) into \((2j) \mod (2n+1)\) into cycle form, then reversing L and R, and erasing all numbers larger than \(n\).
Fortunately, we don't have to work out the solution procedure for ourselves; unfortunately, it is not at all obvious why the solution procedure works, so once we have seen how to apply it, we have to do a bit more work before we can understand it.
The Algorithm
Given an input size \(n\), the algorithm steps are:

Write the doubling permutation (in table form)

Factor the doubling permutation into cycles

Reverse each cycle and remove values larger than n
The result will be the Josephus permutation cycles.
Write the Doubling Permutation
As mentioned above, we will start with the concrete example of \(n = 11, m = 2\).
We start by writing the doubling permutation for the integers \(1 \dots 2n\):
When a number is greater than \(2n+1\), it is reduced \(\mod (2n+1)\), so the next few terms of the doubling permutation written out and reduced modulo \(2n+1\) are:
which reduces modulo \(2n+1\) to:
After 12, we reach 1 again, the starting value, so further doublings will result in repetition of the elements we have so far.
Next, continue the process with the remaining elements. Start with the smallest element not included in the cycle found above, which is 5, and continue the operation of doubling and reducing modulo \(2n+1\). Doing this results in the terms:
which reduces modulo \(2n+1\) to:
Repeating this until the first element repeats yields all of the remaining elements:
Now the final doubling permutation can be written as the product of the two cycles:
We now have the cycles of the doubling permutation! Step 2 finished.
Table Method
Slightly more convenient than writing out the cycles the way we did above is to create a table, with a column for \(j = 1 \dots 2n\), a column for \(2j\), and a column for \((2j) \mod (2n+1)\).
j  2j  2j mod 2n+1

1  2  2
2  4  4
3  6  6
4  8  8
5  10  10
6  12  12
7  14  14
8  16  16
9  18  18
10  20  20
11  22  22
12  24  1
13  26  3
14  28  5
15  30  7
16  32  9
17  34  11
18  36  13
19  38  15
20  40  17
21  42  19
22  44  21
Now the cycles can be constructed by starting with the \(j\) column of the first row in the table (1), reading off the corresponding entry (2) in the \((2j) \mod (2n+1)\) column, adding the pair \(1 \rightarrow 2\) to the cycle, and moving to the row of the table with that corresponding \(j\) value.
The next step takes us to row 2 of the table, where we read off the value 4 from the right column, add the pair \(2 \rightarrow 4\) to the cycle, and keep moving.
Eventually, the procedure will complete the first cycle and take us back to 1. Since there are still rows in the table left, repeat the procedure, starting with the first available row of the table (5), and keep going until all entries of the table are gone. This will yield the same cycles as the process described above, but the bookkeeping is slightly easier:
Reverse the Doubling Permutation
The next step is to reverse the permutation from left to right, which means we step through all cycles from left to right, and step through each cycle from left to right.
Starting with the doubling permutation:
We obtain the reverse:
Trim the Reversed Doubling Permutation
Now we eliminate any numbers from the reversed doubling permutation that are larger than \(n = 11\), to get the trimmed permutation:
This is the final Josephus permutation. The one remaining step is to rewrite the cycles in standard "sorted" order:
Congratulations! You just solved the problem.
Why Does Algorithm D Work?
If at this point you are scratching your head in wonder at the seeming black magic involved in this algorithm, then welcome to The Art of Computer Programming! (Seriously, the exercises are chock full of questions like this one.)
To understand why Algorithm D works, it is instructive to consider a case where \(n\) is a power of two, and make some observations and generalizations to uncover the pattern at work in Algorithm D.
An Important Observation
Start with an important observation: each time we complete one circuit of the circle, we have reduced the number of items in the circle by
If our circle size is even, that means we cut our circle size in half each time through the circle.
The halving operation means the Josephus problem is intimately linked with powers of 2 (or, base 2 logs), so that's where the powers of 2 will come from.
Three Facts
Now we use this observation about halving to state 3 facts about the case:
Fact 1: If we are considering an item at an oddnumbered location in the circle, we have already been through the circle at least once.
(Explanation: The first time through the circle, all of the items in the circle at evennumbered locations will be removed, so only items at oddnumbered locations around the circle are left after the first pass.)
Fact 2: When the circle size is a power of 2, the first item in the circle is the last item removed.
(Explanation: If we have a circle the size of a power of 2, we will always remove the last item in the circle when we complete a circuit (because we will always reach it using twosteps), which means we always skip item 1 if the circle size is a power of 2.)
(Addendum: If a circle size starts as a power of 2, going through the circle once and removing elements will exactly halve the circle size, and preserve the property that the circle size is a power of 2.)
Fact 3: When the circle size is not a power of 2, the size will eventually be reduced to a power of 2. Then the item at the starting position in that round will be the last item removed.
(Explanation: Combining Fact 1 and Fact 2, we can see that eventually we will always reach the special case of the circle size being a power of 2, since with each step, we reduce the circle size by 1. For example, for a circle size of 22, we must carry out 6 removal operations to reach a circle size of 16, at which point the item at the starting position will be the last item to be removed from the circle.)
Now we will use these 3 facts to understand Algorithm D.
Power of Two Example
Let's run through a Josephus solution for an example where \(n\) is a power of two, specifically, \(n = 8, m = 2\).
We start by creating the doubling permutation table, with columns for \(j\), \(2j\), and \((2j) \mod (2n+1)\):
j  2j  2j mod 2n+1

1  2  2
2  4  4
3  6  6
4  8  8
5  10  10
6  12  12
7  14  14
8  16  16
9  18  1
10  20  3
11  22  5
12  24  7
13  26  9
14  28  11
15  30  13
16  32  15
The variable \(j\) takes on \(2n\) values, \(j = 1 \dots 2n\), each of which are doubled and reduced modulo \(2n+1\).
From column 3 we can see why the doubling permutation requires us to consider \(2n\) items instead of \(n\) items: because the first \(n\) items are always even (cuz, uh, we're doubling), and what we want is coverage of odds and evens in the range \(1 \dots n\). This is also why we reduce \(\mod (2n + 1)\), which is guaranteed to be odd  if we reduced \(\mod 2n\) we would run into the same problem of wrapping back garound to the even integers.
If we step through the doubling permutation to factor it
into its cycles, we start at 1 and connect it to 2.
This states that item 2 is removed 1st, so we write
1 > 2
for the doubling permutation cycle, and keep
in mind that it will be reversed to 2 > 1
when we
write the final Josephus permutation cycle.
The next step is to move to 2, and connect it to 4,
so 2 > 4
(also recognizing that this means the
opposite connection 4 > 2
, that item 4 is removed
second, in the final Josephus permutation.)
Now we would expect to see 3 > 6
, since we know that
item 6 is removed 3rd with the same certainty that we know
item 4 is removed 2nd. But the problem is that it doesn't
fit into the existing cycle, so it may belong later in the
cycle or in another cycle altogether. We can't be certain.
This is why we multiply our step size by 2 each time (2, 4, 8, 16), instead of incrementing it by 2 each time (2, 4, 6, 8, 10).
Fact 1 stated that any items at evennumbered positions are eliminated on the first pass  we can see that manifest as we write the doubling cycle. Once we reach 16 (double the size of the circle), the integers begin to loop back around through the odd numbers (also thanks to the choice of the odd modulus \(2n+1\)):
Consider the \(16 \rightarrow 15\) connection. Stepping through the (doubled) circle the first time through, we end on 16. The next doubling takes us 16 more steps forward, but that's 1 less than the size of the circle (again due to the choice of an odd modulus \(2n+1\)), so we end up at \(2n  1\). 15 is larger than the size of the circle, so it will be excluded in the final Josephus cycle.
Continuing to follow this through, we reach 9, which will also be excluded from the Josephus cycle, and 9 returns back to 1 and completes the cycle.
Now we continue the process with the remaining items in
the table, starting with the smallest, which is 3.
Doubling 3, we see the 3 > 6
connection we mentioned
earlier, which we expected to see since it indicates the
known fact that item 6 will be removed 3rd.
Finally, reversing and trimming the values larger than 8:
and writing in sorted order:
Fact 2 tells us that the item in position 1 is the
last to be removed, since the size of the circle \(n = 8 = 2^3\).
We confirm the cycle does contain \(1 \rightarrow 8\)
(meaning, the item in position 1 will be removed
eighth).
Not a Power of Two Example
Now that we've looked at an example where the size of the circle is a power of 2, let's return to the original example that we presented in Part 2 and the beginning of this blog post, namely, \(n = 11, m = 2\).
Here is the doubling permutation table again:
j  2j  2j mod 2n+1

1  2  2
2  4  4
3  6  6
4  8  8
5  10  10
6  12  12
7  14  14
8  16  16
9  18  18
10  20  20
11  22  22
12  24  1
13  26  3
14  28  5
15  30  7
16  32  9
17  34  11
18  36  13
19  38  15
20  40  17
21  42  19
22  44  21
As before, the last column splits into odd and even, with all the evens coming first, so we consider \(2n\) elements to cover all \(n\) odd and even items, and use an odd modulus to ensure we don't wrap back around to the even numbers the second time through.
We follow the entries of the table to factor this permutation into cycles, as above, yielding the two cycles:
and
Finally, we reverse these and eliminate the numbers larger than \(n\), giving the Josephus permutation cycle:
or, rearranging into sorted order:
However, there is a faster way to figure out the last item in the circle that will be removed that only requires us to carry out 3 steps.
The Power of Two Shortcut
Unlike the circle whose size was a power of two, we don't know ahead of time who goes last. However, Fact 3 tells us that eventually our circle will reach the size of a power of two. For this case it only takes 3 steps (11  8). At that point, the circle item at the starting point for the next step will be the last item removed.
Here is the diagram for the third removal operation (left labels show removal indices, right labels show circle position indices):
Fact 3 is in effect here, because after the third item is removed, our starting "cursor" is on item 7 in the circle, and the circle is of size \(2^3\), so from Fact 2 we know that 7 will be the last item removed from the circle.
We can see from the final permutation that \(7 \rightarrow 11\), so indeed, this holds true.
To translate this into mathematical terms, we are carrying out \(r\) operations, where
and at the conclusion of those \(r\) operations, we mark the starting item for the next round. That item will be the last item to be removed.
Algorithm D Summary
To summarize what we covered in this blog post:
This post covers a method for solving the Josephus problem for the special case of a step size of two, \(m = 2\).
We started by covering the recipe for Algorithm D, which uses a doubling permutation to obtain the Josephus permutation in cycle form. We illustrated the three steps of the recipe (populate the table, write the doubling permutation cycles, then filter and reverse) using an example case.
Then we got into the question of why Algorithm D works, by making an observation and building upon it. We deduced several facts that allowed us to come up with a shortcut for the \(m = 2\) case.
Our final solution technique (making use of the abovementioned shortcut) is as follows:

Carry out as many removal operations as needed to make the circle the size of a power of 2 (see formula for \(r\) above)

The starting item for the next round (where our "removal cursor" will start counting) will be the last item removed from the circle.
So if you find yourself being rounded up in some sort of weird mass execution scheme that happens to be a Josephus problem with a step size of 2, the key to surviving is to remember your powers of 2, to carry out \(r = n  2^{ \left\lfloor{ \lg{n} }\right\rfloor }\) removal operations, and to grab the next open spot in the circle.