## All posts in this series:

• One Potato: A description of the problem
• Two potato: The basic equations
• Three potato: An algebraic formula for the last person left
• Four: A neat calculation method using binary numbers
• Five potato: F# functions to check the various formulae
• Six potato: Using a calculation graph to see the pattern
• Seven potato: Generalizing to other intervals
• More: An efficient algorithm for arbitrary intervals
• Online research: In which I discover that it is called the Josephus problem

## Introduction

In previous posts in the series I derived Mathematical formulae (and provided F# scripts) for calculating the last person left in a circle if every second person is asked to leave (continuing until only one person remains).

In the third post in the series I developed an algebraic formula. However algebra often fails to provide insight into the solution. In this post I'd like to provide that insight by taking a very different approach to the problem.

## Finding a pattern

In the second post in the series I derived the basic equations needed to solve the problem:

\begin{align*} f(1) &= 1 \tag{1} \\ f(2n) &= 2.f(n) - 1 \tag{2} \\ f(2n+1) &= 2.f(n) + 1 \tag{3} \end{align*}

Notice how equations 2 and 3 both reference $f(n)$. So knowing $f(n)$ allows you to generate two other numbers: $f(2n)$ and $f(2n+1)$.

Let's generate a graph showing how the values are determined:

There seems to be a pattern here. The output in the first row is 1. The outputs in the second row are 1, 3. The outputs in the 3rd row are 1, 3, 5, 7. It gets a bit harder to see, but if we expand another level down we get 1, 3, 5, 7, 9, 11, 13, 15:

So the inputs in row r go from $2^{r-1}$ to $2^{r}-1$ in increments of 1.

And the outputs in row r seem to go from 1 to $2^{r} - 1$ in increments of 2.

## Insight

A bit further on I will prove this pattern using the principle of induction. But first let's look for the insight into why this pattern is so.

There are two things to show:
• The output of the first number in each row must always be 1
• Consecutive outputs in a row always differ by 2

Why is the output of the first number in a row always 1? In other words why does $f(2^{m}) = 1$?

If the size of the circle is a power of 2, then exactly half the people will be removed during one trip around the circle. And the last person removed will be the person just before person 1. So person 1 is not going to be eliminated in the next round of eliminations either.

And since half of a power of 2 is also a power of 2, this pattern is going to repeat itself. So person 1 is never going to be eliminated. So the answer has to be 1.

Why does each pair of successive nodes in a row always differ by 2?

Subtract equation 2 from equation 3 in the basic equations. The $f(n)$ terms cancel out, showing that: $$f(2n+1) - f(2n) = [2.f(n) + 1] - [2.f(n) - 1] = 2$$

But we also need to consider the other way of pairing successive terms (i.e. an odd followed by an even argument).

In other words, why is $f(2n+2) - f(2n+1) = 2$ for all n, except when $2n+2$ is a power of 2?

We will prove this by induction. Suppose $f(n+1) - f(n) = 2$ in the previous row. Then:

\begin{align*} f(2n+2) &= f(2(n+1)) \\ &= 2.f(n+1) - 1 & \text{by (2)} \\ \\ \text{So: }f(2n+2) - f(2n + 1) & = [2.f(n+1) - 1] - [2.f(n) + 1] & \text{by (3)} \\ &= 2.[f(n+1) - f(n)] - 2 \\ &= 2.[2] - 2 \\ &= 2 \end{align*}
Look at the second row in the graph: $f(3) - f(2) = 2$. Starting at this row, the property of successive outputs differing by 2 is going to ripple down from row to row in the graph.

Now let's prove this formally. We'll do it by induction again, but this time we'll use the row number in the graph as the induction step (or more correctly the row number minus 1).

## Proof by induction

### The inductive step

Suppose that for some integer k, the numbers from $2^k$ to $2^{k+1} - 1$ satisfy:

$f(2^{k} + i) = 2i + 1 \tag{4}$.
Then we wish to show that this holds true for k+1 as well. In other words, the numbers from $2^{k+1}$ to $2^{k+2} - 1$ should satisfy $f(2^{k+1} + j) = 2j + 1$.

To prove this we consider the two cases of odd and even values of j separately:

Proof for even values:

Let $j = 2i$ for any $i \in \left\{ 0, 1, ..., 2^{k}-1 \right\}$.

First note that j will have values in the following range:

$j \in \left\{0, 2, ..., 2^{k+1} - 2 \right\} \tag{5}$

Then:

\begin{align*} f(2^{k+1} + j) &= f(2.2^{k} + 2i) \\ &= f(2[2^{k} + i]) \\ &= 2.f(2^{k} + i) - 1 & \text{by (2)} \\ &= 2.( 2i + 1 ) - 1 & \text{by (4)} \\ &= 2.( j + 1 ) - 1 \\ &= 2j + 1 \end{align*} Derivation for odd values:

Let $j = 2i + 1$ for any $i \in \left\{ 0, 1, ..., 2^{k}-1 \right\}$.

Note that j will have values in the following range:

$j \in \left\{1, 3, ..., 2^{k+1} - 1 \right\} \tag{6}$

Then:

\begin{align*} f(2^{k+1} + j) &= f(2.2^{k} + 2i + 1) \\ &= f(2[2^{k} + i] + 1) \\ &= 2.f(2^{k} + i) + 1 & \text{by (3)} \\ &= 2.( 2i + 1 ) + 1 & \text{by (4)} \\ &= 2.( j ) + 1 \\ &= 2j + 1 \end{align*}

So combining these two cases, we see that $f(2^{k+1} + j) = 2j + 1$ for all $j \in \left\{ 0, 1, ..., 2^{k+1}-1 \right\}$, from (5) and (6).

### The base case

When k = 0:

$f(2^{0}) = f(1) = 1 = 2 . 0 + 1$

So when k = 0 and $i \in \left\{0, ..., 2^{k} - 1\right\} = \left\{0\right\}$, $f(2^{k} + i) = 2i + 1$

### Conclusion

So this is our proof by induction of the following:

For all non-negative integers $k$ and $i$, with $i \in \left\{ 0, 1, ..., 2^{k}-1 \right\}$:

$f(2^{k} + i) = 2i + 1$

But for $n = 2^{k} + i$ this gives: \begin{align*} f(n) &= 2i + 1 \\ &= 2.[i + 2^{k} - 2^{k}] + 1 \\ &= 2.[2^{k} + i] + 1 - 2^{k+1} \\ &= 2n + 1 - 2^{k+1} \\ &= 2n + 1 - 2^{{\lfloor log_2 n \rfloor} + 1} \end{align*} And that is the same as the equation derived in the third post of the series.

## Next time

In the next blog post in the series I will try to generalize the result to intervals other than 2.