# Concrete Mathematics: My Notes on Josephus Problem Induction

The explanations in Concrete Mathematics are very good if a bit terse. So, as before, I will add some extra detail in a blog post to help cement (🥁) my knowledge. I am also shouting it into the void in case it helps someone else.

See the book for full context; I’ll just cover the induction related steps.

# The Recurrences and Closed Form

The recurrence comes in two flavours: odd and even. The complete recurrences are thus:

\begin{align} J(1) &= 1;\\ J(2n) &= 2J(n) - 1, \ \ \ \ \text{for n} \geq 1\\ J(2n+1) &= 2J(n) + 1, \ \ \ \ \text{for n} \geq 1\\ \end{align}

And the closed form is as follows

$J(2^m+\ell)=2\ell+1$

# Even Induction

The book goes on to give an example of the even induction

$J(2^m+\ell)=2J(2^{m-1} + \ell/2) - 1 = 2(2\ell/2 + 1) - 1 = 2\ell + 1$

by (1.8) and the induction hypothesis; this is exactly what we want.

What confused me at first was the leap

$2J(2^{m-1} + \ell/2) - 1 = 2(2\ell/2 + 1) - 1$

This is the key induction step - this is not done via algebraic manipulation. We are substituting the closed form that is equivalent to $J(n)$. Our closed form proposal is actually for $2J(n)-1=2l+1$ so we need to refit it to be for $J(n)$. Thus $l$ is in the previous power of two block (remember $J(n)$ not $2J(n)$) and half as big: $l/2$. So the $J(n)$ closed form is $2l/2 +1$.

So we plug that in and then he next step, however, is done via algebra:

$2(2\ell/2 + 1) - 1 = 2\ell + 1$

Completing the even induction.

# Odd Induction

Credit: I was stuck on this until helped out by the ever helpful Brian M. Scott on Math Stackexchange.

The book leaves the odd induction as an exercise to the reader so this reader thought to undertake it. As a reminder have the closed form

$J(2^m+\ell)=2\ell+1$

And odd recurrence

$J(2n+1) = 2J(n) + 1, \ \ \ \ \text{for n} \geq 1\\$

As with even our closed form is such that $2J(n)+1 = 2\ell+1$ except we are adding one and we also have to make sure that the $\ell$ is odd. So the “conversion process” is as follows; end result highlighted in blue

Starting point:

$2n+1 = 2^m+\ell = 2\ell+1$

First divide by two:

$n+1 = 2^{m-1}+\ell/2 = \frac{2(\ell+1)}{2}$

Then subtract one:

$n = 2^{m-1}+\frac{\ell-1}{2} = \color{blue}{\frac{2(\ell-1)}{2} + 1 }$

We can then substitute in our “n”-ified closed form equivalent into the recurrence (replace the $J()$ recurrent call) to do the induction; highlighted in blue

\begin{align*} J(2^m+\ell)&=2J\left(2^{m-1}+\frac{\ell-1}2\right)+1\\ &=2\left(\color{blue}{\frac{2(\ell-1)}2+1}\right)+1\\ &=2\ell+1\;, \end{align*}

Giving us what we want.

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