# Concrete Mathematics: 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. I will repeat that here and then do my own (simplified) take on thereafter. First Brian’s reply:

Your’re not applying the recurrence for the odd case correctly. Suppose that $2n + 1 = 2^m + \ell$, where $0 \leq \ell \leq 2^m$. The recurrence is $J(2n + 1) = 2J(n) + 1$, and n here is $\frac12(2^m + \ell - 1)$, so

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

as desired

I will do some colour highlighting and ever more verbose explanation for my future self here. The core is we want to convert $2n + 1$ into $n$ hence dividing by two and subtracting one:

$\frac{2(n + 1)}{2} - 1 = n$

We apply these operations during the induction step to turn $2\ell + 1$ into $\frac{2(\ell +1)}{2} - 1$. Notice the division by two and subtraction by one? And now with gusto and in context of the induction step where we “insert” the closed form representing n into the recurrence “framework” (substitution highlighted in blue):

\begin{align} J(2^m + \ell) &= 2\left(\color{blue}\frac{2(\ell + 1)}{2} - 1\color{#3d4144}\right) + 1 \\ &= 2\ell + 1 \end{align}

Giving us what we want.

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