# Concrete Mathematics Chapter 1 Notes

For the first post of this hopefully long series, I have a few notes I wrote down as I was reading Chapter 1. Nothing revolutionary, but it gives me a chance to play with math notation.

# Lines in the Plane

I must admit that my memories of Geometry are far, far away (the subject was not addressed at all when I studied Mathematics at university, and I had no need for Geometry in my work since), so I spent perhaps an unreasonable amount of time to check the validity of the most elementary steps.

It goes without saying that the exercises of a Geometric nature are particularly challenging (as if I needed the extra difficulty).

## Intersecting lines

The notion that one line will add $k$ new regions if it intersects other lines at $k-1$ points is due to the fact that $k-1$ distinct lines define at least $k$ regions (more if they are not all parallel), and one more line that intersects them all will divide these $k$ regions in two.

# Josephus Problem

## $J(5 \cdot 2^m) = 2^{m+1} + 1$

This is based on the fact that $J(10) = 5$ and $J(2n) = 2J(n) -1$.

By induction:

Base case: it is true for $m = 1$: $J(5\cdot 2) = J(10) = 5 = 2^{1+1} + 1$

Recurrence: assuming it is true for $m$,

\begin{aligned} J(5\cdot 2^{m+1}) &= J(2(5\cdot 2^m))\\ &= 2J(5\cdot 2^m) - 1&&\text{as } J(2n) = 2J(n) -1\\ &= 2(2^{m+1}+1) - 1&&\text{induction hypothesis}\\ &= 2\cdot 2^{m+1} + 2 - 1\\ &= 2^{m+2} + 1 \end{aligned}

## $A(2^{m}+l) = 2^{m}$

It took me a while to convince myself that the $l$ was not a problem here. This can be seen by considering $l$ in binary notation, and using $A(2n) = 2A(n)$ and $A(2n+1) = 2A(n)$ to remove the rightmost bit.

That is, with $2^m > l = (b_{m-1}b_{m-2}\cdots b_{1}b_{0})_2$,we have:

\begin{aligned} A(2^{m}+l) &= A(2^{m}+(b_{m-1}b_{m-2}\cdots b_{1}b_{0})_2)\\ &= 2A(2^{m-1}+(b_{m-1}b_{m-2}\cdots b_{1})_2)\\ &= 2^{2}A(2^{m-2}+(b_{m-1}b_{m-2}\cdots b_{2})_2)\\ &= 2^{3}A(2^{m-3}+(b_{m-1}b_{m-2}\cdots b_{3})_2)\\ &= \cdots \end{aligned}

At each iteration, whether $b_i$ is $0$ or $1$, we can ignore it when dividing by $2$. And as $2^m < l$, it takes no more than $m$ steps (removing the $m$ bits $b_0$ to $b_{m-1}$) to reduce $A(2^m+l)$ to $2^mA(1) = 2^m$

$$f \left( ( b_m b_{m-1} \cdots b_1 b_0)_d \right) = \left( \alpha_{b_m} \beta_{b_{m-1}} \beta_{b_{m-2}} \cdots \beta_{b_1} \beta_{b_0} \right)_c$$
is so unnaturally smart and simple that I thought the proof must be missing. But in fact it is indeed trivial, and just as the book states, follows from the rewriting of the argument in base $d$, then recurrence over $m$ (with $m$ the number of digits or the argument in base $d$).