TITLE: Number of Regions in the Plane defined by $n$ Zig-Zag Lines
QUESTION [9 upvotes]: Fellows of Math.SE, I have been scratching my head at a solution to an exercise in Donald Knuth's Concrete Math. Here is the problem:

Here is the solution (I hid it in case someone wants to solve this on their own)

Given $n$ straight lines that define $L_n$ regions, we can replace them by extremely narrow zig-zags with segments sufficiently long that there are nine intersections between each pair of zig-zags. This shows that $ZZ_n = ZZ_{n−1} +9n−8$, for all $n > 0$; consequently $ZZ_n = 9S_n −8n+1 = \frac{9n^2 − 7n}{2} +1$. 

$L_n$ is the number of regions definable by $n$ lines, which is solved as an example earlier in the text. It equals $S_n + 1$, where $S_n = \sum_{k=1}^n k$. I am having difficulty understanding the recurrence solution. I'll hide my question, just to be extra careful I don't spoil this wonderful problem for anyone.

 Where does the "$-8$" come from? Is the recurrence better understood as $ZZ_n = ZZ_{n−1} +9(n-1)+1$? Or does that further convolute the meaning of the recurrence? I figure there must "-8" must have to do with "lost regions" due to the half-lines, but I am having trouble putting my finger on it.

I really love this problem and would love to understand the solution in its entirety! Thank you!

REPLY [2 votes]: I figured a different approach. We all know by now, 

Finding 1: Each new zig-zag will intersect each old zig-zags at $9$ points.
Finding 2: When two zig-zags intersect, they create $8$ finite regions around intersections (see colored regions from the picture of bryanj's answer).

Let's assume each zig-zag is so thin that it can be drawn as a line (two parallel lines are too close to each other) and the segment connecting two parallels is quite long. Now to draw the $n$ zig-zags on a plane, first we will draw $n$ lines on that plane intersecting each other. Then we will replace $n$ lines with $n$ 'thin' zig-zags I was talking about. 
For example, if $n = 3$, we draw 'simple lines' first, and we get 7 regions (See the picture below). It has $L_3 = 7$ regions and $\binom{3}{2} = 3$ intersections. 

Now replacing 'simple lines' with zig-zags, we get, this picture. 

So, $L_3 = 7$ regions are still there. For each of $\binom{3}{2} = 3$ intersections, we get $8$ regions (from finding 2). Total $L_3 + 8\times\binom{3}{2}$ regions. 
Finally, for $n$ zig-zags, $ZZ_n = L_n + 8\times\binom{n}{2}$