The book's solution consists of two parts. I'll go through Part 1 today and see if anybody can solve it, and then I'll go through Part 2 tomorrow.
Part 1 should take you less than 15 minutes to figure out. The number it generates will be used in Part 2 to crank out (in a trivial amount of time) the final answer, which will be the mean average.
Network theory (a.k.a. graph theory) is a branch of mathematics that looks at points joined by arcs. The arcs (at least in our case of the knight, for which moves are reversible) are undirected--that is, they do not have arrows on them. The points are usually termed "vertices" and the arcs are usually termed "edges," so I will go with that terminology.
Represent the 8x8 chessboard squares with vertices, one vertex for each square. Represent a knight move like c5-->e6 by drawing an edge between the vertices that represent c5 and e6. Do this until each of the 64 vertices has its complete set of edges attached to it.
The number needed for tomorrow's Step 2 is the number of edges you find this graph to have.
My source has no diagrams; it just tells what the number of edges is. I wanted to verify the number on my own, so I drew an 8 x 8 array of pencil dots (though a network is abstract enough that I didn't have to draw it in such a squarish way) and started connecting the dots ("vertices" ) with arcs ("edges" ) in ways that the knight moves connect chess squares. [Side note: this website inserted smilies until I put a space between the quote mark and the trailing parenthesis in the previous sentence.]
My plan was to finish all the edges and then put a colored marker dot on each edge, one-by-one, while counting. But my sketch got too messy, and I abandoned this in favor of a more arithmetical approach.
From the point of view of the knight, there are several categories of chessboard squares. One category is the (four) corner squares, because each of them is represented by a node having two edges meeting it. Another category is the (eight) perimeter squares abutting corner squares, because each of them is represented by a node having three edges meeting it.
I figured out all the categories and toted up edges, being careful to account for any edges that might be counted more than once by this method. My result agreed with the book.
Feel free to tell us what you get as your Part 1 result. If you can figure out what Part 2 of the problem is, go ahead and solve for the mean. Otherwise, I'll see you tomorrow.