MathLabs

Problem 2

Consider a 100×100100\times100 table, and identify the cell in row aa and column bb, 1≤a,b≤1001\le a,b\le100, with the ordered pair (a,b)(a,b). Let kk be an integer such that 51≤k≤9951\le k\le99. A kk-knight is a piece that moves one cell vertically or horizontally and kk cells in the other direction; that is, it moves from (a,b)(a,b) to (c,d)(c,d) such that (∣a−c∣,∣b−d∣)(|a-c|,|b-d|) is either (1,k)(1,k) or (k,1)(k,1). The kk-knight starts at cell (1,1)(1,1) and performs several moves. A sequence of moves is a sequence of cells (x0,y0)=(1,1),(x1,y1),…,(xn,yn)(x_0,y_0)=(1,1),(x_1,y_1),\ldots,(x_n,y_n) such that, for all i=1,2,…,ni=1,2,\ldots,n, 1≤xi,yi≤1001\le x_i,y_i\le100 and the kk-knight can move from (xi−1,yi−1)(x_{i-1},y_{i-1}) to (xi,yi)(x_i,y_i). In this case each cell (xi,yi)(x_i,y_i) is said to be reachable. For each kk, find L(k)L(k), the number of reachable cells.
Step 2 of 5: The central square with no moves
S={(x,y):101−k≤x,y≤k},∣S∣=(2k−100)2S=\{(x,y):101-k\le x,y\le k\},\qquad|S|=(2k-100)^2
Detailed analysis

Negating the condition of the previous step, a cell (x,y)(x,y) has no legal move at all exactly when 101−k≤x≤k101-k\le x\le k and 101−k≤y≤k101-k\le y\le k; this is a square SS of side k−(101−k)+1=2k−100k-(101-k)+1=2k-100, so ∣S∣=(2k−100)2|S|=(2k-100)^2. Cells in SS are isolated, while the remaining 1002−(2k−100)2100^2-(2k-100)^2 cells each have at least one move. Since 101−k≥2101-k\ge2 for k≤99k\le99, the starting cell (1,1)(1,1) is never in SS.