This is a sequel to Random knight… don’t expect more than you usually get from sequels. I do not plan to do other chess pieces, although “random pawn” is an intriguing idea.

Queen begins its random walk at d1. After two moves, it can get anywhere on the board.

It is not surprising that the most likely outcome is the return to initial state. It is less obvious why the second most likely position is d3.

After 3 moves, the most likely positions are near the center of the board.

After 4 moves, a concentric-square pattern emerges.

After 10 moves, the distribution is visually identical to the steady-state distribution.

As was noted in the Random knight post, the steady state distribution is the normalized number of moves from a given position. Hence, all probabilities are rational. For the knight they happen to be unit fractions: 1/168, 1/112, 1/84, 1/56, 1/42. The queen probabilities are uglier: 3/208, 23/1456, 25/1456, 27/1456. Random queen is distributed more uniformly than random knight: it spends between 1.44% and 1.85% of its time in any given square. For the knight, the range is from 0.59% to 2.38%.

But the champion in uniform coverage of the chessboard is the rook: when moving randomly, it spends 1/64 of time on every square.

A knight walks randomly on the standard chessboard. What is the proportion of time that it will spend at e4?

The answer (1/42) is not hard to get, but I’ll take the computational approach first (testing Scilab 5.5.0 beta 1, by the way). Begin by placing the knight at b1 and letting it make five random moves. This is the distribution of its position after five moves:

Unsurprisingly, half of the board is black; actually more than half because the knight can’t get to h8 in five moves. the other half isn’t — you can even get to h8 in five moves (want to find all possible ways to do this?).

After ten moves, the colors become more uniform.

After 200 (or any large even number) of moves, the distribution is little changed. But you may notice that it is centrally symmetric, while the previous one isn’t quite.

Let’s repeat the process beginning with two knights at b1 and g1. After five moves of each:

After ten moves:

After a large number of moves (does not matter, even or odd), the variety of colors is greatly reduced:

Indeed, this is the distribution which also describes the proportion of time that the knight (wherever it started) will spend at a given square Q.

Precisely, the proportion of time spent at Q is P(Q)=N(Q)/336 where N(Q) is the number of legal moves from Q. For the sixteen central squares P(Q) = 8/336 = 1/42, while for the corners we get 2/336 = 1/168.

Here is a quick argument to support the above. Let Q1 and Q2 be two squares such that the move from Q1 to Q2 is legal. The proportion of time that the knight makes this move is P(Q1)/N(Q1). Similarly, the time proportion of Q2-Q1 move is P(Q2)/N(Q2). Since the process is symmetric in time (we have a reversible Markov chain), P(Q1)/N(Q1)=P(Q2)/N(Q2). In other words, the ratio P/Q is the same for any two squares that are joined by a legal move; it follows that P/Q is the same for all cells. Finding the coefficient of proportionality is a matter of counting, since the sum of P(Q) over all Q is 1.

The Scilab code I wrote for this post is largely not knight-specific. The function update receives the initial distribution of a chess piece and the set of its legal moves. It computes the distribution after one random move.

function newState=update(state,moves)
[maxK,n]=size(moves)
[n,n]=size(state)
newState=zeros(state)
for i=1:n
for j=1:n
neighborCount=0
contribution=zeros(state)
for k=1:maxK
pos=[i,j]+moves(k,:)
if (min(pos)>=1)&(max(pos)<=n) then
neighborCount=neighborCount+1
contribution(pos(1),pos(2))=state(i,j)
end
end
newState=newState+contribution/neighborCount
end
end
endfunction

This set is given in the function knight which calls update iteratively.

function state=knight(state,iter)
moves=[2 1;1 2;-2 1;1 -2;2 -1;-1 2;-2 -1;-1 -2]
for i=1:iter
state=update(state,moves)
end
endfunction

For example, this is how I plotted the distribution after 10 random moves starting at b1:

initial = zeros(8,8)
initial(8,2)=1
state = knight(initial,10)
isoview(0,0,9,9)
Matplot(state*500)
f = gcf()
f.color_map = hotcolormap(32)

NB: getting the correct color representation of matrices from Matplot requires an up-to-date version of Scilab; in version 5.4.0 Matplot smoothens colors, producing intriguing but less informative images.