Cover time chess board (king)

Consider a random walk of a king on a standard chess board, which at each step moves to a uniformly random permitted square. What’s the exact mean time to visit all squares (cover time), starting from a corner square?

Is there an algebraic solution?

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There is in principle no difficulty in answering this question.
As I point out in my answer here, calculating the expected cover
time of some set $\cal S$ reduces to calculating the expected hitting time of every possible non-empty subset $A$ of $\cal S$:

$$\mathbb{E}(\text{cover time})=\sum_A (-1)^{|A|-1} \mathbb{E}(T_A)$$

These hitting times are defined by $T_A=\inf(n\geq 0: X_n\in A)$.


Look out below! Ignorant of chess rules, I didn’t realize that a king can move diagonally. The calculations below are based on a piece that can only move in four ways: north, south, east, or west.


Just to illustrate, let me show you the solution for a $2\times 2$ chessboard:

a small board

The expected time to cover the other 3 squares ${a,b,c}$ is equal to
$$\mathbb{E}(T_{a})+\mathbb{E}(T_{b})+\mathbb{E}(T_{c})-\mathbb{E}(T_{a,b})-\mathbb{E}(T_{a,c})-\mathbb{E}(T_{b,c})+\mathbb{E}(T_{a,b,c})$$

Standard Markov chain theory uses linear algebra to find these expected hitting times
$$\mathbb{E}(T_{a})=\mathbb{E}(T_{c})=3, \mathbb{E}(T_{b})=4,
\mathbb{E}(T_{a,b})=\mathbb{E}(T_{b,c})=2, \mathbb{E}(T_{a,c})=\mathbb{E}(T_{a,b,c})=1$$

Putting it all together, we find that the expected cover time is $3+3+4-2-2-1+1=6$.

Note that I counted the king’s initial position as already covered. If you
require a return to your starting point you can modify the above technique.

The number of terms in the sum make this method impractical for an $8\times 8$ chessboard, however!


Added: If my calculations are correct, the expected cover time for the $3\times 3$ board is $${140803109038245\over 4517710919176}=31.1669$$

There is an algebraic solution; the mean cover time from any node in a graph is known to be rational and can be found in exponential time. It’s not likely to have a simple form, though. Experimentally, I ran the following code:

import random

def reachable((i,j)):
   ok = lambda(q):(min(q)>=1 and max(q)<=8)
   return filter(ok, [(i-1,j-1), (i-1,j), (i-1,j+1), (i,j-1), (i,j+1), (i+1,j-1), (i+1,j), (i+1,j+1)])

def covertime(sq, rng=random.Random()):
   seen = set([sq])
   path = [sq]
   while len(seen)<64:
      path.append(rng.choice(reachable(path[-1])))
      seen.add(path[-1])
   return len(path)-1

I found that the mean cover time starting from a corner square was about $597$, while the mean cover time starting from a center square was larger, about $621$.