Saturday, June 30, 2012

The White Knight Returns

Back in May, on his excellent blog The Endeavor, John D. Cook posed (and subsequently answered) the following problem: if a knight begins in one corner of an empty chessboard and moves randomly, each time choosing uniformly over the available moves from its current position, what is the mean number of moves until it returns to the corner from which it started?

Several people posed exact or approximate answers in the comments. The approximate answers were based on simulation. The most elegant solution among the comments, posted by 'deinst', modeled the problem as a reversible Markov chain on a graph. John Cook's answer used a rather powerful theorem about a random walk on a graph, which I think boils down to the same logic used by 'deinst'. (Cook also includes Python code for a simulation.)

When I read the problem, my first reaction was also to model it as a Markov chain, but sadly I've forgotten most of what I knew about MC's, including the concept of reversibility. I did, at least, recall the essentials of first passage time. The question reduces to computing the expected first passage time from the corner square to itself. The Wikipedia link for first passage times is not entirely helpful, as it is not specific to Markov chains, but the concept is easy to explain. Let the corner in which the knight starts be position $s$, and let $\mu_j$ denote the expected number of moves required until the knight first returns to position $s$, given that it is currently at position $j$ (i.e., the expected first passage time from $j$ to $s$). The answer to the puzzle is $\mu_s$, the expected first passage time from $s$ to itself. Further, let $M_j$ denote the set of positions that the knight can reach in one move, given that it is currently at position $j$. Letting $p_{jk}$ denote the probability that a knight at position $j$ next moves to position $k$, the first passage times $\mu_\cdot$ satisfy the following system of linear equations:\[ \mu_{j}=1+\sum_{k\in M_{j}\backslash\{s\}}p_{jk}\mu_{k}\quad\forall j. \] In our case, $p_{jk} = 1/|M_j|$ if $k\in M_j$ and 0 otherwise. The solution to this system has $\mu_s=168$ (matching the answers of John Cook and deinst).

Other than a quick refresher on Markov chains, my main interest in this was that I'm teaching myself Python while also trying to get a handle on using Sage. So I wrote a Sage notebook to set up and solve the first passage time equations. Here is the content of that notebook, in case anyone is curious. (Please bear in mind I'm new to Python.)

Define the board

board = [(i, j) for i in range(8) for j in range(8)]

Assign a variable for the expected first passage time (to the lower left corner) from each square of the board

vlist = {b : var('r_' + str(b[0]) + '_' + str(b[1])) for b in board}

List the eight moves a knight can make

moves = [(i, j) for i in [-2, -1, 1, 2] for j in [-2, -1, 1, 2] if abs(i) + abs(j) == 3]

Define a function that takes a position and move and returns the new position if the move is legal, None otherwise

def move(position, change): 
    if (position in board) and (change in moves): 
        landing = (position[0] + change[0], position[1] + change[1])
        return landing if landing in board else None
    else:
        return None

Compute all legal moves from each board position

legal = {p : [move(p, m) for m in moves if not(move(p, m) is None)] for p in board}

Set up the first passage time equations

eqns = [vlist[b] == 1.0 + (1.0/len(legal[b]))*sum([vlist[x] for x in legal[b] if x != (0, 0)]) for b in board]

Solve the equations

sols = solve(eqns, vlist.values(), solution_dict = True)

Verify that the solution is unique

len(sols) == 1

True

Find the mean first passage time from position (0,0) to itself

sols[0][vlist[(0,0)]]

168

One last note: I had originally intended to give the post a Christian Bale buzz by titling it "The Dark Night Returns". Turns out DC Comics beat me to that title. Oh, well.

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