Checkers.jl

finding the size of the minimum dominating set of an n by n grid graph

This is a (very not rigourous) description of what I found.

Note that I did not use the language or apparatus of dominating sets when I first started this. I wanted to know what was the smallest number of checkers (m) that I could use to cover a (n by n) checkerboard where a square is "covered" if it has a checker on it or is directly next to a square with a checker on it (diagonals don't count). Therefore most of this descriptions uses "checkers", "squares", and "holes".

Note that m ≧ n²/5, because each individual checker can cover, at most, 5 squares. Also, m < n²/2, because a board that has a checker on every other square is covered (very inefficiently).

Usefull definitions

The Von Neuman neighborhood (of radius 1) around the i,jth element of matrix x. z here is a special matrix type that returns a default fill value if any of the indices are out of bounds.

neighbors(z, (i, j)) = (z[i-1, j], z[i, j-1], z[i, j+1], z[i+1, j], z[i, j])

A square is "covered" if it has a checker on it (represented by a 1) or is next to a square with a checker on it.

covered(A, (i,j)) = A[i,j] == 1 || 1 in neighbors(A, (i,j))

if all squares on the board are covered, the whole board is also covered

covered(A) = all(covered(A, I) for I in CartesianIndices(A))

A "hole" is a square which is not covered

is_hole(A, I) = !covered(A, I)
is_hole(A) = [is_hole(A, I) for I in CartesianIndices(A)]

The number of holes on a board can be computed by

holes(p) = sum(neighbors(p) .== 0)

The number of squares a square is "covering" can be defined as

neighbors(m::AbstractMatrix, I; fill) = neighbors(Pad(m, fill), I)
covering(A, I) = A[I] * count(==(0), neighbors(A, I, fill=one(eltype(A))))

This uses a special overload of neighbors which counts all squares out of bounds as 1's instead of zeros. Otherwise checkers on the edge would "cover" squares which are outside of the board.

By brute force search

The brute force search is simple and guareteed to give a minimal solution, but performs horribly as n gets large due to the size of the search space increasing factorially.

using Combinatorics
function brute_force(n, M=ceil(Int, n * n / 5):(n*n))
    board = zeros(Bool, n, n)  # pre-allocate a board
    for m in M
        # make an iterator over every combination of indices length m. These is where we'll put the 1s
        combinations = Combinatorics.Combinations(n * n, m)
        for combination in combinations  # most of the allocations happen here
            board .= 0  # Start by clearing the old board, this doesn't allocate
            board[combination] .= 1  # put a 1 on every part of the board specified by that combination of indices.
            if covered(board)  # check if the board is covered. This doesn't allocate somehow
                return (board, m)  # if we find any solution, return early 
            end
        end
    end
    return (Bool[;;], 0)  # if the function gets this far, presumably no solutions exist
end

This function starts at the smallest possible m and only increments m when all boards have been exhausted. This means that the first covered board that this function finds must be a minimal covering board, and therefore the function returns it immediatly. This is helpful because it doesn't require searching through the entire search space.

One issue with this function is that, due to how uses the same memory for every board, it is not clear how to parallelize it.

By Stochastic search.

The way that the brute force search looks for covered boards is far from optimal, checking huge swaths of uncovered boards before reaching any covered ones. Instead, why find a way to take a non-covered board and move it closer to a covered board with the same number of checkers. One way to do this is to find the square with a checker on it which coveres the fewest other squares, put it's checkeron the square with the most un-covered squares (called "holes") around it.

function update!(A)
    (h, holeIndex) = findmax(neighbors(is_hole(A)))
    (c, tokenIndex) = find_min_covering(A)
    @assert A[holeIndex] == 0
    @assert A[tokenIndex] == 1
    A[holeIndex] = 1
    A[tokenIndex] = 0
    return A
end

function find_min_covering(A)
    I = filter(I -> A[I], CartesianIndices(A)) |> collect
    (c, II) = findmin(I -> covering(A, I), I)
    return (c, I[II])
end

Note that find_min_covering only needs to look at the squares with checkers on them.

This update algorithem can be run until the number of holes is 0.

A = zeros(n,n)
A[1:m] .= 1

prev_holes = holes(A)
update!(A)
next_holes = holes(A)
while next_holes > 0
    prev_holes = next_holes
    update!(A)
    next_holes = holes(A)
end

However update! won't always reduce the number of holes. Sometimes it finds a local minimum where it oscilates between multiple boards with a non-zero number of holes. In these case, the board can be shuffled to give the algorithem a new initial condition, and the process can be restarted.

There is no guarentee that a covered n by n board with m checkers exists, so the number of attempts is constrained.

using Random
function stochastic_search(
    n::Int,
    m::Int;
    init=reshape(shuffle(vcat(ones(Bool, m), zeros(Bool, n * n - m))), n, n),
    attempts=10^7,
    (update!)=update!
)
    A = copy(init)
    for n in 1:attempts
        prev_holes = holes(A)
        update!(A)
        next_holes = holes(A)
        while next_holes > 0 && prev_holes > next_holes
            prev_holes = next_holes
            update!(A)
            next_holes = holes(A)
        end
        holes(A) == 0 && return A
        shuffle!(A)
    end
    return Bool[;;]
end

Note that this function can be used to find an upper bound on m, in fact the higher m is the more likely it is to stumble on a solution. However it cannot find a lower bound to m because it does not try all of the boards exhaustivly.

colonelkrud/Checkers.jl

Checkers.jl

Usefull definitions

By brute force search

By Stochastic search.