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#' solve_interval_partition (R version)
#'
#' Solve a for a minimal cost partition of the integers [1,...,nrow(x)] problem where for j>=i x(i,j).
#' is the cost of choosing the partition element [i,...,j].
#' Returned solution is an ordered vector v of length k where: v[1]==1, v[k]==nrow(x)+1, and the
#' partition is of the form [v[i], v[i+1]) (intervals open on the right).
#'
#' @param x NumericMatix, for j>=i x(i,j) is the cost of partition element [i,...,j] (inclusive).
#' @param kmax int, maximum number of steps in solution.
#' @return dynamic program solution.
#'
#' @keywords internal
#'
#' @examples
#'
#' x <- matrix(c(1,1,5,1,1,0,5,0,1), nrow=3)
#' k <- 3
#' solve_interval_partition_R(x, k)
#' solve_interval_partition(x, k)
#'
#' @export
#'
solve_interval_partition_R <- function(x, kmax) {
# for cleaner notation
# solution and x will be indexed from 1 using
# R_INDEX_DELTA
# intermediate arrays will be padded so indexing
# does not need to be shifted
R_INDEX_DELTA = 0L;
R_SIZE_PAD = 0L;
# get shape of problem
n = nrow(x);
if(kmax>n) {
kmax = n;
}
# get some edge-cases
if((kmax<=1)||(n<=1)) {
solution = integer(2);
solution[1 + R_INDEX_DELTA] = 1;
solution[2 + R_INDEX_DELTA] = n+1;
return(solution);
}
# best path cost up to i (row) with exactly k-steps (column)
path_costs = matrix(0.0, n + R_SIZE_PAD, kmax + R_SIZE_PAD);
# how many steps we actually took
k_actual = matrix(0L, n + R_SIZE_PAD, kmax + R_SIZE_PAD);
# how we realized each above cost
prev_step = matrix(0L, n + R_SIZE_PAD, kmax + R_SIZE_PAD);
# fill in initial path and costs tables k = 1 case
for(i in seqi(1, n)) {
prev_step[i, 1] = 1L;
path_costs[i, 1] = x[1 + R_INDEX_DELTA, i + R_INDEX_DELTA];
k_actual[i, 1] = 1L;
}
# refine dynprog table
for(ksteps in seqi(2, kmax)) {
# compute larger paths
for(i in seqi(1, n)) {
# no split case
pick = i;
k_seen = 1;
pick_cost = x[1 + R_INDEX_DELTA, i + R_INDEX_DELTA];
# split cases
for(candidate in seqi(1, i-1)) {
cost = path_costs[candidate, ksteps-1] +
x[candidate + 1 + R_INDEX_DELTA, i + R_INDEX_DELTA];
k_cost = k_actual[candidate, ksteps-1] + 1;
if((cost<=pick_cost) &&
((cost<pick_cost)||(k_cost<k_seen))) {
pick = candidate;
pick_cost = cost;
k_seen = k_cost;
}
}
path_costs[i, ksteps] = pick_cost;
prev_step[i, ksteps] = pick;
k_actual[i, ksteps] = k_seen;
}
}
# now back-chain for solution
k_opt = k_actual[n, kmax];
solution = integer(k_opt+1);
solution[1 + R_INDEX_DELTA] = 1L;
solution[k_opt + 1 + R_INDEX_DELTA] = n+1L;
i_at = n;
k_at = k_opt;
while(k_at>1) {
prev_i = prev_step[i_at, k_at];
solution[k_at + R_INDEX_DELTA] = prev_i + 1;
i_at = prev_i;
k_at = k_at - 1;
}
return(solution);
}
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