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<=kmax 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).

solve_interval_partition(x, kmax)

Arguments

x

NumericMatix, for j>=i x(i,j) is the cost of partition element [i,...,j] (inclusive).

kmax

int, maximum number of segments in solution.

Value

dynamic program solution.

Examples

costs <- matrix(c(1.5, NA ,NA ,1 ,0 , NA, 5, -1, 1), nrow = 3) solve_interval_partition(costs, nrow(costs))
#> [1] 1 2 4