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).
Usage
solve_interval_partition(x, kmax)
Value
dynamic program solution.
Arguments
x
square NumericMatix, for j>=i x(i,j) is the cost of partition element [i,...,j] (inclusive).