solve_interval_partition_R | R Documentation |
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).
solve_interval_partition_R(x, kmax)
x |
NumericMatix, for j>=i x(i,j) is the cost of partition element [i,...,j] (inclusive). |
kmax |
int, maximum number of steps in solution. |
dynamic program solution.
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)
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.