solve_interval_partition_no_k | R Documentation |

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```
solve_interval_partition_no_k(x)
```

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

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).

dynamic program solution.

```
costs <- matrix(c(1.5, NA ,NA ,1 ,0 , NA, 5, -1, 1), nrow = 3)
solve_interval_partition(costs, nrow(costs))
```

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