Module Core.Interval.Int

create l u returns the interval with lower bound l and upper bound u, unless l > u, in which case it returns the empty interval.

convex_hull ts returns an interval whose upper bound is the greatest upper bound of the intervals in the list, and whose lower bound is the least lower bound of the list.

Suppose you had three intervals a, b, and c:

             a:  (   )
             b:    (     )
             c:            ( )

          hull:  (           )

In this case the hull goes from lbound_exn a to ubound_exn c.

bound t x returns None iff is_empty t. If bounds t = Some (a, b), then bound returns Some y where y is the element of t closest to x. I.e.:

        y = a  if x < a
        y = x  if a <= x <= b
        y = b  if x > b

is_superset i1 of_:i2 is whether i1 contains i2. The empty interval is contained in every interval.

map t ~f returns create (f l) (f u) if bounds t = Some (l, u), and empty if t is empty. Note that if f l > f u, the result of map is empty, by the definition of create.

If you think of an interval as a set of points, rather than a pair of its bounds, then map is not the same as the usual mathematical notion of mapping f over that set. For example, map ~f:(fun x -> x * x) maps the interval [-1,1] to [1,1], not to [0,1].

are_disjoint ts returns true iff the intervals in ts are pairwise disjoint.

Returns true iff a given set of intervals would be disjoint if considered as open intervals, e.g., (3,4) and (4,5) would count as disjoint according to this function.

Assuming that ilist1 and ilist2 are lists of disjoint intervals, list_intersect ilist1 ilist2 considers the intersection (intersect i1 i2) of every pair of intervals (i1, i2), with i1 drawn from ilist1 and i2 from ilist2, returning just the non-empty intersections. By construction these intervals will be disjoint, too. For example:

        let i = Interval.create;;
        list_intersect [i 4 7; i 9 15] [i 2 4; i 5 10; i 14 20];;
        [(4, 4), (5, 7), (9, 10), (14, 15)]

Raises an exception if either input list is non-disjoint.

Returns true if the intervals, when considered as half-open intervals, nestle up cleanly one to the next. I.e., if you sort the intervals by the lower bound, then the upper bound of the nth interval is equal to the lower bound of the n+1th interval. The intervals do not need to partition the entire space, they just need to partition their union.