def test_chop_method_w_limit():
### scenario 3
### we want to only chop one period and not both
v1 = [Interval(0, 10, data=0)]
v2 = [Interval(0, 10, data=1)]
v3 = [Interval(0, 4, data=2), Interval(6, 10, data=2)]
intervals = sum((v1, v2, v3), [])
avail_tree = IntervalTree(intervals)
### example booking for v1
avail_tree.chop_intervals_that_envelope_range(3, 7, limit=1)
assert len(avail_tree.items()) == 5
def test_chop_method_scenario2():
### scenario 2
### we want to make sure that we completely removea period that is exactly the same as the search
v1 = Interval(0, 10, data=0)
v2 = [Interval(0, 4, data=1), Interval(6, 10, data=1)]
v2.append(v1)
avail_tree = IntervalTree(v2)
### example booking
avail_tree.chop_intervals_that_envelope_range(0, 10)
assert avail_tree == IntervalTree(
[Interval(0, 4, data=1),
Interval(6, 10, data=1)])
def test_chop_method_scenario1():
### scenario 1
### We want to make sure we only chop the period of availability for v1
v1 = Interval(0, 10, data=0)
v2 = [Interval(0, 4, data=1), Interval(6, 10, data=1)]
v2.append(v1)
avail_tree = IntervalTree(v2)
### example booking
avail_tree.chop_intervals_that_envelope_range(3, 7)
assert avail_tree == IntervalTree([
Interval(0, 3, data=0),
Interval(7, 10, data=0),
Interval(0, 4, data=1),
Interval(6, 10, data=1)
])
