Interval Operations - Edge Cases

Let’s explore how interval operations handle various edge cases:

Adjacent Intervals

When intervals are exactly adjacent (end of one equals start of another), the union operation will merge them into a single interval:

# Create adjacent intervals [1, 2] and [2, 3]
adjacent = Interval(start=[1, 2], end=[2, 3])

# Union will merge them into [1, 3]
merged = adjacent | adjacent
print(merged.start)  # [1]
print(merged.end)    # [3]

Point Intervals

Intervals where start equals end (zero duration) are handled gracefully:

# Create point interval [2, 2]
point = Interval(start=[2], end=[2])

# Intersection with another interval containing that point
other = Interval(start=[1], end=[3])
intersection = point & other
print(intersection.start)  # [2]
print(intersection.end)    # [2]

Empty Intervals

Operations with empty intervals (no time periods) return appropriate results:

# Create an empty interval
empty = Interval(start=[], end=[])

# Any intersection with an empty interval is empty
some_interval = Interval(start=[1], end=[2])
intersection = empty & some_interval
print(len(intersection.start))  # 0

# Union with an empty interval returns the non-empty interval
union = empty | some_interval
print(union.start)  # [1]
print(union.end)    # [2]

Overlapping Input Intervals

If you try to perform operations on intervals that aren’t disjoint or sorted, the operations will raise a ValueError:

# Create overlapping intervals [1, 3] and [2, 4]
overlapping = Interval(start=[1, 2], end=[3, 4])

try:
    # This will raise a ValueError
    overlapping & some_interval
except ValueError as e:
    print(e)  # "left Interval object must be disjoint."

# Fix by making intervals disjoint first
fixed = overlapping | overlapping  # Merges overlapping intervals
result = fixed & some_interval     # Now works correctly

Exact Matches

When interval boundaries exactly match, both intersection and union handle them appropriately:

# Create two identical intervals
a = Interval(start=[1, 5], end=[3, 7])
b = Interval(start=[1, 5], end=[3, 7])

# Intersection returns the same intervals
intersection = a & b
print(intersection.start)  # [1, 5]
print(intersection.end)    # [3, 7]

# Union also returns the same intervals
union = a | b
print(union.start)  # [1, 5]
print(union.end)    # [3, 7]

These edge cases are important to consider when working with intervals, especially in data processing pipelines where unexpected interval patterns might occur.