No. Answers change once from false to true, so one comparison discards half while preserving the boundary.

Imagine a row of false answers followed by true answers. The algorithm is not searching for a value; it is squeezing the single border between those regions.

What the function promises: Return the first index whose value is at least target, or len(values). Input must be nondecreasing.

def lower_bound(values, target):
    left, right = 0, len(values)
    while left < right:
        middle = (left + right) // 2
        if values[middle] >= target:
            right = middle
        else:
            left = middle + 1
    return left

assert lower_bound([1,2,2,4], 2) == 1
assert lower_bound([1,2,2,4], 3) == 3
assert lower_bound([], 3) == 0

Work and storage: O(log n) time and O(1) space.

When to recognize it: Answers change once from false to true, so one comparison discards half while preserving the boundary.

Check before applying it: Off-by-one errors come from mixing interval conventions or searching a predicate that flips more than once.

Ordinary boundary binary search is invalid; find another monotone quantity or inspect the separate regions.