Only with a suitable restriction, such as all remaining candidates being nonnegative. A remaining -1 could complete the target. “Already too large” is not sufficient by itself.

For unique subsets of possibly repeated values, sort first. At one recursion depth, choose only the first occurrence of each equal value. Equal values at deeper depths remain available, allowing [1, 1].

def unique_subsets(values):
    values = sorted(values)
    answers, path = [], []
    def visit(start):
        answers.append(path.copy())
        for i in range(start, len(values)):
            if i > start and values[i] == values[i - 1]:
                continue
            path.append(values[i])
            visit(i + 1)
            path.pop()
    visit(0)
    return answers

assert unique_subsets([1, 1, 2]) == [[], [1], [1, 1], [1, 1, 2], [1, 2], [2]]

The condition i > start means “not the first option at this depth.” Removing it would forbid choosing a second equal value after the first one, incorrectly removing [1, 1].

For optimization, another kind of pruning uses a bound: if even the best possible completion cannot beat the best answer already found, the branch is unnecessary. The bound must be optimistic in the correct direction. A guessed future value is not enough.

Sorting costs O(n log n) here. Search is still exponential in the worst case, and copying outputs still counts. “Uses pruning” does not imply polynomial time.

Analogy limit: pruning invalid completions and choosing one representative of equivalent outputs have different proofs. Neither is the same as a greedy assertion that one locally attractive choice can be committed to globally.

No. The outputs are distinct under the revised identity rule. Value-based deduplication would erase required answers.