Structured Output Learning with Candidate Labels for Local Parts: Supplementary Materials

Theorem 1. Given a structured instance x and arbitrary candidate labeling set Y , no algorithm exists that can always find the most violated label (in Y or not in Y) in poly(|x|) time unless P = N P , where |x| is the length of x. Sketch of the Proof. We prove this theorem by first proving the following lemmas: Lemma 1. We prove that no algorithm exists that can always find the most violated label setting that is in Y where Y could be an arbitrary candidate label set. Lemma 2. We prove that no algorithm exists that can always find the most violated label setting that is in Y/Y where Y could be arbitrary candidate label set. By combining these two lemmas we finish the proof of the theorem. Proof. Lemma 1: Assume that for arbitrary candidate label set Y , an algorithm exists that can find the most violated label setting that is in Y in poly(|x|) time. The value of |Y | can be exponential in |x|, without proper encoding of the candidate label set Y , it would take exp(|x|) time to read Y. So if the algorithm runs in poly(|x|) time, there must exist some kind of encoding of the candidate label set and the given Y is already encoded. Thus we show that even with encoding, the algorithm still cannot run in poly(|x|). Assume that given Y is encoded in the following rules: Rule 1: For all label settings in Y , at least one of the following cases happens: the label of x i is y i ; or the label of x j is y j , or. .. Rule 2: For all label settings in Y , at least one of the following cases happens: the label of x k is y k ; or. .. .. . Now we prove that finding the most violated label setting in Y is NP-hard. More precisely, we prove that the decision version of this problem: determining whether a label setting exists that is in Y is NP-complete.