A Proof of SPNs as Mixture of Trees

Proof. Argue by contradiction that T is not a tree, then there must exist a node v 2 T such that v has more than one parent in T . This means that there exist at least two paths R, p1, . . . , v and R, q1, . . . , v that connect the root of S(T ), which we denote by R, and v. Let t be the last node in R, p1, . . . , v and R, q1, . . . , v such that R, . . . , t are common prefix of these two paths. By construction we know that such t must exist since these two paths start from the same root node R (R will be one candidate of such t). Also, we claim that t 6= v otherwise these two paths overlap with each other, which contradicts the assumption that v has multiple parents. This shows that these two paths can be represented as R, . . . , t, p, . . . , v and R, . . . , t, q, . . . , v where R, . . . , t are the common prefix shared by these two paths and p 6= q since t is the last common node. From the construction process defined in Def. 2, we know that both p and q are children of t in S . Recall that for each sum node in S , Def. 2 takes at most one child, hence we claim that t must be a product node, since both p and q are children of t. Then the paths that t! p v and t! q v indicate that scope(v) ✓ scope(p) ✓ scope(t) and scope(v) ✓ scope(q) ✓ scope(t), leading to ? 6= scope(v) ✓ scope(p) \ scope(q), which is a contradiction of the decomposability of the product node t. Hence as long as S is complete and decomposable, T must be a tree.