Supplementary Materials: Global Connectivity Potentials for Random Field Models

Proof to Lemma 2. First, yi ≥ 0. For each i, we construct |V | affinely independent points in C with yi = 0. Fix i, then one solution is obviously x = 0, the empty subgraph. Next, for all p 6= i, obtain one solution by setting only yp = 1, and for all j 6= p set yj = 0. Clearly, yj = 0 and the |V | − 1 solutions thus obtained are affinely independent. In total we have |V | solutions with yi = 0, thus yi ≥ 0 is facet-defining. Second, yi ≤ 1. Again let i be arbitrary. We construct |V | affinely independent points in C with yi = 1. For this, set yi = 1 and yj = 0 for all j 6= i. This is obviously one solution. Now root a spanning tree in i and set one node k at a time to yk = 1, respecting the order of the spanning tree, i.e. the subgraph selected all nodes j with yj = 1 always remains a connected subgraph of the spanning tree. This constructs |V | − 1 solutions, all affinely independent. Adding the first solution yields |V | solutions in total, completing the proof.