“…For the second phase of the algorithm we now compute the min-cost flow on a graph where we know that we can deliver 2 sat with perfect certainty to B. This updated uncertainty network is depicted in Figure 3 The flow that sends the full residual amount of 2 sat along s, A, B, X, Y, d has a probability of 8 10 • 5 7 • 1 2 = 0.286... while the flow that sends 1 sat along the B, d channel has a probability of 1 2 • 9 10 • 6 7 • 3 4 = 0.289... telling us that for the residual amount we should make another split payment sending one sat along each of the paths q 1 = s, A, B, d and q 2 = s, A, B, X, Y, d.…”