The FordFulkerson Algorithm aka The Labeling Algorithm FordFulkerson

The Ford-Fulkerson Algorithm aka The Labeling Algorithm

Ford-Fulkerson Algorithm begin x : = 0; label node t; while t is labeled do begin unlabel all nodes; pred(j) : = 0 for all j in N; label s; LIST : = {s}; while LIST is not empty and t is not labeled do begin remove a node i from LIST; for all {j in N: (i, j) in A and rij > 0} do if j is unlabeled then pred(j) : = i, label j, add j to LIST; end; if t is labeled then augment flow on path from s to t end;

Labeling Algorithm Example (0, 5) s 2 (0, 2) 4 (0, 4) 1 (0, 6) 3 (0, 5) 6 (0, 7) 5 t

The Residual Network G(x) 2 2 5 s 1 0 6 0 0 0 3 4 4 0 4 5 0 7 5 6 0 t

Iteration 1: LIST = {1}, Labeled = {1} i=1 2 2 5 s 1 0 6 0 0 0 3 4 4 0 4 5 0 0 5 6 7 t

Iteration 1: LIST = {1}, Labeled = {1} • i=1 • LIST = {} • Arc (1, 2) – pred(2) =1 – label 2 – LIST = {2} • Arc (1, 3) – pred (3) = 1 – label 3 – LIST = {2, 3}

Iteration 1: LIST = {2, 3}, Labeled= {1, 2, 3} pred(2) = 1 2 5 s 1 0 3 pred(3) = 1 4 4 0 0 6 0 2 0 4 5 0 7 5 6 0 t

Iteration 1: LIST = {2, 3}, Labeled = {1, 2, 3} • i=2 • LIST = {3} • Arc (2, 4) – pred(4) =2 – label 4 – LIST = {3, 4} • Arc (2, 5) – pred (5) = 2 – label 5 – LIST = {3, 4, 5} • Arc (2, 1) – residual capacity of (2, 1) = 0

Iteration 1: LIST = {3, 4, 5}, Labeled= {1, 2, 3, 4, 5} pred(2) = 1 2 5 s 1 0 3 pred(3) = 1 4 4 0 0 6 0 2 pred(4) = 2 0 4 5 0 7 5 pred(5) = 2 6 0 t

Iteration 1: LIST = {3, 4, 5}, Labeled = {1, 2, 3, 4, 5} • i=3 • LIST = {4, 5} • Arc (3, 5) – 5 is already labeled • Arc (3, 1) – residual capacity of (3, 1) = 0

Iteration 1: LIST = {4, 5}, Labeled = {1, 2, 3, 4, 5} • i=4 • LIST = {5} • Arc (4, 2) – residual capacity of (4, 2) = 0 • Arc (4, 6) – pred(6) =4 – label 6 – LIST = {5, 6}

Iteration 1: LIST = {5, 6}, Labeled= {1, 2, 3, 4, 5, 6} pred(2) = 1 2 5 s 1 0 4 4 0 0 6 0 2 pred(4) = 2 3 pred(3) = 1 0 4 5 0 pred(6) = 4 7 5 pred(5) = 2 6 0 t

Iteration 1: The sink is labeled • Use pred labels to trace back from the sink to the source to find path P – P = 1 -> 2 -> 4 -> 6 • = min {rij: (i, j) in P) = 2 • Send 2 units of flow from to s to t along path P

Flow x After Iteration 1 (2, 5) s 2 (2, 2) 4 (2, 4) (0, 4) 1 (0, 6) 3 (0, 5) 6 (0, 7) 5 v=2 t

The Residual Network G(x) 0 2 3 s 1 2 6 0 2 0 3 2 4 5 0 7 5 6 0 t

Iteration 2: LIST = {1}, Labeled = {1} • i=1 • LIST = {} • Arc (1, 2) – pred(2) =1 – label 2 – LIST = {2} • Arc (1, 3) – pred (3) = 1 – label 3 – LIST = {2, 3}

Iteration 2: LIST = {2, 3}, Labeled={1, 2, 3} p=1 2 3 s 1 2 0 6 0 2 0 3 p=1 2 4 5 0 7 5 6 0 t

Iteration 2: LIST = {2, 3}, Labeled = {1, 2, 3} • i=2 • LIST = {3} • Arc (2, 4) – residual cap (2, 4) = 0 • Arc (2, 5) – pred (5) = 2 – label 5 – LIST = {3, 5} • Arc (2, 1) – residual capacity of (2, 1) = 0

Iteration 2: LIST = {3, 5}, Labeled={1, 2, 3, 5} p=1 2 3 s 1 2 0 6 0 2 0 3 p=1 2 4 5 0 7 5 p=2 6 0 t

Iteration 2: LIST = {3, 5}, Labeled = {1, 2, 3, 5} • i=3 • LIST = {5} • Arc (3, 5) – 5 is already labeled • Arc (3, 1) – residual capacity of (3, 1) = 0 • i=5 • LIST = {} • Arc (5, 2) – residual cap = 0 • Arc (5, 3) – residual cap = 0 • Arc (5, 6) – pred(6) = 5 – label 6 – LIST = {6}

Iteration 2: LIST = {6}, Labeled={1, 2, 3, 5, 6} p=1 2 3 s 1 2 0 6 0 2 0 3 p=1 2 4 5 0 7 5 p=2 6 0 p=5 t

Iteration 2: The sink is labeled • Use pred labels to trace back from the sink to the source to find path P – P = 1 -> 2 -> 5 -> 6 • = min {rij: (i, j) in P) = 3 • Send 3 units of flow from to s to t along path P

Flow x After Iteration 2 (5, 5) s 2 (2, 2) 4 (2, 4) (3, 4) 1 (0, 6) 3 (0, 5) 6 (3, 7) 5 v=5 t

The Residual Network G(x) 0 2 0 s 1 5 6 0 2 3 3 2 4 2 1 5 0 4 5 6 0 t

Iteration 3: LIST = {1}, Labeled = {1} • i=1 • LIST = {} • Arc (1, 2) – residual capacity = 0 • Arc (1, 3) – pred (3) = 1 – label 3 – LIST = {3}

Iteration 3: List = {3}, Labeled = {1, 3} 0 2 0 s 1 5 6 0 2 3 3 p=1 2 4 2 1 5 0 4 5 6 0 t

Iteration 3: LIST = {3}, Labeled = {1, 3} • i=3 • LIST = {} • Arc (3, 1) – residual capacity = 0 • Arc (3, 5) – pred (5) = 3 – label 5 – LIST = {5}

Iteration 3: List = {5}, Labeled = {1, 3, 5} 0 2 0 s 1 5 6 0 2 3 3 p=1 2 4 2 1 5 0 4 5 p=2 6 0 t

Iteration 3: LIST = {5}, Labeled = {1, 3, 5} • i = 5, LIST = {} • Arc (5, 2) – pred(2) = 5 – label 2 – LIST = {2} • Arc (5, 3): residual capacity = 0 • Arc (5, 6) – pred (6) = 5 – label 6 – LIST = {2, 6}

Iteration 3: List = {2, 6}, Labeled = {1, 2, 3, 5, 6} p=5 2 0 s 1 5 0 6 0 2 3 3 p=1 2 4 2 1 5 0 4 5 p=2 6 0 p=5 t

Iteration 3: The sink is labeled • Use pred labels to trace back from the sink to the source to find path P – P = 1 -> 3 -> 5 -> 6 • = min {rij: (i, j) in P) = 4 • Send 4 units of flow from to s to t along path P

Flow x After Iteration 3 (5, 5) s 2 (2, 2) 4 (2, 4) (3, 4) 1 (4, 6) 3 (4, 5) 6 (7, 7) 5 v=9 t

The Residual Network G(x) 0 2 0 s 1 5 2 4 2 3 3 2 4 2 1 1 4 0 5 6 7 t

Iteration 4: LIST = {1}, Labeled = {1} • i=1 • LIST = {} • Arc (1, 2) – residual capacity = 0 • Arc (1, 3) – pred (3) = 1 – label 3 – LIST = {3}

Iteration 4: List = {3}, Labeled = {1, 3} 0 2 0 s 1 5 2 4 2 3 3 p=1 2 4 2 1 1 4 0 5 6 7 t

Iteration 4: LIST = {3}, Labeled = {1, 3} • i=3 • LIST = {} • Arc (3, 1) – 1 is labeled • Arc (3, 5) – pred (5) = 3 – label 5 – LIST = {5}

Iteration 4: List = {5}, Labeled = {1, 3, 5} 0 2 0 s 1 5 2 4 2 3 3 p=1 2 4 2 1 1 4 0 5 p=3 6 7 t

Iteration 4: LIST = {5}, Labeled = {1, 3, 5} • i=5 • LIST = {} • Arc (5, 2) – pred(2) = 5 – label 2 – LIST = {2} • Arc (5, 6) – residual capacity = 0

Iteration 4: List = {2}, Labeled = {1, 2, 3, 5} p=5 2 0 s 1 5 0 2 4 2 3 3 p=1 2 4 2 1 1 4 0 5 p=3 6 7 t

Iteration 4: LIST = {2}, Labeled = {1, 2, 3, 5} • i = 2 LIST = {} • Arc (2, 1) – 1 is already labeled • Arc(2, 4) – residual capacity = 0 • Arc (2, 5) – 5 is already labeled

Iteration 4: List = {} • The sink is not labeled • Algorithm ends with optimal flow x

Correctness • At the end of each iteration, the algorithm either augments the flow or terminates because it can’t label the sink. • Let S be the set of labeled nodes when the algorithm terminates. Let T = N S. • We need to show that when the algorithm terminates v = u[S, T] which implies x is a maximum flow.

Correctness: arcs in (S, T) s • • i rij = 0 rij = uij - xij + xji 0 = uij – xij + xji xij = uij + xji j Since xji must be at least zero and 0 xij uij, it follows that xij = uij.

Correctness: arcs in (T, S) i j s • Suppose xij > 0 • rji = uji – xji + xij • Implies rji > 0 (uji xji) • Implies s can reach i in residual network which means i should have been labeled. • Thus, xij = 0

Correctness Thus, the flows on the forward arcs in [S, T] are at the upper bound and there is no flow on any of the backwards arcs. This means x is a maximum flow by Property 6. 1.

Complexity • Let U = max {(i, j) in A} uij. • If S = {s} and T = N{s}, then u[S, T] is at most n. U. • The maximum flow is at most n. U. • Iteration of the inner while loop is O(m): – Each arc is inspected at most once – Finding is O(n) – Updating the flow on P is O(n) • Complexity is O(n. U x m) = O(nm. U).