The Ford-Fulkerson Algorithm aka The Labeling Algorithm

Ford-Fulkerson Algorithm begin x := 0; label node t; while t is labeled do 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 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

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

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

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

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

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

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

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 1 2 4 3 5 6 s t (2,2) (0,6) (0,5) (2,5) (2,4) (0,4) (0,7) v = 2

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

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

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

Iteration 2: LIST = {3,5}, Labeled = {1,2,3,5} 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) Arc (5,6) pred(6) = 5 label 6 LIST = {6}

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

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 1 2 4 3 5 6 s t (2,2) (0,6) (0,5) (5,5) (2,4) (3,4) (3,7) v = 5

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

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

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

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

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

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 4 2 2 1 6 5 2 3 1 t s 6 4 p=5 3 5 5 p=1 p=2

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 1 2 4 3 5 6 s t (2,2) (4,6) (4,5) (5,5) (2,4) (3,4) (7,7) v = 9

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

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

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

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

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

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

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) j rij = 0 rij = uij - xij + xji 0 = uij – xij + xji xij = uij + xji Since xji must be at least zero and 0  xij  uij, it follows that xij = uij.

Correctness: arcs in (T,S) 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 nU. The maximum flow is at most nU. 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(nU x m) = O(nmU).