Flows sourcesink s t

Flows sourcesink edge-weights = capacities

Flows sourcesink edge-weights = capacities 4 units of flow

Flows sourcesink edge-weights = capacities 6 units of flow 2

Flows sourcesink edge-weights = capacities 7 units of flow 3 1

Optimal?  a larger flow?

Cuts s C C  V s  C t  V-C capacity of a cut cap(C)=  c(u,v) u  C v  V-C for any cut and any flow flow  cut max-flow  min-cut 

Flows edge-weights = capacities 3 FLOW CONSERVATION

Skew symmetry edge-weights = capacities 3 FLOW CONSERVATION x -x

Flows edge-weights = capacities 3 FLOW CONSERVATION x -x  f(u,v) = 0 vVvV

Flow – formal definition FLOW CONSERVATION CAPACITY CONSTRAINTS  f(u,v) = 0 vVvV f(u,v)  c(u,v) SKEW SYMMETRY f(u,v) = - f(v,u)

Max-flow problem INPUT: directed graph G=(V,E), source s  V, sink t  V a capacity c(e) for each edge e  E OUTPUT: maximum flow from s to t

Max-flow problem F  zero-flow while can improve F improve F

Max-flow problem improving can decrease flow on some edges

+1 +1 Augmenting path

+1 Augmenting path

+1 Augmenting path  can improve the flow F  zero-flow while  augmenting path p improve F along p Ford-Fulkerson algorithm

Residual capacity f(u,v) c(u,v) r(u,v) = c(u,v) – f(u,v) makes sense for negative f(u,v)

Residual network

Is there an augmenting path? Is there a path from s to t in the residual network?

Correct ? F  zero-flow while  augmenting path p improve F along p

Correct ? F  zero-flow while  augmenting path p improve F along p Theorem: no augmenting path  max-flow YES

Theorem: no augmenting path  max-flow Proof: no augmenting path  no path from s to t in the residual network st

Theorem: no augmenting path  max-flow Proof: no augmenting path  no path from s to t in the residual network st vertices to which we can get from s

Theorem: no augmenting path  max-flow Proof: no augmenting path  no path from s to t in the residual network st vertices to which we can get from s all edges from C to V-C saturated C

Theorem: no augmenting path  max-flow Theorem 2: max-flow = min-cut

Is there an augmenting path? Is there a path from s to t in the residual network? time O(E) F  zero-flow while  augmenting path p improve F along p time = ?

F  zero-flow while  augmenting path p improve F along p

F  zero-flow while  augmenting path p improve F along p

F  zero-flow while  augmenting path p improve F along p

F  zero-flow while  augmenting path p improve F along p

F  zero-flow while  augmenting path p improve F along p

F  zero-flow while  augmenting path p improve F along p

F  zero-flow while  augmenting path p improve F along p

F  zero-flow while  augmenting path p improve F along p

F  zero-flow while  augmenting path p improve F along p Assume that the capacities are integers. Then the number of augementations is bounded by F * (=max-flow) running time O(E F * )

F  zero-flow while  augmenting path p improve F along p running time O(E F * ) can a good choice of p improve the algorithm?

F  zero-flow while  augmenting path p improve F along p 1. choose path p which increases the flow the most (i.e., p has the maximum bottleneck capacity) 2. choose path p which has the fewest number of edges

1. choose path p which increases the flow the most (i.e., p has the maximum bottleneck capacity) THEOREM: number of augmenting steps  O(E log F * )

Graph G flow f in G Residual network G f f’=flow in G f Claim: h=f+f’ is a flow in G SKEW SYMMETRY: h(u,v)=f(u,v)+f’(u,v)=-f(v,u)-f’(v,u) = - (f(v,u) + f’(v,u)) = - h(v,u) FLOW CONSERVATION:  u h(u,v) =  u (f(u,v)+f’(u,v))=  u f(u,v) +  u f’(u,v) = 0

Graph G flow f in G Residual network G f f’=flow in G f Claim: h=f+f’ is a flow in G CAPACITY CONSTRAINTS: f’(u,v)  c(u,v)-f(u,v)  h(u,v)=f(u,v)+f’(u,v)  c(u,v)

Graph G flow f in G Residual network G f f’=flow in G f Claim: f’=h-g is a flow in G f CAPACITY CONSTRAINTS: f’(u,v)  h(u,v)-f(u,v)  c(u,v)-f(u,v) flow h in G

Claim: any flow f in G is a sum of at most |E| path flows Induction on the number of non-zero edges in f

Claim: any flow f in G is a sum of at most |{e  E;f(e)  0}| path flows Induction on the number of non-zero edges in f G’ with capacities c’(u,v)=max{0,f(u,v)} p = augmenting path from zero flow in G’ p has bottleneck edge  f-p has less non-zero edges than f f = flow f = p + (f-p)

1.choose path p which increases the flow the most (i.e., p has the maximum bottleneck capacity) THEOREM: number of augmenting steps O(E log F * ) Claim: any flow f in G is a sum of at most |E| path flows Graph G flow f in G Residual network G f f’=f * -f max-flow f * in G  path with bottleneck capacity  f’/|E|

1.choose path p which increases the flow the most (i.e., p has the maximum bottleneck capacity) THEOREM: number of augmenting steps O(E log F * ) Graph G flow f in G Residual network G f f’=f * -f max-flow f * in G  path with bottleneck capacity  f’/|E| max-flow in the residual graph decreases by a factor of at least (1-1/|E|)

2. choose path p which has the fewest number of edges THEOREM: number of augmenting steps  O(E V)

2. choose path p which has the fewest number of edges level graph = BFS in the residual network

2. choose path p which has the fewest number of edges level graph = BFS in the residual network

2. choose path p which has the fewest number of edges 1) one such edge saturated 2) no new introduced count edges in the residual graph which are on a path of length d from s to t

Plain Ford-Fulkerson O(E F * ) Maximum bottleneck capacity O( (E+V log V) E log F  ) Fewest edges O(E 2 V)

Max-matching in bipartite graphs

st solution using max-flow all edge weights =1

Max-matching in bipartite graphs solution using max-flow st all edge weights =1 Plain Ford-Fulkerson O(E F * ) Maximum bottleneck capacity O( (E+V log V) E log F  ) Fewest edges O(E 2 V) which one to use?