max-flow (ford-fulkerson) - ecology...
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Max-Flow (Ford-Fulkerson)
Emil Thomas04/25/2019
https://www.cs.princeton.edu/~wayne/kleinberg-tardos/pearson/07MaximumFlow.pdfSlides From
Flow network
A flow network is a tuple G = (V, E, s, t, c).Digraph (V, E) with source s ∈ V and sink t ∈ V.
Capacity c(e) > 0 for each e ∈ E.
Intuition. Material flowing through a transportation network;material originates at source and is sent to sink.
3
s t5
15
1015
16
9
15
6
8 10
154
4 10
10
capacity
assume all nodes are reachable from s
Maximum-flow problem
Def. An st-flow (flow) f is a function that satisfies:
For each e ∈ E : [capacity]
For each v ∈ V – {s, t} : [flow conservation]
8
0 / 4
0 / 4 0 / 15
10 / 10
10 / 105 / 5 vs t
0 / 6
5 / 10
5 / 9
5 / 8
5 / 15
10 / 1010 / 15
10 / 16
inflow at v = 5 + 5 + 0 = 10outflow at v = 10 + 0 = 10
flow capacity
0 / 15
0 � f(e) � c(e)�
e v
f(e) =�
e v
f(e)
Maximum-flow problem
Def. An st-flow (flow) f is a function that satisfies:
For each e ∈ E : [capacity]
For each v ∈ V – {s, t} : [flow conservation]
Def. The value of a flow f is:
9
0 / 4
10 / 10
10 / 105 / 5s t
5 / 10
5 / 9
5 / 8
5 / 15
10 / 1010 / 15
0 / 15
value = 5 + 10 + 10 = 25
0 / 4
0 / 6
10 / 16
0 / 15
0 � f(e) � c(e)
val(f) =�
e s
f(e) ��
e s
f(e)
�
e v
f(e) =�
e v
f(e)
Maximum-flow problem
Def. An st-flow (flow) f is a function that satisfies:
For each e ∈ E : [capacity]
For each v ∈ V – {s, t} : [flow conservation]
Def. The value of a flow f is:
Max-flow problem. Find a flow of maximum value.
10
0 / 4
10 / 10
10 / 105 / 5s
8 / 10
8 / 9
8 / 8
10 / 1013 / 15
0 / 15
value = 10 + 5 + 13 = 28
0 / 4
3 / 6
13 / 16
0 / 15
t
2 / 15
0 � f(e) � c(e)
val(f) =�
e s
f(e) ��
e s
f(e)
�
e v
f(e) =�
e v
f(e)
Toward a max-flow algorithm
Greedy algorithm.
Start with f (e) = 0 for each edge e ∈ E.
Find an s↝t path P where each edge has f (e) < c(e). Augment flow along path P.
Repeat until you get stuck.
12
s t
0 / 20 /
10 0 / 6
0 / 10
0 / 4
0 / 8
0 / 9
flow network G and flow f
0 / 10 0
value of flow
0 / 10
flow capacity
Toward a max-flow algorithm
Greedy algorithm.
Start with f (e) = 0 for each edge e ∈ E.
Find an s↝t path P where each edge has f (e) < c(e). Augment flow along path P.
Repeat until you get stuck.
13
0 / 20 /
10 0 / 6
0 / 10
0 / 4
0 / 8
0 / 90 / 10 0
0 / 10
flow network G and flow f
s t
Toward a max-flow algorithm
Greedy algorithm.
Start with f (e) = 0 for each edge e ∈ E.
Find an s↝t path P where each edge has f (e) < c(e). Augment flow along path P.
Repeat until you get stuck.
14
0 / 20 /
10 0 / 6
0 / 10
0 / 4
0 / 8
0 / 90 / 10 0
0 / 10
+ 8 = 8—
8—
—8
8
flow network G and flow f
s t
+ 2 = 10
Toward a max-flow algorithm
Greedy algorithm.
Start with f (e) = 0 for each edge e ∈ E.
Find an s↝t path P where each edge has f (e) < c(e). Augment flow along path P.
Repeat until you get stuck.
15
0 / 6
0 / 4
8 / 8
0 / 10 8
0 / 100 / 2
8 / 10
8 / 100 / 9
—10 2 —
2—
10—
flow network G and flow f
s t
Toward a max-flow algorithm
Greedy algorithm.
Start with f (e) = 0 for each edge e ∈ E.
Find an s↝t path P where each edge has f (e) < c(e). Augment flow along path P.
Repeat until you get stuck.
16
0 / 4
8 / 8
10
2 / 210 /
10
10 / 10
0 / 6
0 / 10
0 / 10
2 / 9
6 —
8—
6— + 6 = 16
6—
flow network G and flow f
s t
Toward a max-flow algorithm
Greedy algorithm.
Start with f (e) = 0 for each edge e ∈ E.
Find an s↝t path P where each edge has f (e) < c(e). Augment flow along path P.
Repeat until you get stuck.
17
0 / 4
8 / 8
16
2 / 210 /
10
10 / 10
6 / 6
6 / 10
6 / 10
8 / 9
ending flow value = 16
flow network G and flow f
s t
Toward a max-flow algorithm
Greedy algorithm.
Start with f (e) = 0 for each edge e ∈ E.
Find an s↝t path P where each edge has f (e) < c(e). Augment flow along path P.
Repeat until you get stuck.
18
3 / 4
7 / 8
19
0 / 210 /
10
10 / 10
6 / 6
9 / 10
9 / 10
9 / 9
but max-flow value = 19
flow network G and flow f
s t
Q. Why does the greedy algorithm fail?
A. Once greedy algorithm increases flow on an edge, it never decreases it.
Ex. Consider flow network G .
The unique max flow has f *(v, w) = 0.
Greedy algorithm could choose s→v→w→t as first augmenting path.
Bottom line. Need some mechanism to “undo” a bad decision.
Why the greedy algorithm fails
19
s
t
w
v
1
2
2
22
flow network G
Residual network
Original edge. e = (u, v) ∈ E.
Flow f (e). Capacity c(e).
Reverse edge. ereverse = (v, u).
“Undo” flow sent.
Residual capacity.
Residual network. Gf = (V, Ef , s, t, cf ).
Ef = {e : f (e) < c(e)} ∪ {ereverse : f (e) > 0}.
Key property: f ʹ is a flow in Gf iff f + f ʹ is a flow in G.
20
u v
u v
residual capacity
flow
6 / 17
capacity
original flow network G
residual network Gf
11
6
where flow on a reverse edge negates flow on
corresponding forward edge
cf (e) =
�c(e) � f(e) e � E
f(e) e � Ereverse edge
edges with positive residual capacity
Augmenting path
Def. An augmenting path is a simple s↝t path in the residual network Gf . Def. The bottleneck capacity of an augmenting path P is the minimum residual capacity of any edge in P.
Key property. Let f be a flow and let P be an augmenting path in Gf . Then, after calling f ʹ ← AUGMENT( f, c, P), the resulting f ʹ is a flow and val( f ʹ) = val( f ) + bottleneck(Gf, P).
21
AUGMENT( f, c, P) ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
δ ← bottleneck capacity of augmenting path P.
FOREACH edge e ∈ P :
IF (e ∈ E) f (e) ← f (e) + δ.
ELSE f (ereverse) ← f (ereverse) – δ.
RETURN f.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Short Exercise & Discussion of solution in Lab
Ford–Fulkerson algorithm
Ford–Fulkerson augmenting path algorithm.
Start with f (e) = 0 for each edge e ∈ E.
Find an s↝t path P in the residual network Gf .
Augment flow along path P.
Repeat until you get stuck.
23
FORD–FULKERSON(G) _________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
FOREACH edge e ∈ E : f (e) ← 0.
Gf ← residual network of G with respect to flow f.
WHILE (there exists an s↝t path P in Gf )
f ← AUGMENT( f, c, P).
Update Gf.
RETURN f.
augmenting path
Exercise Handout and Survey in Ecampus
• Survey in ecampus->Lab1 makefile
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