Advanced Graph
Homer Lee 2013/10/31
Reference
Slides from Prof. Ya-Yunn Su’s and Prof. Hsueh-I Lu’s course
Today’s goal
Flow networks
Ford-Fulkerson (and Edmonds-Karp)
Bipartite matching
Flow networks
Network
A directed graph G, each of whose edges has a capacity
Two nodes s and t of G
Denoted as (G,s,t)
*
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Flow
A flow of network (G,s,t) is (weighted) subgraph of G satisfying the capacity constraint and the conservation law
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Capacity constraint
Given capacity constraint function c, for all u, v ∈ V, 0 ≤ f(u,v) ≤ c(u,v)
流過edge的flow大小要小於流量限制
常表示為圖的edge的weight
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Conservation Law
For all 𝑢 ∈ 𝑉 − {𝑠, 𝑡} , 𝑣∈𝑉 𝑓 𝑢, 𝑣 = 𝑣∈𝑉 𝑓 𝑣, 𝑢
換句話說，流進來的等於流出去的
上面的sigma是對所有的v，那假如u,v沒有接在一起
怎麼辦?
=> If (u,v)∉ E, f(u,v) = 0
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What’s maximum flow problem?
The value of a flow is denoted as |f|
|f| = 𝑣∈𝑉 𝑓 𝑠, 𝑣 − 𝑣∈𝑉 𝑓 𝑣, 𝑠
Given flow network (G,s,t)
=> Find a flow of maximum value
A famous theorem
Maximum flow Minimum cut
How to solve the problem?
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Ford-Fulkerson
Idea: Iteratively increase the value of the flow
Start with f(u,v) = 0
In each iteration
Find an augmenting path in the residual network 𝐺𝑓
Until no more augmenting paths exist
Residual Network
What if we choose the wrong path?
Just give it a second chance!!
For each edge (u,v) in G,construct 𝐺𝑓
If f(u,v) > 0, 𝐺𝑓 has an edge (v,u) with weight f(u,v)
If c(u,v) ≥ f(u,v) 𝐺𝑓 has an edge (u,v) with weight c(u,v) – f(u,v)
illustration
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Some notes
Why adding an edge if c(u,v) ≥ f(u,v) ?
讓他有回頭的機會
|E𝑓|≤ 2|E|,Why?
Residual Networks上的edges會是原本Flow network 有的edge (with different weight) 或反方向
You can try to prove it by simply using case analysis.
How can residual network help us?
Lemma: Let G = (V,E) be a flow network, and let f be a flow in G. Let 𝐺𝑓 be the residual network of G induced by f. Let g be a flow in 𝐺𝑓. Then
f+g = is a valid flow in G.
Idea: capacity constraint and flow conservation still holds.
Augmenting paths
Given a flow network G and a flow f, an
augmenting path is a simple path from s to t in the residual network 𝐺𝑓
簡單的來說，就是在residual network上找一條可以 走的路
How can Augmenting paths help us?
There exists an augmenting path
=> there exist some potential flow in the path
=> By the capacity constraint, trivially the maximum flow in the path
= min{𝐶𝑓(u,v)|(u,v) is on augmenting path}
How do we know when we have found maximum flow?
From the maximum-flow-minimum-cut theorem, we stop when its residual graph contains no
augmenting graph
2 equivalent things:
1. f is a maximum flow in G
2. The residual network 𝐺𝑓 contains no augmenting path
Let’s prove it!
f is a maximum flow in G => The residual network 𝐺𝑓 contains no augmenting path
=> is simple, use contradiction.
If 𝐺𝑓 still contains augmenting paths, then we can still find 𝑓𝑝 to add to f. Then result in bigger flow |f| + |𝑓𝑝| > |f|
Let’s prove it!!
Goal: f is a maximum flow in G <= The residual network 𝐺𝑓 contains no augmenting path
Equivalent statement: f is not a maximum flow in G => The residual network 𝐺𝑓 contains some
augmenting path
If h is a flow whose value larger than that of f, then g = h – f has to be a positive flow in 𝐺𝑓
Let’s prove it!!!
Goal: f is not a maximum flow in G <= The
residual network 𝐺𝑓 contains some augmenting path
Here provides a sketch of the proof
If h is a flow whose value larger than that of f, then g = h – f has to be a positive flow in 𝐺𝑓
Then why g has to be a positive flow in 𝐺𝑓?
Do the remaining job by yourself!
Pseudo code
Ford-Folkerson (G,s,t){
for each edge (u,v) in G.E (u,v).f = 0
while there exists an augmenting path p in residual network 𝐺𝑓 𝒄𝒇(𝐩) = min(𝒄𝒇(u, v) | (u,v) is on p)
for each edge (u,v) on p //雙向都要考慮 if (u,v) ∈ E, (u,v).f += 𝒄𝒇 𝐩
else, (v,u).f -= 𝒄𝒇 𝐩 }
Example:
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Running time analysis
Initializing part: O(E)
How to find a path in residual network?
BFS or DFS
What time complexity does it take?
|E𝑓| ≤ 2|E|
It takes O(V+E𝑓) = O(E) times
Running time analysis
If the edge capacity are integer, and f be the maximum flow of the network
The for-loop may be executed at most f times(increment by 1 unit at a time)
Each time takes O(E) times
Totally O(E)+O(E)*O(f) = O(Ef)
=> This depends on f, not a good idea
Issues
What is C is not an integer?(each time the
amount that the augmenting path adding has no lower-bound)
How can we improve this bound?
(Edmonds-Karp)
To improve the time complexity
A key observation: Let P be a shortest path from s to t in the residual network 𝐺𝑓. Let g be the flow corresponding to P.
Then, the distance of any nodes v from s in 𝐺𝑓+𝑔 is no less than that in 𝐺𝑓
IDEA: If the above thing holds, it seems that the update times will be bounded
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Proof of the monotonically increased distance
Assume for contradiction that there is a node v whose distance d*(v) in 𝐺𝑓+𝑔 is less than its
distance d(v) in 𝐺𝑓
Let Q be a shortest path from s to v in 𝐺𝑓+𝑔
There has to be some node u on Q such that d*(u) ≥ d(u).(s is such a u.)
So we can assume Q = s~u->v, so d*(v) = d*(u)+1 and (u,v)∈ 𝐺𝑓+𝑔 . such u exists.
u s v
That’s what we said last page Note: d*(u) ≤ d(u)
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Proof-contd.
Claim: (u,v) cannot be an edge in 𝐺𝑓 since the following contradiction:
d(v) ≤ d(u)+1 ≤ d*(u)+1 = d*(v)
Since (u,v) belongs to 𝐺𝑓+𝑔 but not 𝐺𝑓, we know g goes from v to u, but still reach the following contradiction:
d(v) = d(u)-1 ≤ d*(u)-1 = d*(v)-2
