COMP3010: Algorithm Theory and Design
Daniel Sutantyo, Department of Computing, Macquarie University
8.0 - Maximum Flow
Prelude
8.0 - Maximum Flow
- We are going to discuss graph-related algorithms this week:
- Maximum Flow (Ford Fulkerson Method)
- Single-Source Shortest Path (Bellman-Ford Algorithm)
- All-Pairs Shortest Path (Matrix Multiplication Method and Floyd-Warshall Algorithm)
Notation and Terminology
8.0 - Maximum Flow
- A graph \(G\) is characterised by its set of vertices \(V\) and its set of edges \(E\), and we write \(G = \langle V,E \rangle\)
- if \(u\) and \(v\) are two vertices in \(V\), we use \((u,v)\) to denote the edge connecting them, and the degree of a vertex is the number of edges connected to it
- if the graph is weighted, then we use \(w(u,v)\) to denote the weight of the edge connecting \(u\) and \(v\)
- if the graph is directed (digraph) then \((u,v)\) is not the same as \((v,u)\), and they can have different weights as well
- A graph can also be acyclic, i.e. it contains no cycles
- A graph is connected if there is a path from each vertex in the graph to any other vertex
8.0 - Maximum Flow
- A subgraph of \(G = \langle V,E\rangle\) is another graph \(G^\prime = \langle V^\prime,E^\prime\rangle\) formed from a subset of the vertices and edges of \(G\) i.e \(E^\prime \subseteq E\) and \(V^\prime \subseteq V\)
- We also use the term component to refer to a (maximal) connected subgraph
- weakly connected vs strongly connected
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Notation and Terminology
Maximum Flow Problem
8.0 - Maximum Flow
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- Suppose that you want to move materials from one location to another location
- each vertex represents a location
- each edge represents the connections between two locations
Maximum Flow Problem
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- The source vertex produces material and the sink vertex consumes them
- every other vertex, can only propagate the material (so if they receive \(n\), then they have to send out \(n\) as well)
- The weight of each edge signifies the amount of material that can flow through it
Maximum Flow Problem
8.0 - Maximum Flow
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- The source vertex produces material and the sink vertex consumes them
- every other vertex, can only propagate the material (so if they receive \(n\), then they have to send out \(n\) as well)
- The weight of each edge signifies the amount of material that can flow through it
Maximum Flow Problem
8.0 - Maximum Flow
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- In maximum-flow problem, we want to compute the greatest rate at which we can move materials from the source vertex to the sink vertex
- It's not that straightforward: in the above example, the flow out of the source is 29, but you cannot propagate this much material in the network
Maximum Flow Problem
8.0 - Maximum Flow
- It is a relatively common (and complex) problem, but we are going to study a simplified version
- Examples of applications:
- goods distribution
- road networks (traffic control), rail networks
- water distribution, water pipes
- airline scheduling
- It is also well-studied:
- Ford-Fulkerson Algorithm
- Edmonds-Karp Algorithm
- Dinic's Algorithm
- KRT Algorithm
Example
8.0 - Maximum Flow
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Example
8.0 - Maximum Flow
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Here is another flow:
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Definitions
8.0 - Maximum Flow
- A flow network \(G = \langle V,E \rangle\) is a directed graph where:
- each edge \((u,v) \in E\) has a non-negative capacity \(c(u,v)\)
- if \((u,v) \in E\), then \((v,u)\not\in E\), i.e. if you have an edge going in one direction, then you cannot have another edge in the reverse direction
- there is a source vertex \(s\) and a sink vertex \(t\)
- Assumptions (to make our life easier):
- the graph is connected and each node lies in the path between the source and the sink vertices
- \(|E| \ge |V|-1\) because each vertex has degree at least 1
Definitions
8.0 - Maximum Flow
- A flow in \(G\) is a function\(f(u,v) : V \times V \rightarrow \mathbb R\) for \(u,v \in V\) that satisfies the following two properties:
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capacity constraint: for all \(u,v \in V\),
- \(0\le f(u,v) \le c(u,v)\), i.e. the flow through an edge cannot exceed the capacity of that edge
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flow conservation: for all \(u \in V \setminus \{s,t\}\)
- i.e. except for the source and sink vertices, all inflows to a vertex must equal the outflows from that vertex
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capacity constraint: for all \(u,v \in V\),
\[\sum_{v\in V\setminus \{s,t\}}f(v,u) = \sum_{v\in V\setminus\{s,t\}} f(u,v)\]
Definitions
8.0 - Maximum Flow
this is a flow
this is a different flow
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Definitions
8.0 - Maximum Flow
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\(f(s,a) = 11\)
\(f(s,c) = 8\)
\(f(a,b) = 11\)
\(f(b,c) = 3\)
\(f(c,d) = 11\)
\(f(d,b) = 7\)
\(f(b,t) = 15\)
\(f(d,t) = 4\)
the value of this flow is \(19\)
A flow in \(G\) is a function \(f(u,v) : V \times V \rightarrow \mathbb R\) for \(u,v \in V\)
\(f(c,a) = 0\)
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Definitions
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A flow in \(G\) is a function \(f(u,v) : V \times V \rightarrow \mathbb R\) for \(u,v \in V\)
\(f(s,a) = 1\)
\(f(s,c) = 1\)
\(f(a,b) = 1\)
\(f(b,c) = 0\)
\(f(c,d) = 1\)
\(f(d,b) = 0\)
\(f(b,t) = 1\)
\(f(d,t) = 1\)
the value of this flow is \(2\)
\(f(c,a) = 0\)
Definitions
8.0 - Maximum Flow
- A flow network has many flows, and each flow has its own value \(|f|\), defined by:
- \[|f| = \sum_{v \in V} f(s,v)\]
- i.e the total flow out of the source vertex (how much can we send out)
- our goal is to find a flow of maximum value, the maximum flow
- It might be worthwhile for you to stop now and think on how to do this using brute force or also why greedy algorithm won't work
Ford-Fulkerson Method
8.0 - Maximum Flow
- Method:
- start by setting \(f(u,v)\) to \(0\) for all \(u,v \in V\)
- while there is an augmenting path \(p\) in the residual network \(G_f\):
- augment flow \(f\) along \(p\)
- It is not an algorithm, just a method, because it can be implemented in several ways (Edmond-Karp Algorithm is one implementation)
- Let's begin with a run of the method, then we'll go back to these definitions
Ford-Fulkerson Method
8.0 - Maximum Flow
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Ford-Fulkerson Method
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Ford-Fulkerson Method
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- an augmenting path \(p\) is a simple path from \(s\) to \(t\) in the residual network \(G_f\)
Ingredient #1:
Ford-Fulkerson Method
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Ford-Fulkerson Method
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- a residual network of \(G\) is the edges of \(G\) that still have some capacity left, i.e. those that can take additional flow
- we use \(G_f\) to denote the residual network of \(G\)
Ingredient #2:
Ford-Fulkerson Method
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- a residual network of \(G\) is the edges of \(G\) that still have some capacity left, i.e. those that can take additional flow
- we use \(G_f\) to denote the residual network of \(G\)
Ingredient #2:
Ford-Fulkerson Method
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- a residual network of \(G\) is the edges of \(G\) that still have some capacity left, i.e. those that can take additional flow
- we use \(G_f\) to denote the residual network of \(G\)
Ingredient #2:
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Ford-Fulkerson Method
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- a residual network of \(G\) is the edges of \(G\) that still have some capacity left, i.e. those that can take additional flow
- we use \(G_f\) to denote the residual network of \(G\)
Ingredient #2:
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Ford-Fulkerson Method
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- a residual network of \(G\) is the edges of \(G\) that still have some capacity left, i.e. those that can take additional flow
- we use \(G_f\) to denote the residual network of \(G\)
Ingredient #2:
Ford-Fulkerson Method
8.0 - Maximum Flow
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Residual Network
8.0 - Maximum Flow
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- Define the residual capacity \(c_f(u,v)\) by
\[c_f(u,v)=\begin{cases} c(u,v)-f(u,v)& \text{if $(u,v)\in E$,} \\ f(v,u) & \text{if $(v,u)\in E$,}\\ 0 & \text{otherwise} \end{cases} \]
Residual Network
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\[c_f(u,v)=\begin{cases} c(u,v)-f(u,v)& \text{if $(u,v)\in E$,} \\ f(v,u) & \text{if $(v,u)\in E$,}\\ 0 & \text{otherwise} \end{cases} \]
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- Define the residual capacity \(c_f(u,v)\) by
Residual Network
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\[c_f(u,v)=\begin{cases} c(u,v)-f(u,v)& \text{if $(u,v)\in E$,} \\ f(v,u) & \text{if $(v,u)\in E$,}\\ 0 & \text{otherwise} \end{cases} \]
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- Define the residual capacity \(c_f(u,v)\) by
Residual Network
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\[c_f(u,v)=\begin{cases} c(u,v)-f(u,v)& \text{if $(u,v)\in E$,} \\ f(v,u) & \text{if $(v,u)\in E$,}\\ 0 & \text{otherwise} \end{cases} \]
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- Define the residual capacity \(c_f(u,v)\) by
Residual Network
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\[c_f(u,v)=\begin{cases} c(u,v)-f(u,v)& \text{if $(u,v)\in E$,} \\ f(v,u) & \text{if $(v,u)\in E$,}\\ 0 & \text{otherwise} \end{cases} \]
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- Define the residual capacity \(c_f(u,v)\) by
Residual Network
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\[c_f(u,v)=\begin{cases} c(u,v)-f(u,v)& \text{if $(u,v)\in E$,} \\ f(v,u) & \text{if $(v,u)\in E$,}\\ 0 & \text{otherwise} \end{cases} \]
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- Define the residual capacity \(c_f(u,v)\) by
Residual Network
8.0 - Maximum Flow
\[c_f(u,v)=\begin{cases} c(u,v)-f(u,v)& \text{if $(u,v)\in E$,} \\ f(v,u) & \text{if $(v,u)\in E$,}\\ 0 & \text{otherwise} \end{cases} \]
- Define the residual network of \(G\) induced by \(f\) as \(G_f = (V,E_f)\) where
- \(E_f = \{(u,v) \in V \times V : c_f(u,v) > 0\}\)
- notice that a residual network is basically a flow network that is allowed to have both \((u,v)\) and \((v,u)\)
- Define the residual capacity \(c_f(u,v)\) by
Augmenting Path
8.0 - Maximum Flow
- The residual network tells us how to add flow to the original flow network
- The idea is to find an augmenting path in the residual network \(G_f\) and use this augment the flow \(f\) in \(G\)
- An augmenting path \(p\) is a simple path from \(s\) (source) to \(t\) (sink) in the residual network \(G_f\)
8.0 - Maximum Flow
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- The residual network tells us how to add flow to the original flow network
- The idea is to find an augmenting path in the residual network \(G_f\) and use this augment the flow \(f\) in \(G\)
- remember, don't think of a flow as just one path, a flow is a function that tells us how much capacity we are using on each edge
Augmenting Path
8.0 - Maximum Flow
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\(f(s,a) = 1\)
\(f(s,c) = 1\)
\(f(a,t) = 2\)
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\(f(c,a) = 1\)
\(f(c,t) = 0\)
- The residual network tells us how to add flow to the original flow network
- The idea is to find an augmenting path in the residual network \(G_f\) and use this augment the flow \(f\) in \(G\)
- remember, don't think of a flow as just one path, a flow is a function that tells us how much capacity we are using on each edge
Augmenting Path
8.0 - Maximum Flow
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- The residual network tells us how to add flow to the original flow network
- The idea is to find an augmenting path in the residual network \(G_f\) and use this augment the flow \(f\) in \(G\)
- remember, don't think of a flow as just one path, a flow is a function that tells us how much capacity we are using on each edge
Augmenting Path
8.0 - Maximum Flow
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- An augmenting path \(p\) is a simple path from \(s\) (source) to \(t\) (sink) in the residual network \(G_f\)
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Augmenting Path
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- An augmenting path \(p\) is a simple path from \(s\) (source) to \(t\) (sink) in the residual network \(G_f\)
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Augmenting Path
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- Once we found an augmenting path, we simply use it to augment the flow we have so far in \(G\)
- Note that an augmenting path is basically a flow in \(G_f\)
Augmenting Path
8.0 - Maximum Flow
- If \(f\) is a flow in \(G\) and \(f_p\) is a flow in \(G_f\), then define the function \((f \uparrow f_p) : V \times V \rightarrow \mathbb R\) as
\[(f \uparrow f_p)(u,v) = \begin{cases} f(u,v) + f_p(u,v) - f_p(v,u) &\text{if $(u,v) \in E$,} \\ 0 & \text{otherwise.}\end{cases}\]
- The augmenting path is a flow \(f_p\) where the value \(|f_p|\) of the flow is the minimum residual capacity in the path
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Augmenting Path
8.0 - Maximum Flow
- we call the function \((f \uparrow f_p)\) as the augmentation of flow \(f\) by \(f_p\)
- basically we increase the flow on \((u,v)\) by \(f_p(u,v)\), but we decrease it by \(f_p(v,u)\) (we call this a cancellation)
- the function \((f\uparrow f_p)(u,v)\) is a flow in \(G\) with value
\(|f\uparrow f_p| = |f| + |f_p| > |f|\)
(Lemma 26.1, Corollary 26.3 of CLRS, page 717, 720)
- If \(f\) is a flow in \(G\) and \(f_p\) is a flow in \(G_f\), then define the function \((f \uparrow f_p) : V \times V \rightarrow \mathbb R\) as
\[(f \uparrow f_p)(u,v) = \begin{cases} f(u,v) + f_p(u,v) - f_p(v,u) &\text{if $(u,v) \in E$,} \\ 0 & \text{otherwise.}\end{cases}\]
Augmenting Path
8.0 - Maximum Flow
- we call the function \((f \uparrow f_p)\) as the augmentation of flow \(f\) by \(f_p\)
- basically we increase the flow on \((u,v)\) by \(f_p(u,v)\), but we decrease it by \(f_p(v,u)\) (we call this a cancellation)
- the function \((f\uparrow f_p)(u,v)\) is a flow in \(G\) with value
\(|f\uparrow f_p| = |f| + |f_p| > |f|\)
(Lemma 26.1, Corollary 26.3 of CLRS, page 717, 720)
- If \(f\) is a flow in \(G\) and \(f_p\) is a flow in \(G_f\), then define the function \((f \uparrow f_p) : V \times V \rightarrow \mathbb R\) as
\[(f \uparrow f_p)(u,v) = \begin{cases} f(u,v) + f_p(u,v) - f_p(v,u) &\text{if $(u,v) \in E$,} \\ 0 & \text{otherwise.}\end{cases}\]
Augmenting Path
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Ford-Fulkerson Method
8.0 - Maximum Flow
- You don't have to worry about the proof of correctness of the algorithm (it is definitely not simple)
- It is called the Max-Flow Min-Cut Theorem (Theorem 26.6 in CLRS, pg 723)
- What you need to understand is how the method works
- Ford-Fulkerson Method:
- start by setting \(f(u,v)\) to \(0\) for all \(u,v \in V\)
- while there is an augmenting path \(p\) in the residual network \(G_f\):
- augment flow \(f\) along \(p\)
Ford-Fulkerson Method
8.0 - Maximum Flow
- Ford-Fulkerson Method for a flow network \(G = \langle V,E\rangle\):
- set \(f(u,v) = 0\) for all \((u,v)\) in \(E\)
- create the residual network \(G_f\)
- while there exists a simple path \(p\) from source to sink in \(G_f\):
- set \(c_f(p) = \min\{c_f(u,v) : (u,v) \in p\}\)
- for each edge \((u,v)\) in \(p\):
- if \((u,v) \in E\):
- \((u,v) = (u,v) + c_f(p)\)
- else
- \((v,u) = (v,u) - c_f(p)\)
- if \((u,v) \in E\):
Ford-Fulkerson Method
8.0 - Maximum Flow
- One more time, this is just a method, not an algorithm, because we never discussed how to actually find the augmenting path
- Edmond-Karp Algorithm is an implementation that uses BFS to find the augmenting path, and it can be proven that the runtime is \(O(VE^2)\)
- proof of correctness is in CLRS, pg. 727 onwards, but again, this is beyond the scope of this unit
- Dinic's blocking flow algorithm is \(O(V^2E)\) (it also uses BFS)
Example
8.0 - Maximum Flow
a
s
c
t
b
d
source
sink
0/13
0/16
0/12
0/4
0/9
0/7
0/20
0/14
0/4
- We'll do this in the workshop, but you can also give it a try first
COMP3010 - 8.0 - Maximum Flow
By Daniel Sutantyo
COMP3010 - 8.0 - Maximum Flow
- 150
