Combinatorial AlgorithmsMaximum matching
Gale–Shapley algorithm
Assignment problem
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Outline
Definition
A graph G = (V, E) is bipartite if there exists partition V = X ∪ Y with X ∩ Y = ∅ and E ⊆ X × Y.
Bipartite Graph types
weight w(e) .
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Introduction
Example:
There are a set of boys and a set of girls.
Each boy only likes girls and each girl only likes boys.
A common friend wants to match each boy with a girl such that the boy and girl are both happy – but they both will only be happy if the boy likes the girl and the girl likes the boy.
Is it possible for every situation?
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Introduction
We can use a bipartite graph to model this problem
Problem
Yehong
Definition
Matching
A Matching is a subset M ⊆ E such that ∀v ∈ V at most one edge in M is incident upon v
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Maximum matching
A Maximum Matching is matching M such that everyother matching M′ satisfies |M′| ≤ |M|.
Unweighted graph: |M|= the number of edges
Weighted graph: |M|=
Maximum matching
A Matching
Definition
We say that a vertex is matched if it is incident to some edge in M.
Otherwise, the vertex is free
matched
free
Alternating Paths
A path is alternating if its edges alternate between M and E − M.
Augmenting Paths
An alternating path is augmenting if both endpoints are free
Alternating Tree
A tree rooted at some free vertex v in which every path is an alternating path.
Alternating paths
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Property of Augmenting Paths
Replacing the M edges by the E − M ones increments size of the matching
(Path: Y1, X2, Y2, X4, Y4, X5)
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Berge's Theorem: A matching M is maximum iff it has no augmenting path (Proof: Lec01 Page 3)
Maximum matching
At most V times
Maximum matching
Hopcroft-Karp Algorithm
An algorithm to find the maximum matching given a bipartite graph Gordon
Introduction
The Hopcroft-Karp algorithm was published in 1973
It is a matching algorithm that finds a maximum matching in bipartite graphs
The main idea is to augment along a set of vertex-disjoint shortest augment paths simulatenously
The complexity is O(√|V||E|)
In this section, some Theorems and Lemmas from graph theory will be stated without showing the proof.
Definition
We let the set A ⊕ B denote the symmetric difference of the set
A ⊕ B = (A ∪ B) – (A ∩ B)
A maximal set of vertex-disjoint minimum length augmenting path is defined as follows :
It is a set of augmenting path
No two path share a same vertex
If the minimum length augmenting path is of length k, then all paths in S are of length k
If p is an augmenting path not in S, then p shares a vertex with some path p’ in S
Algorithm
Given a graph G = (X ∪ Y),E)
1) Let M = {} ,
3) While S ≠ {}
M = M ⊕ S
Demonstration of algorithm at some stage
Let the dark edges represent the edges in a matching M
Demonstration of algorithm at some stage
Pink edges represent an augmenting path
Deleting them
Another augmenting path
No more paths
Pink edges represent the paths in maximal set S
M ⊕ S Note the before and after
Algorithm
Question : How do we know that this algorithm produces the result that we want ?
Theorem 1 (Berge) : A matching M is maximum if and only if there is no augmenting path with respect to M
This theorem guarantees the correctness of the algorithm
We will now prove that the complexity of the algorithm is O(√|V||E|)
Lemma 2 : A maximal set S of vertex-disjoint minimum length augmenting paths can be found in O(|E|) time
Proof : Let G = (U ∪ V,E) be the graph that we are working on and M be a matching
First , we construct a “tree-like/directed acyclic graph” graph given G
We start with all the free vertices in U at level 0
Starting at level 2k (even) , the vertices at level 2k+1 are obtained by following free edges from edges at level 2k
Starting at level 2k+1 (odd) , the level at 2k+2 are obtained by following matched edges from vertices at level 2k+1
Note that the even levels contain vertices from U and odd levels from V
Continuation of proof of lemma 2
Recall the earlier example : There are 3 levels here
U
V
Continuation of proof of lemma 2
We continue building the tree until all vertices have been visited or until a free vertex is encountered (say t)
Note that in the latter case, the free vertices are encountered at V
Complexity of this portion of building the “tree” is linear to the size of the edges ( similar to BFS)
Continuation of proof of lemma 2
Example :
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Dashed line represent edges while the normal lines represent edges in the matching M
Free-vertex
Continuation of proof of lemma 2
Now we find a maximal set S of vertex disjoint paths in this “tree” that we constructed
We assign a counter to all vertices after level 0
This counter represents the number of edges entering the vertex previous level (think of it like an indegree)
Starting at a free vertex v at level t, we trace a path until we reach a free vertex u at level 0
Continuation of proof of lemma 2
This path is an augmenting path and we add it into S
After which , we add the vertices in this path into a deletion queue
As long as the deletion queue is non empty, we delete the vertex in the queue and from the constructed “tree”
This includes all the edges incident onto it
Recall the earlier example
Continuation of proof of lemma 2
Whenever an edge is deleted , the counter associated with its right endpoint are all decremented
If the counter becomes 0, put the vertex into the deletion queue (there can be no augmenting path from this vertex)
After emptying the deletion queue, if there are still free vertex at level t , it means that an augmenting path must still exist
Continuation of proof of lemma 2
We continue until there are no more free vertex at level t
This entire process takes linear time , since it is proportional to the number of edges deleted
Therefore this part takes O(|E|)
Total time complexity for both parts is O(|E|)
Continuation of proof of lemma 2
Example :
V6
U6
V5
U1
Example :
V3
U3
V1
U2
Theorems and Lemmas cited without proof
Lemma 3 : Let M* be a maximum matching, and let M be any matching in G. If the length of the shortest augmenting path with respect to M is k, then |M*| - |M| ≤ (|V|/k)
Lemma 4 : Let k be the length of the shortest augmenting path with respect to M and let S be a maximal set of shortest disjoint augmenting paths with respect to M, then the length of the shortest augmenting path with respect to M⊕S is larger than k
Theorem 5: The Hopcroft-Karp algorithm finds a maximum matching in a bipartite graph in O(√|V||E|) time
Proof :
Now we run the algorithm for √|V| and let M be matching after running those rounds
Lemma 4 implies that we have that in each phrase of the algorithm, the length of the shortest augmenting path increases by at least 1
Therefore the size of the shortest augmenting path must be at least √|V|
Now from Lemma 3, we have that |M*| - |M| ≤ (|V|/√|V|) = √|V|
In each phrase , we increase the size of the matching by at least 1, so therefore , at most √|V| more phrases needed
Therefore at most 2 √|V| phrases are needed for this entire algorithm.
Therefore with lemma 2, the time complexity of the algorithm is O(√|V| |E|)
Continuation of proof of Theorem 5
Stable Marriage Problem
Wang Wei
Problem definition:
Given n men and n women, each person has a preference list for all members of the opposite sex; Find a one-to-one match M.
If m(a man) and w (a woman) are matched in M, then m is the partner of w, and vice verse.
Blocking pair in a match M: (m, w), m prefers w than his partner in M, and w prefers m than her partner.
Stable match: no blocking pair exist.
Terminology
For each man, try to find a woman, with whom they form a blocking pair; if no such woman exist, then the match is stable.
Complexity: O(n2)
Stability-checking algorithm
For man, propose to every women on his preference list until get engaged;
For woman, wait for proposal, accept if free or prefer the proposer than current partner/fiance; otherwise reject the proposal;
Complexity: O(n2)
Basic Gale-Shapley algorithm
For any given instance of the stable marriage problem, the Gale-Shapley algorithm terminates, and, on termination, the engaged pairs constitute a stable matching.
Termination:
Stability: if m prefers w than his partner , then w must have rejected m, i.e., w prefers her partner to m. (m,w) cannot be a block pairno block pair exists.
If GS not terminate, then at least one man is free
He must be rejected by all women
Once a woman is engaged, she will never be free
To reject a man, the woman must be engaged
All women are engaged
All men are engaged
Theorem 1.
Theorem 2: All possible executions of the Gale-Shapley algorithm(with the men as the proposers) yield the same stable matching, in which, man has the best partner he can have in any stable matching.
Theorem 3: In the man-optimal stable matching, each woman has the worst partner that she can have in any stable matching.
The Hungarian Method
Jinyang
Suppose we have n resources to which we want to assign to n tasks on a one-to-one basis. Suppose we also know the cost(gain) of assigning a given resource to a given task. We wish to nd an optimal assignment–one which minimizes(maximizes) total cost(gain).
Min-Cost or Max-Weight Perfect Matching in Bipartite Graph.
Assignment Problem
Three students collaborate to finish a project. Their efficiency is different.
coding
test
writing
John
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We will formula this problem in a matrix representation. It is easier to illustrate its key idea and how it works.
We will explain how to implement it into algorithm and show its complexity later.
We will use minimum cost form of the problem. For maximum problem, we just reverse the num.
Matrix Representation
Matrix Representation
is the cost of assigning the th resource to the th task.
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Difficulty
If a number is added to or subtracted from all of the entries of any one row or column of a cost matrix, then on optimal assignment for the resulting cost matrix is also an optimal assignment for the original cost matrix.
You have to choose one entry in each row or column any way. So this operation add or reduce the same number for all assignment.
Theorem 1
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When there exist a assignment has a zero cost in a non-negative matrix. This assignment is an optimal assignment.
Theorem 2
The key idea of Hungarian Method is to transform the original matrix to a non-negative matrix which have a zero assignment by add or subtract operation in each row and column.
There will be some slight difference in different implementation.
Hungarian Method
Step 1: Subtract the smallest entry in each row from all the entries of its row.
Step 2: Subtract the smallest entry in each column from all the entries of its column.
This two step is not necessary. But it can reduce the number of iterations later. The only requirement is that it comes to a non-negative matrix.
Hungarian Method
Step 3: Draw lines through appropriate rows and columns so that all the zero entries of the cost matrix are covered and the minimum number of such lines is used.
Step 4: If the minimum number of covering lines is n, an optimal assignment of zeros is possible and we are nished. Else continue step 5.
Step 5: Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to Step 3.
Hungarian Method
Only uncovered point will be subtracted. All uncovered point is not less than the smallest non-zero. So no point will be less than zero.
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Example 2
The lines is a minimum dominating set of all zero point.
Transform the solution of maximum matching to minimum dominating set.
How to Draw Lines
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Mark all rows having no assignments .
0
Find a maximum assignment(maximum match).
$
Find a maximum assignment(maximum match).
$
Find a maximum assignment(maximum match).
Mark all rows having no assignments . Then mark all columns having zeros in marked row(s). Then mark all rows having assignments in marked columns .Repeat this until a closed loop is obtained.
$
How to Draw Lines
There are at most n-1 lines. Suppose there are m row lines and k column lines. The smallest entry is x.
We subtract from the sum of all entries of the matrix. And then add . The sum reduces at least.
The sum value becomes smaller and smaller in each iteration. (complexity will be shown in KM algorithm)
Why Always Stop?
Wang Sheng
1955, Harold Kuhn
Hungarian method was published and was largely based on the earlier works of two Hungarian mathematicians
1957, James Munkres
Munkres observed it is polynomial in O(n4) and since then the algorithm was also known as Kuhn-Munkres algorithm
1960, Edmonds and Karp
The KM algorithm was modified to achieve an O(n3) running time
Evolvement of Hungarian Method
Idea: add/subtract X from all entries of a row/column
Goal: choose 0s from nonnegative matrix
Easy to understand
Easy to analysis and implement
Our Goal
Introduction to KM Algorithm
Matrix vs. Bipartite graph
For each entry Ci,j in matrix, there is an edge in bipartite graph from Xi to Yj with weight equal to Ci,j
In order to be consistent with theorems introduced in the algorithm we consider max-weight matching
Y1
Y2
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X3
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0
A vertex labeling is a function L : V → R
A feasible labeling is one such that L(x)+L(y) ≥ w(x, y), ∀x ∈ X, y ∈ Y
Equality Graphs
Equality Graph is G = (V,EL) where EL = {(x, y) : L(x)+L(y) = w(x, y)}
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Definitions
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If L is feasible and M is a perfect matching in EL then M is a max-weight matching
each node is covered exactly once and L(x)+L(y) ≥ w(x, y) therefore the upper-bound of the weight is the sum of labels
Power of the theorem
1
1
1
Key idea
find a good feasible labeling that remains enough edges in equality graph to ensure perfect matching can be done
Algorithm proposal
Start with any feasible labeling L and some matching M in EL
While M is not perfect, repeat: Find an augmenting path in EL to increase the size of M or if no path exists, improve L to L’ such that EL ⊂ EL’
Inspiration from KM Theorem
Simplest assignment
Maximize L(x) while Minimize L(y) ∀y ∈ Y, L(y) = 0 ∀x ∈ X, L(x) = max{w(x,y)}, y ∈ Y
It is obvious that ∀x ∈ X, y ∈ Y, w(x, y) ≤ L(x)+L(y)
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
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NL(u) = { v : (u,v) ∈ EL }
NL(S) = ∪u ∈ S NL(u)
Set αL = min { L(x) + L(y) – w(x,y) }, x ∈ S, y ∉ T
Update the labels (1) if v ∈ S L’(v) = L(v) - αL (2) if v ∈ T L’(v) = L(v) + αL (3) otherwise L’(v) = L(v)
Then L’ is a feasible labeling
Improving Labeling
αL = min { L(x) + L(y) – w(x,y) }, x ∈ S, y ∉ T
Consider a matrix C with –w(x,y) as its elements
For row x ∈ X / column y ∈ Y, add L(x)/ L(y) to each element
Problem is equivalent to solve the min-cost assignment in C
Edge with L(x) + L(y) = w(x,y) is the 0 element in matrix C
(1) if v ∈ S, L’(v) = L(v) - αL (2) if v ∈ T, L’(v) = L(v) + αL
S : set of uncovered rows T : set of covered columns
αL : the smallest entry not covered by any line
Subtract this entry from each uncovered row, and then add it to each covered column
Equivalence of Graph and Matrix
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Edges in EL’ (1) if v ∈ S L’(v) = L(v) - αL (2) if v ∈ T L’(v) = L(v) + αL (3) otherwise L’(v) = L(v)
If (x,y) ∈ EL for x ∈ S, y ∈ T then (x,y) ∈ EL’
If (x,y) ∈ EL for x ∉ S, y ∉ T then (x,y) ∈ EL’
There is some edge (x,y) ∈ EL’ for x ∈ S, y ∉ T
With good choice of S, we can guarantee there are more edges in new Equality Graph
Effectiveness of Label Update
Generate initial labeling L and matching M in EL
If M is perfect, terminate. Otherwise pick free vertex u ∈ X. Set S = {u}, T = {}.
If NL(S) = T, update labels(forcing NL(S) ≠ T) αL = min { L(x) + L(y) – w(x,y) }, x ∈ S, y ∉ T (1) if v ∈ S L’(v) = L(v) - αL (2) if v ∈ T L’(v) = L(v) + αL (3) otherwise L’(v) = L(v)
If NL(S) ≠ T, pick y ∈ NL(S) – T. If y is free, augmenting u-y and go to 2. If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. Go to 3.
Kuhn-Munkres Algorithm
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
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Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 0
Y3 0
X1 6
X2 8
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 2
Y3 0
X1 4
X2 6
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 2
Y3 0
X1 4
X2 6
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 2
Y3 0
X1 4
X2 6
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 2
Y3 0
X1 4
X2 6
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 2
Y3 0
X1 4
X2 6
X3 4
Pick a free vertex, set S={u} T={}; otherwise stop
If NL(S) = T, update labels (force NL(S) ≠ T)
If NL(S) ≠ T, pick y to be NL(S) – T
If y is free, augment u – y, go to step 2 If y is matched to z, S = S ∪ {z}, T = T ∪ {y}. go to step 3
Y1 0
Y2 2
Y3 0
X1 4
X2 6
X3 4
Example
94
In each phase of algorithm, |M| increases by 1, so there are at most V phases.
∀y ∉ T keep track of slacky = min{L(x)+L(y)-w(x,y)}
In each phase
Initializing all slacks. O(V)
When a vertex moves into S, all slacks need update. O(V) Only |V| vertices can be moved into S. O(V2)
When updating labels, αL = min(slacky). O(V) After getting αL, must update slacky = slacky -αL. O(V) αL can be calculated |V| times per phase. O(V2)
Total time per phase is O(V2)
Total running time is O(V3)
Complexity
95