1. 3.3 Applications of Maximum Flow and Minimum Cut
Bipartite Matchings and Covers
A cover of an undirected graph 𝐺 is a set 𝐶 of nodes such that every edge of G has at least one end in 𝐶.
For any matching 𝑀 and any cover 𝐶, each edge in 𝑀 has an end in 𝐶 and the corresponding nodes in C are all distinct.
⇒ |𝑀|≤|𝐶| (primal-dual relation)
Thm 3.14 (König’s Theorem, 1931) : (König-Egerváry Thm)
For a bipartite graph G,
max { |𝑀|: 𝑀 a matching} = min { |𝐶|: 𝐶 a cover}
Combinatorial Optimization 2016

2. (Transform to max flow problem)Q P Q a A a A b B b B c C c C r s d D d D f F f F h H h H
urp=1
uqs=1 i I i I
upq=
Combinatorial Optimization 2016

3. (Maximum matching and corresponding flow)b B b B c C c C r s d D d D f F f F h H h H
urp=1
uqs=1 i I i I
upq=
node cover
Combinatorial Optimization 2016

4. (Identifying 𝑥-incrementing paths)b B b B c C c C r r s s d D d D f F f F h H h H
urp=1
uqs=1
urp=1
uqs=1 i I i I
upq=
upq=
node reachable from 𝑟
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5. (blue nodes defines cut = flow value)r r s s d D d D f F f F h H h
urp=1
uqs=1 H
urp=1
uqs=1 i I i I
upq=
upq=
: cut arcs
node reachable from 𝑟
Note that no cut arc from blue in 𝑃 → green in 𝑄 (O.w. cut capacity is )
Combinatorial Optimization 2016

6. Let 𝐴 = {blue nodes∖{𝑟}} Cover 𝐶 (pink) =(𝑃\A)∪(𝑄∩𝐴) f F f F h H h
urp=1
uqs=1 H
urp=1
uqs=1 i I i I
upq=
upq=
Let 𝐴 = {blue nodes∖{𝑟}}
Cover 𝐶 (pink) =(𝑃\A)∪(𝑄∩𝐴)
Capacity of cut = |𝑃\A|+|𝑄∩𝐴|=|𝐶|
Combinatorial Optimization 2016

7. min cut: 𝛿′ 𝑟 ∪𝐴 where 𝐴⊆𝑉 no edge from 𝑃∩𝐴 to 𝑄\A Cover 𝐶=(𝑃\A)∪(𝑄∩𝐴) s
𝑃∩𝐴
𝑄∩𝐴
min cut: 𝛿′ 𝑟 ∪𝐴 where 𝐴⊆𝑉
no edge from 𝑃∩𝐴 to 𝑄\A
Cover 𝐶=(𝑃\A)∪(𝑄∩𝐴)
Capacity of cut =|𝑃\A|+|𝑄∩𝐴|=|𝐶|
Running time: 𝑂(𝑚𝑛)
Combinatorial Optimization 2016

8. Reference: Linear Programming, Vasek Chvatal, 1983, Freeman, p336-338.
(Hall’s Theorem)
Reference: Linear Programming, Vasek Chvatal, 1983, Freeman, p
System of distinct representatives
Given a set 𝑄 and a family (𝑆1, 𝑆2, …, 𝑆 𝑚 ) of subsets of 𝑄, a system of distinct representative (SDR) is a set {𝑞1, …, 𝑞 𝑚 } of distinct elements of 𝑄 such that 𝑞𝑖 ∈𝑆𝑖 for 1≤𝑖≤𝑚.
Hall’s Theorem:
The family (𝑆1, 𝑆2, …, 𝑆 𝑚 ) does not have an SDR if and only if there is a set 𝐼 such that 𝑖∈𝐼 𝑆 𝑖 < 𝐼 (20.14)
(The family (𝑆1, 𝑆2, …, 𝑆 𝑚 ) has an SDR if and only if for every subset 𝐼 of {1, …, 𝑚} we have 𝑖∈𝐼 𝑆 𝑖 ≥ 𝐼 .)
Pf) later.
Combinatorial Optimization 2016

9. q1 q1 S1 S1 q2 q2 S2 S2 q3 q3 S3 S3 q4 q4 S4 S4 q5 q5 S5 S5 q6 q6
The family has an SDR iff the corresponding bipartite graph has a matching of size 𝑚.
Combinatorial Optimization 2016

10. Circled nodes have 𝑖∈𝐼 𝑆 𝑖 < 𝐼 , hence no SDR.q1 q1 S1 S1 q2 q2 S2 S2 q3 q3 S3 S3 q4 q4 S4 S4 q5 q5 S5 S5 q6 q6
Circled nodes have 𝑖∈𝐼 𝑆 𝑖 < 𝐼 , hence no SDR.
Combinatorial Optimization 2016

11. Hall’s Theorem:
The family (𝑆1, 𝑆2, …, 𝑆 𝑚 ) does not have an SDR if and only if there is a set 𝐼 such that 𝑖∈𝐼 𝑆 𝑖 < 𝐼 (20.14)
(The family (𝑆1, 𝑆2, …, 𝑆 𝑚 ) has an SDR if and only if for every subset 𝐼 of {1, …, 𝑚} we have 𝑖∈𝐼 𝑆 𝑖 ≥ 𝐼 .)
Pf) ) clear.
) Suppose (𝑆1, 𝑆2, …, 𝑆 𝑚 ) does not have an SDR. Corresponding bipartite graph has no matching of size 𝑚. Let 𝑀 ∗ be the largest matching ( 𝑀 ∗ <𝑚) and 𝐶 ∗ be the smallest cover. By König’s thm 𝑀 ∗ = 𝐶 ∗ <𝑚.
Let 𝐼⊆{1,2,…,𝑚} be such that 𝑖∈𝐼 iff the left node 𝑆 𝑖 does not belong to 𝐶 ∗ . If 𝑖∈𝐼, then each element 𝑞∈ 𝑆 𝑖 , viewed as a right node must be in 𝐶 ∗ . Otherwise 𝐶 ∗ would not be a cover. Hence 𝑖∈𝐼 𝑆 𝑖 is a subset of right nodes in 𝐶 ∗ . Since 𝐶 ∗ has precisely 𝑚− 𝐼 left nodes, it has 𝐶 ∗ − 𝑚− 𝐼 < 𝐼 right nodes. So (20.14) holds. 
Combinatorial Optimization 2016

12. Dilworth’s Theorem Chains and Antichains in Partially Ordered Sets
(Reference: Linear Programming, Vasek Chvatal, 1983, Freeman, p )
A list of pairs 𝑥 → 𝑦 satisfying the following conditions is called a partial order
(i) no 𝑥 → 𝑥
(ii) 𝑥 → 𝑦 and 𝑦 → 𝑧 together imply 𝑥 → 𝑧
A set whose elements appear in a partial order is called a partially ordered set
A subset 𝐶 of a partially ordered set is called a chain if
𝑥 → 𝑦 or 𝑦 → 𝑥 whenever 𝑥, 𝑦∈𝐶 and 𝑥≠𝑦
(The elements can be enumerated as 𝑧1, 𝑧2, …, 𝑧𝑟 in such a way that 𝑧1 → 𝑧2, 𝑧2 → 𝑧3, … , 𝑧 𝑟−1 → 𝑧𝑟 )
A subset 𝐴 of a partially ordered set is called an antichain if
𝑥 → 𝑦 for no 𝑥, 𝑦∈𝐴.
Combinatorial Optimization 2016

13. application: guided tour, airplane assignment
Dilworth’s Theorem:
In every partially ordered set 𝑃, the smallest number of chains whose union is 𝑃 equals the largest size of an antichain
application: guided tour, airplane assignment
pf) text exercise We consider a different approach here.
ex) Set 𝑋={𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔}, tours
partial orders : 𝑎→𝑐, 𝑎→𝑑, 𝑎→𝑓, 𝑎→𝑔, 𝑏→𝑐, 𝑏→𝑔, 𝑑→𝑔, 𝑒→𝑓, 𝑒→𝑔
( need 3 guides, 𝑎→𝑑→𝑔, 𝑏→𝑐, 𝑒→𝑓 )
(Note that 𝑐,𝑔,𝑓 has no partial orders (antichain) )
Construct a bipartite graph by doubling each node 𝑣 as 𝑣′ and 𝑣′′.
There exists an edge 𝑥′𝑦′′ iff 𝑥→𝑦.
Combinatorial Optimization 2016

14. A matching 𝑀 defines 𝑘=𝑛−|𝑀| chains (guides) ex) (i) 𝑎→𝑑→𝑔 (ii) 𝑏→𝑐
- In the matching, each tour has at most one successor and at most one predecessor. Otherwise, M would not be a matching.
By concatenating successive tours, we obtain a set of chains (A chain may consists of a single tour)
- Let 𝑘 be the number of chains and 𝑟𝑖 be the length (number of nodes) of 𝑖-th chain.
Then 𝑖=1 𝑘 𝑟 𝑖 =𝑛 and 𝑖=1 𝑘 ( 𝑟 𝑖 −1) =|𝑀| which imply 𝑘=𝑛−|𝑀|. (Hence min # of guides ≤𝑛− 𝑀 ∗ , 𝑀 ∗ : max matching)
- Converse can be shown too, i.e. 𝑘 chains define 𝑛 – 𝑘 sized matching 𝑀. (Hence 𝑛−𝑘≤ 𝑀 ∗ , i.e., min # of guides ≥𝑛− 𝑀 ∗ , 𝑀 ∗ : max matching)
a’’
b’’ a’
c’’ b’
d’’ c’
e’’ d’
f’’
g’’ e’ f’ g’
Combinatorial Optimization 2016

15. Different reasoning using König’s Theorem:
Let 𝑀 ∗ be a largest matching and 𝐶 ∗ be min cover. Define a set 𝐴 ∗ of trips as follows;
trip 𝑧∈ 𝐴 ∗ iff 𝑧′∉ 𝐶 ∗ and 𝑧′′∉ 𝐶 ∗ .
No 𝑥→𝑦 with 𝑥, 𝑦∈ 𝐴 ∗ is possible. (Otherwise, ∃ 𝑥′𝑦′′ with 𝑥′∉ 𝐶 ∗ , 𝑦′′∉ 𝐶 ∗ , hence 𝑥′𝑦′′ not covered by 𝐶 ∗ , contradiction to 𝐶 ∗ being a cover.)
⇒ no guide can take care of two different trips in 𝐴 ∗ . (e.g., 𝑐→𝑓)
⇒ need at least | 𝐴 ∗ | guides.
But 𝐴 ∗ has at least 𝑛−| 𝐶 ∗ | elements (each node in 𝐶 ∗ makes only one trip ineligible for membership in 𝐴 ∗ )
Hence every schedule must use at least
| 𝐴 ∗ |≥𝑛−| 𝐶 ∗ |=𝑛−| 𝑀 ∗ | different guides.
⇒ The schedule defined by | 𝑀 ∗ |, using only 𝑛−| 𝑀 ∗ | guides, is optimal. (from earlier)
Cover C*
a’’ A*
b’’ a’
c’’ b’
d’’ c’
e’’ d’
f’’
g’’ e’ f’ g’
Combinatorial Optimization 2016

16. Proof of Dilworth’s Theorem:
The partially ordered set 𝑃 may be thought of as a set of tours. Chains are those sets of tours that can be taken care of by one guide. The described procedure gives the smallest 𝑘 of chains. Along with the chains, the procedure also finds an antichain 𝐴 ∗ of size at least 𝑘.
Since every antichain 𝐴 shares at most one element with each of the chains 𝐶1, 𝐶2, …, 𝐶𝑘, it must satisfy |𝐴|≤𝑘≤| 𝐴 ∗ |. So 𝐴 ∗ is a largest antichain and its size is 𝑘 (plug in 𝐴 ∗ for 𝐴). 
Combinatorial Optimization 2016

17. Optimal Closure in a Digraph
Situation: If we want to choose project 𝑣, then we must choose project 𝑤 also. We want to choose the set of projects which gives maximum profit
Model as a problem on a digraph. Use directed arc 𝑣𝑤 if project 𝑤 precedes project 𝑣. Find the maximum benefit set 𝐶⊆𝑉 such that (𝐶)=∅. (closure of 𝐺)
(May be formulated as integer program using constraints 𝑥 𝑣 ≤ 𝑥 𝑤 , ∀ 𝑣𝑤∈𝐸, 𝑥∈ 𝐵 𝑛 .)
Example: open pit mining problem.
Partition the volume to be considered for excavation into small 3-dimensional blocks.
Block 𝑣 produces profit of 𝑏𝑣 (profit of ore in 𝑣 – cost of excavating 𝑣 )
If we need to excavate block 𝑤 to excavate block 𝑣, we include arc 𝑣𝑤 in 𝐺.
Want to find a closed set in 𝐺 which maximizes profit.
Combinatorial Optimization 2016

18. Convert to min-cut problem (max-flow problem)
Divide the nodes into two sets 𝐴 and 𝐵.
𝑣∈𝐴 if 𝑏𝑣≥0 and 𝑣∈𝐵 if 𝑏𝑣<0.
Construct 𝐺′. Add a source 𝑟 and a sink 𝑠 to the graph and add arc 𝑟𝑣 for each 𝑣∈𝐴 and arc 𝑣𝑠 for each 𝑣∈𝐵.
Capacity on the arcs are: 𝑢𝑟𝑣=𝑏𝑣 for each 𝑣∈𝐴 and 𝑢𝑣𝑠=− 𝑏 𝑣 for each 𝑣∈𝐵.
The upper bounds on the original arcs are infinite.
A set 𝐶 of nodes is closed if and only if the cut δ′(𝐶∪ 𝑟 ) has finite capacity.
If the capacity of δ′(𝐶∪ 𝑟 ) is finite, then it equals
𝑣∈𝐴−𝐶 𝑢 𝑟𝑣 + 𝑣∈𝐵∩𝐶 𝑢 𝑣𝑠 = 𝑣∈𝐴−𝐶 𝑏 𝑣 + 𝑣∈𝐵∩𝐶 (−𝑏 𝑣 ) =
𝑣∈𝐴−𝐶 𝑏 𝑣 + 𝑣∈𝐴∩𝐶 𝑏 𝑣 + 𝑣∈𝐴∩𝐶 (− 𝑏 𝑣 ) + 𝑣∈𝐵∩𝐶 (−𝑏 𝑣 ) = 𝑣∈𝐴 𝑏 𝑣 − 𝑣∈𝐶 𝑏 𝑣
Since 𝑣∈𝐴 𝑏 𝑣 is a constant, minimizing the capacity of δ′(𝐶∪ 𝑟 ) amounts to maximizing 𝑣∈𝐶 𝑏 𝑣 .
Combinatorial Optimization 2016

19. 𝑣∈𝐴−𝐶 𝑢 𝑟𝑣 + 𝑣∈𝐵∩𝐶 𝑢 𝑣𝑠 = 𝑣∈𝐴−𝐶 𝑏 𝑣 + 𝑣∈𝐵∩𝐶 (−𝑏 𝑣 ) =
(Example)
𝑣∈𝐴−𝐶 𝑢 𝑟𝑣 + 𝑣∈𝐵∩𝐶 𝑢 𝑣𝑠 = 𝑣∈𝐴−𝐶 𝑏 𝑣 + 𝑣∈𝐵∩𝐶 (−𝑏 𝑣 ) =
𝑣∈𝐴−𝐶 𝑏 𝑣 + 𝑣∈𝐴∩𝐶 𝑏 𝑣 + 𝑣∈𝐴∩𝐶 (− 𝑏 𝑣 ) + 𝑣∈𝐵∩𝐶 (−𝑏 𝑣 ) = 𝑣∈𝐴 𝑏 𝑣 − 𝑣∈𝐶 𝑏 𝑣 2 -7  2  C 7 𝑟 4 4 1 -1  3   3 2 3 𝑠 -3   1 2  -1
Combinatorial Optimization 2016

20. Menger’s Theorems
Ref: Graph Theory with Applications, J. A. Bondy, U. S. R. Murty, 1976, 2008.
Directed version:
Lemma: Let 𝐺 be a digraph in which each arc has unit capacity. Then
(a) The maximum (𝑟, 𝑠) flow is equal to the maximum number 𝑚 of arc disjoint directed (𝑟, 𝑠)-paths in 𝐺; and
(b) The capacity of a minimum (𝑟, 𝑠)-cut is equal to the minimum number 𝑛 of arcs whose deletion destroys all directed (𝑟, 𝑠)-paths in G.
Thm: Let 𝑟, 𝑠 be two nodes of a digraph 𝐺. Then the maximum number of arc-disjoint directed (𝑟, 𝑠)-paths in G is equal to the minimum number of arcs whose deletion destroys all directed (𝑟, 𝑠)-paths in G.
(pf) From above Lemma and max-flow and min-cut theorem.
Combinatorial Optimization 2016

21. Thm: Let 𝑟, 𝑠 be two nodes of a graph 𝐺
Thm: Let 𝑟, 𝑠 be two nodes of a graph 𝐺. Then the maximum number of edge-disjoint (𝑟, 𝑠)-paths in 𝐺 is equal to the minimum number of edges whose deletion destroys all (𝑟, 𝑠)-paths in G.
(pf) Replace each edge of G by two oppositely oriented arcs and apply the previous results. 
Combinatorial Optimization 2016

22. (node connectivity) Directed version:
Thm: Let 𝑟, 𝑠 be two nodes of a digraph 𝐺, such that 𝑟 is not joined to 𝑠. Then the maximum number of node-disjoint directed (𝑟, 𝑠)-paths in 𝐺 is equal to the minimum number of nodes whose deletion destroys all directed (𝑟, 𝑠)-paths in 𝐺.
(pf) Construct 𝐺′ as follows ( except 𝑟 and 𝑠). Then directed (𝑟,𝑠)-paths in 𝐺′ are arc-disjoint if and only if the corresponding paths in 𝐺 are node-disjoint.   𝑣 𝑣′
𝑣′′ 1   
Combinatorial Optimization 2016

23. Thm: Let 𝑟 and 𝑠 be two nonadjacent nodes of a graph 𝐺
Thm: Let 𝑟 and 𝑠 be two nonadjacent nodes of a graph 𝐺. Then the maximum number of node-disjoint (𝑟, 𝑠)-paths in 𝐺 is equal to the minimum number of nodes whose deletion destroys all (𝑟, 𝑠)-paths.
(pf) Replace each edge of G by two oppositely oriented arcs and use the previous transformation (reduction). 
Recall from connectivity results earlier. Thm 9.2: If 𝐺 has at least one pair of nonadjacent vertices, 𝜅(𝐺) = min {𝑝(𝑢,𝑣): 𝑢,𝑣∈𝑉, 𝑢≠𝑣, 𝑢𝑣∉𝐸}.
Min cut can be obtained in polynomial time. Hence the (edge, node) connectivity of a graph can be found in polynomial time.
Combinatorial Optimization 2016