1 A cover of an undirected graph is a set of nodes such that every edge of G has at least one end in For any matching and any cover each edge in has an end in and the corresponding nodes in C are all distinct ID: 778746
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Slide1
Combinatorial Optimization 2014
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A cover of an undirected graph is a set of nodes such that every edge of G has at least one end in . For any matching and any cover , each edge in has an end in and the corresponding nodes in C are all distinct. (primal-dual relation)Thm 3.14 (König’s Theorem, 1931) : For a bipartite graph G, max { : a matching} = min { : a cover}
3.3 Applications of Maximum Flow and Minimum Cut
Bipartite
Matchings
and Covers
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(Transform to max flow problem)
P
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(Maximum matching and corresponding flow)
node cover
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(Identifying
-augmenting path)
node reachable from
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(blue nodes defines cut = flow value)node reachable from
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cut arcs
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Note that no cut arc from blue in
→ green in
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Cover
(pink)
Capacity of cut
Let
{blue nodes
}
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min cut: where no edge from to Cover Capacity of cut Running time: rs
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Reference: Linear Programming, Vasek Chvatal, 1983, Freeman, p336-338.System of distinct representatives Given a set and a family of subsets of , a system of distinct representative (SDR) is a set of distinct elements of such that for .Hall’s Theorem: The family
has an SDR if and only if for every subset
of we have
.
(The family
does not have an SDR if and only if there is a set
such that
.
)
Pf) see reference.
(Hall’s Theorem)
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The family has
an
SDR
iff
the corresponding bipartite graph has a matching of size
.
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S1S2
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Circled nodes have
, hence no SDR.
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Chains and Antichains in Partially Ordered Sets (Reference: Linear Programming, Vasek Chvatal, 1983, Freeman, p338-341.) A list of pairs satisfying the following conditions is called a partial order (i) no (ii) and together imply A set whose elements appear in a partial order is called a partially ordered set A subset of a partially ordered set is called a chain if
or
whenever and (The elements can be enumerated as
in such a way that
,
, … ,
)
A subset
of a partially ordered set is called an
antichain
if
for no
.
Dilworth’s Theorem
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Dilworth’s Theorem: In every partially ordered set , the smallest number of chains whose union is equals the largest size of an antichain application: guided tour, airplane assignment pf) text exercise 3.37. We consider a different approach here. ex) Set , tourspartial order :
(
need 3 guides,
)
Construct
a bipartite graph by doubling each node
as
and
.
There
exists an edge
iff
.
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a’b’c’
d’
e’
g’’
f’
f’’
g’
e’’
d’’
c’’
b’’
a’’
A matching
defines
chains (guides)
ex) (
i
)
(ii)
(iii)
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
be the number of chains and
be the length (number of nodes) of
-th
chain.
Then
and
which imply
. (Hence min # of guides
,
: max matching)
Converse can be shown too, i.e.
chains define
sized matching
. (
Hence min # of guides
,
: max matching)
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a’b’c’
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Different reasoning using
K
önig’s
Theorem
:
Let
be a largest matching and
be min cover. Define a set
of trips as follows;
trip
iff
and
.
No
with
is possible. (Otherwise,
with
,
, hence
not covered by
, contradiction to
being a
cover.)
no guide can take care of two different trips in
. (e.g.,
)
need at least
guides.
But
has at least
elements (each node in
makes
only
one
trip
ineligible for membership in
)
Hence every schedule must use at least
different guides. The schedule defined by
, using only guides, is optimal. (from earlier)
Cover C
*
A*
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Proof of Dilworth’s Theorem: The partially ordered set 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 of chains. Along with the chains, the procedure also finds an antichain of size at least . Since every antichain shares at most one element with each of the chains , it must satisfy . So is a largest antichain and its size is (plug in for ).
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Situation: If we want to choose project , then we must choose project also. We want to choose the set of projects which gives maximum profitModel as a problem on a digraph. Use directed arc if project precedes project . Find the maximum benefit set such that . (closure of )Example: open pit mining problem. Partition the volume to be considered for excavation into small 3-dimensional blocks. Block produces profit of (profit of ore in – cost of excavating ) If we need to excavate block to excavate block , we include arc in . Want to find a closed set in which maximizes profit.
Optimal Closure in a Digraph
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Convert to min-cut problem (max-flow problem) Divide the nodes into two sets and . if and if . Add a source and a sink to the graph and add arc for each and arc for each . Capacity on the arcs are: for each and for each . The upper bounds on the original arcs are infinite. A set of nodes is closed if and only if the cut has finite capacity.If the capacity of is finite, then it equals
Since
is a constant, m
inimizing the capacity of
amounts
to maximizing
.
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(Example)4
-1
2
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C
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Ref: Graph Theory with Applications, J. A. Bondy, U. S. R. Murty, 1976, 2008.Directed version:Lemma: Let be a digraph in which each arc has unit capacity. Then (a) The maximum flow is equal to the maximum number of arc disjoint directed -paths in ; and (b) The capacity of a minimum -cut is equal to the minimum number of arcs whose deletion destroys all directed -paths in G.Thm: Let , be two nodes of a digraph
. Then the maximum number of arc-disjoint directed
-paths in G is equal to the minimum number of arcs whose deletion destroys all directed -paths in G. (pf)
From above Lemma and max-flow and min-cut theorem.
Menger’s
Theorems
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Thm: Let , be two nodes of a graph . Then the maximum number of edge-disjoint -paths in is equal to the minimum number of edges whose deletion destroys all -paths in G. (pf) Replace each edge of G by two oppositely oriented arcs and apply the previous results.
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(node connectivity)Directed version:Thm: Let , be two nodes of a digraph , such that is not joined to . Then the maximum number of node-disjoint directed -paths in is equal to the minimum number of nodes whose deletion destroys all directed -paths in . (pf) Construct as follows ( except and )
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Thm: Let and be two nonadjacent nodes of a graph . Then the maximum number of node-disjoint -paths in is equal to the minimum number of nodes whose deletion destroys all -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.