Set problems
Vertex Cover, Dominating set,
Clique, Independent set
This lecture
• Several basic graph problems:
– Finding subsets of vertices or edges with simple
property – as small or as large as possible
– Relations
• Approximation Algorithms
• Special cases and a brief introduction to
some important graph classes
2
Set problems
1
Stating the problems
Vertex cover
• Set of vertices W  V with for all {x,y} 
E: x  W or y  W.
• Vertex Cover problem:
– Given G, find vertex cover of minimum size
4
Set problems
Dominating Set
• Set of vertices W such that for all v  V: v
 W or v has a neighbor w with w W.
• Dominating Set problem: find dominating
set of minimum size
• E.g.: facility location problems
5
Set problems
Clique and Independent Set
• Clique: Set of vertices W with for all v, w 
W: {v,w}  E.
• Independent set: Set of vertices W with for
all v, w  W: {v,w}  E.
• Clique problem: find largest clique.
• Independent set problem: find largest
independent set
6
Set problems
Relations
• The following are equivalent
– G has an independent set with at least k vertices
– The complement of G has a clique with at least
k vertices
– G has a vertex cover with at most n-k vertices
7
Set problems
NP-complete
• Each of the following problems is NP-complete:
– Decide if given graph G has vertex cover of size at
most a given integer K
– Decide if given graph G has dominating set of size at
most a given integer K
– Decide if given graph G has independent set of size at
least a given integer K
– Decide if given graph G has clique of size at least a
given integer K
• Proofs omitted in this lecture
8
Set problems
2
On approximation 1:
Clique and Independent Set
Approximation
• Clique, IS are very hard to approximate
• Ratio 2 approximation algorithm for Vertex Cover
– H = G, S = 
– repeat until H is empty
• Select an edge {v,w} . Put v and w in S, and remove v, w,
and all adjacent edges from H
• Proof of ratio: for each selected two vertices, at
least one belongs to optimal solution
10
Set problems
Warning
• The vertex cover approximation does NOT
give a ratio for an approximation for Clique
or Independent Set
– E.g., take a graph with 1000 vertices and a vertex cover
of size 498.
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Set problems
3
Approximating Set Cover and
Dominating Set
Approximating Dominating Set
on general graphs
• Can be done with ratio O( log n)
• Here: algorithm for generalization: set covering
problem
• Given: set X, collection C of subsets of X
• Find: smallest number of sets from C such that its
union is X.
– From DS to Set Covering: For each vertex v, take a set
Sv in the collection, with Sv = {v}  N(v).
13
Set problems
Greedy approximation of Set
Covering problem
•
•
•
•
U=X
A=
while U 
do
– Select an S from C with maximum | S  U|
– A = A  {S}
–U=U–S
• Output A
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Set problems
Ratio
• H(n) = Si=1…n 1/i
• Theorem. Greedy algorithm gives a number
of sets that is at most H( max{|S| | S  C})
times the optimal number of sets.
• Corollary: Greedy algorithm has ratio
O(log n).
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Set problems
Proof (1)
• Assign a cost 1 to each selected set.
• Spread this cost evenly over the newly
covered elements, cx.
• Say A* is optimal solution, A solution by
algorithm.
• |A| = Sx  X cx  SS  A* Sx  S cx
• We’ll proof: Sx  S cx  H(|S|)
• That proves the theorem.
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Set problems
Sx  S cx  H(|S|)
• Consider some S in C.
• Write ui = number of elements of S that still are in U
after i sets have been taken = |S – (S1  …  Si)|. u0
= |S|.
• Suppose uk = 0; uk-1 >0.
• ui-1 – ui > 0 elements from S were covered for the first
time by Si.
 Sx  S cx = Si=1…k (ui-1 – ui)* 1/|Si –(S1  …  Si-1)| 
Si=1…k (ui-1 – ui)/ ui-1
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Set problems
Proof continues
 Sx  S cx  Si=1…k (ui-1 – ui)/ ui-1
 Si=1…k (H(ui-1) – H(ui)) = H(u0) – H(uk) =
H(|S|) – 0.
• Use H(b) – H(a)  (b – a)/b
• Now we showed: Sx  S cx  H(|S|)
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Set problems
Comments
• (I)LP methods work also: log n ratio
• O(log n) is the best one can hope for unless
P=NP
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Set problems
4
Bipartite graphs
Vertex cover in bipartite graphs
• In a bipartite graph: the minimum size of a vertex
cover equals the maximum size of a matching
– If we have a matching: each edge must have an
endpoint in the vertex cover
– In the other direction:
• Let X be a vertex cover. Take flow model. Look at
the edges from s to vertices in X, and from vertices
in X to t. These form a cut. Now use the min-cut
max-flow theorem.
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Set problems
Bipartite graphs
• We saw: min vertex cover in bipartite graph
equals maximum matching
• Corollary: Independent Set in bipartite
graphs is polynomial time computable
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Set problems
Special cases for Independent Set
• For many types of graphs (e.g., planar) IS
stays NP-complete when restricted to these
graphs.
• Some special cases are polynomial
– Trees, bounded treewidth
– Bipartite graphs
– Interval graphs, circular arc graphs
23
Set problems
5
Interval graphs and generalizations
Interval graphs
• Each vertex represents an interval
• Edge between two vertices when their intervals
have non-empty intersection
• Independent set on interval graphs models roombooking problem
25
Set problems
Greedy algorithms
on interval graphs
• Greedy algorithm for IS on interval graphs is
optimal:
– Repeat:
• Select vertex v with leftmost right-endpoint of
interval
• Remove v and its neighbors
• Clique is trivial on interval graphs
26
Set problems
Circular arc graphs
• Like interval graphs, but now `intervals on a
circle’ (model of repeating time…)
• If we remove a vertex and its neighbors from an
circular arc graph, then we get an interval graph
• Algorithm, e.g.:
– for all vertices v do
• Compute the maximum independent set of G minus v and its
neighbors
– If the largest i.s. is found for v*, output that set with v*
added.
27
Set problems
Perfect graphs
• A graph G is perfect if for each induced subgraph
H of G: the size of the maximum clique in H
equals the chromatic number of H (nb. of colors
needed to `vertex color H’).
• Strong perfect graph theorem (conjectured Berge
1966, proven Chudnowsky et al., 2002).
– G is perfect, if and only if G does not contain an
induced subgraph that is a cycle without additional
edges of odd length at least four, and the complement of
G also does not contain such a cycle. (`No odd hole or
antihole’)
28
Set problems
Coloring and independent set in
perfect graphs
• Polynomial time algorithm for finding the
maximum independent set and the coloring
number of a perfect graph
– Grötschel, Lovasz, Schrijver (1981)
– Uses Linear Programming methods
29
Set problems
6
Approximating facility location
Facility location
• In a town, we want to place a number of
facilities (e.g., fire stations, hospitals, …),
such that the maximum distance to a facility
is small
• Models as a graph problem
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Set problems
Facility location problem
• Given: complete graph G, distances d
between vertices fulfilling triangle
inequality, integer k
• Question: find a set S of k vertices, such
that the maximum over all vertices v of
minw S d(v,w)
is as small as possible
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Set problems
Restating the problem
• Find the smallest r and a set S such that S is
a dominating set of size at most k for the
graph Gr formed by the edges of length at
most r.
• OPT is smallest r.
• G2 is square of a graph: edge in G2 when
there is a path of length 2
33
Set problems
A Lemma
• Let I be an independent set in H2. Then |I| is
at most the minimum size of a dominating
set in H.
• Proof: Let D be a minimum dominating set
of H. For each v  D, N(v) {v} contains at
most one vertex from I.
34
Set problems
Approximation algorithm for
facility location
• Sort the values of the lengths of the edges in
G, say d1  …  dm
• for i = 1 to m do
–
–
–
–
35
Compute Gx with x = di
Compute Gx2
Find a maximal independent set I in Gx2
If it has size at most k, then return I; stop.
Set problems
Analysis
• The optimal solution has cost at least x=di
– For y<x: the minimum size of a dominating set
in Gy is k+1 or more (use lemma)
• Ratio of algorithm is 2
– Each maximal independent set is a domination
set, so I is a dominating set in Gx2.
– Each vertex has a path of length at most 2 to a
vertex in I in Gx: so, I is a k-center with `cost’ at
most 2x.
36
Set problems
Graphs that are not complete
• Given: graph G, distance for each edge d(e),
integer k.
• Question: find a set S of k vertices, such that the
maximum over all vertices of the length of the
shortest path to a vertex in S is as small as
possible.
 Can be translated to previous problem: make
complete graph with d(v,w) the length of the
shortest path from v to w
37
Set problems
7
Conclusions
Conclusions
• Several techniques for approximation
problems like vertex cover, dominating set,
independent set
• Work often for other problems too (with
modifications)
39
Set problems
Omitted parts
• After this some omitted parts in 2010
– NP-completeness
– Unit disk graphs
40
Set problems
Unit disk graphs
• Vertices are disks on
plane with radius 1
• Edge when disks
intersect
• Model radio
transmitters and
inferences or
connections
44
Set problems
More on unit disk graphs
• Finding the intersection model is hard (NPhard; not yet known to be in NP).
• Approximation algorithms for Independent
Set, Dominating Set, …
– Greedy algorithm is a 5-approximation
– Simple modification gives 3 approximation
(next)
45
Set problems
3-approximation for Independent
set on Unit Disk graphs
• W=V
• S=
• repeat until W = 
Leftmost Greedy
Algorithm
– Choose a vertex v from W whose center is as
much as possible to the left
– S = S  {v}
– W = W – N(v) – {v}
• Return S
46
Set problems
Ratio 3
• The maximum
independent set is
never larger than 3
times the set, given by
the leftmost greedy
algorithm.
– If we choose v, then
N(v) has an
independent set of size
at most 3
47
Between successive
nonintersecting
neighbors, there is an
angle that is larger than
p/3
Set problems
A PTAS
• For every e>0, there is a polynomial time
algorithm for Independent set that gives an
independent set of size at least (1-e)*OPT.
– Uses geometric separation technique
PTAS =
Polynomial Time
Approximation
Scheme
48
Set problems
Bounded regions
• Suppose all disks are in a region of size r by r.
Then the maximum size of an independent set in
this graph is at most 4r2/p2.
– Each disk takes space p2/4.
– The disks in the solution do not overlap.
• Then, the maximum independent set problem can
be solved in O(nO(r*r)) time.
– Check each subset of size at most 4r2/p2. Take the
largest independent one of these.
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Set problems
Removing strips and getting
small area subgraphs
• Remove, at steps of r+1, for a 1-wide
horizontal strip all disks intersecting the
strip.
• Remove, at steps of r+1, for a 1-wide
vertical strip all disks intersecting the strip.
• On the remaining graph, each connected
component fits in an r by r area
50
Set problems
All disks intersecting
a green strip are
removed. We end up
with the graph,
formed by the disks
fitting in the white
area
Each connected component
fits in one white area
51
For fixed r, we can
solve the IS problem
on the remaining
graph in polynomial
time
Set problems
52
Set problems
Graphs Gi,j
• Remove horizontal strips at spaces i,
i+(r+1), i + 2*(r+1), … (I.e., start at i.)
• Remove vertical strips at spaces j, j+(r+1), j
+ 2*(r+1), … (I.e., start at j.)
• Let Gi,j = (Vi,j,Ei,j) be the resulting graph
• Algorithm: try all (r+1)2 choices for (i,j),
for each solve IS optimally for Gi,j and
output the largest IS from these.
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Set problems
Quality of heuristic (1)
• Let X be an independent set of maximum
size.
• Let v  X. There are at most 4(r+1) graphs
Gi,j that do not contain v.
– v can be removed by a horizontal strip (upside,
downside intersects) or a vertical strip (left side,
right side intersects) .
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Set problems
Quality of heuristic (2)
• Total size over all m.i.s.’s over all Gi,j is at
least the sum over all |Vi,j  X|  (r+1)2 *|X|
- 4(r+1)*|X|
• There must be at least one Gi,j that contains
an independent set of size at least ((r+1)2 –
4)/(r+1)2 * |X|
– Maximum is at least average
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Set problems
Setting r
• Let r be the smallest integer such that
(r  1) 2  4(r  1) r  3

 1 e
2
(r  1)
r 1
• For fixed e, this is a constant – gives a
polynomial time algorithm: PTAS
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Set problems
Variants
•
•
•
•
Other type of intersecting objects
Other type of surfaces
Other problems
Example: Dominating set.
– Instead of removing strips, we duplicate
vertices in strips giving them to the left, and
right area
– Leads also to PTAS
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Set problems