1. Vertex Cover, Dominating set, Clique, Independent set
Set problems
Vertex Cover, Dominating set,
Clique, Independent set

2. Basic problems Several basic graph problems: Relations
Finding subsets of vertices or edges with simple property – as small or as large as possible
Coloring (next time)
Relations
Approximation Algorithms
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3. Vertex cover
Set of vertices W Í V with for all {x,y} Î E: x Î W or y Î W.
Vertex Cover problem:
Given G, find vertex cover of minimum size
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4. Dominating Set
Set of vertices W such that for all v Î V: v Î W or v has a neighbor w with w ÎW.
Dominating Set problem: find dominating set of minimum size
E.g.: facility location problems
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5. Clique and Independent Set
Clique: Set of vertices W with for all v, w Î W: {v,w} Î E.
Independent set: Set of vertices W with for all v, w Î W: {v,w} Ï E.
Clique problem: find largest clique.
Independent set problem: find largest independent set
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6. 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
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7. Approximation Clique, IS are very hard to approximate
Ratio 2 approximation algorithm for Vertex Cover
H = G, S = Æ
repeat until H is empty
Select an edge {v,w} Î E. Put v and w in S, and remove v, w, and all adjacent edges from H
Does not translate to approx. algorithm for Clique or Independent Set
E.g., take a graph with 1000 vertices and a vertex cover of size 498.
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8. 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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9. 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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10. 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
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11. Interval graphs Each vertex represents an interval
Edge between two vertices when their intervals have non-empty intersection
Independent set on interval graphs models room-booking problem
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12. 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
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13. Circular arc graphs
Like interval graphs, but now `intervals on a circle’.
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.
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14. 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’)
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15. 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
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16. Unit disk graphs Vertices are disks on plane with radius 1
Edge when disks intersect
Model radio transmitters and inferences or connections
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17. More on unit disk graphs
Finding the intersection model is hard (NP-hard; 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)
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18. 3-approximation for Independent set on Unit Disk graphs
W = V
S = Æ
repeat until W = Æ
Choose a vertex v from W whose center is as much as possible to the left
S = S È {v}
W = W – N(v) – {v}
Return S
Leftmost Greedy
Algorithm
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19. 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
Between successive
nonintersecting neighbors, there is an
angle that is larger than
p/3
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20. Polynomial Time Approximation
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
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21. 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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22. 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
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23. Each connected component fits in one white area
All disks intersecting a green strip are removed. We end up with the graph, formed by the disks fitting in the white area
For fixed r, we can solve the IS problem on the remaining graph in polynomial time
Each connected component
fits in one white area
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24. Set problems

25. 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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26. Quality of heuristic (1)
Let X be an independent set of maximum size.
Let v Î 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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27. 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 Ç X| ³ (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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28. Setting r Let r be the smallest integer such that
For fixed e, this is a constant – gives a polynomial time algorithm: PTAS
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29. 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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30. 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} È N(v).
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31. Greedy approximation of Set Covering problem
U = X
A = Æ
while U ¹ Æ do
Select an S from C with maximum | S Ç U|
A = A È {S}
U = U – S
Output A
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32. Ratio
H(n) = Si=1…n 1/i
Theorem. Greedy algorithm gives a number of sets that is at most H( max{|S| | S Î C}) times the optimal number of sets.
Corollary: Greedy algorithm has ratio O(log n).
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33. 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 Î X cx £ SS Î A* Sx Î S cx
We’ll proof: Sx Î S cx £ H(|S|)
That proves the theorem.
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34. Sx Î 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 È … È 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 Î S cx = Si=1…k (ui-1 – ui)* 1/(|Si –((S1 È … È Si-1)| £ Si=1…k (ui-1 – ui)/ ui-1
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35. Proof continues
Sx Î 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) ³ (b – a)/b
Now we showed: Sx Î S cx £ H(|S|)
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36. 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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37. 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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38. 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 ÎS d(v,w) is as small as possible
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39. 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
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40. 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 Î D, N(v)È {v} contains at most one vertex from I.
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41. 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
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.
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42. 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.
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43. 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
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44. Conclusions
Several techniques for approximation problems like vertex cover, dominating set, independent set
Work often for other problems too (with modifications)
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