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Enumeration
Branch an Bound
Greedy
Approximation
◦ PTAS
◦ K-Approximation
◦ No Approximation
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A problem L has a polynomial-time approximation
scheme (PTAS) if it has a polynomial-time
(1+ε)-approximation algorithm, for any fixed ε >0 (this value
can appear in the running time).
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For example, there is a PTAS for finding the maximum
independent set problem on planar graphs.
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An independent set is a set of vertices in a graph, no
two of which are adjacent.
An maximal independent set is an independent set
that is not a subset of any other independent set.
maximum
independent sets
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The input is an undirected graph, and the output is a
maximum independent set in the graph.
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It is a NP-hard problem and it is also hard to
approximate, and the decision problem is NPComplete.
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Fortunately, there is a PTAS for finding the maximum
independent set problem on planar graphs.
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A planar graph is a graph that can be embedded in the
plane, i.e., it can be drawn on the plane in such a way
that its edges intersect only at their endpoints.
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(a) A Planar Graph.
(b) A Graph Which Is Not Planar.
Figure 9-41 Planar Graphs.
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The unbounded faces are called exterior faces and all
other faces are called interior faces.
exterior face
interior face
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We can use faces to mark the level of each node.
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A graph is k-outerplanar if it has no nodes with level greater
than k.
Figure 9-43 An Example of 2-Outerplanar Graph.
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Given an arbitrary planar graph G, we can
decompose it into a set of k-outerplanar graphs.
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For a k-outerplanar graph, an optimal solution for the
maximum independent set problem can be found in
O(8kn) time through the dynamic programming
approach where n is the number of vertices.
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A Planar Graph which Has 9 Levels.
The Graph Obtained by Removing
Nodes in levels 3, 6 and 9.
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Step 1.
For all i = 0, 1, ... , k, do
◦ (1.1) Let Gi be the graph obtained by deleting all nodes with levels
congruent to i (mod k + 1). The remaining subgraphs are all kouterplanar graphs.
◦ (1.2) For each k-outerplanar graph, find its maximum independent
set. Let Si denote the union of these solutions.
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Step 2.
Among S0 , S1 , ... , Sk , choose the Sj with the
maximum size and let it be our approximation solution
SAPX .
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The time-complexity of our approximation algorithm is
obviously O(8kkn).
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1
k 1
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Thus there is at least one r, such that at most
of
vertices in SOPT are at a level which is congruent to r
(mod k + 1).
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This means that the solution Sr obtained by deleting the
nodes in class r from SOPT will have at least
1
k
|SOPT| (1 - k  1 ) = k  1 |SOPT| nodes.
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k
k 1
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Therefore, |Sr| 
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According to our algorithm,
◦ |SAPX|  |Sr| 
or
 e=

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S OPT  S APX
S OPT
|SOPT| .
k
k 1
|SOPT|

1
k 1
Thus if we set k = 1 / E  -1, then the above formula
becomes
1
1
e
=
 E.
1
/
E
k 1
 
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This shows that for every given error bound E, we have
a corresponding k to guarantee that the approximation
solution differs from the optimum one within this error
ratio.
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