Algorithm Design

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Chapter 10 Extending the Limits of Tractability
Vt
No edge
~ g the choice of U
breaks all communication[
between desc~’ndants |
~.(and with the parent).
Parent(t)
No edge
Figure 10.8 The subproblem it(U) in the subgraph Gr in the optimal solution to this
subproblem, we consider independent sets Si in the descendant subgraphs Gti, subject
to the constraint that Sin Vt = U n Vtr
Let S be the maximum-weight independent set in G t subject to the requirement
that S N V t = U; that is, w(S) = it(U). The key is to understand how this
set S looks when intersected with each of the subgraphs Gti, as suggested in
Figure 10.8. We let St denote the intersection of S with the nodes of Gt~.
(10.17) S t is a maximum-weight independent set of Gtg, subject to the
straint that S t ~ Vt = U C~ Vt,.
10.4 Tree Decompositions of Graphs
Proof. Suppose there were an independent set S~ of Gti with the property that
S~ C] V t = U N Vti and w(S~) > w(S~). Then consider the set S’ = (S-Si) u S~.
Clearly w(S’) > w(S). Also, it is easy to check that S’ ~ V t = U.
We claim that S’ is an independent set in G; this will contradict our choice
of S as the maximum-weight independent set in Gt subject to S ~ V t = U. For
suppose S’ is not independent, and let e = (u, v) be an edge with both ends in S’.
It cannot be that u and v both belong to S, or that they both belong to S~, since
these are both .independent sets. Thus we must have u ~ S-S[ and v e S~-S,
from which it follows that u is not a node of Gt~ while u ~ Gti - (V t ~ Vti)" But
then, by (10.14), there cannot be an edge joining u and v. []
Statement (10.17) is exactly what we need to design a recurrence relation
for our subproblems. It says that the ~nformation needed to compute ft(U)
is implicit in the values already computed for the subtrees. Specifically, for
each child t~, we need simply determine the value of ~e maximum-weight
independent set St of Gt~ , subject to the constraint that Si ~ V t _-_ U ~ Vti. This
constraint does not completely determine what S t ~ Vt~ should be; rather, it
says that it can be any independent set Uz c__ Vt~ such that U~ ~ V t = U ~ Vtg.
Thus the weight of the optimal S~ is equal to
max{ft~(U~) : Ui ~ V t = U ~ Vti and U~ c_ Vt~ is independent}.
Finally, the value of f~ (U) is simply w(U) plus these maxima added over the d
children of t--except that to avoid overcoundng the nodes in U, we exclude
them from the contribution of the children. Thus we have
(10.18) The value "
of it(U) is given by the following recurrence:
d
it(U) = w(U) + ~ max{ft~(U~ ) _ w(U~ ~ U)"
U~ ~ V t = U ~ Vt~ and U~ c__ Vt~ is independent}.
The overall algorithm now just builds up the values of all the subproblems
from the leaves of T upward.
To find a maximum-weight independent set of G,
given a tree decomposition (T, [Vt} ) of G:
Modify the tree decomposition if necessary so
Root T at a node r
For each node ~ of T in post-order
If t is a leaf then
it is nonredundant
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