Unformatted text preview: pendencies between comparators.
The shortest path is the one with the most edges, or the most dependencies between comparators.
Constructing our dependency graph requires ¢´Ö ¾ µ time because we consider all pairs of comparators once. The graph
is a dag since we only have edges ´Ú Ú µ if
. Thus, we can run
Ç´Ö · Ö ¾ µ Thus, we have a total
DAG -S HORTEST-PATHS, which has a running time of ¢´Î · µ
running time of ¢´Ö ¾ µ.
Deﬁne a comparison network as standard if Õ Õ for Õ ½¾ Ö. (b) Give an algorithm to convert a comparison network with Ò inputs and Ö comparators into an equivalent
standard comparison network with Ò inputs and Ö comparators.
Solution: We iterate over all the comparators in order. At a comparator ´ Õ Õ µ, if Õ
Õ , we do not
perform any action and continue with the next comparator. If Õ
Õ , then we need to standardize the
comparator by changing it to ´ Õ Õ µ. Note that now the output that used to be on Õ is on Õ and viceversa. Thus, we also need to reverse Õ and Õ in all subsequent comparators. That is to say, we consider
all comparators ´ Õ· Õ· µ for ½
Ö Õ. If Õ·
Õ , then we change the comparator to be
´ Õ Õ · µ. If Õ ·
Õ , then we change the comparator to be ´ Õ Õ · µ. We perform similar actions for
Õ and Õ ·
Õ . We then continue on to the next comparator.
The proof of correctness is just a simple induction argument over the number of comparators iterated over.
To be equivalent, we simply need that after ﬂipping the Õ Ø co...
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- Spring '07
- Graph Theory, Shortest path problem, µ, Comparator, comparators