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Subsections
In this section, we show that IP-opt is correct by proving
the following
Theorem 28822
The algorithm IP-opt delivers an optimal partition for an
instance of IP if there is a partition that satisfies
(5.7).
The proof is by induction over the loop variable i for which
is the invariant of the for-loop in IP-opt where
and
Using the facts
- that c(qi) is always forced to be minimal through the
min function
- and that Pi is the optimal partition for the segment
we can conclude that c(qn) is the number of overlaps in an
optimal partition as specified by IP and that an optimal partition
is given by Pn.
If there is no partition that satisfies constraint (5.7)
then there must be some qi such that for all
j<i. This causes the set J to be empty and thus the algorithm to
report this fact and then to stop. In the remainder of the proof we
assume that there is an optimal partition and consequently that J
is non-empty for all .
- Base case:
- i=1
The algorithm provides
with . Thus which causes the sum
in (5.11) to evaluate to 0 which matches with (*).
Thus the assumption holds.
- Hypothesis:
- Assume that (5.11) holds for all
(**).
- Inductive Step:
- i=x+1
Let be the minimising point
with j < i = x+1 as guaranteed by the algorithm. Then
Therefore (5.11) holds for i=x+1.
Next: Alternative: Reducing IP to
Up: Optimal Partitioning
Previous: Example
Thomas Zurek