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Running
time recurrence
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T(n) =
2T(n/2) + n log n
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and
T(1) = 1
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This
does not solve to
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T(n) =
O(n log n)
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Final
trick: Presorting
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Sort
the set of points by y-
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coordinate
before we start
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Whenever
we split a point
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set,
we can run through the
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list
sorted by y-coordinate
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and
create a new list for
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each
part, sorted by y-
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coordinate
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Recurrence
becomes
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T(n) =
2T(n/2) + n
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and
T(1) = 1
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Solution:
T(n) = O(n log n)
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Its
possible to show that the
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closest
pair can be found in
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O(n
log n) time for any
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dimension
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