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Unformatted text preview: e use a notation called bigO.
We
big
So order n2 is written O(n2). Applied Programming So we need to look at best/worst case
So
best/worst
performance. All three of the sorting algorithms discussed so far
All
are O(n2).
are Is there some initial arrangement of the data which
Is
makes the algorithm perform better or worse than
O(n2)?
O(n This doesn't mean that they're all equal. Selection Sort has constant performance. They differ in the constant multiplier. Thus, if the data is already sorted, it will take just
Thus,
as long as if the data is in reverse order.
as But, they all act the same as n increases. Programming
Applied Applied Programming For Bubble Sort, one pass through the data is all
For
that is needed to recognise this.
No exchanges are required. So it becomes O(n).
No
But, if the data is completely reversed, the full set
But,
of comparisons and all the moves will be needed,
so worst case is O(n2).
so
On average, less than n(n1)/2 comparisons will
1)/2
be needed.
be
Applied Programming
Applied Can we improve on O(n2)? Applied Programming Insertion Sort's best case is sorted.
Each new entry is compared to the last of the
Each
already sorted data. So O(n).
already
The worst case is again reversed data.
This is O(n2).
The average is about n(n1)/4, still O(n2).
So Selection seems to be worst, followed by
So
Bubble and Insertion. This is only on comparisons
– not moves.
Applied Programming...
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 Spring '14
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