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PseudocodeThe simplest form of cocktail sort goes through the whole list each time:
procedure cocktailSort( A : list of sortable items ) defined as:
do
swapped := false
for each i in 0 to length( A ) - 2 do:
if A[ i ] > A[ i + 1 ] then // test whether the two elements are in the wrong order
swap( A[ i ], A[ i + 1 ] ) // let the two elements change places
swapped := true
end if
end for
if swapped = false then
// we can exit the outer loop here if no swaps occurred.
break do-while loop
end if
swapped := false
for each i in length( A ) - 2 to 0 do:
if A[ i ] > A[ i + 1 ] then
swap( A[ i ], A[ i + 1 ] )
swapped := true
end if
end for
while swapped // if no elements have been swapped, then the list is sorted
end procedure
The first rightward pass will shift the largest element to its correct place at the end, and the following leftward pass will shift the smallest element to its correct place at the beginning. The second complete pass will shift the second largest and second smallest elements to their correct places, and so on. After i passes, the first i and the last i elements in the list are in their correct positions, and do not need to be checked. By shortening the part of the list that is sorted each time, the number of operations can be halved (see bubble sort). procedure oddEvenSort( A : list of sortable items ) defined as: // `begin` and `end` marks the first and last index to check begin := -1 end := length( A ) - 2 do swapped := false // increases `begin` because the elements before `begin` are in correct order begin := begin + 1 for each i in begin to end do: if A[ i ] > A[ i + 1 ] then swap( A[ i ], A[ i + 1 ] ) swapped := true end if end for if swapped = false then break do-while loop end if swapped := false // decreases `end` because the elements after `end` are in correct order end := end - 1 for each i in end to begin do: if A[ i ] > A[ i + 1 ] then swap( A[ i ], A[ i + 1 ] ) swapped := true end if end for while swapped end procedure Differences from bubble sortCocktail sort is a slight variation of bubble sort. It differs in that instead of repeatedly passing through the list from bottom to top, it passes alternately from bottom to top and then from top to bottom. It can achieve slightly better performance than a standard bubble sort. The reason for this is that bubble sort only passes through the list in one direction and therefore can only move items backward one step each iteration. An example of a list that proves this point is the list (2,3,4,5,1), which would only need to go through one pass of cocktail sort to become sorted, but if using an ascending bubble sort would take four passes. However one cocktail sort pass should be counted as two bubble sort passes. Typically cocktail sort is less than two times faster than bubble sort. ComplexityThe complexity of cocktail sort in big O notation is O(n2) for both the worst case and the average case, but it becomes closer to O(n) if the list is mostly ordered before applying the sorting algorithm, for example, if every element is at a position that differs at most k (k ≥ 1) from the position it is going to end up in, the complexity of cocktail sort becomes O(k * n). External links
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This article is from Wikipedia. All text is available under the terms of the GNU Free Documentation License.
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