# Batcher odd–even mergesort

(Redirected from Batcher odd-even mergesort)
Class Visualization of the odd–even mergesort network with eight inputs Sorting algorithm Array ${\displaystyle O(\log ^{2}(n))}$ parallel time ${\displaystyle O(\log ^{2}(n))}$ parallel time ${\displaystyle O(\log ^{2}(n))}$ parallel time ${\displaystyle O(n\log ^{2}(n))}$ non-parallel time

Batcher's odd–even mergesort is a generic construction devised by Ken Batcher for sorting networks of size O(n (log n)2) and depth O((log n)2), where n is the number of items to be sorted. Although it is not asymptotically optimal, Knuth concluded in 1998, with respect to the AKS network that "Batcher's method is much better, unless n exceeds the total memory capacity of all computers on earth!"[1]

It is popularized by the second GPU Gems book,[2] as an easy way of doing reasonably efficient sorts on graphics-processing hardware.

## Example code

The following is an implementation of odd–even mergesort algorithm in Python. The input is a list x of length a power of 2. The output is a list sorted in ascending order.

```def oddeven_merge(lo: int, hi: int, r: int):
step = r * 2
if step < hi - lo:
yield from oddeven_merge(lo, hi, step)
yield from oddeven_merge(lo + r, hi, step)
yield from [(i, i + r) for i in range(lo + r, hi - r, step)]
else:
yield (lo, lo + r)

def oddeven_merge_sort_range(lo: int, hi: int):
""" sort the part of x with indices between lo and hi.

Note: endpoints (lo and hi) are included.
"""
if (hi - lo) >= 1:
# if there is more than one element, split the input
# down the middle and first sort the first and second
# half, followed by merging them.
mid = lo + ((hi - lo) // 2)
yield from oddeven_merge_sort_range(lo, mid)
yield from oddeven_merge_sort_range(mid + 1, hi)
yield from oddeven_merge(lo, hi, 1)

def oddeven_merge_sort(length: int):
""" "length" is the length of the list to be sorted.
Returns a list of pairs of indices starting with 0 """
yield from oddeven_merge_sort_range(0, length - 1)

def compare_and_swap(x, a, b) -> None:
if x[a] > x[b]:
x[a], x[b] = x[b], x[a]
```
```>>> data = [2, 4, 3, 5, 6, 1, 7, 8]
>>> pairs_to_compare = list(oddeven_merge_sort(len(data)))
>>> pairs_to_compare
[(0, 1), (2, 3), (0, 2), (1, 3), (1, 2), (4, 5), (6, 7), (4, 6), (5, 7), (5, 6), (0, 4), (2, 6), (2, 4), (1, 5), (3, 7), (3, 5), (1, 2), (3, 4), (5, 6)]
>>> for i in pairs_to_compare: compare_and_swap(data, *i)
>>> data
[1, 2, 3, 4, 5, 6, 7, 8]
```

More concise and non-recursive calculation of partner node is possible. Here is a Scala implementation to get the partner of an index at each step:[3]

```  def partner(index: Int, merge: Int, step: Int): Int = {
if (step == 1)
index ^ (1 << (merge - 1))
else {
val (scale, box) = (1 << (merge - step), 1 << step)
val sn = index / scale - (index / scale / box) * box

if (sn == 0 || sn == box - 1) index // no exchange at this level
else if (sn % 2 == 0) index - scale else index + scale
}
}
```