http://www.geeksforgeeks.org/bitonic-sort/
Bitonic Sort is a classic parallel algorithm for sorting.
- Bitonic sort does O(n Log n2) comparisons.
- The number of comparisons done by Bitonic sort are more than popular sorting algorithms like Merge Sort [ does O(nLogn) comparisons], but Bitonice sort is better for parallel implementation because we always compare elements in predefined sequence and the sequence of comparison doesn’t depend on data. Therefore it is suitable for implementation in hardware and parallel processor array.
- A sequence, sorted in increasing order is considered Bitonic with the decreasing part as empty. Similarly, decreasing order sequence is considered Bitonic with the increasing part as empty.
- A rotation of Bitonic Sequence is also bitonic.
How to form a Bitonic Sequence from a random input?
We start by forming 4-element bitonic sequences from consecutive 2-element sequence. Consider 4-element in sequence x0, x1, x2, x3. We sort x0 and x1 in ascending order and x2 and x3 in descending order. We then concatenate the two pairs to form a 4 element bitonic sequence.
Next, we take two 4 element bitonic sequences, sorting one in ascending order, the other in descending order (using the Bitonic Sort which we will discuss below), and so on, until we obtain the bitonic sequence.
We start by forming 4-element bitonic sequences from consecutive 2-element sequence. Consider 4-element in sequence x0, x1, x2, x3. We sort x0 and x1 in ascending order and x2 and x3 in descending order. We then concatenate the two pairs to form a 4 element bitonic sequence.
Next, we take two 4 element bitonic sequences, sorting one in ascending order, the other in descending order (using the Bitonic Sort which we will discuss below), and so on, until we obtain the bitonic sequence.
To form a sorted sequence of length n from two sorted sequences of length n/2, log(n) comparisons are required (for example: log(8) = 3 when sequence size. Therefore, The number of comparisons T(n) of the entire sorting is given by:
T(n) = log(n) + T(n/2)
The solution of this recurrence equation is
T(n) = log(n) + log(n)-1 + log(n)-2 + … + 1 = log(n) · (log(n)+1) / 2
As, each stage of the sorting network consists of n/2 comparators. Therefore total Θ(n log n2) comparators.
/* The parameter dir indicates the sorting direction,
ASCENDING or DESCENDING; if (a[i] > a[j]) agrees
with the direction, then a[i] and a[j] are
interchanged. */
void
compAndSwap(
int
a[],
int
i,
int
j,
int
dir)
{
if
( (a[i] > a[j] && dir ==
1
) ||
(a[i] < a[j] && dir ==
0
))
{
// Swapping elements
int
temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}
/* It recursively sorts a bitonic sequence in ascending
order, if dir = 1, and in descending order otherwise
(means dir=0). The sequence to be sorted starts at
index position low, the parameter cnt is the number
of elements to be sorted.*/
void
bitonicMerge(
int
a[],
int
low,
int
cnt,
int
dir)
{
if
(cnt>
1
)
{
int
k = cnt/
2
;
for
(
int
i=low; i<low+k; i++)
compAndSwap(a,i, i+k, dir);
bitonicMerge(a,low, k, dir);
bitonicMerge(a,low+k, k, dir);
}
}
/* This funcion first produces a bitonic sequence by
recursively sorting its two halves in opposite sorting
orders, and then calls bitonicMerge to make them in
the same order */
void
bitonicSort(
int
a[],
int
low,
int
cnt,
int
dir)
{
if
(cnt>
1
)
{
int
k = cnt/
2
;
// sort in ascending order since dir here is 1
bitonicSort(a, low, k,
1
);
// sort in descending order since dir here is 0
bitonicSort(a,low+k, k,
0
);
// Will merge wole sequence in ascending order
// since dir=1.
bitonicMerge(a, low, cnt, dir);
}
}
/*Caller of bitonicSort for sorting the entire array
of length N in ASCENDING order */
void
sort(
int
a[],
int
N,
int
up)
{
bitonicSort(a,
0
, N, up);
}