Bitonic sorter

A bitonic sorter for ${\displaystyle k=1}$ ${\displaystyle (n=2)}$ is simply a comparator.^[3] This is illustrated by the given box layout, in which X and Y represent the inputs, while H and L represent the higher and lower outputs, respectively.

With the sorter for ${\displaystyle k=1}$, we can recursively create a sorter of higher order. For example, consider the following ${\displaystyle k=2}$ ${\displaystyle (n=4)}$ bitonic sorter.^[3]

The bitonic sorter consists of two layers: a recombination layer, which recombines the bitonic inputs into two new bitonic sequences that are each half as long as the original sequence, and a bitonic sort layer consisting of two bitonic sorters of order ${\displaystyle k-1}$, each of which sorts one of the two bitonic sequences produced by the previous layer. This structure may be extended recursively for higher values of ${\displaystyle k}$ by ensuring that each comparator always accepts one input from each of the two halves of the bitonic sequence it is meant to help sort. The following illustration depicts these connections schematically.^[3]

As you can see, the first half of the input sequence is compared against the last half of the input sequence. Comparing each element of the (green) subsequence with the element of the other (orange) subsequence at the respective index produces two bitonic subsequences. These two bitonic series (blue and red, respectively) can then be fed into the next lower-order bitonic sorter. This can be done because all elements in the red sequence are guaranteed to be higher than all elements in the blue series. ^[3]

Ken Batcher provided some mathematical proof sketch in his paper.^[3] With out loss of generality the bitonic input sequence is assumed to be ${\displaystyle a_{1}\leq a_{2}\leq \dots \leq a_{j-1}\leq a_{j}\geq a_{j+1}\geq \dots \geq a_{2n}}$ with ${\displaystyle 1\leq j\leq 2n}$. With out loss of generality the sequence can be reversed therefore, we can assume ${\displaystyle n\leq j\leq 2n}$.

Case 1: If ${\displaystyle a_{n}\leq a_{2n}}$ then every element of the two subsequences are smaller. In this case ${\displaystyle d_{i}=a_{i}}$ and ${\displaystyle e_{i}=a_{n+i}}$ with ${\displaystyle 1\leq i\leq n}$ and therefore ${\displaystyle d_{i}}$ and ${\displaystyle e_{i}}$ are trivially bitonic.

Case 2: Otherwise there exists a ${\displaystyle k}$ such that the element ${\displaystyle a_{k}}$ of the first sub-sequence is bigger then ${\displaystyle a_{k}}$ of the second sub-sequence while it is the opposite for ${\displaystyle a_{k+1}}$. This means that ${\displaystyle a_{k}\leq a_{k+n}}$ and ${\displaystyle a_{k+1}>a_{k+n+1}}$ are true for a specific ${\displaystyle k}$. Therefore we now know that:

1. For ${\displaystyle 1\leq i\leq k}$ the sequences are ${\displaystyle d_{i}=a_{i}}$ and ${\displaystyle e_{i}=a_{n+i}}$

2. For ${\displaystyle k<i\leq 2n}$ the sequences are defined as the opposite of 1, with ${\displaystyle d_{i}=a_{n+i}}$ and ${\displaystyle e_{i}=a_{i}}$

In the original paper he then claims the following inequalities result from those definitions:^[3]

Following from 1:

• For ${\displaystyle 1\leq i\leq k}$ that ${\displaystyle d_{i}\leq d_{i+1}}$
• For ${\displaystyle j-n\leq i\leq k}$ that ${\displaystyle e_{i}\geq e_{i+1}}$
• For ${\displaystyle 1\leq i\leq j-n}$ that ${\displaystyle e_{i}\leq e_{i+1}}$

Following from 2:

• For ${\displaystyle k<i\leq 2n}$ that ${\displaystyle d_{i}\geq d_{i+1}}$
• For ${\displaystyle k<i\leq 2n}$ that ${\displaystyle e_{i}\leq e_{i+1}}$

From both: ${\displaystyle e_{n}\leq e_{1}}$

From the paper claims follows that the sequences ${\displaystyle d_{i}}$ and ${\displaystyle e_{i}}$ are in fact bitonic.^[3]

A bitonic sorting network is created by using several bitonic sorters. These bitonic sorters are recursively used to create two monotonic sequences, one decreasing and one increasing, which are then put into the next stage. This creates a bitonic series for the next stage, which can then use this bitonic series as a monotonic series for the next stage. Consider the following example for an ${\displaystyle n=4}$ bitonic sort network.^[3]

The bitonic sorting network for ${\displaystyle k=2}$ can be created by using a ${\displaystyle k=2}$ bitonic sorter and two ${\displaystyle k=k_{prev}-1}$ sorters. The two sorters create a decreasingly or increas- ingly sorted sequence in order to create a bitonic input for the bitonic sorter. Bitonic sorting networks of a lower order are mostly used for the two pre-sorters; therefore, a recursive definition of a bitonic sorting network from bitonic sorters can be described. In the above example, the two bitonic sorting networks are ${\displaystyle k=1}$ networks; hence, they are just a comparator.^[3] The following figure shows the overall scheme.

This overall scheme requires the sorter to have an input of sequence that is a power of two. There are, however, possibilities to mitigate this by, for example, using sentinel values.

The following pseudocode describes the sorting process. In the code, a is the array to be sorted, low is the index of the first item in the sub-array to be sorted, k and count is the number items in the sub-array that are being sorted in this function call. direction is a boolean value that determines whether the sub-array is being sorted into ascending / descending order.

The function call bitonicSort(a, 0, n, 1) is used to sort a, where n is the number of items in a .

function bitonicMerge(a, low, count, direction) is
    if count > 1 THEN
        k ← count / 2
        // Compare and swap elements across the halves
        for i ← low to low + k - 1 do
            // determine if two elements of a are out of order in relation to the direction of sorting. 
            if (direction == 1 AND a[i] > a[i + k]) OR (direction == 0 AND a[i] < a[i + k]) THEN
                swap a[i] with a[i + k]

        // Recursively merge both halves
        bitonicMerge(a, low, k, direction)
        bitonicMerge(a, low + k, k, direction)

// This only works when input size is a power of 2.
function bitonicSort(a, low, count, direction) is
    if count > 1 THEN
        k ← count / 2

        // Sort first/second half into ascending/descending order
        bitonicSort(a, low, k, 1)
        bitonicSort(a, low + k, k, 0)

        // Merge entire sequence in desired order
        bitonicMerge(a, low, count, direction)

In this section we assume that our sorter has ${\displaystyle n=2^{k}}$ input elements as previously.

Each recursion in a bitonic sorting network adds a sorter of order ${\displaystyle k_{next}=k_{prev}-1}$, which consists of bitonic ${\displaystyle k_{next}}$ sorter and the next recursion. As both sub-sorters can be done in parallel, only one level is added for each level in both sub-sorters. Each bitonic sorter has, therefore, one recombination layer and a lower-order bitonic sorter for its recursion. This results in ${\displaystyle k}$ levels per bitonic sorter. Therefore we can describe this construction's levels as the following sum: ${\displaystyle \sum _{i=1}^{k}i}$.

This sum can be reduced using the Gauss sum formula ${\displaystyle \sum _{i=1}^{k}i={\dfrac {1}{2}}k(k+1)}$

Therefore, the number of levels in which each comparison can be done in parallel is given by ${\displaystyle {\dfrac {1}{2}}k(k+1)}$.^[3] Which gives us ${\displaystyle {\mathcal {O}}(k^{2}+k)={\mathcal {O}}(k^{2})={\mathcal {O}}((\log _{2}n)^{2})}$ assuming ${\displaystyle n}$ comparisons can be performed in parallel.

Although the absolute number of comparisons is typically higher than Batcher's odd-even sort, many of the consecutive operations in a bitonic sort retain a locality of reference, making implementations more cache-friendly and typically more efficient in practice.^[3]

• Batcher odd–even mergesort
• Pairwise sorting network