Block sort

As previously stated, the outer loop of a block sort is identical to a bottom-up merge sort. However, it benefits from the variant that ensures each A and B subarray are the same size to within one item:

   BlockSort(array)
       power_of_two = FloorPowerOfTwo(array.size)
       scale = array.size/power_of_two // 1.0 ≤ scale < 2.0
      
       // insertion sort 16–31 items at a time
       for (merge = 0; merge < power_of_two; merge += 16)
           start = merge * scale
           end = start + 16 * scale
           InsertionSort(array, [start, end))
      
       for (length = 16; length < power_of_two; length += length)
           for (merge = 0; merge < power_of_two; merge += length * 2)
               start = merge * scale
               mid = (merge + length) * scale
               end = (merge + length * 2) * scale
              
               if (array[end − 1] < array[start])
                   // the two ranges are in reverse order, so a rotation is enough to merge them
                   Rotate(array, mid − start, [start, end))
               else if (array[mid − 1] > array[mid])
                   Merge(array, A = [start, mid), B = [mid, end))
               // else the ranges are already correctly ordered

Fixed-point math may also be used, by representing the scale factor as a fraction integer_part + numerator/denominator:

   power_of_two = FloorPowerOfTwo(array.size)
   denominator = power_of_two/16
   numerator_step = array.size % denominator
   integer_step = floor(array.size/denominator)
  
   // insertion sort 16–31 items at a time
  
   while (integer_step < array.size)
       integer_part = numerator = 0
       while (integer_part < array.size)
           // get the ranges for A and B
           start = integer_part
          
           integer_part += integer_step
           numerator += numerator_step
           if (numerator ≥ denominator)
               numerator −= denominator
               integer_part++
          
           mid = integer_part
          
           integer_part += integer_step
           numerator += numerator_step
           if (numerator ≥ denominator)
               numerator −= denominator
               integer_part++
          
           end = integer_part
          
           if (array[end − 1] < array[start])
               Rotate(array, mid − start, [start, end))
           else if (array[mid − 1] > array[mid])
               Merge(array, A = [start, mid), B = [mid, end))
      
       integer_step += integer_step
       numerator_step += numerator_step
       if (numerator_step ≥ denominator)
           numerator_step −= denominator
           integer_step++

The buffer extraction process for block sort.

The two internal buffers needed for each level of the merge step are created by moving the first 2√A first instances of their values within an A subarray to the start of A. First it iterates over the elements in A and counts off the unique values it needs, then it applies array rotations to move those unique values to the start.[6] If A did not contain enough unique values to fill the two buffers (of size √A each), B can be used just as well. In this case it moves the last instance of each value to the end of B, with that part of B not being included during the merges.

while (integer_step < array.size)
    block_size = √integer_step
    buffer_size = integer_step/block_size + 1
    [extract two buffers of size 'buffer_size' each]

If B does not contain enough unique values either, it pulls out the largest number of unique values it could find, then adjusts the size of the A and B blocks such that the number of resulting A blocks is less than or equal to the number of unique items pulled out for the buffer. Only one buffer will be used in this case – the second buffer won't exist.

buffer_size = [number of unique values found]
block_size = integer_step/buffer_size + 1
  
integer_part = numerator = 0
while (integer_part < array.size)
    [get the ranges for A and B]
    [adjust A and B to not include the ranges used by the buffers]

Tagging the A blocks using values from the first internal buffer. Note that the first A block and last B block are unevenly sized.

Once the one or two internal buffers have been created, it begins merging each A and B subarray for this level of the merge sort. To do so, it divides each A and B subarray into evenly sized blocks of the size calculated in the previous step, where the first A block and last B block are unevenly sized if needed. It then loops over each of the evenly sized A blocks and swaps the second value with a corresponding value from the first of the two internal buffers. This is known as tagging the blocks.

// blockA is the range of the remaining A blocks,
// and firstA is the unevenly sized first A block
blockA = [A.start, A.end)
firstA = [A.start, A.start + |blockA| % block_size)
  
// swap the second value of each A block with the value in buffer1
for (index = 0, indexA = firstA.end + 1; indexA < blockA.end; indexA += block_size)
    Swap(array[buffer1.start + index], array[indexA])
    index++
  
lastA = firstA
blockB = [B.start, B.start + minimum(block_size, |B|))
blockA.start += |firstA|

Two A blocks rolling through the B blocks. Once the first A block is dropped behind, the unevenly sized A block is locally merged with the B values that follow it.

After defining and tagging the A blocks in this manner, the A blocks are rolled through the B blocks by block swapping the first evenly sized A block with the next B block. This process repeats until the first value of the A block with the smallest tag value is less than or equal to the last value of the B block that was just swapped with an A block.

At that point, the minimum A block (the A block with the smallest tag value) is swapped to the start of the rolling A blocks and the tagged value is restored with its original value from the first buffer. This is known as dropping a block behind, as it will no longer be rolled along with the remaining A blocks. That A block is then inserted into the previous B block, first by using a binary search on B to find the index where the first value of A is less than or equal to the value at that index of B, and then by rotating A into B at that index.

   minA = blockA.start
   indexA = 0
  
   while (true)
       // if there's a previous B block and the first value of the minimum A block is ≤
       // the last value of the previous B block, then drop that minimum A block behind.
       // or if there are no B blocks left then keep dropping the remaining A blocks.
       if ((|lastB| > 0 and array[lastB.end - 1] ≥ array[minA]) or |blockB| = 0)
           // figure out where to split the previous B block, and rotate it at the split
           B_split = BinaryFirst(array, array[minA], lastB)
           B_remaining = lastB.end - B_split
          
           // swap the minimum A block to the beginning of the rolling A blocks
           BlockSwap(array, blockA.start, minA, block_size)
          
           // restore the second value for the A block
           Swap(array[blockA.start + 1], array[buffer1.start + indexA])
           indexA++
          
           // rotate the A block into the previous B block
           Rotate(array, blockA.start - B_split, [B_split, blockA.start + block_size))
          
           // locally merge the previous A block with the B values that follow it,
           // using the second internal buffer as swap space (if it exists)
           if (|buffer2| > 0)
               MergeInternal(array, lastA, [lastA.end, B_split), buffer2)
           else
               MergeInPlace(array, lastA, [lastA.end, B_split))
          
           // update the range for the remaining A blocks,
           // and the range remaining from the B block after it was split
           lastA = [blockA.start - B_remaining, blockA.start - B_remaining + block_size)
           lastB = [lastA.end, lastA.end + B_remaining)
          
           // if there are no more A blocks remaining, this step is finished
           blockA.start = blockA.start + block_size
           if (|blockA| = 0)
               break
          
           minA = [new minimum A block] (see below)
       else if (|blockB| < block_size)
           // move the last B block, which is unevenly sized,
           // to before the remaining A blocks, by using a rotation
           Rotate(array, blockB.start - blockA.start, [blockA.start, blockB.end))
          
           lastB = [blockA.start, blockA.start + |blockB|)
           blockA.start += |blockB|
           blockA.end += |blockB|
           minA += |blockB|
           blockB.end = blockB.start
       else
           // roll the leftmost A block to the end by swapping it with the next B block
           BlockSwap(array, blockA.start, blockB.start, block_size)
           lastB = [blockA.start, blockA.start + block_size)
           if (minA = blockA.start)
               minA = blockA.end
          
           blockA.start += block_size
           blockA.end += block_size
           blockB.start += block_size
          
           // this is equivalent to minimum(blockB.end + block_size, B.end),
           // but that has the potential to overflow
           if (blockB.end > B.end - block_size)
               blockB.end = B.end
           else
               blockB.end += block_size
  
   // merge the last A block with the remaining B values
   if (|buffer2| > 0)
       MergeInternal(array, lastA, [lastA.end, B.end), buffer2)
   else
       MergeInPlace(array, lastA, [lastA.end, B.end))

One optimization that can be applied during this step is the floating-hole technique.[7] When the minimum A block is dropped behind and needs to be rotated into the previous B block, after which its contents are swapped into the second internal buffer for the local merges, it would be faster to swap the A block to the buffer beforehand, and to take advantage of the fact that the contents of that buffer do not need to retain any order. So rather than rotating the second buffer (which used to be the A block before the block swap) into the previous B block at position index, the values in the B block after index can simply be block swapped with the last items of the buffer.

The floating hole in this case refers to the contents of the second internal buffer floating around the array, and acting as a hole in the sense that the items do not need to retain their order.

Once the A block has been rotated into the B block, the previous A block is then merged with the B values that follow it, using the second buffer as swap space. When the first A block is dropped behind this refers to the unevenly sized A block at the start, when the second A block is dropped behind it means the first A block, and so forth.

MergeInternal(array, A, B, buffer)
    // block swap the values in A with those in 'buffer'
    BlockSwap(array, A.start, buffer.start, |A|)

    A_count = 0, B_count = 0, insert = 0
    while (A_count < |A| and B_count < |B|)
        if (array[buffer.start + A_count] ≤ array[B.start + B_count])
            Swap(array[A.start + insert], array[buffer.start + A_count])
            A_count++
        else
            Swap(array[A.start + insert], array[B.start + B_count])
            B_count++
        insert++

    // block swap the remaining part of the buffer with the remaining part of the array
    BlockSwap(array, buffer.start + A_count, A.start + insert, |A| - A_count)

If the second buffer does not exist, a strictly in-place merge operation must be performed, such as a rotation-based version of the Hwang and Lin algorithm,[7][8] the Dudzinski and Dydek algorithm,[9] or a repeated binary search and rotate.

MergeInPlace(array, A, B)
    while (|A| > 0 and |B| > 0)
        // find the first place in B where the first item in A needs to be inserted
        mid = BinaryFirst(array, array[A.start], B)

        // rotate A into place
        amount = mid - A.end
        Rotate(array, amount, [A.start, mid))

        // calculate the new A and B ranges
        B = [mid, B.end)
        A = [A.start + amount, mid)
        A.start = BinaryLast(array, array[A.start], A)

After dropping the minimum A block and merging the previous A block with the B values that follow it, the new minimum A block must be found within the blocks that are still being rolled through the array. This is handled by running a linear search through those A blocks and comparing the tag values to find the smallest one.

minA = blockA.start
for (findA = minA + block_size; findA < blockA.end - 1; findA += block_size)
    if (array[findA + 1] < array[minA + 1])
        minA = findA

These remaining A blocks then continue rolling through the array and being dropped and inserted where they belong. This process repeats until all of the A blocks have been dropped and rotated into the previous B block.

Once the last remaining A block has been dropped behind and inserted into B where it belongs, it should be merged with the remaining B values that follow it. This completes the merge process for that particular pair of A and B subarrays. However, it must then repeat the process for the remaining A and B subarrays for the current level of the merge sort.

Note that the internal buffers can be reused for every set of A and B subarrays for this level of the merge sort, and do not need to be re-extracted or modified in any way.

After all of the A and B subarrays have been merged, the one or two internal buffers are still left over. The first internal buffer was used for tagging the A blocks, and its contents are still in the same order as before, but the second internal buffer may have had its contents rearranged when it was used as swap space for the merges. This means the contents of the second buffer will need to be sorted using a different algorithm, such as insertion sort. The two buffers must then be redistributed back into the array using the opposite process that was used to create them.

After repeating these steps for every level of the bottom-up merge sort, the block sort is completed.

Block sort begins by performing insertion sort on groups of 16–31 items in the array. Insertion sort is an ${\displaystyle O(n^{2})}$ operation, so this leads to anywhere from ${\displaystyle O(16^{2}\times n/16)}$ to ${\displaystyle O(31^{2}\times n/31)}$, which is ${\displaystyle O(n)}$ once the constant factors are omitted. It must also apply an insertion sort on the second internal buffer after each level of merging is completed. However, as this buffer was limited to ${\displaystyle {\sqrt {A}}}$ in size, the ${\displaystyle O{\sqrt {n}}^{2}}$ operation also ends up being ${\displaystyle O(n)}$.

Next it must extract two internal buffers for each level of the merge sort. It does so by iterating over the items in the A and B subarrays and incrementing a counter whenever the value changes, and upon finding enough values it rotates them to the start of A or the end of B. In the worst case this will end up searching the entire array before finding ${\displaystyle {\sqrt {A}}}$ non-contiguous unique values, which requires ${\displaystyle O(n)}$ comparisons and ${\displaystyle {\sqrt {A}}}$ rotations for ${\displaystyle {\sqrt {A}}}$ values. This resolves to ${\displaystyle O(n+{\sqrt {n}}\times {\sqrt {n}})}$, or ${\displaystyle O(n)}$.

When none of the A or B subarrays contained ${\displaystyle {\sqrt {A}}}$ unique values to create the internal buffers, a normally suboptimal in-place merge operation is performed where it repeatedly binary searches and rotates A into B. However, the known lack of unique values within any of the subarrays places a hard limit on the number of binary searches and rotations that will be performed during this step, which is again ${\displaystyle {\sqrt {A}}}$ items rotated up to ${\displaystyle {\sqrt {A}}}$ times, or ${\displaystyle O(n)}$. The size of each block is also adjusted to be smaller in the case where it found ${\displaystyle {\sqrt {A}}}$ unique values but not 2${\displaystyle {\sqrt {A}}}$, which further limits the number of unique values contained within any A or B block.

Tagging the A blocks is performed ${\displaystyle {\sqrt {A}}}$ times for each A subarray, then the A blocks are rolled through and inserted into the B blocks up to ${\displaystyle {\sqrt {A}}}$ times. The local merges retain the same ${\displaystyle O(n)}$ complexity of a standard merge, albeit with more assignments since the values must be swapped rather than copied. The linear search for finding the new minimum A block iterates over ${\displaystyle {\sqrt {A}}}$ blocks ${\displaystyle {\sqrt {A}}}$ times. And the buffer redistribution process is identical to the buffer extraction but in reverse, and therefore has the same ${\displaystyle O(n)}$ complexity.

After omitting all but the highest complexity and considering that there are ${\displaystyle \log {n}}$ levels in the outer merge loop, this leads to a final asymptotic complexity of ${\displaystyle O(n\log {n})}$ for the worst and average cases. For the best case, where the data is already in order, the merge step performs ${\displaystyle n/16}$ comparisons for the first level, then ${\displaystyle n/32}$, ${\displaystyle n/64}$, ${\displaystyle n/128}$, etc. This is a well-known mathematical series which resolves to ${\displaystyle O(n)}$.

As block sort is non-recursive and does not require the use of dynamic allocations, this leads to constant stack and heap space. It uses O(1) auxiliary memory in a transdichotomous model, which accepts that the O(log n) bits needed to keep track of the ranges for A and B cannot be any greater than 32 or 64 on 32-bit or 64-bit computing systems, respectively, and therefore simplifies to O(1) space for any array that can feasibly be allocated.

Although items in the array are moved out of order during a block sort, each operation is fully reversible and will have restored the original order of equivalent items by its completion.

Stability requires the first instance of each value in an array before sorting to still be the first instance of that value after sorting. Block sort moves these first instances to the start of the array to create the two internal buffers, but when all of the merges are completed for the current level of the block sort, those values are distributed back to the first sorted position within the array. This maintains stability.

Before rolling the A blocks through the B blocks, each A block has its second value swapped with a value from the first buffer. At that point the A blocks are moved out of order to roll through the B blocks. However, once it finds where it should insert the smallest A block into the previous B block, that smallest A block is moved back to the start of the A blocks and its second value is restored. By the time all of the A blocks have been inserted, the A blocks will be in order again and the first buffer will contain its original values in the original order.

Using the second buffer as swap space when merging an A block with some B values causes the contents of that buffer to be rearranged. However, as the algorithm already ensured the buffer only contains unique values, sorting the contents of the buffer is sufficient to restore their original stable order.

Block sort is an adaptive sort on two levels: first, it skips merging A and B subarrays that are already in order. Next, when A and B need to be merged and are broken into evenly sized blocks, the A blocks are only rolled through B as far as is necessary, and each block is only merged with the B values immediately following it. The more ordered the data originally was, the fewer B values there will be that need to be merged into A.