I was wondering, is the bitonic sort algorithm stable? I searched the original paper, wikipedia and some tutorials, could not find it.

It seems to me that it should be, as it is composed of merge / sort steps, however was unable to find answer anywhere.

The reason why I'm asking - I was comparing this particular implementation of bitonic sort to the sort implemented in the C++ standard library, for array length 9 it requires 28 comparison / swap operations, while the standard library sort (which is unstable) requires 25. The three extra cswaps do not seem enough to make the sort stable.

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    $\begingroup$ Bitonic sort is a sorting network. Can a sorting network be not stable? How do you define for a sorting network to be stable? $\endgroup$ Oct 22, 2014 at 11:44
  • $\begingroup$ By the way, I cannot understand the motivation you stated in the last paragraph. You compared the number of comparisons needed for n=9 between bitonic sort and some unstable sort, and you found that bitonic sort required more comparisons. How is this related to whether bitonic sort is stable or not? $\endgroup$ Oct 22, 2014 at 11:54
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    $\begingroup$ @TsuyoshiIto: I also thought that sorting networks would be automatically stable (as a corollary of the 0-1 principle), but it seems that this is not the case. See hoytech.github.io/sorting-networks pages 27–28. $\endgroup$ Oct 22, 2014 at 16:58
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    $\begingroup$ @Jukka Suomela: Thanks for the link! I stand corrected. $\endgroup$ Oct 22, 2014 at 21:41
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    $\begingroup$ @theswine: In The Art of Computer Programming Vol. 3 Sorting and Searching Knuth explains that we pay for the uniformity of sorting networks with an increased number of comparisons compared to the optimum. $\endgroup$ Sep 2, 2015 at 9:58

2 Answers 2


No, bitonic sort is not stable.

For this post I will denote numbers as 2;0 where only the part before the ; is used for comparison and the part behind ; to mark the initial position. Comparison-exchanges are denoted by arrows where the head points at the desired location of the greater value.

As written in the link that @JukkaSuomela posted a stable sorting network needs to avoid swaps of equal values.

When swapping equal values, the bitonic sorter for two values is already unstable:

0;0 ----- 0;1
0;1 ----- 0;0

Of course, this can be fixed when we don't swap equal values:

0;0 ----- 0;0
0;1 ----- 0;1

However, it could happen that the order of two equal elements is swapped without them being compared to each other.

This is exactly the case in this example of a bitonic sorter for 4 values:

1;0 ------ 1;0 ------ 0;2 ------ 0;2
      ^          |          |
      |          |          v
1;1 ------ 1;1 --|--- 1;1 ------ 1;1
0;2 ------ 0;2 ---|-- 1;0 ------ 1;0
      |           |         |
      v           v         v
2;3 ------ 2;3 ------ 2;3 ------ 2;3

Although we were careful not to swap elements that compared equal (upper left comparison), the merging pass swapped the order of 1;1 and 1;0 which cannot be corrected later on.

This counterexample proves that bitonic sort cannot be stable.

  • $\begingroup$ I am afraid the second stage is wrong since a merger network reverses its second input to make the whole concatenated input bitonic. Therefore, the second stage should compare between element 1 and element 4, and element 2 and 3. $\endgroup$ Feb 13, 2023 at 9:01

It could become stable, but in an inefficient way.

See Is there any efficient Network stable sort (not bubble sort)? which demonstrates it


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