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As discussed in the Lounge room on Stack Overflow:

if you can't implement the Quicksort algorithm given in any language you have minimal knowledge of, you might want to consider a different profession. @sbi

but SBI also noted that maybe BrainF*** was an exception.

So, here's the puzzle/challenge: implement QuickSort in BrainF***. The implementation must

  • be interpreted by this and/or by the interpreter(s) here (for large scripts)
  • implement the algorithm as described on Wikipedia - if possible as an in-place sort
  • sort the following list of integers: [0,4,6,4,2,3,9,2,3,6,5,3] and print the result
share|improve this question
Searching around a bit I'm able to find one implementation, but it's 6kB (and compiled from Haskell). – Peter Taylor May 3 '11 at 23:02
@Peter actually the brainfuck implementation is 474.2 K inside the archive - which is a bit bigger than I expected (and too big for the on-line interpreter). Maybe I should change the target interpreter.. (but I would love to see something hand-written) – Ronald May 4 '11 at 0:01
Actually, I got an error on that one: an underflow at byte 484109 (using the bcci interpreter) – Ronald May 4 '11 at 0:12
I bet I could do bubble sort instead and nobody looking at the code would know the difference... – Peter Olson May 4 '11 at 16:37
@Peter Of The Corn: We would discover a bubble-sort by the bad performance. – user unknown Jul 13 '11 at 16:57
up vote 47 down vote accepted

BrainF* (697 bytes)


Below is an annotated version. In order to keep track of what was supposed to be happening while developing it, I used a comment notation that looks like this: |a|b=0|c=A0|@d|A0|A1|```|

|a| represents a named cell
|b=X| means we know the cell has value X, where X can be a constant or a variable name
|@d|  means the data pointer is in this cell
|A0|A1|```| is variable length array. (using ``` for ... because . is a command)

The memory is laid out with a left-growing stack of partitions to process on the left, a scratch space in the center, and the array being sorted to the right. Array indexing is handled by moving a "data bus" containing the index and working space through the array. So for instance a 3-wide bus of |i|data|0|A0|A1|A2, will become |A0|i-1|data|0|A1|A2 after shifting by one. The partitioning is performed by keeping the bus between the high and low elements.
Here's the full version:

Get input
>>>>>>>> ,[>,]                      |A0|A1|```|An|@0|
Count items
<[ [>>>+<<<-]>[<+>-]<+ <]  |@0|n|0|0|A0|A1|```
Make 8wide data bus w/ stack on left
>[<<<<<<<<+>>>>>>>>-]  ```|K1=n|K0=0|Z=0|a|b|c|d|e|@f|g|X=0|A0|A1|```
K1 and K0 represent the first index to process (I) and one past the last (J)
Check if still partitions to process
  Copy K1 to a&c via Z
  [>>+>+>>+<<<<<-]>>[<<+>>-] ```|K1=J|K0=I|@Z=0|a=J|b|c=J|d|e|f|g|X=0|A0|A1|```
  Copy K0 to b&d via Z
  <[>+>>+>>+<<<<<-]>[<+>-] ```|K1|K0|@Z=0|a=J|b=I|c=J|d=I|e|f|g|X=0|A0|A1|```
  Check if J minus I LE 1 : Subtract d from c
  >>>>[-<->]                    |a=J|b=I|c=JminusI|@d=0|e|f|g|
  d= c==0; e = c==1
  +<[>- >+<<-[>>-<<[-]]]        |a=J|b=I|@c=0|d=c==0|e=c==1|f|g|
  if d or e is 1 then J minus I LE 1: partition empty
  >[<+>-]>[<<+>>-]<+<      |a=J|b=I|@c=isEmpty|d=1|e=0|f|g|
  If Partition Empty;
  [->-                      |a=J|b=I|@c=0|d=0|c=0|f|g|
    pop K0: Zero it and copy the remaining stack right one; inc new K0
    <<[-]<[-]<<[-]<[[>+<-]<]>>[>]<+    ``|K1|@Z=0|a=J|b=I|c=0|d=0|e|f|g|
  >>>>]>[-                   Z|a=J|b=I|c=isEmpty=0|@d=0|e|f|g|X|A0|A1
    Move Bus right I plus 1 frames; leaving first element to left
    <<+[ -[>+<-]<-[>+<-]>>>>>>>>      (dec J as we move)
      [<<<<<<<<+>>>>>>>>-]<<<<<< ]      Z|Ai|a=J|@b=0|c=0|d|e|f|g|X|Aq
    first element becomes pivot Ap; store in b
    <<[>>+<<-]            Z|@0|a=J|b=Ap|c=0|d|e|f|g|X|Aq
    While there are more elements (J GT 0);
    >[                    Z|0|@a=J|b=Ap|c=0|d|e|f|g|X|Aq
      copy Ap to e via c
      >[>+>>+<<<-]>[<+>-]  Z|0|a=J|b=Ap|@c=0|d=0|e=Ap|f|g|X=0|Aq
       copy Aq to g via X
      >>>>>>[<+<+>>-]<[>+<-] |c|d=0|e=Ap|f|g=Aq|@X=0|Aq
      Test Aq LT Ap:  while e; mark f; clear it if g 
      <<<[ >+>[<-]<[<]           |@d=0|e|f=gLTe|g|
        if f: set d and e to 1; dec e and g 
      set g to 1; if d: set f 
      >>[-]+<<< [->>+<<]
      If Aq LT Ap move Aq across Bus
      >>[->- <<<<<[>+<-] <[>+<-] >>>>>>>>
        [<<<<<<<<+>>>>>>>>-] <<]  Z|0|Aq|a=J|b=Ap|c|d|e|@f=0|g=0|X=0|Ar
      Else Swap AQ w/ Aj: Build a 3wide shuttle holding J and Aq                
      >[[-] <<<<<<[>>+>>>>>+<<<<<<<-]>>[<<+>>-] |@c=0|d|e|f=0|g=0|X=J|Aq|Ar|```
      If J then dec J
        & While J shuttle right
        [>>[<<<+>>>-]<[>+<-]<-[>+<-]>] |a=J|b=Ap|c|d|e|f|Ar|```|Aj|g=0|@X=0|Aq|
        Leave Aq out there and bring Aj back
        <<[ [>>+<<-] < ]              |a=J|b=Ap|c|d|e|@f=0|g|X=0|Ar|```|Aj|Aq|
    Either bus moved or last element swapped; reduce J in either case
    <<<<<<-]                 |Aq|@a=0|b=Ap|c|d|e|f|g|X|Ar|```|
    Insert Ap To right of bus
    >[>>>>>>+<<<<<<-]        |Aq|a=0|@b=0|c|d|e|f|g|Ap|Ar|```|
    Move the bus back to original location tracking pivot location
    <<[ [>>>>>>>+<<<<<<<-]>[<+>-]<+ <]     
    <[ [>>>>>>>>+<<<<<<<<-]>>[<+>-]<+ <<] |K1|K0|@Z=0|a=0|b=p|c|d|e|f|g|X|Ar|```
    if p is not 0:  put new partition on stack between K0 and K1:
    >+>[<-                                 |K1|K0|Z=0|@a=pEQ0|b=p|
      move K0 to Z; search for last K
      <<[>+<-] <[<]                           |@0|Kn|```|K1|0|Z=K0|a=0|b=p| 
      shift left until return to 0 at K0;
      >[ [<+>-] >]                            |Kn|```|K1|0|@0|Z=K0|a=0|b=p|
      put p one left of there making it K1; restore K0 from Z;
      >>>[<<<<+>>>>-]<<[<+>-]                 |Kn|```|K2|K1=p|K0|@Z=0|a=0|b=0|
    else increment K0 (special case when first partition empty) 
    >>]<[- <<+>>]              
  >>>]  End if !empty
<<<<<<] End If Partitions remaining   @K1=0|K0=0|Z=0|a|b|c|d|e|f|g|X=0|A0|A1|```
Print the Results
share|improve this answer
+1 for actually building this in Brainf*ck! – ChristopheD Jan 19 '12 at 23:03
I was working on a similar solution but couldn't quite get it working. Awesome idea to do the partitioning that way. I was pulling out one element at a time and replacing it, and it got quite cumbersome quite quickly. I was also 1.5k into it, so you destroyed me on efficiency too. – captncraig Jan 20 '12 at 17:01
Everything in BF gets cumbersome quite quickly :) Even seemingly simple things like how to do an efficient if (i<j) {} else {} took several tries to get right. And the edge cases are killers. I don't know how many times I thought "just this one little thing left..." and then discovered a test case which caused another several hours work to handle. I think I can reduce it by a few dozen characters, but I'm not sure I want to put in the effort. – AShelly Jan 20 '12 at 19:18
One word: wow! I honestly didn't think it was humanly possible. I'm going to run a few inputs through it just to see how it works :-) – Ronald Jan 23 '12 at 0:05
Epic! Just epic! – vsz Jun 29 '12 at 18:04

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