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About Code-Bowling:

In golf, you try to get the lowest score (smallest application, most elegant, etc). In Bowling, you try to get the highest score. So if you follow, the goal of a Code-Bowling challenge is to make the biggest, most bastardized, hardest to maintain piece of code that still meets the requirements of the challenge. However, there's no point in making source longer just for the sake of it. It needs to seem like that added length was from design and not just padding.

The Challenge:

Create a program that sorts a list of numbers in ascending order.

Example:

Input: 1, 4, 7, 2, 5

Output: 1, 2, 4, 5, 7

Code: Obviously this wouldn't be a good answer, since there aren't many WTFs in there

function doSort(array $input) {
    sort($input);
    return $input;
}

Rules:

There are no real rules. As long as the program functions, have at it!

Remember: This is code-bowling, not golf. The objective is to make the worst, most bastardized code that you can! Bonus points for code that looks good, but is actually deceptively evil...

share|improve this question
    
I wish I still had the calendar queue code I wrote once... –  Peter Taylor Feb 17 '11 at 20:21
12  
sort(sort(sort(sort(sort(sort(myarray)))))) Guarantees perfect sorting! –  muntoo Mar 12 '11 at 22:55

31 Answers 31

Sorting by user input. What can be worse?

function sort(array) {
    newArray = new Array
    notSorted = true;
    while (notSorted) {
        sortfail = false;
        i = -1
        for (;;) {
            if ((++i + 1) == array.length) break;
            if (confirm("is " + array[i] + " less than " + array[i + 1])) {
                newArray[i] = array[i]
                newArray[i + 1] = array[i + 1]
            } else {
                newArray[i] = array[i + 1]
                newArray[i + 1] = array[i]
                array[i] = newArray[i]
                array[i + 1] = newArray[i + 1]
                sortfail = true
            }
        }
        array = newArray;
        if (!sortfail) notSorted = false
    }
    return array
}

console.log(sort([2,1,3]))

See live example

share|improve this answer
    
You can do better than that... :-D –  ircmaxell Feb 3 '11 at 22:43
    
Better and still sort correctly? That's a pain. This is algorithmically horrible. I don't need to make the code that much more horrible. –  Raynos Feb 3 '11 at 22:46
5  
User input is often wrong. Users may be drunk, or high, and there may also be adversaries who give you the wrong order on purpose. Should randomly shuffle and ask again until you have a fair number of answers for the same pairing. –  Paul Mar 15 '11 at 10:26
    
@Paul the lack of defence on user input is one of the major advantages of this code bowl. I guess having completely useless validation on user input would be better then none. –  Raynos Mar 15 '11 at 12:36
    
not sure about the manual element of this. –  Steve Robbins Dec 12 '11 at 19:52
class Array
  def sort
    self.permutation.min
  end
end

Ah, the elegance of Ruby... self.permutation results in an enumerator. No harm done, yet. The innocent looking .min however sucks this enumerator into an array. This array's size explodes when the number of elements rises. Not to mention this ruines a fine pre-existing sort.

share|improve this answer
2  
It's code bowling if you haven't seen it before – you must write the longest code! ;) –  nyuszika7h Feb 15 '11 at 6:42
2  
I'm going for the 'good looking but deceptively evil' bonus :) –  steenslag Feb 15 '11 at 22:46
1  
I'd say that this looks perfectly fine to any SQL programmer. Now the compiler just has to figure that what you wanted was just a simple sort. –  eBusiness Mar 3 '11 at 16:30

Bogosort!

import random

def is_sorted(seq):
    for x, y in zip(seq[:-1], seq[1:]):
        if x > y:
            return False
    return True

def sort(seq):
    while not is_sorted(seq):
        random.shuffle(seq)
share|improve this answer
2  
+1, that is beautiful! –  Ben Jan 1 '12 at 19:30
    
damnit you beat me to it... –  arrdem Dec 21 '12 at 21:27

Perl Bubble Sort

push(@m,'0');$a=<>;push(@m,$a);
$a=pop@m;push(@c,$a);
push(@m,'p');$a=pop@m;$b=pop@m;$hash{$a}=$b;
$a=pop@c;push(@c,$a);for(1..$a){$a=<>;push(@m,$a);
push(@m,'p');$a=pop@m;push(@m,$hash{$a});
$a=pop@m;$b=pop@m;$hash{$a}=$b;
push(@m,'p');$a=pop@m;push(@m,$hash{$a});
push(@m,'1');$a=pop@m;$b=pop@m;push(@m,$b+$a);
push(@m,'p');$a=pop@m;$b=pop@m;$hash{$a}=$b;
}$a=pop@c;push(@m,$a);
push(@m,'2');$a=pop@m;$b=pop@m;for(1..$a){push(@m,$b);};
$a=pop@m;$b=pop@m;push(@m,$b*$a);
$a=pop@m;push(@c,$a);
$a=pop@c;push(@c,$a);for(1..$a){push(@m,'1');$a=pop@m;push(@m,-$a);
push(@m,'p');$a=pop@m;$b=pop@m;$hash{$a}=$b;
$a=pop@c;push(@c,$a);for(1..$a){push(@m,'p');$a=pop@m;push(@m,$hash{$a});
push(@m,'1');$a=pop@m;$b=pop@m;push(@m,$b+$a);
push(@m,'p');$a=pop@m;$b=pop@m;$hash{$a}=$b;
push(@m,'p');$a=pop@m;push(@m,$hash{$a});
$a=pop@m;push(@m,$hash{$a});
push(@m,'p');$a=pop@m;push(@m,$hash{$a});
push(@m,'1');$a=pop@m;push(@m,-$a);
$a=pop@m;$b=pop@m;push(@m,$b+$a);
$a=pop@m;push(@m,$hash{$a});
$a=pop@m;$b=pop@m;if($b < $a){push(@c,1)}else{push(@c,0)};
$a=pop@c;push(@c,$a);for(1..$a){push(@m,'p');$a=pop@m;push(@m,$hash{$a});
$a=pop@m;push(@m,$hash{$a});
push(@m,'t');$a=pop@m;$b=pop@m;$hash{$a}=$b;
push(@m,'p');$a=pop@m;push(@m,$hash{$a});
push(@m,'1');$a=pop@m;push(@m,-$a);
$a=pop@m;$b=pop@m;push(@m,$b+$a);
$a=pop@m;push(@m,$hash{$a});
push(@m,'p');$a=pop@m;push(@m,$hash{$a});
$a=pop@m;$b=pop@m;$hash{$a}=$b;
push(@m,'t');$a=pop@m;push(@m,$hash{$a});
push(@m,'p');$a=pop@m;push(@m,$hash{$a});
push(@m,'1');$a=pop@m;push(@m,-$a);
$a=pop@m;$b=pop@m;push(@m,$b+$a);
$a=pop@m;$b=pop@m;$hash{$a}=$b;
}$a=pop@c;push(@m,$a);
$a=pop@m;
}}push(@m,'0');push(@m,'p');$a=pop@m;$b=pop@m;$hash{$a}=$b;
$a=pop@c;push(@c,$a);for(1..$a){push(@m,'p');$a=pop@m;push(@m,$hash{$a});
$a=pop@m;push(@m,$hash{$a});
$a=pop@m;print$a;
push(@m,'p');$a=pop@m;push(@m,$hash{$a});
push(@m,'1');$a=pop@m;$b=pop@m;push(@m,$b+$a);
push(@m,'p');$a=pop@m;$b=pop@m;$hash{$a}=$b;
}

I was going for the "every single line looks like the same piece of line noise" look. The first line of input (on STDIN) tells the program how many numbers there are, while the next N number of lines contain one number that needs to be sorted.

share|improve this answer
1  
stack-based perl :) –  jozefg Dec 22 '12 at 0:36

Generic, multithreaded bogosort in Java

46 seconds to sort 4 numbers

I thought it would be elegant with support for generics. Also, multithreading is always nice, so I use that instead of randomization: This program generates one thread for each number to be sorted. Each thread tries to insert its element into an array object, and when all elements have been inserted, the program will check if the array is sorted. If not, try again. Off course this insertion needs to be synchronized.

ElementInserter

This will be the class that we use for the threads. This class holds one element and tries to insert it into its sortedArray:

class ElementInserter<E extends Comparable<? super E>> implements Runnable {
    E element;
    SortedArray<? super E> target;

    ElementInserter(E e, SortedArray<? super E> a) {
        element = e;
        target = a;
    }

    @Override
    public void run() {
        target.insert(element);
    }
}

sortedArray

Sports a straight-forward insert method. When last element has been inserted, it checks if the array is sorted.

class SortedArray<E extends Comparable<? super E>> {
    boolean sorted; int i = 0; E[] a;

    SortedArray(E[] e) { a = e; }

    synchronized public void insert(E e) {
        while (!sorted) {
            a[i++] = e;

            if (i == a.length) {
                sorted = true;

                for (int j = 1; j < i; ++j)
                    if (a[j-1].compareTo(a[j]) > 0) {
                        sorted = false;
                        i = 0;
                        notifyAll();
                    }

                if (sorted) {
                    for (int j = 0; j < a.length; ++j) System.out.print(a[j]+" ");
                    System.out.println();
                    notifyAll(); // allow all threads to terminate
                }

            } else
                try { wait(); }
                catch(InterruptedException x) { }
        }
    }
}

Main method

Parses command line arguments as Integers, creates a new array and a new thread for each Integer:

    public static void main (String[] args) {
        Integer[] i = new Integer[args.length];
        SortedArray<Integer> c = new SortedArray<Integer>(i);

        for (String s : args)
            new Thread(new ElementInserter<Integer>(Integer.parseInt(s), c)).start();
    }

Test run

Execution time will depend on the original order of the elements and your scheduler implementation. This is on a dual-core 2.9 GHz Intel i7 MacBook Pro:

$ time java Main 3 1 2 4
1 2 3 4 

real    0m46.307s
user    0m14.629s
sys     0m26.207s
share|improve this answer
1  
Haha +1 I got a kick out of this –  arshajii Dec 22 '12 at 13:11

The F# "I hate functional programming" :

let numbers = System.Console.ReadLine().Split(',')
let mutable check = 0
let mutable sorted = ""

while not (sorted.Split(',').Length = numbers.Length) do
    for i in [check..check + 1000] do
        for number in numbers do
            if number = i.ToString() then
                if sorted.Length > 0 then
                    sorted <- sorted + "," 
                sorted <- sorted + number
    check <- check + 1000

printfn "%s" sorted
share|improve this answer
1  
+1: That's just foul! Even for an imperative algorithm. –  cfern Feb 17 '11 at 8:45

Though I can't take credit for this Java code, Smoothsort is a good example of the tradeoff between readability and performance:

static final int LP[] = { 1, 1, 3, 5, 9, 15, 25, 41, 67, 109,
  177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891,
  35421, 57313, 92735, 150049, 242785, 392835, 635621, 1028457,
  1664079, 2692537, 4356617, 7049155, 11405773, 18454929, 29860703,
  48315633, 78176337, 126491971, 204668309, 331160281, 535828591,
  866988873 // the next number is > 31 bits.
};

public static <C extends Comparable<? super C>> void sort(C[] m,
    int lo, int hi) {
  int head = lo; // the offset of the first element of the prefix into m

  int p = 1; // the bitmap of the current standard concatenation >> pshift
  int pshift = 1;

  while (head < hi) {
    if ((p & 3) == 3) {
      sift(m, pshift, head);
      p >>>= 2;
      pshift += 2;
    } else {
      // adding a new block of length 1
      if (LP[pshift - 1] >= hi - head) {
        // this block is its final size.
        trinkle(m, p, pshift, head, false);
      } else {
        // this block will get merged. Just make it trusty.
        sift(m, pshift, head);
      }

      if (pshift == 1) {
        // LP[1] is being used, so we add use LP[0]
        p <<= 1;
        pshift--;
      } else {
        // shift out to position 1, add LP[1]
        p <<= (pshift - 1);
        pshift = 1;
      }
    }
    p |= 1;
    head++;
  }

  trinkle(m, p, pshift, head, false);

  while (pshift != 1 || p != 1) {
    if (pshift <= 1) {
      // block of length 1. No fiddling needed
      int trail = Integer.numberOfTrailingZeros(p & ~1);
      p >>>= trail;
      pshift += trail;
    } else {
      p <<= 2;
      p ^= 7;
      pshift -= 2;

      trinkle(m, p >>> 1, pshift + 1, head - LP[pshift] - 1, true);
      trinkle(m, p, pshift, head - 1, true);
    }

    head--;
  }
}

private static <C extends Comparable<? super C>> void sift(C[] m, int pshift,
    int head) {   
  C val = m[head];

  while (pshift > 1) {
    int rt = head - 1;
    int lf = head - 1 - LP[pshift - 2];

    if (val.compareTo(m[lf]) >= 0 && val.compareTo(m[rt]) >= 0)
      break;
    if (m[lf].compareTo(m[rt]) >= 0) {
      m[head] = m[lf];
      head = lf;
      pshift -= 1;
    } else {
      m[head] = m[rt];
      head = rt;
      pshift -= 2;
    }
  }  

  m[head] = val;
}

private static <C extends Comparable<? super C>> void trinkle(C[] m, int p,
    int pshift, int head, boolean isTrusty) {

  C val = m[head];

  while (p != 1) {
    int stepson = head - LP[pshift];

    if (m[stepson].compareTo(val) <= 0)
      break; // current node is greater than head. Sift.

    if (!isTrusty && pshift > 1) {
      int rt = head - 1;
      int lf = head - 1 - LP[pshift - 2];
      if (m[rt].compareTo(m[stepson]) >= 0
          || m[lf].compareTo(m[stepson]) >= 0)
        break;
    }

    m[head] = m[stepson];

    head = stepson;
    int trail = Integer.numberOfTrailingZeros(p & ~1);
    p >>>= trail;
    pshift += trail;
    isTrusty = false;
  }

  if (!isTrusty) {
    m[head] = val;
    sift(m, pshift, head);
  }
}

/* insert some basic static void main here... 
   my Java's too rusty to do it from the top of my head 
*/

(note: some comments removed for effect and to make it shorter; source taken from the wikipedia page linked above)

share|improve this answer
    
Could you please add the language you used for this to the post? –  Nathan Osman Feb 14 '11 at 7:04

A Cobol subprogram to sort a table of integers, guaranteed to have a higher WTF/minute than any other language. For performance purposes, the QuickSort algorithm is used:

Identification Division.                                        
Program-ID. QwikSort is recursive.                              
Environment Division.                                           
Data Division.                                                  
Working-Storage Section.                                        
01 QwikSort-Working-Storage.                                    
 05 Swap-Space           Pic X(80) Value spaces.               
 05 Pivot         Binary Pic S9(8).                            
 05 I             Binary Pic S9(8).                            

Local-Storage Section.                                          
01 QwikSort-Local-Storage.                                      
 05 Lo            Binary Pic S9(8) Value 0.                    
 05 Hi            Binary Pic S9(8) Value 0.                    

Linkage Section.                                                

01 The-Table-Area.                                              
 05 The-Table     occurs 0 to 200000 depending on High-Element.
  10 The-Key     Binary Pic S9(8).                            

01 Low-Element     Binary Pic S9(8).                            
01 High-Element    Binary Pic S9(8).                            

Procedure Division using The-Table-Area                         
                         Low-Element                            
                         High-Element.                          

  Compute Lo  = Low-Element                                   
  Compute Hi  = High-Element                                  

  If ( High-Element > Low-Element )                           
     Perform Select-Pivot          

* ----- Loop through table until indices cross                   
       Perform until Hi < Lo                                    

* -------- Locate an item that should not be in less partition    
          Perform varying Lo from Lo by 1                        
            until (( Lo >= High-Element )                        
               or  ( The-Key of The-Table ( Lo ) >= Pivot ))     
          End-Perform                                            


* -------- Locate an item that should not be in greater partition 
          Perform varying Hi from Hi by -1                       
            until (( Hi <= Low-Element )                         
               or  ( The-Key of The-Table ( Hi ) <= Pivot ))     
          End-Perform                                            

* -------- Exchange the two and keep looking                      
          If ( Lo <= Hi )                                        
             Perform Swap-Elements                               
             Compute Lo = Lo + 1                                 
             Compute Hi = Hi - 1                                 
          End-If                                                 

    End-Perform                                               

    Perform Qsort-Less-Partition                              

    Perform Qsort-Greater-Partition                           

 End-If                                                       

 Goback.                                                      

*--------------------------------------------------------------*  
* Select the pivot using median of three rule.                    
*--------------------------------------------------------------*  
 Select-Pivot.                                                    

    Compute Pivot = ( Lo + Hi ) / 2                              
    Compute Pivot = Function Median (                            
            The-Key of The-Table ( Lo )                          
            The-Key of The-Table ( Pivot )                       
            The-Key of The-Table ( Hi )                          
         )                                                       
    End-Compute                                                  

    Exit.                                                        

*--------------------------------------------------------------*  
* Exchange elements in the wrong partition                        
*--------------------------------------------------------------*  
 Swap-Elements.                                                   
    Move The-Table ( Lo ) to Swap-Space                          
    Move The-Table ( Hi ) to The-Table ( Lo )                    
    Move Swap-Space       to The-Table ( Hi )                    
    Exit.                   

*--------------------------------------------------------------*  
* Sort the less sub-partition                                     
*  -- Optimization opportunity for the user: if the partition     
*    size is sufficiently small you might want to apply a simple  
*    sort like insertion or bubble, or perhaps a bose-nelson      
*    network to order the last few and save the overhead of       
*    another recursive call.                                      
*--------------------------------------------------------------*  
 Qsort-Less-Partition.                                            
    If ( Low-Element < Hi )                                      
       Call 'QwikSort' Using                                     
          The-Table-Area                                         
          Low-Element                                            
          Hi                                                     
       End-Call                                                  
    End-If 
    Exit.                                                        

*--------------------------------------------------------------*  
* Sort the greater sub-partition                                  
*  -- Optimization opportunity for the user: same as left part.   
*--------------------------------------------------------------*  
 Qsort-Greater-Partition.                                         
    If ( Lo < High-Element )                                     
       Call 'QwikSort' Using                                     
          The-Table-Area                                         
          Lo                                                     
          High-Element                                           
       End-Call                                                  
    End-If                              
    Exit.                                                        

 End Program QwikSort.
share|improve this answer
    
Is there a reason why it's QwikSort instead of QuickSort? (i.e. length restriction or you can't use the letter u in a name?) Because that would make it even more fun :P –  marinus Dec 21 '12 at 0:25
    
@marinus It's probably a linker restriction. –  Craig Trader Sep 30 at 3:48
var inp = [1, 4, 7, 2, 5],
    loops = 0,
    trysort;

while(++loops) {
    trysort = new Array(inp.length);

    for(var i = 0, len = inp.length; i < len; i++) {
        trysort[i] = inp[~~(Math.random()*inp.length)];
    }

    if( trysort.join(',') === '1,2,4,5,7' )
        break;
}
alert('done! Only ' + loops + ' iterations! ' + trysort);

http://www.jsfiddle.net/sAFMC/

share|improve this answer
    
hard coded in the array tut tut. –  Raynos Feb 3 '11 at 22:59
    
Bogosort for the win! \o/ –  Chris Jester-Young Feb 3 '11 at 23:05
    
nice. i like it. –  Vaibhav Feb 3 '11 at 23:09

Python quicksort using lambda

Something I wrote on my blog:

qsort = lambda seq: [] if not seq else qsort(filter(lambda n: n<=seq[0], seq[1:]))+[seq[0]]+qsort(filter(lambda n: n>seq[0], seq[1:]))

Here's the actual blog post

share|improve this answer
1  
Nice solution! The shortest known quicksort in Python is only 79 bytes long (cortesy legendary codegolfer Mark Byers): news.e-scribe.com/314#581 –  hallvabo Dec 22 '12 at 10:10
    
That's a very esoteric solution. Blew my mind! –  dougk Dec 26 '12 at 19:27

Ruby

#get user input from standard input and store it in a variable
var = gets
#make user input into a ruby array in a string
array_string = '[' + var + ']'
#evaluate this ruby array and store the array in a variable
array = eval(array_string)
#convert from strings to integers
#initialize loop variable as 0
i = 0
#loop until the loop variable is no longer small enough
#to be a valid index of the array
until !(i < array.length)
    #convert to a number then ensure it is an integer
    array[i] = array[i].to_f.floor
    #increment i
    i = i + 1
end

sorted = false

#infinite loop
while true
    #if it hasn't been sorted
    if ((!sorted) == true)
        #try to sort it
        #initialize loop variable as 0
        j = 0
        #loop while i is less than or equal to the length of the array
        while (i <= array.length)
            random = rand # get a random number
            #this gets a zero or one
            random = (random + random.round).floor
            #turn random into a boolean
            if random == 0
                random = true
            end
            if random == 1
                random = false
            end
            #don't do anything if this is the last one
            if j == (array.length - 1)
                #do nothing
            else
                if random == true
                    #true means keep these 2 elements in the same order
                end
                if random == false
                    #false means swap with next element
                    #initialize a variable to store the current element
                    swapvar = array[j]
                    #initialize a variable to store the next element
                    swapvar2 = array[j+1]
                    #set the current element to what was the next element
                    array[j] = swapvar2
                    #set the next element to what was in this one
                    array[j+1] = swapvar
                end
            end
            #increment j
            j = j + 1
            #loop kept looping, fixed now
            if j >= array.length
                break
            end
        end
        #now test if it is sorted
        #lets say it is
        sorted = true
        #now see if we are wrong
        #initialise a loop variable to zero
        k = 0
        while (array.length > k)
            #get current value from array and store it in a variable
            current = array[k]
            #get current value from array and store it in a variable
            next_ = array[k + 1]
            if !(array[k +1].nil? == true)#stop errors
                #if sorted correctly
                if next_ < current || next_ == current
                    #sorted stil== true
                else
                    sorted = false #not sorted, try again
                end
            end
            #increment loop variable
            k = k + 1
        end
            #DEBUG
        #p array
    else
        break #break out of infinite loop
    end
end


#it is now sorted in descending order
#we want ascending so lets flip it

#initialise a loop variable to zero
l = 0

final_array = Array.new([])

while (l <= (array.length - 1))
    final_array[array.length - l - 1] = array[l]
    #increment loop variable
    l += 1
end

#now we need to make it into a string
str = ''

#first map each element to a string
#to do this lets extend Array, then use this for our array
class Array
    def map #map each element to a string
        #initialise a loop variable to zero
        @m = 0
        arr = []
        while @m < self.length
            arr += [self[@m].to_s]
            #increment loop variable
            @m = @m + 1
        end
        return arr
    end
end

final_array = Array.new(final_array).map()
#DEBUG
#p final_array

#now add the elements to a string, separated by commas and spaces
#initialise a loop variable to zero
n = 0

while n < array.length
    #increment loop variable
    #add item to string
    str = str + final_array[n]
    #if not last item
    if !((array.length - 1 )== n)
        #add comma
        str += ','
    end
    #if not last item
    if !((array.length - 1 )== n)
        #add space
        str += ' '
    end
    n = n + 1
end

#now print the string
puts str
share|improve this answer

C++ (4409)

#include <iostream>
#include <cstdlib>
#include <cctype>
#include <string>
#include <sstream>


std::string itostr(int number)
{
   std::stringstream ss;
   ss << number;
   return(ss.str());
}


class CStackLIFO
{
private:
    int* stack;
    size_t len;
public:
    CStackLIFO();
    CStackLIFO(int);
    ~CStackLIFO();
    void init();
    void init(int);
    void push(int);
    int pop();
    int pop(int&);
    void destroy(size_t);
    int get(size_t);
    size_t get_len();
};


void CStackLIFO::init()
{
    len = 0;
    stack = NULL;
}


void CStackLIFO::init(int val)
{
    len = 1;
    stack = new int[len];
    stack[0] = val;
}


CStackLIFO::CStackLIFO()
{
    init();
}


CStackLIFO::CStackLIFO(int val)
{
    init(val);
}


CStackLIFO::~CStackLIFO()
{
    len = 0;
    delete[] stack;
    stack = NULL;
}


void CStackLIFO::push(int val = NULL)
{
    if((stack == NULL) || (len == 0))
    {
        this->init(val);
    }
    else
    {
        int* buf = stack;
        len++;
        stack = new int [len];

        for(size_t i = 0; i < (len - 1); i++)
        {
            stack[i] = buf[i];
        }

        stack[len - 1] = val;
        delete[] buf;
    }
}


int CStackLIFO::pop()
{
    int val;
    int* buf = stack;
    len--;
    stack = new int [len];

    for(size_t i = 0; i < len; i++)
    {
        stack[i] = buf[i];
    }

    val = buf[len];
    delete[] buf;

    return(val);
}


int CStackLIFO::pop(int &out)
{
    if(len <= 0)
        return(false);

    int* buf = stack;
    len--;
    stack = new int [len];

    for(size_t i = 0; i < len; i++)
    {
        stack[i] = buf[i];
    }

    out = buf[len];
    delete[] buf;

    return(true);
}


void CStackLIFO::destroy(size_t idx)
{
    int* buf = stack;
    len--;
    stack = new int [len];
    for(size_t i = 0; i < idx; i++)
    {
        stack[i] = buf[i];
    }
    for(size_t i = idx; i < len; i++)
    {
        stack[i] = buf[i + 1];
    }
    delete[] buf;
}


int CStackLIFO::get(size_t idx)
{
    return(stack[idx]);
}


size_t CStackLIFO::get_len()
{
    return(len);
}


class CSort
{
private:
    void removeWhitespace();
public:
    std::string in;
    void sort();
    void display();
};


void CSort::display()
{
    std::cout << in << std::endl;
    return;
}


void CSort::removeWhitespace()
{
    for(size_t i = 0; i < in.size(); i++)
    {
        if(in[i] == ' ')
        {
            in = in.substr(0, i) + in.substr(i+1);
        }
    }
}


void CSort::sort()
{
    removeWhitespace();

    CStackLIFO nums;
    CStackLIFO s;

    // build stack
    for(size_t i = 0, offset = 0; i < in.size(); i++)
    {
        for(offset = 0; (i + offset) < in.size(); offset++)
        {
            if(!(isdigit(in[i + offset]) ||
                 (in[i + offset] == '-')))
                break;
        }

        if(offset > 0)
        {
            nums.push(atoi(in.substr(i, offset).c_str()));
        }

        i += offset;
    }

    // sort
    for(size_t lowest = 0; 0 < nums.get_len(); lowest = 0)
    {
        for(size_t i = 0; i < nums.get_len(); i++)
        {
            if(nums.get(i) < nums.get(lowest))
                lowest = i;
        }

        s.push(nums.get(lowest));
        nums.destroy(lowest);
    }    

    // convert to string
    in = "";
    for(size_t i = 0; i < s.get_len(); i++)
    {
        // Convert num to string
        in += itostr(s.get(i));
        in += ", ";
    }
    if(in.size() > 2)
        in = in.substr(0, in.size() - 2);
}


int main(int argc, char* argv[])
{
    while(1)
    {
        CSort unsorted;
        unsorted.in = "";

        while(1)
        {
            std::string userin = "";
            std::cin >> userin;

            if(userin.compare("end") == 0)
                break;

            unsorted.in += userin + ", ";
        }

        unsorted.in = unsorted.in.substr(0, unsorted.in.size() - 2);

        CSort sorted;
        sorted.in = unsorted.in;
        sorted.sort();

        std::cout << "Input:" << std::endl;
        unsorted.display();
        std::cout << "Output:" << std::endl;
        sorted.display();
    }

    return(0);
}

Some things "bad" about this program:

  • Really stringy. :) I could have simply inputted the numbers directly, instead of parsing the string for numbers.

  • Uses excessive classes. I could have used the built-in ones, too, but a bad coder would simply reinvent the wheel.

  • The loop for the sort is terribly inefficient. I think it's the slowest way possible, without making it look like I actually tried to make it slow. Actually, the actual "sort" part of the code is only like 11 lines, including the curly braces and line breaks.

share|improve this answer

This one is O(n*n!)

from itertools import permutations, izip, islice
L=[10,9,8,7,6,5,4,3,2,1]
for l in permutations(L):
    if all([i<j for i,j in izip(L, islice(L,1,None))]):
        break
print l

It iterates all the permutations of the list and tests whether they are sorted. It is so awful that sorting just 10 items takes 17 seconds

$ time python badsort.py 
(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

real    0m17.563s
user    0m17.537s
sys     0m0.020s
share|improve this answer

Ruby Metasort

#I know Ruby has a built-in method to order an array,
#but I can't remember what it is.  Oh well, metaprogramming to the rescue!
#I'll just try all of the available methods, and see which one works.

#check whether an array is ordered
def is_in_order? arr
  arr.is_a?(Array) && (0...(arr.size-1)).all? {|i| arr[i]<arr[i+1]}
end

Array.instance_methods.each do |meth|
  begin
    possibly_ordered_array = ARGV.map(&:to_i).send(meth)
    #have to check both that the new array is ordered, and that it's still the same size,
    #so that methods that change the array's elements don't create false positives.
    if is_in_order?(possibly_ordered_array) && possibly_ordered_array.size == ARGV.size
      puts possibly_ordered_array.join(", ")
      exit 0 #success!
    end
  rescue
    #method needed an argument or something
  end
end

exit 1 #could not order the array
share|improve this answer

JavaScript (with animation!). 8172 chars. Few hours for 6 numbers.

We like lottery, right? Similar to Bogosort by dan04, but using physics and animation...

(function(){var e={},v=!1,G=/abc/.test(function(){abc})?/\b_super\b/:/.*/,H=function(c,a,b){return function(){var l=this._super,s;this._super=b[c];try{s=a.apply(this,arguments)}finally{this._super=l}return s}};e.Class=function(){};e.Class.extend=function a(b){var l=this.prototype;v=!0;var s=new this;v=!1;for(var c in b)s[c]="function"===typeof b[c]&&"function"===typeof l[c]&&G.test(b[c])?H(c,b[c],l):b[c];b=function(){!v&&this.init&&this.init.apply(this,arguments)};b.prototype=s;b.prototype.constructor=
b;b.extend=a;return b};e.Events={};var k=e.Events,z,A,w=Object.prototype.toString,I=document.createEvent,r=Date.now;k.EventListener=e.Class.extend({handleEvent:function(){},init:function(a){this.handleEvent=a}},"EventListener");z=k.EventListener;k.Event=e.Class.extend({timeStamp:0,type:"",target:null,cancelable:!1,defaultPrevented:!1,preventDefault:function(){this.cancelable&&(this.defaultPrevented=!0)},initEvent:function(a,b){this.type=a;this.target=b;this.timeStamp=r()}},"Event");A=k.Event;var t=
Object.create(null);e.createEventType=function(a,b){t[a]=A.extend(b,a);return t[a]};e.createEvent=function(a){return null!=t[a]?(a=new t[a],a.timeStamp=r(),a):I(a)};k.EventTarget=e.Class.extend({_listeners:{},addEventListener:function(a,b,l){var c=this._listeners[a];if(!(null!=c&&function(){for(var a="object"===typeof l?l:null,d,e=0,f=c.length;e<f;e++)if(d=this.listeners[e],d.listener.handleEvent===b&&d.scope===a)return!0;return!1}())){var d={listener:"object"===typeof b&&b.handleEvent?b:new z(b),
scope:"object"===typeof l?l:null};"[object Array]"===w.call(c)?c.push(d):this._listeners[a]=[d]}},removeEventListener:function(a,b,c){a=this._listeners[a];if("[object Array]"===w.call(a)){c="object"===typeof c?c:null;for(var d=0,e=a.length;d<e;d++)if(a[d].listener.handleEvent===b&&a[d].scope===c){a.splice(d,1);break}}},dispatchEvent:function(a){a.target||(a.target=this);var b=0,c=this._listeners[a.type],d=c.length,e;if("[object Array]"===w.call(c))for(;b<d;b++)e=c[b].listener.handleEvent,e.call(c[b].scope,
a);return a.defaultPrevented}},"EventTarget");for(var B=0,k=["ms","moz","webkit","o"],h=0;h<k.length&&!window.requestAnimationFrame;++h)window.requestAnimationFrame=window[k[h]+"RequestAnimationFrame"],window.cancelAnimationFrame=window[k[h]+"CancelAnimationFrame"]||window[k[h]+"CancelRequestAnimationFrame"];window.requestAnimationFrame||(window.requestAnimationFrame=function(a){var b=r(),c=Math.max(0,16-(b-B)),d=window.setTimeout(function(){a(b+c)},c);B=b+c;return d});window.cancelAnimationFrame||
(window.cancelAnimationFrame=function(a){clearTimeout(a)});var x=window.clearInterval,C=window.setInterval,D=e.Class.extend({fn:function(){},args:[],thisArg:window,sleepTime:0,init:function(a,b,c){this.method=a||function(){};this.thisArg=b||window;this.args=c||[]}},"FunctionToCall"),r=Date.now,u=Object.create(null);u.animation=0;u.steady=1;u.as_fast_as_possible=2;e.Timer=e.Events.EventTarget.extend({ANIMATION:0,STEADY:1,AS_FAST_AS_POSSIBLE:2,TimerEvent:e.createEventType("TimerEvent",{FPS:0,targetFPS:0,
delta:0,initEvent:function(a,b){if(null==b.getFPS||null==b.targetFPS||null==b.dt)throw notATimerException;this.FPS=b.getFPS();this.targetFPS=b.targetFPS;this.delta=b.dt;this._super(a,b)}}),_clearIntervalFlag:!1,_recursionCounter:0,_tickHandlersFromMethod:Object.create(null),running:!1,paused:!1,mode:0,autoStartStop:!0,setAutoStartStop:function(a){this.autoStartStop=a},prevTime:0,dt:0,maxDelta:0.05,setMaxDelta:function(a){this.maxDelta=a},sysTimerId:-1,targetFPS:50,setFPS:function(a){this.targetFPS=
a;if(this.running&&this.mode===this.STEADY){x(this.sysTimerId);var b=this;this.sysTimerId=C(function(){b.tick()},1E3/a)}},getFPS:function(){return 1/this.dt},queue:[],addToQueue:function(a,b,c){var d=new D(a,b,c),e;this.queue.push(d);var f=function J(f){0<d.sleepTime?d.sleepTime-=f.delta:(e=a.apply(b,[f].concat(c)),"number"===typeof e&&0<e?d.sleepTime=e:!0!==e&&(this.removeEventListener("tick",J,this),this.queue.splice(this.queue.indexOf(d),1)))};null==this._tickHandlersFromMethod[a]?this._tickHandlersFromMethod[a]=
[f]:this._tickHandlersFromMethod[a].push(f);this.addEventListener("tick",f,this);this.autoStartStop&&!this.running&&this.startLoop()},removeFromQueue:function(a){var b=this._tickHandlersFromMethod[a];a=this.queue.indexOf(a);if(-1!==a){for(var c=0,d=b.length;c<d;c++)this.removeEventListener("tick",b[c],this);return this.queue.splice(a,1)}},getQueue:function(){return this.queue.slice(0)},pause:function(){this._clearIntervalFlag=this.paused=!0;this.dt=0},resume:function(){this.paused=!1;this.startLoop()},
startLoop:function(){this.prevTime=r();this.running=!0;if(this.mode===this.steady){var a=this;this.sysTimerId=C(function(){a.tick()},1E3/this.targetFPS)}else this.tick()},tick:function(){if(this._clearIntervalFlag&&-1!==this.sysTimer)x(this.sysTimerId),this.sysTimerId=-1,this._clearIntervalFlag=!1;else if(!this.paused){var a=r();this.dt=(a-this.prevTime)/1E3;this.dt>this.maxDelta&&(this.dt=this.maxDelta);if(0===this.dt&&4>this._recursionCounter)this._recursionCounter++,this.tick();else{0<this._recursionCounter&&
(this._recursionCounter=0);var b=e.createEvent("TimerEvent");b.initEvent("tick",this);this.dispatchEvent(b);this.prevTime=a;a=!(this.autoStartStop&&0>=this.queue.length);if(this.mode===this.AS_FAST_AS_POSSIBLE&&a){var c=this;setTimeout(function(){c.tick()},0)}if(this.mode===this.ANIMATION&&a){var d=this;requestAnimationFrame(function(){d.tick()})}a||(this.running=!1,this.mode===this.STEADY&&-1!==this.sysTimerId&&(x(this.sysTimerId),this.sysTimerId=-1))}}},init:function(a){this.mode="number"===typeof a?
a:u[a];if(null==this.mode||0>this.mode||2<this.mode)this.mode=0}},"Timer");e.Timer.ANIMATION=0;e.Timer.STEADY=1;e.Timer.AS_FAST_AS_POSSIBLE=2;e.Timer.FunctionToCall=D;var j=[2,3,5,8,4,6],m=new (e.Class.extend({displace:0,velocity:0,drag:0.01,tickInterval:10,update:function(a){this.displace+=this.velocity*a.delta;this.velocity*=1-this.drag},init:function(){}})),q=j.slice(0),k=new e.Timer(0),h=document.createElement("div");h.style.width=23*j.length-28+"px";h.style.height="22px";h.style.overflowX="hidden";
h.style.overflowY="hidden";h.style.border="1px #555 solid";h.style.fontFamily='Consolas, "Courier New"';for(var c=[],f=0;f<j.length;f++)c[f]=document.createElement("div"),c[f].style.width="20px",c[f].style.height="20px",c[f].style.display="inline-block",c[f].style.position="relative",c[f].style.left=f+1===j.length?"-22px":"0px",c[f].style.top=f+1===j.length?"-22px":"0px",c[f].style.border="1px #AAA solid",c[f].style.textAlign="center",c[f].innerHTML=j[f],h.appendChild(c[f]);var g=0,d=1,n,p,E=0,F=
0,y=0;m.velocity=1E3*Math.random()+500;document.getElementsByTagName("body")[0].appendChild(h);k.addToQueue(function(a){if(5>m.velocity)if(0===g){for(a=0;a<c.length;a++)if((m.displace+22+22*a)%(22*j.length)-22<11*c.length&&(m.displace+22+22*a)%(22*j.length)>11*c.length){n=c[a];E=parseFloat(n.style.left);break}for(a=0;a<c.length;a++)if((m.displace+22+22*a)%(22*j.length)<11*c.length&&(m.displace+22+22*a)%(22*j.length)+22>11*c.length){p=c[a];F=parseFloat(p.style.left);break}y=p.offsetLeft-n.offsetLeft;
g=1}else if(1===g)n.style.backgroundColor="rgba(255,255,0,"+Math.round(1-d)+")",p.style.backgroundColor="rgba(255,255,0,"+Math.round(1-d)+")",d-=10*a.delta,0>d&&(g=2);else if(2==g)n.style.backgroundColor="rgba(255,255,0,"+Math.round(1-d)+")",p.style.backgroundColor="rgba(255,255,0,"+Math.round(1-d)+")",d+=10*a.delta,1<d&&(g=3,n.style.backgroundColor="rgba(255,255,255,0)",p.style.backgroundColor="rgba(255,255,255,0)",d=0);else if(3==g)n.style.left=E+d*y+"px",p.style.left=F-d*y+"px",d+=2*a.delta,1<
d&&(g=4);else if(4==g){if(d-=a.delta,0.5>d){a=c.indexOf(n);var b=c.indexOf(p),e;e=n.innerHTML;c[a].innerHTML=p.innerHTML;c[b].innerHTML=e;e=q[a];q.splice(a,1,q[b]);q.splice(b,1,e);b=-1/0;for(a=0;a<q.length;a++){if(q[a]<b){g=5;break}b=q[a]}g=5===g?5:6}}else if(5===g)d=1,m.velocity=1E3*Math.random()+500,g=0;else{if(6===g)return d=1,g=0,document.getElementsByTagName("body")[0].innerHTML+=q,!1}else{m.update(a);for(a=0;a<c.length;a++)b=(m.displace+22+22*a)%(22*j.length),a+1===j.length&&(b+=22*(j.length-
1)),c[a].style.left=b-22*a-22+"px"}return!0})})();​

However, I used a bit of Google Closure Compiler to, er... I don't know. But it looks uglier, right?

It's like one of those lottory wheels, and each time it stops it swaps to numbers in the middle. You can have a play here: http://jsfiddle.net/VkJUE/5/ (to understand what I mean)

It does work, a bit, except that it may take hours for 6 numbers. However, I tested it on 3 numbers and it works fine!

share|improve this answer

Python implementation of Funnel Sort. This algorithm can actually have good cache performance when implemented properly, which in this case it is most certainly not (it does, however, sort correctly).

class Funnel(object):
  """ In place funnel sorting algorithm. """
  def __init__(self, left_child, right_child):
    self.left_child = left_child
    self.right_child = right_child
    self.lookahead = (False, -1)

  def get_next(self):
    if self.left_child.has_next() and self.right_child.has_next():
      l, r = self.left_child.get_next(), self.right_child.get_next()
      if l < r:
        self.right_child.put_back(r)
        return l
      else:
        self.left_child.put_back(l)
        return r
    elif self.left_child.has_next():
      return self.left_child.get_next()
    elif self.right_child.has_next():
      return self.right_child.get_next()
    else:
      raise Exception("Out of elements.")

  def has_next():
    if self.lookahead[0]: 
      self.lookahead = (False, self.lookahead[1])
      return self.lookahead[1]
    return self.left_child.has_next() or self.right_child.has_next()

  def put_back(self, item):
    self.lookahead = (True, item)

class Funnel_Base(Funnel):
  def __init__(self, buffer):
    self.buf = buffer
    self.lookahead = (False, -1)

  def get_next(self):
    if self.lookahead[0]:
      self.lookahead = (False, self.lookahead[1])
      return self.lookahead[1]
    return self.buf.pop(0)

  def has_next():
    return len(self.buf) > 0

def make_funnel(funnels):
  while len(funnels) > 1: funnels.append(Funnel(funnels.pop(0), funnels.pop(0)))
  return funnels[0]

def insertion_sort(array):
  def swap(i, j):
    t = array[i]
    array[i] = array[j]
    array[j] = t
  for i in xrange(1,len(array)):
    while i > 0 and  array[i] < array[i-1]:
      swap(i, i-1)
      i -= 1

def funnel_sort_internal(array):
  if len(array) < 100:
    # sort subsections with insertion sort and make base funnels
    insertion_sort(array)
    return Funnel_Base(array)
  else:
    K = int(len(array)**(1.0/3))
    new_funnels = [funnel_sort_internal(array[base:min(base+K, len(array))])
                   for base in xrange(0, len(array), K)]
    return make_funnel(funnels)

def funnel_sort(array):
  cp = array[:]
  srtd = funnel_sort_internal(cp)
  for i in xrange(len(array)): array[i] = cp[i]
share|improve this answer

Sorts a list of 32-bit integers. It is actually quite efficient for most ordinary cases:

#include <assert.h>
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>

static unsigned char *bmp[131072];
static size_t total, printed;

static void set(int n);
static void print_page(unsigned char *page, int offset);
static void print_number(int n);

int main(void)
{
    int i, n;

    assert(INT_MIN == -2147483647 - 1 && INT_MAX == 2147483647);

    while (scanf("%d,*", &n) > 0) {
        set(n);
        total++;
    }

    for (i = 0; i < 131072; i++)
        if (bmp[i] != NULL)
            print_page(bmp[i], INT_MIN + i * 32768);
    putchar('\n');

    return 0;
}

static void set(int n)
{
    unsigned int i = n - INT_MIN;

    if (bmp[i >> 15] == NULL)
        bmp[i >> 15] = calloc(4096, sizeof(unsigned char));

    bmp[i >> 15][(i >> 3) & 4095] |= (unsigned int)1 << (i & 7);
}

static void print_page(unsigned char *page, int offset)
{
    int i, j;

    for (i = 0; i < 4096; i++, offset += 8)
        if (page[i])
            for (j = 0; j < 8; j++)
                if (page[i] & (1 << j))
                    print_number(offset + j);
}

static void print_number(int n)
{
    printf("%d%s", n, ++printed < total ? ", " : "");
}

Example:

$ echo '1, 4, -3, 7, -2147483648, 2147483647, 2, 5' | ./bitmap-sort
-2147483648, -3, 1, 2, 4, 5, 7, 2147483647
share|improve this answer
<?php
    /*  THIS SOFTWARE IS PROVIDED "AS IS" AND ANY
        EXPRESSED OR IMPLIED WARRANTIES, INCLUDING,
        BUT NOT LIMITED TO, THE IMPLIED WARRANTIES
        OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
        PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE
        REGENTS OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT,
        INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
        CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED
        TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES;
        LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
        INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF
        LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY,
        OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING
        IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF
        ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.          */

    // Make the function
    function sort_numbers($nums) {
        // Remove duplicates from array
        array_unique($nums);
        // Filter out everything that's not a number or a numeric string
        foreach ($nums as $val) {
            if (!is_numeric($val)) {
                unset($array[$val]);
            }
        }
        // Do some random computations
        class SorterFunction {
            public $timestamp = 0;
            public $rand_num = 0;
            public function sort_a_numeric_array($num_arr) {
                // Sort the array
                sort($num_arr, SORT_NUMERIC);
                return $num_arr;
            }
            public function __construct() {
            $this->timestamp = time();
            $this->rand_num = rand() % 100000;
        }
    }
    // Create a new SorterFunction class
    $MySorterFunction = new SorterFunction();
    // Print out the sorted array's elements using var_dump
        $MySorterFunction->sort_a_numeric_array($nums);
        var_dump($nums);
    }
    $my_nums = array(1, 4, 7, 2, 5);
    sort_numbers($my_nums);
?>
share|improve this answer
    
That won't even work, since sort() works by reference and doesn't even return anything... –  ircmaxell Feb 4 '11 at 19:53
    
@ircmaxell: Fixed, thanks. –  nyuszika7h Feb 4 '11 at 20:44
    
sigh. (although it is a wtf, so it technically counts). But it still doesn't work (wrong sort change) –  ircmaxell Feb 4 '11 at 20:47
1  
+1 for license, but it could have covered more lines. –  user475 Feb 14 '11 at 18:26
1  
@Tim See the update. ;) –  nyuszika7h Feb 14 '11 at 20:11

C++, not sure if there is name for this sort, but here goes <ducks for cover/>

#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>

int main(void)
{
  std::istream_iterator<int> in(std::cin), end;
  std::vector<int> input(in, end);

  std::cout << "sorting.." << std::endl;
  std::copy(input.begin(), input.end(), std::ostream_iterator<int>(std::cout, " "));
  std::cout << std::endl;

  // this is the smart bit.. ;)
  while (std::next_permutation (input.begin(), input.end()));

  std::cout << "sorted.." << std::endl;
  std::copy(input.begin(), input.end(), std::ostream_iterator<int>(std::cout, " "));
  std::cout << std::endl;

  return 0;
}
share|improve this answer
    
It's similar to BogoSort, with the difference that your worst case complexity is the average case complexity of Bogosort. –  saeedn Dec 11 '11 at 12:51

In Ruby, a visual representation of a "Spaghetti Sort", designed to be run in an 80 column terminal:

$CONSOLE_WIDTH = 78
def sort a
  factor = $CONSOLE_WIDTH.to_f/a.max
  puts "Strands:"
  puts (strands=a.map{|v|['-'*(v*factor).to_i,v*factor]}).map{|s|s.first}
  results=[];
  len=0
  n = 1
  until strands.empty? do
   #puts "\nITERATION #{n}\n\n"
    n+=1
    $CONSOLE_WIDTH.downto(0) do |i|
      len = i
      contacts = strands.find_all{|s|s.first[i]!=nil}
      break unless contacts.empty?
    end
    puts "\nResults:"
    puts results.map{|s|s.first}
    puts "\nStrands:"
    puts strands.map{|s|s.first.ljust(len+1)+"|"}
    longest = strands.sort_by{|s|s.last}.max
    strands.delete_at strands.index(longest)
    results << longest
  end
  puts "\nResults:"
  puts results.map{|s|s.first}
  return results.map{|s|s.last/factor}

end

Usage: p sort (1..10).map{rand(100)}

share|improve this answer

PL\SQL - 109 lines

This is a response to @eBusiness's comment in @steenslag's answer. It serves no useful purpose.

It has 4 stages:

  1. Take a comma separated list of numbers and work out how many commas are in it.
  2. Use the number of commas to split the string positionally based on the commas.
  3. Loop through the known number of values and find the minimum value each time - deleting the previous minimum as we go.
  4. Output them.

 create or replace procedure sort ( Plist_of_numbers varchar2 ) is

   type t_num_array is table of number index by binary_integer;

   t_sorted t_num_array;
   t_unsorted t_num_array;

   i binary_integer := 0;
   -- Can you tell this one was an afterthought?
   x binary_integer := 0;

   no_of_commas integer := 0;

   v_list_of_numbers varchar2(32000);
   v_min_value number;
   v_min_index binary_integer;

begin

   -- Best to be overly careful about these things.
   if trim(Plist_of_numbers) is null then
      raise_application_error(-20000,'You need to give me something to work with.');
   end if;

   -- Firstly we have to make sure that there is a trailing comma
   -- in Plist_of_numbers for this to work.
   if substr(Plist_of_numbers,-1) = ',' then
      v_list_of_numbers := Plist_of_numbers;
   else
      v_list_of_numbers := Plist_of_numbers || ',';
   end if;


   /* SQL ( being SQL ) we had to pass in a string of numbers
      because we'd need to create a type outside the function which
      wouldn-t really be in the spirit of things so, let's unpack
      our string. In order to do this ( bowling ) we need to know how
      many commas are in our sting.
   */

   for j in 1 .. length(v_list_of_numbers) loop

      if substr(v_list_of_numbers,j,1) = ',' then
         no_of_commas := no_of_commas + 1;
      end if;

   end loop;


   -- Next we unpack our list into t_unsorted;
   for j in 1 .. ( no_of_commas - 1 ) loop

      i := i + 1;
      t_unsorted(i) := to_number( substr( v_list_of_numbers
                                        , instr(v_list_of_numbers, ',', 1, j)
                                        , instr(v_list_of_numbers, ',', 1, j + 1)
                                           )
                                  );

   end loop;


   -- Next the actual sorting.
   -- Loop through the known number of elements in the array.
   for j in 1 .. i loop

      -- Then the array each time to find the minimum.
      -- As we-re deleting stuff in the middle here we have to be careful;
      k := t_unsorted.first;
      v_min_value := null;

      while k < t_unsorted.last loop

         if v_min_value is null then 
            v_min_value := t_unsorted(k);
            v_min_index := k;
         elsif t_unsorted(k) < v_min_value then
            v_min_value := t_unsorted(k);
            v_min_index := k;
         end if;

         k := t_unsorted.next;

      end loop;

      -- Now we-ve found the next minimum value put it into the 
      -- t_sorted array
      x := x + 1;          
      t_sorted(x) := v_min_value;
      -- and delete our min value from the unsorted array.
      t_unsorted.delete(v_min_index);

   end loop;


   -- Lastly show that everything worked.
   for j in 1 .. x loop

      dbms_output.put_line( t_sorted(j) );

      -- it-s also nice to show the wider world what-s happening 
      -- but let-s not do it too often, server load etc.
      if mod(j,10) = 0 then
          dbms_application_info.set_module('finding min', 'total: ' || j );
      end if;

   end loop;

end sort;

As you can see it's ridiculous... Bad things include:

  • Doing this in SQL

That's pretty much it actually, it should be extremely quick.

share|improve this answer

Both the choice of language and algorithm need to be explained.

This algorithm is called slow sort. I intend to beat the crap out of bogosort (test random permutations until it's sorted) because, while algorithmically terribly inefficient, it's implementation is too simple and you cannot garantee its slowness.

I code slow sort in Scheme because Scheme's main goal is to be simple, which makes this a greater challenge. An interesting feature of Scheme is its unbeatable extensibility; in fact, implementations of the language are commonly (often completely) implemented in Scheme itself. Even better: all you need is abstractions (lambdas) and applications. Applications are evaluated in prefix notation:

(function arg1 arg2 ...)

... is just syntactic sugar for:

(apply function (list arg1 arg2 ...))

... which applies a function to lists of arguments.

In order to make a strike out of this code bowling, I have to redefine the max function to recursively split the list until it can compare itself. The min function is also redefined recursively using max by removing out all the maximums until one number remains. Finally, the sort is redefined by successively appending the minimums.

Slow sort is based on "multiply and surrender", as opposed to "divide and conquer". It works by recursively pulling out the maximums until you are left with the minimum, appending the resulting minimum each time to the solution and restarting over until all minimums have been appended in turn. While totally inefficient, my implementation reuses computations as much as possible because: 1) It is needed for the algorithm 2) You might want this sort to end some day...

(define sort
  (lambda ns
    (define max
      (lambda (ns)
        (let ([len (length ns)])
          (case len
            [(2) (receive (a b) (car+cdr ns)
                   (if (> a b) a b))]
            [(1) (car ns)]
            [else (receive (a b)
                    (split-at ns
                      (floor (/ len 2)))
                    (let ([ma (max a)]
                          [mb (max b)])
                      (if (> ma mb) ma mb)))]))))
    (define min
      (lambda (ns)
        (if (null? (cdr ns)) (car ns)
            (min (remove ns (max ns))))))
    (if (every number? ns)
        (let sort ([ns ns] [sorted (list)])
          (if (null? ns) sorted
              (let ([m (min ns)])
                (sort (remove ns m)
                  (append ns m)))))
    (error "These are not all numbers: " ns))))
share|improve this answer

An enterprisey solution (in pseudocode):

create sql connection
create temporary table: one autoincrementing ID field and one value field
convert array into xml
post xml to sql database
use stored procedure to create records from xml
run sql command "select * from temptable sort by value"
iterate records, adding to array
drop temporary table
close connection
share|improve this answer

Java exaggerated mess/heaviness

That was fun! Feels weird to post something like this.

import java.util.LinkedList;

public class WTF {

    // XXX We all know linkedlists are awesome for random access!
    public static LinkedList<Integer> sort(LinkedList<Integer> array) {
        // XXX Instead of sorting in place, we create another array entirely!
        LinkedList<Integer> result = new LinkedList<Integer>();

        // while array is not all null
        // XXX No comments!
        boolean arrayAllNull = true;
        for (int a = 0; a < array.size(); a = a + 1) {
            if (array.get(a) != null) {
                arrayAllNull = false;
            }
        }

        while (arrayAllNull != true) {

            // XXX We could use Integer.MAX_VALUE, but lets assumes something instead
            Integer i = 999999999;
            // find lowest
            for (int a = 0; a < array.size(); a = a + 1) {
                if (array.get(a) != null) {
                    if (array.get(a) < i) {
                        i = array.get(a);
                    }
                }
            }

            // replace lowest by null
            // XXX Yup, another loop! :)
            // XXX We're breaking the original array, but don't tell anyone.
            for (int a = 0; a < array.size(); a = a + 1) {
                if (array.get(a) != null) {
                    if (array.get(a) == i) {
                        array.set(a, null);
                    }
                }
            }

            result.add(i);

            // XXX Lets repeat that test once again, functions are bad.
            arrayAllNull = true;
            for (int a = 0; a < array.size(); a = a + 1) {
                if (array.get(a) != null) {
                    arrayAllNull = false;
                }
            }

        }

        return result;
    }

    public static void main(String[] args) {
        int[] numbers = new int[]{1, 4, 7, 2, 5};
        LinkedList<Integer> nums = new LinkedList<Integer>();
        for (int i : numbers)
            nums.add(i);
        System.out.println(sort(nums));
    }

}
share|improve this answer

Python

Only works if all the items are less than 9e99 :)

>>> L=[10,9,8,7,6,5,4,3,2,1]
>>> 
>>> print [[min(L),L.__setitem__(L.index(min(L)),9e99)][0] for i in L]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Each time through the loop, extract the smallest element and replace it with 9e99

share|improve this answer

Python (634)

If the list is sorted, print it. Otherwise, swap two elements that are in the wrong order, then generate a python script that handles the new order. After running that, delete the created script.

list_to_sort = input()
n = 0
n += 1
import sys, os
for i in range(len(list_to_sort)-1):
    if list_to_sort[i]>list_to_sort[i+1]:
        (list_to_sort[i],list_to_sort[i+1]) = (list_to_sort[i+1],list_to_sort[i])
        source_file = open(sys.argv[0],'r')
        temp_filename = 'temp'+repr(n)+'.py'
        temp_file = open(temp_filename,'w')
        temp_file.write('list_to_sort = ' + repr(list_to_sort) + '\n')
        temp_file.write('n = ' + repr(n) + '\n')
        for line in source_file.readlines()[2:]:
            temp_file.write(line)
        source_file.close()
        temp_file.close()
        os.system('python ' + temp_filename)
        os.remove(temp_filename)
        sys.exit(0)
print list_to_sort
share|improve this answer

Bogobogosort in Python

import random
import copy

def isSorted(list):
    newlist = copy.deepcopy(list)
    while True:
        newlist[:len(newlist)-1] = bogobogosort(newlist[:len(newlist)-1])
        if newlist[len(newlist)-2] < newlist[len(newlist)-1]:
            break
        random.shuffle(newlist)
    return newlist == list

def bogobogosort(list):
    ret = copy.deepcopy(list)
    if (len(ret) > 1):
        while not isSorted(ret):
            random.shuffle(ret)
    return ret

Algorithm invented by David Morgan-Mar.

Warning: don't try this with any list larger than 5 elements. Even 5 is very slow.

share|improve this answer

Python

My solution. Includes error handling, and a feature that preserves whitespace when sorting numbers, i.e., whitespace stays in place; numbers move.

e.g.,

Please tell me your favorite list of numbers.
I'll sort them for you.  
7, 8, 6, 7, 1, 3, 5
1, 3, 5, 6, 7, 7, 8

9,  6  ,8, 43, 90, 13   , 54, 3223, 4
4,  6  ,8, 9, 13, 43   , 54, 90, 3223

Code:

#!/usr/bin/env python
import sys
import re

class SillyUserException(ValueError):
    def __init__(self, reason, input):
        self.reason = reason
        self.input = input

    def __str__(self):
        return '''
"No man is exempt from saying silly things; the mischief is to say them deliberately."
 - Michel de Montaigne, The Complete Essays
 Your offense was {}.  You should know that it's {}.
         '''.format(self.input, self.reason)

class Item(object):
    def __init__(self, number, prespace, postspace):
        self.number = number
        self.prespace = prespace
        self.postspace = postspace

    def __str__(self):
        return self.prespace + str(self.number) + self.postspace

class ListOfNumbers(object):
    def __init__(self, string):
        self.array = self.parse(string)

    def __str__(self):
        outs = ''
        for i in self.array:
            outs += '{}{}{},'.format(i.prespace, i.number, i.postspace)
        return outs[:-1]

    def parse(self, string):
        array = []

        for item in string.split(','):
            try:
                x = int(item)
            except ValueError:
                raise SillyUserException("not a number", item)

            whitespace = re.findall("\s+", item)
            pre = ''
            post = ''
            if len(whitespace) > 0:
                pre = whitespace[0]
            if len(whitespace) > 1:
                post = whitespace[-1]

            array.append(Item(x, pre, post))

        def swap(i, j):
            tmp = array[i].number
            array[i].number = array[j].number
            array[j].number = tmp

        def min(arr, offset):
            m = arr[0]
            index = 0
            for i, e in enumerate(arr):
                if e.number < m.number:
                    m = e
                    index = i
            return index + offset


        for i, x in enumerate(array):
            m = min(array[i:], i)
            if m != i:
                swap(m, i)

        return array


if __name__ == "__main__":
    sys.stdout.writelines('\n'.join([
        "Please tell me your favorite list of numbers.",
        "I'll sort them for you.  ",
        ""
    ]))

    sys.stdout.flush()

    def read_lines():
        while True:
            line = raw_input()
            yield line

    try:
        for line in read_lines():
            if not line:
                raise SillyUserException('polite to speak when asked a question.', "<nothing>")
            if line.strip() == '42':
                raise SillyUserException('the answer to life, the universe, and everything',
                    'using and overly powerful value')
            if ',' not in line:
                raise SillyUserException("not a list", line)


            print ListOfNumbers(line)

    except SillyUserException as sillyness:
        print sillyness
share|improve this answer

Perl

not so long but self-explained:

if (my @list = (2,5,4,1)) {
    do { @list = sort @list and print @list} if("you want to sort it")
}
share|improve this answer

Radix Sort in Haskell

I have tried to do some optimization using Data.Sequence but it probably needs some more around assingBuckets method.

import qualified Data.Sequence as Seq
import qualified Data.Foldable as Fold
import System.Environment

digit :: Int -> Int -> Int
digit x pos = if pos >= 0 && pos < xlength
                then read [xstr !! (xlength - 1 - pos)] 
                else 0
    where xlength = length xstr
          xstr = show x

assignBuckets :: [Int] -> Int -> [[Int]]
assignBuckets xs dig = map Fold.toList . Fold.toList $ 
                            assign xs dig
                              (Seq.fromList $ take 10 $ repeat (Seq.empty :: Seq.Seq Int))
    where assign :: [Int] -> Int -> Seq.Seq (Seq.Seq Int) -> Seq.Seq (Seq.Seq Int)
          assign []     _   seq = seq
          assign (x:xs) dig seq = assign xs dig $ 
                                    Seq.update idx
                                      ((Seq.index seq idx) Seq.|> x) seq
           where idx = digit x dig

radixSort     :: [Int] -> [Int]
radixSort []   = []
radixSort xs   = sort xs 0 (length . show . maximum $ xs)
    where sort xs n m = if n >= m
                         then xs
                         else sort buckets (n+1) m
           where buckets = (concat $ assignBuckets xs n) 

main = do
  args <- getArgs
  print . radixSort . map (read :: String -> Int) $ args

This sort (if done properly I haven't checked the asymptotic runtime 100%) should run i O(kN) where k is the number of digits in the largest number which is good if k is constant.

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