Sorting algorithms/Bead sort: Difference between revisions

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In this task, the goal is to sort an array of positive integers using the Bead Sort Algorithm.

Algorithm has O(S), where S is the sum of the integers in the input set: Each bead is moved individually. This is the case when bead sort is implemented without a mechanism to assist in finding empty spaces below the beads, such as in software implementations.

C

A rather straightforward implementation; since we do not use dynamic matrix, we have to know the maximum value in the array in advance. Using no sparse matrix means the matrix needs MAX*MAX times the size of an integer bytes to be stored.

<lang c>#include <stdio.h>

include <stdlib.h>
include <stdbool.h>
include <string.h>

int *bead_sort(int *a, size_t len) {

 size_t i, j, k;
 bool fallen;
 int *t, *r = NULL;
 int max = a[0];

 for(i = 1; i < len; i++) 
 {
   if ( a[i] < 0 ) return NULL;  // we can't sort nums < 0
   if ( max < a[i] ) max = a[i];
 }
 
 t = malloc(max*max*sizeof(int));
 if ( t == NULL ) return NULL;
 memset(t, 0, max*max*sizeof(int));

 r = malloc(len*sizeof(int));
 memset(r, 0, len*sizeof(int));
 if (r != NULL) 
 {
   // "split" numbers into "beads" (units)
   for(i = 0; i < len; i++)
   {
     for(j = 0; j < a[i]; j++) t[i*max + j]++;
   }

   // make them fall down
   do
   {
     fallen = false;
     for(i = 0; i < max-1; i++)
     {

for(j = 0; j < max; j++) { if ( t[i*max + j] == 1 && t[(i+1)*max + j] == 0 ) { fallen = true; t[i*max + j] = 0; t[(i+1)*max + j] = 1; } }

     }
   } while(fallen);

if defined(SHOW_BEADS)

   for(i = 0; i < max; i++)
   {
     for(j = 0; j < max; j++)
     {

printf("%d ", t[i*max + j]);

     }
     printf("\n");
   }

endif

   // count beads
   k = 0;
   for(i = 0; i < max; i++)
   {
     if ( t[(max - i - 1)*max + 0] == 0 ) break;
     for(j = 0; j < max; j++)
     {

int v = t[(max - i - 1)*max + j]; if ( v == 0 ) break; r[k] += v;

     }
     k++;
   }
   
 }
 free(t);
 return r;

}

int main() {

 int values[] = {5, 3, 1, 7, 4, 1, 1, 20};
 size_t i, len = sizeof(values)/sizeof(int);

 int *r = bead_sort(values, len);
 if ( r == NULL ) return EXIT_FAILURE;

 for(i = 0; i < len; i++)
 {
   printf("%d ", r[i]);
 }
 putchar('\n');

 free(r);
 return EXIT_SUCCESS;

}</lang>

C++

<lang cpp>//this algorithm only works with positive, whole numbers. //O(2n) time complexity where n is the summation of the whole list to be sorted. //O(3n) space complexity.

include<iostream>
include<vector>

using namespace std;

void distribute( int dist, vector<int> &List)//*beads* go down into different buckets using gravity (addition). {

   if (dist > List.size() )
       List.resize(dist,0); //resize if too big for current vector

   for (int i=0; i < dist; i++)
       List[i] = List[i]+1;

}

int main() {

   vector<int> list;
   vector<int> list2;
   int myints[] = {5,3,1,7,4,1,1};
   vector<int> fifth (myints, myints + sizeof(myints) / sizeof(int) );

   cout << "#1 Beads falling down: ";
   for (int i=0; i < fifth.size(); i++)
       distribute (fifth[i], list);
   cout << endl;

   cout <<endl<< "Beads on their sides: ";
   for (int i=0; i < list.size(); i++)
       cout << " " << list[i];
   cout << endl;

   //second part

   cout << "#2 Beads right side up: ";
   for (int i=0; i < list.size(); i++)
       distribute (list[i], list2);
   cout << endl;

   cout <<endl<< "Sorted list/array";
   for (int i=0; i < list2.size(); i++)
       cout << " " << list2[i];
   cout << endl;
   return 0;

}</lang>

Output:

Beads falling down:

Beads on their sides: 7 4 4 3 2 1 1 Beads right side up:

Sorted list/array 7 5 4 3 1 1 1

Positive, Negative, and Zeros (all integers)

Some lines in this example are too long (more than 80 characters). Please fix the code if it's possible and remove this message.

<lang Cpp>//combination of both positive and negative bead sort (with zeros) //positive bead sort = O(1/2n) where n is the sumation of all positive integers //negative bead sort = O(1/2|n|) where n is the absolute value of the summation of all negative integers //count zeros and insert = O(z) where z is number of zeros //so all in all, the bead sort is still (O(S) where is is the summation of negative and positive bead sort algorithms //space complexity is now O(5n) where 1 array is set and the others are expandable. If lists were used, it could //probably be faster and better for insertion later.

//Feel free to improve upon coding if you will. Otherwise, it proves correctness

include<iostream>
include<vector>

using namespace std;

void distribute_neg( int dist, vector<int> &List)//in theory makes *beads* go down into different buckets using gravity. { dist=-dist; //resets to positive number for implamentation

if (dist > List.size() ) List.resize(dist,0);//can be done differently but *meh*

for (int i=0; i < dist; i++) List[i]=List[i]-1; } //end of distribute negative

void distribute_pos( int dist, vector<int> &List)//in theory makes *beads* go down into different buckets using gravity. { if (dist > List.size() ) List.resize(dist,0);

for (int i=0; i < dist; i++) List[i]=List[i]+1; } //end of distribute positive

void sort(vector<int> &List){ int i; int zeros=0; vector<int> list; vector<int> list_pos; vector<int> sorted; vector<int> sorted_pos; cout << "#1 Beads falling down: "; for (i=0; i < List.size(); i++) if (List[i] < 0) distribute_neg (List[i], list); else if (List[i] > 0) distribute_pos(List[i], list_pos); else zeros++;

cout << endl;

cout <<endl<< "Beads on their sides neg: "; for (i=0; i < list.size(); i++) cout << " " << list[i]; cout << endl;

cout <<endl<< "Beads on their sides positive: "; for (i=0; i < list_pos.size(); i++) cout << " " << list_pos[i]; cout << endl; //second part

cout << "#2 Beads right side up: "; for (i=0; i < list.size(); i++) distribute_neg (list[i], sorted);

for (i=0; i < list_pos.size(); i++) distribute_pos(list_pos[i], sorted_pos); cout << endl;

cout << endl;

cout <<endl<< "Sorted list/array neg"; for (i=0; i < sorted.size(); i++) cout << " " << sorted[i]; cout << endl;

cout <<endl<< "Sorted list/array pos"; for (i=0; i < sorted_pos.size(); i++) cout << " " << sorted_pos[i]; cout << endl;

//combine two at end. //In theory, a list for both positive and negative structures would give better performance at the end, //combining the two lists in O(1) time. You may chose to do so if you wish. The same goes with zeros.

while (zeros > 0) { sorted_pos.push_back(0); zeros--; }

i=sorted.size()-1; while (i >= 0) { sorted_pos.push_back(sorted[i]); i--; }

cout <<endl<< "Sorted list/array"; for (i=0; i < sorted_pos.size(); i++) cout << " " << sorted_pos[i]; cout << endl; }

int main(){ int myints[] = {-1, -4, -3, 1, 4, 3, 0}; vector<int> here_be_dragons (myints, myints + sizeof(myints) / sizeof(int) ); sort(here_be_dragons); return 0;

}</lang>

Clojure

Translation of: Haskell

<lang Clojure>(defn transpose [xs]

 (loop [transposed [], remaining xs]    
   (if (empty? remaining)
     transposed
     (recur 
       (conj transposed (map #(first %) remaining))
       (filter #(not-empty %) (map #(rest %) remaining)))) ))

(defn bead-sort [xs]

 (map #(reduce + %) 
   (transpose 
     (transpose (map #(replicate % 1) xs)))))

(println (bead-sort [5 2 4 1 3 3 9])) </lang>

Output:

(9 5 4 3 3 2 1)

D

Translation of: Python

<lang d>import std.stdio, std.algorithm, std.range;

auto beadSort(int[] l) {

   auto columns(R)(R m) {
       int[][] r;
       foreach (i; 0 .. reduce!max(map!walkLength(m))) {
           r.length += 1;
           foreach (sub; m)
               if (sub.length > i)
                   r[$-1] ~= 0;
       }
       return r;
   }
   auto m = map!"new int[a]"(l);
   return map!walkLength(columns(columns(m)));

}

void main() {

   writeln(beadSort([5, 3, 1, 7, 4, 1, 1]));

}</lang> Output (D V.2.048):

[7, 5, 4, 3, 1, 1, 1]

Fortran

Works with: Fortran version 2003

Works with: Fortran version 95

removing the iso_fortran_env as explained in code

This implementation suffers the same problems of the C implementation: if the maximum value in the array to be sorted is very huge, likely there will be not enough free memory to complete the task. Nonetheless, if the Fortran implementation would use "silently" sparse arrays and a compact representation for "sequences" of equal values in an array, then this very same code would run fine even with large integers.

<lang fortran>program BeadSortTest

 use iso_fortran_env 
 ! for ERROR_UNIT; to make this a F95 code,
 ! remove prev. line and declare ERROR_UNIT as an
 ! integer parameter matching the unit associated with
 ! standard error

 integer, dimension(7) :: a = (/ 7, 3, 5, 1, 2, 1, 20 /)

 call beadsort(a)
 print *, a

contains

 subroutine beadsort(a)
   integer, dimension(:), intent(inout) :: a

   integer, dimension(maxval(a), maxval(a)) :: t
   integer, dimension(maxval(a)) :: s
   integer :: i, m

   m = maxval(a)
   
   if ( any(a < 0) ) then
      write(ERROR_UNIT,*) "can't sort"
      return
   end if

   t = 0
   forall(i=1:size(a)) t(i, 1:a(i)) = 1  ! set up abacus
   forall(i=1:m)             ! let beads "fall"; instead of
      s(i) = sum(t(:, i))    ! moving them one by one, we just
      t(:, i) = 0            ! count how many should be at bottom,
      t(1:s(i), i) = 1       ! and then "reset" and set only those
   end forall
   
   forall(i=1:size(a)) a(i) = sum(t(i,:))
   
 end subroutine beadsort

end program BeadSortTest</lang>

Haskell

<lang haskell>import Data.List

beadSort :: [Int] -> [Int] beadSort = map sum. transpose. transpose. map (flip replicate 1)</lang> Example; <lang haskell>*Main> beadSort [2,4,1,3,3] [4,3,3,2,1]</lang>

Icon and Unicon

Icon

The program below handles integers and not just whole numbers. As are so many others, the solution is limited by the lack of sparse array or list compression.

Some lines in this example are too long (more than 80 characters). Please fix the code if it's possible and remove this message.

<lang Icon>procedure main() #: demonstrate various ways to sort a list and string

  write("Sorting Demo using ",image(beadsort))
     writes("  on list : ")
     writex(UL := [3, 14, 1, 5, 9, 2, 6, 3])
     displaysort(beadsort,copy(UL))

end

procedure beadsort(X) #: return sorted list ascending(or descending) local base,i,j,x # handles negatives and zeros, may also reduce storage

  poles := list(max!X-(base := min!X -1),0)                       # set up poles, we will track sums not individual beads
  every x := !X do {                                              # each item in the list
     if integer(x) ~= x then runerr(101,x)                        # ... must be an integer
     every poles[1 to x - base] +:= 1                             # ... beads "fall" into the sum for that pole 
     }

  every (X[j := *X to 1 by -1] := base) & (i := 1 to *poles) do   # read from the bottom of the poles
    if poles[i] > 0 then {                                        # if there's a bead on the pole ... 
       poles[i] -:= 1                                             # ... remove it

X[j] +:= 1 # ... and add it in place

    }
  return X

end</lang>

Note: This example relies on the supporting procedures 'writex' in Bubble Sort.

This example is in need of improvement:

Need to confirm if min/max are native in Icon. They are available in the IPL

Abbreviated sample output:

Sorting Demo using procedure beadsort
  on list : [ 3 14 1 5 9 2 6 3 ]
    with op = &null:         [ 1 2 3 3 5 6 9 14 ]   (0 ms)

Unicon

The Icon solution works in Unicon.

J

Generally, this task should be accomplished in J using \:~. Here we take an approach that's more comparable with the other examples on this page.

Example use:

<lang> bead bead 2 4 1 3 3 4 3 3 2 1

  bead bead 5 3 1 7 4 1 1

7 5 4 3 1 1 1</lang>

Extending to deal with sequences of arbitrary integers:

<lang j>bball=: ] (] + [: bead^:2 -) <./ - 1:</lang>

Example use:

<lang> bball 2 0 _1 3 1 _2 _3 0 3 2 1 0 0 _1 _2 _3</lang>

Octave

Translation of: Fortran

<lang octave>function sorted = beadsort(a)

 sorted = a;
 m = max(a);
 if ( any(a < 0) )
   error("can't sort");
 endif
 t = zeros(m, m);
 for i = 1:numel(a)
   t(i, 1:a(i)) = 1;
 endfor
 for i = 1:m
   s = sum(t(:, i));
   t(:, i) = 0;
   t(1:s, i) = 1;
 endfor
 for i = 1:numel(a)
   sorted(i) = sum(t(i, :));
 endfor

endfunction

beadsort([5, 7, 1, 3, 1, 1, 20])</lang>

Perl

Instead of storing the bead matrix explicitly, I choose to store just the number of beads in each row and column, compacting on the fly. At all times, the sum of the row widths is equal to the sum column heights.

<lang perl>sub beadsort {

   my @data = @_;

   my @columns;
   my @rows;

   for my $datum (@data) {
       for my $column ( 0 .. $datum-1 ) {
           ++ $rows[ $columns[$column]++ ];
       }
   }

   return reverse @rows;

}

beadsort 5, 7, 1, 3, 1, 1, 20; </lang>

Perl 6

Translation of: Haskell

<lang perl6>use List::Utils;

sub beadsort(@l) {

   (transpose(transpose(map {[1 xx $_]}, @l))).map(*.elems);

}

my @list = 2,1,3,5; say beadsort(@list).perl;</lang>

Output:

(5, 3, 2, 1)

PureBasic

<lang PureBasic>#MAXNUM=100

Dim MyData(Random(15)+5) Global Dim Abacus(0,0)

Declare BeadSort(Array InData(1)) Declare PresentData(Array InData(1))

If OpenConsole()

 Define i
 ;- Generate a random array
 For i=0 To ArraySize(MyData())
   MyData(i)=Random(#MAXNUM)
 Next i
 PresentData(MyData())
 ;
 ;- Sort the array
 BeadSort(MyData())
 PresentData(MyData())
 ;
 Print("Press ENTER to exit"): Input()

EndIf

Procedure LetFallDown(x)

 Protected y=ArraySize(Abacus(),2)-1
 Protected ylim=y
 While y>=0
   If Abacus(x,y) And Not Abacus(x,y+1)
     Swap Abacus(x,y), Abacus(x,y+1)
     If y<ylim: y+1: Continue: EndIf
   Else
     y-1
   EndIf
 Wend

EndProcedure

Procedure BeadSort(Array n(1))

 Protected i, j, k
 NewList T()
 Dim Abacus(#MAXNUM,ArraySize(N()))
 ;- Set up the abacus
 For i=0 To ArraySize(Abacus(),2)
   For j=1 To N(i)
     Abacus(j,i)=#True
   Next
 Next
 ;- sort it in threads to simulate free beads falling down
 For i=0 To #MAXNUM
   AddElement(T()): T()=CreateThread(@LetFallDown(),i)
 Next
 ForEach T()
   WaitThread(T())
 Next
 ;- send it back to a normal array
 For j=0 To ArraySize(Abacus(),2)
   k=0
   For i=0 To ArraySize(Abacus())
     k+Abacus(i,j)
   Next
   N(j)=k
 Next

EndProcedure

Procedure PresentData(Array InData(1))

 Protected n, m, sum
 PrintN(#CRLF$+"The array is;")
 For n=0 To ArraySize(InData())
   m=InData(n): sum+m
   Print(Str(m)+" ")
 Next
 PrintN(#CRLF$+"And its sum= "+Str(sum))

EndProcedure</lang>

The array is;
4 38 100 25 69 69 16 8 59 71 53 33
And its sum= 545

The array is;
4 8 16 25 33 38 53 59 69 69 71 100
And its sum= 545

Python

Translation of: Haskell

<lang python>def beadsort(l):

 return map(len, columns(columns([[1] * e for e in l])))

def columns(l):

 try:
   from itertools import zip_longest
 except:
   zip_longest = lambda *args: map(None, *args)
 return [filter(None, x) for x in zip_longest(*l)]

Demonstration code:

beadsort([5,3,1,7,4,1,1])</lang>

Output:

=> [7, 5, 4, 3, 1, 1, 1]

Ruby

Translation of: Haskell

<lang ruby>class Array def beadsort self.map {|e| [1] * e}.columns.columns.map {|e| e.length} end

def columns y = self.length x = self.map {|l| l.length}.max

Array.new(x) do |row| Array.new(y) { |column| self[column][row] }.compact # Remove nulls. end end end

Demonstration code:

[5,3,1,7,4,1,1].beadsort</lang>

Output:

=> [7, 5, 4, 3, 1, 1, 1]

Tcl

<lang tcl>package require Tcl 8.5

proc beadsort numList {

   # Special case: empty list is empty when sorted.
   if {![llength $numList]} return
   # Set up the abacus...
   foreach n $numList {

for {set i 0} {$i<$n} {incr i} { dict incr vals $i }

   }
   # Make the beads fall...
   foreach n [dict values $vals] {

for {set i 0} {$i<$n} {incr i} { dict incr result $i }

   }
   # And the result is...
   dict values $result

}

Demonstration code

puts [beadsort {5 3 1 7 4 1 1}]</lang> Output:

7 5 4 3 1 1 1