You are encouraged to solve this task according to the task description, using any language you may know.

Sorting Algorithm
This is a sorting algorithm.   It may be applied to a set of data in order to sort it.     For comparing various sorts, see compare sorts.   For other sorting algorithms,   see sorting algorithms,   or:

O(n logn) sorts

O(n log2n) sorts
Shell Sort

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>

1. include <stdlib.h>
2. include <stdbool.h>
3. 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);
```
```   for(i = 0; i < max; i++)
{
for(j = 0; j < max; j++)
{
```

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

```     }
printf("\n");
}
```
1. 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.

1. include<iostream>
2. 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 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)

<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

```
1. include<iostream>
2. 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>
```

## 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.

``` 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
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,:))

```

## J

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>

## Octave

Translation of: Fortran

``` 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>

## Ruby

<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

1. Demonstration code:

Output:

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

## Tcl

<lang tcl>package require Tcl 8.5

```   # 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 }

```   }
foreach n [dict values \$vals] {
```

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

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

}

1. Demonstration code

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

`7 5 4 3 1 1 1`