Sorting algorithms/Cycle sort
From the the Wikipedia entry on cycle sorting:
| This page uses content from Wikipedia. The original article was at Cycle_sort. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) |
- Cycle sort is an in-place, unstable sorting algorithm, a comparison sort that is theoretically optimal in terms of the total number of writes to the original array, unlike any other in-place sorting algorithm.
- It is based on the idea that the permutation to be sorted can be factored into cycles, which can individually be rotated to give a sorted result.
- Unlike nearly every other sort, items are never written elsewhere in the array simply to push them out of the way of the action.
- Each value is either written zero times, if it's already in its correct position, or written one time to its correct position.
- This matches the minimal number of overwrites required for a completed in-place sort.
- Minimizing the number of writes is useful when making writes to some huge data set is very expensive, such as with EEPROMs like Flash memory where each write reduces the lifespan of the memory.
- See also
- Youtube Visualization and audibilization of Cycle Sort algorithm.
360 Assembly
The program uses ASM structured macros and two ASSIST macros to keep the code as short as possible. <lang 360asm>* Cycle sort 26/06/2016 CYCLESRT CSECT
USING CYCLESRT,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) prolog
ST R13,4(R15) "
ST R15,8(R13) "
LR R13,R15 "
LA RJ,1 jcycle=1
L R2,N n
BCTR R2,0 n-1
ST R2,NM1 nm1=n-1
DO WHILE=(C,RJ,LE,NM1) do jcycle=1 to n-1
LR R1,RJ jcycle
SLA R1,2 .
L RM,A-4(R1) item=a(jcycle)
LR RK,RJ kpos=jcycle /*find{*/
LA RI,1(RJ) i=jcycle+1
DO WHILE=(C,RI,LE,N) do i=jcycle+1 to n
LR R1,RI i
SLA R1,2 .
L R2,A-4(R1) a(i)
IF CR,R2,LT,RM THEN if a(i)<item then
LA RK,1(RK) kpos=kpos+1
ENDIF , end if
LA RI,1(RI) i=i+1
ENDDO , end do /*}*/
IF CR,RK,NE,RJ THEN if kpos^=jcycle then ======
LR R1,RK kpos /*put{*/
SLA R1,2 .
LA R2,A-4(R1) @a(kpos)
DO WHILE=(C,RM,EQ,0(R2)) do while item=a(kpos)
LA RK,1(RK) kpos=kpos+1
LA R2,4(R2) @a(kpos)=@a(kpos)+4
ENDDO , end do
LR R1,RK kpos
SLA R1,2 .
LA R2,A-4(R1) @a(kpos)
L RT,0(R2) temp=a(kpos)
ST RM,0(R2) a(kpos)=item
LR RM,RT item=temp
L R2,WRITES writes
LA R2,1(R2) writes+1
ST R2,WRITES writes=writes+1 /*}*/
DO WHILE=(CR,RK,NE,RJ) do while(kpos^=jcycle) -----
LR RK,RJ kpos=jcycle /*find{*/
LA RI,1(RJ) i=jcycle+1
DO WHILE=(C,RI,LE,N) do i=jcycle+1 to n
LR R1,RI i
SLA R1,2 .
L R2,A-4(R1) a(i)
IF CR,R2,LT,RM THEN if a(i)<item then
LA RK,1(RK) kpos=kpos+1
ENDIF , end if
LA RI,1(RI) i=i+1
ENDDO , end do /*}*/
LR R1,RK kpos /*put{*/
SLA R1,2 .
LA R2,A-4(R1) @a(kpos)
DO WHILE=(C,RM,EQ,0(R2)) do while item=a(kpos)
LA RK,1(RK) kpos=kpos+1
LA R2,4(R2) @a(kpos)=@a(kpos)+4
ENDDO , end do
LR R1,RK kpos
SLA R1,2 .
LA R2,A-4(R1) @a(kpos)
L RT,0(R2) temp=a(kpos)
ST RM,0(R2) a(kpos)=item
LR RM,RT item=temp
L R2,WRITES writes
LA R2,1(R2) writes+1
ST R2,WRITES writes=writes+1 /*}*/
ENDDO , end while ------------------
ENDIF , end if =====================
LA RJ,1(RJ) jcycle=jcycle+1
ENDDO , end do jcycle
LA R3,PG pgi=0
LA RI,1 i=1
DO WHILE=(C,RI,LE,N) do i=1 to n
LR R1,RI i
SLA R1,2 .
L R2,A-4(R1) a(i)
XDECO R2,XDEC edit a(i)
MVC 0(4,R3),XDEC+8 output a(i)
LA R3,4(R3) pgi=pgi+4
LA RI,1(RI) i=i+1
ENDDO , end do
XPRNT PG,L'PG print buffer
L R1,WRITES writes
XDECO R1,XDEC edit writes
MVC XDEC(7),=CL7'writes='
XPRNT XDEC,L'XDEC print buffer
L R13,4(0,R13) epilog
LM R14,R12,12(R13) "
XR R15,R15 "
BR R14 exit
A DC F'4',F'65',F'2',F'-31',F'0',F'99',F'2',F'83',F'782',F'1'
DC F'45',F'82',F'69',F'82',F'104',F'58',F'88',F'112',F'89',F'74'
N DC A((N-A)/L'A) number of items of a NM1 DS F n-1 PG DC CL80' ' buffer XDEC DS CL12 temp for xdeco WRITES DC F'0' number of writes
YREGS
RI EQU 6 i RJ EQU 7 jcycle RK EQU 8 kpos RT EQU 9 temp RM EQU 10 item
END CYCLESRT</lang>
- Output:
-31 0 1 2 2 4 45 58 65 69 74 82 82 83 88 89 99 104 112 782
C
<lang C>
- include <stdio.h>
- include <stdlib.h>
int cycleSort(int * list, size_t l_len); void show_array(int * array, size_t a_len);
/*
* Sort an array in place and return the number of writes. */
int cycleSort(int * list, size_t l_len) {
int writes = 0;
/* Loop through the array to find cycles to rotate. */
for (int cycleStart = 0; cycleStart < l_len - 1; ++cycleStart)
{
int item = list[cycleStart];
int swap_tmp;
/* Find where to put the item. */
int pos = cycleStart;
for (int i = cycleStart + 1; i < l_len; ++i)
{
if (list[i] < item)
{
++pos;
}
}
/* If the item is already there, this is not a cycle. */
if (pos == cycleStart)
{
continue;
}
/* Otherwise, put the item there or right after any duplicates. */
while (item == list[pos])
{
++pos;
}
swap_tmp = list[pos];
list[pos] = item;
item = swap_tmp;
++writes;
/* Rotate the rest of the cycle. */
while (pos != cycleStart)
{
/* Find where to put the item. */
pos = cycleStart;
for (int i = cycleStart + 1; i < l_len; ++i)
{
if (list[i] < item)
{
++pos;
}
}
/* Put the item there or right after any duplicates. */
while (item == list[pos])
{
++pos;
}
swap_tmp = list[pos];
list[pos] = item;
item = swap_tmp;
++writes;
}
}
return writes;
}
int main(int argc, char ** argv) {
int arr[] = { 0, 1, 2, 2, 2, 2, 1, 9, 3, 5, 5, 8, 4, 7, 0, 6, };
int arr_k = sizeof(arr) / sizeof(arr[0]);
int writes;
show_array(arr, arr_k);
writes = cycleSort(arr, arr_k);
show_array(arr, arr_k);
printf("writes: %d\n", writes);
return 0;
}
void show_array(int * array, size_t a_len) {
for (int ix = 0; ix < a_len; ++ix)
{
printf("%d ", array[ix]);
}
putchar('\n');
return;
} </lang>
- Output:
0 1 2 2 2 2 1 9 3 5 5 8 4 7 0 6 0 0 1 1 2 2 2 2 3 4 5 5 6 7 8 9 writes: 10
C++
Based on example code on Wikipedia <lang Cpp>
- include <time.h>
- include <iostream>
- include <vector>
using namespace std;
class cSort { public:
void doIt( vector<unsigned> s )
{
sq = s; display(); c_sort(); cout << "writes: " << wr << endl; display();
}
private:
void display()
{
copy( sq.begin(), sq.end(), ostream_iterator<unsigned>( std::cout, " " ) ); cout << endl;
}
void c_sort()
{
wr = 0; unsigned it, p, vlen = static_cast<unsigned>( sq.size() ); for( unsigned c = 0; c < vlen - 1; c++ ) { it = sq[c]; p = c; for( unsigned d = c + 1; d < vlen; d++ ) if( sq[d] < it ) p++;
if( c == p ) continue;
doSwap( p, it );
while( c != p ) { p = c; for( unsigned e = c + 1; e < vlen; e++ ) if( sq[e] < it ) p++;
doSwap( p, it ); } }
}
void doSwap( unsigned& p, unsigned& it )
{
while( sq[p] == it ) p++; swap( it, sq[p] ); wr++;
} vector<unsigned> sq; unsigned wr;
};
int main(int argc, char ** argv) {
srand( static_cast<unsigned>( time( NULL ) ) ); vector<unsigned> s; for( int x = 0; x < 20; x++ )
s.push_back( rand() % 100 + 21 );
cSort c; c.doIt( s ); return 0;
} </lang>
- Output:
38 119 38 33 33 28 24 101 108 120 99 59 69 24 117 22 90 94 78 75 writes: 19 22 24 24 28 33 33 38 38 59 69 75 78 90 94 99 101 108 117 119 120
D
This version doesn't use Phobos algorithms beside 'swap'. Algorithms can be used to find where to put the item1 and elsewhere.
<lang d>import std.stdio, std.algorithm;
/// Sort an array in place and return the number of writes. uint cycleSort(T)(T[] data) pure nothrow @safe @nogc {
typeof(return) nWrites = 0;
// Loop through the data to find cycles to rotate.
foreach (immutable cycleStart, item1; data) {
// Find where to put the item1.
size_t pos = cycleStart;
foreach (item2; data[cycleStart + 1 .. $])
if (item2 < item1)
pos++;
// If the item1 is already there, this is not a cycle.
if (pos == cycleStart)
continue;
// Otherwise, put the item1 there or right after any duplicates.
while (item1 == data[pos])
pos++;
data[pos].swap(item1);
nWrites++;
// Rotate the rest of the cycle.
while (pos != cycleStart) {
// Find where to put the item1.
pos = cycleStart;
foreach (item2; data[cycleStart + 1 .. $])
if (item2 < item1)
pos++;
// Put the item1 there or right after any duplicates.
while (item1 == data[pos])
pos++;
data[pos].swap(item1);
nWrites++;
}
}
return nWrites;
}
void main() {
immutable x = [0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6]; auto xs = x.dup; immutable nWrites = xs.cycleSort;
if (!xs.isSorted) {
"Wrong order!".writeln;
} else {
writeln(x, "\nIs correctly sorted using cycleSort to:");
writefln("%s\nusing %d writes.", xs, nWrites);
}
}</lang>
- Output:
[0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6] Is correctly sorted using cycleSort to: [0, 0, 1, 1, 2, 2, 2, 2, 3.5, 4, 5, 6, 7, 8, 9] using 10 writes.
Elixir
<lang elixir>defmodule Sort do
def cycleSort(list) do
tuple = List.to_tuple(list)
# Loop through the array to find cycles to rotate.
{data,writes} = Enum.reduce(0 .. tuple_size(tuple)-2, {tuple,0}, fn cycleStart,{data,writes} ->
item = elem(data, cycleStart)
pos = find_pos(data, cycleStart, item)
if pos == cycleStart do
# If the item is already there, this is not a cycle.
{data, writes}
else
# Otherwise, put the item there or right after any duplicates.
{data, item} = swap(data, pos, item)
rotate(data, cycleStart, item, writes+1)
end
end)
{Tuple.to_list(data), writes}
end
# Rotate the rest of the cycle.
defp rotate(data, cycleStart, item, writes) do
pos = find_pos(data, cycleStart, item)
{data, item} = swap(data, pos, item)
if pos==cycleStart, do: {data, writes+1},
else: rotate(data, cycleStart, item, writes+1)
end
# Find where to put the item.
defp find_pos(data, cycleStart, item) do
cycleStart + Enum.count(cycleStart+1..tuple_size(data)-1, &elem(data, &1) < item)
end
# Put the item there or right after any duplicates.
defp swap(data, pos, item) when elem(data, pos)==item, do: swap(data, pos+1, item)
defp swap(data, pos, item) do
{put_elem(data, pos, item), elem(data, pos)}
end
end
IO.inspect a = [0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6] {b, writes} = Sort.cycleSort(a) IO.puts "writes : #{writes}" IO.inspect b</lang>
- Output:
[0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6] writes : 10 [0, 0, 1, 1, 2, 2, 2, 2, 3.5, 4, 5, 6, 7, 8, 9]
FreeBASIC
Uses algorithm in Wikipedia article: <lang freebasic>' FB 1.05.0 Win64
' sort an array in place and return the number of writes Function cycleSort(array() As Integer) As Integer
Dim length As Integer = UBound(array) - LBound(array) + 1 If Length = 0 Then Return 0 Dim As Integer item, position, writes = 0
' loop through the array to find cycles to rotate For cycleStart As Integer = LBound(array) To UBound(array) - 1 item = array(cycleStart)
' find where to put the item
position = cycleStart
For i As Integer = cycleStart + 1 To UBound(array)
If array(i) < item Then position += 1
Next i
' If the item is already there, this is not a cycle
If position = cycleStart Then Continue For
' Otherwise, put the item there or right after any duplicates
While item = array(position)
position += 1
Wend
Swap array(position), item
writes += 1
'rotate the rest of the cycle
While position <> cycleStart
' Find where to put the item
position = cycleStart
For i As Integer = cycleStart + 1 To UBound(array)
If array(i) < item Then position += 1
Next i
' Put the item there or right after any duplicates
While item = array(position)
position += 1
Wend
Swap array(position), item
writes +=1
Wend
Next cycleStart
Return writes
End Function
Sub printArray(array() As Integer)
For i As Integer = LBound(array) To UBound(array) Print Str(array(i)); " "; Next Print
End Sub
Dim array(1 To 16) As Integer = {0, 1, 2, 2, 2, 2, 1, 9, 3, 5, 5, 8, 4, 7, 0, 6} printArray(array()) Dim writes As Integer = cycleSort(array()) Print "After sorting with"; writes; " writes :" printArray(array()) Print Dim array2(1 To 20) As Integer = {38, 119, 38, 33, 33, 28, 24, 101, 108, 120, 99, 59, 69, 24, 117, 22, 90, 94, 78, 75} printArray(array2()) writes = cycleSort(array2()) Print "After sorting with"; writes; " writes :" printArray(array2()) Print Print "Press any key to quit" Sleep</lang>
- Output:
0 1 2 2 2 2 1 9 3 5 5 8 4 7 0 6 After sorting with 10 writes : 0 0 1 1 2 2 2 2 3 4 5 5 6 7 8 9 38 119 38 33 33 28 24 101 108 120 99 59 69 24 117 22 90 94 78 75 After sorting with 19 writes : 22 24 24 28 33 33 38 38 59 69 75 78 90 94 99 101 108 117 119 120
Go
This implementation was translated from the example code on Wikipedia.
<lang go>package main
import ( "fmt" "math/rand" "time" )
func cyclesort(ints []int) int { writes := 0
for cyclestart := 0; cyclestart < len(ints)-1; cyclestart++ { item := ints[cyclestart]
pos := cyclestart
for i := cyclestart + 1; i < len(ints); i++ { if ints[i] < item { pos++ } }
if pos == cyclestart { continue }
for item == ints[pos] { pos++ }
ints[pos], item = item, ints[pos]
writes++
for pos != cyclestart { pos = cyclestart for i := cyclestart + 1; i < len(ints); i++ { if ints[i] < item { pos++ } }
for item == ints[pos] { pos++ }
ints[pos], item = item, ints[pos] writes++ } }
return writes }
func main() { rand.Seed(time.Now().Unix())
ints := rand.Perm(10)
fmt.Println(ints) fmt.Printf("writes %d\n", cyclesort(ints)) fmt.Println(ints) }</lang>
- Output:
[1 9 3 5 8 4 7 0 6 2] writes 10 [0 1 2 3 4 5 6 7 8 9]
Note: output may be different due to the random numbers used.
J
J's sort is natively a single write sort, but it assigns the whole array at once. It would be trivial do the writes one at a time, and to avoid updating values which are not changed:
<lang J>noncyc=:3 :0
writes=. 0
for_item. /:~y do.
if. item ~: item_index{y do.
writes=. writes+1
y=.item item_index} y
end.
end.
smoutput (":writes),' writes'
y
)</lang>
- Example use:
<lang J> noncyc 9 8 15 17 4 0 1 2 17 9 3 12 11 12 19 15 3 9 16 9 17 writes 0 1 2 3 3 4 8 9 9 9 9 11 12 12 15 15 16 17 17 19</lang>
Meanwhile, if we just wanted the "value at a time swapping" mechanism, an idiomatic approach might look something like this:
<lang j>cyc0=:3 :0
c=. (#~ 1 < #@>)C./:/: y
writes=. 0
for_box. c do.
inds=. >box
v=. ({:inds) { y
for_ind. inds do.
writes=. writes+1
t=. ind{ y
y=. v ind} y
v=. t
end.
end.
smoutput (":writes),' writes'
y
)</lang>
- Example use:
<lang J> cyc0 9 8 15 17 4 0 1 2 17 9 3 12 11 12 19 15 3 9 16 9 18 writes 0 1 2 3 3 4 8 9 9 9 9 11 12 12 15 15 16 17 17 19</lang>
This gives us an extra write, because we're using a generic cycle abstraction.
Also that's still a bit different from the wikipedia algorithm. We might model the wikipedia algorithm like this:
<lang J>cyc1=:3 :0
writes=. 0
for_index. i.(#y)-1 do.
item=. index{y
adj=. item+/ .>(1+index)}.y
if. 0<adj do.
pos=. index+adj
while. item=pos{y do. pos=.pos+1 end.
writes=. writes+1
t=. pos{y
y=. item pos} y
item=. t
while. pos ~: index do.
pos=. index+item+/ .>(1+index)}.y
while. item=pos{y do. pos=.pos+1 end.
writes=. writes+1
t=. pos{y
y=. item pos} y
item=. t
end.
end.
end.
smoutput (":writes),' writes'
y
)</lang>
- Example use:
<lang J> cyc1 9 8 15 17 4 0 1 2 17 9 3 12 11 12 19 15 3 9 16 9 17 writes 0 1 2 3 3 4 8 9 9 9 9 11 12 12 15 15 16 17 17 19</lang>
Note that we've saved a write in this case, by following the wikipedia algorithm.
Java
<lang java>import java.util.Arrays;
public class CycleSort {
public static void main(String[] args) {
int[] arr = {5, 0, 1, 2, 2, 3, 5, 1, 1, 0, 5, 6, 9, 8, 0, 1};
System.out.println(Arrays.toString(arr));
int writes = cycleSort(arr);
System.out.println(Arrays.toString(arr));
System.out.println("writes: " + writes);
}
static int cycleSort(int[] a) {
int writes = 0;
for (int cycleStart = 0; cycleStart < a.length - 1; cycleStart++) {
int val = a[cycleStart];
// count the number of values that are smaller than val
// since cycleStart
int pos = cycleStart;
for (int i = cycleStart + 1; i < a.length; i++)
if (a[i] < val)
pos++;
// there aren't any
if (pos == cycleStart)
continue;
// skip duplicates
while (val == a[pos])
pos++;
// put val into final position
int tmp = a[pos];
a[pos] = val;
val = tmp;
writes++;
// repeat as long as we can find values to swap
// otherwise start new cycle
while (pos != cycleStart) {
pos = cycleStart;
for (int i = cycleStart + 1; i < a.length; i++)
if (a[i] < val)
pos++;
while (val == a[pos])
pos++;
tmp = a[pos];
a[pos] = val;
val = tmp;
writes++;
}
}
return writes;
}
}</lang>
- Output:
[5, 0, 1, 2, 2, 3, 5, 1, 1, 0, 5, 6, 9, 8, 0, 1] [0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 5, 5, 5, 6, 8, 9] writes: 14
Kotlin
Translation of the algorithm in the Wikipedia article: <lang scala>// version 1.1.0
/** Sort an array in place and return the number of writes */ fun <T : Comparable<T>> cycleSort(array: Array<T>): Int {
var writes = 0
// Loop through the array to find cycles to rotate.
for (cycleStart in 0 until array.size - 1) {
var item = array[cycleStart]
// Find where to put the item.
var pos = cycleStart
for (i in cycleStart + 1 until array.size) if (array[i] < item) pos++
// If the item is already there, this is not a cycle.
if (pos == cycleStart) continue
// Otherwise, put the item there or right after any duplicates.
while (item == array[pos]) pos++
val temp = array[pos]
array[pos] = item
item = temp
writes++
// Rotate the rest of the cycle.
while (pos != cycleStart) {
// Find where to put the item.
pos = cycleStart
for (i in cycleStart + 1 until array.size) if (array[i] < item) pos++
// Otherwise, put the item there or right after any duplicates.
while (item == array[pos]) pos++
val temp2 = array[pos]
array[pos] = item
item = temp2
writes++
}
}
return writes
}
fun <T : Comparable<T>> printResults(array: Array<T>) {
println(array.asList())
val writes = cycleSort(array)
println("After sorting with $writes writes:")
println(array.asList())
println()
}
fun main(args: Array<String>) {
val array = arrayOf(0, 1, 2, 2, 2, 2, 1, 9, 3, 5, 5, 8, 4, 7, 0, 6)
printResults(array)
val array2 = arrayOf(5, 0, 1, 2, 2, 3, 5, 1, 1, 0, 5, 6, 9, 8, 0, 1)
printResults(array2)
val array3 = "the quick brown fox jumps over the lazy dog".split(' ').toTypedArray()
printResults(array3)
val array4 = "sphinx of black quartz judge my vow".replace(" ", "").toCharArray().distinct().toTypedArray()
printResults(array4)
}</lang>
- Output:
[0, 1, 2, 2, 2, 2, 1, 9, 3, 5, 5, 8, 4, 7, 0, 6] After sorting with 10 writes: [0, 0, 1, 1, 2, 2, 2, 2, 3, 4, 5, 5, 6, 7, 8, 9] [5, 0, 1, 2, 2, 3, 5, 1, 1, 0, 5, 6, 9, 8, 0, 1] After sorting with 14 writes: [0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 5, 5, 5, 6, 8, 9] [the, quick, brown, fox, jumps, over, the, lazy, dog] After sorting with 8 writes: [brown, dog, fox, jumps, lazy, over, quick, the, the] [s, p, h, i, n, x, o, f, b, l, a, c, k, q, u, r, t, z, j, d, g, e, m, y, v, w] After sorting with 26 writes: [a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z]
NetRexx
Direct translation of the Wikipedia entry example <lang NetRexx>/* Rexx */ options replace format comments java crossref symbols nobinary
runSample(arg) return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -- Sort an array in place and return the number of writes. method cycleSort(array = Rexx[]) public static
writes = 0
-- Loop through the array to find cycles to rotate. loop cycleStart = 0 to array.length - 1 - 1 item = array[cycleStart]
-- Find where to put the item.
pos = cycleStart
loop i = cycleStart + 1 to array.length - 1
if array[i] < item then
pos = pos + 1
end i
-- If the item is already there, this is not a cycle.
if pos == cycleStart then
iterate
-- Otherwise, put the item there or right after any duplicates.
loop while item == array[pos]
pos = pos + 1
end
swap_tmp = array[pos]
array[pos] = item
item = swap_tmp
writes = writes + 1
-- Rotate the rest of the cycle. loop while pos \= cycleStart
-- Find where to put the item.
pos = cycleStart
loop i = cycleStart + 1 to array.length - 1
if array[i] < item then
pos = pos + 1
end i
-- Put the item there or right after any duplicates.
loop while item == array[pos]
pos = pos + 1
end
swap_tmp = array[pos]
array[pos] = item
item = swap_tmp
writes = writes + 1
end
end cycleStart return writes
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(arg) public static
samples = ArrayList() samples.add([1, 9, 3, 5, 8, 4, 7, 0, 6, 2]) samples.add([0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6]) samples.add(['Greygill Hole', 'Ogof Draenen', 'Ogof Ffynnon Ddu', 'Malham Tarn Pot']) samples.add([-3.14 ,3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8, 4]) samples.add(['George Washington: Virginia', 'John Adams: Massachusetts', 'Thomas Jefferson: Virginia', 'James Madison: Virginia', 'James Monroe: Virginia'])
list = Rexx[] loop i_ = 0 to samples.size() - 1 list = Rexx[] samples.get(i_) say 'Input list ' Arrays.asList(list) writes = cycleSort(list) say 'Sorted list' Arrays.asList(list) say 'Total number of writes:' writes say end i_ return
</lang>
- Output:
Input list [1, 9, 3, 5, 8, 4, 7, 0, 6, 2] Sorted list [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] Total number of writes: 10 Input list [0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6] Sorted list [0, 0, 1, 1, 2, 2, 2, 2, 3.5, 4, 5, 6, 7, 8, 9] Total number of writes: 10 Input list [Greygill Hole, Ogof Draenen, Ogof Ffynnon Ddu, Malham Tarn Pot] Sorted list [Greygill Hole, Malham Tarn Pot, Ogof Draenen, Ogof Ffynnon Ddu] Total number of writes: 3 Input list [-3.14, 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8, 4] Sorted list [-3.14, 0, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9] Total number of writes: 34 Input list [George Washington: Virginia, John Adams: Massachusetts, Thomas Jefferson: Virginia, James Madison: Virginia, James Monroe: Virginia] Sorted list [George Washington: Virginia, James Madison: Virginia, James Monroe: Virginia, John Adams: Massachusetts, Thomas Jefferson: Virginia] Total number of writes: 4
Objeck
<lang objeck>class Test {
function : Main(args : String[]) ~ Nil {
arr := [5, 0, 1, 2, 2, 3, 5, 1, 1, 0, 5, 6, 9, 8, 0, 1];
arr->ToString()->PrintLine();
writes := CycleSort(arr);
"writes: {$writes}"->PrintLine();
arr->ToString()->PrintLine();
}
function : CycleSort(a : Int[]) ~ Int {
writes := 0;
for(cycleStart := 0; cycleStart < a->Size() - 1; cycleStart+=1;) {
val := a[cycleStart];
pos := cycleStart;
for(i := cycleStart + 1; i < a->Size(); i+=1;) {
if(a[i] < val) {
pos++;
};
};
if(pos <> cycleStart) {
while(val = a[pos]) {
pos+=1;
};
tmp := a[pos];
a[pos] := val;
val := tmp;
writes+=1;
while(pos <> cycleStart) {
pos := cycleStart;
for(i := cycleStart + 1; i < a->Size(); i+=1;) {
if(a[i] < val) {
pos+=1;
};
};
while(val = a[pos]) {
pos++;
};
tmp := a[pos];
a[pos] := val;
val := tmp;
writes++;
};
};
};
return writes;
}
}</lang>
[5,0,1,2,2,3,5,1,1,0,5,6,9,8,0,1] writes: 14 [0,0,0,1,1,1,1,2,2,3,5,5,5,6,8,9]
ooRexx
<lang oorexx>/*REXX program demonstrates a cycle sort on a list of numbers**********
- 13.06.2014 Walter Pachl
- Modified from Rexx Version 2
- ooRexx allows to pass a stemmed variable by reference
- swapping variables uses a temporary instead of the parse.
- /
a.1='George Washington Virginia' a.2='John Adams Massachusetts' a.3='Thomas Jefferson Virginia' a.4='James Madison Virginia' a.5='James Monroe Virginia' n=5 Call show 'Unsorted list: ' w=sortcycle(a.,n) Say 'sorted' Call show 'Sorted list' Say ' ' Say 'This took' w 'writes.' Exit
sortcycle: Procedure
Use Arg a.,n
writes=0
Do c=1 For n
x=a.c
p=c
x=a.c
Do j=c+1 To n
If a.j<x Then
p=p+1
End
If p==c Then
Iterate
Do While x==a.p
p=p+1
End
t=x
x=a.p
a.p=t
writes=writes+1
Do While p\==c
p=c
Do k=c+1 To n
If a.k<x Then
p=p+1
End
Do While x==a.p
p=p+1
End
t=x
x=a.p
a.p=t
writes=writes+1
End
End
Return writes
show:
Parse Arg hdr Say ' ' Say hdr Do i=1 To n Say format(i,2) a.i End Return</lang>
- Output:
Unsorted list: 1 George Washington Virginia 2 John Adams Massachusetts 3 Thomas Jefferson Virginia 4 James Madison Virginia 5 James Monroe Virginia sorted Sorted list 1 George Washington Virginia 2 James Madison Virginia 3 James Monroe Virginia 4 John Adams Massachusetts 5 Thomas Jefferson Virginia This took 4 writes.
Perl
This is based on the Wikipedia pseudocode. <lang perl>use strict; use warnings;
sub cycleSort(@) { my ($array) = @_; my $writes = 0;
my @alreadysorted;
# For each index except the last: for my $start ( 0 .. $#$array - 1 ) { next if $alreadysorted[$start]; my $item = $array->[$start]; # If there are N items less than $item, then we # must move $item N items rightward. my $pos = $start + grep $array->[$_] lt $item, $start + 1 .. $#$array; # If the item is where it should be, continue. next if $pos == $start; # If $item is one of several repetitions, move it to the right # of the last repeat. ++$pos while $item eq $array->[ $pos ]; # Store $item at $pos, where it belongs, and fetch the # value that had been at $pos, and put it in $item. ($array->[ $pos ], $item) = ($item, $array->[ $pos ]); ++$writes;
# Whatever $item is now, it certainly doesn't belong at $pos; do { # Find the correct $pos, $pos = $start + grep $array->[$_] lt $item, $start+1 .. $#$array; ++$pos while $item eq $array->[ $pos ]; # Swap the value there with $item, ($array->[ $pos ] , $item ) = ($item, $array->[ $pos ]); # And mark $pos as having the correct value in it.. $alreadysorted[ $pos ] = 1; ++$writes; # The loop ends after we have just written an item to $start } while $pos != $start; } $writes; }
use List::Util 'shuffle'; my @test = shuffle( ('a'..'z') x 2 ); print "Before sorting: @test\n"; print "There were ", cycleSort( \@test ), " writes\n"; print "After sorting: @test\n"; </lang>
- Output:
Before sorting: a t d b f g y l t p w c r r x i y j k i z q e v a f o q j u x k m h s u v z g m b o l e n h p n c s w d There were 50 writes After sorting: a a b b c c d d e e f f g g h h i i j j k k l l m m n n o o p p q q r r s s t t u u v v w w x x y y z z
Perl 6
<lang perl6>sub cycle_sort ( @nums ) {
my $writes = 0;
# Loop through the array to find cycles to rotate.
for @nums.kv -> $cycle_start, $item is copy {
# Find where to put the item.
my $pos = $cycle_start
+ @nums[ $cycle_start ^.. * ].grep: * < $item;
# If the item is already there, this is not a cycle.
next if $pos == $cycle_start;
# Otherwise, put the item there or right after any duplicates.
$pos++ while $item == @nums[$pos];
( @nums[$pos], $item ) .= reverse;
$writes++;
# Rotate the rest of the cycle.
while $pos != $cycle_start {
# Find where to put the item.
$pos = $cycle_start
+ @nums[ $cycle_start ^.. * ].grep: * < $item;
# Put the item there or right after any duplicates.
$pos++ while $item == @nums[$pos];
( @nums[$pos], $item ) .= reverse;
$writes++;
}
}
return $writes;
}
my @a = <0 1 2 2 2 2 1 9 3.5 5 8 4 7 0 6>;
say @a; say 'writes ', cycle_sort(@a); say @a; </lang>
- Output:
0 1 2 2 2 2 1 9 3.5 5 8 4 7 0 6 writes 10 0 0 1 1 2 2 2 2 3.5 4 5 6 7 8 9
Python
The Wikipedia algorithm pseudocode is very nearly Python. The main changes needed were to change the name array to vector to stop it obscuring a built-in name, and iterating over an enumerated collection rather than using explicit indices.
<lang python>def cycleSort(vector):
"Sort a vector in place and return the number of writes."
writes = 0
# Loop through the vector to find cycles to rotate.
for cycleStart, item in enumerate(vector):
# Find where to put the item.
pos = cycleStart
for item2 in vector[cycleStart + 1:]:
if item2 < item:
pos += 1
# If the item is already there, this is not a cycle.
if pos == cycleStart:
continue
# Otherwise, put the item there or right after any duplicates.
while item == vector[pos]:
pos += 1
vector[pos], item = item, vector[pos]
writes += 1
# Rotate the rest of the cycle.
while pos != cycleStart:
# Find where to put the item.
pos = cycleStart
for item2 in vector[cycleStart + 1:]:
if item2 < item:
pos += 1
# Put the item there or right after any duplicates.
while item == vector[pos]:
pos += 1
vector[pos], item = item, vector[pos]
writes += 1
return writes
if __name__ == '__main__':
x = [0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6]
xcopy = x[::]
writes = cycleSort(xcopy)
if xcopy != sorted(x):
print('Wrong order!')
else:
print('%r\nIs correctly sorted using cycleSort to'
'\n%r\nUsing %i writes.' % (x, xcopy, writes))</lang>
- Output:
[0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6] Is correctly sorted using cycleSort to [0, 0, 1, 1, 2, 2, 2, 2, 3.5, 4, 5, 6, 7, 8, 9] Using 10 writes.
Racket
<lang racket>#lang racket/base (require racket/match)
- Sort an array in place and return the number of writes.
(define (cycle-sort! v < =?)
(define v-len (vector-length v))
(for/sum ; Loop through the array to find cycles to rotate.
((cycle-start (in-range 0 (sub1 v-len))))
(define item (vector-ref v cycle-start))
(define (find-insertion-point) ; Find where to put the item.
(+ cycle-start
(for/sum
((i (in-range (add1 cycle-start) v-len))
#:when (< (vector-ref v i) item)) 1)))
;; Put the item there or right after any duplicates
(define (insert-after-duplicates pos)
(match (vector-ref v pos)
[(== item =?) (insert-after-duplicates (add1 pos))]
[tmp (vector-set! v pos item) ; / swap
(set! item tmp) ; \ [this is my only write point]
pos]))
(define i-p (find-insertion-point))
(if (= i-p cycle-start)
0 ; If the item is already there, this is not a cycle.
(let loop ; Rotate the rest of the cycle.
((e-p (insert-after-duplicates i-p))
(W 1 #| we've already written once |#))
(if (= e-p cycle-start)
W
(loop (insert-after-duplicates (find-insertion-point))
(add1 W))))))) ; we've written again!
(module+ main
;; This will be random with duplicates (define A (list->vector (build-list 30 (λ (i) (random 20))))) A (cycle-sort! A < =) A (define B #(1 1 1 1 1 1)) B (cycle-sort! B < =))</lang>
- Output:
'#(7 17 5 16 14 9 18 10 1 4 10 1 9 3 3 0 1 18 16 12 9 14 14 12 19 2 12 15 16 8) 28 '#(0 1 1 1 2 3 3 4 5 7 8 9 9 9 10 10 12 12 12 14 14 14 15 16 16 16 17 18 18 19) '#(1 1 1 1 1 1) 0
REXX
version 1
<lang rexx>/* REXX ***************************************************************
- 12.06.2014 Walter Pachl translated from Wikipedia's code
- 20.06.2014 WP corrected (courtesy Alan Sampson)
- 30.05.2017 WP fixed for Classic Rexx (courtesy GS)
- /
list='1 9 3 5 8 4 7 0 6 2' n=words(list) Do i=0 To n-1
array.i=word(list,i+1) End
Say list writes=cyclesort() Say 'writes='writes ol= Do i=0 To n-1
ol=ol array.i End
Say strip(ol) Exit
cycleSort: procedure expose array. n
writes = 0 /* Loop through the array to find cycles to rotate. */ do cycleStart=0 to n-2 item = array.cycleStart
/* Find where to put the item. */
pos = cycleStart
Do i=cycleStart+1 to n-1
if array.i < item Then
pos=pos+1
End
/* If the item is already there, this is not a cycle. */
if pos == cycleStart Then
Iterate
/* Otherwise, put the item there or right after any duplicates. */
Do while item == array.pos
pos=pos+1
End
Parse Value array.pos item With item array.pos
writes=writes+1
/* Rotate the rest of the cycle. */ Do while pos <> cycleStart
/* Find where to put the item. */
pos = cycleStart
Do i=cycleStart + 1 to n-1
if array.i < item Then
pos=pos+1
End
/* Put the item there or right after any duplicates. */
Do while item == array.pos
pos=pos+1
End
Parse Value array.pos item With item array.pos
writes=writes+1
End
End
return writes</lang>
- Output:
1 9 3 5 8 4 7 0 6 2 writes=10 0 1 2 3 4 5 6 7 8 9
version 2
This REXX version demonstrates the use of negative numbers and non-integer values in the list.
As a default, the program uses (for the input list) some digits of pi, which for practical purposes, appear random. <lang rexx>/*REXX program demonstrates a cycle sort on a list of items. */ parse arg z /* [↓] not specified? Use "pi" digits.*/ if z= then z=-3.14 3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0 2 8 8 4 say 'unsorted list: ' z /*show the original unsorted numbers. */ w=sortCycle(z) /*W: the number of writes done in sort*/ say 'and took' w "writes." /*show number of writes done in sort. */ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ sortCycle: procedure expose @.; parse arg y; #=words(y); writes=0
do i=1 for #; @.i=word(y,i); end /*i*/ /*put each of the items ───► @. array.*/
/* [↓] find a "cycle" to rotate. */
do c=1 for #; x=@.c; p=c /*X is the item being sorted. */
do j=c+1 to #; if @.j<x then p=p+1; end /*determine where to put X. */
if p==c then iterate /*Is it there? No, this ain't a cycle.*/
do while x==@.p; p=p+1; end /*put X right after any duplicate. */
parse value @.p x with x @.p /*swap the two values: @.p and X.*/
writes=writes+1 /*bump counter for the number of writes*/
do while p\==c; p=c /*rotate the rest of the "cycle". */
do k=c+1 to #; if @.k<x then p=p+1; end /*k*/
do while x==@.p; p=p+1; end /*put X here or right after dups. */
parse value @.p x with x @.p /*swap the two values: @.p and X.*/
writes=writes+1 /*bump the counter for number of writes*/
end /*while p\==c*/
end /*c*/
/* [↓] display the sorted list. */
_=@.1; do j=2 to #; _=_ @.j; end; say ' sorted list: ' _ return writes</lang> output when using the default input:
unsorted list: -3.14 3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0 2 8 8 4 sorted list: -3.14 0 1 1 2 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 7 7 8 8 8 8 8 9 9 9 9 and took 34 writes.
output when using the input of: FM Stereo has been around since 1961.
unsorted list: FM Stereo has been around since 1961. sorted list: 1961. FM Stereo around been has since and took 7 writes.
Note (for the above output). This REXX program was executed on an ASCII machine.
On an ASCII machine, the order of sorting is numbers, uppercase letters, lowercase letters.
On an EBCDIC machine, the order of sorting is lowercase letters, uppercase letters, numbers.
Other (special) characters may also be in a different order.
version 3
This version uses a faster (but a more cryptic) version of incrementing 1 (one) to P within two do loops. <lang rexx>/*REXX program demonstrates a cycle sort on a list of items. */ parse arg z /* [↓] not specified? Use "pi" digits.*/ if z= then z=-3.14 3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0 2 8 8 4 say 'unsorted list: ' z /*show the original unsorted numbers. */ w=sortCycle(z) /*W: the number of writes done in sort*/ say 'and took' w "writes." /*show number of writes done in sort. */ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ sortCycle: procedure expose @.; parse arg y; #=words(y); writes=0
do i=1 for #; @.i=word(y,i); end /*i*/ /*put each of the items ───► @. array.*/
/* [↓] find a "cycle" to rotate. */
do c=1 for #; x=@.c; p=c /*X is the item being sorted. */
do j=c+1 to #; if @.j<x then p=p+1; end /*determine where to put X. */
if p==c then iterate /*Is it there? No, this ain't a cycle.*/
do while x==@.p; p=p+1; end /*put X right after any duplicate. */
parse value @.p x with x @.p /*swap the two values: @.p and X.*/
writes=writes+1 /*bump counter for the number of writes*/
do while p\==c; p=c /*rotate the rest of the "cycle". */
do k=c+1 to #; if @.k<x then p=p+1; end /*k*/
do p=p while x==@.p; end /*put X here or right after dups.*/
parse value @.p x with x @.p /*swap the two values: @.p and X.*/
writes=writes+1 /*bump the counter for number of writes*/
end /*while p\==c*/
end /*c*/
/* [↓] display the sorted list. */
_=@.1; do j=2 to #; _=_ @.j; end; say ' sorted list: ' _ return writes</lang> output is identical to the 2nd version.
version 4
This version uses a subroutine to perform the task of handling an (sorted) item placement (possibly after duplicates). <lang rexx>/*REXX program demonstrates a cycle sort on a list of items. */ parse arg z /* [↓] not specified? Use "pi" digits.*/ if z= then z=-3.14 3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0 2 8 8 4 say 'unsorted list: ' z /*show the original unsorted numbers. */ w=sortCycle(z) /*W: the number of writes done in sort*/ say 'and took' w "writes." /*show number of writes done in sort. */ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ sortCycle: procedure expose @.; parse arg y; #=words(y); w=0
do i=1 for #; @.i=word(y,i); end /*i*/ /*put each of the items ───► @. array.*/
do c=1 for #; x=@.c; p=c /*X is the item being sorted. */
do j=c+1 to #; if @.j<x then p=p+1; end /*j*/ /*where to put X.*/
if p==c then iterate /*Is it there? Then this ain't a cycle*/
call .swap /*put X here or right after dups.*/
do while p\==c; p=c /*rotate the rest of the "cycle". */
do k=c+1 to #; if @.k<x then p=p+1; end /*k*/
call .swap /*put X here or right after dups.*/
end /*while p\==c*/
end /*c*/
/* [↓] display the sorted list to term*/
_=@.1; do j=2 to #; _=_ @.j; end; say ' sorted list: ' _ return w /* [↓] find where to put X into @ */ .swap: do p=p while x==@.p; end; parse value @.p x with x @.p; w=w+1; return</lang> output is identical to the 2nd version.
Ruby
Direct translation of the pseudocode on the Wikipedia. <lang ruby>def cycleSort!(array)
writes = 0
# Loop through the array to find cycles to rotate.
for cycleStart in 0 .. array.size-2
item = array[cycleStart]
# Find where to put the item.
pos = cycleStart
for i in cycleStart+1 ... array.size
pos += 1 if array[i] < item
end
# If the item is already there, this is not a cycle.
next if pos == cycleStart
# Otherwise, put the item there or right after any duplicates.
pos += 1 while item == array[pos]
array[pos], item = item, array[pos]
writes += 1
# Rotate the rest of the cycle.
while pos != cycleStart
# Find where to put the item.
pos = cycleStart
for i in cycleStart+1 ... array.size
pos += 1 if array[i] < item
end
# Put the item there or right after any duplicates.
pos += 1 while item == array[pos]
array[pos], item = item, array[pos]
writes += 1
end
end
writes
end
p a = [0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6] puts "writes : #{cycleSort!(a)}" p a</lang>
- Output:
[0, 1, 2, 2, 2, 2, 1, 9, 3.5, 5, 8, 4, 7, 0, 6] writes : 10 [0, 0, 1, 1, 2, 2, 2, 2, 3.5, 4, 5, 6, 7, 8, 9]
Scala
Translation of Java version <lang scala>
def cycleSort(a: Array[Int]): (Array[Int], Int) = {
var writes = 0
for (cycleStart <- 0 until a.length - 1) {
var value = a(cycleStart)
// count the number of values that are smaller than value since cycleStart
var pos = cycleStart
for (i <- cycleStart + 1 until a.length)
if (a(i) < value) pos += 1
// skip if there aren't any
if (pos != cycleStart) {
// skip duplicates
while (a(pos) == value) pos += 1
// put val into final position
val tmp = a(pos)
a(pos) = value
value = tmp
writes += 1
// repeat as long as we can find values to swap
// otherwise start new cycle
while (pos != cycleStart) {
pos = cycleStart
for (i <- cycleStart + 1 until a.length)
if (a(i) < value) pos += 1
while (a(pos) == value) pos += 1
val tmp = a(pos)
a(pos) = value
value = tmp
writes += 1
}
}
}
(a, writes)
}
</lang>
Sidef
<lang ruby>func cycle_sort (array) {
var (writes=0, pos=0)
func f(i, Ref item, bool=false) {
pos = (i + array.ft(i+1).count{ _ < *item })
return(false) if (bool && pos==i)
while (*item == array[pos]) { ++pos }
(array[pos], *item) = (*item, array[pos])
++writes
return true
}
array.each_kv { |i, item|
f(i, \item, true) || next
while (pos != i) {
f(i, \item)
}
}
return writes
}
var a = %n(0 1 2 2 2 2 1 9 3.5 5 8 4 7 0 6)
say a.join(' ') say ('writes ', cycle_sort(a)) say a.join(' ')</lang>
- Output:
0 1 2 2 2 2 1 9 3.5 5 8 4 7 0 6 writes 10 0 0 1 1 2 2 2 2 3.5 4 5 6 7 8 9
Tcl
Direct translation of the pseudocode on the Wikipedia page <lang tcl>proc cycleSort {listVar} {
upvar 1 $listVar array set writes 0
# Loop through the array to find cycles to rotate.
for {set cycleStart 0} {$cycleStart < [llength $array]} {incr cycleStart} {
set item [lindex $array $cycleStart]
# Find where to put the item. set pos $cycleStart for {set i [expr {$pos + 1}]} {$i < [llength $array]} {incr i} { incr pos [expr {[lindex $array $i] < $item}] }
# If the item is already there, this is not a cycle. if {$pos == $cycleStart} continue
# Otherwise, put the item there or right after any duplicates. while {$item == [lindex $array $pos]} { incr pos } set tmp [lindex $array $pos] lset array $pos $item set item $tmp incr writes
# Rotate the rest of the cycle. while {$pos != $cycleStart} { # Find where to put the item. set pos $cycleStart
for {set i [expr {$cycleStart + 1}]} {$i < [llength $array]} {incr i} { incr pos [expr {[lindex $array $i] < $item}] }
# Put the item there or right after any duplicates. while {$item == [lindex $array $pos]} { incr pos } set tmp [lindex $array $pos] lset array $pos $item set item $tmp incr writes }
} return $writes
}</lang> Demonstrating: <lang tcl>set example {0 1 2 2 2 2 1 9 3.5 5 8 4 7 0 6} puts "Data was: $example" set writes [cycleSort example] puts "Data is now: $example" if {$example eq [lsort -real $example]} {
puts "\twhich is correctly sorted"
} else {
puts "\twhich is the wrong order!"
} puts "Writes required: $writes"</lang>
- Output:
Data was: 0 1 2 2 2 2 1 9 3.5 5 8 4 7 0 6 Data is now: 0 0 1 1 2 2 2 2 3.5 4 5 6 7 8 9 which is correctly sorted Writes required: 10