Sorting algorithms/Pancake sort

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Task
Sorting algorithms/Pancake sort
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 other sorting algorithms, see Category:Sorting Algorithms, or:
O(n logn) Sorts
Heapsort | Mergesort | Quicksort
O(n log2n) Sorts
Shell Sort
O(n2) Sorts
Bubble sort | Cocktail sort | Comb sort | Gnome sort | Insertion sort | Selection sort | Strand sort
Other Sorts
Bead sort | Bogosort | Counting sort | Pancake sort | Permutation sort | Radix sort | Sleep sort | Stooge sort

Sort an array of integers (of any convenient size) into ascending order using Pancake sorting. In short, instead of individual elements being sorted, the only operation allowed is to "flip" one end of the list, like so:

Before:
6 7 8 9 2 5 3 4 1
After:
9 8 7 6 2 5 3 4 1

Only one end of the list can be flipped; this should be the low end, but the high end is okay if it's easier to code or works better, but it must be the same end for the entire solution. (The end flipped can't be arbitrarily changed.)

Show both the initial, unsorted list and the final sorted list. (Intermediate steps during sorting are optional.) Optimizations are optional (but recommended).

For more information on pancake sorting, see the Wikipedia entry.

See also:

Contents

[edit] Ada

with Ada.Text_IO;
procedure Pancake_Sort is
generic
type Element_Type is private;
type Index_Type is range <>;
type Array_Type is array (Index_Type range <>) of Element_Type;
with function ">" (Left, Right : Element_Type) return Boolean is <>;
procedure Pancake_Sort (Data: in out Array_Type);
 
procedure Pancake_Sort (Data: in out Array_Type) is
procedure Flip (Up_To : in Index_Type) is
Temp : constant Array_Type := Data (Data'First .. Up_To);
begin
for I in Temp'Range loop
Data (I) := Temp (Temp'First + Up_To - I);
end loop;
end Flip;
Max_Index : Index_Type;
begin
for I in reverse Data'First + 1 .. Data'Last loop
Max_Index := Data'First;
for A in Data'First + 1 .. I loop
if Data(A) > Data (Max_Index) then
Max_Index := A;
end if;
end loop;
if Max_Index /= I then
if Max_Index > Data'First then
Flip (Max_Index);
end if;
Flip (I);
end if;
end loop;
end Pancake_Sort;
 
type Integer_Array is array (Positive range <>) of Integer;
procedure Int_Pancake_Sort is new Pancake_Sort (Integer, Positive, Integer_Array);
Test_Array : Integer_Array := (3, 14, 1, 5, 9, 2, 6, 3);
begin
Int_Pancake_Sort (Test_Array);
for I in Test_Array'Range loop
Ada.Text_IO.Put (Integer'Image (Test_Array (I)));
end loop;
Ada.Text_IO.New_Line;
end Pancake_Sort;

Output:

 1 2 3 3 5 6 9 14

[edit] AutoHotkey

;---------------------------------------------------------------------------
Loop { ; test loop
;---------------------------------------------------------------------------
Loop, % Data0 := 10
Random, Data%A_Index%, 10, 99
Unsorted := Array2List("Data")
PancakeSort("Data")
Sorted := Array2List("Data")
MsgBox, 1, Pancake Sort, %Unsorted%`n%Sorted%
IfMsgBox, Cancel, Break
}
 
 
 
;---------------------------------------------------------------------------
PancakeSort(Array) { ; implementation of pancake sort algorithm
;---------------------------------------------------------------------------
Loop, % %Array%0 - 1 {
m := 0
Loop, % s := %Array%0 - A_Index + 1
If (m <= %Array%%A_Index%)
m := %Array%%A_Index%, p := A_Index
If (p < s) && (p > 1)
Flip(Array, p)
If (p < s)
Flip(Array, s)
}
}
 
 
 
;---------------------------------------------------------------------------
Flip(Array, n) { ; flip the first n members of Array
;---------------------------------------------------------------------------
Loop, % x := n // 2 {
i := n - A_Index + 1
%Array%%i% := (%Array%%A_Index% "", %Array%%A_Index% := %Array%%i%)
}
}
 
 
 
;---------------------------------------------------------------------------
Array2List(Array) { ; returns a space separated list from an array
;---------------------------------------------------------------------------
Loop, % %Array%0
List .= (A_Index = 1 ? "" : " ") %Array%%A_Index%
Return, List
}
 

[edit] BASIC

[edit] Text

Works with: QBasic
RANDOMIZE TIMER
 
DIM nums(9) AS INTEGER
DIM L0 AS INTEGER, L1 AS INTEGER, n AS INTEGER
 
'initial values
FOR L0 = 0 TO 9
nums(L0) = L0
NEXT
'scramble
FOR L0 = 9 TO 1 STEP -1
n = INT(RND * (L0)) + 1
IF n <> L0 THEN SWAP nums(n), nums(L0)
NEXT
'display initial condition
FOR L0 = 0 TO 9
PRINT nums(L0);
NEXT
PRINT
 
FOR L1 = 9 TO 1 STEP -1
n = 0
FOR L0 = 1 TO L1
IF nums(n) < nums(L0) THEN n = L0
NEXT
 
IF (n < L1) THEN
IF (n > 0) THEN
FOR L0 = 0 TO (n \ 2)
SWAP nums(L0), nums(n - L0)
NEXT
FOR L0 = 0 TO 9
PRINT nums(L0);
NEXT
PRINT
END IF
FOR L0 = 0 TO (L1 \ 2)
SWAP nums(L0), nums(L1 - L0)
NEXT
 
FOR L0 = 0 TO 9
PRINT nums(L0);
NEXT
PRINT
END IF
NEXT

Sample output:

0  4  6  1  8  7  2  5  3  9
8  1  6  4  0  7  2  5  3  9
3  5  2  7  0  4  6  1  8  9
7  2  5  3  0  4  6  1  8  9
1  6  4  0  3  5  2  7  8  9
6  1  4  0  3  5  2  7  8  9
2  5  3  0  4  1  6  7  8  9
5  2  3  0  4  1  6  7  8  9
1  4  0  3  2  5  6  7  8  9
4  1  0  3  2  5  6  7  8  9
2  3  0  1  4  5  6  7  8  9
3  2  0  1  4  5  6  7  8  9
1  0  2  3  4  5  6  7  8  9
0  1  2  3  4  5  6  7  8  9

[edit] Graphics

This is a graphical variation of the above.

RANDOMIZE TIMER
 
CONST delay = .1 'controls display speed
 
DIM nums(14) AS INTEGER
DIM L0 AS INTEGER, L1 AS INTEGER, n AS INTEGER, ttmp AS SINGLE
 
'initial values
FOR L0 = 0 TO 14
nums(L0) = L0
NEXT
'scramble
FOR L0 = 14 TO 1 STEP -1
n = INT(RND * (L0)) + 1
IF n <> L0 THEN SWAP nums(n), nums(L0)
NEXT
 
'display initial condition
CLS
GOSUB displayer
 
FOR L1 = 14 TO 1 STEP -1
n = 0
FOR L0 = 1 TO L1
IF nums(n) < nums(L0) THEN n = L0
NEXT
 
IF (n < L1) THEN
IF (n > 0) THEN
FOR L0 = 0 TO (n \ 2)
SWAP nums(L0), nums(n - L0)
NEXT
GOSUB displayer
END IF
FOR L0 = 0 TO (L1 \ 2)
SWAP nums(L0), nums(L1 - L0)
NEXT
 
GOSUB displayer
END IF
NEXT
 
COLOR 7
END
 
displayer:
LOCATE 1, 1
FOR L0 = 0 TO 14
COLOR nums(L0) + 1
IF nums(L0) < 10 THEN PRINT " ";
PRINT RTRIM$(LTRIM$(STR$(nums(L0)))); STRING$(nums(L0), 219); SPACE$(20)
NEXT
ttmp = TIMER
DO WHILE TIMER < ttmp + delay: LOOP
RETURN

Sample output:

Pancake.gif

[edit] BBC BASIC

      DIM test(9)
test() = 4, 65, 2, -31, 0, 99, 2, 83, 782, 1
PROCpancakesort(test())
FOR i% = 0 TO 9
PRINT test(i%) ;
NEXT
PRINT
END
 
DEF PROCpancakesort(a())
LOCAL i%, j%, m%
FOR i% = DIM(a(),1)+1 TO 2 STEP -1
m% = 0
FOR j% = 1 TO i%-1
IF a(j%) > a(m%) m% = j%
NEXT
m% += 1
IF m% < i% THEN
IF m% > 1 PROCflip(a(), m%)
PROCflip(a(), i%)
ENDIF
NEXT
ENDPROC
 
DEF PROCflip(a(), n%)
IF n% < 2 ENDPROC
LOCAL i%
n% -= 1
FOR i% = 0 TO n% DIV 2
SWAP a(i%), a(n%-i%)
NEXT
ENDPROC

Output:

       -31         0         1         2         2         4        65        83        99       782

[edit] C

The function that sorts:

int pancake_sort(int *list, unsigned int length)
{
//If it's less than 2 long, just return it as sorting isn't really needed...
if(length<2)
return 0;
 
int i,a,max_num_pos,moves;
moves=0;
 
for(i=length;i>1;i--)
{
//Find position of the max number in pos(0) to pos(i)
max_num_pos=0;
for(a=0;a<i;a++)
{
if(list[a]>list[max_num_pos])
max_num_pos=a;
}
 
if(max_num_pos==i-1)
//It's where it need to be, skip
continue;
 
 
//Get the found max number to the beginning of the list (unless it already is)
if(max_num_pos)
{
moves++;
do_flip(list, length, max_num_pos+1);
}
 
 
//And then move it to the end of the range we're working with (pos(0) to pos(i))
moves++;
do_flip(list, length, i);
 
//Then everything above list[i-1] is sorted and don't need to be touched
 
}
 
return moves;
}

Where do_flip() is a simple function to flip a part of an array:

void do_flip(int *list, int length, int num)
{
int swap;
int i=0;
for(i;i<--num;i++)
{
swap=list[i];
list[i]=list[num];
list[num]=swap;
}
}

Testing the function:

int main(int argc, char **argv)
{
//Just need some random numbers. I chose <100
int list[9];
int i;
srand(time(NULL));
for(i=0;i<9;i++)
list[i]=rand()%100;
 
 
//Print list, run code and print it again displaying number of moves
printf("\nOriginal: ");
print_array(list, 9);
 
int moves = pancake_sort(list, 9, 1);
 
printf("\nSorted: ");
print_array(list, 9);
printf(" - with a total of %d moves\n", moves);
}

[edit] C++

#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
 
// pancake sort template (calls predicate to determine order)
template <typename BidIt, typename Pred>
void pancake_sort(BidIt first, BidIt last, Pred order)
{
if (std::distance(first, last) < 2) return; // no sort needed
 
for (; first != last; --last)
{
BidIt mid = std::max_element(first, last, order);
if (mid == last - 1)
{
continue; // no flips needed
}
if (first != mid)
{
std::reverse(first, mid + 1); // flip element to front
}
std::reverse(first, last); // flip front to final position
}
}
 
// pancake sort template (ascending order)
template <typename BidIt>
void pancake_sort(BidIt first, BidIt last)
{
pancake_sort(first, last, std::less<typename std::iterator_traits<BidIt>::value_type>());
}
 
int main()
{
std::vector<int> data;
for (int i = 0; i < 20; ++i)
{
data.push_back(i); // generate test data
}
std::random_shuffle(data.begin(), data.end()); // scramble data
 
std::copy(data.begin(), data.end(), std::ostream_iterator<int>(std::cout, " "));
std::cout << "\n";
 
pancake_sort(data.begin(), data.end()); // ascending pancake sort
 
std::copy(data.begin(), data.end(), std::ostream_iterator<int>(std::cout, " "));
std::cout << "\n";
}
Output:
4 10 11 15 14 16 17 1 6 9 3 7 19 2 0 12 5 18 13 8 
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 

[edit] C#

 
public static class PancakeSorter
{
public static void Sort<T>(List<T> list) where T : IComparable
{
if (list.Count < 2)
{
return;
}
int i, a, max_num_pos;
for (i = list.Count; i > 1; i--)
{
max_num_pos = 0;
for (a = 0; a < i; a++)
{
if (list[a].CompareTo(list[max_num_pos]) > 0)
{
max_num_pos = a;
}
}
if (max_num_pos == i - 1)
{
continue;
}
if (max_num_pos > 0)
{
Flip(list, list.Count, max_num_pos + 1);
}
Flip(list, list.Count, i);
}
return;
}
private static void Flip<T>(List<T> list, int length, int num)
{
for (int i = 0; i < --num; i++)
{
T swap = list[i];
list[i] = list[num];
list[num] = swap;
}
}
}
 

[edit] Common Lisp

(defun pancake-sort (seq)
"A destructive version of Pancake Sort that works with either lists or arrays of numbers."
(defun flip (lst index)
(setf (subseq lst 0 index) (reverse (subseq lst 0 index))))
(loop with lst = (coerce seq 'list)
for i from (length lst) downto 2
for index = (position (apply #'max (subseq lst 0 i)) lst)
do (unless (= index 0)
(flip lst (1+ index)))
(flip lst i)
finally (return (coerce lst (type-of seq)))))

Output:

CL-USER> (pancake-sort '(6 7 8 9 2 5 3 4 1))  ;list
(1 2 3 4 5 6 7 8 9)
CL-USER> (pancake-sort #(6 7 8 9 2 5 3 4 1)) ;array
#(1 2 3 4 5 6 7 8 9)
CL-USER> (pancake-sort #(6.5 7.5 8 9 2 5 3 4 1.0)) ;array with integer and floating point values
#(1.0 2 3 4 5 6.5 7.5 8 9)

[edit] D

Translation of: Python
import std.stdio, std.algorithm;
 
void pancakeSort(bool tutor=false, T)(T[] data) {
foreach_reverse (i; 2 .. data.length + 1) {
immutable maxIndex = i - data[0 .. i].minPos!q{a > b}().length;
if (maxIndex + 1 != i) {
if (maxIndex != 0) {
static if (tutor)
writeln("With: ", data, " doflip ", maxIndex + 1);
data[0 .. maxIndex + 1].reverse();
}
 
static if (tutor)
writeln("With: ", data, " doflip ", i);
data[0 .. i].reverse();
}
}
}
 
void main() {
char[] data = "769248135".dup;
pancakeSort!true(data);
writeln(data);
}
Output:
With: 769248135 doflip 3
With: 967248135 doflip 9
With: 531842769 doflip 4
With: 813542769 doflip 8
With: 672453189 doflip 2
With: 762453189 doflip 7
With: 135426789 doflip 3
With: 531426789 doflip 5
With: 241356789 doflip 2
With: 421356789 doflip 4
With: 312456789 doflip 3
With: 213456789 doflip 2
123456789

[edit] Euphoria

function flip(sequence s, integer n)
object temp
for i = 1 to n/2 do
temp = s[i]
s[i] = s[n-i+1]
s[n-i+1] = temp
end for
return s
end function
 
function pancake_sort(sequence s)
integer m
for i = length(s) to 2 by -1 do
m = 1
for j = 2 to i do
if compare(s[j], s[m]) > 0 then
m = j
end if
end for
 
if m < i then
if m > 1 then
s = flip(s,m)
end if
s = flip(s,i)
end if
end for
return s
end function
 
constant s = rand(repeat(100,10))
 
? s
? pancake_sort(s)

Output:

{24,32,100,15,8,34,50,79,46,52}
{8,15,24,32,34,46,50,52,79,100}

[edit] F#

open System
 
let show data = data |> Array.iter (printf "%d ") ; printfn ""
let split (data: int[]) pos = data.[0..pos], data.[(pos+1)..]
 
let flip items pos =
let lower, upper = split items pos
Array.append (Array.rev lower) upper
 
let pancakeSort items =
let rec loop data limit =
if limit <= 0 then data
else
let lower, upper = split data limit
let indexOfMax = lower |> Array.findIndex ((=) (Array.max lower))
let partialSort = Array.append (flip lower indexOfMax |> Array.rev) upper
loop partialSort (limit-1)
 
loop items ((Array.length items)-1)

Usage: pancakeSort [|31; 41; 59; 26; 53; 58; 97; 93; 23; 84|] |> show

Output:

  23 26 31 41 53 58 59 84 93 97

[edit] Fortran

Works with: Fortran version 90 and later
program Pancake_Demo
implicit none
 
integer :: list(8) = (/ 1, 4, 7, 2, 5, 8, 3, 6 /)
 
call Pancake_sort(list)
 
contains
 
subroutine Pancake_sort(a)
 
integer, intent(in out) :: a(:)
integer :: i, maxpos
 
write(*,*) a
do i = size(a), 2, -1
 
! Find position of max number between index 1 and i
maxpos = maxloc(a(1:i), 1)
 
! is it in the correct position already?
if (maxpos == i) cycle
 
! is it at the beginning of the array? If not flip array section so it is
if (maxpos /= 1) then
a(1:maxpos) = a(maxpos:1:-1)
write(*,*) a
end if
 
! Flip array section to get max number to correct position
a(1:i) = a(i:1:-1)
write(*,*) a
end do
 
end subroutine
 
end program Pancake_Demo

Output:

            1           4           7           2           5           8           3           6
            8           5           2           7           4           1           3           6
            6           3           1           4           7           2           5           8
            7           4           1           3           6           2           5           8
            5           2           6           3           1           4           7           8
            6           2           5           3           1           4           7           8
            4           1           3           5           2           6           7           8
            5           3           1           4           2           6           7           8
            2           4           1           3           5           6           7           8
            4           2           1           3           5           6           7           8
            3           1           2           4           5           6           7           8
            2           1           3           4           5           6           7           8
            1           2           3           4           5           6           7           8

[edit] Go

package main
 
import "fmt"
 
func main() {
list := pancake{31, 41, 59, 26, 53, 58, 97, 93, 23, 84}
fmt.Println("unsorted:", list)
 
list.sort()
fmt.Println("sorted! ", list)
}
 
type pancake []int
 
func (a pancake) sort() {
for uns := len(a) - 1; uns > 0; uns-- {
// find largest in unsorted range
lx, lg := 0, a[0]
for i := 1; i <= uns; i++ {
if a[i] > lg {
lx, lg = i, a[i]
}
}
// move to final position in two flips
a.flip(lx)
a.flip(uns)
}
}
 
func (a pancake) flip(r int) {
for l := 0; l < r; l, r = l+1, r-1 {
a[l], a[r] = a[r], a[l]
}
}

Output:

unsorted: [31 41 59 26 53 58 97 93 23 84]
sorted!   [23 26 31 41 53 58 59 84 93 97]

[edit] Groovy

This formulation of the pancake sort achieves stability by picking the last index (rather than, say, the first) in the remaining sublist that matches the max value of the remaining sublist. Performance is enhanced somewhat by not flipping if the flipPoint is already at the end of the remaining sublist.

def makeSwap = { a, i, j = i+1 -> print "."; a[[j,i]] = a[[i,j]] }
 
def flip = { list, n -> (0..<((n+1)/2)).each { makeSwap(list, it, n-it) } }
 
def pancakeSort = { list ->
def n = list.size()
(1..<n).reverse().each { i ->
def max = list[0..i].max()
def flipPoint = (i..0).find{ list[it] == max }
if (flipPoint != i) {
flip(list, flipPoint)
flip(list, i)
}
}
list
}

Test:

println (pancakeSort([23,76,99,58,97,57,35,89,51,38,95,92,24,46,31,24,14,12,57,78,4]))
println (pancakeSort([88,18,31,44,4,0,8,81,14,78,20,76,84,33,73,75,82,5,62,70,12,7,1]))
println ()
println (pancakeSort([10, 10.0, 10.00, 1]))
println (pancakeSort([10, 10.00, 10.0, 1]))
println (pancakeSort([10.0, 10, 10.00, 1]))
println (pancakeSort([10.0, 10.00, 10, 1]))
println (pancakeSort([10.00, 10, 10.0, 1]))
println (pancakeSort([10.00, 10.0, 10, 1]))

The use of decimals and integers that compare as equal demonstrates, but of course not prove, that the sort is stable.

Output:

..........................................................................................................................................[4, 12, 14, 23, 24, 24, 31, 35, 38, 46, 51, 57, 57, 58, 76, 78, 89, 92, 95, 97, 99]
............................................................................................................................................................................................................[0, 1, 4, 5, 7, 8, 12, 14, 18, 20, 31, 33, 44, 62, 70, 73, 75, 76, 78, 81, 82, 84, 88]

...[1, 10, 10.0, 10.00]
...[1, 10, 10.00, 10.0]
...[1, 10.0, 10, 10.00]
...[1, 10.0, 10.00, 10]
...[1, 10.00, 10, 10.0]
...[1, 10.00, 10.0, 10]

[edit] Haskell

import Data.List
import Control.Arrow
import Control.Monad
import Data.Maybe
 
dblflipIt :: (Ord a) => [a] -> [a]
dblflipIt = uncurry ((reverse.).(++)). first reverse
. ap (flip splitAt) (succ. fromJust. (elemIndex =<< maximum))
 
dopancakeSort :: (Ord a) => [a] -> [a]
dopancakeSort xs = dopcs (xs,[]) where
dopcs ([],rs) = rs
dopcs ([x],rs) = x:rs
dopcs (xs,rs) = dopcs $ (init &&& (:rs).last ) $ dblflipIt xs

Example:

*Main>  dopancakeSort [3,2,1,0,2]
[0,1,2,2,3]

[edit] Icon and Unicon

procedure main()                                        #: demonstrate various ways to sort a list and string 
demosort(pancakesort,[3, 14, 1, 5, 9, 2, 6, 3],"qwerty")
pancakeflip := pancakeflipshow # replace pancakeflip procedure with a variant that displays each flip
pancakesort([3, 14, 1, 5, 9, 2, 6, 3])
end
 
procedure pancakesort(X,op) #: return sorted list ascending(or descending)
local i,m
 
op := sortop(op,X) # select how and what we sort
 
every i := *X to 2 by -1 do { # work back to front
m := 1
every j := 2 to i do
if op(X[m],X[j]) then m := j # find X that belongs @i high (or low)
if i ~= m then { # not already in-place
X := pancakeflip(X,m) # . bring max (min) to front
X := pancakeflip(X,i) # . unsorted portion of stack
}
}
return X
end
 
procedure pancakeflip(X,tail) #: return X[1:tail] flipped
local i
 
i := 0
tail := integer(\tail|*X) + 1 | runerr(101,tail)
while X[(i +:= 1) < (tail -:= 1)] :=: X[i] # flip
return X
end
 
procedure pancakeflipshow(X,tail) #: return X[1:tail] flipped (and display)
local i
 
i := 0
tail := integer(\tail|*X) + 1 | runerr(101,tail)
while X[(i +:= 1) < (tail -:= 1)] :=: X[i] # flip
every writes(" ["|right(!X,4)|" ]\n") # show X
return X
end

Note: This example relies on the supporting procedures 'sortop', and 'demosort' in Bubble Sort. The full demosort exercises the named sort of a list with op = "numeric", "string", ">>" (lexically gt, descending),">" (numerically gt, descending), a custom comparator, and also a string.

Abbreviated sample output:
Sorting Demo using procedure pancakesort
  on list : [ 3 14 1 5 9 2 6 3 ]
    with op = &null:         [ 1 2 3 3 5 6 9 14 ]   (0 ms)
  ...
  on string : "qwerty"
    with op = &null:         "eqrtwy"   (0 ms)
The output below shows the flipping:
     [  14   3   1   5   9   2   6   3 ]
     [   3   6   2   9   5   1   3  14 ]
     [   9   2   6   3   5   1   3  14 ]
     [   3   1   5   3   6   2   9  14 ]
     [   6   3   5   1   3   2   9  14 ]
     [   2   3   1   5   3   6   9  14 ]
     [   5   1   3   2   3   6   9  14 ]
     [   3   2   3   1   5   6   9  14 ]
     [   3   2   3   1   5   6   9  14 ]
     [   1   3   2   3   5   6   9  14 ]
     [   3   1   2   3   5   6   9  14 ]
     [   2   1   3   3   5   6   9  14 ]
     [   2   1   3   3   5   6   9  14 ]
     [   1   2   3   3   5   6   9  14 ]

[edit] 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.
flip=: C.~ C.@i.@-
unsorted=: #~ 1 , [: >./\. 2 >/\ ]
FlDown=: flip 1 + (i. >./)@unsorted
FlipUp=: flip 1 >. [:+/>./\&|.@(< {.)
 
pancake=: FlipUp@FlDown^:_

Example use:

   (,:pancake) ?~9
1 0 8 7 4 6 3 5 2
0 1 2 3 4 5 6 7 8

See the discussion page for illustrations of the other words.

[edit] Java

 
public class PancakeSort
{
int[] heap;
 
public String toString() {
String info = "";
for (int x: heap)
info += x + " ";
return info;
}
 
public void flip(int n) {
for (int i = 0; i < (n+1) / 2; ++i) {
int tmp = heap[i];
heap[i] = heap[n-i];
heap[n-i] = tmp;
}
System.out.println("flip(0.." + n + "): " + toString());
}
 
public int[] minmax(int n) {
int xm, xM;
xm = xM = heap[0];
int posm = 0, posM = 0;
 
for (int i = 1; i < n; ++i) {
if (heap[i] < xm) {
xm = heap[i];
posm = i;
}
else if (heap[i] > xM) {
xM = heap[i];
posM = i;
}
}
return new int[] {posm, posM};
}
 
public void sort(int n, int dir) {
if (n == 0) return;
 
int[] mM = minmax(n);
int bestXPos = mM[dir];
int altXPos = mM[1-dir];
boolean flipped = false;
 
if (bestXPos == n-1) {
--n;
}
else if (bestXPos == 0) {
flip(n-1);
--n;
}
else if (altXPos == n-1) {
dir = 1-dir;
--n;
flipped = true;
}
else {
flip(bestXPos);
}
sort(n, dir);
 
if (flipped) {
flip(n);
}
}
 
PancakeSort(int[] numbers) {
heap = numbers;
sort(numbers.length, 1);
}
 
public static void main(String[] args) {
int[] numbers = new int[args.length];
for (int i = 0; i < args.length; ++i)
numbers[i] = Integer.valueOf(args[i]);
 
PancakeSort pancakes = new PancakeSort(numbers);
System.out.println(pancakes);
}
}

Example:

$ java PancakeSort  1 2 5 4 3 10 9 8 7
flip(0..5): 10 3 4 5 2 1 9 8 7
flip(0..8): 7 8 9 1 2 5 4 3 10
flip(0..2): 9 8 7 1 2 5 4 3 10
flip(0..7): 3 4 5 2 1 7 8 9 10
flip(0..2): 5 4 3 2 1 7 8 9 10
flip(0..4): 1 2 3 4 5 7 8 9 10
1 2 3 4 5 7 8 9 10
 
$ java PancakeSort 6 7 2 1 8 9 5 3 4
flip(0..5): 9 8 1 2 7 6 5 3 4
flip(0..8): 4 3 5 6 7 2 1 8 9
flip(0..1): 3 4 5 6 7 2 1 8 9
flip(0..4): 7 6 5 4 3 2 1 8 9
flip(0..6): 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9

[edit] JavaScript

Array.prototype.pancake_sort = function () {
for (var i = this.length - 1; i >= 1; i--) {
// find the index of the largest element not yet sorted
var max_idx = 0;
var max = this[0];
for (var j = 1; j <= i; j++) {
if (this[j] > max) {
max = this[j];
max_idx = j;
}
}
 
if (max_idx == i)
continue; // element already in place
 
var new_slice;
 
// flip this max element to index 0
if (max_idx > 0) {
new_slice = this.slice(0, max_idx+1).reverse();
for (var j = 0; j <= max_idx; j++)
this[j] = new_slice[j];
}
 
// then flip the max element to its place
new_slice = this.slice(0, i+1).reverse();
for (var j = 0; j <= i; j++)
this[j] = new_slice[j];
}
return this;
}
ary = [7,6,5,9,8,4,3,1,2,0]
sorted = ary.concat().pancake_sort();

[edit] Mathematica

LMaxPosition[ a_, n_ ] := Part[Position[a[[;;n]],Max[a[[;;n]]]],1,1]
 
SetAttributes[Flip,HoldFirst]; Flip[a_] := Set[a,Reverse[a]]
 
pancakeSort[a_] : = For[n = Length[a], n > 1, n--,
If[LMaxPosition[a,n] < n,
Flip[a[[;;LMaxPosition[a,n]]]]; Print[a];
Flip[a[[;;n]]]; Print[a];
];
];
(* each major sort step is printed in example usage *)
pancakeSort[{6, 7, 8, 9, 2, 5, 3, 4, 1}]

{9,8,7,6,2,5,3,4,1}
{1,4,3,5,2,6,7,8,9}
{5,3,4,1,2,6,7,8,9}
{2,1,4,3,5,6,7,8,9}
{4,1,2,3,5,6,7,8,9}
{3,2,1,4,5,6,7,8,9}
{3,2,1,4,5,6,7,8,9}
{1,2,3,4,5,6,7,8,9}

[edit] MATLAB / Octave

function list = pancakeSort(list)
 
for i = (numel(list):-1:2)
 
minElem = list(i);
minIndex = i;
 
%Find the min element in the current subset of the list
for j = (i:-1:1)
if list(j) <= minElem
minElem = list(j);
minIndex = j;
end
end
 
%If the element is already in the correct position don't flip
if i ~= minIndex
 
%First flip flips the min element in the stack to the top
list(minIndex:-1:1) = list(1:minIndex);
 
%Second flip flips the min element into the correct position in
%the stack
list(i:-1:1) = list(1:i);
 
end
end %for
end %pancakeSort

Sample Usage:

>> pancakeSort([4 3 1 5 6 2])
 
ans =
 
6 5 4 3 2 1

[edit] NetRexx

Sorts integers, decimal numbers and strings because they're all the same to NetRexx.

/* NetRexx */
options replace format comments java crossref symbols nobinary
 
import java.util.List
 
runSample(arg)
return
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method pancakeSort(tlist = List, debug = (1 == 0)) private static returns List
 
if tlist.size() > 1 then do
loop i_ = tlist.size() by -1 while i_ > 1
maxPos = 0
loop a_ = 0 while a_ < i_
if Rexx tlist.get(a_) > Rexx tlist.get(maxPos) then maxPos = a_
end a_
if maxPos = i_ - 1 then iterate i_
if maxPos > 0 then pancakeFlip(tlist, maxPos + 1, debug)
pancakeFlip(tlist, i_, debug)
end i_
end
return tlist
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method pancakeFlip(tlist = List, offset, debug = (1 == 0)) private static returns List
z_ = offset - 1
pl = 3
if debug then do
plx = offset.length()
if plx > pl then pl = plx
say ' flip{1-'offset.right(pl, 0)'} Before:' tlist
end
loop i_ = 0 while i_ < z_
Collections.swap(tlist, i_, z_)
z_ = z_ - 1
end i_
if debug then do
say ' flip{1-'offset.right(pl, 0)'} After:' tlist
end
return tlist
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method runSample(arg) private static
 
isTrue = (1 == 1)
isFalse = \isTrue
 
parse arg debug .
if '-debug'.abbrev(debug.lower(), 2) then debug = isTrue
else debug = isFalse
 
lists = sampleData()
loop il = 1 to lists[0]
clist = words2list(lists[il])
say ' Input:' clist
say 'Output:' pancakeSort(clist, debug)
say
end il
 
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method sampleData() private static
lists = ''
i_ = 0
i_ = i_ + 1; lists[0] = i_; lists[i_] = '1 4 3 5 2 9 8 7 6'
i_ = i_ + 1; lists[0] = i_; lists[i_] = '10 -9 8 -7 6 -5 4 -3 2 -1 0 -10 9 -8 7 -6 5 -4 3 -2 1'
i_ = i_ + 1; lists[0] = i_; lists[i_] = '88 18 31 44 4 0 8 81 14 78 20 76 84 33 73 75 82 5 62 70 12 7 1'
i_ = i_ + 1; lists[0] = i_; lists[i_] = '10 10.0 10.00 1 -10.0 10. -1'
i_ = i_ + 1; lists[0] = i_; lists[i_] = 'To be or not to be that is the question'
i_ = i_ + 1; lists[0] = i_; lists[i_] = '1'
return lists
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method words2list(wordlist) private static returns List
 
clist = ArrayList()
loop w_ = 1 to wordlist.words()
clist.add(wordlist.word(w_))
end w_
 
return clist
 
Output:
 Input: [1, 4, 3, 5, 2, 9, 8, 7, 6]
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9]

 Input: [10, -9, 8, -7, 6, -5, 4, -3, 2, -1, 0, -10, 9, -8, 7, -6, 5, -4, 3, -2, 1]
Output: [-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

 Input: [88, 18, 31, 44, 4, 0, 8, 81, 14, 78, 20, 76, 84, 33, 73, 75, 82, 5, 62, 70, 12, 7, 1]
Output: [0, 1, 4, 5, 7, 8, 12, 14, 18, 20, 31, 33, 44, 62, 70, 73, 75, 76, 78, 81, 82, 84, 88]

 Input: [10, 10.0, 10.00, 1, -10.0, 10., -1]
Output: [-10.0, -1, 1, 10.00, 10.0, 10., 10]

 Input: [To, be, or, not, to, be, that, is, the, question]
Output: [be, be, is, not, or, question, that, the, to, To]

 Input: [1]
Output: [1]

[edit] Nimrod

import algorithm
 
proc pancakeSort[T](list: var openarray[T]) =
var length = list.len
if length < 2: return
 
var moves = 0
 
for i in countdown(length, 2):
var maxNumPos = 0
for a in 0 .. <i:
if list[a] > list[maxNumPos]:
maxNumPos = a
 
if maxNumPos == i - 1: continue
 
if maxNumPos > 0:
inc moves
reverse(list, 0, maxNumPos)
 
inc moves
reverse(list, 0, i - 1)
 
var a = @[4, 65, 2, -31, 0, 99, 2, 83, 782]
pancakeSort a
echo a

Output:

@[-31, 0, 2, 2, 4, 65, 83, 99, 782]

[edit] OCaml

let rec sorted = function
| [] -> (true)
| x::y::_ when x > y -> (false)
| x::xs -> sorted xs
 
let rev_until_max li =
let rec aux acc greater prefix suffix = function
| x::xs when x > greater -> aux (x::acc) x acc xs xs
| x::xs -> aux (x::acc) greater prefix suffix xs
| [] -> (greater, (prefix @ suffix))
in
aux [] min_int [] li li
 
let pancake_sort li =
let rec aux i li suffix =
let greater, li = rev_until_max li in
let suffix = greater :: suffix
and li = List.rev li in
if sorted li
then (li @ suffix), i
else aux (succ i) li suffix
in
aux 0 li []
 
let print_list li =
List.iter (Printf.printf " %d") li;
print_newline()
 
let make_rand_list n bound =
let rec aux acc i =
if i >= n then (acc)
else aux ((Random.int bound)::acc) (succ i)
in
aux [] 0
 
let () =
Random.self_init();
let li = make_rand_list 8 100 in
print_list li;
let res, n = pancake_sort li in
print_list res;
Printf.printf " sorted in %d loops\n" n;
;;

[edit] PARI/GP

pancakeSort(v)={
my(top=#v);
while(top>1,
my(mx=1,t);
for(i=2,top,if(v[i]>v[mx], mx=i));
if(mx==top, top--; next);
for(i=1,mx\2,
t=v[i];
v[i]=v[mx+1-i];
v[mx+1-i]=t
);
for(i=1,top\2,
t=v[i];
v[i]=v[top+1-i];
v[top+1-i]=t
);
top--
);
v
};

[edit] Pascal

Program PancakeSort (output);
 
procedure flip(var b: array of integer; last: integer);
 
var
swap, i: integer;
 
begin
for i := low(b) to (last - low(b) - 1) div 2 do
begin
swap := b[i];
b[i] := b[last-(i-low(b))];
b[last-(i-low(b))] := swap;
end;
end;
 
procedure PancakeSort(var a: array of integer);
 
var
i, j, maxpos: integer;
 
begin
for i := high(a) downto low(a) do
begin
// Find position of max number between beginning and i
maxpos := i;
for j := low(a) to i - 1 do
if a[j] > a[maxpos] then
maxpos := j;
 
// is it in the correct position already?
if maxpos = i then
continue;
 
// is it at the beginning of the array? If not flip array section so it is
if maxpos <> low(a) then
flip(a, maxpos);
 
// Flip array section to get max number to correct position
flip(a, i);
end;
end;
 
var
data: array of integer;
i: integer;
 
begin
setlength(data, 8);
Randomize;
writeln('The data before sorting:');
for i := low(data) to high(data) do
begin
data[i] := Random(high(data));
write(data[i]:4);
end;
writeln;
PancakeSort(data);
writeln('The data after sorting:');
for i := low(data) to high(data) do
begin
write(data[i]:4);
end;
writeln;
end.

Output:

:>./PancakeSort
The data before sorting:
   3   1   3   2   4   0   2   6
The data after sorting:
   0   1   2   2   3   3   4   6

[edit] Perl

sub pancake {
my @x = @_;
for my $idx (0 .. $#x - 1) {
my $min = $idx;
$x[$min] > $x[$_] and $min = $_ for $idx + 1 .. $#x;
 
next if $x[$min] == $x[$idx];
 
@x[$min .. $#x] = reverse @x[$min .. $#x] if $x[$min] != $x[-1];
@x[$idx .. $#x] = reverse @x[$idx .. $#x];
}
@x;
}
 
my @a = map (int rand(100), 1 .. 10);
print "Before @a\n";
@a = pancake(@a);
print "After @a\n";
 

Sample output:

Before 57 37 35 35 22 58 70 53 77 15
After  15 22 35 35 37 53 57 58 70 77

[edit] Perl 6

sub pancake_sort ( @a is copy ) {
my $endpoint = @a.end;
while $endpoint > 0 and not [<] @a {
my $max_i = [0..$endpoint].max: { @a[$_] };
my $max = @a[$max_i];
if @a[$endpoint] == $max {
$endpoint-- while @a[$endpoint] == $max;
next;
}
# @a[$endpoint] is not $max, so it needs flipping;
# Flip twice if max is not already at the top.
@a[0..$max_i] .= reverse if $max_i != 0;
@a[0..$endpoint] .= reverse;
$endpoint--;
}
return @a;
}
my @data = 6, 7, 2, 1, 8, 9, 5, 3, 4;
say 'input = ' ~ @data;
say 'output = ' ~ @data.&pancake_sort;
 
Output:
input  = 6 7 2 1 8 9 5 3 4
output = 1 2 3 4 5 6 7 8 9

[edit] PL/I

 
pancake_sort: procedure options (main); /* 23 April 2009 */
declare a(10) fixed, (i, n, loc) fixed binary;
 
a(1) = 3; a(2) = 9; a(3) = 2; a(4) = 7; a(5) = 10;
a(6) = 1; a(7) = 8; a(8) = 5; a(9) = 4; a(10) = 6;
 
n = hbound(A,1);
put skip edit (A) (f(5));
do i = 1 to n-1;
loc = max(A, n);
call flip (A, loc);
call flip (A, n);
n = n - 1;
put skip edit (A) (f(5));
end;
 
max: procedure (A, k) returns (fixed binary);
declare A(*) fixed, k fixed binary;
declare (i, maximum, loc) fixed binary;
maximum = A(1); loc = 1;
do i = 2 to k;
if A(i) > maximum then do; maximum = A(i); loc = i; end;
end;
return (loc);
end max;
 
flip: procedure (A, k);
declare A(*) fixed, k fixed binary;
declare (i, t) fixed binary;
do i = 1 to (k+1)/2;
t = A(i); A(i) = A(k-i+1); A(k-i+1) = t;
end;
end flip;
 
end pancake_sort;
 

Output:

 
3 9 2 7 10 1 8 5 4 6
6 4 5 8 1 3 9 2 7 10
7 2 6 4 5 8 1 3 9 10
3 1 7 2 6 4 5 8 9 10
5 4 6 2 3 1 7 8 9 10
1 3 2 5 4 6 7 8 9 10
4 1 3 2 5 6 7 8 9 10
2 3 1 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
 

[edit] PicoLisp

(de pancake (Lst)
(prog1 (flip Lst (index (apply max Lst) Lst))
(for (L @ (cdr (setq Lst (cdr L))) (cdr L))
(con L (flip Lst (index (apply max Lst) Lst))) ) ) )

Output:

: (trace 'flip)
-> flip

: (pancake (6 7 2 1 8 9 5 3 4))
 flip : (6 7 2 1 8 9 5 3 4) 6
 flip = (9 8 1 2 7 6 5 3 4)
 flip : (8 1 2 7 6 5 3 4) 1
 flip = (8 1 2 7 6 5 3 4)
 flip : (1 2 7 6 5 3 4) 3
 flip = (7 2 1 6 5 3 4)
 flip : (2 1 6 5 3 4) 3
 flip = (6 1 2 5 3 4)
 flip : (1 2 5 3 4) 3
 flip = (5 2 1 3 4)
 flip : (2 1 3 4) 4
 flip = (4 3 1 2)
 flip : (3 1 2) 1
 flip = (3 1 2)
 flip : (1 2) 2
 flip = (2 1)
-> (9 8 7 6 5 4 3 2 1)

[edit] PowerShell

Function FlipPancake( [Object[]] $indata, $index = 1 )
{
$data=$indata.Clone()
$datal = $data.length - 1
if( $index -gt 0 )
{
if( $datal -gt $index )
{
$first = $data[ $index..0 ]
$last = $data[ ( $index + 1 )..$datal ]
$data = $first + $last
} else {
$data = $data[ $index..0 ]
}
}
$data
}
 
Function MaxIdx( [Object[]] $data )
{
$data | ForEach-Object { $max = $data[ 0 ]; $i = 0; $maxi = 0 } { if( $_ -gt $max ) { $max = $_; $maxi = $i }; $i++ } { $maxi }
}
 
Function PancakeSort( [Object[]] $data, $index = 0 )
{
"unsorted - $data"
$datal = $data.length - 1
if( $datal -gt 0 )
{
for( $i = $datal; $i -gt 0; $i-- )
{
$data = FlipPancake ( FlipPancake $data ( MaxIdx $data[ 0..$i ] ) ) $i
}
}
"sorted - $data"
}
 
$l = 100; PancakeSort ( 1..$l | ForEach-Object { $Rand = New-Object Random }{ $Rand.Next( 0, $l - 1 ) } )

[edit] PureBasic

If OpenConsole()
Define i, j, k, Loops
Dim Pile(9)
;--------------------------------------------------------------
;- Create a Random Pile()
For i=1 To 9 ;- Initiate the Pile
Pile(i)=i
Next
For i=9 To 1 Step -1 ;- Do a Fisher-Yates shuffle
Swap Pile(i),Pile(Random(i-1)+1)
Next
Print("Random Pile()  :")
For i=1 To 9
Print(" "+Str(Pile(i)))
Next
;--------------------------------------------------------------
;- Start Sorting
For i=9 To 2 Step -1
If Pile(i)<>i ;- Only Flip it if the current cake need Swapping
Loops+1
j=0
Repeat ;- find place of Pancake(i) in the Pile()
j+1
Until Pile(j)=i
 
For k=1 To (j/2) ;- Flip it up
Swap Pile(k),Pile(j-k+1)
Next
For k=1 To i/2 ;- Flip in place
Swap Pile(k),Pile(i-k+1)
Next
 
EndIf
Next
 
Print(#CRLF$+"Resulting Pile() :")
For i=1 To 9
Print(" "+str(Pile(i)))
Next
Print(#CRLF$+"All done in "+str(Loops)+" loops.")
Print(#CRLF$+#CRLF$+"Press ENTER to quit."): Input()
CloseConsole()
EndIf

Output can look like

Original Pile()  : 9 4 1 8 6 3 2 5 7
Resulting Pile() : 1 2 3 4 5 6 7 8 9
All done in 6 loops.

Press ENTER to quit.

[edit] Python

The function:

tutor = False
 
def pancakesort(data):
if len(data) <= 1:
return data
if tutor: print()
for size in range(len(data), 1, -1):
maxindex = max(range(size), key=data.__getitem__)
if maxindex+1 != size:
# This indexed max needs moving
if maxindex != 0:
# Flip the max item to the left
if tutor: print('With: %r doflip  %i'
 % ( ' '.join(str(x) for x in data), maxindex+1 ))
data[:maxindex+1] = reversed(data[:maxindex+1])
# Flip it into its final position
if tutor: print('With: %r doflip %i'
 % ( ' '.join(str(x) for x in data), size ))
data[:size] = reversed(data[:size])
if tutor: print()

A test:

if __name__ == '__main__':
import random
 
tutor = True
data = list('123456789')
while data == sorted(data):
random.shuffle(data)
print('Original List: %r' % ' '.join(data))
pancakesort(data)
print('Pancake Sorted List: %r' % ' '.join(data))

Sample output:

Original List: '6 7 2 1 8 9 5 3 4'

With: '6 7 2 1 8 9 5 3 4' doflip  6
With: '9 8 1 2 7 6 5 3 4'  doflip 9
With: '4 3 5 6 7 2 1 8 9' doflip  5
With: '7 6 5 3 4 2 1 8 9'  doflip 7
With: '1 2 4 3 5 6 7 8 9' doflip  3
With: '4 2 1 3 5 6 7 8 9'  doflip 4
With: '3 1 2 4 5 6 7 8 9'  doflip 3
With: '2 1 3 4 5 6 7 8 9'  doflip 2

Pancake Sorted List: '1 2 3 4 5 6 7 8 9'

[edit] Racket

 
#lang racket
 
(define (pancake-sort l)
(define (flip l n) (append (reverse (take l n)) (drop l n)))
(for/fold ([l l]) ([i (in-range (length l) 1 -1)])
(let* ([i2 (cdr (for/fold ([m #f]) ([x l] [j i])
(if (and m (<= x (car m))) m (cons x j))))]
[l (if (zero? i2) l (flip l (add1 i2)))])
(flip l i))))
 
(pancake-sort (shuffle (range 0 10)))
;; => '(0 1 2 3 4 5 6 7 8 9)
 

[edit] REXX

/*REXX program sorts/shows an array using the  pancake  sort  algorithm.*/
call gen@ /*generate elements in the array.*/
call show@ 'before sort' /*show the BEFORE array elements.*/
call pancakeSort #items /*invoke the pancake sort. Yummy.*/
call show@ ' after sort' /*show the AFTER array elements.*/
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────PANCAKESORT subroutine──────────────*/
pancakeSort: procedure expose @.; parse arg n
 
do n-1 /*perform this loop N-1 times.*/
 !=@.1;  ?=1
do j=2 to n
if @.j<=! then iterate
 !=@.j;  ?=j
end /*j*/
call flip ?; call flip n
n=n-1
end /*n-1*/
return
/*──────────────────────────────────FLIP subroutine─────────────────────*/
flip: procedure expose @.; parse arg w
do i=1 for (w+1)%2 /*In REXX,  % is integer divide.*/
kmip1=w-i+1; _=@.i; @.i=@.kmip1; @.kmip1=_
end /*i*/
return
/*──────────────────────────────────GEN@ subroutine─────────────────────*/
gen@: @.='' /*assign a default value. */
/*Generate some bread primes which are primes of the*/
/*form: (p-3)÷2 and 2∙p+3 where p is a prime.*/
/*Bread primes are related to sandwich & meat primes.*/
 
@.1= 2  ; @.17= 113  ; @.33= -55
@.2= 17  ; @.18= 461  ; @.34= -21
@.3= 5  ; @.19= 137  ; @.35= -1
@.4= 29  ; @.20= 557  ; @.36= -8
@.5= 7  ; @.21= 167  ; @.37= -8
@.6= 37  ; @.22= 677  ; @.38= -21
@.7= 13  ; @.23= 173  ; @.39= -55
@.8= 61  ; @.24= 701  ; @.40= 0
@.9= 43  ; @.25= 797  ; @.41= 0
@.10= 181  ; @.26= 1117
@.11= 47  ; @.27= 307 /*The non-positive numbers*/
@.12= 197  ; @.28= 1237 /*above are some negative */
@.13= 67  ; @.29= 1597 /*Fibonacci numbers, some */
@.14= 277  ; @.30= 463 /*of which where specified*/
@.15= 97  ; @.31= 1861 /*twice just 'cause I felt*/
@.16= 397  ; @.32= 467 /*like it. Also, added */
/*two zeroes, just 'cause.*/
do #items=1 while @.#items\==''; end /*find how many entries in array.*/
 
#items=#items-1 /*adjust the #items slightly. */
return
/*──────────────────────────────────SHOW@ subroutine────────────────────*/
show@: widthH=length(#items) /*the maximum width of any line. */
do k=1 for #items
say 'element' right(k,widthH) arg(1)':' @.k
end /*k*/
say copies('─',79) /*show an eyeball separator line.*/
return

output

element  1 before sort: 2
element  2 before sort: 17
element  3 before sort: 5
element  4 before sort: 29
element  5 before sort: 7
element  6 before sort: 37
element  7 before sort: 13
element  8 before sort: 61
element  9 before sort: 43
element 10 before sort: 181
element 11 before sort: 47
element 12 before sort: 197
element 13 before sort: 67
element 14 before sort: 277
element 15 before sort: 97
element 16 before sort: 397
element 17 before sort: 113
element 18 before sort: 461
element 19 before sort: 137
element 20 before sort: 557
element 21 before sort: 167
element 22 before sort: 677
element 23 before sort: 173
element 24 before sort: 701
element 25 before sort: 797
element 26 before sort: 1117
element 27 before sort: 307
element 28 before sort: 1237
element 29 before sort: 1597
element 30 before sort: 463
element 31 before sort: 1861
element 32 before sort: 467
element 33 before sort: -55
element 34 before sort: -21
element 35 before sort: -1
element 36 before sort: -8
element 37 before sort: -8
element 38 before sort: -21
element 39 before sort: -55
element 40 before sort: 0
element 41 before sort: 0
───────────────────────────────────────────────────────────────────────────────
element  1  after sort: -55
element  2  after sort: -55
element  3  after sort: -21
element  4  after sort: -21
element  5  after sort: -8
element  6  after sort: -8
element  7  after sort: -1
element  8  after sort: 0
element  9  after sort: 0
element 10  after sort: 2
element 11  after sort: 5
element 12  after sort: 7
element 13  after sort: 13
element 14  after sort: 17
element 15  after sort: 29
element 16  after sort: 37
element 17  after sort: 43
element 18  after sort: 47
element 19  after sort: 61
element 20  after sort: 67
element 21  after sort: 97
element 22  after sort: 113
element 23  after sort: 137
element 24  after sort: 167
element 25  after sort: 173
element 26  after sort: 181
element 27  after sort: 197
element 28  after sort: 277
element 29  after sort: 307
element 30  after sort: 397
element 31  after sort: 461
element 32  after sort: 463
element 33  after sort: 467
element 34  after sort: 557
element 35  after sort: 677
element 36  after sort: 701
element 37  after sort: 797
element 38  after sort: 1117
element 39  after sort: 1237
element 40  after sort: 1597
element 41  after sort: 1861
───────────────────────────────────────────────────────────────────────────────

[edit] Ruby

class Array
def pancake_sort!
num_flips = 0
(self.size-1).downto(1) do |end_idx|
max = self[0..end_idx].max
max_idx = self[0..end_idx].index(max)
next if max_idx == end_idx
 
if max_idx > 0
self[0..max_idx] = self[0..max_idx].reverse
p [num_flips += 1, self] if $DEBUG
end
 
self[0..end_idx] = self[0..end_idx].reverse
p [num_flips += 1, self] if $DEBUG
end
self
end
end
 
p a = (1..9).to_a.shuffle
p a.pancake_sort!

sample output:

$ ruby -d sorting_pancake.rb
[7, 3, 6, 8, 2, 4, 5, 1, 9]
[1, [8, 6, 3, 7, 2, 4, 5, 1, 9]]
[2, [1, 5, 4, 2, 7, 3, 6, 8, 9]]
[3, [7, 2, 4, 5, 1, 3, 6, 8, 9]]
[4, [6, 3, 1, 5, 4, 2, 7, 8, 9]]
[5, [2, 4, 5, 1, 3, 6, 7, 8, 9]]
[6, [5, 4, 2, 1, 3, 6, 7, 8, 9]]
[7, [3, 1, 2, 4, 5, 6, 7, 8, 9]]
[8, [2, 1, 3, 4, 5, 6, 7, 8, 9]]
[9, [1, 2, 3, 4, 5, 6, 7, 8, 9]]
[1, 2, 3, 4, 5, 6, 7, 8, 9]

[edit] Tcl

package require Tcl 8.5
# Some simple helper procedures
proc flip {nlist n} {
concat [lreverse [lrange $nlist 0 $n]] [lrange $nlist $n+1 end]
}
proc findmax {nlist limit} {
lsearch -exact $nlist [tcl::mathfunc::max {*}[lrange $nlist 0 $limit]]
}
 
# Simple-minded pancake sort algorithm
proc pancakeSort {nlist {debug ""}} {
for {set i [llength $nlist]} {[incr i -1] > 0} {} {
set j [findmax $nlist $i]
if {$i != $j} {
if {$j} {
set nlist [flip $nlist $j]
if {$debug eq "debug"} {puts [incr flips]>>$nlist}
}
set nlist [flip $nlist $i]
if {$debug eq "debug"} {puts [incr flips]>>$nlist}
}
}
return $nlist
}

Demonstrate (with debug mode enabled so it prints intermediate states):

puts [pancakeSort {27916 5928 23535 14711 32184 14621 21093 14422 29844 11093} debug]

Output:

1>>32184 14711 23535 5928 27916 14621 21093 14422 29844 11093
2>>11093 29844 14422 21093 14621 27916 5928 23535 14711 32184
3>>29844 11093 14422 21093 14621 27916 5928 23535 14711 32184
4>>14711 23535 5928 27916 14621 21093 14422 11093 29844 32184
5>>27916 5928 23535 14711 14621 21093 14422 11093 29844 32184
6>>11093 14422 21093 14621 14711 23535 5928 27916 29844 32184
7>>23535 14711 14621 21093 14422 11093 5928 27916 29844 32184
8>>5928 11093 14422 21093 14621 14711 23535 27916 29844 32184
9>>21093 14422 11093 5928 14621 14711 23535 27916 29844 32184
10>>14711 14621 5928 11093 14422 21093 23535 27916 29844 32184
11>>14422 11093 5928 14621 14711 21093 23535 27916 29844 32184
12>>5928 11093 14422 14621 14711 21093 23535 27916 29844 32184
5928 11093 14422 14621 14711 21093 23535 27916 29844 32184

As you can see, it took 12 flips.

[edit] VBA

 
 
'pancake sort
'uses two auxiliary routines "printarray" and "flip"

Public Sub printarray(A)
For i = LBound(A) To UBound(A)
Debug.Print A(i),
Next
Debug.Print
End Sub
 
Public Sub Flip(ByRef A, p1, p2, trace)
'flip first elements of A (p1 to p2)
If trace Then Debug.Print "we'll flip the first "; p2 - p1 + 1; "elements of the array"
Cut = Int((p2 - p1 + 1) / 2)
For i = 0 To Cut - 1
'flip position i and (n - i + 1)
temp = A(i)
A(i) = A(p2 - i)
A(p2 - i) = temp
Next
End Sub
 
Public Sub pancakesort(ByRef A(), Optional trace As Boolean = False)
'sort A into ascending order using pancake sort

lb = LBound(A)
ub = UBound(A)
Length = ub - lb + 1
If Length <= 1 Then 'no need to sort
Exit Sub
End If
 
For i = ub To lb + 1 Step -1
'find position of max. element in subarray A(lowerbound to i)
P = lb
Maximum = A(P)
For j = lb + 1 To i
If A(j) > Maximum Then
P = j
Maximum = A(j)
End If
Next j
'check if maximum is already at end - then we don't need to flip
If P < i Then
'flip the first part of the array up to the maximum so it is at the head - skip if it is already there
If P > 1 Then
Flip A, lb, P, trace
If trace Then printarray A
End If
'now flip again so that it is in its final position
Flip A, lb, i, trace
If trace Then printarray A
End If
Next i
End Sub
 
'test routine
Public Sub TestPancake(Optional trace As Boolean = False)
Dim A()
A = Array(5, 7, 8, 3, 1, 10, 9, 23, 50, 0)
Debug.Print "Initial array:"
printarray A
pancakesort A, trace
Debug.Print "Final array:"
printarray A
End Sub
 

Sample output:

testpancake True
Initial array:
 5             7             8             3             1             10            9             23            50            0            
we'll flip the first  9 elements of the array
 50            23            9             10            1             3             8             7             5             0            
we'll flip the first  10 elements of the array
 0             5             7             8             3             1             10            9             23            50           
we'll flip the first  7 elements of the array
 10            1             3             8             7             5             0             9             23            50           
we'll flip the first  8 elements of the array
 9             0             5             7             8             3             1             10            23            50           
we'll flip the first  7 elements of the array
 1             3             8             7             5             0             9             10            23            50           
we'll flip the first  3 elements of the array
 8             3             1             7             5             0             9             10            23            50           
we'll flip the first  6 elements of the array
 0             5             7             1             3             8             9             10            23            50           
we'll flip the first  3 elements of the array
 7             5             0             1             3             8             9             10            23            50           
we'll flip the first  5 elements of the array
 3             1             0             5             7             8             9             10            23            50           
we'll flip the first  3 elements of the array
 0             1             3             5             7             8             9             10            23            50           
Final array:
 0             1             3             5             7             8             9             10            23            50           
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