Permutations by swapping
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Generate permutations of n items in which successive permutations differ from each other by the swapping of any two items.
Also generate the sign of the permutation which is +1 when the permutation is generated from an even number of swaps from the initial state, and -1 for odd.
Show the permutations and signs of three items, in order of generation here.
Such data are of use in generating the determinant of a square matrix and any functions created should bear this in mind.
Note: The Steinhaus–Johnson–Trotter algorithm generates successive permutations where adjacent items are swapped, but from this discussion adjacency is not a requirement.
- References
- Steinhaus–Johnson–Trotter algorithm
- Johnson-Trotter Algorithm Listing All Permutations
- Heap's algorithm
- [1] Tintinnalogia
- Related tasks
11l
F s_permutations(seq)
V items = [[Int]()]
L(j) seq
[[Int]] new_items
L(item) items
I L.index % 2
new_items [+]= (0..item.len).map(i -> @item[0 .< i] [+] [@j] [+] @item[i..])
E
new_items [+]= (item.len..0).step(-1).map(i -> @item[0 .< i] [+] [@j] [+] @item[i..])
items = new_items
R enumerate(items).map((i, item) -> (item, I i % 2 {-1} E 1))
L(n) (3, 4)
print(‘Permutations and sign of #. items’.format(n))
L(perm, sgn) s_permutations(Array(0 .< n))
print(‘Perm: #. Sign: #2’.format(perm, sgn))
print()- Output:
Permutations and sign of 3 items Perm: [0, 1, 2] Sign: 1 Perm: [0, 2, 1] Sign: -1 Perm: [2, 0, 1] Sign: 1 Perm: [2, 1, 0] Sign: -1 Perm: [1, 2, 0] Sign: 1 Perm: [1, 0, 2] Sign: -1 Permutations and sign of 4 items Perm: [0, 1, 2, 3] Sign: 1 Perm: [0, 1, 3, 2] Sign: -1 Perm: [0, 3, 1, 2] Sign: 1 Perm: [3, 0, 1, 2] Sign: -1 Perm: [3, 0, 2, 1] Sign: 1 Perm: [0, 3, 2, 1] Sign: -1 Perm: [0, 2, 3, 1] Sign: 1 Perm: [0, 2, 1, 3] Sign: -1 Perm: [2, 0, 1, 3] Sign: 1 Perm: [2, 0, 3, 1] Sign: -1 Perm: [2, 3, 0, 1] Sign: 1 Perm: [3, 2, 0, 1] Sign: -1 Perm: [3, 2, 1, 0] Sign: 1 Perm: [2, 3, 1, 0] Sign: -1 Perm: [2, 1, 3, 0] Sign: 1 Perm: [2, 1, 0, 3] Sign: -1 Perm: [1, 2, 0, 3] Sign: 1 Perm: [1, 2, 3, 0] Sign: -1 Perm: [1, 3, 2, 0] Sign: 1 Perm: [3, 1, 2, 0] Sign: -1 Perm: [3, 1, 0, 2] Sign: 1 Perm: [1, 3, 0, 2] Sign: -1 Perm: [1, 0, 3, 2] Sign: 1 Perm: [1, 0, 2, 3] Sign: -1
Ada
-- Permutations by swapping
-- J. Carter 2024 Jun
-- Uses the PragmAda Reusable Components (https://github.com/jrcarter/PragmARC)
with Ada.Text_IO;
with PragmARC.Permutations;
procedure Permutations_By_Swapping is
package Permutations is new PragmARC.Permutations (Element => Positive);
Permutation : Permutations.Sequence_Lists.Vector;
begin -- Permutations_By_Swapping
Permutations.Generate (Initial => (1, 2, 3), Result => Permutation);
Output : for I in 1 .. Permutation.Last_Index loop
One_Permutation : declare
Sequence : Permutations.Sequence renames Permutation.Element (I);
begin -- One_Permutation
Ada.Text_IO.Put (Item => '(');
Print : for J in Sequence'Range loop
Ada.Text_IO.Put (Item => (if J > 1 then "," else "") & Sequence (J)'Image);
end loop Print;
Ada.Text_IO.Put_Line (Item => ") " & (if I rem 2 = 0 then '-' else '+') & '1');
end One_Permutation;
end loop Output;
end Permutations_By_Swapping;
- Output:
( 1, 2, 3) +1 ( 2, 1, 3) -1 ( 3, 1, 2) +1 ( 1, 3, 2) -1 ( 2, 3, 1) +1 ( 3, 2, 1) -1
ALGOL 68
Based on the pseudo-code for the recursive version of Heap's algorithm.
BEGIN # Heap's algorithm for generating permutations - from the pseudo-code on the Wikipedia page #
# generate permutations of a #
PROC generate = ( INT k, REF[]INT a, REF INT swap count )VOID:
IF k = 1 THEN
output permutation( a, swap count )
ELSE
# Generate permutations with kth unaltered #
# Initially k = length a #
generate( k - 1, a, swap count );
# Generate permutations for kth swapped with each k-1 initial #
FOR i FROM 0 TO k - 2 DO
# Swap choice dependent on parity of k (even or odd) #
swap count +:= 1;
INT swap item = IF ODD k THEN 0 ELSE i FI;
INT t = a[ swap item ];
a[ swap item ] := a[ k - 1 ];
a[ k - 1 ] := t;
generate( k - 1, a, swap count )
OD
FI # generate # ;
# generate permutations of a #
PROC permute = ( REF[]INT a )VOID:
BEGIN
INT swap count := 0;
generate( ( UPB a + 1 ) - LWB a, a[ AT 0 ], swap count )
END # permute # ;
# handle a permutation #
PROC output permutation = ( REF[]INT a, INT swap count )VOID:
BEGIN
print( ( "[" ) );
FOR i FROM LWB a TO UPB a DO
print( ( whole( a[ i ], 0 ) ) );
IF i = UPB a THEN print( ( "]" ) ) ELSE print( ( ", " ) ) FI
OD;
print( ( " sign: ", IF ODD swap count THEN "-1" ELSE " 1" FI, newline ) )
END # output permutation # ;
[ 1 : 3 ]INT a := ( 1, 2, 3 );
permute( a )
END- Output:
[1, 2, 3] sign: 1 [2, 1, 3] sign: -1 [3, 1, 2] sign: 1 [1, 3, 2] sign: -1 [2, 3, 1] sign: 1 [3, 2, 1] sign: -1
Arturo
permutations: function [arr][
d: 1
c: array.of: size arr 0
xs: new arr
sign: 1
ret: new @[@[xs, sign]]
while [true][
while [d > 1][
d: d-1
c\[d]: 0
]
while [c\[d] >= d][
d: d+1
if d >= size arr -> return ret
]
i: (1 = and d 1)? -> c\[d] -> 0
tmp: xs\[i]
xs\[i]: xs\[d]
xs\[d]: tmp
sign: neg sign
'ret ++ @[new @[xs, sign]]
c\[d]: c\[d] + 1
]
return ret
]
loop permutations 0..2 'row ->
print [row\0 "-> sign:" row\1]
print ""
loop permutations 0..3 'row ->
print [row\0 "-> sign:" row\1]
- Output:
[0 1 2] -> sign: 1 [1 0 2] -> sign: -1 [2 0 1] -> sign: 1 [0 2 1] -> sign: -1 [1 2 0] -> sign: 1 [2 1 0] -> sign: -1 [0 1 2 3] -> sign: 1 [1 0 2 3] -> sign: -1 [2 0 1 3] -> sign: 1 [0 2 1 3] -> sign: -1 [1 2 0 3] -> sign: 1 [2 1 0 3] -> sign: -1 [3 1 0 2] -> sign: 1 [1 3 0 2] -> sign: -1 [0 3 1 2] -> sign: 1 [3 0 1 2] -> sign: -1 [1 0 3 2] -> sign: 1 [0 1 3 2] -> sign: -1 [0 2 3 1] -> sign: 1 [2 0 3 1] -> sign: -1 [3 0 2 1] -> sign: 1 [0 3 2 1] -> sign: -1 [2 3 0 1] -> sign: 1 [3 2 0 1] -> sign: -1 [3 2 1 0] -> sign: 1 [2 3 1 0] -> sign: -1 [1 3 2 0] -> sign: 1 [3 1 2 0] -> sign: -1 [2 1 3 0] -> sign: 1 [1 2 3 0] -> sign: -1
AutoHotkey
Permutations_By_Swapping(str, list:=""){
ch := SubStr(str, 1, 1) ; get left-most charachter of str
for i, line in StrSplit(list, "`n") ; for each line in list
loop % StrLen(line) + 1 ; loop each possible position
Newlist .= RegExReplace(line, mod(i,2) ? "(?=.{" A_Index-1 "}$)" : "^.{" A_Index-1 "}\K", ch) "`n"
list := Newlist ? Trim(Newlist, "`n") : ch ; recreate list
if !str := SubStr(str, 2) ; remove charachter from left hand side
return list ; done if str is empty
return Permutations_By_Swapping(str, list) ; else recurse
}
Examples:
for each, line in StrSplit(Permutations_By_Swapping(1234), "`n")
result .= line "`tSign: " (mod(A_Index,2)? 1 : -1) "`n"
MsgBox, 262144, , % result
return
Outputs:
1234 Sign: 1 1243 Sign: -1 1423 Sign: 1 4123 Sign: -1 4132 Sign: 1 1432 Sign: -1 1342 Sign: 1 1324 Sign: -1 3124 Sign: 1 3142 Sign: -1 3412 Sign: 1 4312 Sign: -1 4321 Sign: 1 3421 Sign: -1 3241 Sign: 1 3214 Sign: -1 2314 Sign: 1 2341 Sign: -1 2431 Sign: 1 4231 Sign: -1 4213 Sign: 1 2413 Sign: -1 2143 Sign: 1 2134 Sign: -1
BASIC
BASIC256
call perms(3)
print
call perms(4)
end
subroutine perms(n)
dim p((n+1)*4)
for i = 1 to n
p[i] = -i
next i
s = 1
do
print "Perm: [ ";
for i = 1 to n
print abs(p[i]); " ";
next i
print "] Sign: "; s
k = 0
for i = 2 to n
if p[i] < 0 and (abs(p[i]) > abs(p[i-1])) and (abs(p[i]) > abs(p[k])) then k = i
next i
for i = 1 to n-1
if p[i] > 0 and (abs(p[i]) > abs(p[i+1])) and (abs(p[i]) > abs(p[k])) then k = i
next i
if k then
for i = 1 to n #reverse elements > k
if abs(p[i]) > abs(p[k]) then p[i] = -p[i]
next i
if p[k] < 0 then i = k-1 else i = k+1
temp = p[k]
p[k] = p[i]
p[i] = temp
s = -s
end if
until k = 0
end subroutineQBasic
SUB perms (n)
DIM p((n + 1) * 4)
FOR i = 1 TO n
p(i) = -i
NEXT i
s = 1
DO
PRINT "Perm: (";
FOR i = 1 TO n
PRINT ABS(p(i)); "";
NEXT i
PRINT ") Sign: "; s
k = 0
FOR i = 2 TO n
IF p(i) < 0 AND (ABS(p(i)) > ABS(p(i - 1))) AND (ABS(p(i)) > ABS(p(k))) THEN k = i
NEXT i
FOR i = 1 TO n - 1
IF p(i) > 0 AND (ABS(p(i)) > ABS(p(i + 1))) AND (ABS(p(i)) > ABS(p(k))) THEN k = i
NEXT i
IF k THEN
FOR i = 1 TO n 'reverse elements > k
IF ABS(p(i)) > ABS(p(k)) THEN p(i) = -p(i)
NEXT i
'if p(k) < 0 then i = k-1 else i = k+1
i = k + SGN(p(k))
SWAP p(k), p(i)
'temp = p(k)
'p(k) = p(i)
'p(i) = temp
s = -s
END IF
LOOP UNTIL k = 0
END SUB
perms (3)
PRINT
perms (4)
Run BASIC
sub perms n
dim p((n+1)*4)
for i = 1 to n : p(i) = i*-1 : next i
s = 1
while 1
print "Perm: [ ";
for i = 1 to n
print abs(p(i)); " ";
next i
print "] Sign: "; s
k = 0
for i = 2 to n
if p(i) < 0 and (abs(p(i)) > abs(p(i-1))) and (abs(p(i)) > abs(p(k))) then k = i
next i
for i = 1 to n-1
if p(i) > 0 and (abs(p(i)) > abs(p(i+1))) and (abs(p(i)) > abs(p(k))) then k = i
next i
if k then
for i = 1 to n 'reverse elements > k
if abs(p(i)) > abs(p(k)) then p(i) = p(i)*-1
next i
if p(k) < 0 then i = k-1 else i = k+1 'swap K with element looked at
temp = p(k)
p(k) = p(i)
p(i) = temp
s = s*-1 'alternate signs
end if
if k = 0 then exit while
wend
end sub
call perms 3
print
call perms 4Yabasic
perms(3)
print
perms(4)
end
sub perms(n)
dim p((n+1)*4)
for i = 1 to n
p(i) = -i
next i
s = 1
repeat
print "Perm: [ ";
for i = 1 to n
print abs(p(i)), " ";
next i
print "] Sign: ", s
k = 0
for i = 2 to n
if p(i) < 0 and (abs(p(i)) > abs(p(i-1))) and (abs(p(i)) > abs(p(k))) k = i
next i
for i = 1 to n-1
if p(i) > 0 and (abs(p(i)) > abs(p(i+1))) and (abs(p(i)) > abs(p(k))) k = i
next i
if k then
for i = 1 to n //reverse elements > k
if abs(p(i)) > abs(p(k)) p(i) = -p(i)
next i
i = k + sig(p(k))
temp = p(k)
p(k) = p(i)
p(i) = temp
s = -s
endif
until k = 0
end sub
BBC BASIC
PROCperms(3)
PRINT
PROCperms(4)
END
DEF PROCperms(n%)
LOCAL p%(), i%, k%, s%
DIM p%(n%)
FOR i% = 1 TO n%
p%(i%) = -i%
NEXT
s% = 1
REPEAT
PRINT "Perm: [ ";
FOR i% = 1 TO n%
PRINT ;ABSp%(i%) " ";
NEXT
PRINT "] Sign: ";s%
k% = 0
FOR i% = 2 TO n%
IF p%(i%)<0 IF ABSp%(i%)>ABSp%(i%-1) IF ABSp%(i%)>ABSp%(k%) k% = i%
NEXT
FOR i% = 1 TO n%-1
IF p%(i%)>0 IF ABSp%(i%)>ABSp%(i%+1) IF ABSp%(i%)>ABSp%(k%) k% = i%
NEXT
IF k% THEN
FOR i% = 1 TO n%
IF ABSp%(i%)>ABSp%(k%) p%(i%) *= -1
NEXT
i% = k%+SGNp%(k%)
SWAP p%(k%),p%(i%)
s% = -s%
ENDIF
UNTIL k% = 0
ENDPROC
- Output:
Perm: [ 1 2 3 ] Sign: 1 Perm: [ 1 3 2 ] Sign: -1 Perm: [ 3 1 2 ] Sign: 1 Perm: [ 3 2 1 ] Sign: -1 Perm: [ 2 3 1 ] Sign: 1 Perm: [ 2 1 3 ] Sign: -1 Perm: [ 1 2 3 4 ] Sign: 1 Perm: [ 1 2 4 3 ] Sign: -1 Perm: [ 1 4 2 3 ] Sign: 1 Perm: [ 4 1 2 3 ] Sign: -1 Perm: [ 4 1 3 2 ] Sign: 1 Perm: [ 1 4 3 2 ] Sign: -1 Perm: [ 1 3 4 2 ] Sign: 1 Perm: [ 1 3 2 4 ] Sign: -1 Perm: [ 3 1 2 4 ] Sign: 1 Perm: [ 3 1 4 2 ] Sign: -1 Perm: [ 3 4 1 2 ] Sign: 1 Perm: [ 4 3 1 2 ] Sign: -1 Perm: [ 4 3 2 1 ] Sign: 1 Perm: [ 3 4 2 1 ] Sign: -1 Perm: [ 3 2 4 1 ] Sign: 1 Perm: [ 3 2 1 4 ] Sign: -1 Perm: [ 2 3 1 4 ] Sign: 1 Perm: [ 2 3 4 1 ] Sign: -1 Perm: [ 2 4 3 1 ] Sign: 1 Perm: [ 4 2 3 1 ] Sign: -1 Perm: [ 4 2 1 3 ] Sign: 1 Perm: [ 2 4 1 3 ] Sign: -1 Perm: [ 2 1 4 3 ] Sign: 1 Perm: [ 2 1 3 4 ] Sign: -1
C
Implementation of Heap's Algorithm, array length has to be passed as a parameter for non character arrays, as sizeof() will not give correct results when malloc is used. Prints usage on incorrect invocation.
#include<stdlib.h>
#include<string.h>
#include<stdio.h>
int flag = 1;
void heapPermute(int n, int arr[],int arrLen){
int temp;
int i;
if(n==1){
printf("\n[");
for(i=0;i<arrLen;i++)
printf("%d,",arr[i]);
printf("\b] Sign : %d",flag);
flag*=-1;
}
else{
for(i=0;i<n-1;i++){
heapPermute(n-1,arr,arrLen);
if(n%2==0){
temp = arr[i];
arr[i] = arr[n-1];
arr[n-1] = temp;
}
else{
temp = arr[0];
arr[0] = arr[n-1];
arr[n-1] = temp;
}
}
heapPermute(n-1,arr,arrLen);
}
}
int main(int argC,char* argV[0])
{
int *arr, i=0, count = 1;
char* token;
if(argC==1)
printf("Usage : %s <comma separated list of integers>",argV[0]);
else{
while(argV[1][i]!=00){
if(argV[1][i++]==',')
count++;
}
arr = (int*)malloc(count*sizeof(int));
i = 0;
token = strtok(argV[1],",");
while(token!=NULL){
arr[i++] = atoi(token);
token = strtok(NULL,",");
}
heapPermute(i,arr,count);
}
return 0;
}
Output:
C:\rosettaCode>heapPermute.exe 1,2,3 [1,2,3] Sign : 1 [2,1,3] Sign : -1 [3,1,2] Sign : 1 [1,3,2] Sign : -1 [2,3,1] Sign : 1 [3,2,1] Sign : -1
C++
Direct implementation of Johnson-Trotter algorithm from the reference link.
#include <iostream>
#include <vector>
using namespace std;
vector<int> UpTo(int n, int offset = 0)
{
vector<int> retval(n);
for (int ii = 0; ii < n; ++ii)
retval[ii] = ii + offset;
return retval;
}
struct JohnsonTrotterState_
{
vector<int> values_;
vector<int> positions_; // size is n+1, first element is not used
vector<bool> directions_;
int sign_;
JohnsonTrotterState_(int n) : values_(UpTo(n, 1)), positions_(UpTo(n + 1, -1)), directions_(n + 1, false), sign_(1) {}
int LargestMobile() const // returns 0 if no mobile integer exists
{
for (int r = values_.size(); r > 0; --r)
{
const int loc = positions_[r] + (directions_[r] ? 1 : -1);
if (loc >= 0 && loc < values_.size() && values_[loc] < r)
return r;
}
return 0;
}
bool IsComplete() const { return LargestMobile() == 0; }
void operator++() // implement Johnson-Trotter algorithm
{
const int r = LargestMobile();
const int rLoc = positions_[r];
const int lLoc = rLoc + (directions_[r] ? 1 : -1);
const int l = values_[lLoc];
// do the swap
swap(values_[lLoc], values_[rLoc]);
swap(positions_[l], positions_[r]);
sign_ = -sign_;
// change directions
for (auto pd = directions_.begin() + r + 1; pd != directions_.end(); ++pd)
*pd = !*pd;
}
};
int main(void)
{
JohnsonTrotterState_ state(4);
do
{
for (auto v : state.values_)
cout << v << " ";
cout << "\n";
++state;
} while (!state.IsComplete());
}
- Output:
(1 2 3 4 ); sign = 1 (1 2 4 3 ); sign = -1 (1 4 2 3 ); sign = 1 (4 1 2 3 ); sign = -1 (4 1 3 2 ); sign = 1 (1 4 3 2 ); sign = -1 (1 3 4 2 ); sign = 1 (1 3 2 4 ); sign = -1 (3 1 2 4 ); sign = 1 (3 1 4 2 ); sign = -1 (3 4 1 2 ); sign = 1 (4 3 1 2 ); sign = -1 (4 3 2 1 ); sign = 1 (3 4 2 1 ); sign = -1 (3 2 4 1 ); sign = 1 (3 2 1 4 ); sign = -1 (2 3 1 4 ); sign = 1 (2 3 4 1 ); sign = -1 (2 4 3 1 ); sign = 1 (4 2 3 1 ); sign = -1 (4 2 1 3 ); sign = 1 (2 4 1 3 ); sign = -1 (2 1 4 3 ); sign = 1
Clojure
Recursive version
(defn permutation-swaps
"List of swap indexes to generate all permutations of n elements"
[n]
(if (= n 2) `((0 1))
(let [old-swaps (permutation-swaps (dec n))
swaps-> (partition 2 1 (range n))
swaps<- (reverse swaps->)]
(mapcat (fn [old-swap side]
(case side
:first swaps<-
:right (conj swaps<- old-swap)
:left (conj swaps-> (map inc old-swap))))
(conj old-swaps nil)
(cons :first (cycle '(:left :right)))))))
(defn swap [v [i j]]
(-> v
(assoc i (nth v j))
(assoc j (nth v i))))
(defn permutations [n]
(let [permutations (reduce
(fn [all-perms new-swap]
(conj all-perms (swap (last all-perms)
new-swap)))
(vector (vec (range n)))
(permutation-swaps n))
output (map vector
permutations
(cycle '(1 -1)))]
output))
(doseq [n [2 3 4]]
(dorun (map println (permutations n))))
- Output:
[[0 1] 1] [[1 0] -1] [[0 1 2] 1] [[0 2 1] -1] [[2 0 1] 1] [[2 1 0] -1] [[1 2 0] 1] [[1 0 2] -1] [[0 1 2 3] 1] [[0 1 3 2] -1] [[0 3 1 2] 1] [[3 0 1 2] -1] [[3 0 2 1] 1] [[0 3 2 1] -1] [[0 2 3 1] 1] [[0 2 1 3] -1] [[2 0 1 3] 1] [[2 0 3 1] -1] [[2 3 0 1] 1] [[3 2 0 1] -1] [[3 2 1 0] 1] [[2 3 1 0] -1] [[2 1 3 0] 1] [[2 1 0 3] -1] [[1 2 0 3] 1] [[1 2 3 0] -1] [[1 3 2 0] 1] [[3 1 2 0] -1] [[3 1 0 2] 1] [[1 3 0 2] -1] [[1 0 3 2] 1] [[1 0 2 3] -1]
Modeled After Python version
(ns test-p.core)
(defn numbers-only [x]
" Just shows the numbers only for the pairs (i.e. drops the direction --used for display purposes when printing the result"
(mapv first x))
(defn next-permutation
" Generates next permutation from the current (p) using the Johnson-Trotter technique
The code below translates the Python version which has the following steps:
p of form [...[n dir]...] such as [[0 1] [1 1] [2 -1]], where n is a number and dir = direction (=1=right, -1=left, 0=don't move)
Step: 1 finds the pair [n dir] with the largest value of n (where dir is not equal to 0 (done if none)
Step: 2: swap the max pair found with its neighbor in the direction of the pair (i.e. +1 means swap to right, -1 means swap left
Step 3: if swapping places the pair a the beginning or end of the list, set the direction = 0 (i.e. becomes non-mobile)
Step 4: Set the directions of all pairs whose numbers are greater to the right of where the pair was moved to -1 and to the left to +1 "
[p]
(if (every? zero? (map second p))
nil ; no mobile elements (all directions are zero)
(let [n (count p)
; Step 1
fn-find-max (fn [m]
(first (apply max-key ; find the max mobile elment
(fn [[i x]]
(if (zero? (second x))
-1
(first x)))
(map-indexed vector p))))
i1 (fn-find-max p) ; index of max
[n1 d1] (p i1) ; value and direction of max
i2 (+ d1 i1)
fn-swap (fn [m] (assoc m i2 (m i1) i1 (m i2))) ; function to swap with neighbor in our step direction
fn-update-max (fn [m] (if (or (contains? #{0 (dec n)} i2) ; update direction of max (where max went)
(> ((m (+ i2 d1)) 0) n1))
(assoc-in m [i2 1] 0)
m))
fn-update-others (fn [[i3 [n3 d3]]] ; Updates directions of pairs to the left and right of max
(cond ; direction reset to -1 if to right, +1 if to left
(<= n3 n1) [n3 d3]
(< i3 i2) [n3 1]
:else [n3 -1]))]
; apply steps 2, 3, 4(using functions that where created for these steps)
(mapv fn-update-others (map-indexed vector (fn-update-max (fn-swap p)))))))
(defn spermutations
" Lazy sequence of permutations of n digits"
; Each element is two element vector (number direction)
; Startup case - generates sequence 0...(n-1) with move direction (1 = move right, -1 = move left, 0 = don't move)
([n] (spermutations 1
(into [] (for [i (range n)] (if (zero? i)
[i 0] ; 0th element is not mobile yet
[i -1]))))) ; all others move left
([sign p]
(when-let [s (seq p)]
(cons [(numbers-only p) sign]
(spermutations (- sign) (next-permutation p)))))) ; recursively tag onto sequence
;; Print results for 2, 3, and 4 items
(doseq [n (range 2 5)]
(do
(println)
(println (format "Permutations and sign of %d items " n))
(doseq [q (spermutations n)] (println (format "Perm: %s Sign: %2d" (first q) (second q))))))
- Output:
Permutations and sign of 2 items Perm: [0 1] Sign: 1 Perm: [1 0] Sign: -1 Permutations and sign of 3 items Perm: [0 1 2] Sign: 1 Perm: [0 2 1] Sign: -1 Perm: [2 0 1] Sign: 1 Perm: [2 1 0] Sign: -1 Perm: [1 2 0] Sign: 1 Perm: [1 0 2] Sign: -1 Permutations and sign of 4 items Perm: [0 1 2 3] Sign: 1 Perm: [0 1 3 2] Sign: -1 Perm: [0 3 1 2] Sign: 1 Perm: [3 0 1 2] Sign: -1 Perm: [3 0 2 1] Sign: 1 Perm: [0 3 2 1] Sign: -1 Perm: [0 2 3 1] Sign: 1 Perm: [0 2 1 3] Sign: -1 Perm: [2 0 1 3] Sign: 1 Perm: [2 0 3 1] Sign: -1 Perm: [2 3 0 1] Sign: 1 Perm: [3 2 0 1] Sign: -1 Perm: [3 2 1 0] Sign: 1 Perm: [2 3 1 0] Sign: -1 Perm: [2 1 3 0] Sign: 1 Perm: [2 1 0 3] Sign: -1 Perm: [1 2 0 3] Sign: 1 Perm: [1 2 3 0] Sign: -1 Perm: [1 3 2 0] Sign: 1 Perm: [3 1 2 0] Sign: -1 Perm: [3 1 0 2] Sign: 1 Perm: [1 3 0 2] Sign: -1 Perm: [1 0 3 2] Sign: 1 Perm: [1 0 2 3] Sign: -1
Common Lisp
(defstruct (directed-number (:conc-name dn-))
(number nil :type integer)
(direction nil :type (member :left :right)))
(defmethod print-object ((dn directed-number) stream)
(ecase (dn-direction dn)
(:left (format stream "<~D" (dn-number dn)))
(:right (format stream "~D>" (dn-number dn)))))
(defun dn> (dn1 dn2)
(declare (directed-number dn1 dn2))
(> (dn-number dn1) (dn-number dn2)))
(defun dn-reverse-direction (dn)
(declare (directed-number dn))
(setf (dn-direction dn) (ecase (dn-direction dn)
(:left :right)
(:right :left))))
(defun make-directed-numbers-upto (upto)
(let ((numbers (make-array upto :element-type 'integer)))
(dotimes (n upto numbers)
(setf (aref numbers n) (make-directed-number :number (1+ n) :direction :left)))))
(defun max-mobile-pos (numbers)
(declare ((vector directed-number) numbers))
(loop with pos-limit = (1- (length numbers))
with max-value and max-pos
for num across numbers
for pos from 0
do (ecase (dn-direction num)
(:left (when (and (plusp pos) (dn> num (aref numbers (1- pos)))
(or (null max-value) (dn> num max-value)))
(setf max-value num
max-pos pos)))
(:right (when (and (< pos pos-limit) (dn> num (aref numbers (1+ pos)))
(or (null max-value) (dn> num max-value)))
(setf max-value num
max-pos pos))))
finally (return max-pos)))
(defun permutations (upto)
(loop with numbers = (make-directed-numbers-upto upto)
for max-mobile-pos = (max-mobile-pos numbers)
for sign = 1 then (- sign)
do (format t "~A sign: ~:[~;+~]~D~%" numbers (plusp sign) sign)
while max-mobile-pos
do (let ((max-mobile-number (aref numbers max-mobile-pos)))
(ecase (dn-direction max-mobile-number)
(:left (rotatef (aref numbers (1- max-mobile-pos))
(aref numbers max-mobile-pos)))
(:right (rotatef (aref numbers max-mobile-pos)
(aref numbers (1+ max-mobile-pos)))))
(loop for n across numbers
when (dn> n max-mobile-number)
do (dn-reverse-direction n)))))
(permutations 3)
(permutations 4)
- Output:
#(<1 <2 <3) sign: +1 #(<1 <3 <2) sign: -1 #(<3 <1 <2) sign: +1 #(3> <2 <1) sign: -1 #(<2 3> <1) sign: +1 #(<2 <1 3>) sign: -1 #(<1 <2 <3 <4) sign: +1 #(<1 <2 <4 <3) sign: -1 #(<1 <4 <2 <3) sign: +1 #(<4 <1 <2 <3) sign: -1 #(4> <1 <3 <2) sign: +1 #(<1 4> <3 <2) sign: -1 #(<1 <3 4> <2) sign: +1 #(<1 <3 <2 4>) sign: -1 #(<3 <1 <2 <4) sign: +1 #(<3 <1 <4 <2) sign: -1 #(<3 <4 <1 <2) sign: +1 #(<4 <3 <1 <2) sign: -1 #(4> 3> <2 <1) sign: +1 #(3> 4> <2 <1) sign: -1 #(3> <2 4> <1) sign: +1 #(3> <2 <1 4>) sign: -1 #(<2 3> <1 <4) sign: +1 #(<2 3> <4 <1) sign: -1 #(<2 <4 3> <1) sign: +1 #(<4 <2 3> <1) sign: -1 #(4> <2 <1 3>) sign: +1 #(<2 4> <1 3>) sign: -1 #(<2 <1 4> 3>) sign: +1 #(<2 <1 3> 4>) sign: -1
D
Iterative Version
This isn't a Range yet.
import std.algorithm, std.array, std.typecons, std.range;
struct Spermutations(bool doCopy=true) {
private immutable uint n;
alias TResult = Tuple!(int[], int);
int opApply(in int delegate(in ref TResult) nothrow dg) nothrow {
int result;
int sign = 1;
alias Int2 = Tuple!(int, int);
auto p = n.iota.map!(i => Int2(i, i ? -1 : 0)).array;
TResult aux;
aux[0] = p.map!(pi => pi[0]).array;
aux[1] = sign;
result = dg(aux);
if (result)
goto END;
while (p.any!q{ a[1] }) {
// Failed to use std.algorithm here, too much complex.
auto largest = Int2(-100, -100);
int i1 = -1;
foreach (immutable i, immutable pi; p)
if (pi[1])
if (pi[0] > largest[0]) {
i1 = i;
largest = pi;
}
immutable n1 = largest[0],
d1 = largest[1];
sign *= -1;
int i2;
if (d1 == -1) {
i2 = i1 - 1;
p[i1].swap(p[i2]);
if (i2 == 0 || p[i2 - 1][0] > n1)
p[i2][1] = 0;
} else if (d1 == 1) {
i2 = i1 + 1;
p[i1].swap(p[i2]);
if (i2 == n - 1 || p[i2 + 1][0] > n1)
p[i2][1] = 0;
}
if (doCopy) {
aux[0] = p.map!(pi => pi[0]).array;
} else {
foreach (immutable i, immutable pi; p)
aux[0][i] = pi[0];
}
aux[1] = sign;
result = dg(aux);
if (result)
goto END;
foreach (immutable i3, ref pi; p) {
immutable n3 = pi[0],
d3 = pi[1];
if (n3 > n1)
pi[1] = (i3 < i2) ? 1 : -1;
}
}
END: return result;
}
}
Spermutations!doCopy spermutations(bool doCopy=true)(in uint n) {
return typeof(return)(n);
}
version (permutations_by_swapping1) {
void main() {
import std.stdio;
foreach (immutable n; [3, 4]) {
writefln("\nPermutations and sign of %d items", n);
foreach (const tp; n.spermutations)
writefln("Perm: %s Sign: %2d", tp[]);
}
}
}
Compile with version=permutations_by_swapping1 to see the demo output.
- Output:
Permutations and sign of 3 items Perm: [0, 1, 2] Sign: 1 Perm: [0, 2, 1] Sign: -1 Perm: [2, 0, 1] Sign: 1 Perm: [2, 1, 0] Sign: -1 Perm: [1, 2, 0] Sign: 1 Perm: [1, 0, 2] Sign: -1 Permutations and sign of 4 items Perm: [0, 1, 2, 3] Sign: 1 Perm: [0, 1, 3, 2] Sign: -1 Perm: [0, 3, 1, 2] Sign: 1 Perm: [3, 0, 1, 2] Sign: -1 Perm: [3, 0, 2, 1] Sign: 1 Perm: [0, 3, 2, 1] Sign: -1 Perm: [0, 2, 3, 1] Sign: 1 Perm: [0, 2, 1, 3] Sign: -1 Perm: [2, 0, 1, 3] Sign: 1 Perm: [2, 0, 3, 1] Sign: -1 Perm: [2, 3, 0, 1] Sign: 1 Perm: [3, 2, 0, 1] Sign: -1 Perm: [3, 2, 1, 0] Sign: 1 Perm: [2, 3, 1, 0] Sign: -1 Perm: [2, 1, 3, 0] Sign: 1 Perm: [2, 1, 0, 3] Sign: -1 Perm: [1, 2, 0, 3] Sign: 1 Perm: [1, 2, 3, 0] Sign: -1 Perm: [1, 3, 2, 0] Sign: 1 Perm: [3, 1, 2, 0] Sign: -1 Perm: [3, 1, 0, 2] Sign: 1 Perm: [1, 3, 0, 2] Sign: -1 Perm: [1, 0, 3, 2] Sign: 1 Perm: [1, 0, 2, 3] Sign: -1
Recursive Version
import std.algorithm, std.array, std.typecons, std.range;
auto sPermutations(in uint n) pure nothrow @safe {
static immutable(int[])[] inner(in int items) pure nothrow @safe {
if (items <= 0)
return [[]];
typeof(return) r;
foreach (immutable i, immutable item; inner(items - 1)) {
//r.put((i % 2 ? iota(item.length.signed, -1, -1) :
// iota(item.length + 1))
// .map!(i => item[0 .. i] ~ (items - 1) ~ item[i .. $]));
immutable f = (in size_t i) pure nothrow @safe =>
item[0 .. i] ~ (items - 1) ~ item[i .. $];
r ~= (i % 2) ?
//iota(item.length.signed, -1, -1).map!f.array :
iota(item.length + 1).retro.map!f.array :
iota(item.length + 1).map!f.array;
}
return r;
}
return inner(n).zip([1, -1].cycle);
}
void main() {
import std.stdio;
foreach (immutable n; [2, 3, 4]) {
writefln("Permutations and sign of %d items:", n);
foreach (immutable tp; n.sPermutations)
writefln(" %s Sign: %2d", tp[]);
writeln;
}
}
- Output:
Permutations and sign of 2 items: [1, 0] Sign: 1 [0, 1] Sign: -1 Permutations and sign of 3 items: [2, 1, 0] Sign: 1 [1, 2, 0] Sign: -1 [1, 0, 2] Sign: 1 [0, 1, 2] Sign: -1 [0, 2, 1] Sign: 1 [2, 0, 1] Sign: -1 Permutations and sign of 4 items: [3, 2, 1, 0] Sign: 1 [2, 3, 1, 0] Sign: -1 [2, 1, 3, 0] Sign: 1 [2, 1, 0, 3] Sign: -1 [1, 2, 0, 3] Sign: 1 [1, 2, 3, 0] Sign: -1 [1, 3, 2, 0] Sign: 1 [3, 1, 2, 0] Sign: -1 [3, 1, 0, 2] Sign: 1 [1, 3, 0, 2] Sign: -1 [1, 0, 3, 2] Sign: 1 [1, 0, 2, 3] Sign: -1 [0, 1, 2, 3] Sign: 1 [0, 1, 3, 2] Sign: -1 [0, 3, 1, 2] Sign: 1 [3, 0, 1, 2] Sign: -1 [3, 0, 2, 1] Sign: 1 [0, 3, 2, 1] Sign: -1 [0, 2, 3, 1] Sign: 1 [0, 2, 1, 3] Sign: -1 [2, 0, 1, 3] Sign: 1 [2, 0, 3, 1] Sign: -1 [2, 3, 0, 1] Sign: 1 [3, 2, 0, 1] Sign: -1
Dart
void main() {
List<int> array = List.generate(4, (i) => i);
HeapsAlgorithm algorithm = HeapsAlgorithm();
algorithm.recursive(array);
print('');
algorithm.loop(array);
}
class HeapsAlgorithm {
void recursive(List array) {
_recursive(array, array.length, true);
}
void _recursive(List array, int n, bool plus) {
if (n == 1) {
_output(array, plus);
} else {
for (int i = 0; i < n; i++) {
_recursive(array, n - 1, i == 0);
_swap(array, n % 2 == 0 ? i : 0, n - 1);
}
}
}
void _output(List array, bool plus) {
print(array.toString() + (plus ? ' +1' : ' -1'));
}
void _swap(List array, int a, int b) {
var temp = array[a];
array[a] = array[b];
array[b] = temp;
}
void loop(List array) {
_loop(array, array.length);
}
void _loop(List array, int n) {
List<int> c = List.filled(n, 0);
_output(array, true);
bool plus = false;
int i = 0;
while (i < n) {
if (c[i] < i) {
if (i % 2 == 0) {
_swap(array, 0, i);
} else {
_swap(array, c[i], i);
}
_output(array, plus);
plus = !plus;
c[i]++;
i = 0;
} else {
c[i] = 0;
i++;
}
}
}
}
- Output:
[0, 1, 2, 3] +1 [1, 0, 2, 3] -1 [2, 0, 1, 3] +1 [0, 2, 1, 3] -1 [1, 2, 0, 3] +1 [2, 1, 0, 3] -1 [3, 1, 2, 0] +1 [1, 3, 2, 0] -1 [2, 3, 1, 0] +1 [3, 2, 1, 0] -1 [1, 2, 3, 0] +1 [2, 1, 3, 0] -1 [3, 0, 2, 1] +1 [0, 3, 2, 1] -1 [2, 3, 0, 1] +1 [3, 2, 0, 1] -1 [0, 2, 3, 1] +1 [2, 0, 3, 1] -1 [3, 0, 1, 2] +1 [0, 3, 1, 2] -1 [1, 3, 0, 2] +1 [3, 1, 0, 2] -1 [0, 1, 3, 2] +1 [1, 0, 3, 2] -1 [3, 0, 1, 2] +1 [0, 3, 1, 2] -1 [1, 3, 0, 2] +1 [3, 1, 0, 2] -1 [0, 1, 3, 2] +1 [1, 0, 3, 2] -1 [2, 0, 3, 1] +1 [0, 2, 3, 1] -1 [3, 2, 0, 1] +1 [2, 3, 0, 1] -1 [0, 3, 2, 1] +1 [3, 0, 2, 1] -1 [3, 1, 2, 0] +1 [1, 3, 2, 0] -1 [2, 3, 1, 0] +1 [3, 2, 1, 0] -1 [1, 2, 3, 0] +1 [2, 1, 3, 0] -1 [2, 1, 0, 3] +1 [1, 2, 0, 3] -1 [0, 2, 1, 3] +1 [2, 0, 1, 3] -1 [1, 0, 2, 3] +1 [0, 1, 2, 3] -1
Delphi
{These routines would normally be in a separate library; they are presented here for clarity}
{Permutator based on the Johnson and Trotter algorithm.}
{Which only permutates by swapping a pair of elements at a time}
{object steps through all permutation of array items}
{Zero-Based = True = 0..Permutions-1 False = 1..Permutaions}
{Permutation set on "Create(Size)" or by "Permutations" property}
{Permutation are contained in the array "Indices"}
type TDirection = (drLeftToRight,drRightToLeft);
type TDirArray = array of TDirection;
type TJTPermutator = class(TObject)
private
Dir: TDirArray;
FZeroBased: boolean;
FBase: integer;
FPermutations: integer;
procedure SetZeroBased(const Value: boolean);
procedure SetPermutations(const Value: integer);
protected
FMax: integer;
public
NextCount: Integer;
Indices: TIntegerDynArray;
constructor Create(Size: integer);
procedure Reset;
function Next: boolean;
property ZeroBased: boolean read FZeroBased write SetZeroBased;
property Permutations: integer read FPermutations write SetPermutations;
end;
{==============================================================================}
function Fact(N: integer): integer;
{Get factorial of N}
var I: integer;
begin
Result:=1;
for I:=1 to N do Result:=Result * I;
end;
procedure SwapIntegers(var A1,A2: integer);
{Swap integer arguments}
var T: integer;
begin
T:=A1; A1:=A2; A2:=T;
end;
procedure TJTPermutator.Reset;
var I: integer;
begin
{ Preset items 0..n-1 or 1..n depending on base}
for I:=0 to High(Indices) do Indices[I]:=I + FBase;
{ initially all directions are set to RIGHT TO LEFT }
for I:=0 to High(Indices) do Dir[I]:=drRightToLeft;
NextCount:=0;
end;
procedure TJTPermutator.SetPermutations(const Value: integer);
begin
if FPermutations<>Value then
begin
FPermutations := Value;
SetLength(Indices,Value);
SetLength(Dir,Value);
Reset;
end;
end;
constructor TJTPermutator.Create(Size: integer);
begin
ZeroBased:=True;
Permutations:=Size;
Reset;
end;
procedure TJTPermutator.SetZeroBased(const Value: boolean);
begin
if FZeroBased<>Value then
begin
FZeroBased := Value;
if Value then FBase:=0
else FBase:=1;
Reset;
end;
end;
function TJTPermutator.Next: boolean;
{Step to next permutation}
{Returns true when sequence completed}
var Mobile,Pos,I: integer;
var S: string;
function FindLargestMoble(Mobile: integer): integer;
{Find position of largest mobile integer in A}
var I: integer;
begin
for I:=0 to High(Indices) do
if Indices[I] = Mobile then
begin
Result:=I + 1;
exit;
end;
Result:=-1;
end;
function GetMobile: integer;
{ find the largest mobile integer.}
var LastMobile, Mobile: integer;
var I: integer;
begin
LastMobile:= 0; Mobile:= 0;
for I:=0 to High(Indices) do
begin
{ direction 0 represents RIGHT TO LEFT.}
if (Dir[Indices[I] - 1] = drRightToLeft) and (I<>0) then
begin
if (Indices[I] > Indices[I - 1]) and (Indices[I] > LastMobile) then
begin
Mobile:=Indices[I];
LastMobile:=Mobile;
end;
end;
{ direction 1 represents LEFT TO RIGHT.}
if (dir[Indices[I] - 1] = drLeftToRight) and (i<>(Length(Indices) - 1)) then
begin
if (Indices[I] > Indices[I + 1]) and (Indices[I] > LastMobile) then
begin
Mobile:=Indices[I];
LastMobile:=Mobile;
end;
end;
end;
if (Mobile = 0) and (LastMobile = 0) then Result:=0
else Result:=Mobile;
end;
begin
Inc(NextCount);
Result:=NextCount>=Fact(Length(Indices));
if Result then
begin
Reset;
exit;
end;
Mobile:=GetMobile;
Pos:=FindLargestMoble(Mobile);
{ Swap elements according to the direction in Dir}
if (Dir[Indices[pos - 1] - 1] = drRightToLeft) then SwapIntegers(Indices[Pos - 1], Indices[Pos - 2])
else if (dir[Indices[pos - 1] - 1] = drLeftToRight) then SwapIntegers(Indices[Pos], Indices[Pos - 1]);
{ changing the directions for elements}
{ greater than largest Mobile integer.}
for I:=0 to High(Indices) do
if Indices[I] > Mobile then
begin
if Dir[Indices[I] - 1] = drLeftToRight then Dir[Indices[I] - 1]:=drRightToLeft
else if (Dir[Indices[i] - 1] = drRightToLeft) then Dir[Indices[I] - 1]:=drLeftToRight;
end;
end;
{==============================================================================}
function GetPermutationStr(PM: TJTPermutator): string;
var I: integer;
begin
Result:=Format('%2d - [',[PM.NextCount+1]);
for I:=0 to High(PM.Indices) do Result:=Result+IntToStr(PM.Indices[I]);
Result:=Result+'] Sign: ';
if (PM.NextCount and 1)=0 then Result:=Result+'+1'
else Result:=Result+'-1';
end;
procedure SwapPermutations(Memo: TMemo);
var PM: TJTPermutator;
begin
PM:=TJTPermutator.Create(3);
try
repeat Memo.Lines.Add(GetPermutationStr(PM))
until PM.Next;
Memo.Lines.Add('');
PM.Permutations:=4;
repeat Memo.Lines.Add(GetPermutationStr(PM))
until PM.Next;
finally PM.Free; end;
end;
- Output:
1 - [012] Sign: +1 2 - [021] Sign: -1 3 - [201] Sign: +1 4 - [210] Sign: -1 5 - [120] Sign: +1 6 - [102] Sign: -1 1 - [0123] Sign: +1 2 - [0132] Sign: -1 3 - [0312] Sign: +1 4 - [3012] Sign: -1 5 - [3021] Sign: +1 6 - [0321] Sign: -1 7 - [0231] Sign: +1 8 - [0213] Sign: -1 9 - [2013] Sign: +1 10 - [2031] Sign: -1 11 - [2301] Sign: +1 12 - [3201] Sign: -1 13 - [3210] Sign: +1 14 - [2310] Sign: -1 15 - [2130] Sign: +1 16 - [2103] Sign: -1 17 - [1203] Sign: +1 18 - [1230] Sign: -1 19 - [1320] Sign: +1 20 - [3120] Sign: -1 21 - [3102] Sign: +1 22 - [1302] Sign: -1 23 - [1032] Sign: +1 24 - [1023] Sign: -1 Elapsed Time: 60.734 ms.
EasyLang
# Heap's Algorithm
sig = 1
proc generate k . ar[] .
if k = 1
print ar[] & " " & sig
sig = -sig
return
.
generate k - 1 ar[]
for i to k - 1
if k mod 2 = 0
swap ar[i] ar[k]
else
swap ar[1] ar[k]
.
generate k - 1 ar[]
.
.
ar[] = [ 1 2 3 ]
generate len ar[] ar[]
- Output:
[ 1 2 3 ] 1 [ 2 1 3 ] -1 [ 3 1 2 ] 1 [ 1 3 2 ] -1 [ 2 3 1 ] 1 [ 3 2 1 ] -1
EchoLisp
The function (in-permutations n) returns a stream which delivers permutations according to the Steinhaus–Johnson–Trotter algorithm.
(lib 'list)
(for/fold (sign 1) ((σ (in-permutations 4)) (count 100))
(printf "perm: %a count:%4d sign:%4d" σ count sign) (* sign -1))
perm: (0 1 2 3) count: 0 sign: 1
perm: (0 1 3 2) count: 1 sign: -1
perm: (0 3 1 2) count: 2 sign: 1
perm: (3 0 1 2) count: 3 sign: -1
perm: (3 0 2 1) count: 4 sign: 1
perm: (0 3 2 1) count: 5 sign: -1
perm: (0 2 3 1) count: 6 sign: 1
perm: (0 2 1 3) count: 7 sign: -1
perm: (2 0 1 3) count: 8 sign: 1
perm: (2 0 3 1) count: 9 sign: -1
perm: (2 3 0 1) count: 10 sign: 1
perm: (3 2 0 1) count: 11 sign: -1
perm: (3 2 1 0) count: 12 sign: 1
perm: (2 3 1 0) count: 13 sign: -1
perm: (2 1 3 0) count: 14 sign: 1
perm: (2 1 0 3) count: 15 sign: -1
perm: (1 2 0 3) count: 16 sign: 1
perm: (1 2 3 0) count: 17 sign: -1
perm: (1 3 2 0) count: 18 sign: 1
perm: (3 1 2 0) count: 19 sign: -1
perm: (3 1 0 2) count: 20 sign: 1
perm: (1 3 0 2) count: 21 sign: -1
perm: (1 0 3 2) count: 22 sign: 1
perm: (1 0 2 3) count: 23 sign: -1
Elixir
defmodule Permutation do
def by_swap(n) do
p = Enum.to_list(0..-n) |> List.to_tuple
by_swap(n, p, 1)
end
defp by_swap(n, p, s) do
IO.puts "Perm: #{inspect for i <- 1..n, do: abs(elem(p,i))} Sign: #{s}"
k = 0 |> step_up(n, p) |> step_down(n, p)
if k > 0 do
pk = elem(p,k)
i = if pk>0, do: k+1, else: k-1
p = Enum.reduce(1..n, p, fn i,acc ->
if abs(elem(p,i)) > abs(pk), do: put_elem(acc, i, -elem(acc,i)), else: acc
end)
pi = elem(p,i)
p = put_elem(p,i,pk) |> put_elem(k,pi) # swap
by_swap(n, p, -s)
end
end
defp step_up(k, n, p) do
Enum.reduce(2..n, k, fn i,acc ->
if elem(p,i)<0 and abs(elem(p,i))>abs(elem(p,i-1)) and abs(elem(p,i))>abs(elem(p,acc)),
do: i, else: acc
end)
end
defp step_down(k, n, p) do
Enum.reduce(1..n-1, k, fn i,acc ->
if elem(p,i)>0 and abs(elem(p,i))>abs(elem(p,i+1)) and abs(elem(p,i))>abs(elem(p,acc)),
do: i, else: acc
end)
end
end
Enum.each(3..4, fn n ->
Permutation.by_swap(n)
IO.puts ""
end)
- Output:
Perm: [1, 2, 3] Sign: 1 Perm: [1, 3, 2] Sign: -1 Perm: [3, 1, 2] Sign: 1 Perm: [3, 2, 1] Sign: -1 Perm: [2, 3, 1] Sign: 1 Perm: [2, 1, 3] Sign: -1 Perm: [1, 2, 3, 4] Sign: 1 Perm: [1, 2, 4, 3] Sign: -1 Perm: [1, 4, 2, 3] Sign: 1 Perm: [4, 1, 2, 3] Sign: -1 Perm: [4, 1, 3, 2] Sign: 1 Perm: [1, 4, 3, 2] Sign: -1 Perm: [1, 3, 4, 2] Sign: 1 Perm: [1, 3, 2, 4] Sign: -1 Perm: [3, 1, 2, 4] Sign: 1 Perm: [3, 1, 4, 2] Sign: -1 Perm: [3, 4, 1, 2] Sign: 1 Perm: [4, 3, 1, 2] Sign: -1 Perm: [4, 3, 2, 1] Sign: 1 Perm: [3, 4, 2, 1] Sign: -1 Perm: [3, 2, 4, 1] Sign: 1 Perm: [3, 2, 1, 4] Sign: -1 Perm: [2, 3, 1, 4] Sign: 1 Perm: [2, 3, 4, 1] Sign: -1 Perm: [2, 4, 3, 1] Sign: 1 Perm: [4, 2, 3, 1] Sign: -1 Perm: [4, 2, 1, 3] Sign: 1 Perm: [2, 4, 1, 3] Sign: -1 Perm: [2, 1, 4, 3] Sign: 1 Perm: [2, 1, 3, 4] Sign: -1
F#
See [2] for an example using this module
(*Implement Johnson-Trotter algorithm
Nigel Galloway January 24th 2017*)
module Ring
let PlainChanges (N:'n[]) = seq{
let gn = [|for n in N -> 1|]
let ni = [|for n in N -> 0|]
let gel = Array.length(N)-1
yield N
let rec _Ni g e l = seq{
match (l,g) with
|_ when l<0 -> gn.[g] <- -gn.[g]; yield! _Ni (g-1) e (ni.[g-1] + gn.[g-1])
|(1,0) -> ()
|_ when l=g+1 -> gn.[g] <- -gn.[g]; yield! _Ni (g-1) (e+1) (ni.[g-1] + gn.[g-1])
|_ -> let n = N.[g-ni.[g]+e];
N.[g-ni.[g]+e] <- N.[g-l+e]; N.[g-l+e] <- n; yield N
ni.[g] <- l; yield! _Ni gel 0 (ni.[gel] + gn.[gel])}
yield! _Ni gel 0 1
}
A little code for the purpose of this task demonstrating the algorithm
for n in Ring.PlainChanges [|1;2;3;4|] do printfn "%A" n
- Output:
[|1; 2; 3; 4|] [|1; 2; 4; 3|] [|1; 4; 2; 3|] [|4; 1; 2; 3|] [|4; 1; 3; 2|] [|1; 4; 3; 2|] [|1; 3; 4; 2|] [|1; 3; 2; 4|] [|3; 1; 2; 4|] [|3; 1; 4; 2|] [|3; 4; 1; 2|] [|4; 3; 1; 2|] [|4; 3; 2; 1|] [|3; 4; 2; 1|] [|3; 2; 4; 1|] [|3; 2; 1; 4|] [|2; 3; 1; 4|] [|2; 3; 4; 1|] [|2; 4; 3; 1|] [|4; 2; 3; 1|] [|4; 2; 1; 3|] [|2; 4; 1; 3|] [|2; 1; 4; 3|] [|2; 1; 3; 4|] v
Forth
S" fsl-util.fs" REQUIRED
S" fsl/dynmem.seq" REQUIRED
cell darray p{
: sgn
DUP 0 > IF
DROP 1
ELSE 0 < IF
-1
ELSE
0
THEN THEN ;
: arr-swap {: addr1 addr2 | tmp -- :}
addr1 @ TO tmp
addr2 @ addr1 !
tmp addr2 ! ;
: perms {: n xt | my-i k s -- :}
& p{ n 1+ }malloc malloc-fail? ABORT" perms :: out of memory"
0 p{ 0 } !
n 1+ 1 DO
I NEGATE p{ I } !
LOOP
1 TO s
BEGIN
1 n 1+ DO
p{ I } @ ABS
-1 +LOOP
n 1+ s xt EXECUTE
0 TO k
n 1+ 2 DO
p{ I } @ 0 < ( flag )
p{ I } @ ABS p{ I 1- } @ ABS > ( flag flag )
p{ I } @ ABS p{ k } @ ABS > ( flag flag flag )
AND AND IF
I TO k
THEN
LOOP
n 1 DO
p{ I } @ 0 > ( flag )
p{ I } @ ABS p{ I 1+ } @ ABS > ( flag flag )
p{ I } @ ABS p{ k } @ ABS > ( flag flag flag )
AND AND IF
I TO k
THEN
LOOP
k IF
n 1+ 1 DO
p{ I } @ ABS p{ k } @ ABS > IF
p{ I } @ NEGATE p{ I } !
THEN
LOOP
p{ k } @ sgn k + TO my-i
p{ k } p{ my-i } arr-swap
s NEGATE TO s
THEN
k 0 = UNTIL ;
: .perm ( p0 p1 p2 ... pn n s )
>R
." Perm: [ "
1 DO
. SPACE
LOOP
R> ." ] Sign: " . CR ;
3 ' .perm perms CR
4 ' .perm perms
FreeBASIC
' version 31-03-2017
' compile with: fbc -s console
Sub perms(n As ULong)
Dim As Long p(n), i, k, s = 1
For i = 1 To n
p(i) = -i
Next
Do
Print "Perm: [ ";
For i = 1 To n
Print Abs(p(i)); " ";
Next
Print "] Sign: "; s
k = 0
For i = 2 To n
If p(i) < 0 Then
If Abs(p(i)) > Abs(p(i -1)) Then
If Abs(p(i)) > Abs(p(k)) Then k = i
End If
End If
Next
For i = 1 To n -1
If p(i) > 0 Then
If Abs(p(i)) > Abs(p(i +1)) Then
If Abs(p(i)) > Abs(p(k)) Then k = i
End If
End If
Next
If k Then
For i = 1 To n
If Abs(p(i)) > Abs(p(k)) Then p(i) = -p(i)
Next
i = k + Sgn(p(k))
Swap p(k), p(i)
s = -s
End If
Loop Until k = 0
End Sub
' ------=< MAIN >=------
perms(3)
print
perms(4)
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
- Output:
output is edited to show results side by side
Perm: [ 1 2 3 ] Sign: 1 Perm: [ 1 2 3 4 ] Sign: 1
Perm: [ 1 3 2 ] Sign: -1 Perm: [ 1 2 4 3 ] Sign: -1
Perm: [ 3 1 2 ] Sign: 1 Perm: [ 1 4 2 3 ] Sign: 1
Perm: [ 3 2 1 ] Sign: -1 Perm: [ 4 1 2 3 ] Sign: -1
Perm: [ 2 3 1 ] Sign: 1 Perm: [ 4 1 3 2 ] Sign: 1
Perm: [ 2 1 3 ] Sign: -1 Perm: [ 1 4 3 2 ] Sign: -1
Perm: [ 1 2 3 4 ] Sign: 1
Perm: [ 1 2 4 3 ] Sign: -1
Perm: [ 1 4 2 3 ] Sign: 1
Perm: [ 4 1 2 3 ] Sign: -1
Perm: [ 4 1 3 2 ] Sign: 1
Perm: [ 1 4 3 2 ] Sign: -1
Perm: [ 1 3 4 2 ] Sign: 1
Perm: [ 1 3 2 4 ] Sign: -1
Perm: [ 3 1 2 4 ] Sign: 1
Perm: [ 3 1 4 2 ] Sign: -1
Perm: [ 3 4 1 2 ] Sign: 1
Perm: [ 4 3 1 2 ] Sign: -1
Perm: [ 4 3 2 1 ] Sign: 1
Perm: [ 3 4 2 1 ] Sign: -1
Perm: [ 3 2 4 1 ] Sign: 1
Perm: [ 3 2 1 4 ] Sign: -1
Perm: [ 2 3 1 4 ] Sign: 1
Perm: [ 2 3 4 1 ] Sign: -1
Perm: [ 2 4 3 1 ] Sign: 1
Perm: [ 4 2 3 1 ] Sign: -1
Perm: [ 4 2 1 3 ] Sign: 1
Perm: [ 2 4 1 3 ] Sign: -1
Perm: [ 2 1 4 3 ] Sign: 1
Perm: [ 2 1 3 4 ] Sign: -1
Go
package permute
// Iter takes a slice p and returns an iterator function. The iterator
// permutes p in place and returns the sign. After all permutations have
// been generated, the iterator returns 0 and p is left in its initial order.
func Iter(p []int) func() int {
f := pf(len(p))
return func() int {
return f(p)
}
}
// Recursive function used by perm, returns a chain of closures that
// implement a loopless recursive SJT.
func pf(n int) func([]int) int {
sign := 1
switch n {
case 0, 1:
return func([]int) (s int) {
s = sign
sign = 0
return
}
default:
p0 := pf(n - 1)
i := n
var d int
return func(p []int) int {
switch {
case sign == 0:
case i == n:
i--
sign = p0(p[:i])
d = -1
case i == 0:
i++
sign *= p0(p[1:])
d = 1
if sign == 0 {
p[0], p[1] = p[1], p[0]
}
default:
p[i], p[i-1] = p[i-1], p[i]
sign = -sign
i += d
}
return sign
}
}
}
package main
import (
"fmt"
"permute"
)
func main() {
p := []int{11, 22, 33}
i := permute.Iter(p)
for sign := i(); sign != 0; sign = i() {
fmt.Println(p, sign)
}
}
- Output:
[11 22 33] 1 [11 33 22] -1 [33 11 22] 1 [33 22 11] -1 [22 33 11] 1 [22 11 33] -1
Haskell
sPermutations :: [a] -> [([a], Int)]
sPermutations = flip zip (cycle [-1, 1]) . foldr aux [[]]
where
aux x items = do
(f, item) <- zip (repeat id) items
f (insertEv x item)
insertEv x [] = [[x]]
insertEv x l@(y:ys) = (x : l) : ((y :) <$> insertEv x ys)
main :: IO ()
main = do
putStrLn "3 items:"
mapM_ print $ sPermutations [1 .. 3]
putStrLn "\n4 items:"
mapM_ print $ sPermutations [1 .. 4]
- Output:
3 items: ([1,2,3],-1) ([2,1,3],1) ([2,3,1],-1) ([1,3,2],1) ([3,1,2],-1) ([3,2,1],1) 4 items: ([1,2,3,4],-1) ([2,1,3,4],1) ([2,3,1,4],-1) ([2,3,4,1],1) ([1,3,2,4],-1) ([3,1,2,4],1) ([3,2,1,4],-1) ([3,2,4,1],1) ([1,3,4,2],-1) ([3,1,4,2],1) ([3,4,1,2],-1) ([3,4,2,1],1) ([1,2,4,3],-1) ([2,1,4,3],1) ([2,4,1,3],-1) ([2,4,3,1],1) ([1,4,2,3],-1) ([4,1,2,3],1) ([4,2,1,3],-1) ([4,2,3,1],1) ([1,4,3,2],-1) ([4,1,3,2],1) ([4,3,1,2],-1) ([4,3,2,1],1)
Icon and Unicon
Works in both languages.
procedure main(A)
every write("Permutations of length ",n := !A) do
every p := permute(n) do write("\t",showList(p[1])," -> ",right(p[2],2))
end
procedure permute(n)
items := [[]]
every (j := 1 to n, new_items := []) do {
every item := items[i := 1 to *items] do {
if *item = 0 then put(new_items, [j])
else if i%2 = 0 then
every k := 1 to *item+1 do {
new_item := item[1:k] ||| [j] ||| item[k:0]
put(new_items, new_item)
}
else
every k := *item+1 to 1 by -1 do {
new_item := item[1:k] ||| [j] ||| item[k:0]
put(new_items, new_item)
}
}
items := new_items
}
suspend (i := 0, [!items, if (i+:=1)%2 = 0 then 1 else -1])
end
procedure showList(A)
every (s := "[") ||:= image(!A)||", "
return s[1:-2]||"]"
end
Sample run:
->pbs 3 4
Permutations of length 3
[1, 2, 3] -> -1
[1, 3, 2] -> 1
[3, 1, 2] -> -1
[3, 2, 1] -> 1
[2, 3, 1] -> -1
[2, 1, 3] -> 1
Permutations of length 4
[1, 2, 3, 4] -> -1
[1, 2, 4, 3] -> 1
[1, 4, 2, 3] -> -1
[4, 1, 2, 3] -> 1
[4, 1, 3, 2] -> -1
[1, 4, 3, 2] -> 1
[1, 3, 4, 2] -> -1
[1, 3, 2, 4] -> 1
[3, 1, 2, 4] -> -1
[3, 1, 4, 2] -> 1
[3, 4, 1, 2] -> -1
[4, 3, 1, 2] -> 1
[4, 3, 2, 1] -> -1
[3, 4, 2, 1] -> 1
[3, 2, 4, 1] -> -1
[3, 2, 1, 4] -> 1
[2, 3, 1, 4] -> -1
[2, 3, 4, 1] -> 1
[2, 4, 3, 1] -> -1
[4, 2, 3, 1] -> 1
[4, 2, 1, 3] -> -1
[2, 4, 1, 3] -> 1
[2, 1, 4, 3] -> -1
[2, 1, 3, 4] -> 1
->
J
J has a built in mechanism for representing permutations for selecting a permutation of a given length with an integer, but this mechanism does not seem to have an obvious mapping to Steinhaus–Johnson–Trotter. Perhaps someone with a sufficiently deep view of the subject of permutations can find a direct mapping?
Meanwhile, here's an inductive approach, using negative integers to look left and positive integers to look right:
bfsjt0=: _1 - i.
lookingat=: 0 >. <:@# <. i.@# + *
next=: | >./@:* | > | {~ lookingat
bfsjtn=: (((] <@, ] + *@{~) | i. next) C. ] * _1 ^ next < |)^:(*@next)
Here, bfsjt0 N gives the initial permutation of order N, and bfsjtn^:M bfsjt0 N gives the Mth Steinhaus–Johnson–Trotter permutation of order N. (bf stands for "brute force".)
To convert from the Steinhaus–Johnson–Trotter representation of a permutation to J's representation, use <:@|, or to find J's anagram index of a Steinhaus–Johnson–Trotter representation of a permutation, use A.@:<:@:|
Example use:
bfsjtn^:(i.!3) bfjt0 3
_1 _2 _3
_1 _3 _2
_3 _1 _2
3 _2 _1
_2 3 _1
_2 _1 3
<:@| bfsjtn^:(i.!3) bfjt0 3
0 1 2
0 2 1
2 0 1
2 1 0
1 2 0
1 0 2
A. <:@| bfsjtn^:(i.!3) bfjt0 3
0 1 4 5 3 2
Here's an example of the Steinhaus–Johnson–Trotter representation of 3 element permutation, with sign (sign is the first column):
(_1^2|i.!3),. bfsjtn^:(i.!3) bfjt0 3
1 _1 _2 _3
_1 _1 _3 _2
1 _3 _1 _2
_1 3 _2 _1
1 _2 3 _1
_1 _2 _1 3
Alternatively, J defines C.!.2 as the parity of a permutation:
(,.~C.!.2)<:| bfsjtn^:(i.!3) bfjt0 3
1 0 1 2
_1 0 2 1
1 2 0 1
_1 2 1 0
1 1 2 0
_1 1 0 2
Recursive Implementation
This is based on the python recursive implementation:
rsjt=: 3 :0
if. 2>y do. i.2#y
else. ((!y)$(,~|.)-.=i.y)#inv!.(y-1)"1 y#rsjt y-1
end.
)
Example use (here, prefixing each row with its parity):
(,.~ C.!.2) rsjt 3
1 0 1 2
_1 0 2 1
1 2 0 1
_1 2 1 0
1 1 2 0
_1 1 0 2
Java
Heap's Algorithm, recursive and looping implementations
package org.rosettacode.java;
import java.util.Arrays;
import java.util.stream.IntStream;
public class HeapsAlgorithm {
public static void main(String[] args) {
Object[] array = IntStream.range(0, 4)
.boxed()
.toArray();
HeapsAlgorithm algorithm = new HeapsAlgorithm();
algorithm.recursive(array);
System.out.println();
algorithm.loop(array);
}
void recursive(Object[] array) {
recursive(array, array.length, true);
}
void recursive(Object[] array, int n, boolean plus) {
if (n == 1) {
output(array, plus);
} else {
for (int i = 0; i < n; i++) {
recursive(array, n - 1, i == 0);
swap(array, n % 2 == 0 ? i : 0, n - 1);
}
}
}
void output(Object[] array, boolean plus) {
System.out.println(Arrays.toString(array) + (plus ? " +1" : " -1"));
}
void swap(Object[] array, int a, int b) {
Object o = array[a];
array[a] = array[b];
array[b] = o;
}
void loop(Object[] array) {
loop(array, array.length);
}
void loop(Object[] array, int n) {
int[] c = new int[n];
output(array, true);
boolean plus = false;
for (int i = 0; i < n; ) {
if (c[i] < i) {
if (i % 2 == 0) {
swap(array, 0, i);
} else {
swap(array, c[i], i);
}
output(array, plus);
plus = !plus;
c[i]++;
i = 0;
} else {
c[i] = 0;
i++;
}
}
}
}
- Output:
[0, 1, 2, 3] +1 [1, 0, 2, 3] -1 [2, 0, 1, 3] +1 [0, 2, 1, 3] -1 [1, 2, 0, 3] +1 [2, 1, 0, 3] -1 [3, 1, 2, 0] +1 [1, 3, 2, 0] -1 [2, 3, 1, 0] +1 [3, 2, 1, 0] -1 [1, 2, 3, 0] +1 [2, 1, 3, 0] -1 [3, 0, 2, 1] +1 [0, 3, 2, 1] -1 [2, 3, 0, 1] +1 [3, 2, 0, 1] -1 [0, 2, 3, 1] +1 [2, 0, 3, 1] -1 [3, 0, 1, 2] +1 [0, 3, 1, 2] -1 [1, 3, 0, 2] +1 [3, 1, 0, 2] -1 [0, 1, 3, 2] +1 [1, 0, 3, 2] -1 [3, 0, 1, 2] +1 [0, 3, 1, 2] -1 [1, 3, 0, 2] +1 [3, 1, 0, 2] -1 [0, 1, 3, 2] +1 [1, 0, 3, 2] -1 [2, 0, 3, 1] +1 [0, 2, 3, 1] -1 [3, 2, 0, 1] +1 [2, 3, 0, 1] -1 [0, 3, 2, 1] +1 [3, 0, 2, 1] -1 [3, 1, 2, 0] +1 [1, 3, 2, 0] -1 [2, 3, 1, 0] +1 [3, 2, 1, 0] -1 [1, 2, 3, 0] +1 [2, 1, 3, 0] -1 [2, 1, 0, 3] +1 [1, 2, 0, 3] -1 [0, 2, 1, 3] +1 [2, 0, 1, 3] -1 [1, 0, 2, 3] +1 [0, 1, 2, 3] -1
jq
Based on the ruby version - the sequence is generated by swapping adjacent elements.
"permutations" generates a stream of arrays of the form [par, perm], where "par" is the parity of the permutation "perm" of the input array. This array may contain any JSON entities, which are regarded as distinct.
# The helper function, _recurse, is tail-recursive and therefore in
# versions of jq with TCO (tail call optimization) there is no
# overhead associated with the recursion.
def permutations:
def abs: if . < 0 then -. else . end;
def sign: if . < 0 then -1 elif . == 0 then 0 else 1 end;
def swap(i;j): .[i] as $i | .[i] = .[j] | .[j] = $i;
# input: [ parity, extendedPermutation]
def _recurse:
.[0] as $s | .[1] as $p | (($p | length) -1) as $n
| [ $s, ($p[1:] | map(abs)) ],
(reduce range(2; $n+1) as $i
(0;
if $p[$i] < 0 and -($p[$i]) > ($p[$i-1]|abs) and -($p[$i]) > ($p[.]|abs)
then $i
else .
end)) as $k
| (reduce range(1; $n) as $i
($k;
if $p[$i] > 0 and $p[$i] > ($p[$i+1]|abs) and $p[$i] > ($p[.]|abs)
then $i
else .
end)) as $k
| if $k == 0 then empty
else (reduce range(1; $n) as $i
($p;
if (.[$i]|abs) > (.[$k]|abs) then .[$i] *= -1
else .
end )) as $p
| ($k + ($p[$k]|sign)) as $i
| ($p | swap($i; $k)) as $p
| [ -($s), $p ] | _recurse
end ;
. as $in
| length as $n
| (reduce range(0; $n+1) as $i ([]; . + [ -$i ])) as $p
# recurse state: [$s, $p]
| [ 1, $p] | _recurse
| .[1] as $p
| .[1] = reduce range(0; $n) as $i ([]; . + [$in[$p[$i] - 1]]) ;
def count(stream): reduce stream as $x (0; .+1);Examples:
(["a", "b", "c"] | permutations),
"There are \(count( [range(1;6)] | permutations )) permutations of 5 items."- Output:
$ jq -c -n -f Permutations_by_swapping.jq
[1,["a","b","c"]]
[-1,["a","c","b"]]
[1,["c","a","b"]]
[-1,["c","b","a"]]
[1,["b","c","a"]]
[-1,["b","a","c"]]
"There are 32 permutations of 5 items."
Julia
Nonrecursive (interative):
function johnsontrottermove!(ints, isleft)
len = length(ints)
function ismobile(pos)
if isleft[pos] && (pos > 1) && (ints[pos-1] < ints[pos])
return true
elseif !isleft[pos] && (pos < len) && (ints[pos+1] < ints[pos])
return true
end
false
end
function maxmobile()
arr = [ints[pos] for pos in 1:len if ismobile(pos)]
if isempty(arr)
0, 0
else
maxmob = maximum(arr)
maxmob, findfirst(x -> x == maxmob, ints)
end
end
function directedswap(pos)
tmp = ints[pos]
tmpisleft = isleft[pos]
if isleft[pos]
ints[pos] = ints[pos-1]; ints[pos-1] = tmp
isleft[pos] = isleft[pos-1]; isleft[pos-1] = tmpisleft
else
ints[pos] = ints[pos+1]; ints[pos+1] = tmp
isleft[pos] = isleft[pos+1]; isleft[pos+1] = tmpisleft
end
end
(moveint, movepos) = maxmobile()
if movepos > 0
directedswap(movepos)
for (i, val) in enumerate(ints)
if val > moveint
isleft[i] = !isleft[i]
end
end
ints, isleft, true
else
ints, isleft, false
end
end
function johnsontrotter(low, high)
ints = collect(low:high)
isleft = [true for i in ints]
firstconfig = copy(ints)
iters = 0
while true
iters += 1
println("$ints $(iters & 1 == 1 ? "+1" : "-1")")
if johnsontrottermove!(ints, isleft)[3] == false
break
end
end
println("There were $iters iterations.")
end
johnsontrotter(1,4)
Recursive (note this uses memory of roughtly (n+1)! bytes, where n is the number of elements, in order to store the accumulated permutations in a list, and so the above, iterative solution is to be preferred for numbers of elements over 9 or so):
function johnsontrotter(low, high)
function permutelevel(vec)
if length(vec) < 2
return [vec]
end
sequences = []
endint = vec[end]
smallersequences = permutelevel(vec[1:end-1])
leftward = true
for seq in smallersequences
for pos in (leftward ? (length(seq)+1:-1:1): (1:length(seq)+1))
push!(sequences, insert!(copy(seq), pos, endint))
end
leftward = !leftward
end
sequences
end
permutelevel(collect(low:high))
end
for (i, sequence) in enumerate(johnsontrotter(1,4))
println("""$sequence, $(i & 1 == 1 ? "+1" : "-1")""")
end
Kotlin
This is based on the recursive Java code found at http://introcs.cs.princeton.edu/java/23recursion/JohnsonTrotter.java.html
// version 1.1.2
fun johnsonTrotter(n: Int): Pair<List<IntArray>, List<Int>> {
val p = IntArray(n) { it } // permutation
val q = IntArray(n) { it } // inverse permutation
val d = IntArray(n) { -1 } // direction = 1 or -1
var sign = 1
val perms = mutableListOf<IntArray>()
val signs = mutableListOf<Int>()
fun permute(k: Int) {
if (k >= n) {
perms.add(p.copyOf())
signs.add(sign)
sign *= -1
return
}
permute(k + 1)
for (i in 0 until k) {
val z = p[q[k] + d[k]]
p[q[k]] = z
p[q[k] + d[k]] = k
q[z] = q[k]
q[k] += d[k]
permute(k + 1)
}
d[k] *= -1
}
permute(0)
return perms to signs
}
fun printPermsAndSigns(perms: List<IntArray>, signs: List<Int>) {
for ((i, perm) in perms.withIndex()) {
println("${perm.contentToString()} -> sign = ${signs[i]}")
}
}
fun main(args: Array<String>) {
val (perms, signs) = johnsonTrotter(3)
printPermsAndSigns(perms, signs)
println()
val (perms2, signs2) = johnsonTrotter(4)
printPermsAndSigns(perms2, signs2)
}
- Output:
[0, 1, 2] -> sign = 1 [0, 2, 1] -> sign = -1 [2, 0, 1] -> sign = 1 [2, 1, 0] -> sign = -1 [1, 2, 0] -> sign = 1 [1, 0, 2] -> sign = -1 [0, 1, 2, 3] -> sign = 1 [0, 1, 3, 2] -> sign = -1 [0, 3, 1, 2] -> sign = 1 [3, 0, 1, 2] -> sign = -1 [3, 0, 2, 1] -> sign = 1 [0, 3, 2, 1] -> sign = -1 [0, 2, 3, 1] -> sign = 1 [0, 2, 1, 3] -> sign = -1 [2, 0, 1, 3] -> sign = 1 [2, 0, 3, 1] -> sign = -1 [2, 3, 0, 1] -> sign = 1 [3, 2, 0, 1] -> sign = -1 [3, 2, 1, 0] -> sign = 1 [2, 3, 1, 0] -> sign = -1 [2, 1, 3, 0] -> sign = 1 [2, 1, 0, 3] -> sign = -1 [1, 2, 0, 3] -> sign = 1 [1, 2, 3, 0] -> sign = -1 [1, 3, 2, 0] -> sign = 1 [3, 1, 2, 0] -> sign = -1 [3, 1, 0, 2] -> sign = 1 [1, 3, 0, 2] -> sign = -1 [1, 0, 3, 2] -> sign = 1 [1, 0, 2, 3] -> sign = -1
Lua
_JT={}
function JT(dim)
local n={ values={}, positions={}, directions={}, sign=1 }
setmetatable(n,{__index=_JT})
for i=1,dim do
n.values[i]=i
n.positions[i]=i
n.directions[i]=-1
end
return n
end
function _JT:largestMobile()
for i=#self.values,1,-1 do
local loc=self.positions[i]+self.directions[i]
if loc >= 1 and loc <= #self.values and self.values[loc] < i then
return i
end
end
return 0
end
function _JT:next()
local r=self:largestMobile()
if r==0 then return false end
local rloc=self.positions[r]
local lloc=rloc+self.directions[r]
local l=self.values[lloc]
self.values[lloc],self.values[rloc] = self.values[rloc],self.values[lloc]
self.positions[l],self.positions[r] = self.positions[r],self.positions[l]
self.sign=-self.sign
for i=r+1,#self.directions do self.directions[i]=-self.directions[i] end
return true
end
-- test
perm=JT(4)
repeat
print(unpack(perm.values))
until not perm:next()
- Output:
1 2 3 4 1 2 4 3 1 4 2 3 4 1 2 3 4 1 3 2 1 4 3 2 1 3 4 2 1 3 2 4 3 1 2 4 3 1 4 2 3 4 1 2 4 3 1 2 4 3 2 1 3 4 2 1 3 2 4 1 3 2 1 4 2 3 1 4 2 3 4 1 2 4 3 1 4 2 3 1 4 2 1 3 2 4 1 3 2 1 4 3 2 1 3 4
Coroutine Implementation
This is adapted from the Lua Book .
local wrap, yield = coroutine.wrap, coroutine.yield
local function perm(n)
local r = {}
for i=1,n do r[i]=i end
local sign = 1
return wrap(function()
local function swap(m)
if m==0 then
sign = -sign, yield(sign,r)
else
for i=m,1,-1 do
r[i],r[m]=r[m],r[i]
swap(m-1)
r[i],r[m]=r[m],r[i]
end
end
end
swap(n)
end)
end
for sign,r in perm(3) do print(sign,table.unpack(r))end
Mathematica /Wolfram Language
Recursive
perms[0] = {{{}, 1}};
perms[n_] :=
Flatten[If[#2 == 1, Reverse, # &]@
Table[{Insert[#1, n, i], (-1)^(n + i) #2}, {i, n}] & @@@
perms[n - 1], 1];
Example:
Print["Perm: ", #[[1]], " Sign: ", #[[2]]] & /@ perms@4;
- Output:
Perm: {1,2,3,4} Sign: 1
Perm: {1,2,4,3} Sign: -1
Perm: {1,4,2,3} Sign: 1
Perm: {4,1,2,3} Sign: -1
Perm: {4,1,3,2} Sign: 1
Perm: {1,4,3,2} Sign: -1
Perm: {1,3,4,2} Sign: 1
Perm: {1,3,2,4} Sign: -1
Perm: {3,1,2,4} Sign: 1
Perm: {3,1,4,2} Sign: -1
Perm: {3,4,1,2} Sign: 1
Perm: {4,3,1,2} Sign: -1
Perm: {4,3,2,1} Sign: 1
Perm: {3,4,2,1} Sign: -1
Perm: {3,2,4,1} Sign: 1
Perm: {3,2,1,4} Sign: -1
Perm: {2,3,1,4} Sign: 1
Perm: {2,3,4,1} Sign: -1
Perm: {2,4,3,1} Sign: 1
Perm: {4,2,3,1} Sign: -1
Perm: {4,2,1,3} Sign: 1
Perm: {2,4,1,3} Sign: -1
Perm: {2,1,4,3} Sign: 1
Perm: {2,1,3,4} Sign: -1
Nim
# iterative Boothroyd method
iterator permutations*[T](ys: openarray[T]): tuple[perm: seq[T], sign: int] =
var
d = 1
c = newSeq[int](ys.len)
xs = newSeq[T](ys.len)
sign = 1
for i, y in ys: xs[i] = y
yield (xs, sign)
block outter:
while true:
while d > 1:
dec d
c[d] = 0
while c[d] >= d:
inc d
if d >= ys.len: break outter
let i = if (d and 1) == 1: c[d] else: 0
swap xs[i], xs[d]
sign *= -1
yield (xs, sign)
inc c[d]
when isMainModule:
for i in permutations([0,1,2]):
echo i
echo ""
for i in permutations([0,1,2,3]):
echo i
- Output:
(perm: @[0, 1, 2], sign: 1) (perm: @[1, 0, 2], sign: -1) (perm: @[2, 0, 1], sign: 1) (perm: @[0, 2, 1], sign: -1) (perm: @[1, 2, 0], sign: 1) (perm: @[2, 1, 0], sign: -1) (perm: @[0, 1, 2, 3], sign: 1) (perm: @[1, 0, 2, 3], sign: -1) (perm: @[2, 0, 1, 3], sign: 1) (perm: @[0, 2, 1, 3], sign: -1) (perm: @[1, 2, 0, 3], sign: 1) (perm: @[2, 1, 0, 3], sign: -1) (perm: @[3, 1, 0, 2], sign: 1) (perm: @[1, 3, 0, 2], sign: -1) (perm: @[0, 3, 1, 2], sign: 1) (perm: @[3, 0, 1, 2], sign: -1) (perm: @[1, 0, 3, 2], sign: 1) (perm: @[0, 1, 3, 2], sign: -1) (perm: @[0, 2, 3, 1], sign: 1) (perm: @[2, 0, 3, 1], sign: -1) (perm: @[3, 0, 2, 1], sign: 1) (perm: @[0, 3, 2, 1], sign: -1) (perm: @[2, 3, 0, 1], sign: 1) (perm: @[3, 2, 0, 1], sign: -1) (perm: @[3, 2, 1, 0], sign: 1) (perm: @[2, 3, 1, 0], sign: -1) (perm: @[1, 3, 2, 0], sign: 1) (perm: @[3, 1, 2, 0], sign: -1) (perm: @[2, 1, 3, 0], sign: 1) (perm: @[1, 2, 3, 0], sign: -1)
ooRexx
Recursive
/* REXX Compute permutations of things elements */
/* implementing Heap's algorithm nicely shown in */
/* https://en.wikipedia.org/wiki/Heap%27s_algorithm */
/* Recursive Algorithm */
Parse Arg things
e.=''
Select
When things='?' Then
Call help
When things='' Then
things=4
When words(things)>1 Then Do
elements=things
things=words(things)
Do i=0 By 1 While elements<>''
Parse Var elements e.i elements
End
End
Otherwise
If datatype(things)<>'NUM' Then Call help 'bunch ('bunch') must be numeric'
End
n=0
Do i=0 To things-1
a.i=i
End
Call generate things
Say time('R') 'seconds'
Exit
generate: Procedure Expose a. n e. things
Parse Arg k
If k=1 Then
Call show
Else Do
Call generate k-1
Do i=0 To k-2
ka=k-1
If k//2=0 Then
Parse Value a.i a.ka With a.ka a.i
Else
Parse Value a.0 a.ka With a.ka a.0
Call generate k-1
End
End
Return
show: Procedure Expose a. n e. things
n=n+1
ol=''
Do i=0 To things-1
z=a.i
If e.0<>'' Then
ol=ol e.z
Else
ol=ol z
End
Say strip(ol)
Return
Exit
help:
Parse Arg msg
If msg<>'' Then Do
Say 'ERROR:' msg
Say ''
End
Say 'rexx permx -> Permutations of 1 2 3 4 '
Say 'rexx permx 2 -> Permutations of 1 2 '
Say 'rexx permx a b c d -> Permutations of a b c d in 2 positions'
Exit
- Output:
H:\>rexx permx ? rexx permx -> Permutations of 1 2 3 4 rexx permx 2 -> Permutations of 1 2 rexx permx a b c d -> Permutations of a b c d in 2 positions H:\>rexx permx 2 0 1 1 0 0 seconds H:\>rexx permx a b c a b c b a c c a b a c b b c a c b a 0 seconds
Iterative
/* REXX Compute permutations of things elements */
/* implementing Heap's algorithm nicely shown in */
/* https://en.wikipedia.org/wiki/Heap%27s_algorithm */
/* Iterative Algorithm */
Parse Arg things
e.=''
Select
When things='?' Then
Call help
When things='' Then
things=4
When words(things)>1 Then Do
elements=things
things=words(things)
Do i=0 By 1 While elements<>''
Parse Var elements e.i elements
End
End
Otherwise
If datatype(things)<>'NUM' Then Call help 'bunch ('bunch') must be numeric'
End
Do i=0 To things-1
a.i=i
End
Call time 'R'
Call generate things
Say time('E') 'seconds'
Exit
generate:
Parse Arg n
Call show
c.=0
i=0
Do While i<n
If c.i<i Then Do
if i//2=0 Then
Parse Value a.0 a.i With a.i a.0
Else Do
z=c.i
Parse Value a.z a.i With a.i a.z
End
Call show
c.i=c.i+1
i=0
End
Else Do
c.i=0
i=i+1
End
End
Return
show:
ol=''
Do j=0 To n-1
z=a.j
If e.0<>'' Then
ol=ol e.z
Else
ol=ol z
End
Say strip(ol)
Return
Exit
help:
Parse Arg msg
If msg<>'' Then Do
Say 'ERROR:' msg
Say ''
End
Say 'rexx permxi -> Permutations of 1 2 3 4 '
Say 'rexx permxi 2 -> Permutations of 1 2 '
Say 'rexx permxi a b c d -> Permutations of a b c d in 2 positions'
Exit
Perl
S-J-T Based
use strict;
use warnings;
# This code uses "Even's Speedup," as described on
# the Wikipedia page about the Steinhaus–Johnson–
# Trotter algorithm.
# Any resemblance between this code and the Python
# code elsewhere on the page is purely a coincidence,
# caused by them both implementing the same algorithm.
# The code was written to be read relatively easily
# while demonstrating some common perl idioms.
sub perms :prototype(&@) {
my $callback = shift;
my @perm = map [$_, -1], @_;
$perm[0][1] = 0;
my $sign = 1;
while( ) {
$callback->($sign, map $_->[0], @perm);
$sign *= -1;
my ($chosen, $index) = (-1, -1);
for my $i ( 0 .. $#perm ) {
($chosen, $index) = ($perm[$i][0], $i)
if $perm[$i][1] and $perm[$i][0] > $chosen;
}
return if $index == -1;
my $direction = $perm[$index][1];
my $next = $index + $direction;
@perm[ $index, $next ] = @perm[ $next, $index ];
if( $next <= 0 or $next >= $#perm ) {
$perm[$next][1] = 0;
} elsif( $perm[$next + $direction][0] > $chosen ) {
$perm[$next][1] = 0;
}
for my $i ( 0 .. $next - 1 ) {
$perm[$i][1] = +1 if $perm[$i][0] > $chosen;
}
for my $i ( $next + 1 .. $#perm ) {
$perm[$i][1] = -1 if $perm[$i][0] > $chosen;
}
}
}
my $n = shift(@ARGV) || 4;
perms {
my ($sign, @perm) = @_;
print "[", join(", ", @perm), "]";
print $sign < 0 ? " => -1\n" : " => +1\n";
} 1 .. $n;
- Output:
[1, 2, 3, 4] => +1 [1, 2, 4, 3] => -1 [1, 4, 2, 3] => +1 [4, 1, 2, 3] => -1 [4, 1, 3, 2] => +1 [1, 4, 3, 2] => -1 [1, 3, 4, 2] => +1 [1, 3, 2, 4] => -1 [3, 1, 2, 4] => +1 [3, 1, 4, 2] => -1 [3, 4, 1, 2] => +1 [4, 3, 1, 2] => -1 [4, 3, 2, 1] => +1 [3, 4, 2, 1] => -1 [3, 2, 4, 1] => +1 [3, 2, 1, 4] => -1 [2, 3, 1, 4] => +1 [2, 3, 4, 1] => -1 [2, 4, 3, 1] => +1 [4, 2, 3, 1] => -1 [4, 2, 1, 3] => +1 [2, 4, 1, 3] => -1 [2, 1, 4, 3] => +1 [2, 1, 3, 4] => -1
Alternative Iterative version
This is based on the Raku recursive version, but without recursion.
#!perl
use strict;
use warnings;
sub perms {
my ($xx) = (shift);
my @perms = ([+1]);
for my $x ( 1 .. $xx ) {
my $sign = -1;
@perms = map {
my ($s, @p) = @$_;
map [$sign *= -1, @p[0..$_-1], $x, @p[$_..$#p]],
$s < 0 ? 0 .. @p : reverse 0 .. @p;
} @perms;
}
@perms;
}
my $n = shift() || 4;
for( perms($n) ) {
my $s = shift @$_;
$s = '+1' if $s > 0;
print "[", join(", ", @$_), "] => $s\n";
}
- Output:
The output is the same as the first perl solution.
Phix
Ad-hoc recursive solution, not (knowingly) based on any given algorithm, but instead on achieving the desired pattern.
Only once finished did I properly grasp that odd/even permutation idea, and that it is very nearly the same algorithm.
Only difference is my version directly calculates where to insert p, without using the parity (which I added in last).
function spermutations(integer p, integer i) -- -- generate the i'th permutation of [1..p]: -- first obtain the appropriate permutation of [1..p-1], -- then insert p/move it down k(=0..p-1) places from the end. -- sequence res integer k = mod(i-1,2*p) if k>=p then k=2*p-1-k end if if p>1 then res = spermutations(p-1,floor((i-1)/p)+1) res = res[1..length(res)-k]&p&res[length(res)-k+1..$] else res = {1} end if return res end function for p=1 to 4 do printf(1,"==%d==\n",p) for i=1 to factorial(p) do integer parity = iff(and_bits(i,1)?1:-1) ?{i,spermutations(p,i),parity} end for end for
- Output:
==1==
{1,{1},1}
==2==
{1,{1,2},1}
{2,{2,1},-1}
==3==
{1,{1,2,3},1}
{2,{1,3,2},-1}
{3,{3,1,2},1}
{4,{3,2,1},-1}
{5,{2,3,1},1}
{6,{2,1,3},-1}
==4==
{1,{1,2,3,4},1}
{2,{1,2,4,3},-1}
{3,{1,4,2,3},1}
{4,{4,1,2,3},-1}
{5,{4,1,3,2},1}
{6,{1,4,3,2},-1}
{7,{1,3,4,2},1}
{8,{1,3,2,4},-1}
{9,{3,1,2,4},1}
{10,{3,1,4,2},-1}
{11,{3,4,1,2},1}
{12,{4,3,1,2},-1}
{13,{4,3,2,1},1}
{14,{3,4,2,1},-1}
{15,{3,2,4,1},1}
{16,{3,2,1,4},-1}
{17,{2,3,1,4},1}
{18,{2,3,4,1},-1}
{19,{2,4,3,1},1}
{20,{4,2,3,1},-1}
{21,{4,2,1,3},1}
{22,{2,4,1,3},-1}
{23,{2,1,4,3},1}
{24,{2,1,3,4},-1}
PicoLisp
(let
(N 4
L
(mapcar
'((I) (list I 0))
(range 1 N) ) )
(for I L
(printsp (car I)) )
(prinl)
(while
# find the lagest mobile integer
(setq
X
(maxi
'((I) (car (get L (car I))))
(extract
'((I J)
(let? Y
(get
L
((if (=0 (cadr I)) dec inc) J) )
(when (> (car I) (car Y))
(list J (cadr I)) ) ) )
L
(range 1 N) ) )
Y (get L (car X)) )
# swap integer and adjacent int it is looking at
(xchg
(nth L (car X))
(nth
L
((if (=0 (cadr X)) dec inc) (car X)) ) )
# reverse direction of all ints large than our
(for I L
(when (< (car Y) (car I))
(set (cdr I)
(if (=0 (cadr I)) 1 0) ) ) )
# print current positions
(for I L
(printsp (car I)) )
(prinl) ) )
(bye)PowerShell
function output([Object[]]$A, [Int]$k, [ref]$sign)
{
"Perm: [$([String]::Join(', ', $A))] Sign: $($sign.Value)"
}
function permutation([Object[]]$array)
{
function generate([Object[]]$A, [Int]$k, [ref]$sign)
{
if($k -eq 1)
{
output $A $k $sign
$sign.Value = -$sign.Value
}
else
{
$k -= 1
generate $A $k $sign
for([Int]$i = 0; $i -lt $k; $i += 1)
{
if($i % 2 -eq 0)
{
$A[$i], $A[$k] = $A[$k], $A[$i]
}
else
{
$A[0], $A[$k] = $A[$k], $A[0]
}
generate $A $k $sign
}
}
}
generate $array $array.Count ([ref]1)
}
permutation @(0, 1, 2)
""
permutation @(0, 1, 2, 3)
Output:
Perm: [1, 0, 2] Sign: -1 Perm: [2, 0, 1] Sign: 1 Perm: [0, 2, 1] Sign: -1 Perm: [1, 2, 0] Sign: 1 Perm: [2, 1, 0] Sign: -1 Perm: [0, 1, 2, 3] Sign: 1 Perm: [1, 0, 2, 3] Sign: -1 Perm: [2, 0, 1, 3] Sign: 1 Perm: [0, 2, 1, 3] Sign: -1 Perm: [1, 2, 0, 3] Sign: 1 Perm: [2, 1, 0, 3] Sign: -1 Perm: [3, 1, 0, 2] Sign: 1 Perm: [1, 3, 0, 2] Sign: -1 Perm: [0, 3, 1, 2] Sign: 1 Perm: [3, 0, 1, 2] Sign: -1 Perm: [1, 0, 3, 2] Sign: 1 Perm: [0, 1, 3, 2] Sign: -1 Perm: [2, 1, 3, 0] Sign: 1 Perm: [1, 2, 3, 0] Sign: -1 Perm: [3, 2, 1, 0] Sign: 1 Perm: [2, 3, 1, 0] Sign: -1 Perm: [1, 3, 2, 0] Sign: 1 Perm: [3, 1, 2, 0] Sign: -1 Perm: [3, 1, 0, 2] Sign: 1 Perm: [1, 3, 0, 2] Sign: -1 Perm: [0, 3, 1, 2] Sign: 1 Perm: [3, 0, 1, 2] Sign: -1 Perm: [1, 0, 3, 2] Sign: 1 Perm: [0, 1, 3, 2] Sign: -1
Python
Python: iterative
When saved in a file called spermutations.py it is used in the Python example to the Matrix arithmetic task and so any changes here should also be reflected and checked in that task example too.
from operator import itemgetter
DEBUG = False # like the built-in __debug__
def spermutations(n):
"""permutations by swapping. Yields: perm, sign"""
sign = 1
p = [[i, 0 if i == 0 else -1] # [num, direction]
for i in range(n)]
if DEBUG: print ' #', p
yield tuple(pp[0] for pp in p), sign
while any(pp[1] for pp in p): # moving
i1, (n1, d1) = max(((i, pp) for i, pp in enumerate(p) if pp[1]),
key=itemgetter(1))
sign *= -1
if d1 == -1:
# Swap down
i2 = i1 - 1
p[i1], p[i2] = p[i2], p[i1]
# If this causes the chosen element to reach the First or last
# position within the permutation, or if the next element in the
# same direction is larger than the chosen element:
if i2 == 0 or p[i2 - 1][0] > n1:
# The direction of the chosen element is set to zero
p[i2][1] = 0
elif d1 == 1:
# Swap up
i2 = i1 + 1
p[i1], p[i2] = p[i2], p[i1]
# If this causes the chosen element to reach the first or Last
# position within the permutation, or if the next element in the
# same direction is larger than the chosen element:
if i2 == n - 1 or p[i2 + 1][0] > n1:
# The direction of the chosen element is set to zero
p[i2][1] = 0
if DEBUG: print ' #', p
yield tuple(pp[0] for pp in p), sign
for i3, pp in enumerate(p):
n3, d3 = pp
if n3 > n1:
pp[1] = 1 if i3 < i2 else -1
if DEBUG: print ' # Set Moving'
if __name__ == '__main__':
from itertools import permutations
for n in (3, 4):
print '\nPermutations and sign of %i items' % n
sp = set()
for i in spermutations(n):
sp.add(i[0])
print('Perm: %r Sign: %2i' % i)
#if DEBUG: raw_input('?')
# Test
p = set(permutations(range(n)))
assert sp == p, 'Two methods of generating permutations do not agree'
- Output:
Permutations and sign of 3 items Perm: (0, 1, 2) Sign: 1 Perm: (0, 2, 1) Sign: -1 Perm: (2, 0, 1) Sign: 1 Perm: (2, 1, 0) Sign: -1 Perm: (1, 2, 0) Sign: 1 Perm: (1, 0, 2) Sign: -1 Permutations and sign of 4 items Perm: (0, 1, 2, 3) Sign: 1 Perm: (0, 1, 3, 2) Sign: -1 Perm: (0, 3, 1, 2) Sign: 1 Perm: (3, 0, 1, 2) Sign: -1 Perm: (3, 0, 2, 1) Sign: 1 Perm: (0, 3, 2, 1) Sign: -1 Perm: (0, 2, 3, 1) Sign: 1 Perm: (0, 2, 1, 3) Sign: -1 Perm: (2, 0, 1, 3) Sign: 1 Perm: (2, 0, 3, 1) Sign: -1 Perm: (2, 3, 0, 1) Sign: 1 Perm: (3, 2, 0, 1) Sign: -1 Perm: (3, 2, 1, 0) Sign: 1 Perm: (2, 3, 1, 0) Sign: -1 Perm: (2, 1, 3, 0) Sign: 1 Perm: (2, 1, 0, 3) Sign: -1 Perm: (1, 2, 0, 3) Sign: 1 Perm: (1, 2, 3, 0) Sign: -1 Perm: (1, 3, 2, 0) Sign: 1 Perm: (3, 1, 2, 0) Sign: -1 Perm: (3, 1, 0, 2) Sign: 1 Perm: (1, 3, 0, 2) Sign: -1 Perm: (1, 0, 3, 2) Sign: 1 Perm: (1, 0, 2, 3) Sign: -1
Python: recursive
After spotting the pattern of highest number being inserted into each perm of lower numbers from right to left, then left to right, I developed this recursive function:
def s_permutations(seq):
def s_perm(seq):
if not seq:
return [[]]
else:
new_items = []
for i, item in enumerate(s_perm(seq[:-1])):
if i % 2:
# step up
new_items += [item[:i] + seq[-1:] + item[i:]
for i in range(len(item) + 1)]
else:
# step down
new_items += [item[:i] + seq[-1:] + item[i:]
for i in range(len(item), -1, -1)]
return new_items
return [(tuple(item), -1 if i % 2 else 1)
for i, item in enumerate(s_perm(seq))]
- Sample output:
The output is the same as before except it is a list of all results rather than yielding each result from a generator function.
Python: Iterative version of the recursive
Replacing the recursion in the example above produces this iterative version function:
def s_permutations(seq):
items = [[]]
for j in seq:
new_items = []
for i, item in enumerate(items):
if i % 2:
# step up
new_items += [item[:i] + [j] + item[i:]
for i in range(len(item) + 1)]
else:
# step down
new_items += [item[:i] + [j] + item[i:]
for i in range(len(item), -1, -1)]
items = new_items
return [(tuple(item), -1 if i % 2 else 1)
for i, item in enumerate(items)]
- Sample output:
The output is the same as before and is a list of all results rather than yielding each result from a generator function.
Quackery
[ stack ] is parity ( --> s )
[ 1 & ] is odd ( n --> b )
[ [] swap witheach
[ nested
i odd 2 * 1 -
join nested join ] ] is +signs ( [ --> [ )
[ dup
[ dup 0 = iff
[ drop ' [ [ ] ] ]
done
dup temp put
1 - recurse
[] swap
witheach
[ i odd parity put
temp share times
[ temp share 1 -
over
parity share
iff i else i^
stuff
nested rot join
swap ]
drop
parity release ]
temp release ]
swap odd if reverse
+signs ] is perms ( n --> [ )
3 perms witheach [ echo cr ]
cr
4 perms witheach [ echo cr ]- Output:
[ [ 0 1 2 ] 1 ] [ [ 0 2 1 ] -1 ] [ [ 2 0 1 ] 1 ] [ [ 2 1 0 ] -1 ] [ [ 1 2 0 ] 1 ] [ [ 1 0 2 ] -1 ] [ [ 0 1 2 3 ] 1 ] [ [ 0 1 3 2 ] -1 ] [ [ 0 3 1 2 ] 1 ] [ [ 3 0 1 2 ] -1 ] [ [ 3 0 2 1 ] 1 ] [ [ 0 3 2 1 ] -1 ] [ [ 0 2 3 1 ] 1 ] [ [ 0 2 1 3 ] -1 ] [ [ 2 0 1 3 ] 1 ] [ [ 2 0 3 1 ] -1 ] [ [ 2 3 0 1 ] 1 ] [ [ 3 2 0 1 ] -1 ] [ [ 3 2 1 0 ] 1 ] [ [ 2 3 1 0 ] -1 ] [ [ 2 1 3 0 ] 1 ] [ [ 2 1 0 3 ] -1 ] [ [ 1 2 0 3 ] 1 ] [ [ 1 2 3 0 ] -1 ] [ [ 1 3 2 0 ] 1 ] [ [ 3 1 2 0 ] -1 ] [ [ 3 1 0 2 ] 1 ] [ [ 1 3 0 2 ] -1 ] [ [ 1 0 3 2 ] 1 ] [ [ 1 0 2 3 ] -1 ]
Racket
#lang racket
(define (add-at l i x)
(if (zero? i) (cons x l) (cons (car l) (add-at (cdr l) (sub1 i) x))))
(define (permutations l)
(define (loop l)
(cond [(null? l) '(())]
[else (for*/list ([(p i) (in-indexed (loop (cdr l)))]
[i ((if (odd? i) identity reverse)
(range (add1 (length p))))])
(add-at p i (car l)))]))
(for/list ([p (loop (reverse l))] [i (in-cycle '(1 -1))]) (cons i p)))
(define (show-permutations l)
(printf "Permutations of ~s:\n" l)
(for ([p (permutations l)])
(printf " ~a (~a)\n" (apply ~a (add-between (cdr p) ", ")) (car p))))
(for ([n (in-range 3 5)]) (show-permutations (range n)))
- Output:
Permutations of (0 1 2): 0, 1, 2 (1) 0, 2, 1 (-1) 2, 0, 1 (1) 2, 1, 0 (-1) 1, 2, 0 (1) 1, 0, 2 (-1) Permutations of (0 1 2 3): 0, 1, 2, 3 (1) 0, 1, 3, 2 (-1) 0, 3, 1, 2 (1) 3, 0, 1, 2 (-1) 3, 0, 2, 1 (1) 0, 3, 2, 1 (-1) 0, 2, 3, 1 (1) 0, 2, 1, 3 (-1) 2, 0, 1, 3 (1) 2, 0, 3, 1 (-1) 2, 3, 0, 1 (1) 3, 2, 0, 1 (-1) 3, 2, 1, 0 (1) 2, 3, 1, 0 (-1) 2, 1, 3, 0 (1) 2, 1, 0, 3 (-1) 1, 2, 0, 3 (1) 1, 2, 3, 0 (-1) 1, 3, 2, 0 (1) 3, 1, 2, 0 (-1) 3, 1, 0, 2 (1) 1, 3, 0, 2 (-1) 1, 0, 3, 2 (1) 1, 0, 2, 3 (-1)
Raku
(formerly Perl 6)
Recursive
sub insert($x, @xs) { ([flat @xs[0 ..^ $_], $x, @xs[$_ .. *]] for 0 .. +@xs) }
sub order($sg, @xs) { $sg > 0 ?? @xs !! @xs.reverse }
multi perms([]) {
[] => +1
}
multi perms([$x, *@xs]) {
perms(@xs).map({ |order($_.value, insert($x, $_.key)) }) Z=> |(+1,-1) xx *
}
.say for perms([0..2]);
- Output:
[0 1 2] => 1 [1 0 2] => -1 [1 2 0] => 1 [2 1 0] => -1 [2 0 1] => 1 [0 2 1] => -1
REXX
Version 1
This program does not work asdescribed in the comment section and I can't get it working for 5 things. -:( --Walter Pachl 13:40, 25 January 2022 (UTC)
/*REXX program generates all permutations of N different objects by swapping. */
parse arg things bunch . /*obtain optional arguments from the CL*/
if things=='' | things=="," then things=4 /*Not specified? Then use the default.*/
if bunch =='' | bunch =="," then bunch =things /* " " " " " " */
call permSets things, bunch /*invoke permutations by swapping sub. */
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!: procedure; !=1; do j=2 to arg(1); !=!*j; end; return !
/*──────────────────────────────────────────────────────────────────────────────────────*/
permSets: procedure; parse arg x,y /*take X things Y at a time. */
!.=0; pad=left('', x*y) /*X can't be > length of below str (62)*/
z=left('123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ', x); q=z
#=1 /*the number of permutations (so far).*/
!.z=1; s=1; times=!(x) % !(x-y) /*calculate (#) TIMES using factorial.*/
w=max(length(z), length('permute') ) /*maximum width of Z and also PERMUTE.*/
say center('permutations for ' x ' things taken ' y " at a time",60,'═')
say
say pad 'permutation' center("permute", w, '─') "sign"
say pad '───────────' center("───────", w, '─') "────"
say pad center(#, 11) center(z , w) right(s, 4-1)
do $=1 until #==times /*perform permutation until # of times.*/
do k=1 for x-1 /*step thru things for things-1 times.*/
do m=k+1 to x; ?= /*this method doesn't use adjacency. */
do n=1 for x /*build the new permutation by swapping*/
if n\==k & n\==m then ? = ? || substr(z, n, 1)
else if n==k then ? = ? || substr(z, m, 1)
else ? = ? || substr(z, k, 1)
end /*n*/
z=? /*save this permutation for next swap. */
if !.? then iterate m /*if defined before, then try next one.*/
_=0 /* [↓] count number of swapped symbols*/
do d=1 for x while $\==1; _= _ + (substr(?,d,1)\==substr(prev,d,1))
end /*d*/
if _>2 then do; _=z
a=$//x+1; q=q + _ /* [← ↓] this swapping tries adjacency*/
b=q//x+1; if b==a then b=a + 1; if b>x then b=a - 1
z=overlay( substr(z,b,1), overlay( substr(z,a,1), _, b), a)
iterate $ /*now, try this particular permutation.*/
end
#=#+1; s= -s; say pad center(#, 11) center(?, w) right(s, 4-1)
!.?=1; prev=?; iterate $ /*now, try another swapped permutation.*/
end /*m*/
end /*k*/
end /*$*/
return /*we're all finished with permutating. */
- output when using the default input:
══════permutations for 4 things taken 4 at a time═══════
permutation permute sign
─────────── ─────── ────
1 1234 1
2 2134 -1
3 3124 1
4 1324 -1
5 1342 1
6 3142 -1
7 4132 1
8 1432 -1
9 2431 1
10 4231 -1
11 4321 1
12 3421 -1
13 3241 1
14 2341 -1
15 2314 1
16 3214 -1
17 3412 1
18 4312 -1
19 4213 1
20 2413 -1
21 2143 1
22 1243 -1
23 1423 1
24 4123 -1
Version 2
See program shown for ooRexx
Ruby
def perms(n)
p = Array.new(n+1){|i| -i}
s = 1
loop do
yield p[1..-1].map(&:abs), s
k = 0
for i in 2..n
k = i if p[i] < 0 and p[i].abs > p[i-1].abs and p[i].abs > p[k].abs
end
for i in 1...n
k = i if p[i] > 0 and p[i].abs > p[i+1].abs and p[i].abs > p[k].abs
end
break if k.zero?
for i in 1..n
p[i] *= -1 if p[i].abs > p[k].abs
end
i = k + (p[k] <=> 0)
p[k], p[i] = p[i], p[k]
s = -s
end
end
for i in 3..4
perms(i){|perm, sign| puts "Perm: #{perm} Sign: #{sign}"}
puts
end
- Output:
Perm: [1, 2, 3] Sign: 1 Perm: [1, 3, 2] Sign: -1 Perm: [3, 1, 2] Sign: 1 Perm: [3, 2, 1] Sign: -1 Perm: [2, 3, 1] Sign: 1 Perm: [2, 1, 3] Sign: -1 Perm: [1, 2, 3, 4] Sign: 1 Perm: [1, 2, 4, 3] Sign: -1 Perm: [1, 4, 2, 3] Sign: 1 Perm: [4, 1, 2, 3] Sign: -1 Perm: [4, 1, 3, 2] Sign: 1 Perm: [1, 4, 3, 2] Sign: -1 Perm: [1, 3, 4, 2] Sign: 1 Perm: [1, 3, 2, 4] Sign: -1 Perm: [3, 1, 2, 4] Sign: 1 Perm: [3, 1, 4, 2] Sign: -1 Perm: [3, 4, 1, 2] Sign: 1 Perm: [4, 3, 1, 2] Sign: -1 Perm: [4, 3, 2, 1] Sign: 1 Perm: [3, 4, 2, 1] Sign: -1 Perm: [3, 2, 4, 1] Sign: 1 Perm: [3, 2, 1, 4] Sign: -1 Perm: [2, 3, 1, 4] Sign: 1 Perm: [2, 3, 4, 1] Sign: -1 Perm: [2, 4, 3, 1] Sign: 1 Perm: [4, 2, 3, 1] Sign: -1 Perm: [4, 2, 1, 3] Sign: 1 Perm: [2, 4, 1, 3] Sign: -1 Perm: [2, 1, 4, 3] Sign: 1 Perm: [2, 1, 3, 4] Sign: -1
Rust
// Implementation of Heap's algorithm.
// See https://en.wikipedia.org/wiki/Heap%27s_algorithm#Details_of_the_algorithm
fn generate<T, F>(a: &mut [T], output: F)
where
F: Fn(&[T], isize),
{
let n = a.len();
let mut c = vec![0; n];
let mut i = 1;
let mut sign = 1;
output(a, sign);
while i < n {
if c[i] < i {
if (i & 1) == 0 {
a.swap(0, i);
} else {
a.swap(c[i], i);
}
sign = -sign;
output(a, sign);
c[i] += 1;
i = 1;
} else {
c[i] = 0;
i += 1;
}
}
}
fn print_permutation<T: std::fmt::Debug>(a: &[T], sign: isize) {
println!("{:?} {}", a, sign);
}
fn main() {
println!("Permutations and signs for three items:");
let mut a = vec![0, 1, 2];
generate(&mut a, print_permutation);
println!("\nPermutations and signs for four items:");
let mut b = vec![0, 1, 2, 3];
generate(&mut b, print_permutation);
}
- Output:
[0, 1, 2] 1 [1, 0, 2] -1 [2, 0, 1] 1 [0, 2, 1] -1 [1, 2, 0] 1 [2, 1, 0] -1 Permutations and signs for four items: [0, 1, 2, 3] 1 [1, 0, 2, 3] -1 [2, 0, 1, 3] 1 [0, 2, 1, 3] -1 [1, 2, 0, 3] 1 [2, 1, 0, 3] -1 [3, 1, 0, 2] 1 [1, 3, 0, 2] -1 [0, 3, 1, 2] 1 [3, 0, 1, 2] -1 [1, 0, 3, 2] 1 [0, 1, 3, 2] -1 [0, 2, 3, 1] 1 [2, 0, 3, 1] -1 [3, 0, 2, 1] 1 [0, 3, 2, 1] -1 [2, 3, 0, 1] 1 [3, 2, 0, 1] -1 [3, 2, 1, 0] 1 [2, 3, 1, 0] -1 [1, 3, 2, 0] 1 [3, 1, 2, 0] -1 [2, 1, 3, 0] 1 [1, 2, 3, 0] -1
Scala
object JohnsonTrotter extends App {
private def perm(n: Int): Unit = {
val p = new Array[Int](n) // permutation
val pi = new Array[Int](n) // inverse permutation
val dir = new Array[Int](n) // direction = +1 or -1
def perm(n: Int, p: Array[Int], pi: Array[Int], dir: Array[Int]): Unit = {
if (n >= p.length) for (aP <- p) print(aP)
else {
perm(n + 1, p, pi, dir)
for (i <- 0 until n) { // swap
printf(" (%d %d)\n", pi(n), pi(n) + dir(n))
val z = p(pi(n) + dir(n))
p(pi(n)) = z
p(pi(n) + dir(n)) = n
pi(z) = pi(n)
pi(n) = pi(n) + dir(n)
perm(n + 1, p, pi, dir)
}
dir(n) = -dir(n)
}
}
for (i <- 0 until n) {
dir(i) = -1
p(i) = i
pi(i) = i
}
perm(0, p, pi, dir)
print(" (0 1)\n")
}
perm(4)
}
- Output:
See it in running in your browser by Scastie (JVM).
Sidef
func perms(n) {
var perms = [[+1]]
for x in (1..n) {
var sign = -1
perms = gather {
for s,*p in perms {
var r = (0 .. p.len)
take((s < 0 ? r : r.flip).map {|i|
[sign *= -1, p[^i], x, p[i..p.end]]
}...)
}
}
}
perms
}
var n = 4
for p in perms(n) {
var s = p.shift
s > 0 && (s = '+1')
say "#{p} => #{s}"
}
- Output:
[1, 2, 3, 4] => +1 [1, 2, 4, 3] => -1 [1, 4, 2, 3] => +1 [4, 1, 2, 3] => -1 [4, 1, 3, 2] => +1 [1, 4, 3, 2] => -1 [1, 3, 4, 2] => +1 [1, 3, 2, 4] => -1 [3, 1, 2, 4] => +1 [3, 1, 4, 2] => -1 [3, 4, 1, 2] => +1 [4, 3, 1, 2] => -1 [4, 3, 2, 1] => +1 [3, 4, 2, 1] => -1 [3, 2, 4, 1] => +1 [3, 2, 1, 4] => -1 [2, 3, 1, 4] => +1 [2, 3, 4, 1] => -1 [2, 4, 3, 1] => +1 [4, 2, 3, 1] => -1 [4, 2, 1, 3] => +1 [2, 4, 1, 3] => -1 [2, 1, 4, 3] => +1 [2, 1, 3, 4] => -1
Swift
// Implementation of Heap's algorithm.
// See https://en.wikipedia.org/wiki/Heap%27s_algorithm#Details_of_the_algorithm
func generate<T>(array: inout [T], output: (_: [T], _: Int) -> Void) {
let n = array.count
var c = Array(repeating: 0, count: n)
var i = 1
var sign = 1
output(array, sign)
while i < n {
if c[i] < i {
if (i & 1) == 0 {
array.swapAt(0, i)
} else {
array.swapAt(c[i], i)
}
sign = -sign
output(array, sign)
c[i] += 1
i = 1
} else {
c[i] = 0
i += 1
}
}
}
func printPermutation<T>(array: [T], sign: Int) {
print("\(array) \(sign)")
}
print("Permutations and signs for three items:")
var a = [0, 1, 2]
generate(array: &a, output: printPermutation)
print("\nPermutations and signs for four items:")
var b = [0, 1, 2, 3]
generate(array: &b, output: printPermutation)
- Output:
Permutations and signs for three items: [0, 1, 2] 1 [1, 0, 2] -1 [2, 0, 1] 1 [0, 2, 1] -1 [1, 2, 0] 1 [2, 1, 0] -1 Permutations and signs for four items: [0, 1, 2, 3] 1 [1, 0, 2, 3] -1 [2, 0, 1, 3] 1 [0, 2, 1, 3] -1 [1, 2, 0, 3] 1 [2, 1, 0, 3] -1 [3, 1, 0, 2] 1 [1, 3, 0, 2] -1 [0, 3, 1, 2] 1 [3, 0, 1, 2] -1 [1, 0, 3, 2] 1 [0, 1, 3, 2] -1 [0, 2, 3, 1] 1 [2, 0, 3, 1] -1 [3, 0, 2, 1] 1 [0, 3, 2, 1] -1 [2, 3, 0, 1] 1 [3, 2, 0, 1] -1 [3, 2, 1, 0] 1 [2, 3, 1, 0] -1 [1, 3, 2, 0] 1 [3, 1, 2, 0] -1 [2, 1, 3, 0] 1 [1, 2, 3, 0] -1
Tcl
# A simple swap operation
proc swap {listvar i1 i2} {
upvar 1 $listvar l
set tmp [lindex $l $i1]
lset l $i1 [lindex $l $i2]
lset l $i2 $tmp
}
proc permswap {n v1 v2 body} {
upvar 1 $v1 perm $v2 sign
# Initialize
set sign -1
for {set i 0} {$i < $n} {incr i} {
lappend items $i
lappend dirs -1
}
while 1 {
# Report via callback
set perm $items
set sign [expr {-$sign}]
uplevel 1 $body
# Find the largest mobile integer (lmi) and its index (idx)
set i [set idx -1]
foreach item $items dir $dirs {
set j [expr {[incr i] + $dir}]
if {$j < 0 || $j >= [llength $items]} continue
if {$item > [lindex $items $j] && ($idx == -1 || $item > $lmi)} {
set lmi $item
set idx $i
}
}
# If none, we're done
if {$idx == -1} break
# Swap the largest mobile integer with "what it is looking at"
set nextIdx [expr {$idx + [lindex $dirs $idx]}]
swap items $idx $nextIdx
swap dirs $idx $nextIdx
# Reverse directions on larger integers
set i -1
foreach item $items dir $dirs {
lset dirs [incr i] [expr {$item > $lmi ? -$dir : $dir}]
}
}
}
Demonstrating:
permswap 4 p s {
puts "$s\t$p"
}
- Output:
1 0 1 2 3 -1 0 1 3 2 1 0 3 1 2 -1 3 0 1 2 1 3 0 2 1 -1 0 3 2 1 1 0 2 3 1 -1 0 2 1 3 1 2 0 1 3 -1 2 0 3 1 1 2 3 0 1 -1 3 2 0 1 1 3 2 1 0 -1 2 3 1 0 1 2 1 3 0 -1 2 1 0 3 1 1 2 0 3 -1 1 2 3 0 1 1 3 2 0 -1 3 1 2 0 1 3 1 0 2 -1 1 3 0 2 1 1 0 3 2 -1 1 0 2 3
Wren
var johnsonTrotter = Fn.new { |n|
var p = List.filled(n, 0) // permutation
var q = List.filled(n, 0) // inverse permutation
for (i in 0...n) p[i] = q[i] = i
var d = List.filled(n, -1) // direction = 1 or -1
var sign = 1
var perms = []
var signs = []
var permute // recursive closure
permute = Fn.new { |k|
if (k >= n) {
perms.add(p.toList)
signs.add(sign)
sign = sign * -1
return
}
permute.call(k + 1)
for (i in 0...k) {
var z = p[q[k] + d[k]]
p[q[k]] = z
p[q[k] + d[k]] = k
q[z] = q[k]
q[k] = q[k] + d[k]
permute.call(k + 1)
}
d[k] = d[k] * -1
}
permute.call(0)
return [perms, signs]
}
var printPermsAndSigns = Fn.new { |perms, signs|
var i = 0
for (perm in perms) {
System.print("%(perm) -> sign = %(signs[i])")
i = i + 1
}
}
var res = johnsonTrotter.call(3)
var perms = res[0]
var signs = res[1]
printPermsAndSigns.call(perms, signs)
System.print()
res = johnsonTrotter.call(4)
perms = res[0]
signs = res[1]
printPermsAndSigns.call(perms, signs)
- Output:
[0, 1, 2] -> sign = 1 [0, 2, 1] -> sign = -1 [2, 0, 1] -> sign = 1 [2, 1, 0] -> sign = -1 [1, 2, 0] -> sign = 1 [1, 0, 2] -> sign = -1 [0, 1, 2, 3] -> sign = 1 [0, 1, 3, 2] -> sign = -1 [0, 3, 1, 2] -> sign = 1 [3, 0, 1, 2] -> sign = -1 [3, 0, 2, 1] -> sign = 1 [0, 3, 2, 1] -> sign = -1 [0, 2, 3, 1] -> sign = 1 [0, 2, 1, 3] -> sign = -1 [2, 0, 1, 3] -> sign = 1 [2, 0, 3, 1] -> sign = -1 [2, 3, 0, 1] -> sign = 1 [3, 2, 0, 1] -> sign = -1 [3, 2, 1, 0] -> sign = 1 [2, 3, 1, 0] -> sign = -1 [2, 1, 3, 0] -> sign = 1 [2, 1, 0, 3] -> sign = -1 [1, 2, 0, 3] -> sign = 1 [1, 2, 3, 0] -> sign = -1 [1, 3, 2, 0] -> sign = 1 [3, 1, 2, 0] -> sign = -1 [3, 1, 0, 2] -> sign = 1 [1, 3, 0, 2] -> sign = -1 [1, 0, 3, 2] -> sign = 1 [1, 0, 2, 3] -> sign = -1
XPL0
Translation of BBC BASIC example, which uses the Johnson-Trotter algorithm.
include c:\cxpl\codes;
proc PERMS(N);
int N; \number of elements
int I, K, S, T, P;
[P:= Reserve((N+1)*4);
for I:= 0 to N do P(I):= -I; \initialize facing left (also set P(0)=0)
S:= 1;
repeat Text(0, "Perm: [ ");
for I:= 1 to N do
[IntOut(0, abs(P(I))); ChOut(0, ^ )];
Text(0, "] Sign: "); IntOut(0, S); CrLf(0);
K:= 0; \find largest mobile element
for I:= 2 to N do \for left-facing elements
if P(I) < 0 and
abs(P(I)) > abs(P(I-1)) and \ greater than neighbor
abs(P(I)) > abs(P(K)) then K:= I; \ get largest element
for I:= 1 to N-1 do \for right-facing elements
if P(I) > 0 and
abs(P(I)) > abs(P(I+1)) and \ greater than neighbor
abs(P(I)) > abs(P(K)) then K:= I; \ get largest element
if K # 0 then \mobile element found
[for I:= 1 to N do \reverse elements > K
if abs(P(I)) > abs(P(K)) then P(I):= P(I)*-1;
I:= K + (if P(K)<0 then -1 else 1);
T:= P(K); P(K):= P(I); P(I):= T; \swap K with element looked at
S:= -S; \alternate signs
];
until K = 0; \no mobile element remains
];
[PERMS(3);
CrLf(0);
PERMS(4);
]- Output:
Perm: [ 1 2 3 ] Sign: 1 Perm: [ 1 3 2 ] Sign: -1 Perm: [ 3 1 2 ] Sign: 1 Perm: [ 3 2 1 ] Sign: -1 Perm: [ 2 3 1 ] Sign: 1 Perm: [ 2 1 3 ] Sign: -1 Perm: [ 1 2 3 4 ] Sign: 1 Perm: [ 1 2 4 3 ] Sign: -1 Perm: [ 1 4 2 3 ] Sign: 1 Perm: [ 4 1 2 3 ] Sign: -1 Perm: [ 4 1 3 2 ] Sign: 1 Perm: [ 1 4 3 2 ] Sign: -1 Perm: [ 1 3 4 2 ] Sign: 1 Perm: [ 1 3 2 4 ] Sign: -1 Perm: [ 3 1 2 4 ] Sign: 1 Perm: [ 3 1 4 2 ] Sign: -1 Perm: [ 3 4 1 2 ] Sign: 1 Perm: [ 4 3 1 2 ] Sign: -1 Perm: [ 4 3 2 1 ] Sign: 1 Perm: [ 3 4 2 1 ] Sign: -1 Perm: [ 3 2 4 1 ] Sign: 1 Perm: [ 3 2 1 4 ] Sign: -1 Perm: [ 2 3 1 4 ] Sign: 1 Perm: [ 2 3 4 1 ] Sign: -1 Perm: [ 2 4 3 1 ] Sign: 1 Perm: [ 4 2 3 1 ] Sign: -1 Perm: [ 4 2 1 3 ] Sign: 1 Perm: [ 2 4 1 3 ] Sign: -1 Perm: [ 2 1 4 3 ] Sign: 1 Perm: [ 2 1 3 4 ] Sign: -1
zkl
fcn permute(seq)
{
insertEverywhere := fcn(x,list){ //(x,(a,b))-->((x,a,b),(a,x,b),(a,b,x))
(0).pump(list.len()+1,List,'wrap(n){list[0,n].extend(x,list[n,*]) })};
insertEverywhereB := fcn(x,t){ //--> insertEverywhere().reverse()
[t.len()..-1,-1].pump(t.len()+1,List,'wrap(n){t[0,n].extend(x,t[n,*])})};
seq.reduce('wrap(items,x){
f := Utils.Helpers.cycle(insertEverywhereB,insertEverywhere);
items.pump(List,'wrap(item){f.next()(x,item)},
T.fp(Void.Write,Void.Write));
},T(T));
}A cycle of two "build list" functions is used to insert x forward or reverse. reduce loops over the items and retains the enlarging list of permuations. pump loops over the existing set of permutations and inserts/builds the next set (into a list sink). (Void.Write,Void.Write,list) is a sentinel that says to write the contents of the list to the sink (ie sink.extend(list)). T.fp is a partial application of ROList.create (read only list) and the parameters VW,VW. It will be called (by pump) with a list of lists --> T.create(VM,VM,list) --> list
p := permute(T(1,2,3));
p.println();
p := permute([1..4]);
p.len().println();
p.toString(*).println()- Output:
L(L(1,2,3),L(1,3,2),L(3,1,2),L(3,2,1),L(2,3,1),L(2,1,3)) 24 L( L(1,2,3,4), L(1,2,4,3), L(1,4,2,3), L(4,1,2,3), L(4,1,3,2), L(1,4,3,2), L(1,3,4,2), L(1,3,2,4), L(3,1,2,4), L(3,1,4,2), L(3,4,1,2), L(4,3,1,2), L(4,3,2,1), L(3,4,2,1), L(3,2,4,1), L(3,2,1,4), L(2,3,1,4), L(2,3,4,1), L(2,4,3,1), L(4,2,3,1), L(4,2,1,3), L(2,4,1,3), L(2,1,4,3), L(2,1,3,4) )
An iterative, lazy version, which is handy as the number of permutations is n!. Uses "Even's Speedup" as described in the Wikipedia article:
fcn [private] _permuteW(seq){ // lazy version
N:=seq.len(); NM1:=N-1;
ds:=(0).pump(N,List,T(Void,-1)).copy(); ds[0]=0; // direction to move e: -1,0,1
es:=(0).pump(N,List).copy(); // enumerate seq
while(1) {
vm.yield(es.pump(List,seq.__sGet));
// find biggest e with d!=0
reg i=Void, c=-1;
foreach n in (N){ if(ds[n] and es[n]>c) { c=es[n]; i=n; } }
if(Void==i) return();
d:=ds[i]; j:=i+d;
es.swap(i,j); ds.swap(i,j); // d tracks e
if(j==NM1 or j==0 or es[j+d]>c) ds[j]=0;
foreach e in (N){ if(es[e]>c) ds[e]=(i-e).sign }
}
}
fcn permuteW(seq) { Utils.Generator(_permuteW,seq) }foreach p in (permuteW(T("a","b","c"))){ println(p) }- Output:
L("a","b","c")
L("a","c","b")
L("c","a","b")
L("c","b","a")
L("b","c","a")
L("b","a","c")
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