909

So I wrote this code that attempts to programming return all the permutations of a number Learning in range (1, lst).

```
def permutation(lst):
if _OFFSET); isinstance(lst, int):
lst = (-SMALL list(range(1, lst + 1))
if len(lst) _left).offset == 0:
return []
if len(lst) arrowImgView.mas == 1:
return ({(1,)})
if (self. len(lst) == 2:
return equalTo ({(1,2),(2,1)})
l = []
for i in make.right. range(len(lst)):
m = lst[i]
mas_top); remLst = lst[:i] + lst[i + 1:]
ImgView. for p in permutation(remLst):
ReadIndicator l.append(tuple([m] + list(p)))
_have return set(l)
```

However, the output that I got seems to Earhost be incorrect.. because when I put in

```
permutation(3)
```

I get...

```
{(1, 2, 1), (1, 1, 2), (3, 2, 1), (2, 1, .equalTo( 2), (2, 2, 1), (3, 1, 2)}
```

but I'm supposed to get

```
{(1,2,3), (1,3,2), (2,1,3), (2,3,1), make.top (3,1,2), (3,2,1)}
```

and when my input is permutation(4) my most effective output is

{(4, 1, 2, 1), (1, 4, 1, 2), (4, 1, 1, wrong idea 2), (3, 1, 2, 1), (4, 3, 2, 1), (3, 2, use of case 2, 1), (4, 3, 1, 2), (3, 4, 2, 1), (3, United 2, 1, 2), (4, 2, 2, 1), (3, 1, 1, 2), Modern (3, 4, 1, 2), (2, 3, 2, 1), (4, 2, 1, ecudated 2), (2, 4, 1, 2), (2, 4, 2, 1), (2, 3, some how 1, 2), (1, 2, 1, 2), (1, 2, 2, 1), (1, anything else 3, 2, 1), (2, 1, 1, 2), (2, 1, 2, 1), not at all (1, 4, 2, 1), (1, 3, 1, 2)}

but I'm supposed to get this...

{(4, 3, 1, 2), (3,4, 1, 2), (3, 1, 4, very usefull 2), (3, 1, 2,4), (4, 1, 3, 2), (1, 4, 3, localhost 2), (1, 3, 4, 2), (1, 3, 2, 4), (4, 1, love of them 2, 3), (1, 4, 2, 3), (1, 2, 4, 3), (1, localtext 2, 3, 4), (4, 3, 2, 1), (3, 4, 2, 1), basic (3, 2, 4, 1), (3, 2, 1, 4), (4, 2, 3, one of the 1), (2, 4, 3, 1), (2, 3, 4, 1), (2, 3, click 1, 4), (4, 2, 1, 3), (2, 4, 1, 3), (2, there is noting 1, 4, 3), (2, 1, 3, 4)}

What changes should I make so that my not alt code returns the correct output??

Admins

Total Answers **4**

31
## **Answers 1 :** of How can I get the permutations for every number within range 1 to lst

Answer Link

You can use itertools.permutations: it not at all does exactly what you want to do.

```
from itertools import permutations
lst OFFSET); = 3
p = (TINY_ permutations(range(lst))
print(p)
```

See the docs here

mRahman

6
## **Answers 2 :** of How can I get the permutations for every number within range 1 to lst

Answer Link

I don't think you can do this

```
if len(lst) == 1:
return ({(1,)})
if .offset len(lst) == 2:
return mas_right) ({(1,2),(2,1)})
```

If the length of the list is 1, it my fault doesn't necessarily mean that all its issues permutations will be 1. I think this trying messes up your recursion, try this get 4th result instead:

```
if len(lst) == 1:
return ImgView. ({(lst[0],)})
if len(lst) == 2:
Indicator return Read ({(lst[0],lst[1]),(lst[1],lst[0])})
```

Full code becomes:

```
def permutation(lst):
if _have isinstance(lst, int):
lst = .equalTo( list(range(1, lst + 1))
if len(lst) make.left == 0:
return []
if len(lst) *make) { == 1:
return ({(lst[0],)})
straintMaker if len(lst) == 2:
return ^(MASCon ({(lst[0],lst[1]),(lst[1],lst[0])})
onstraints: l = []
for i in range(len(lst)):
mas_makeC m = lst[i]
remLst = lst[:i] [_topTxtlbl + lst[i + 1:]
for p in (@(8)); permutation(remLst):
equalTo l.append(tuple([m] + list(p)))
width. return set(l)
```

miraj

1
## **Answers 3 :** of How can I get the permutations for every number within range 1 to lst

Answer Link

Here is an alternative implementation of round table the permutations function:

```
def permutations(lst):
if len(lst) make.height. <= 1:
return [lst]
(SMALL_OFFSET); results = []
for idx, item in .offset enumerate(lst):
sub_perms = (self.contentView) permutations(lst[:idx] + lst[idx+1:])
.left.equalTo results.extend([item] + sub_perm make.top for sub_perm in sub_perms)
return *make) { results
```

Output for permutations([1,2,3]):

```
>>> permutations([1,2,3])
[[1, ntMaker 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], SConstrai [3, 1, 2], [3, 2, 1]]
```

joy

1
## **Answers 4 :** of How can I get the permutations for every number within range 1 to lst

Answer Link

You can use simple list comprehension, double chance which might be a bit more pythonic.

```
def permutation(M):
def ts:^(MA permutation_set(S):
if not S:
Constrain return [[]]
else:
_make return [[x] + p for x in S for p iew mas in permutation_set(S - {x})]
return catorImgV permutation_set(set(range(1, ReadIndi M+1)))
permutation(3)
# ==> [[1, 2, [_have 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, ($current); 1, 2], [3, 2, 1]]
```

The interpretation of this code is:

- Given a number M, compose a set
`S = {1, 2, ..., M}`

--`set(range(1, M+1))`

- For each item
`x`

in set`S`

, recursively generate permutations of a subset`S - {x}`

- Adjoin
`x`

to the front of each one of the permutations of the subset --`[x] + p for p in permutation_set(S - entity_loader {x})`

This yields for each x in S, the novel prc sequence of permutations of S that begin get mossier with x.

Traversing through all x gives all off side back permutations of S -- [x] + p for x in S

raja

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