apply Permutation cycles to a letters list

permutation

asked 2024-11-19 16:44:05 +0100

ortollj
1355 ●29 ●69 ●106

updated 2024-11-22 09:02:16 +0100

Hi

Sorry for this very basic question, but how to apply the Permutation function to a list of letters ? I think it should exist a simplest way than the the cooking below. I searched the web, without success !

def applyGen(ge,dic) :   
    return [ge[dic.get(e)] for e in dic.keys() ]

def decrement_dic(dic): 
    return {k - 1: v - 1 for k, v in dic.items()} 

lettersL=['a','b','c','d','e','f','g','h']

test=Permutation((1,5,8,4),(2,6,7,3))
dicD=decrement_dic(test.dict())
applyGen(lettersL,dicD)

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answered 2024-11-19 17:17:37 +0100

FrédéricC

5524 ●4 ●46 ●123

updated 2024-11-19 18:28:45 +0100

Like this

sage: L = ['a','b','c','d','e','f','g','h']
sage: test = Permutation((1,5,8,4),(2,6,7,3))
sage: [L[test(i + 1) - 1] for i in range(len(L))]
['e', 'b', 'c', 'a', 'h', 'f', 'g', 'd']

Note that Python numbering starts from 0.

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ok thank you @FrédéricC

ortollj ( 2024-11-19 18:26:40 +0100 )edit

Btw, one can replace test(i + 1) with test[i].

Max Alekseyev ( 2024-11-20 03:27:17 +0100 )edit

@Max Alekseyev What do you mean ?

ortollj ( 2024-11-22 09:04:50 +0100 )edit

1

I mean that [L[test[i] - 1] for i in range(len(L))] will work equally well.

Max Alekseyev ( 2024-11-22 14:30:58 +0100 )edit

Ok @Max Alekseyev

ortollj ( 2024-11-22 15:17:25 +0100 )edit

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answered 2024-12-27 14:50:46 +0100

dan_fulea
5685 ●5 ●44 ●97

Here is an alternative solution that explicitly uses the set $X={1,2,\dots,8}$ where the permutation is acting on, and a dictionary dic that passes from $X$ to the letters. The dictionary is a replacement for the increment/decrement that converts between pythonic indexing and the natural set permutations act on...

letters = list('abcdefgh')
X = list(range(1, len(letters) + 1))
dic = dict( zip(X, letters) )

test_pi = Permutation((1,5,8,4), (2,6,7,3))
[dic[test_pi(x)] for x in X]

We obtain:

['e', 'b', 'c', 'a', 'h', 'f', 'g', 'd']
sage: ''.join(_)
'ebcahfgd'

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answered 2024-11-22 08:34:23 +0100

ortollj
1355 ●29 ●69 ●106

updated 2024-11-26 05:59:46 +0100

I propose to improve @FrédéricC's solution as below to avoid errors in certain cases :

tuplesL=[ ((1,5,8,4),(2,6,7,3)), ((4,5)),((1,2,3)),((1,2,3),(4,5)),((1,2,3),(4,5),(7,8))]
for t in tuplesL :
    print('permutation tuples= ',t)
    try :
        test=Permutation(t)
        print('FredericC solution OK : ',[L[test(i + 1) - 1] for i in range(len(L))])
    except :
        print('error from FredericC solution')
    P=Permutation(t)
    Pletters= [L[P(i + 1) - 1] for i in range(len(P.dict()))] +[L[i] for i in range(len(P.dict()),len(L))]
    print(Pletters)

But unfortunately this solution does not allow singletons in list of tuples ! like Wolfram alpha do

see Max Alekseyev solution but unfortunately it does not accept element which appears more than once in the input like Wolfram does

As this post has become a bit messy, final correct answer from Max Alekseyev to the question below:

import string
import sage.combinat.permutation as permutation
nl=12 # letters number
L=list(string.ascii_lowercase[0:nl])
tuplesL=[ ((1,5,8,4),(2,6,7,3)), ((4,5),),((1,2,3),),((1,2,3),(4,5),),((1,2,3),(4,5),(7,8)),((1, 2 ,6),(3, 5),),
((1, 2, 3, 4),(1, 3)(2, 4),)]
for t in tuplesL :
    print('permutation tuples= ',t)
    try :
        #test=permutation.from_cycles(len(L),t)
        test = prod(map(lambda x: permutation.from_cycles(len(L),[x]),t))
        print('Max Alekseyev solution OK : ',[L[test[i] - 1] for i in range(len(L))])
    except :
        print('error ')

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But that doesn't accept singletons is a minor problem

ortollj ( 2024-11-22 08:58:48 +0100 )edit

1

First, to specify tuples with single cycles, you need to have an extra "," - like ((4,5),) and ((1,2,3),). Second, to construct permutations with implicut presence of singletons you can replace test=Permutation(t) with

import sage.combinat.permutation as permutation
test = permutation.from_cycles(len(L),t)

Max Alekseyev ( 2024-11-22 14:41:32 +0100 )edit

Thank you @Max Alekseyev

import string
import sage.combinat.permutation as permutation
nl=12 # letters number
L=list(string.ascii_lowercase[0:nl])
tuplesL=[ ((1,5,8,4),(2,6,7,3)), ((4,5),),((1,2,3),),((1,2,3),(4,5),),((1,2,3),(4,5),(7,8)),((1, 2 ,6),(3, 5),)]
for t in tuplesL :
    print('permutation tuples= ',t)
    try :
        test=permutation.from_cycles(len(L),t)
        print('Max Alekseyev solution OK : ',[L[test[i] - 1] for i in range(len(L))])
    except :
        print('error from Max Alekseyev solution solution')

is it possible to transform Max Alekseyev solution as THE solution of this question ?

ortollj ( 2024-11-22 15:43:52 +0100 )edit

Just edit your own answer.

Max Alekseyev ( 2024-11-22 16:42:28 +0100 )edit

I wanted to copy Max Alekseyev's solution into the incomplete solution I gave.So first, I tried to transform my answer into a comment using the menu --> more --> repost as a question comment, but nothing happens.

ortollj ( 2024-11-22 17:07:25 +0100 )edit

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