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# Permutation String 2 2Bits
Hi everyone... Can someone help me?

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1. def lexico_permute_string(s):
2.     ''' Generate all permutations in lexicographic order of string `s`
3.
4.         This algorithm, due to Narayana Pandita, is from
5.         https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
6.
7.         To produce the next permutation in lexicographic order of sequence `a`
8.
9.         1. Find the largest index j such that a[j] < a[j + 1]. If no such index exists,
10.         the permutation is the last permutation.
11.         2. Find the largest index k greater than j such that a[j] < a[k].
12.         3. Swap the value of a[j] with that of a[k].
13.         4. Reverse the sequence from a[j + 1] up to and including the final element a[n].
14.     '''
15.
16.     a = sorted(s)
17.     n = len(a) - 1
18.     while True:
19.         yield ''.join(a)
20.
21.         #1. Find the largest index j such that a[j] < a[j + 1]
22.         for j in range(n-1, -1, -1):
23.             if a[j] < a[j + 1]:
24.                 break
25.         else:
26.             return
27.
28.         #2. Find the largest index k greater than j such that a[j] < a[k]
29.         v = a[j]
30.         for k in range(n, j, -1):
31.             if v < a[k]:
32.                 break
33.
34.         #3. Swap the value of a[j] with that of a[k].
35.         a[j], a[k] = a[k], a[j]
36.
37.         #4. Reverse the tail of the sequence
38.         a[j+1:] = a[j+1:][::-1]
39.
40. for s in lexico_permute_string('data):
OUTPUT :
aatd
atda
daat
data
dtaa
tdaa

CAN OUTPUT TO :
data
data
daat
dtaa
dtaa
daat
data
atda
aatd
tdaa
tdaa
aatd
atda
Jun 15 '21 #1
3 3142 SioSio
272 256MB
I don't know what to help, but the syntax is wrong.
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1. for s in lexico_permute_string('data):
Jun 15 '21 #2
gieforce
2 2Bits
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1. def lexico_permute_string(s):
2.     ''' Generate all permutations in lexicographic order of string `s`
3.
4.         This algorithm, due to Narayana Pandita, is from
5.         https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
6.
7.         To produce the next permutation in lexicographic order of sequence `a`
8.
9.         1. Find the largest index j such that a[j] < a[j + 1]. If no such index exists,
10.         the permutation is the last permutation.
11.         2. Find the largest index k greater than j such that a[j] < a[k].
12.         3. Swap the value of a[j] with that of a[k].
13.         4. Reverse the sequence from a[j + 1] up to and including the final element a[n].
14.     '''
15.
16.     a = sorted(s)
17.     n = len(a) - 1
18.     while True:
19.         yield ''.join(a)
20.
21.         #1. Find the largest index j such that a[j] < a[j + 1]
22.         for j in range(n-1, -1, -1):
23.             if a[j] < a[j + 1]:
24.                 break
25.         else:
26.             return
27.
28.         #2. Find the largest index k greater than j such that a[j] < a[k]
29.         v = a[j]
30.         for k in range(n, j, -1):
31.             if v < a[k]:
32.                 break
33.
34.         #3. Swap the value of a[j] with that of a[k].
35.         a[j], a[k] = a[k], a[j]
36.
37.         #4. Reverse the tail of the sequence
38.         a[j+1:] = a[j+1:][::-1]
39.
40. for s in lexico_permute_string('data'):
Sorry...
Jun 15 '21 #3
SioSio
272 256MB
Does that mean including duplicate characters?
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1. import itertools
2. all = itertools.permutations('data', 4)
3. for x in all:
4.     print(''.join(x))
Jun 16 '21 #4

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