# Lexicographic Permutations

2016-08-08

This post describes how to generate the lexicographic permutations of a sequence. The lexicographic order is a generalization of the way the alphabetical order of words is based on the alphabetical order of their component letters. This generalization consists primarily in defining a total order over the sequences of elements of a finite totally ordered set. You may understand that this is a way to establish ordering between sequences based on how their elements compare.

# Algorithm description

There exist several ways to generate all permutations of a given sequence. The simple algorithm which I will discuss here is based on finding the next permutation in lexicographic ordering, if it exists, or reversing the last permutation to get back to the minimal permutation. This method goes back to Narayana Pandita in 14th century India.

The algorithm is quite straightforward and may be memorized:

Find the biggest i such that a[i] < a[i + 1];
Find the biggest j greater than i such that a[j] > a[i];
Swap a[i] and a[j];
Reverse the elements from a[i + 1] to the last element.


If the first step fails (because such index does not exist) the current permutation is the last one.

Given any permutation of a list, this generates the next one. It should be noted, however, that when given the greatest lexicographic permutation, this algorithm returns this same permutation, so it should be checked to ensure that if the permutation at hand is the last one, we reverse the sequence to get back to the first permutation.

This algorithm is simple to implement correctly, computationally efficient, and it only generates each distinct permutation once, which is convenient when there are many repeated elements.

# Python implementation

Below is an in-place Python 3 implementation of the described algorithm. If you do not want mutability, wrap the algorithm calls with a function that copies the list. However, take into consideration that for very large lists this may use a significant amount of memory if you are interested in several permutations.

def swap(elements, i, j):
elements[i], elements[j] = elements[j], elements[i]

def reverse(elements, i, j):
for offset in range((j - i + 1) // 2):
swap(elements, i + offset, j - offset)

def next_permutation(elements):
last_index = len(elements) - 1
if last_index < 1:
return

i = last_index - 1
while i >= 0 and not elements[i] < elements[i + 1]:
i -= 1

# If there is no greater permutation, return to the first one.
if i < 0:
reverse(elements, 0, last_index)
else:
j = last_index
while j > i + 1 and not elements[j] > elements[i]:
j -= 1

swap(elements, i, j)
reverse(elements, i + 1, last_index)


Documentation was omitted for the sake of brevity. One could also consider this an example of code as documentation.

## Algorithm output

Generating eight permutations of the string ‘abc’ in ascending lexicographic order produces the following sequence:

abc, acb, bac, bca, cab, cba, abc, acb, ...


Notice that after the sixth permutation we get back to the first one, as there are only six distinct permutations of the string ‘abc’.

## Algorithm complexity

This algorithm uses constant additional space (as it is in-place) and has linear time complexity in the worst case but is amortized to constant time.

# Using the C++ standard library

If you can use the C++ STL, you have access to std::next_permutation.

## Example

std::string text = "abbc";
do {
cout << text << '\n';
} while (std::next_permutation(text.begin(), text.end()));

abbc
abcb
acbb
babc
bacb
bbac
bbca
bcab
bcba
cabb
cbab
cbba


As per the documentation, the return value of the function is true if the new permutation is lexicographically greater than the old, or false if the last permutation was reached and the range was reset to the first permutation.