Python Program for Generating Lyndon Words of Length n

In this problem, we will find all Lyndon words using the array's alphanumeric characters.

Before we start, let’s understand the definition of the Lyndon word.

• All words are Lyndon words which are strictly lexicographically smaller than all of their rotations.

Here are examples of Lyndon words.

• ab − The ‘ab’ is strictly lexicographically smaller than all of its permutations which is ‘ba’.

• 89 − The rotations of ‘89’ is ‘98’, which is strictly lexicographically larger than ‘89’.

• abc − The rotations of ‘abc’ are ‘bca’ and ‘cab’, which are strictly greater than ‘abc’.

Here are examples of non−Lyndon words.

• aaa − The aaa is the non−Lyndon word, as all rotations of ‘aaa’ is the same.

• bca − The ‘bca’ is non−Lyndon words as ‘abc’ its rotation is smaller than it,

Problem statement − We have given a chars array of length K containing the alphanumeric characters. Also, we have given n containing the positive integer. The task is that we need to find all Lyndon words of length n using the alphanumeric characters given in the array.

Sample examples

Input

chars = ['1', '3', '2'], n = 3

Output

112, 113, 122, 123, 132, 133, 223, 233

Explanation − It has generated all Lydon words of length 3 using the array characters.

Input

 n = 2, chars = ['1', '0']

Output

01

Explanation − The ‘01’ is the only Lyndon word that we can make using 0 and 1.

Input

 n = 2, chars = ['c', 'a', 'd']

Output

ac, ad, cd

Explanation − It has generated Lyndon words of length 2 using the a, c, and d characters.

Approach 1

We have a special algorithm to generate the Lyndon words called Duval’s algorithm.

Algorithm

Step 1 − Define the value of ‘n’ representing the Lyndon word’s length and chars array containing the characters to use while creating Lyndon words.

Step 2 − Sort the list.

Step 3 − Initialize the ‘indexes’ list with −1.

Step 4 − Make iterations until the indexes list is not null.

Step 5 − Increase the last element of the ‘indexes’ list by 1.

Step 6 − If list_size is equal to n, print the list values.

Step 7 − Append indexes to the list to make its length equal to n.

Step 8 − If the last element of the list is equal to the last index of the array, remove it from the list.

Example

Let’s understand the example with sample input.

• The sorted list will be [‘a’, ‘c’, ‘d’].

• The indexes list will be updated from [−1] to [0] in the first iteration. After that, it will equal the length of indexes to 2, and it will become [0, 0].

• In the second iteration, the list will be updated to [0, 1], and we found the first Lyndon word, ' ac’.

• The list will become [0, 2] in the third iteration, and the second Lyndon word is ‘ad’. Also, remove the last element from the list as it is equal to array_len −1.

• In the fourth iteration, the list will become [1]. After that, it will be updated [1, 1].

• The list will become [1, 2] in the next iteration,, and we found the third Lyndon wor, ' ‘cd’.

# Input
n = 2
chars = ['c', 'a', 'd']

# sort the list
initial_size = len(chars)
chars.sort()

# Initializing the list
indexes = [-1]

print("The Lyndon words of length {} is".format(n))

# Making iterations
while indexes:
    # Add 1 to the last element of the list
    indexes[-1] += 1
    list_size = len(indexes)

# If the list contains n characters, it means we found a Lyndon word
    if list_size == n:
        print(''.join(chars[p] for p in indexes))

    # Make the list size equal to n by adding characters
    while len(indexes) < n:
        indexes.append(indexes[-list_size])

    while indexes and indexes[-1] == initial_size - 1:
        indexes.pop()

Output

The Lyndon words of length 2 is
ac
ad
cd

Time complexity− O(nlogn), as we need to sort the ‘chars’ list initially.

Space complexity − O(n), as we store n indexes in the list.

Duval’s algorithm is the most effective approach to generate Lyndon words of length n. However, we have customized the approach to use only array characters.