Python Program to find all Palindromic Bitlists in length

Last Updated : 05 Sep, 2024

In this article, we will learn to generate all Palindromic Bitlists of a given length using Backtracking in Python. Backtracking can be defined as a general algorithmic technique that considers searching every possible combination in order to solve a computational problem. Whereas, Bitlists are lists of bits(0 and 1). A bit list that is a palindrome is called Palindromic Bitlists. To generate all Palindromic Bitlists of length N, the best way is to generate all possible bit lists of length N/2 and then form palindromic Bitlist using this list of length N/2. Here we will discuss 2 cases:

Case 1: When the length of the list is odd

Let's consider the value of N = 3. So, N/2 = 1 (floor value).

All possible bit lists of length 1 are [0], [1]. In this case, we can not do exact same thing which we have done for the even length case because doing that thing will give a bit list of length 2 and not 3. Here we will add [0] or [1] in mid of bit list. Consider [0], its reverse will be [0]. To generate bit lists we have to combine 3 lists that are generated bit lists of length 1, [0], Or [1], the reverse of the generated list. Here every bit list will produce 2-bit lists. For [0] there will be [0] + [0] +[0] and [0] + [1] +[0].

Example: The value of N is 5

Python

# list to store all the possible solutions fianl_ans = []  # this function generates all possible bitlists  # of length N//2 and then saves the final answer  # in the list   def generate_palindromic(n, is_even, lst=[]):     # n==0 is the base condition and after      # reaching here we don't go forward     if n == 0:         if is_even:             # even case             # final_ans = generated list + reverse              # of generated list             fianl_ans.append((lst + lst                               [-1::-1]))         else:             # odd case             # final_ans_1 = generated list +              # [0] + reverse of generated list             fianl_ans.append(lst + [0] +                               lst[-1::-1])             # final_ans_1 = generated list              # + [1] + reverse of generated list             fianl_ans.append(lst + [1] +                               lst[-1::-1])         return     # store 0 in the list to provide the value of      # this step to the next step     lst.append(0)     # call recursive function and decrease      # the value of n by 1     generate_palindromic(n-1, is_even, lst)     # remove 0 from the list (Backtrack)     lst.pop()     # store 1 in the list to provide the value      # of this step to the next step     lst.append(1)     # call recursive function and decrease      # the value of n by 1     generate_palindromic(n-1, is_even, lst)     # remove 1 from the list (Backtrack)     lst.pop()   def check(n):     if n % 2 == 0:         # if n is even then pass 1 to the          # recursive function         generate_palindromic(n/2, 1)     else:         # if n is odd then pass 0 to the          # recursive function         generate_palindromic(n//2, 0)   if __name__ == "__main__":     # calling function to generate all      # bitlists of length 5     check(5)     # printing final answers     for i in fianl_ans:         print(i)

Output:

Use Backtracking to find all Palindromic Bitlists of a given length in Python

Case 2: When the length of the list is even

Let's consider the value of N = 4. So, N/2 = 2.

All possible bit lists of length 2 are [0,0], [0,1], [1,0], [1,1]. Consider [0,1], its reverse will be [1,0]. Combining both of them we will get [0,1,1,0] which is a Palindromic bit list of length 4. In this way, we can generate all Palindromic Bitlists of length N where N is even.

Example: The value of N is 4

Python

# list to store all the possible solutions fianl_ans = []  # this function generates all possible bitlists  # of length N//2 and then saves the final answer in the list   def generate_palindromic(n, is_even, lst=[]):     # n==0 is the base condition and after      # reaching here we don't go forward     if n == 0:         if is_even:             # even case             # final_ans = generated list + reverse              # of generated list             fianl_ans.append((lst +                                lst[-1::-1]))         else:             # odd case             # final_ans_1 = generated list +              # [0] + reverse of generated list             fianl_ans.append(lst + [0] +                               lst[-1::-1])             # final_ans_1 = generated list +               # [1] + reverse of generated list             fianl_ans.append(lst + [1] +                               lst[-1::-1])         return     # store 0 in the list to provide the value of      # this step to the next step     lst.append(0)     # call recursive function and decrease the      # value of n by 1     generate_palindromic(n-1, is_even, lst)     # remove 0 from the list (Backtrack)     lst.pop()     # store 1 in the list to provide the      # value of this step to the next step     lst.append(1)     # call recursive function and decrease      # the value of n by 1     generate_palindromic(n-1, is_even, lst)     # remove 1 from the list (Backtrack)     lst.pop()   def check(n):     if n % 2 == 0:         # if n is even then pass 1 to          # the recursive function         generate_palindromic(n/2, 1)     else:         # if n is odd then pass 0 to          # the recursive function         generate_palindromic(n//2, 0)   if __name__ == "__main__":     # calling function to generate all      # bitlists of length 5     check(4)     # printing final answers     for i in fianl_ans:         print(i)

Output:

The time complexity of both Case 1 and Case 2 of the Python program to find all Palindromic Bitlists of a given length using backtracking is O(2^(N/2)), where N is the length of the bitlist.The space complexity of both Case 1 and Case 2 algorithms is O(2^(N/2)), where N is the length of the palindromic bitlist.

This is because the algorithms generate all possible palindromic bitlists of length N, which is equal to generating all possible bitlists of length N/2 and then combining them in a palindromic manner. Therefore, the space required to store all possible bitlists of length N/2 is O(2^(N/2)), which is the maximum space required by both the algorithms.

In both cases, the bitlists are generated recursively and stored in a list, which is then appended to the final_ans list. The maximum size of this list would be O(2^(N/2)), as there are 2^(N/2) possible bitlists of length N/2. Hence, the space complexity of both algorithms is O(2^(N/2)).