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
Python3
# list to store all the possible solutionsfianl_ans = []Â
# this function generates all possible bitlists # of length N//2 and then saves the final answer # in the listÂ
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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()Â
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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)Â
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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:
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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
Python3
# list to store all the possible solutionsfianl_ans = []Â
# this function generates all possible bitlists # of length N//2 and then saves the final answer in the listÂ
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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()Â
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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)Â
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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:
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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)).
