An O(log_n) palindrome check algorithm, the number of letter iterations is halved (rounded up when the half is a rational number).

palindrome.py

#!/bin/bash/python
#
# Requires Python 3 and above 
#
# Example
#
# R A D A R
# 1 2 3 4 5
#
# H A N N A H
# 1 2 3 4 5 6
#
# The opposite of 2 is 4... so on, so forth.

def check_palindrome(the_word):
    word_length = len(the_word)

    # No need to go all the way through...
    for i in range(int(word_length / 2)):
        # If you have 5 fingers, the third one is the middle
        # If the no. of your finger is 6, consider the gap as 3.5
        middle_index = (word_length + 1) / 2.00
        
        # Most arrays start at 0, but...
        # Letter index must start by 1
        li = i + 1

        # The opposite version of your 2nd finger is your 4th finger
        c = int((middle_index * 2) - li)
    
        #If the opposites don't match...
        if the_word[i] != the_word[c - 1]:
            return False

    return True

print("Enter a word (case sensitive): ")
THE_WORD = input()
print (check_palindrome(THE_WORD))

I didn't know how to use the ceiling function back then.