How many bit strings of length n are palindromes? Hint: Consider two cases n is even and n is odd. Note a palindrome is a “string” of letters or numbers which read the same “frontwards” and backwards”. Examples: 1101011, 10111101 are palindromes.
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To determine the number of bit strings of length n that are palindromes, we need to consider two cases: n being even and n being odd.
Case 1: n is even
In this case, the bit string will have an equal number of bits on both sides, which means the left half must be the reverse of the right half. For example, in the given examples, "1101011" and "10111101," you can observe this property.
To count the number of palindromes of length n when n is even, we can start by choosing bits for the left half of the string. Since the left half determines the right half, we only need to consider choosing bits for the left half.
Since a palindrome reads the same frontwards and backwards, if we choose the bits for the left half, it will automatically determine the bits for the right half. Therefore, we only need to select bits for half of the string.
For each bit, we have two choices: 0 or 1. So, for the left half of the string, we have 2^(n/2) possible combinations.
Therefore, the number of bit strings of length n that are palindromes when n is even is 2^(n/2).
Case 2: n is odd
When n is odd, the bit string will have a middle bit, and the left half should be the reverse of the right half, excluding the middle bit. For instance, in the example "10111101," you can observe this pattern.
Similar to the previous case, we need to choose bits for the left half of the string, excluding the middle bit. Again, for each bit, we have two choices: 0 or 1. Hence, for the left half, we have 2^((n-1)/2) possible combinations.
Since there are two choices for the middle bit, the total number of palindromic bit strings of length n when n is odd is 2^((n-1)/2) * 2.
In summary:
- When n is even: Number of palindromic bit strings = 2^(n/2)
- When n is odd: Number of palindromic bit strings = 2^((n-1)/2) * 2