91Ó°ÊÓ

Strings are sequences of characters. In our exercise, the set \(\Sigma = \{0, 1, 2, 3\}\) represents the alphabet from which these characters are drawn. An important aspect of strings is their length, denoted as \(n\) in this context, which tells us how many characters are in the string.
For example, a string of length 3 could be '012' or '333'.

In counting problems like this one, we're often interested in strings that meet certain criteria. Here, we're concerned with strings that contain an **odd number of 1's.**

• The total number of possible strings of length \(n\) is calculated by raising the size of \(\Sigma\) (which is 4) to the power of \(n\), resulting in \(4^n\).
• Understanding this total helps us form parts of a recurrence relation since any string of a given length can be broken down into smaller problems.

A recurrence relation is a mathematical way to define a sequence of numbers, specifying each term as a function of the preceding ones.
In this problem, we are interested in the sequence \(a_n\), which counts the number of strings of length \(n\) with an odd number of 1's.

By looking at the progression from previous steps, we need to find a way to relate \(a_{n}\) with \(a_{n-1}\).

• We start with the base case where \(a_1 = 1\). This is because only the string '1' from \(\Sigma^{1}\) meets the condition.
• To move forward, we consider strings of length \(n\) and relate them to those of length \(n-1\).
• The key transformation is recognizing that the relation \(a_n = 4^{n-1} - a_{n-1}\) accurately depicts this progression.

This relation makes it possible to compute \(a_n\) using the previously calculated \(a_{n-1}\), providing an efficient way to build larger terms from smaller ones.

Counting strings with conditions involves identifying strings that meet specific criteria, like containing an odd number of a certain character—in this case, '1'.
We break this down by understanding how each addition to a string affects its character count.

Consider a string of length \(n-1\).

• If we add '1', it changes the parity of the number of 1's (odd becomes even, and vice versa).
• If we add '0', '2', or '3', the parity remains the same.

This insight is crucial because it leads to the understanding that we can split all strings into two categories: those with an odd number of 1's and those with an even number of 1's.

Hence, we can deduce:
- For every odd string of length \(n-1\), we form an even string of length \(n\) by adding '1'.
- Conversely, for every even string of length \(n-1\), adding '1' creates an odd string.
This balanced approach ensures that our recurrence relation \(a_n = 4^{n-1} - a_{n-1}\) correctly accounts for the entirety of \(\Sigma^n\), linking back to how many strings meet our condition of having an odd number of 1's.