16.4. Ternary Strings with Occurrences

Let $s_n$ denote my functions which is the number of valid sequences ending with $n$ for my recurrence condition. A trivial recurrence condition for $s_n$ is the following: $s_n = s_{n-1} + s_{n-2} + \dots + s_1$. This is because the second last value in the sequence, namely $m$, can be any number from {1, 2, \dots, n-1}. If $m = 1$, we find out all possible sequences starting and ending with 1 and append it with $n$. How many such sequences can you have? You can have $s_{n-1}$ such sequences. If $m = 2$, find out all valid sequences starting with 1, ending with 2, which is a strictly increasing sequence, and append $n$ at the end. You can have another $s_{n-1}$ such sequences. This continues up to $m = n - 1$.

In this chunk, we are discussing the recurrence relation of valid sequences that lead to ternary strings with specific occurrences. The function $s_n$ counts the total number of valid sequences where the last character is $n$. To find this count, we consider all sequences where the second last character can be any value from 1 to $n-1$. This produces a recursive structure where all valid sequences of length increased by 1 depend on the previous valid sequences. Thus, we derive that $s_n$ equals the sum of all possible preceding sequences that lead to valid sequences ending in $n$. This creates a way to compute the count incrementally.

Think of this process as building a tower of blocks. Each block can only be placed on top of a block that is smaller in height. Suppose you have blocks numbered from 1 to n. The number of ways to build a tower such that the last block is n depends on how many towers you can build where the last block is less than n (1 through n-1). Each valid tower that ends in 1, 2, ..., (n-1) can be extended by adding block n on top.

Now, we want a more compact recurrence condition. Let us derive an alternate recurrence equation. Imagine you have a valid strictly increasing sequence starting with 1, ending with $n$, and second last value is $m$. The first category considers all sequences where $m = n - 1$. For each sequence like this, when we add $n$ at the end, we create valid sequences starting with 1 and ending in n. The second category considers all values where $m$ can be 1, 2, or $n-2$. For these, we also append $n$. Thus, all such sequences represent different categories that are disjoint.

In this chunk, the intent is to create a simplified form of the recurrence relation. By identifying two distinct categories of valid sequences (ending with $n-1$ or any number between 1 and $n-2$), we reinterpret the sequences for easier computation. Each category can be calculated separately but contributes to the overall count. This leads to a new recurrence relation $s_n = 2 * s_{n-1}$. The relationship shows a dependency that only relies on the immediate previous term, making it easier to compute in comparison to the previous relation which was dependent on multiple earlier terms.

Imagine a system where you have to choose your team members based on their performance in previous rounds. If you know the best available teams from the earlier rounds (let's say last round is like 'previous term'), and you can double the number of teams by just replacing only those who did not perform well. Thus the overall performance hinges solely on how well your last performed group did, simplifying the decision-making process.

Despite the fact that the equation is of degree 1, we need 2 initial conditions. For $n = 1$, there is only 1 sequence which is strictly increasing starting and ending with 1. For $n = 2$, there is also just 1 sequence. Therefore, it's necessary to specify these conditions.

This chunk illustrates the importance of initial conditions when working with recurrence relations, even when they are simplified. Initial conditions serve as starting points from which we can generate all other terms in the sequence. For both cases $n = 1$ and $n = 2$, specifying that there is only one valid sequence each helps in establishing the true foundation for the calculations of future sequences from those terms.

Think of starting a project. Initially, you might have just one way to start, for example, writing your first line, which is your foundation. Similarly, before you can develop your project further (like creating teams or launching stages), you need to outline how the first stages will look. Without these initial bases (like your 1 and 2 sequences), none of the advanced features (later sequences) would make sense or work properly.

In the context of ternary strings of length $n$ that contain an occurrence of "00", "11", or "22", we denote $s$ as the number of such ternary strings. The recurrence relation satisfies $s = 2s + 3\cdot 2^{n-1}$. Here, category 1 consists of strings of length $n-1$ that already have the condition satisfied. In category 2, we derive strings without occurrences of the specified pairs and add a character to fulfill the conditions.

In this final chunk, we delve into the counting of ternary strings specifically focusing on those that contain specific occurrences of pairs. By subdividing the strings into categories, we clarify how many ternary strings satisfy the conditions while also deriving a proper equation to calculate them. This recurrence relationship focuses on the known count of strings that meet the specified conditions and segregates those that don't (and subsequently transforming them), much in the same way we identified prior with valid sequences.

Consider making a salad using three types of ingredients: greens, vegetables, and toppings (0, 1, 2). You want your salad to meet specific tasty combinations (like having two types of greens next to each other, or two toppings next). The total number of ways to combine these ingredients into a salad that meets those requirements can be counted based on how many combinations you already know that are satisfying those conditions. Then you can tweak or add ingredients to make new salads while ensuring they meet the tasty requirements.

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