16.5. Onto Functions

Let $f$ denote my functions which is the number of valid sequences ending with $n$.

In this section, we define a function $f$ that counts the number of valid sequences that end with a specific integer, $n$. These sequences must adhere to specific rules, such that the integers in the sequence are in strictly increasing order and range from 1 up to $n$. The concept of valid sequences is foundational for the recurrence relations we will explore later.

Imagine you are tasked with listing all the distinct ways you can arrange a set of building blocks numbered 1 through $n$ in a way that each arrangement starts with 1 and ends with $n$, while ensuring that each block is only used once, and that the numbers are in ascending order.

A trivial recurrence condition for $f$ is the following. I can say that $f(n) = f(n-1) + f(n-2) + ... + f(1)$.

Here, we establish a recurrence relation for the function $f(n)$. The number of valid sequences ending with $n$ can be calculated by summing the number of valid sequences that end with all integers less than $n$. This includes sequences that end with 1, 2, ..., and so forth up to $n-1$. This illustrates how each valid sequence can build upon previous sequences.

Think of this as a way of counting paths in a game where each step is determined by choosing to move to a lower-numbered block. The total number of ways you can reach block $n$ is equal to the total ways to reach each of the blocks leading up to it.

We want a more compact recurrence condition. The degree of this equation is $n - 1$.

While we have an initial recurrence formula to work with, it includes multiple terms and can be complex. We aim to derive a more compact version, which simplifies solving our recurrence relation by reducing the dependence on multiple previous terms. Specifically, we wish to decrease the degree from $n - 1$ to 1, allowing us to work with just one preceding value of $f$.

Consider simplifying a recipe that requires multiple previous steps. If you can reduce the recipe to only needing the result of the last step, it becomes easier to follow. Similarly, we want our function to reduce complexity so we can compute it more efficiently.

Now there can be two categories. Category 1 where $m = n - 1$ and Category 2 where $m$ can be anything from the set 1, 2, ..., to $n - 2$.

In deriving a more manageable recurrence condition, we break down the valid sequences into two distinct categories based on their last integer. In the first category, the second last number is fixed as $n - 1$, while in the second category it can be any number less than $n - 1$. The sequences in these categories are entirely separate, which aid in ensuring that our counts don't overlap when deriving the new relation.

Imagine you are organizing a race with two separate tracks. One track allows participants only to finish a lap just before the end, while the other allows more flexibility. By analyzing the race's outcome based on these two tracks, you can better predict overall finishing times.

Thus, the alternate recurrence equation will be $f(n) = 2 \cdot f(n-1)$.

After identifying the distinct categories of valid sequences, we combine the counts to establish a final, more compact recurrence relation of degree 1. This means the number of sequences ending in $n$ depends solely on the number ending with $n-1$. This significantly simplifies calculations and reflections on the sequence's growth.

Think of this like doubling the number of ways a single performer can present in front of an audience when given additional rehearsal time. If they can choose to repeat their last performance or adjust it slightly, the variety of outcomes grows predictably based on just the last performance.

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