21.1.3. Second Problem: Sequences of 1s and -1s

So, we recall from the last lecture that we discussed two problems whose solutions are the Catalan numbers. The first problem was that of coming up with the number of strings with n pairs of opening and closing parenthesis where a string is called valid, if whenever we scan the string from left to right then at any point of time or at any position in the string the number of; each instance of an opening parenthesis has a closing parenthesis. And the second problem that we saw in the last lecture is that of coming up with a number of sequences consisting of n number of 1s and n number of –1s such that if we scan the string from the first position to the last position then each partial sum should be greater than equal to 0.

In this section, we discuss two problems that lead to the solution known as Catalan numbers. The first problem deals with counting valid sequences of parentheses, where a valid string means it must have matching pairs of opening and closing parentheses at every point. The second problem involves counting specific sequences of numbers, namely 1s and -1s, that maintain a non-negative sum when read sequentially from start to finish.

Imagine a game where you are trying to balance weights on a seesaw. Each '1' represents adding a weight to one side, while each '-1' represents removing a weight. A sequence maintaining a non-negative sum is like ensuring the seesaw never tips over to one side; it must always remain balanced or tipped towards the side where the weights are heavier.

And we saw a bijection between the set of sequences consisting of n 1s and n –1s where each partial sum is greater than equal to 0 and the set of all strings all valid strings with n pairs of opening and closing parenthesis.

A bijection refers to a one-to-one correspondence between two sets. In this case, we are finding a relationship between valid sequences of 1s and -1s that do not drop into negative sums and valid strings of parentheses. Each valid arrangement of parentheses can be translated into a corresponding arrangement of 1s and -1s and vice versa.

Think of matching socks in a drawer. If you have pairs (like parentheses), pulling out one sock means you can only pull out its pair without leaving a sock unmatched. This resembles how we track the number of 1s and -1s; every time we add one, we need to ensure there’s a corresponding one to keep the balance intact.

So, this is the statement which we want to prove. We want to prove that the number of sequences consisting of n 1s and n –1s, where in each sequence the partial sum at any position is greater C(2n, n) than equal to 0 is . So, for that the proof strategy will be the following.

To prove that there are exactly C(2n, n) sequences meeting these criteria, we will use a systematic strategy. We will first define A as all the possible sequences of n 1s and n -1s without any restriction. Next, we will determine a subset of those sequences, B, that violate the non-negativity condition and then subtract the count of these 'bad' sequences from total sequences to arrive at the final count of valid sequences.

Consider organizing a race between two teams: Team A and Team B. Team A completes 10 laps (1s), and Team B completes 10 laps (-1s). If we let the laps completed be free of any rules, Team A can finish in any order (A's total count), but we then need to subtract instances where Team B overtakes (B's count of bad instances) to determine the actual races where Team A stays ahead.

So, now for the rest of our discussion our focus will be to find out the cardinality of the set of bad sequences and what is a bad sequence? A sequence is a bad sequence if it consists of n number of 1s, n number of –1s such that in such sequence there is at least one occurrence of a partial negative sum.

A bad sequence is defined specifically as one that, despite having the necessary number of 1s and -1s, crosses below zero at some point, indicating an imbalance. The task is to analyze the set of these bad sequences to understand their prevalence in the total set of sequences.

Imagine a baker trying to balance the perfect recipe. Each ingredient represents a unit (1s and -1s). If adding too much of one ingredient dips the total weight below a certain threshold, that would be considered a bad mix, and we'd want to figure out how many such mixes exist in all possible combinations.

So, what we will do is we will introduce a very nice method called as reflection method or why it is called reflection method will be clear very soon. So, we will find the cardinality of the set B using the reflection method.

The reflection method is a technique employed to count the number of bad sequences. By identifying the first point where a sequence drops below zero, we can then 'reflect' the remainder of the sequence. This reflection allows us to create a new sequence that can be counted more easily. Ultimately, we'll be able to establish a clear link between the bad sequences and their valid counterparts.

Think of using a mirror to study your reflection while learning dance steps. When you make a mistake on your left foot (the point dropping below zero), you can look in the mirror to see how to correct that step on the right foot (the reflected sequence). This method helps keep track of steps in a coherent way!

Now what we will show is that the cardinality of the set B is C(2n, n+1) and if we subtract the cardinality of B from the cardinality of A then we will get our required answer.

After identifying and understanding the bad sequences, we'll count them using combinatorial formulas. The count of the bad sequences (B) can be shown to equal C(2n, n + 1). The idea is to show how many sequences would be formed with an extra '1' or reduced '-1', leading to a new total. By subtracting this from the total sequences (A), we arrive at the valid combinations we seek.

Imagine again the teams racing, where Team A normally gains an advantage. By figuring out how many times Team B has gained an excess lap and accounting for them, we can ultimately calculate the number of valid winning sequences of Team A after all disqualifications are subtracted.

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