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Chapter 10: Problem 11

An alphabet \(\Sigma\) consists of the four numeric characters \(1,2,3,4\), and the seven alphabetic characters \(a, b, c, d, e, f, g\). Find and solve a recurrence relation for the number of words of length \(n\) (in \(\Sigma^{*}\) ) where there are no consecutive (identical or distinct) alphabetic characters.

Short Answer

Expert verified
The recurrence relation for the number of words of length \(n\) without consecutive alphabetic characters is: \(A_{n} = 4 * A_{n-1} + 28 * A_{n-2}\), with initial conditions \(A_1 = 11\) and \(A_2 = 112\).

Step by step solution

01

Identifying Patterns

Let's define \(A_{n}\) as the number of words of length \(n\) without consecutive alphabetic characters. Since the numeric characters (1,2,3,4) can be adjacent without restriction, they can form words of any length. Notice that a word of length \(n\) in \(\Sigma^{*}\) can be formed by either adding a numeric character to a word of length \(n-1\), or by adding a numeric character and an alphabetic character to a word of length \(n-2\). Therefore, we can find that the number of words can be expressed as a sum of previous terms, leading to the recurrence relation.
02

Forming the Recurrence Relation

Since we can add one numeric character to a string of length \(n-1\), or a numeric and alphabetic character to a string of length \(n-2\), we find that: \(A_{n} = 4 * A_{n-1} + 28 * A_{n-2}\), since there are 4 numeric characters and 7 alphabetic characters to choose from in the alphabet.
03

Solving the Recurrence Relation

The solution to this recurrence relation will depend on the initial conditions. Since the sequence starts with single-character words, we can say that: \(A_{1} = 11\) (4 numeric characters, 7 alphabetic characters) and \(A_{2} = 112\) (when forming a two-character string, we can combine any 2 characters without restrictions), these being all the possible strings of length 1 and 2. With this initial setup, we can now calculate \(A_n\) for any \(n > 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alphabet
The term "alphabet" in discrete mathematics refers to a set of symbols or characters used to construct strings or sequences. Typically, alphabets are finite, meaning they contain a specific number of characters.
In the exercise above, the alphabet \( \Sigma \) consists of four numeric characters \( 1, 2, 3, 4 \) and seven alphabetic characters \( a, b, c, d, e, f, g \). This means our alphabet contains a total of \(4 + 7 = 11\) distinct characters.
When we refer to creating words of length \( n \) from this alphabet, we mean that we can form sequences of \( n \) characters, each chosen from the alphabet \( \Sigma \).
  • Numeric characters can be placed next to each other without restriction.
  • Alphabetic characters, however, cannot be consecutive, whether they are identical or distinct.
Understanding the structure and rules of the alphabet is crucial in solving problems related to sequence formation, like this one, where specific conditions affect word creation.
Combinatorics
Combinatorics is an area of mathematics concerned with counting, arrangement, and combination of sets. In this exercise, combinatorics is applied to determine how different sequences or "words" can be formed from the given alphabet, considering certain restrictions.
We use combinatorial methods to find the recurrence relation, which mathematically expresses how to compute the number of valid words of length \( n \).
  • A recurrence relation allows us to express the terms in a sequence based on previous terms, thus making it easier to compute the number of combinations for larger values of \( n \).
  • For instance, the recurrence relation \( A_n = 4 \times A_{n-1} + 28 \times A_{n-2} \) arises from considering the ways numeric and alphabetic characters can combine.
Solving such problems typically involves identifying patterns in how sequences can grow as new characters are added while considering restrictions like non-consecutive alphabetic characters.
Discrete Mathematics
Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. This includes combinatorics, graph theory, and set theory, among others. In the context of this exercise, discrete math is applied to understand the structure and count of sequences formed from a finite alphabet under certain rules.
We have a word formation problem where we need to ensure that no two alphabetic characters are consecutive. This is a typical discrete math problem where understanding constraints and logical structure is key.
  • The initial conditions \( A_1 = 11 \) and \( A_2 = 112 \) provide a starting point for solving the recurrence relation.
  • The challenge lies in using these conditions along with the recurrence formula to predict the number of sequences for larger \( n\).
Ultimately, this exercise demonstrates how discrete mathematics principles can solve seemingly complex combinatorial problems by breaking them down into simpler, countable units.

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Most popular questions from this chapter

Solve the following recurrence relations. (No final answer should involve complex numbers.) a) \(a_{n}=5 a_{n-1}+6 a_{n-2}, \quad n \geq 2, \quad a_{0}=1, \quad a_{1}=3\) b) \(2 a_{n+2}-11 a_{n+1}+5 a_{n}=0, \quad n \geq 0, \quad a_{0}=2, \quad a_{1}=-8\) c) \(3 a_{n+1}=2 a_{n}+a_{n-1}, \quad n \geq 1, \quad a_{0}=7, \quad a_{1}=3\) d) \(a_{n+2}+a_{n}=0, \quad n \geq 0, \quad a_{0}=0, \quad a_{1}=3\) e) \(a_{n+2}+4 a_{n}=0, \quad n \geq 0, \quad a_{0}=a_{1}=1\) f) \(a_{n}-6 a_{n-1}+9 a_{n-2}=0, \quad n \geq 2, \quad a_{0}=5, \quad a_{1}=12\) g) \(a_{n}+2 a_{n-1}+2 a_{n-2}=0, \quad n \geq 2, \quad a_{0}=1, a_{1}=3\)

For \(n \geq 0, b_{n}=\left(\frac{1}{n+1}\right)\left(\begin{array}{c}2 n \\\ n\end{array}\right)\) is the \(n\)th Catalan number. a) Show that for all \(n \geq 0, b_{n+1}=\frac{2(2 n+1)}{(n+2)} b_{n}\). b) Use the result of part (a) to write a computer program (or develop an algorithm) that calculates the first 15 Catalan numbers.

Consider a tennis tournament for \(n\) players, where \(n=2^{k}, k \in \mathbf{Z}^{+}\). In the first round \(n / 2\) matches are played, and the \(n / 2\) winners advance to round 2 , where \(n / 4\) matches are played. This halving process continues until a winner is determined. a) For \(n=2^{k}, k \in \mathbf{Z}^{+}\), let \(f(n)\) count the total number of matches played in the tournament. Find and solve a recurrence relation for \(f(n)\) of the form $$ \begin{aligned} &f(1)=d \\ &f(n)=a f(n / 2)+c, \quad n=2,4,8, \ldots, \end{aligned} $$ where \(a, c\), and \(d\) are constants. b) Show that your answer in part (a) also solves the recurrence relation $$ \begin{aligned} &f(1)=d \\ &f(n)=f(n / 2)+(n / 2), \quad n=2,4,8, \ldots \end{aligned} $$

Find and solve a recurrence relation for the number of ways to park motorcycles and compact cars in a row of \(n\) spaces if each cycle requires one space and each compact needs two. (All cycles are identical in appearance, as are the cars, and we want to use up all the \(n\) spaces.)

$$ \text { a) For } n \geq 0, \text { let } B_{n} \text { denote the number of parti- } $$tions of \(\\{1,2,3, \ldots, n\\}\). Set \(B_{0}=1\) for the partitions of \(\emptyset\). Verify that for all \(n \geq 0\), $$ B_{n+1}=\sum_{i=0}^{n}\left(\begin{array}{c} n \\ n-i \end{array}\right) B_{i}=\sum_{i=0}^{n}\left(\begin{array}{c} n \\ i \end{array}\right) B_{i} $$ [The numbers \(B_{i}, i \geq 0\), are referred to as the Bell numbers after Eric Temple Bell (18831960).] b) How are the Bell numbers related to the Stirling numbers of the second kind?
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