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Chapter 6: Problem 1

Let \(\mathbf{\Sigma}=\\{a, b, c, d, e\\} .\) (a) What is \(\left|\mathbf{\Sigma}^{2}\right| ?\left|\mathbf{\Sigma}^{3}\right| ?\) (b) How many strings in \(\mathbf{\Sigma}^{*}\) have length at most five?

Short Answer

Expert verified
The sizes of \( \mathbf{\Sigma}^{2}\) and \( \mathbf{\Sigma}^{3}\) are 25 and 125 respectively. There are 3906 strings in \( \mathbf{\Sigma}^{*}\) with a length of at most five.

Step by step solution

01

Find the size of \( \mathbf{\Sigma}^{2}\) and \( \mathbf{\Sigma}^{3}\)

Use the product principle. The size of \( \mathbf{\Sigma}^{2}\) is \( |\mathbf{\Sigma}|^{2}\), and the size of \( \mathbf{\Sigma}^{3}\) is \( |\mathbf{\Sigma}|^{3}\), where \( |\mathbf{\Sigma}|\) is the size of the alphabet, which is 5 in this case. Hence, \( \left|\mathbf{\Sigma}^{2}\right| =5^{2} = 25\), and \( \left|\mathbf{\Sigma}^{3}\right| =5^{3} = 125\).
02

Determine the number of strings of at most length five

To find the total amount of the strings of length from 0 to 5, we can use the sum rule and realize that the number of strings of certain length 'n' is \( |\mathbf{\Sigma}|^{n}\). So, the total number of strings of at most length five from symbol set \(\mathbf{\Sigma}\) is \(1 + |\mathbf{\Sigma}| + |\mathbf{\Sigma}|^{2} + |\mathbf{\Sigma}|^{3} + |\mathbf{\Sigma}|^{4} + |\mathbf{\Sigma}|^{5} = 1 + 5 + 25 + 125 + 625 + 3125 = 3906\). Note: '1' is for the empty string that also included in \(\mathbf{\Sigma}^{*}\).

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Key Concepts

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

Alphabet
An alphabet in combinatorics refers to a set of symbols or characters that are used to construct strings. In our exercise, the alphabet is represented by the Greek letter \(\mathbf{\Sigma}\), and it consists of the five characters \(\{a, b, c, d, e\}\).
These characters are the building blocks for forming different strings of varying lengths.

When dealing with problems involving alphabets, it is crucial to understand how these symbols can be combined to create sequences. The size of the alphabet, which is the total number of unique symbols it has, plays a significant role in determining how many different strings or combinations can be made.
Strings
Strings are sequences of characters derived from a given alphabet. In this context, strings are formed using the alphabet \(\{a, b, c, d, e\}\). The exercises often ask about the number of possible strings of a certain length.
For example, a string of length 2, such as "ab" or "de", is formed by picking any two characters from the alphabet. The strings can vary greatly in terms of their length, which adds to the calculation of their possible numbers.

Strings play a critical role in combinatorics as they are outcomes that we can count, and they demonstrate how symbols can be systematically arranged.
Product Principle
The product principle is a fundamental rule in combinatorics used to calculate the total number of outcomes when events are independent and sequential. In the context of our exercise, it is used to calculate the number of strings of a given length.
Let's say we need to find the number of strings of length 3 formed from our alphabet \(\{a, b, c, d, e\}\). For each position in the string, there are 5 possible choices, since there are 5 characters in the alphabet. Hence, the total number of strings of length 3 is calculated as \(|\mathbf{\Sigma}|^3\).
This gives us \(5^3 = 125\) possible strings, showcasing how the product principle efficiently gives us the count by multiplying the number of choices for each stage in the sequence.
Sum Rule
The sum rule in combinatorics is used to determine the total number of outcomes for events that cannot occur simultaneously. It is especially useful for calculating total outcomes over different scenarios.
In the exercise, the sum rule helps find the number of strings with lengths at most 5. The idea is to use the sum rule to add together the number of strings of each length from 0 to 5.
  • The count for strings of length 0 (empty string) is \(1\).
  • For length 1, the product principle gives us \(5\).
  • For length 2, it's \(25\) (calculated as \(5^2\)).
  • For length 3, it's \(125\) (calculated as \(5^3\)).
  • For length 4, it's \(625\) (calculated as \(5^4\)).
  • Finally, for length 5, it's \(3125\) (calculated as \(5^5\)).
Adding these up as per the sum rule, we get a total of \(1 + 5 + 25 + 125 + 625 + 3125 = 3906\) strings, capturing all strings with lengths up to five.

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Most popular questions from this chapter

For the alphabet \(\mathbf{\Sigma}=\\{0,1\\}\), let \(A, B, C \subseteq \mathbf{\Sigma}^{*}\) be the following languages: $$ \begin{aligned} &A=\\{0,1,00,11,000,111,0000,1111\\}, \\ &B=\left\\{w \in \mathbf{\Sigma}^{*} \mid 2 \leq\|w\|\right\\}, \\ &C=\\{w \in \mathbf{\Sigma} * \mid 2 \geq\|w\|\\} . \end{aligned} $$ Determine the following subsets (languages) of \(\mathbf{\Sigma}^{*}\). a) \(A \cap B\) b) \(A-B\) c) \(A \triangle B\) d) \(A \cap C\) e) \(B \cap C\) f) \(B \cup C\) g) \(\overline{(A \cap C)}\) h) \(\bar{A} \cap \bar{C}\) i) \(\bar{A} \cap B\)

Show that it is not possible to construct a finite state machine that recognizes precisely those sequences in the language \(A=\left\\{0^{\prime} 1^{\prime} \mid i, j \in \mathbf{Z}^{+}, i>j\right\\} .\) (Here the alphabet for \(A\) is \(\mathbf{\Sigma}=\\{0,1\\} .)\)

Let \(\mathbf{\Sigma}\) be the alphabet \(\\{0,1\\}\), and let \(A \subseteq \mathbf{\Sigma}^{*}\) be the language defined recursively as follows: 1) The symbols 0,1 are both in \(A\)-this is the base for our definition; and, 2) For any word \(x\) in \(A\), the word \(0 x 1\) is also in \(A\)-this constitutes the recursive process. a) Find four different words-two of length 3 and two of length 5 -in \(A\). b) Use the given recursive definition to show that 0001111 is in \(A\). c) Explain why 00001111 is not in \(A\).

Let \(\mathbf{\Sigma}\) be an alphabet. Let \(x_{i} \in \mathbf{\Sigma}\) for \(1 \leq i \leq 100\) (where \(x_{i} \neq x_{j}\) for any \(1 \leq i

Provide a recursive definition for each of the following languages \(A \subseteq \mathbf{\Sigma}^{*}\) where \(\mathbf{\Sigma}=\\{0,1\\} .\) a) \(x \in A\) if (and only if) the number of 0 's in \(x\) is even. b) \(x \in A\) if (and only if) there is only one 0 in \(x\). c) \(x \in A\) if (and only if) all of the 1 's in \(x\) precede all of the 0 's.
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