Example 3.5 For E = {0,1}, give a regular expression r such that L (r) = {w E £* : w has at least one pair of consecutive zeros}. %3D One can arrive at an answer by reasoning something like this: Every string in L (r) must contain 00 somewhere, but what comes before and what goes after is completely arbitrary. An arbitrary string on {0,1} can be denoted by (0+ 1)*. Putting these observations together, we arrive at the solution r = (0+1)* 00 (0+1)* .

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
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Does the expression ((0 + 1) (0 + 1)∗)∗00 (0 + 1)∗ denote the language in Example 3.5?

**Example 3.5**

For Σ = {0,1}, give a regular expression r such that

\[ L(r) = \{w \in \Sigma^* : w \text{ has at least one pair of consecutive zeros}\}. \]

One can arrive at an answer by reasoning something like this: Every string in \( L(r) \) must contain 00 somewhere, but what comes before and what goes after is completely arbitrary. An arbitrary string on \(\{0,1\}\) can be denoted by \((0 + 1)^*\). Putting these observations together, we arrive at the solution

\[ r = (0 + 1)^* \, 00 \, (0 + 1)^* .\]
Transcribed Image Text:**Example 3.5** For Σ = {0,1}, give a regular expression r such that \[ L(r) = \{w \in \Sigma^* : w \text{ has at least one pair of consecutive zeros}\}. \] One can arrive at an answer by reasoning something like this: Every string in \( L(r) \) must contain 00 somewhere, but what comes before and what goes after is completely arbitrary. An arbitrary string on \(\{0,1\}\) can be denoted by \((0 + 1)^*\). Putting these observations together, we arrive at the solution \[ r = (0 + 1)^* \, 00 \, (0 + 1)^* .\]
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