Define a language that is not a regular language using a context-free grammar that is not ambiguous. The alphabet of the language must contain at least five characters. Describe in English the strings of the language. Choose any string that is in the language and demonstrate that it is in the language by showing the left-most derivation that derives it. Draw the parse tree that corresponds to the left-most derivation. Give an example of one string that is not in the language.
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To define a language that is not a regular language using a context-free grammar (CFG) that is not ambiguous, we can consider the language L = {a^n b^n c^n | n ≥ 0}. This language consists of strings with an equal number of "a"s, "b"s, and "c"s, where the number of characters of each type is the same.
Now, let's construct a CFG for this language:
1. Start with the production rule: S → A
2. Add the following production rules:
- A → aBc
- B → aBc | ε (epsilon represents an empty string)
Using this CFG, we can describe the strings in the language as follows:
- The string "abc" is in the language because it follows the pattern a^n b^n c^n, where n = 1. The left-most derivation would be:
S → A → aBc → abc
The parse tree for the derivation of the string "abc" would be as follows:
S
|
A
/ \
a B
/ \
a c
Now, an example of a string that is not in the language would be "aabbcc". This string violates the requirement of having equal numbers of "a"s, "b"s, and "c"s, thus making it not part of the language L.
To define a language that is not regular using a context-free grammar, one way is to consider a language that requires counting or matching nested elements. Let's define a language L using the following context-free grammar:
1. Start symbol: S
2. Terminal symbols: a, b, c, d, e
3. Non-terminal symbols: A, B, C, D, E
4. Production rules:
- S -> AA | AB
- A -> aAb | B
- B -> cB | e
In this grammar, the alphabet of the language consists of the characters a, b, c, d, and e.
Description of the strings in the language:
- Any string in L will have an equal number of a's and b's in any order, followed by any number of c's, and possibly ending with an e.
Example string in the language: abbccddee
Left-most derivation for the string:
S -> AA -> AAB -> AbbB -> abbB -> abbCcB -> abbccB -> abbccdBe -> abbccddee
Parse tree for the left-most derivation:
```
S
/ \
A A
/ \ \
A B B
| | |
a b c
| |
b B
|
c
|
c
|
d
|
B
|
e
```
Example of a string not in the language: abcde
Explanation: The string abcde has an unequal number of a's and b's, violating the production rule A -> aAb | B, which requires an equal number of a's and b's. Therefore, it is not in the language L.