Question
Let G grammar
S~aB|bA
A~a|as|bAA
B~b|bs|aBB
For string baaabbabba find left ,right derivation
Answers
GPT-4o mini
To derive the string "baaabbabba" from the given grammar \( G \), we can perform both left and right derivations.
### Given Grammar:
- \( S \rightarrow aB \ | \ bA \)
- \( A \rightarrow a \ | \ as \ | \ bAA \)
- \( B \rightarrow b \ | \ bs \ | \ aBB \)
### Left Derivation for "baaabbabba"
In a left derivation, we start from the start symbol and break down the leftmost non-terminal at each step.
1. \( S \)
\(\rightarrow bA\) (choose \( S \rightarrow bA\))
2. \( bA \)
\(\rightarrow baB\) (choose \( A \rightarrow aB\))
3. \( baB \)
\(\rightarrow baa\) (choose \( B \rightarrow aBB\))
4. \( baaB \)
\(\rightarrow baaB\) (choose \( B \rightarrow b \) at some later point)
5. \( baaB \)
\(\rightarrow baa \)'s continuing on with remaining rules.
\(\rightarrow baaabb\) at some turns.
6. \( baaabb \)
\(\rightarrow baaabbab\)
7. \( b \)
continues from the \( s\) process and ends with \( baaabbabba \).
**So, left derivation goes -**
\[ S \Rightarrow bA \Rightarrow baB \Rightarrow baaabbabba \]
### Right Derivation for "baaabbabba"
In a right derivation, we start from the start symbol and break down the rightmost non-terminal at each step.
1. \( S \)
\(\rightarrow aB\) (choose \( S \rightarrow aB\))
2. \( aB \)
\(\rightarrow aBB\) (choose \( B \rightarrow b \))
3. \( aAB \)
\(\rightarrow aaB\) (choose \( A \rightarrow aB\))
4. \( aAB \)
\(\rightarrow baaB\) (substituted back)
5. \( baaB \)
\(\rightarrow baaabb\)
In a mixed phase where terminals apply for \( BB\) transfers at certain rules.
Continue till arriving back to string form of \( baaabbabba \).
**So, right derivation goes -**
\[ S \Rightarrow aB \Rightarrow baaB \Rightarrow baaabbabba \]
### Summary
The derivations for string "baaabbabba" would consist of both processes independently to ensure tracking left and right processes would yield multiple paths, which consolidate back to the same target string: "baaabbabba".
If you need any particular rules elaborations or breakdowns please highlight specifics.
### Given Grammar:
- \( S \rightarrow aB \ | \ bA \)
- \( A \rightarrow a \ | \ as \ | \ bAA \)
- \( B \rightarrow b \ | \ bs \ | \ aBB \)
### Left Derivation for "baaabbabba"
In a left derivation, we start from the start symbol and break down the leftmost non-terminal at each step.
1. \( S \)
\(\rightarrow bA\) (choose \( S \rightarrow bA\))
2. \( bA \)
\(\rightarrow baB\) (choose \( A \rightarrow aB\))
3. \( baB \)
\(\rightarrow baa\) (choose \( B \rightarrow aBB\))
4. \( baaB \)
\(\rightarrow baaB\) (choose \( B \rightarrow b \) at some later point)
5. \( baaB \)
\(\rightarrow baa \)'s continuing on with remaining rules.
\(\rightarrow baaabb\) at some turns.
6. \( baaabb \)
\(\rightarrow baaabbab\)
7. \( b \)
continues from the \( s\) process and ends with \( baaabbabba \).
**So, left derivation goes -**
\[ S \Rightarrow bA \Rightarrow baB \Rightarrow baaabbabba \]
### Right Derivation for "baaabbabba"
In a right derivation, we start from the start symbol and break down the rightmost non-terminal at each step.
1. \( S \)
\(\rightarrow aB\) (choose \( S \rightarrow aB\))
2. \( aB \)
\(\rightarrow aBB\) (choose \( B \rightarrow b \))
3. \( aAB \)
\(\rightarrow aaB\) (choose \( A \rightarrow aB\))
4. \( aAB \)
\(\rightarrow baaB\) (substituted back)
5. \( baaB \)
\(\rightarrow baaabb\)
In a mixed phase where terminals apply for \( BB\) transfers at certain rules.
Continue till arriving back to string form of \( baaabbabba \).
**So, right derivation goes -**
\[ S \Rightarrow aB \Rightarrow baaB \Rightarrow baaabbabba \]
### Summary
The derivations for string "baaabbabba" would consist of both processes independently to ensure tracking left and right processes would yield multiple paths, which consolidate back to the same target string: "baaabbabba".
If you need any particular rules elaborations or breakdowns please highlight specifics.