Cfg to pda equivalnce 3
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The document summarizes how to construct a context-free grammar G that generates the same language as a given finite state automaton. It explains that if the transition function of the automaton replaces the stack symbol X with a string like YZ when reading input a, then the grammar must generate productions that erase Y and Z in sequence. It also notes that if the transition replaces X with a string of k symbols, the grammar must generate a family of productions with k-1 intermediate states.
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![Next simplest case: 隆(p, a, X) contains (r, Y)
for some state r and symbol Y.
G has production [pXq] -> a[rYq].
We can erase X and go from p to q by reading a
(entering state r and replacing the X by Y) and
then reading some w that gets P from r to q while
erasing the Y.
Note: [pXq] =>* aw whenever [rYq]
=>* w.
2](https://image.slidesharecdn.com/cfgtopdaequivalnce3-190130105154/85/Cfg-to-pda-equivalnce-3-2-320.jpg)
![Since we do not know state s, we must
generate a family of productions:
[pXq] -> a[rYs][sZq]
for all states s.
[pXq] =>* awx whenever [rYs] =>* w and
[sZq] =>* x.
5](https://image.slidesharecdn.com/cfgtopdaequivalnce3-190130105154/85/Cfg-to-pda-equivalnce-3-5-320.jpg)
![Suppose 隆(p, a, X) contains (r, Y1,Yk) for
some state r and k > 3.
Generate family of productions
[pXq] ->
a[rY1s1][s1Y2s2][sk-2Yk-1sk-1][sk-1Ykq]
6](https://image.slidesharecdn.com/cfgtopdaequivalnce3-190130105154/85/Cfg-to-pda-equivalnce-3-6-320.jpg)
![We can prove that (q0, w, Z0)*(p, 竜, 竜) if and
only if [q0Z0p] =>* w.
Proof is in text; it is two easy inductions.
But state p can be anything.
Thus, add to G another variable S, the start
symbol, and add productions S ->
[q0Z0p] for each state p.
7](https://image.slidesharecdn.com/cfgtopdaequivalnce3-190130105154/85/Cfg-to-pda-equivalnce-3-7-320.jpg)
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- 1. Prof.Neeraj Bhargava Abhishek Kumar Department of Computer Science School of Engineering & System Sciences, MDS, University Ajmer, Rajasthan, India 1
- 2. Next simplest case: 隆(p, a, X) contains (r, Y) for some state r and symbol Y. G has production [pXq] -> a[rYq]. We can erase X and go from p to q by reading a (entering state r and replacing the X by Y) and then reading some w that gets P from r to q while erasing the Y. Note: [pXq] =>* aw whenever [rYq] =>* w. 2
- 3. Third simplest case: 隆(p, a, X) contains (r, YZ) for some state r and symbols Y and Z. Now, P has replaced X by YZ. To have the net effect of erasing X, P must erase Y, going from state r to some state s, and then erase Z, going from s to q. 3
- 4. 4 p X a w x r Y Z w x s Z x q
- 5. Since we do not know state s, we must generate a family of productions: [pXq] -> a[rYs][sZq] for all states s. [pXq] =>* awx whenever [rYs] =>* w and [sZq] =>* x. 5
- 6. Suppose 隆(p, a, X) contains (r, Y1,Yk) for some state r and k > 3. Generate family of productions [pXq] -> a[rY1s1][s1Y2s2][sk-2Yk-1sk-1][sk-1Ykq] 6
- 7. We can prove that (q0, w, Z0)*(p, 竜, 竜) if and only if [q0Z0p] =>* w. Proof is in text; it is two easy inductions. But state p can be anything. Thus, add to G another variable S, the start symbol, and add productions S -> [q0Z0p] for each state p. 7