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# Whether <math>L(R) \subseteq L(G)</math>? | # Whether <math>L(R) \subseteq L(G)</math>? | ||
# Whether <math>L(G)</math> is a CFL? (trivial) | # Whether <math>L(G)</math> is a CFL? (trivial) | ||
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| Grammar | <math>w \in L(G)</math> | <math>L(G) = \phi</math> | <math>L(G) = \Sigma^*</math> | <math>L(G_1) \subseteq L(G_2)</math> | <math>L(G_1) = L(G_2)</math> | <math>L(G_1) \cap L(G_2) = \phi</math> | <math>L(G)</math> is regular? | $L(G)$ is finite? |
|---|---|---|---|---|---|---|---|---|
| Regular Grammar | D | D | D | D | D | D | D | D |
| Det. Context Free | D | D | D | UD | D | UD | D | D |
| Context Free | D | D | UD | UD | UD | UD | UD | D |
| Context Sensitive | D | UD | UD | UD | UD | UD | UD | UD |
| Recursive | D | UD | UD | UD | UD | UD | UD | UD |
| Recursively Enumerable | UD | UD | UD | UD | UD | UD | UD | UD |
Checking if <math>L(CFG)</math> is finite is decidable because we just need to see if <math>L(CFG)</math> contains any string with length between <math>n</math> and <math>2n-1</math>, where <math>n</math> is the pumping lemma constant. If so, <math>L(CFG)</math> is infinite otherwise its finite.
The following problems are undecidable:
But whether <math>L(G) \subseteq L(R)</math> is decidable. (We can test if <math>L(G) \cap compl(L(R))</math> is <math>\phi</math>)
The following problems are decidable:
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