Difference between revisions of "Grammar: Decidable and Undecidable Problems"

(→‎For CFGs G, G1 and G2 and regular set R)
(→‎For CFGs G, G1 and G2 and regular set R)
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===For CFGs G, G1 and G2 and regular set R===
 
===For CFGs G, G1 and G2 and regular set R===
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The following problems are undecidable:
  
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1.) the compliment of <math>(L(G1))^\complement</math> is a CFL
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# <math>(L(G1))^\complement</math> is a CFL
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      2.) L(G1) intersected with L(G2) is a CFL
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# L(G1) \cap L(G2) is a CFL
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3.) L(G1) = R
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# L(G1) \cap L(G2) is empty
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It is undecidable whether an arbitrary CFG is ambiguous
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# L(G1) = R
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    It is undecidable for arbitrary CFG's G1 and G2 whether L(G1) intersected with L(G2) is empty
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# Whether G is ambiguous

Revision as of 18:31, 26 February 2014

Grammar: Decidable and Undecidable Problems
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) is finite</math>
Regular Grammar D D D D D D D
Det. Context Free D D D UD ? UD D
Context Free D D UD UD UD UD D
Context Sensitive D UD UD UD UD UD UD
Recursive D UD UD UD UD UD UD
Recursively Enumerable D UD UD UD UD UD UD


Other Undecidable Problems

For CFGs G, G1 and G2 and regular set R

The following problems are undecidable:

  1. <math>(L(G1))^\complement</math> is a CFL
  2. L(G1) \cap L(G2) is a CFL
  3. L(G1) \cap L(G2) is empty
  4. L(G1) = R
  5. Whether G is ambiguous
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Grammar: Decidable and Undecidable Problems
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) is finite</math>
Regular Grammar D D D D D D D
Det. Context Free D D D UD ? UD D
Context Free D D UD UD UD UD D
Context Sensitive D UD UD UD UD UD UD
Recursive D UD UD UD UD UD UD
Recursively Enumerable D UD UD UD UD UD UD


Other Undecidable Problems[edit]

For CFGs G, G1 and G2 and regular set R[edit]

1.) the compliment of <math>(L(G1))^\complement</math> is a CFL 

     2.) L(G1) intersected with L(G2) is a CFL
3.) L(G1) = R

It is undecidable whether an arbitrary CFG is ambiguous 

   It is undecidable for arbitrary CFG's G1 and G2 whether L(G1) intersected with L(G2) is empty