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

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{{alert|Please don't byheart this table. This is just to check your understanding |alert-danger}}
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|+ align="top"|Grammar: Decidable and Undecidable Problems
 
|+ align="top"|Grammar: Decidable and Undecidable Problems
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! <math>L(G_1) = 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_1) \cap L(G_2) = \phi</math>
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! <math>L(G)</math> is finite
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! <math>L(G)</math> is regular?
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! $L(G)$ is finite?
 
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|Regular Grammar
 
|Regular Grammar
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| {{D}}
 
| {{D}}
 
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| {{D}}
 
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| {{D}}
 
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|Context Sensitive  
 
|Context Sensitive  
 
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| {{D}}
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| {{UD}}
 
| {{UD}}
 
| {{UD}}
 
| {{UD}}
 
| {{UD}}
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|Recursive
 
|Recursive
 
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|Recursively Enumerable
 
|Recursively Enumerable
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Checking if  L(CFG) is finite is decidable because we just need to see if L(CFG) contains any string with length between n and 2n-1 where n is the pumping lemma constant. If so,  L(CFG)  is infinite otherwise its finite.
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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.
  
 
==Other Undecidable Problems ==
 
==Other Undecidable Problems ==
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[http://www.cis.upenn.edu/~jean/gbooks/PCPh04.pdf Proofs]
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===For arbitrary CFGs <math>G</math>, <math>G_1</math> and <math>G_2</math> and an arbitrary regular expression <math>R</math>===
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The following problems are '''undecidable''':
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# Whether <math>(L(G_1))^\complement</math> is a CFL?
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# Whether <math>L(G_1) \cap L(G2)</math> is a CFL? (undecidable for DCFG also)
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# Whether <math>L(G_1) \cap L(G2)</math> is empty? (undecidable for DCFG also)
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# Whether <math>L(G) = L(R)</math>?
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# Whether <math>L(R) \subseteq L(G)</math>?
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# Whether <math>G</math> is ambiguous?
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# Whether <math>L(G)</math> is a DCFL?
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# Whether <math>L(G)</math> is a regular language?
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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>)
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===For arbitrary DCFGs <math>G</math>, <math>G_1</math> and <math>G_2</math> and an arbitrary regular expression <math>R</math>===
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The following problems are '''decidable''':
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# Whether <math>(L(G_1))^\complement</math> is a DCFL? (trivial)
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# Whether <math>L(G) = L(R)</math>?
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# Whether <math>L(G) \subseteq (R)</math>?
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# Whether <math>L(R) \subseteq L(G)</math>?
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# Whether <math>L(G)</math> is a CFL? (trivial)
  
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===For arbitrary CFGs G, G1 and G2 and an arbitrary regular set R===
 
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The following problems are undecidable:
 
  
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# Whether <math>(L(G1))^\complement</math> is a CFL
 
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# Whether <math>L(G1) \cap L(G2)</math> is a CFL
 
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# Whether <math>L(G1) \cap L(G2)</math> is empty
 
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# Whether <math>L(G) = R</math>
 
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# Whether <math>L(G) \subseteq R</math>
 
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# Whether <math>G</math> is ambiguous
 
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# Whether <math>L(G)</math> is a DCFL
 
  
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But whether <math>R \subseteq L(G)</math> is decidable
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{{Template:FBD}}
  
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===For arbitrary DCFGs G, G1 and G2 and an arbitrary regular set R===
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[[Category: Automata Theory Notes]]
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The following problems are decidable:
 
  
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# Whether <math>(L(G1))^\complement</math> is a DCFL
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[[Category: Compact Notes for Reference of Understanding]]
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# Whether <math>L(G) = R</math>
 
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# Whether <math>L(G) \subseteq R</math>
 
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# Whether <math>R \subseteq L(G)</math>
 
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# Whether <math>L(G)</math> is a CFL
 

Latest revision as of 16:01, 9 January 2016

Heads Up! Please don't byheart this table. This is just to check your understanding
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)</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.

Other Undecidable Problems

Proofs

For arbitrary CFGs <math>G</math>, <math>G_1</math> and <math>G_2</math> and an arbitrary regular expression <math>R</math>

The following problems are undecidable:

  1. Whether <math>(L(G_1))^\complement</math> is a CFL?
  2. Whether <math>L(G_1) \cap L(G2)</math> is a CFL? (undecidable for DCFG also)
  3. Whether <math>L(G_1) \cap L(G2)</math> is empty? (undecidable for DCFG also)
  4. Whether <math>L(G) = L(R)</math>?
  5. Whether <math>L(R) \subseteq L(G)</math>?
  6. Whether <math>G</math> is ambiguous?
  7. Whether <math>L(G)</math> is a DCFL?
  8. Whether <math>L(G)</math> is a regular language?

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>)

For arbitrary DCFGs <math>G</math>, <math>G_1</math> and <math>G_2</math> and an arbitrary regular expression <math>R</math>

The following problems are decidable:

  1. Whether <math>(L(G_1))^\complement</math> is a DCFL? (trivial)
  2. Whether <math>L(G) = L(R)</math>?
  3. Whether <math>L(G) \subseteq (R)</math>?
  4. Whether <math>L(R) \subseteq L(G)</math>?
  5. Whether <math>L(G)</math> is a CFL? (trivial)





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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)</math> is finite
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

Checking if L(CFG) is finite is decidable because we just need to see if L(CFG) contains any string with length between n and 2n-1 where n is the pumping lemma constant. If so, L(CFG) is infinite otherwise its finite.

Other Undecidable Problems[edit]

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

The following problems are undecidable:

  1. Whether <math>(L(G1))^\complement</math> is a CFL
  2. Whether <math>L(G1) \cap L(G2)</math> is a CFL
  3. Whether <math>L(G1) \cap L(G2)</math> is empty
  4. Whether <math>L(G) = R</math>
  5. Whether <math>L(G) \subseteq R</math>
  6. Whether <math>G</math> is ambiguous
  7. Whether <math>L(G)</math> is a DCFL

But whether <math>R \subseteq L(G)</math> is decidable

For arbitrary DCFGs G, G1 and G2 and an arbitrary regular set R[edit]

The following problems are decidable:

  1. Whether <math>(L(G1))^\complement</math> is a DCFL
  2. Whether <math>L(G) = R</math>
  3. Whether <math>L(G) \subseteq R</math>
  4. Whether <math>R \subseteq L(G)</math>
  5. Whether <math>L(G)</math> is a CFL