Problem | Description | Computability | Notes
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AFA | (i,j) ⊆ N × N s.t. FAi accepts j | decidable | run DFA |
EFA | i ⊆ N s.t. L(FAi) = ∅ | decidable | non-det guess element in complement |
EQFA | (i,j) ⊆ N × N s.t. L(FAi) =L(FAj) | decidable | emptiness of symmetric difference |
ACFG | (i,j) ⊆ N × N s.t. PDAi accepts j | decidable | Chomsky Normal Form |
ECFG | i ⊆ N s.t. L(CFGi) = ∅ | decidable | marking algorithm |
X | x ∉ X, where X is decidable | decidable |
co-EQCFG | (i,j) ∉ EQCFG | RE | ∃ x s.t. (x,i) ∈ ACFG etc |
EQCFG | (i,j) ⊆ N × N s.t. L(PDAi) = L(PDAj) | non-RE | because not recursive |
ATM | (i,j) ⊆ N × N s.t. Mi(j) halts accepts | RE | run j on Mi |
co-ATM | (i,j) ∉ HALT | non-RE | because not recursive |
HALT | (i,j) ⊆ N × N s.t. Mi(j) halts | RE | run j on Mi |
co-HALT | (i,j) ∉ HALT | non-RE | because not recursive |
ETM | i ⊆ N s.t. L(Mi) = ∅ | non-RE | Rice's theorm |
co-ETM | i ∉ETM | RE | ∃ x s.t. (x,i) ∈ ATM |
EQTM | (i,j) ⊆ N × N s.t. L(Mi)=L(Mj) | non-RE
| ɸi,j(x)=(x!=j)?F:Mi(j), solve EQ(ɸ,∅) |
co-EQTM | (i,j) ∉ EQTM non-RE
| ɸi,j(x)=(x!=j)?T:Mi(j), solve NEQ(ɸ,Ʃ*) | |