CSE 105 THEORY OF COMPUTATION Fall 2016 - - PowerPoint PPT Presentation

CSE 105

THEORY OF COMPUTATION

Fall 2016 http://cseweb.ucsd.edu/classes/fa16/cse105-abc/

T

• day’s lecture
• Examples of undecidable languages
• Examples of decidable languages
• Reductions
• Reading: Chapter 5

Previously …

• Diag = {<M> | M is a TM s.t. <M> is not in L(M) }

–

Undecidable (in fact, not even RE)

• DD = {<M,<M>> | M is a TM s.t. <M> is not in L(M) }

–

Just like Diag: Undecidavle (in fact, not even RE)

–

Proof: Assume DD is decidable. It follows that Diag is decidable. Contradiction!

• ATM = { <M,w> | M is a TM such that M(w) accepts }

–

ATM is RE, but not Decidable. Therefore, also not coRE.

–

Proof: Assume ATM is decidable. It follows that MM is decidable. Contradiction!

ATM is undecidable (Proof details)

• Assume for contradiction ATM is decidable
• MM = {<M,<M>> | M is a TM} is decidable
• MM - ATM is decidable because the class of decidable

languages is closed under set difgerence.

• Notice: MM – ATM

= { <M,<M>> | M is a TM} - {<M,w> | w ∈ L(M) } = { <M,<M>> | M is a TM and <M> ∉ L(M) } = DD

• So, DD is decidable. Contradiction!

Some example languages

• Acceptance problem:

–

ADFA = { <M,w> | M is a DFA and M(w) accepts }

–

ATM = { <M,w> | M is a TM and M(w) accepts }

• Emptyness problem:

–

EDFA = {<M> | M is a DFA and L(M) is the empty set }

–

ETM = {<M> | M is a TM and L(M) is the empty set }

• Equivalence problem:

–

EQDFA = { <M,M’> | M and M’ are DFAs and L(M) = L(M’) }

Some example languages

• Acceptance problem:

–

ADFA = { <M,w> | M is a DFA and M(w) accepts } DECIDABLE

–

ATM = { <M,w> | M is a TM and M(w) accepts } UNDECIDABLE

• Emptyness problem:

–

EDFA = {<M> | M is a DFA and L(M) is the empty set } DECIDABLE

–

ETM = {<M> | M is a TM and L(M) is the empty set } UNDECIDABLE

• Equivalence problem:

–

EQDFA = { <M,M’> | M and M’ are DFAs and L(M) = L(M’) } DECIDABLE

EDFA is decidable

• PEDFA(w) =
• 1. Parse input w as <Q,Σ,δ,s,F>. If parse fails, then reject.
• 2. Let X = {s}
• 3. For all q in X and a in Σ

a) Let q’ = δ(q,a) b) If q’ is not in X, then X ← X U {q’} and restart the loop at 3)

• 4. If X intersects F, then accept, else reject.

EDFA is decidable

• PEDFA(w) =
• 1. Parse input w as <Q,Σ,δ,s,F>. If parse fails, then reject.
• 2. Let X = {s}
• 3. For all q in X and a in Σ

a) Let q’ = δ(q,a) b) If q’ is not in X, then X ← X U {q’} and restart the loop at 3)

• 4. If X intersects F, then accept, else reject.

EDFA

EDFA is decidable

• PEDFA(w) =
• 1. Parse input w as <Q,Σ,δ,s,F>. If parse fails, then reject.
• 2. Let X = {s}
• 3. For all q in X and a in Σ

a) Let q’ = δ(q,a) b) If q’ is not in X, then X ← X U {q’} and restart the loop at 3)

• 4. If X intersects F, then accept, else reject.

What is the language of PEDFA? A) {<P,w> | P is a DFA and P(w) accepts} B) {<P> | P is a DFA} C) {<P> | P is a DFA and L(P) = Ø } D) {<P> | P is a DFA and L(P) = {ε} } E) None of the above

EDFA is decidable

• PEDFA(w) =
• 1. Parse input w as <Q,Σ,δ,s,F>. If parse fails, then reject.
• 2. Let X = {s}
• 3. For all q in X and a in Σ

a) Let q’ = δ(q,a) b) If q’ is not in X, then X ← X U {q’} and restart the loop at 3)

• 4. If X intersects F, then reject

reject, else accept accept

EQDFA is decidable

• PEQDFA(<M,M’>)

1) Check if both M and M’ are DFA. If not, then reject. 2) Use closure properties of regular languages to build a DFA M’’ for L(M’’) = (L(M) – L(M’)) U (L(M’) - L(M))

–

Run PEDFA(<M’’>). If PEDFA accepts, then accept, else reject.

• Notice L(M)=L(M’) if and only if L(M’’) is empty.

Summary

• DD is decidable → Diag is decidable

–

Since Diag is undecidable, then DD is also undecidable

• ATM is decidable → DD is decidable

–

Since DD is undecidable, then ATM is undecidable

• EDFA is decidable → EQDFA is decidable

–

Since EDFA is decidable, then EQDFA is decidable

Proving implications

• Claim: If A is decidable, then B is decidable
• Proof:

–

Assume A is decidable

–

Show that B is also decidable

• Proof methods:

–

Using closure properties of decidable languages: transform A into B

–

Let M be a decider for A. Use M to build a decider for B

• Notice: We do not need to know if A is decidable

Reductions

• “If A is decidable then B is decidable”

–

Can be proved using closure properties, or assuming a decider for A

–

Makes sense even when A or B are undecidable

• When decidability of B is unknown

–

This reduces the problem of solving B to the (possibly easier) problem of solving A

• Proof is called a “reduction” from B to A

–

Oftern written B < A

Reductions

• “If A is decidable then B is decidable”

–

Can be proved using closure properties, or assuming a decider for A

–

Makes sense even when A or B are undecidable

• When decidability of B is unknown

–

This reduces the problem of solving B to the (possibly easier) problem of solving A

• Proof is called a “reduction” from B to A

–

Oftern written B B < < A A Assume B<A B<A and A is decidable. What can you conclude? A) B is also decidable B) B is a subset of A C) B is undecidable D) B may be decidable or undecidable E) I don’t know

Reductions

• “If A is decidable then B is decidable”

–

Can be proved using closure properties, or assuming a decider for A

–

Makes sense even when A or B are undecidable

• When decidability of B is unknown

–

This reduces the problem of solving B to the (possibly easier) problem of solving A

• Proof is called a “reduction” from B to A

–

Oftern written B B < < A A Assume A<B A<B and A is decidable. What can you conclude? A) B is also decidable B) B is a subset of A C) B is undecidable D) B may be decidable or undecidable E) I don’t know

Reductions

• “If A is decidable then B is decidable”

–

Can be proved using closure properties, or assuming a decider for A

–

Makes sense even when A or B are undecidable

• When decidability of B is unknown

–

This reduces the problem of solving B to the (possibly easier) problem of solving A

• Proof is called a “reduction” from B to A

–

Oftern written B B < < A A Assume A<B A<B and A is undecidable. What can you conclude? A) B is also decidable B) B is a subset of A C) B is undecidable D) B may be decidable or undecidable E) I don’t know

Reductions

• “If A is decidable then B is decidable”

–

Can be proved using closure properties, or assuming a decider for A

–

Makes sense even when A or B are undecidable

• When decidability of B is unknown

–

This reduces the problem of solving B to the (possibly easier) problem of solving A

• Proof is called a “reduction” from B to A

–

Oftern written B B < < A A Assume B<A B<A and A is undecidable. What can you conclude? A) B is also decidable B) B is a subset of A C) B is undecidable D) B may be decidable or undecidable E) I don’t know

Some problems on CFG

• EQCFG = {<G1,G2> | G1,G2 CFG s.t. L(G1) = L(G2) }
• SUBCFG = {<G1,G2> | G1,G2 CFG s.t. L(G1) ⊆ L(G2) }
• SUPCFG = {<G1,G2> | G1,G2 CFG s.t. L(G1) ⊇ L(G2) }
• Can you give reductions between any two of these

problems? In what direction?

–

EQCFG < SUBCFG ?

–

SUBCFG < SUPCFG ?

–

SUPCFG < EQCFG ?

Next Time

• We have class on Wednesday
• Thursday and Friday: Thanksgiving!
• Haskell 4 due tonight
• HW7 due T

ue Nov 29

• Last week of classes after Thanksgiving