CS 461 – Nov. 9

• Chomsky hierarchy of language classes– Review– Let’s find a language outside the TM world!– Hints: languages and TM are countable, but the set

of all languages is uncountable

• Undecidability– We have seen that ATM is undecidable.– Let’s look at some more undecidable problems.

TM properties

• 2 kinds of TMs 2 kinds of languages.– Turing-recognizable (a.k.a. recursively enumerable)

• Example: ATM

– Decidable (a.k.a. recursive)• Example: 0*

• If L is decidable, then L’ is decidable.• If L and L’ are both Turing-recognizable,

then L is decidable. (since either L or L’ must accept)

• Therefore, the complement of ATM is not even Turing recognizable.

Decidable?

for n = 3 .. maxint:

for each combo of (x, y, z) starting at 1:

if (xn + yn == zn)

print (“hello, world”);

• Are we able to tell if this program will print “hello, world” or not?– Only since 1993 . For 300 years we didn’t know.– Solving this problem in general (for any program)

must be hard… but what if it could be done?

“hello, world” tester

• Suppose H can decide the question.• Then we can create similar algorithm H2.

– Returns “hello, world” instead of “no”.– Let’s run H2 on itself!

H

program & inputPrint “yes”

Print “no”

H2

H2Print “yes”

Print “hello, world”

Regular?

• REG = “Is there an algorithm to determine if a TM recognizes a regular language?”

• Proof by contradiction, p. 191• Assume REG is decidable.

– Then, TM “R” that can tell if an input <TM> recognizes a regular language.

– We’ll use R to solve the ATM problem by creating a decider S.

– In other words:

Machine R solves REG -- which implies existence of ….

Machine S which solves ATM.

– Do we understand the I/O of R and S?

Regular, continued

• Here is how to build machine S, which will “solve” ATM.

• Input to S is <T, w>.

• Create a new TM “M2”– M2 sees if its input, x, is of the form 0n1n. If so, M2 accepts x.– If x is not of the form 0n1n,

• M2 runs T on w.• If T accepts w, M2 accepts x.

– Thus, L(M2) is either Σ*, if T accepts w, or 0n1n.– The only way for M2 to accept Σ* is for T to accept w.

• Feed M2 into R.– If R says that M2 accepts a regular language, we can conclude T

accepts w, so S can say “yes”, i.e. accept. Otherwise S rejects.– But S can’t exist. So we have a contradiction. REG undecidable.

Rice’s Theorem

• Any question Q about a TM language is undecidable.• Proof by contradiction:

• Assume Q is decidable. Then TM “QD” that – always halts, and – answers Q for any input TM it receives.

• Using QD, we can build a machine AD to decide ATM.

• (But we know that AD can’t exist, therefore QD can’t exist. That is the contradiction.)

Building TM “AD”• AD will do the following:

– Its input is <T, w>.– Build a TM “Q1” that satisfies Q.– Build a TM “M2” whose input is x.

• M2 runs T on w. If T loops/rejects, M2 will loop/reject. • If T accepts w, next run Q1 on x. M2 accepts if Q1 accepts.• Note that L(M2) = L(Q1) if T accepts w. L(M2) = if T does

not accept w.• In other words, if T accepts, then M2 satisfies Q.

– Feed M2 into QD. If QD accepts, AD accepts. Otherwise AD rejects.

• Thus: QD is a decider AD is a decider.

Glossary to proof

• Q = some question or property about a TM language• QD = a TM that can decide Q• Q1 = a TM for which Q is true, built inside AD

• ATM = the acceptance question for TM’s, already known to be undecidable

• AD = a hypothetical decider for ATM

• T = the TM input to AD• w = the corresponding word also input to AD• M2 = a second TM built inside AD

• Rice’s theorem is a little complex because 5 TM’s!

Examples

• Let Q = “Is the language infinite?” or ….• “Does the language include 001?”• “Is the language finite?”• “Is the language nonregular?”

• Can also prove by contradiction:– Is it decidable to determine if one TM language is the

subset of another?