CSE 105 theory of computation Proving

slide1. CSE 105 theory of computation Proving undecidability by diagonalization

Fall 2025

https://cseweb.ucsd.edu/classes/fa25/cse105-a/<br>
slide2. Today's learning goals Sipser Ch 4.1, 4.2 Trace high-level descriptions of algorithms for computational problems.
Use counting arguments to prove the existence of unrecognizable (undecidable) languages.
Use diagonalization in a proof of undecidability.<br>
slide3. Encoding input for TMs Sipser p. 159 By definition, TM inputs are strings

To define TM M:
"On input w …
..
..
… For inputs that aren't strings,
we have to encode the object
(represent it as a string) first Notation:
<O> is the string that represents (encodes) the object O
<O1, …, On> is the single string that represents the tuple of objects O1, …, On<br>
slide4. Encoding inputs Payoff: problems we care about can be reframed as languages of strings

e.g. "Recognize whether a string is a palindrome."
{ w | w in {0,1}* and w = wR }
e.g. "Check whether a string is accepted by a DFA."
{ <B,w> | B is a DFA over Σ, w in Σ*, and w is in L(B) }
e.g. "Check whether the language of a PDA is infinite."
{ <A> | A is a PDA and L(A) is infinite}<br>
slide5. Encoding inputs Payoff: problems we care about can be reframed as languages of strings

e.g. "Recognize whether a string is a palindrome."
{ w | w in {0,1}* and w = wR }
This set is regular and decidable.
This set is regular and not decidable
This set is nonregular and decidable
This set is nonregular and not decidable.
None of the above<br>
slide6. Computational problems A computational problem is decidable iff the language encoding the problem instances is decidable<br>
slide7. Computational problems Sample computational problems and their encodings:
ADFA "Check whether a string is accepted by a DFA."
{ <B,w> | B is a DFA over Σ, w in Σ*, and w is in L(B) }

EDFA "Check whether the language of a DFA is empty."
{ <A> | A is a DFA over Σ, L(A) is empty }

EQDFA "Check whether the languages of two DFAs are equal."
{ <A, B> | A and B are DFA over Σ, L(A) = L(B)}

FACT: all of these problems are decidable!<br>
slide8. Computational problems Sample computational problems and their encodings:
APDA "Check whether a string is accepted by a PDA."
{ <B,w> | B is a PDA over Σ, w in Σ*, and w is in L(B) }

EPDA "Check whether the language of a PDA is empty."
{ <A> | A is a PDA over Σ, L(A) is empty }

EQPDA "Check whether the languages of two PDAs are equal."
{ <A, B> | A and B are PDA over Σ, L(A) = L(B)}

FACT: some of these problems are decidable, and some are not!<br>
slide9. Computational problems Sample computational problems and their encodings:
ATM "Check whether a string is accepted by a TM."
{ <B,w> | B is a TM over Σ, w in Σ*, and w is in L(B) }

ETM "Check whether the language of a TM is empty."
{ <A> | A is a TM over Σ, L(A) is empty }

EQTM "Check whether the languages of two TMs are equal."
{ <A, B> | A and B are TM over Σ, L(A) = L(B)}

FACT: all of these problems are undecidable!<br>
slide10. Undecidable? There are many ways to prove that a problem is decidable.
How do we find (and prove) that a problem is not decidable?<br>
slide11. Before we proved the Pumping Lemma …
We proved there was a set that was not regular because All sets of strings Counting arguments All Regular
Sets Countable Uncountable<br>
slide12. Why is the set of Turing-recognizable languages countable?
It's equal to the set of all TMs, which we showed is countable.
It's a subset of the set of all TMs, which we showed is countable.
Each Turing-recognizable language is associated with a TM, so there can be no more Turing-recognizable languages than TMs.
More than one of the above.
I don't know. All sets of strings Counting arguments All Turing-recognizable sets Countable Uncountable<br>
slide13. Satisfied? Maybe not …

What's a specific example of a language that is not Turing-recognizable? or not Turing-decidable?

Idea: consider a set that, were it to be Turing-decidable, would have to "talk" about itself, and contradict itself!<br>
slide14. ATM Recall ADFA = {<B,w> | B is a DFA and w is in L(B) }
Decider for this set simulates arbitrary DFA

ATM = {<M,w> | M is a TM and w is in L(M) }
Decider for this set simulates arbitrary TMs ???<br>
slide15. ATM ATM = {<M,w> | M is a TM and w is in L(M) }

Define the TM N = "On input <M,w>:
Simulate M on w.
If M accepts, accept. If M rejects, reject."<br>
slide16. ATM ATM = {<M,w> | M is a TM and w is in L(M) }

Define the TM N = "On input <M,w>:
Simulate M on w.
If M accepts, accept. If M rejects, reject." What is L(N)?<br>
slide17. ATM ATM = {<M,w> | M is a TM and w is in L(M) }

Define the TM N = "On input <M,w>:
Simulate M on w.
If M accepts, accept. If M rejects, reject." Does N decide ATM?<br>
slide18. ATM ATM = {<M,w> | M is a TM and w is in L(M) }

Define the TM N = "On input <M,w>:
Simulate M on w.
If M accepts, accept. If M rejects, reject."

Conclusion: ATM is Turing-recognizable.
Is it decidable?<br>
slide19. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that it is.

Call MATM the decider for ATM:

For every TM M and every string w,
Computation of MATM on <M,w> halts and accepts if w is in L(M).
Computation of MATM on <M,w> halts and rejects if w is not in L(M). MATM ≠ N<br>
slide20. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that MATM decides ATM

Define the TM D = "On input <M>:
Run MATM on <M, <M>>.
If MATM accepts, reject; if MATM rejects, accept."<br>
slide21. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that MATM decides ATM

Define the TM D = "On input <M>:
Run MATM on <M, <M>>.
If MATM accepts, reject; if MATM rejects, accept." Which of the following computations halt?
Computation of D on <X>
Computation of D on <Y> where Y is TM with L(Y) =Σ*
Computation of D on <D>
All of the above.<br>
slide22. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that MATM decides ATM

Define the TM D = "On input <M>:
Run MATM on <M, <M>>.
If MATM accepts, reject; if MATM rejects, accept."

Consider running D on input <D>. Because D is a decider:
either computation halts and accepts …
or computation halts and rejects …<br>
slide23. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that MATM decides ATM

Define the TM D = "On input <M>:
Run MATM on <M, <M>>.
If MATM accepts, reject; if MATM rejects, accept."

If D(<D>) accepts, then MATM(<D>,<D>) rejects, which means D(<D>) should have rejected
If D(<D>) rejects, then MATM(<D>,<D>) accepts, which means D(<D>) should have accepted
Either way, we reached a contradiction<br>
slide24. Diagonalization proof: ATM not decidable Sipser 4.11 Assume, towards a contradiction, that MATM decides ATM

Define the TM D = "On input <M>:
Run MATM on <M, <M>>.
If MATM accepts, reject; if MATM rejects, accept."

Consider running D on input <D>. Because D is a decider:
either computation halts and accepts …
or computation halts and rejects … Diagonalization???

Self-reference

"Is <D> an element of L(D)?"<br>
slide25. Why is this called diagonalization?<br>
slide26. Why is this called diagonalization? Rows of table describe all recognizable languages
Any other row (=language) cannot be recognizable
We will construct such a language<br>
slide27. Why is this called diagonalization? Next idea: specialize only to inputs <Mi>

Suffices to construct row (=language) that differs from any row on some of these inputs<br>
slide28. Why is this called diagonalization? Consider the diagonal: running Mi on <Mi>; then flip it

L = {<M>: M doesn’t accept <M>}
L cannot be recognized by any Mi – they disagree on input <Mi>
Since we consider all TMs, L must be un-recognizable<br>
slide29. Another perspective Define language: L = {<M>: M doesn’t accept <M>}

Assume towards a contradiction some TM M decides L
There are two options:
If <M> in L, then M(<M>) accepts, but then <M> not in L
If <M> not in L, then M(<M>) rejects, but then <M> in L

Conclusion: no TM can decide L<br>
slide30. ATM Recognizable
Not decidable

Fact: A language is decidable iff it and its complement are both recgonizable.

Corollary: The complement of ATM is unrecognizable
ATMC= {<M,w> | M is a TM and w is not in L(M) }<br>
slide31. Decidable vs. undecidable Which of the following languages is undecidable?
INFINITEDFA = { <A> | A is a DFA and L(A) is an infinite language}

STM = { <M> | M is a TM and M has exactly 7 states}

RevDFA = { <B> | B is a DFA and for all strings w, B accepts w iff it accepts wR}
RecTM = { <X> | X is a TM and L(X) is recognizable}
DecTM = { <Y> | Y is a TM and L(Y) is decidable}<br>
slide32. So far<br>
slide33. Do we have to diagonalize? Next time: comparing difficulty of problems.<br>