1. Remaining Topics Decidability Concept 4.1 The Halting Problem 4.2
P vs. NP and 7.3
NP-completeness & Cook-Levin Theorem

2. Review: Turing Machines in a nutshell
Church-Turing Thesis
Turing Machine equal Notion of an Algorithm
Turing Machine
Most simple machine possible
Computational power of
modern computer and
high-level language
Not particularly efficient in a practical sense

3. Review: Turing Machines in a nutshell
Most simply model possible…
Adding another tape can improve
efficiency (time or computational “speed”)
but not computational “power” (ability to solve a problem).
Every multi-tape TM has an equivalent single-tape TM (Theorem 3.13, p.149)
Similar to an Automata Non-determinism does NOT add any “computational power” (Theorem 3.15, p.150)

4. Review: Turing Machines in a nutshell
Computational power of
modern computer and
high-level language
Every operation and statement in a high level language can be implemented with a Turing Machine (TM)
Just as statements can be combined So can TMs (HW5 illustrates this)

5. Decidability 4.1 Is there an algorithm that can decide if
An item is in a set.
A string is in a language
A formula is a member of a theory
These are all variations of the same concept, i.e., the concept of decidability

6. Decidability in Languages
We will concentrate on this:
Algorithms for deciding if a string is in a language
But, the strings and languages are going to represent deeper problems
ADFA = {<B,w> | B is a DFA that accepts input string w}

7. ADFA ADFA = {<B,w> | B is a DFA that accepts input string w}
B is the encoding of a DFA
Remember that you can encode a DFA as follows:
B = (Q,Σ, δ, qstart, F)
We are literally encoding the machine and the input (w) as a string “<({1,2,3},{a,b},{(1,2,a),(1,3,b)},1,{3}),abc>”

8. ADFA ADFA = {<B,w> | B is a DFA that accepts input string w}
Testing whether a DFA accepts an input w is the same as the problem of testing whether the string <B,w> is a member of the language ADFA
Just as A = {w | w = (11)*} would accept the set {ε, 11, 1111, , …}
ADFA would enumerate all the <B,w>’s such that w is accepted by the encoded B.

9. ADFA is decidable
ADFA = {<B,w> | B is a DFA that accepts input string w}
What does this mean in plain English?
How can we prove it?

10. ADFA is decidable
ADFA = {<B,w> | B is a DFA that accepts input string w}
What does this mean in plain English?
“An algorithm exists that can accept strings that adhere to the definition of ADFA and reject string that don’t”

11. ADFA is decidable
ADFA = {<B,w> | B is a DFA that accepts input string w}
How can we prove it?
Proof is on p.167

12. ANFA is decidable
ANFA = {<B,w> | B is a NFA that accepts input string w}
How do we know this to be true?
Hint: How are NFAs and DFAs different?

13. AREX is decidable
AREX = {<R,w> | R is a Regular Expression that generates the string w}
How do we know this to be true?
Hint: How are Regular Expressions and DFAs related?

14. EDFA is decidable EDFA = {<A> | A is a DFA and L(A) is empty}
Prove it
Hint: Just as Turing Machine can “simulate” a DFA it can also determine if a state is unreachable.

15. EQDFA is decidable
EQDFA = {<A,B> | A and B are DFAs and L(A) = L(B)}
How do we know this to be true?
Hint: Symmetric Difference formula

16. Decidability of Regular Languages
Deciding if
a language is Regular or not
If given DFA, NFA or REX
a Regular language is empty
two Regular languages are equal

17. Decidability of Context Free
Deciding if
a language is Context Free or not (Theorem 4.7)
If given CFG
a Context Free language is empty (Theorem 4.8)
two Context Free languages are equal

18. Classes of languages Turing-recognized Decidable Context-Free (ACFG)
Regular (ADFA)

19. The Halting Problem Will an algorithm halt on a given input.
Intuition:
Can you ever be sure that a loop is infinite?
It might just terminate in a few minutes, hours, years, millenniums, etc.
Sometimes you can make such a determination:
while (x > 0) {x=1;}
But is it always possible to make such a determination?

20. Infinite Looping DFA: PDA: TM:
by definition, upon consuming the input, the machine rejects unless it is in an accept state.
Looping is simply not an option by definition.
PDA:
very, very hard to make deterministic PDA’, but it can be done.
Once the input is consumed, empty transitions can move to a reject/accept state.
Every CF language has a PDA that will halt (not loop).
TM:
Just like a high-level language TMs can loop forever.
Intuition: you don’t consume the input, you can move on the tape infinitely, and the states can have a loop with no accept or reject.

21. ATM ATM= {<M,w> | M is a TM and M accepts w} Simple intuition:
U = “on input <M,w> simulate M on w”
If M accepts, U accepts
If M rejects, U rejects
Simple intuition:
M could be a Turing Machine that loops forever on certain input.
If M loops forever, U cannot be a decider for ATM

22. Is ATM decidable? ATM= {<M,w> | M is a TM and M accepts w}
U = “on input <M,w> simulate M on w”
If M accepts, U accepts
If M rejects, U rejects
BUT! Perhaps there is a way to implement M such that we can detect the infinite loop?
Upon infinite loop detection, U rejects.
U could still be a decider for ATM

23. The Halting Problem
H(<M,w>) = if M accepts w accept if M rejects w reject

24. The Halting Problem is Undecidable
Proof: First, consider the machine/algorithm D:
D = “on input <M>, where M is a TM:
Run H on input <M,<M>>
Output the opposite of what H outputs; that is; if H accepts, reject and if H rejects, accept.”
Recall H:
H(<M,w>) = if M accepts w accept if M rejects w reject

25. D is a crazy Decider Algorithm
D is implemented with a Turing Machine
D(<M>) = if D does not accept <M>, accept if D accepts <M>, reject
What happens if we run D with its own Turing Machine description?
D(<D>) = if D does not accept <D>, accept if D accepts <D>, reject

26. A paradox emerges
D(<D>) = if D does not accept <D>, accept if D accepts <D>, reject
If D accepts, how can D(<D>) reject?
We assumed that H could decide ATM because it could ‘somehow detect an infinite loop”
Think of H as a deterministic decider if a Turing Machine loops
Then, we use H to build D (the crazy decider)
Here we assume H can stop D from looping infinitely
Then, we run D on its own encoding, which creates a paradox.

27. Paradox resolved Either H or D cannot exist. Which one?
D is a TM machine that can simulate another Turing Machine, which has been elegantly proven.
Intuition: Consider a program that can take another program and simulate its execution.
Program, Algorithm, and Turing Machine are all synonymous (Church-Turing Thesis)
Compilers
Virtual Machines

28. Significance of Turing Machines
Turing Machines are the “tool” we used to prove that the Halting Problem is un-decidable.
In other words, no algorithm exists to determine if a general algorithm will halt or not.
Note: There are some algorithms where its easy to show/prove that it will halt, but we are interested in the general case (any/all algorithms).

29. Un-decidable Languages…
…there are many, but this is the interesting one:
ATM= {<M,w> | M is a TM and M accepts w}
Obviously, this language can’t be generated by a REX or CFG.
So, a NFA, DFA, and PDA can’t be used as a decider to accept/reject strings
But, even a Turing machine cannot act as a decider.
It may be able to decide some input on some machines, but not all.
There are strings in ATM that will cause the decider to loop infinitely. Specifically <D,<D>> and likely other strings.

30. Significance of ATM
A formal language that cannot be decided by Turing Machine.
We can define this language’s concept
But we cannot create an algorithm (TM) to determine if a string is in this language or not.
ATM
Turing-Decidable
Context-Free (ACFG)
Regular (ADFA)

31. Decidable vs. Recognizable
Turing Decidable Languages
Language such that some TM will accept all of its strings
And, reject strings in the language’s compliment
Halts on all input
Turing Recognizable Languages
Language such that some TM will accept all of its strings
But, might not halt on strings in the language’s compliment
Its it looping infinitely or will it accept?
We don’t know.

32. ATM is Turing Recognizable
ATM= {<M,w> | M is a TM and M accepts w}
U = “on input <M,w> simulate M on w”
If M accepts, U accepts
If M rejects, U rejects
U will always halt if M halts. If M doesn’t halt on w than M doesn’t accept w, so <M,w> isn’t in the language. By its very definition U will always halt on strings in ATM.
The un-decidability is when U has been looping for 10 million years,
we really don’t know
Is it eventually going to be an accepted w or
an infinite loop caused by a rejected w. This is why infinity is trouble.

33. Time Complexity TM’s are a formal way to describe algorithms
Some problems don’t have algorithms that will always halt, i.e., determining if a string is in ATM.
Algorithms that do halt can still take a long time.
How long is long?

34. General Time Unit
With Turing Machines we can define a unit of time to be the execution time of one TM transition.
With more practical machines, a time unit could be a CPU clock cycle, which might execute one machine-level instruction.
Some machines can execute 1 billion instructions per second, so the time unit would be 1/ seconds.

35. Time as a function of input size
N is the size of the input
f(N) is the number of time unit to solve the problem.
The running time of algorithms can be expressed as functions:
f(N) = 2N + 5;
Two loops of size N and 5 setup instructions
Or, on loop of size N with two instruction inside and 5 instructions outside the loop.

36. Constants don’t matter
For really big problems, constants don’t matter
f(N) = 2N is the same as g(N) = 100N
While 100 days seems like forever compared to 2 days, parallel computation and faster computers can eventually make up the difference (we hope).
For big problems,
f(N) = N4 is much different than g(N) = N2
A faster computer may not help, why?

37. 1.15 Days vs. 11 billion years N N^2 days N^4 years 1 1.15741E-1416
5.07E-16 3 9
E-13 81
2.57E-15
E-13
256
8.12E-15 10
100
E-12
10000
3.17E-13 20
400
E-12
160000
5.07E-12 50
2500
E-11
1.98E-10
1000
E-08
1E+12
3.17E-05
E-06
1E+16
3.17E-01
100000
1E+20
3.17E+03
1E+24
3.17E+07
1E+14
1E+28
3.17E+11

38. Big-O Review Constants don’t matter Only the leading exponent matters
Why?

39. 1.157407 days vs. 1.157419 days N N^2 days N^2 + 100N 1 1.15741E-14
101
E-12 2 4
E-14
204
E-12 3 9
E-13
309
E-12 16
E-13
416
E-12 10
100
E-12
1100
E-11 20
400
E-12
2400
E-11 50
2500
E-11
7500
E-11
1000
E-08
E-08
10000
E-06
E-06
100000
1E+12
1.0001E+12
1E+14
E+14

40. Why do we only care about big N’s
Same reason I would worry about a $10,000 bill in my wallet but not a penny.
Same reason I would worry about a trip to Mars but not a trip to Menands.

41. Real Algorithm  TM Decider
Prepare for “hand-waving magic:”
Any algorithm that can be programmed can be reduced into a language problem.
A = {<p,i,o> | p is the encoding/description of a problem, i is the input, and o is the correct output.}
Deciding if a string is in L is the same thing as solving the problem.
The TM that decides A solves problem p.

42. The class P
The class of languages that can be decided in polynomial time.
Corresponds, the set of problems that can be solved in polynomial time.
Polynomial is O(nk)
What are some problem in P that you have studied?

43. Did you know?
Every context free language is in P

44. The class NP Non-deterministically Polynomial.
One way to think of this is NOT Polynomial.
Or, exponential
Or N!
But that is not the whole story.