Remaining Topics Decidability Concept 4.1 The Halting Problem 4.2 P vs. NP 7.2 and 7.3 NP-completeness & Cook-Levin Theorem 7.4

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

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)

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)

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

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}

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>”

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, 111111, …} ADFA would enumerate all the <B,w>’s such that w is accepted by the encoded B.

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?

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”

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

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?

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?

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.

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

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

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

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

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?

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.

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

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

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

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

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

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.

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

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).

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.

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)

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.

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.

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?

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 1/100000000 seconds.

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.

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?

1.15 Days vs. 11 billion years N N^2 days N^4 years 1 1.15741E-14 16 5.07E-16 3 9 1.04167E-13 81 2.57E-15 1.85185E-13 256 8.12E-15 10 100 1.15741E-12 10000 3.17E-13 20 400 4.62963E-12 160000 5.07E-12 50 2500 2.89352E-11 6250000 1.98E-10 1000 1000000 1.15741E-08 1E+12 3.17E-05 100000000 1.15741E-06 1E+16 3.17E-01 100000 10000000000 0.000115741 1E+20 3.17E+03 0.011574074 1E+24 3.17E+07 10000000 1E+14 1.157407407 1E+28 3.17E+11

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

1.157407 days vs. 1.157419 days N N^2 days N^2 + 100N 1 1.15741E-14 101 1.16898E-12 2 4 4.62963E-14 204 2.36111E-12 3 9 1.04167E-13 309 3.57639E-12 16 1.85185E-13 416 4.81481E-12 10 100 1.15741E-12 1100 1.27315E-11 20 400 4.62963E-12 2400 2.77778E-11 50 2500 2.89352E-11 7500 8.68056E-11 1000 1000000 1.15741E-08 1100000 1.27315E-08 10000 100000000 1.15741E-06 101000000 1.16898E-06 100000 10000000000 0.000115741 10010000000 0.000115856 1E+12 0.011574074 1.0001E+12 0.011575231 10000000 1E+14 1.157407407 1.00001E+14 1.157418981

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.

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.

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?

Did you know? Every context free language is in P

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.