Toán học - Chapter 13: Modeling computation

Modeling ComputationChapter 13Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.Chapter SummaryLanguages and GrammarsFinite-State Machines with OutputFinite-State Machines with No OutputLanguage RecognitionTuring MachinesLanguages and GrammarsSection 13.1Section SummaryPhrase-Structure GrammarsTypes of Phrase-Structure GrammarsDerivation TreesBackus-Naur FormIntroductionSyntax (form of a sentence) vs. semantics (meaning of a sentence)The sentence the frog writes neatly is a valid sentence according to the rules of English grammar. That is, it is syntactically correct, even though it’s nonsensical (unless we are talking about a fantasy world).The sequence of words swims quickly mathematics is not a valid sentence according to the rules of English grammar.GrammarsThe rules that specify the syntactically correct sentences of a natural language such as English are complex. Instead of studying natural languages, we can define formal languages that have well-defined rules of syntax. These rules of syntax are important both in linguistics (the study of natural languages) and in the study of programming languages. An Example GrammarAn example sequence of replacements:noun phrase verb phrasearticle adjective noun verb phrasearticle adjective noun verb adverbthe adjective noun verb adverbthe large noun verb adverbthe large rabbit verb adverbthe large rabbit hops adverbthe large rabbit hops quicklya sentence is made up of a noun phrase followed by a verb phrase;a noun phrase is made up of an article followed by an adjective followed by a noun, or a noun phrase is made up of an article followed by a noun;a verb phrase is made up of a verb followed by an adverb, or a verb phrase is made up of a verb;an article is a, oran article is the;an adjective is large, oran adjective is hungry;a noun is rabbit, ora noun is mathematician;a verb is eats, or a verb is hops;an adverb is quickly, oran adverb is wildly.We use these rules to form valid sentences by making a series of replacements until no more rules can be used.Some additional valid sentences are:a hungry mathematician eats wildly,a large mathematician hops, the rabbit eats quickly, etc.But note that the following is not valid:the quickly eats mathematicianPhrase-Structure GrammarsA vocabulary (or alphabet) V is a finite, nonempty set of elements called symbols. A word (or sentence) over V is a string of finite length of elements of V. The empty string or null string, denoted by λ, is the string containing no symbols.The set of all words over V is denoted by V*. A language over V is a subset of V*.The elements of V that can not be replaced by other symbols are called terminals, e.g., a, the, and rabbit in the example grammar. Those that can be replaced by other symbols are called nonterminals, e.g., sentence, noun phrase, etc.The rules that specify when we can replace a string V* with another string are called productions of the grammar. We denote by z0 → z1 the production that specifies that z0 can be replaced by z1 within a string. Phrase-Structure Grammars (cont.)A phrase-structure grammar G =(V, T, S, P) consists of a vocabulary V, a subset T of V consisting of terminal symbols, a start symbol S from V, and a finite set of productions P. The set N = V −T is the set of nonterminal symbols. Every production in P must contain at least one nonterminal on its left side. Example (Grammar 1): Let G =(V, T, S, P), where V = {a, b, A, B, S}, T = {a,b}, S is the start symbol, and P = {S →Aba, A →BB, B →ab, AB →b}.Derivations Language Generation Types of Phrase Structure GrammarsPhrase-structure grammars are classified by the types of allowable productions.Type 2 grammars are called context-free grammars. A language generated by a context-free grammar is called a context-free language.Type 3 grammars are called context-sensitive grammars (or a regular grammar). A language generated by a context-sensitive grammar is called a context-sensitive language (or a regular language).Avram Noam Chomsky(Born 1928)Derivation TreesWe can represent a derivation in the language generated by a context-free grammar by an ordered rooted tree, called a derivation, or parse tree. The root of the tree represents the start symbol.The internal vertices represent the nonterminal symbols that arise in the derivation.The leaves represent the terminal symbols that arise.If the production A →w, where w is a word, arises in the derivation, the vertex that represents A has as children vertices that represent each symbol in w, in order from left to right. A derivation tree for the derivation of the hungry rabbit eats quickly, given the grammar described earlier. Backus-Naur FormBackus-Naur form (BNF) is sometimes used to specify a type 2 grammar. It is often used to specify the syntactic rules of computer languages.The productions of a type 2 grammar have a single nonterminal symbol on their left-hand side. All the productions with the same nonterminal symbol on the left-hand side are combined into one statement using the symbol ::= instead of →. Additionally,, all nonterminal symbols are enclosed in brackets (〈〉), and the right-hand side of productions are spearated by bars.For example, the productions A →Aa, A →a, and A →AB are written as 〈A〉 ::= 〈A〉a | a | 〈A〉 〈B〉.John Backus(1924-2007)Peter Naur(Born 1928)BNF and ALGOL 60In the programming language ALGOL 60 an identifier consists of a string of alphanumeric characters and must begin with a letter.The BNF description of allowable identifiers is: 〈identifier〉 ::=〈letter〉 | 〈identifier〉〈letter〉 | 〈identifier〉〈digit〉 〈letter 〉 ::= a | b | ⋯ | y | z 〈digit〉 ::= 0 | 1 | ⋯ | 8 | 9x99a is a valid identifier since the first rule can be used to replace 〈identifier〉 by 〈identifier〉〈letter〉 , the second rule to obtain 〈identifier〉 a, the first rule twice to obtain 〈identifier〉〈digit〉〈digit〉 a, the third rule twice to obtain 〈identifier〉99a, the first rule to obtain 〈letter〉99a, and finally the second rule to obtain x99a.Finite-State Machines with OutputSection 13.2Section SummaryFinite-State Machines (FSMs) with OutputsTypes of Finite-State Machines with Outputs (not currently included in overheads)IntroductionMany kinds of machines, including computers, can be modeled using a structure called a finite-state machine (or finite automaton).A finite-state machine consists of a finite set of states, a designated start state, an input alphabet, and a transition function that assigns a next state to every (state, input) pairAs we will see in Sections 13.2 − 13.4, some types of finite-state machines produce output, while for other types of finite-state machines that do not produce output some states are designated as accepting states. Finite-state machines are used in many diverse applications, including spell-checking programs, grammar checking, indexing, searching large bodies of text, speech recognition, XML, HTML, and network protocolsAn Example of a Finite-State Machine with OutputA vending machine accepts nickels (5 cents) , dimes (10 cents) , and quarters (25 cents). When 30 cents or more has been deposited, the machine returns the amount over 30 cents. The customer can then press an orange button to receive a container of orange juice or a red button to receive a container of apple juice.The machine can be in any of the states si, i = 0, , 6, where si is the state where the machine has received 5i cents. The machine starts in state s0, with 0 cents received. The possible inputs are 5 cents, 10 cents, 25 cents, the orange button (O), and the red button (R). The possible outputs are nothing (n), 5 cents, 15 cents, 20 cents, 25 cents, an orange juice, and an apple juice. An Example (cont.)We represent this vending machine using a directed graph with labeled edges, where each state is represented by a circle, edges represent transitions, and edges are labeled with the input and output for that transition.We will trace the transitions and outputs of the vending machine when a student puts in a dime followed by a quarter, receives 5 cents back, and then pushes the orange button and receives an orange juice. The machine starts in state s0. The first input is 10 cents, which changes the state to s2 and gives no output. After the second input of 25 cents, the state changes to s6 and gives 5 cents as output.The last input is the orange button, which changes the state back to s0 and gives an orange juice as output. FSMs with OutputsA finite-state machine M =(S, I, O, f, g, s0) consists of a finite set S of states, a finite input alphabet I, a finite output alphabet O, a transition function f that assigns to each state and input pair a new state, an output function g that assigns to each state and input pair an output, and an initial state s0 .A state table is used to represent the values of the transition function f and the output function g for all (state, input).Alternatively, a finite-state machine can be represented by a state diagram, which is a directed graph with labeled edges. Each state is represented by a circle, and arrows labeled with the input and output pair represent the transitions.The state table and state diagram both represent the finite state machine with S = {s0 ,s1 ,s2 ,s3}, I = {0, 1}, and O = {0, 1}. Unit-delay MachineAn important element in many electronic devices is a unit-delay machine, which produces as output the input string delayed by a specified amount of time, i.e., padded with an initial string of 0s. How can a finite-state machine be constructed that delays an input string by one unit of time, that is, produces as output the bit string 0x1x2xk-1 given the input bit string x1x2xk-1?A delay machine can be constructed that has 0 or 1 as possible inputs. The machine has the start state s0. The transition from s0 produces an output of 0. The machine is in state s1 if the previous input was a 1 and it produces 1 as output for its next transition, and in state s2 if the previous input was 0 and it produces an output of 0 for its next transition. Addition MachineWe will construct a finite-state machine that adds two positive integers using their binary expansions.Recall the conventional procedure to add (xnx1x0)2 and (yny1y0)2 .First, the bits x0 and y0 are added, producing a sum bit z0 and a carry bit c0. Next the bits x1 and y1 are added together with the carry bit c0. This gives a sum bit z1 and a carry bit c1. The procedure continues until the nth stage, where xn, yn and the previous carry cn-1 are added to produce the sum bit zn and the carry bit cn, which is equal to the sum bit zn+1.We can construct a finite state machine that uses just two states. The start state s0 is used to remember that the previous carry is 0. The other state s1 is used to remember that the previous carry is 1. (For simplicity, we assume that both xn and yn are 0.)The inputs are pairs of bits. The transitions and the outputs are constructed from the sum of the two bits in the input and the carry represented by the state. For example, when the machine is in state s1 and receives 01 as input, the next state is s1 and the output is 0, because the sum 0 + 1 + 1 =(10)2. Finite-State Machines with No OutputSection 13.3Section SummarySet of StringsFinite-State AutomataLanguage Recognition by Finite-State MachinesDesigning Finite-State AutomataEquivalent Finite-State Automata (not currently included in overheads)Nondeterministic Finite-State AutomataSet of Strings Stephen Cole Kleene (1909-1994)Finite-State Automata (FSA)A finite-state automaton M = (S, I, f, s0, F) consists of a finite set S of states, a finite input alphabet I, a transition function f that assigns a next state to every pair of state and input (so that f: S × I → S), an initial or start state s0, and a subset F of S consisting of final (or accepting) states.FSAs can be represented using either state tables or state diagrams, in which final states are indicated with a double circle.The state diagram for the FSA M = (S, I, f, s0, F), where S = {s0, s1, s2,s3}, I = {0, 1}, F = {s0,s3}, and the transition diagram is in Table 1, is shown here. Language Recognition by FSAsA string x is said to be recognized (or accepted) by the machine M = (S, I, f, s0, F) if it takes the initial state s0 to a final state, that is, f(s0, x). The language recognized (or accepted) by M, denoted by L(M), is the set of all strings that are recognized by M. Two finite-state automata are called equivalent if they recognize the same language.The only final state of M1 is s0. The strings that take s0 to itself consist of zero or more consecutive 1s. Hence, L(M1) = {1n | n = 0, 1, 2, .}. The only final state of M2 is s2. The strings that take s0 to s2 are 1 and 01 . Hence , L(M2) = {1, 01}.The final state of M3 are s0 and s3. The strings that take s0 to itself are λ, 0, 00, 000, . The strings that take s0 to s3 are a string of zero or more consecutive 0s, followed by 10, followed by any string. Hence, L(M3) = {0n,0n10x | n = 0, 1, 2, ., and x is any string} Language Recognition by FSAs (cont.)Example: Construct a FSA that recognizes the set of bit strings that begin with two 0s.Solution: Besides the start state s0, we include a nonfinal state s1; we move to s1 from s0 if the first bit is a 0. Next, we add a final state s2, which we move to from s1, if the second bit is a 0. We stay in this state no matter what the succeeding bits (if any) are.We need a nonfinal state s3, so that we can move to it from s0 if the first bit is a 1 and from s1 if the second bit is a 1. Language Recognition by FSA (cont.)Example: Construct a FSA that recognizes the set of bit strings that contain two consecutive 0s. Solution: Besides the start state s0, we include a nonfinal state s1, which tells us that the last input bit seen is a 0, but either the bit before was a 1, or this is the initial bit. Next, we add a final state s2, which we move to from s1, if the next bit after a 0 is also 0. We stay in this state no matter what the succeeding bits (if any) are.We return from s0 or s1 , if a 1 follows a 0 in the string, before we come to two consecutive 0s. NDFSAA nondeterministic finite-state automaton M = (S, I, f, s0, F) consists of a finite set S of states, a finite input alphabet I, a transition function f that assigns a set of states to every pair of state and input (so that f: S × I → P(S)), an initial or start state s0, and a subset F of S consisting of final (or accepting) states.We can represent a nondeterministic finite-state automaton using a state table where we give a list of possible next states for each pair of a state and an input value.We construct a state diagram for a nondeterministic automaton by including an edge from each state to all possible next states, labeling edges with the input or inputs that lead to this transition.We use the abbreviation NDFSA for a nondeterministic finite-state automaton and DFSA for a deterministic finite-state automata when we needed to distinguish between NDFSA and DFSA.Example: Find the state diagram for the NDFSA with the state table shown in Table 2. The final states are s2 and s3.Solution: Finding a DFSA Equivalent to a NFSA Finding an Equivalent DFSA (cont.)Example: Find a DFSA that recognizes the same language as the NFSA:Solution: Following the steps of the procedure described on the previous slide, we obtain the DFSA shown here. Language RecognitionSection 13.4Section SummaryRegular ExpressionsKleene’s TheoremRegular Sets and Regular GrammarsMore Powerful Types of MachinesRegular ExpressionsThe regular expressions over a set I are defined recursively by:Each regular expression represents a set specified by these rules: Sets represented by regular expressions are called regular sets. the symbol  is a regular expression;the symbol λ is a regular expression;the symbol x is a regular expression whenever x ∈ I;the symbols (AB), (A ∪ B), and A* are regular expressions whenever A and B are regular expressions.  represents the empty set, that is, the set with no strings;λ represents the set {λ}, which is the set containing the empty string;x represents the set {x} containing the string with one symbol x;(AB) represents the concatenation of the sets represented by A and by B; (A ∪ B) represents the union of the sets represented by A and by B; A* represents the Kleene closure of the set represented by A. Regular Expressions (cont.)Example: What are the strings in the regular sets specified by the regular expressions 10*, (10)*, 0 ∪ 01, 0(0 ∪ 1)*, and (0*1)*?Solution:Finite-State Automata, Regular Sets, and Regular GrammarsIn 1956 Kleene established the connection between regular sets and sets recognized by a FSA.He showed that a set is regular if and only if it is recognized by a FSA. This result is known as Kleene's theorem.See the text for the lengthy proof of this theorem.There is a close connection between regular grammars and regular sets.Specifically, a set is generated by a regular grammar if and only if it is a regular set.See the text for a proof.We will give an example of a set that is not regular later in this section by finding a set that is not recognized by an FSA.A Set Not Recognized by a FSAThe set {0n1n | n = 0, 1, 2, } of all strings consisting of a block of 0s followed by a block of an equal number of 1s is not regular.To show that this set is not regular, suppose that this set was regular. Then there would be a NDFSA M = (S, I, f, s0, F) recognizing it. Let N be the number of states in this machine, i.e., N = |S|. M must recognize 0N1N since it is made up of a block of 0s followed by a block of an equal number of 1s. Let s0, s1, s2,,s2N be the sequence of states obtained starting at s0 and using the symbols of 0N1N as input. So, s1 =f(s0 ,0), s2 =f(s1 ,0), , sN =f(sN-1 ,0), sN+1 =f(sN ,1), ,s2N = f(s2N-1 ,1), and s2N is a final state.Because there are only N states, by the pigeonhole principle at least two of the first N + 1 states s0, s2,,sN must be the same.Suppose that si and sj are identical states with 0 ≤ i < j ≤ N. This means that f(si ,0t)= sj, where t = j − i. Hence, there is a loop leading from si back to itself, using 0 a total of t times. continuedA Set Not Recognized by a FSA (cont.)Now consider the input string 0N0t1N = 0N+t1N . The string is not of the correct form and so, it is not recognized by M.Consequently, f(s0 ,0N+t1N) can not be a final state. However, when we use the string 0N+t1N as input, we end up in the same state as before, namely, s2N. The reason is that we go through the loop one more time. This contradiction shows that {0n1n | n = 0, 1, 2, } is not regular.More Powerful Types of MachinesThe main limitation of finite-state automata is their finite amount of memory. This has led to the development of more powerful types of machines.Pushdown Automaton (PDA): includes a stack, which provides unlimited memory. We can use a PDA to recognize {0n1n | n = 0, 1, 2, }, but no PDA recognizes the set {0n1n2n | n = 0, 1, 2, }.Linear Bounded Automaton (LBA): More powerful than pushdown automata. We can use a LBA to recognize {0n1n2n | n = 0, 1, 2, }, but there are languages generated by phrase-structure grammars that cannot be recognized by a LBA.Turing Machine (TM): Yet more powerful machines (to be studied in the next section) which can recognize all languages generated by phrase-structure grammars.Turing MachinesSection 13.5Section SummaryDefinition of Turing MachinesUsing Turing Machines to Recognize SetsComputing Functions with Turing Machines Different Types of Turing Machines (not currently included in overheads)The Church-Turing ThesisComputational Complexity, Computability, and DecidabilityIntroductionInformally, a Turing machine consists of a control unit, which at any step is in one of finitely many different states, together with a tape, infinite in both directions, which is divided into cells. Turing machines have read and write capabilities on the tape as the control unit moves back and forth along this tape, changing states depending on the tape symbol read. Turing machines are more powerful than finite-state machines because they include additional memory capability. Turing machines are the most general models of computation; essentially they can do whatever a computer can do. Alan Mathison Turing (1912-1954)Definition of Turing Machines (TM)A Turing machine T = (S, I, f, s0) consists of a finite set S of states, an alphabet I containing the blank symbol B, a partial function f from S × I to S × I ×{R,L}, and a starting state s0.For some (state, symbol) pairs the partial function f may be undefined, but for a pair for which it is defined, there is a unique (state, symbol, direction) triple associated to this pair. The five-tuples corresponding to the partial function in the definition of a TM are called the transition rules of the machine.At each step, the control unit reads the current tape symbol x. If the control unit is in state s and if the partial function f is defined for the pair (s, x) with f(s, x) = (s′, x′, d), the control unit:enters the state s′,writes the symbol x′ in the current cell, erasing x, andmoves right one cell if d = R or moves left one cell if d = L.This step is written as the five-tuple (s, x, s′, x′, d). Turing machines are defined by specifying a set of such five-tuples. If the partial function f is undefined for the pair (s, x) then T will halt. At the beginning of its operation a TM is assumed to be in the initial state s0 and to be positioned over the leftmost nonblank symbol on the tape. This is the initial position of the machine.A TM in OperationExample: What is the final tape when the TM T defined by the seven five-tuples (s0, 0, s0, 0, R), (s0, 1, s1, 1, R), (s0, B, s3, B, R), (s1, 0, s0, 0, R), (s1, 1, s2, 0, L), (s1, B, s3, B, R), and (s2, 1, s3, 0, R) is run on the tape shown here in (a)?Solution: The transitions of this TM are shown to the right. The final tape is shown in (g).Using TM to Recognize SetsLet V be a subset of an alphabet I. A TM T = (S, I, f, s0) recognizes a string x in V* if and only if T, starting in the initial position when x is written on the tape, halts in a final state. T is said to recognize a subset A of V* if x is recognized by T if and only if x belongs to A.Note that to recognize a subset A of V* we can use symbols not in V. This means that the input alphabet I may include symbols not in V. We will see that these extra symbols are used as markers.A TM operating on a tape containing the symbols of a string x in consecutive cells, does not recognize x if it does not halt or halts in a state that is not final. Using TMs to Recognize Sets (cont.)In Section 13.4 we showed that there is no DFA that recognizes the set the set {0n1n | n ≥ 1}. We will now construct a TM that recongizes this set.We use an auxiliary tape symbol M as a marker, and specify that V = {0, 1} and I = {0, 1, M}. Our TM has one final state, s6 . The TM successively replaces a 0 at the leftmost position of the string with an M and a 1 at the rightmost position of the string with an M, sweeping back and forth, terminating in a final state if and only if the string consists of a block of 0s followed by a block of the same number of 1s.The five-tuples are (s0, 0, s1, M, R), (s1, 0, s1, 0, R), (s1, 1, s1, 1, R), (s1, M, s2, M, L), (s1, B, s2, B, L), (s2, 1, s3, M, L), (s3, 1, s3, 1, L), (s3, 0, s4, 0, L), (s3, M, s5, M, R), (s4, 0, s4, 0, L), (s4, M, s0, M, R), and (s5, M, s6, M, R).For example, the string 000111 would successively become M00111, M0011M, MM011M, MM01MM, MMM1MM, MMMMMM as the machine operates until it halts. Computing Functions with TMsA Turing machine can be used to compute the values of a partial function.Suppose that the TM T, when given the string x as input, halts with the string y on its tape. We can then define T(x) = y.To consider a TM as a computer of functions from the set of k-tuples of nonnegative integers to the set of nonnegative integers, we use the unary representation of integers.A nonnegative integer n is represented by a string of n + 1 1s. So, 0 is represented by 1, 5 by 111111, etc.To represent an input that is a k-tuple of integers, we represent each integer in the k-tuple separately and separate these representations using asterisks. For example, (2,0,1,3) is represented by 111*1*11*1111. Constructing a Turing machine that computes a particular function can be extremely complicated. Fortunately, the ability of Turing machines to compute functions is of theoretical, rather than practical, interest.Computing Functions with TMs (cont.)To construct a TM T that computes the function f(n1,n2) = n1 + n2, we first represent the pair (n1,n2) by a string of n1 + 1 1s followed by an asterisk, followed by n2 + 1 1s. The machine starts at the leftmost 1 of the input string, and proceeds to erase this 1. If the next character is an asterisk, n1 = 0. In this case, it replaces the asterisk with a blank and halts.Otherwise, it erases the next 1, and then passes over the remaining 1s, until it comes to the asterisk. The asterisk is then replaced by a 1. The five-tuples defining this Turing machine are: (s0, 1, s1, B, R), (s1, ∗, s3, B, R), (s1, 1, s2, B, R), (s2, 1, s2, 1, R), (s2, ∗, s3, 1, R).The Church-Turing ThesisThe Church-Turing Thesis says that given any problem that can be solved with an effective algorithm, there is a TM that can solve this problem. It is called a thesis rather than a theorem because the concept of solvability by an effective algorithm is informal and imprecise, as opposed to the concept of solvability by a TM, which is formal and precise.Decidability and ComplexityA decision problem asks whether statements from a particular class of statements are true. Decision problems are also known as yes-or-no problems. Consider the question for a particular integer n, “Is n prime?” The answer is "yes" or "no.“The halting problem is the decision problem that asks whether a Turing machine T eventually halts when given an input string x. When there is an effective algorithm that decides whether instances of a decision problem are true, we say that this problem is solvable or decidable.In Section 3.5, an algorithm was given for determining whether a positive integer n is prime by checking whether it is divisible by primes not exceeding its square root. However, if no effective algorithm exists for solving a problem, then we say the problem is unsolvable or undecidable.The halting problem is an unsolvable decision problem (proved in Section 3.1). That is, no TM exists that, when given an encoding of a TM T and its input string x as input, can determine whether T eventually halts when started with x written on its tape. A function that can be computed by a TM is called computable and a function that cannot be computed by a TM is called uncomputable. The busy beaver function, which when given a positive integer n gives the maximum number of 1s that a TM with n states and alphabet {1,B} may print on an initially blank tape is uncomputable (see Exercise 35 in Section 13.5).The Classes P and NPIn a nondeterministic Turing machine (NDTM), the restriction that no two transition rules begin with the same pair (s, x) is eliminated. Hence, there may be more than one transition rule beginning with each (state, tape symbol) pair, so that there may be a choice as to which rule to use at each step. A NDTM T recognizes a string x if and only if there exists some sequence of transitions of T that ends in a final state when the machine starts in the initial position with x written on the tape. The Classes P and NP (cont.)A decision problem is in P, the class of polynomial-time problems, if it can be solved by a deterministic Turing machine in polynomial time in terms of the size of its input. That is, a decision problem is in P if there is a deterministic Turing machine T that solves the decision problem and a polynomial p(n) such that for all integers n, T halts in a final state after no more than p(n) transitions whenever the input to T is a string of length n. A decision problem is in NP, the class of nondeterministic polynomial-time problems, if it can be solved by a nondeterministic Turing machine in polynomial time in terms of the size of its input. That is, a decision problem is in NP if there is a nondeterministic Turing machine T that solves the problem and a polynomial p(n) such that for all integers n, T halts for every choice of transitions after no more than p(n) transitions whenever the input to T is a string of length n. Problems in P are called tractable, whereas problems not in P are called intractable. The Classes P and NP (cont.)For a problem to be in NP, it is necessary only that there be a NDTM that when given a true statement from the set of statements addressed by the problem, can verify its truth in polynomial time by making the correct guess at each step. The problem of determining whether a given graph has a Hamilton circuit is an NP problem, because a NDTM can easily verify that a simple circuit in a graph passes through each vertex exactly once.It can do this by making a series of correct guesses corresponding to successively adding edges to form the circuit. Wrapping Everything up with a Millenium ProblemBecause every DTM can also be a considered to be a NDTM, where each (state, tape symbol) pair occurs in exactly one transition rule, P ⊆ NP.The most famous open question in theoretical CS, and one of the millennium problems with a $1,000,000 prize, is whether every problem in NP is also in P, that is, whether P = NP.There is an important class of problems, known as NP-complete problems, where a problem is in this class if it is the class NP and if this problem was also in the class P, then every problem in NP must also be in P. That is, a problem is NP-complete if the existence of a polynomial-time algorithm for solving it implies the existence of a polynomial-time algorithm for every problem in NP.We have studied several problems that can be shown to be NP-complete in this text, including determining whether a simple graph has a Hamilton circuit and determining whether a proposition in n variables is a tautology.This concludes our introduction to discrete mathematics, but it should come as no surprise that there is a lot more to learn!