Lecture 16: Modeling Computation Languages and Grammars Finite-State Machines with Output Finite-State Machines with No Output

Transcription

1 Lecture 16: Modeling Computation Languages and Grammars Finite-State Machines with Output Finite-State Machines with No Output

2 Natural Language vs. Formal Language

3 Rules of English Grammar verifying a sentence

4 Phrase Structure Grammar

5 An Example

6 Another Example

7 Example

8 Example

9 Example

10 Example

11 Example

12 Types of Phrase Structure Grammars

13 Introduction to Grammars A grammar is is used to generate or evaluate members of a formal language. the context-free grammar G 1 below, generates the language {0 n 1 n for n>=0}. A -> 0A1 A -> φ (the empty string) A grammar is a set of production rules as shown above. A production rule is comprised of a variable, a right-arrow, and a string of variables and terminals. The variables are used to control the arrangement of possible substitutions and the terminals are symbols from the alphabet of the language being generated or recognized. By convention, variables are represented by uppercase letters. One of the variables is designated as the start variable, usually the left-hand symbol of the first rule in the grammar.

14 Applying Production Rules There are usually choices for applying production rules, which means that in order to create a particular member of a language we need to specify an order of application of the rules of the grammar, called a derivation. For example, the following is a derivation to show that is a member of the language defined by G1. start A rule 1 0A1 rule 1 00A11 rule 1 000A111 rule A1111 rule

15 The Language of the Grammar All the member strings that can be generated through derivations using G 1 are collectively referred to as the language of the grammar G 1 or L(G 1 ). We say that a context-free grammar is one in which candidate strings are accepted or rejected as members of L(G 1 ) based strictly on their structures and not based on the context in which they appear. The syntax of a programming language is (mostly) definable with a context-free grammar. For example, the reserved words of a programming language are parsed and recognized as reserved words regardless of their context. (Can you think of an example in a programming language in which context is important?)

16 Formal Definition of a Context-Free Grammar A context-free grammar is a 4-tuple (V,Σ,R,S), where 1. V is a finite set of variables 2. Σ is a finite set, disjoint from V, called the terminals 3. R is a finite set of rules, with each rule being a variable and a string of variables and terminals, and 4. S is an element of V called the start variable. If u, v, and w are strings of variables and terminals, and A -> w is a rule of the grammar, we say that uav yields uwv, written uav -> uwv. Write u -> * v if u=v or if a sequence u 1,u 2,..., u k exists for k>=0 and, u -> u 1 -> u 2 ->... -> u k -> v.

17 An Example A Practical Example - The following context-free grammar can be used to derive (or recognize) infix arithmetic expressions in x,y and z. The start variable is S and the termials are+, -, *, /, (, ), x, y, andz. S -> T+S T-S T T -> T*T T/T (S) x y z Use the grammar above to show that(x+y)/(x-y)*z is a valid in-fix expression. S -> T -> T/T -> T/T*T ->... -> (S)/(S)*T ->... -> (T+S)/(T-S)*T ->... -> (x+y)/(x-y)*z

18 The Problem with Grammars How do we go about demonstrating that a particular string is not derivable with a given contextfree grammar? For example how would we use the grammar above to show that (x+y)*/z is not a valid in-fix expression? The problem we are dealing with is that we have no means to choose which rules to apply or in which order to apply them. If the same grammar generates the same string in more than one way, we say that the grammar is ambiguous. Using a grammar to generate arbitrary parts of a candidate string in a random order can lead to an exponential number of alternate paths to test. What we need is a systematic method for applying the rules to show that a particular string is or is not derivable.

19 Chomsky Normal Form A context-free grammar is said to be in Chomsky normal form (CNF) when every rule has the form A -> BC A -> a where a is any terminal and A, B, and C are any variables except B and C are not the start variable. A rule that replaces the start variable with nothing (i.e. S -> e ) is permitted. We can show that "Any context-free language is generated by a context-free grammar in Chomsky normal form." Perhaps more important to us than the proof itself is the proof idea which we can use to convert a context-free grammar into a CNF grammar.

20 Converting a Grammar into CNF CNF grammars are very useful since they simplify derivations making them more easily implemented as computer algorithms. The method of conversion of a context-free grammar into CNF is detailed below: 1. Create a new start variable. 2. Eliminate all ε rules of the form A -> ε. 3. Eliminate all unit rules of the form A -> B. 4. Convert the remaining rules into the form A -> BC or A -> a

21 An Example Example: Lets work through an example of this conversion process. In the following, rules being eliminated will be shown in blue, rules being added will be shown in red. Given the grammar, we apply Step 1, creating a new start variable S 0, Now we need to eliminate rules of the form A->ε. In order to do this without changing the grammar, we must also create replacement rules to permit the same derivations that were possible when the epsilon rules were present. For example, removing B -> ε requires that we add a rule S -> a to cover the derivation S-> ab -> a that would have been possible with the rule B -> ε. Similarly we need to include the rule A -> ε to account for the A -> B -> ε derivation.

22 We still need to remove the A -> ε rule. Next we remove all unit rules. Removing S 0 -> S we must add S 0 -> SA AS ASA ab a. We can remove S -> S without adding any other rules.

23 When we remove A -> B we must add the rule A -> b. Removing A -> S we must add A -> SA AS ASA ab a. This gives us, Finally we must make sure that each rule is in one of the two forms allowed in CNF. Specifically we can have only rules that replace a single variable with a pair of variables or rules that replace a single variable with a single terminal.

24 Therefore we must modify the rules, S0 -> ASA, S -> ASA, S -> ab, A -> ASA, and A -> ab. We create new variables and new rules to ensure that the grammar derivation is not changed. The original grammar is shown on the left and the CNF of our grammar is on the right.

25 Derivation using the Original Grammar The CNF grammar is larger and less readable than our original grammar, so what is the advantage of the CNF grammar? We may learn more if we attempt to derive candidate strings using each form.

26 Derivation using the CNF Grammar The value of the CNF grammar begins to be revealed in these sample derivations. Although the CNF grammar has a larger rule set, its derivations are significantly shorter.

27 Moore Machine Moore Machine - A Moore machine is a 6-tuple (Q, Σ, Ω, δ, q 0, Φ) where, (1) Q is a finite set of states. (2) Σ is a finite set of symbols called the input alphabet. (3) Ω is a finite set of symbols called the output alphabet. (4) δ: Q x Σ-> Q is the transition function (5) q 0 is an element of Q called the start state, and (6) Φ: Q -> Ω is the output function mapping the current state to the output.

28 Example Moore Machine Each of the six elements of this example Moore machine are given below, Q = {1, 2, 3} δ = ([1,X]->2, [1,Y]->3, [1,Z]->1, [2,X]->3, [2,Y]->1, [2,Z]->2, [3,X]->1, [3,Y]->3, [3,Z]->3) Σ = { X, Y, Z} q 0 = 1 Ω = {A, B} Φ = (1->A, 2->B, 3->A)

29 Mealy Machine Mealy Machine - A Mealy machine is a 6-tuple (Q, Σ, Ω, δ, q 0, Τ) where, (1) Q is a finite set of states. (2) Σ is a finite set of symbols called the input alphabet. (3) Ω is a finite set of symbols called the output alphabet. (4) δ: Q x Σ-> Q is the transition function (5) q 0 is an element of Q called the start state, and (6) Τ: Q x Σ-> Ω is the output function mapping the current transition to the output.

30 Moore Machine Substring Recognizer Now let's look at how we build an FSM to find all occurrences of a particular substring in an input binary string. The first issue we need to resolve is whether we wish to allow overlap in the substrings. We can use the substring 1101 to illustrate substring overlap. Permitting overlap means that one or more bits can be part of more than one instance of a substring. Consider the input string, If we permit overlap, we find that the string contains six instances of the substring However, if we do not permit overlap, this string contains only three instances of the substring.

31 Moore Machine: Recognizer for '1101' with Overlap

32 Mealy Machine: Recognizer for "1101" with Overlap As a final example of bit string recognizers, we will build a Mealy machine to find the substring 1101 with overlap. Recall that a Mealy machine generates its output on the transition rather than upon entering a state. This may seem like a trivial difference, but it can sometimes permit us to reduce the number of states required. Upon reading the final bit in the substring sequence, this machine outputs a 1, otherwise it outputs a 0 each time it reads an input bit. This Mealy machine has one less state than the Moore machine designed to solve the same problem. Mealy machines will always require the same or fewer states to solve the same problem as Moore machines.

33 Finite State Machines Finite State Machines (FSMs) are theoretical models of computational systems. Finite state machines, also called finite automata, are used in the study of the theory of computation. R = right L = left U = up D = down

34 Formal Definition Definition of a Finite State Machine - Formally, a finite Automaton (FA) is defined as a 5-tuple (Q,Σ, δ, q 0, F) where, (1) Q is a finite set of states. (2) Σ is a finite set of symbols or the alphabet. (3) δ: Q x Σ-> Q is the transition function (4) q 0 is an element of Q called the start state, and (5) F is a subset of Q called the set of accept states.

35 An Example Q = {1, 2, 3, 4, 5, 6} q 0 = 1 (start state) Σ = {U, D, L, R} F = {6}

36 An FSM to Recognize (Accept) Integers Valid Integers Not an Integer Hello There

37 How to Implement an FSM in a Computer Program

38 Example Binary String Recognizer accepts strings containing at least three 1's The double circle indicates the accept state, the start state is labeled (0), and the transitions between states are labeled according the input symbol currently being read. The binary strings below, are accepted or rejected as indicated, reject accept 0111 accept reject

39 Example Binary String Recognizer accepts strings containing three consecutive 1's Applying this FSM to the sample binary strings listed below gives the indicated results reject accept reject accept

40 Building Finite State Machines Consider an FSM that accepts (or recognizes) strings containing the substring "1101". We create a start state and a state for each bit in the substring for which we are searching.

41 Accepting Binary Encoded Values divisible by four

42 Dual Parity Tester We can check the parity of 1's or 0's or both in a binary value using an FSM. In this example, we want an FSM that accepts binary strings with an even number of 0's and an even number of 1's. The start state is an accept state since the empty string has zero 0's and zero 1's (and zero is even).

43 Computational Limits of Finite State Machines Counting 1's and 0's - Consider the design of an FSM that accepts all binary strings that have an equal number of 1's and 0's. At first this may seem to be a relatively simple problem; however, we need to deal with the possibility that the count of the number of 1's or 0's could grow without bound. We can produce a partial FSM for recognizing binary strings with the same number of 1's and 0's, but a consecutive number of 1's or 0's could occur which would exceed any finite number of states in the FSM.

44 FSM for Recognizing Binary Palindromes In our example problem, we want to design an FSM that accepts all binary strings that have the same sequence of 1's and 0's when read either left-toright or right-to-left. Again, we are limited by the fact that an input string can be arbitrarily long. In this case however, the FSM is impractical even for strings of finite length. For binary strings of maximum length Nmax, our palindromic binary string recognizer uses 2 Nmax+1-1 states

45 Describe the Language Recognized by this FSM