1

Computer Computer Language TheoryLanguage Theory

Chapter 2: Context-Free Chapter 2: Context-Free LanguagesLanguages

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OverviewOverview In Chapter 1 we introduced two equivalent In Chapter 1 we introduced two equivalent

methods for describing a language: Finite methods for describing a language: Finite Automata and Regular ExpressionsAutomata and Regular Expressions

In this chapter we do something analogousIn this chapter we do something analogous We introduce context free grammars (CFGs)We introduce context free grammars (CFGs) We introduce push-down automata (PDA)We introduce push-down automata (PDA)

PDAs recognize CFGsPDAs recognize CFGs In my view the order is reversed from before since In my view the order is reversed from before since

the PDA is introduced secondthe PDA is introduced second We even have another pumping lemma (Yeah!)We even have another pumping lemma (Yeah!)

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Why Context Free Why Context Free GrammarsGrammars

They were first used to study human They were first used to study human languageslanguages You may have even seen something like You may have even seen something like

them beforethem before They are definitely used for “real” They are definitely used for “real”

computer languages (C, C++, etc.)computer languages (C, C++, etc.) They define the languageThey define the language A parser uses the grammar to parse the A parser uses the grammar to parse the

inputinput Of course you can also parse EnglishOf course you can also parse English

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Section 2.1Section 2.1

Context-Free GrammarsContext-Free Grammars

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A Context-Free GrammarA Context-Free Grammar

Here is an example grammar G1Here is an example grammar G1A A 0A1 0A1

A A B B

B B # #

A grammar has substitution rules or A grammar has substitution rules or productionsproductions

Each rule has a variable and arrow and a Each rule has a variable and arrow and a combination of variables and terminal symbolscombination of variables and terminal symbols

We will capitalize symbols but not terminalsWe will capitalize symbols but not terminals A special variable is the start variableA special variable is the start variable

Usually on the left-hand side of topmost ruleUsually on the left-hand side of topmost rule Here the variables are A and B and the terminals Here the variables are A and B and the terminals

are 0, 1, #are 0, 1, #

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Using the GrammarUsing the Grammar

Use the grammar to generate a Use the grammar to generate a language by replacing variables using language by replacing variables using the rules in the grammarthe rules in the grammar Start with the start variableStart with the start variable Give me some strings that grammar G1 Give me some strings that grammar G1

generates?generates? One answer: 000#111One answer: 000#111 The sequence of steps is the derivationThe sequence of steps is the derivation For this example the derivation is:For this example the derivation is:

A A 0A1 0A1 00A11 00A11 000A111 000A111 000B111 000B111 000#111000#111

You can also represent this with a parse treeYou can also represent this with a parse tree

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The Language of The Language of Grammar G1Grammar G1

All strings generated by G1 form the All strings generated by G1 form the languagelanguage We write it L(G1)We write it L(G1) What is the language of G1? What is the language of G1?

L(G1) = {0L(G1) = {0nn#1#1nn| n ≥0}| n ≥0} This should look familiar. Can we This should look familiar. Can we

generate this with a FA?generate this with a FA?

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An Example English An Example English GrammarGrammar

Page 101 of the text has a simplified Page 101 of the text has a simplified English grammarEnglish grammar Follow the derivation for “a boy sees”Follow the derivation for “a boy sees” Can you do this without looking at the Can you do this without looking at the

solution?solution?

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Formal Definition of a Formal Definition of a CFGCFG

A CFG is a 4-tuple (V, A CFG is a 4-tuple (V, , R, S) , R, S) wherewhere

1.1. V is a finite set called the variablesV is a finite set called the variables

2.2. is a finite set, disjoint from V, called is a finite set, disjoint from V, called the terminalsthe terminals

3.3. R is a finite set of rules, with each rule R is a finite set of rules, with each rule being a variable and a string of being a variable and a string of variables and terminals, andvariables and terminals, and

4.4. S S V is the start variable V is the start variable

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ExampleExample

Grammar G3 = ({S}, {a,b}, R, S), Grammar G3 = ({S}, {a,b}, R, S), where the set of rules R is:where the set of rules R is:S S aSb | SS | aSb | SS | εε What does this generate:What does this generate:

abab, aaabbb, aababbabab, aaabbb, aababb If you view a as “(“ and b as “)” then you get If you view a as “(“ and b as “)” then you get

all strings of properly nested parenthesesall strings of properly nested parentheses Note they consider ()() to be okayNote they consider ()() to be okay I think the key property here is that at any point I think the key property here is that at any point

in the string you have at least as many a’s to the in the string you have at least as many a’s to the left of it as b’sleft of it as b’s

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Another ExampleAnother Example

Consider example 2.4 in text on Consider example 2.4 in text on page 103page 103

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Designing CFGsDesigning CFGs Like designing FA, some creativity is requiredLike designing FA, some creativity is required

It is probably even harder with CFGs since they are It is probably even harder with CFGs since they are more expressive than FA (we will show that soon)more expressive than FA (we will show that soon)

Here are some guidelinesHere are some guidelines If the CFL is the union of simpler CFLs, design grammars If the CFL is the union of simpler CFLs, design grammars

for the simpler ones and then combinefor the simpler ones and then combine For example, S For example, S G1 | G2 | G3 G1 | G2 | G3

If the language is regular, then can design a CFG that If the language is regular, then can design a CFG that mimics a DFAmimics a DFA

Make a variable Ri for every state qiMake a variable Ri for every state qi If If δδ(qi, a) = qj, then add Ri (qi, a) = qj, then add Ri aRjaRj Add Ri Add Ri εε if i is an accept state if i is an accept state Make R0 the start variable where q0 is the start state of the Make R0 the start variable where q0 is the start state of the

DFADFA Assuming this really works, what did we just show?Assuming this really works, what did we just show?

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Designing CFGs Designing CFGs continuedcontinued

A final guideline: A final guideline: Certain CFLs contain strings that are Certain CFLs contain strings that are

linked in the sense that a machine for linked in the sense that a machine for recognizing this language would need recognizing this language would need to remember an unbounded amount of to remember an unbounded amount of information about one substring to information about one substring to “verify” the other substring.“verify” the other substring. This is sometimes trivial with a CFGThis is sometimes trivial with a CFG Example: 0Example: 0nn11nn

S S 0S1 | 0S1 | εε

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AmbiguityAmbiguity

Sometimes a grammar can generate the Sometimes a grammar can generate the same string in multiple wayssame string in multiple ways If a grammar generates even a single string If a grammar generates even a single string

in multiple ways the grammar is ambiguousin multiple ways the grammar is ambiguous Example:Example:

EXPR EXPR EXPR + EXPR | EXPR × EXPR |(EXPR) | a EXPR + EXPR | EXPR × EXPR |(EXPR) | a This generates the string a+a × a ambiguouslyThis generates the string a+a × a ambiguously Try it: generate two parse treesTry it: generate two parse trees Using your extensive knowledge of arithmetic, Using your extensive knowledge of arithmetic,

insert parentheses to shows what each parse tree insert parentheses to shows what each parse tree really representsreally represents

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An English ExampleAn English Example

Grammar G2 on page 101 Grammar G2 on page 101 ambiguously generates ambiguously generates the girl the girl touches the boy with the flowertouches the boy with the flower

Using your extensive knowledge of Using your extensive knowledge of English, what are the two meanings English, what are the two meanings of this phraseof this phrase

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Definition of AmbiguityDefinition of Ambiguity

A grammar generates a string A grammar generates a string ambiguously if there are two different ambiguously if there are two different parse treesparse trees Two derivations may differ in the order that Two derivations may differ in the order that

the rules are applied, but if they generate the rules are applied, but if they generate the same parse tree, it is not really the same parse tree, it is not really ambiguousambiguous

Definitions: Definitions: A derivation is a A derivation is a leftmost derivationleftmost derivation if at every if at every

step the leftmost remaining variable is replacedstep the leftmost remaining variable is replaced A string A string ww is derived is derived ambiguouslyambiguously in a CFG G if in a CFG G if

it has two or more different leftmost derivations.it has two or more different leftmost derivations.

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Chomsky Normal FormChomsky Normal Form

It is often convenient to convert a It is often convenient to convert a CFG into a simplified formCFG into a simplified form

A CFG is in Chomsky normal form if A CFG is in Chomsky normal form if every rule is of the form:every rule is of the form:

A A BC BCA A a aWhere a is any terminal and A, B, and C are Where a is any terminal and A, B, and C are

any variables–except B and C may not be any variables–except B and C may not be the start variable. The start variable can the start variable. The start variable can also go to also go to εε

Any CFL can be generated by a CFG in Any CFL can be generated by a CFG in Chomsky normal formChomsky normal form

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Converting CFG to Chomsky Converting CFG to Chomsky Normal FormNormal Form

Here are the steps:Here are the steps: Add rule SAdd rule S00 S, where S was original start variable S, where S was original start variable Remove Remove εε-rules. Remove A -rules. Remove A εε and for each and for each

occurrence of A add a new rule with A deleted.occurrence of A add a new rule with A deleted. If we have R If we have R uAvAw, we get: uAvAw, we get:

R R uvAw | uAvw | uvw uvAw | uAvw | uvw

Handle all unit rulesHandle all unit rules If we had A If we had A B, then whenever a rule B B, then whenever a rule B u exists, we add A u exists, we add A

u. u.

Replace rules A Replace rules A u u11uu22uu33… u… ukk with: with: A A u u11AA11, A, A11 u u22AA22, A, A22 u u33AA33 … A … Ak-2k-2 u uk-1k-1uukk

You will have a HW question like thisYou will have a HW question like this Prior to doing it, go over example 2.10 in the textbook (page Prior to doing it, go over example 2.10 in the textbook (page

108)108)

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Section 2.2Section 2.2

Pushdown AutomataPushdown Automata

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Pushdown Automata Pushdown Automata (PDA)(PDA)

Similar to NFAs but have an extra Similar to NFAs but have an extra component called a component called a stackstack The stack provides extra memory that is The stack provides extra memory that is

separate from the controlseparate from the control Allows PDA to recognize non-regular Allows PDA to recognize non-regular

languageslanguages Equivalent in power/expressiveness to a Equivalent in power/expressiveness to a

CFGCFG Some languages easily described by Some languages easily described by

generators others by recognizersgenerators others by recognizers Nondeterministic PDA’s Nondeterministic PDA’s notnot equivalent to equivalent to

deterministic ones but NPDA = CFGdeterministic ones but NPDA = CFG

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Schematic of a FASchematic of a FA

The state control represents the The state control represents the states and transition functionstates and transition function

Tape contains the input stringTape contains the input string Arrow represents the input head and Arrow represents the input head and

points to the next symbol to be readpoints to the next symbol to be read

Statecontrol

a a b b

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Schematic of a PDASchematic of a PDA

The PDA adds a stackThe PDA adds a stack Can write to the stack and read them back laterCan write to the stack and read them back later Write to the top (push) and rest “push down” orWrite to the top (push) and rest “push down” or Can remove from the top (pop) and other symbols move upCan remove from the top (pop) and other symbols move up A stack is a LIFO (Last In First Out) and size is A stack is a LIFO (Last In First Out) and size is notnot bounded bounded

Statecontrol

a a b b

x

y

z

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PDA and Language 0PDA and Language 0nn11nn

Can a PDA recognize this?Can a PDA recognize this? Yes, because size of stack is not boundedYes, because size of stack is not bounded Describe the PDA that recognizes this languageDescribe the PDA that recognizes this language

Read symbols from input. Push each 0 onto the Read symbols from input. Push each 0 onto the stack.stack.

As soon as a 1’s are seen, starting popping one 0 for As soon as a 1’s are seen, starting popping one 0 for each 1each 1

If finish reading the input and have no 0’s on stack, If finish reading the input and have no 0’s on stack, then accept the input stringthen accept the input string

If stack is empty and 1s remain or if stack becomes If stack is empty and 1s remain or if stack becomes empty and still 1’s in string, rejectempty and still 1’s in string, reject

If at any time see a 0 after seeing a 1, then rejectIf at any time see a 0 after seeing a 1, then reject

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Formal Definition of a Formal Definition of a PDAPDA

The formal definition of a PDA is The formal definition of a PDA is similar to that of a FA but now we similar to that of a FA but now we have a stackhave a stack Stack alphabet may be different from Stack alphabet may be different from

input alphabetinput alphabet Stack alphabet represented by Stack alphabet represented by ΓΓ

Transition function key part of definitionTransition function key part of definition Domain of transition function is Q × Domain of transition function is Q × εε × × ΓεΓε The current state, next input symbol and top The current state, next input symbol and top

stack symbol determine the next movestack symbol determine the next move

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Definition of PDADefinition of PDA A pushdown automata is a 6-tuple (Q, A pushdown automata is a 6-tuple (Q, , , ΓΓ, ,

δδ, q, q00, F), where Q, , F), where Q, , , ΓΓ, and F are finite , and F are finite setssets

1.1. Q is the set of statesQ is the set of states2.2. is the input alphabetis the input alphabet3.3. ΓΓ is the stack alphabet is the stack alphabet4.4. δ δ : Q × : Q × εε × × ΓεΓε P(Q × P(Q × ΓεΓε) is transition function ) is transition function

5.5. qq00 Q is the start state, and Q is the start state, and6.6. F F Q is the set of accept states Q is the set of accept states

Note that at any step the PDA may enter a Note that at any step the PDA may enter a new state and possibly write a symbol on top new state and possibly write a symbol on top of the stackof the stack

This definition allows nondeterminism since This definition allows nondeterminism since δδ can can return a setreturn a set

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How Does a PDA How Does a PDA Compute?Compute?

The following 3 conditions must be The following 3 conditions must be satisfied for a string to be accepted:satisfied for a string to be accepted:

1.1. M must start in the start state with an empty M must start in the start state with an empty stackstack

2.2. M must move according to the transition functionM must move according to the transition function

3.3. At the end of the input, M must be in an accept At the end of the input, M must be in an accept statestate

To make it easy to test for an empty stack, a To make it easy to test for an empty stack, a $ is initially pushed onto the stack$ is initially pushed onto the stack

If you see a $ at the top of the stack, you know it If you see a $ at the top of the stack, you know it is emptyis empty

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NotationNotation

We write a,b We write a,b c to mean: c to mean: when the machine is reading an a from the when the machine is reading an a from the

inputinput it may replace the b on the top of the stack with cit may replace the b on the top of the stack with c

Any of a, b, or c can be Any of a, b, or c can be εε If a is If a is εε then can make stack change without then can make stack change without

reading an input symbolreading an input symbol If b is If b is εε then no need to pop a symbol (just push c) then no need to pop a symbol (just push c) If c is If c is εε then no new symbol is written (just pop b) then no new symbol is written (just pop b)

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0, ε 0

1, 0 ε

ε, ε $

ε, $ ε

Example 1: a PDA for Example 1: a PDA for 00nn11nn

Formally describe PDA that accepts {0Formally describe PDA that accepts {0nn11nn|n |n ≥0}≥0} Let M1 be (Q, Let M1 be (Q, , , ΓΓ, , δδ, q, q00, F), where, F), where

Q = {qQ = {q11, q, q22, q, q33, q, q44} } = {0, 1} = {0, 1}

ΓΓ = {0, $} F = {q = {0, $} F = {q11, q, q44}}

q1

q4

q2

q3

1, 0 ε

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Example 2: PDA for aExample 2: PDA for aiibbjjcckk, , i=j or i=ki=j or i=k

Come up with a PDA that recognizes Come up with a PDA that recognizes the language {athe language {aiibbjjcckk| i, j, k ≥0 and i=j | i, j, k ≥0 and i=j or i=k}or i=k} Come up with an informal description, as Come up with an informal description, as

we did initially for 0we did initially for 0nn11nn

Can you do it without using non-determinism?Can you do it without using non-determinism? NoNo

With non-determinism?With non-determinism? Easy, similar to 0Easy, similar to 0nn11n n except that guess whether to except that guess whether to

match a’s with b’s or c’s. See Figure 2.17 page 114match a’s with b’s or c’s. See Figure 2.17 page 114 Since NPDA ≠ PDA, create a PDA if possibleSince NPDA ≠ PDA, create a PDA if possible

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Example 3: PDA for Example 3: PDA for {ww{wwRR}}

Come up with a PDA for the following Come up with a PDA for the following language: {wwlanguage: {wwRR| w | w {0,1}* } {0,1}* } Recall that wRecall that wRR is the reverse of w so this is is the reverse of w so this is

the language of palindromesthe language of palindromes Can you informally describe the PDA? Can Can you informally describe the PDA? Can

you come up with a deterministic one?you come up with a deterministic one? NoNo

Can you come up with a non-deterministic Can you come up with a non-deterministic one?one?

Yes, push symbols that are read onto the stack Yes, push symbols that are read onto the stack and at some point nondeterministically guess that and at some point nondeterministically guess that you are in the middle of the string and then pop you are in the middle of the string and then pop off stack value as they match the input (if no off stack value as they match the input (if no match, then reject)match, then reject)

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Diagram of PDA for Diagram of PDA for {ww{wwRR}}

0, ε 01, ε 1

0, 0 ε1, 1 ε

ε, ε $

ε, $ ε

q1

q4

q2

q3

ε, ε ε

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Equivalence with CFGsEquivalence with CFGs Theorem: A language is context free if and only if Theorem: A language is context free if and only if

some pushdown automaton recognizes itsome pushdown automaton recognizes it Lemma: if a language L is context free then some PDA Lemma: if a language L is context free then some PDA

recognizes it (we won’t bother doing other direction)recognizes it (we won’t bother doing other direction) Proof idea: We show how to take a CFG that generates L and Proof idea: We show how to take a CFG that generates L and

convert it into an equivalent PDA Pconvert it into an equivalent PDA P Thus P accepts a string only if the CFG can derive itThus P accepts a string only if the CFG can derive it Each main step of the PDA involves an application of one rule in Each main step of the PDA involves an application of one rule in

the CFGthe CFG The stack contains the intermediate strings generated by the CFGThe stack contains the intermediate strings generated by the CFG Since the CFG may have a choice of rules to apply, the PDA must Since the CFG may have a choice of rules to apply, the PDA must

use its non-determinismuse its non-determinism One issue: since the PDA can only access the top of the stack, any One issue: since the PDA can only access the top of the stack, any

terminal symbols pushed onto the top of the stack must be terminal symbols pushed onto the top of the stack must be checked against the input string immediately.checked against the input string immediately.

If the terminal symbol matches the next input character, then If the terminal symbol matches the next input character, then advance input stringadvance input string

If the terminal symbol does not match, then terminate that pathIf the terminal symbol does not match, then terminate that path

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Informal Description of P Informal Description of P

Place marker symbol $ and the start variable on Place marker symbol $ and the start variable on the stackthe stack

Repeat foreverRepeat forever If the top of the stack is a variable A, If the top of the stack is a variable A,

nondeterministically select one of the rules for A and nondeterministically select one of the rules for A and substitute A by the string on the right-hand side of substitute A by the string on the right-hand side of the rulethe rule

If the top of stack is a terminal symbol a, read the If the top of stack is a terminal symbol a, read the next symbol from the input and compare it to a. If next symbol from the input and compare it to a. If they match, repeat. If they do not match, reject this they match, repeat. If they do not match, reject this branch.branch.

If the top of stack is the symbol $, enter the accept If the top of stack is the symbol $, enter the accept state. Doing so accepts the input if it has all been state. Doing so accepts the input if it has all been read. read.

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ExampleExample

Look at Example 2.25 on page 118 for an Look at Example 2.25 on page 118 for an example of constructing a PDA P1 from a example of constructing a PDA P1 from a CFG GCFG G Note that the top path that branches to the right Note that the top path that branches to the right

will take S and replace it with aTbwill take S and replace it with aTb It first pushes b, then T, then aIt first pushes b, then T, then a

Note the path below that replaces T with TaNote the path below that replaces T with Ta It replaces T with a then pops T on top of thatIt replaces T with a then pops T on top of that

Your task:Your task: Show how this PDA accepts the string aab, which has the Show how this PDA accepts the string aab, which has the

following derivation:following derivation: S S aTb aTb aTab aTab aab aab

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Example continuedExample continued In the following, the left is the top of stackIn the following, the left is the top of stack

We start with S$We start with S$ We take the top branch to the right and we get the We take the top branch to the right and we get the

following as we go thru each state:following as we go thru each state: S$ S$ b$ b$ Tb$ Tb$ aTb$ aTb$

We read a and use rule a,a We read a and use rule a,a εε to pop it to get Tb$ to pop it to get Tb$ We next take the 2We next take the 2ndnd branch going to right: branch going to right:

Tb$ Tb$ ab$ ab$ Tab$ Tab$ We next use rule We next use rule εε,T ,T εε to pop T to get ab$ to pop T to get ab$ Then we pop a then pop b at which point we have $Then we pop a then pop b at which point we have $ Everything read so acceptEverything read so accept

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Relationship of Regular Relationship of Regular Languages & CFLsLanguages & CFLs

We know that CFGs define CFLsWe know that CFGs define CFLs We now know that a PDA recognizes the same We now know that a PDA recognizes the same

class of languages and hence recognizes CFLsclass of languages and hence recognizes CFLs We know that every PDA is a FA that just We know that every PDA is a FA that just

ignores the stackignores the stack Thus PDAs recognize regular languagesThus PDAs recognize regular languages Thus the class of CFLs contains regular Thus the class of CFLs contains regular

languageslanguages But since we know that a FA is not as powerful But since we know that a FA is not as powerful

as a PDA (e.g., 0as a PDA (e.g., 0nn11nn) we can say more) we can say more CFLs and regular languages are not equivalentCFLs and regular languages are not equivalent

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Relationship between CFLs Relationship between CFLs and Regular Languagesand Regular Languages

Regular languages

Context Free Languages

38

Section 2.3Section 2.3

Non-Context Free LanguagesNon-Context Free Languages

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Non Context Free Non Context Free LanguagesLanguages

Just like there are languages that are Just like there are languages that are not regular, there are languages that not regular, there are languages that are not context freeare not context free This means that they cannot be generated This means that they cannot be generated

by a CFGby a CFG This means that they cannot be generated This means that they cannot be generated

by a PDAby a PDA Just your luck! There is also a pumping Just your luck! There is also a pumping

lemma to prove that a language is not lemma to prove that a language is not context free!context free!

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Pumping Lemma for Pumping Lemma for CFGsCFGs

If A is a context-free language then If A is a context-free language then there is a pumping length p where, there is a pumping length p where, if s is any string in A with length ≥ if s is any string in A with length ≥ p, then s may be divided into five p, then s may be divided into five pieces x = uvxyz satisfying 3 pieces x = uvxyz satisfying 3 conditions:conditions:

1.1. For each i ≥0, uvFor each i ≥0, uviixyxyiiz z AA

2.2. |vy| > 0, and|vy| > 0, and

3.3. |vxy| ≤ p|vxy| ≤ p

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Proof IdeaProof Idea

For regular languages we applied the For regular languages we applied the pigeonhole principle to the number of states to pigeonhole principle to the number of states to show that a state had to be repeated.show that a state had to be repeated. Here we apply the same principle to the number of Here we apply the same principle to the number of

variables in the CFG to show that some variable will variables in the CFG to show that some variable will need to be repeated given a sufficiently long stringneed to be repeated given a sufficiently long string

We will call this variable R and assume it can derive XWe will call this variable R and assume it can derive X I don’t find the diagram on the next slide as obvious I don’t find the diagram on the next slide as obvious

as the various texts suggest. However, if you first as the various texts suggest. However, if you first put the CFG into a specific form, then it is a bit put the CFG into a specific form, then it is a bit clearer.clearer.

Mainly just be sure you can apply the pumping lemmaMainly just be sure you can apply the pumping lemma

42

Proof Idea ContinuedProof Idea Continued

T T uRz R uRz R x R x R vRy vRySince we can keep applying rule RSince we can keep applying rule R vRy, we can derive vRy, we can derive

uvuviixyxyiiz which therefore must also belong to the languagesz which therefore must also belong to the languages

T

R

R

u v x y z

43

Example IExample I Use the pumping lemma to show that the language Use the pumping lemma to show that the language

B = {aB = {annbbnnccnn | n ≥ 0} (Ex 2.36) | n ≥ 0} (Ex 2.36) What string should we pick?What string should we pick? Select the string s = aSelect the string s = appbbppccpp. S . S B and |s| > p B and |s| > p Condition 2 says that v and y cannot both be emptyCondition 2 says that v and y cannot both be empty Using the books reasoning we can break things into two Using the books reasoning we can break things into two

cases (note that |vxy| ≤ p). What should they be or how cases (note that |vxy| ≤ p). What should they be or how would you proceed?would you proceed?

v and y each only contain one type of symbol (one symbol is v and y each only contain one type of symbol (one symbol is left out)left out)

uvuv22xyxy22z cannot contain equal number of a’s, b’s and c’sz cannot contain equal number of a’s, b’s and c’s Either v or y contains more than one type of symbolEither v or y contains more than one type of symbol

Pumping will violate the separation of a’s, b’s and c’sPumping will violate the separation of a’s, b’s and c’s We used this reasoning before for regular languagesWe used this reasoning before for regular languages

Using my reasoningUsing my reasoning Since |vxy| ≤ p v and y contain at most 2 symbols and hence Since |vxy| ≤ p v and y contain at most 2 symbols and hence

at least one is left out when pump up (technically we should at least one is left out when pump up (technically we should say at least one symbols is in v or y so that pumping break say at least one symbols is in v or y so that pumping break the equality)the equality)

44

Example IIExample II Let D = {ww | w Let D = {ww | w {0,1}*} (Ex 2.38) {0,1}*} (Ex 2.38)

What string should we pick?What string should we pick? A possible choice is to choose s = {0A possible choice is to choose s = {0pp1010pp1}1}

But this can be pumped– try generating uvxyz so it can be But this can be pumped– try generating uvxyz so it can be pumpedpumped

Hint: we need to straddle the middle for this to workHint: we need to straddle the middle for this to work Solution: u=0Solution: u=0p-1p-1, v=0, x=1, y=0, z=0, v=0, x=1, y=0, z=0p-1p-111 Check it. Does uvCheck it. Does uv22xyxy22z z D? Does uwz D? Does uwz D? D?

Choose s = {0Choose s = {0pp11pp00pp11pp}} First note that vxy must straddle midpoint. Otherwise First note that vxy must straddle midpoint. Otherwise

pumping makes 1st half ≠ 2nd halfpumping makes 1st half ≠ 2nd half Book says another way: if vxy on left, pumping it moves a 1 Book says another way: if vxy on left, pumping it moves a 1

into second half and if on right, moves 0 into first halfinto second half and if on right, moves 0 into first half since |vxy| < p, it is all 1’s in left case and all 0’s in right since |vxy| < p, it is all 1’s in left case and all 0’s in right

casecase If does straddle midpoint, if pump down, then not of form If does straddle midpoint, if pump down, then not of form

ww since neither 1’s or 0’s match in each halfww since neither 1’s or 0’s match in each half

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Summary of Pumping Lemma Summary of Pumping Lemma for CFGsfor CFGs

Just remember the basics and the Just remember the basics and the conditionsconditions You should memorize them for both You should memorize them for both

pumping lemmas, but think about what pumping lemmas, but think about what they meanthey mean

I may give you the pumping lemma I may give you the pumping lemma definitions on the exam, but perhaps definitions on the exam, but perhaps not. You should remember both not. You should remember both pumping lemmas and the conditions pumping lemmas and the conditions They should not be too hard to rememberThey should not be too hard to remember