CSCI 3130: Formal languages and automata theory Tutorial 4 Chin

CSCI 3130: Formal languagesand automata theory

Tutorial 4

Chin

Reminder

• Homework 3 is due on next Thursday.

• You can get back homework 1 after this tutorial.

Context Free Grammar

A → 0A1A → BB → #

A 0A1 00A11 000A111

000B111 000#111

derivation = a sequence of productions that results a string

productions

variablesterminals

start variable


• Regular languages are context free• Context free languages NOT necessarily regular

• CFG describes the recursive structure of languages


• Design a CFG that represents the following.• = {a, b}

L1 = {wywR : y, w ∈ *}

L2 = {ai bj : i < j}


• L1 = {wywR : y, w ∈ *}• Consider a simpler CFG L1’ = {wwR : w ∈ *}

• e.g. z = aabaabaa• First character and the last character are the same

z = aabaabaa = ax1a

x1 = abaaba = ax2a

x2 = baab = bx3b

x3 = aa = ax4a

S → aSa | bSb ? How to stop? base case?S → aSa | bSb |

recursive


• L1 = {wywR : y, w ∈ *}• Consider a simpler CFG L1’ = {wwR : w ∈ *}

S → aSa | bSb | How to turn it into a CFG for L1?

e.g. aabaaababaa? S → aSa | bSb | < change this

How to write a CFG for {y : y ∈ *}? T → aT | bT | Replace by T → aT | bT |

S → aSa | bSb | TT → aT | bT |


• L2 = {ai bj : i < j}• Consider a simpler CFG L2’ = {aibj : i = j}

• e.g. z = aaaabbbb• First character must be a and the last character must be b

z = aaaabbbb = ax1b

x1 = aaabbb = ax2b

x2 = aabb = ax3b

x3 = ab = ax4b

S → aSb ? Base case again…S → aSb |

recursive


• L2 = {ai bj : i < j}• Consider a simpler CFG L2’ = {wwR : w ∈ *}

S → aSb | How to turn it into a CFG for L2? need at least 1 more b in bb…b

e.g. aaaabbbbbb? S → aSb | < change this. insert b’s in front of bb…b

How to write a CFG for {bi : i > 0}? T → bT | bReplace by T → bT | b

S → aSb | | TT → bT | b

Parse Trees

• Representation of derivationse.g. S → AB | ba

A → aA | aB → b

Derivation of aab• S → AB → aAB → aaB → aab

S

B

b

a

Aa

A

Ambiguity

• A CFG is ambiguous if the deviation of some string has two different parse trees.

• Removing ambiguity is impossible for some CFG.

Ambiguity

• Show that the following CFG is ambiguous.E → E * E | E / E | (A) | NA → A + A | NN → 6 | 1 | 2

• Find a string in the CFG and show that it has two different parse trees.

Ambiguity

E → E * E | E / E | (A) | NA → A + A | NN → 6 | 1 | 2

•6 / 2 * (1 + 2)

* E

+A A

1

N

2

N

E

/E E

6

N

2

N

A

E

( )

E

E

*

6

N

E /

E E

2

N

+A A

1

N

2

N

A( )

Parsing - Preprocessing

1. Eliminate productions2. Eliminate unit productions

• Do the steps in order (1 then 2)


1. Eliminate productions i. Identify nullable variables (N )Let Q be the queue containing the nullable variables.Repeat the following:

If X , push X into QIf X YZ…W and XZ…W are all in Q, push X into Q

If start variable S nullable, add S’ → S |

*


1. Eliminate productions i. Identify nullable variables (N )e.g. S → XY

X → Y →

Q = (X, Y, S) but not (S, X, Y) S’ → SS → XYX → Y →

*

Repeat the following:If X , push X into QIf X YZ…W and XZ…W are all in Q, push X into Q


1. Eliminate productionsii. Remove nullable variablesRepeat the following until Q is empty:

Let N be the first element in Qa) For each X → N, add X → b) Remove all N → Remove N from Q

Caution!: S → NTN becomes S → NTN | NT | TN | T



• Eliminate productions for the following CFGS → ASA | aBA → B | SB → b |


S → ASA | aBA → B | SB → b |

•B → , Q = (B)•A → B and B is in Q, Q = (B, A)

i. Identify nullable variables (N )Let Q be the queue containing the nullable variables.Repeat the following:

If X , push X into QIf X YZ…W and XZ…W are all in Q, push X into Q

*


ii. Remove nullable variablesRepeat the following until Q is empty:

Let N be the first element in Qa) For each X → N, add X → b) Remove all N → Remove N from Q.


ii. Remove nullable variablesRepeat the following until Q is empty:

Let N be the first element in Qa) For each X → N, add X → b) Remove all N → Remove N from Q.




i. If there is a cycle of unit productionsA → B → ... → C → A

delete it and replace everything with A


ii. Replace every chainA → B → ... → C →

by A → , B → , ... , C →


CYK Algorithm

1. Eliminate productions2. Eliminate unit productions

• CYK Algorithm1. More preprocessing – Chomsky normal form2. Dynamic Programming

CYK Algorithm - Preprocessing

1. Chomsky normal formevery production has the form

A → BC or A → aAllow S → for start variable

A → BCDEC → c

A → BXX → CDEC → c

A → BXX → CYY → DEC → c

A → BcDE





Parsing

• CYK algorithmDynamic Programming (taught in CSCI3160)Let s = s1s2s3s4…sn be a string

Let s(i, j) be the substring si…sj of s

e.g. s = abcde, s(2,4) = bcd

s(i, j) can be construct by s(i, k) + s(k + 1, j) for some k.e.g. s = abcde

s = “a” + “bcde”, “ab” + “cde”, …, “abcd” + “e”

Parsing

• CYK algorithmDynamic Programming (taught in CSCI3160)s(i, j) can be construct by s(i, k) + s(k + 1, j) for some k.

Main Idea:If A derives s(i, k) and B derives s(k + 1, j), and S → AB. Then S derives s(i, j)e.g. A derives “ab”, B derives “cde”, S → AB.Then S derives “abcde”

Cocke-Younger-Kasami algorithm

• Use the CYK algorithm to parse abbab

for the CFGS → ASA | aBA → B | SB → b |


ba a bb

-- - ---- ---- ---

-


ba a bb

ABSAU SAU ABAB-- ---- ---

-

S,A,U can derive aA,B can derive b


ba a bb

ABSAU SAU ABAB-SAX SAXSA-- ---

-

If we can derive“a”(SAU) and “b”(AB), thenwe can derive “ab”.

SAU derives “a”, AB derives “b”.Look for variables that produceSA, SB, AA, AB, UA, or UBS → SA, A → SA, X → SA

similarly for “bb”, “ba”, “ab”


ba a bb

ABSAU SAU ABAB-SAX SAXSA

SASAX SAX--

-

If we can derive“ab”(SAX) and “b”(AB), or“a”(SAU) and “bb”(-), then we can derive “abb”.

1. SAU derives “a”, nothing derives “b”

2. SAX derives “ab”, AB derives “b”.Look for variables that produceSA, SB, AA, AB, XA, or XBS → SA, A → SA, X → SA

similarly for “bba” and “bab”


ba a bb


SASAX SAXSAXSAX

-

If we can derive“a”(SAU) and “bba”(SA), or“ab”(SAX) and “ba”(SA), or“abb”(SAX) and “a”(SAU), thenwe can derive “abba”.

SAX derives “ab”, SA derives “ba”.Look for variables that produceSS, SA, AA, AS, XS, or XAS → SA, A → SA, X → SA

similarly for “bbab”


ba a bb


SASAX SAXSAXSAX

SAX

If we can derive“a”(SAU) and “bbab”(SAX), or“ab”(SAX) and “bab”(SAX), or“abb”(SAX) and “ab”(SAX), thenwe can derive “abbab”.

SAU derives “a”, AB derives “bbab”.Look for variables that produceSA, SB, AA, AB, UA, or UBS → SA, A → SA, X → SA

If S is on the top left cell, thenx can be derived.

Starting from S in the top left, there must be some k such that

s(i, j) = s(i, k) + s(k+1, j)

Then there must be some l1, l2

such thats(i, k) = s(i, l1) + s(l1+1, k)

s(i, k) = s(k+1, l2) + s(l2+1, j)

Do the rest recursively.

Parse tree reconstruction

ba a bb


SASAX SAXSAXSAX

SAX


s(i, j) = s(i, k) + s(k+1, j)



s(i, k) = s(k+1, l2) + s(l2+1, j)



ba a bb


SASAX SAXSAXSAX

SAX


s(i, j) = s(i, k) + s(k+1, j)



s(i, k) = s(k+1, l2) + s(l2+1, j)



ba a bb


SASAX SAXSAXSAX

SAX

End

• Questions?

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CSCI 3130: Formal languages and automata theory Tutorial 4 Chin