Homework Assignment 4 Solutions

CSCI 2670 Introduction to Theory of Computing, Fall 2016

Due Tuesday October 18, 2016

This homework assignment is about context-free grammars, Chomsky nor-mal form, and pushdown automata.

1. Design a context-free (or regular) grammar to describe the language recognized bythe following NFA.

Give a derivation and parsing tree for string aababa.

Solutions:

• The regular grammar for the language recognized by the NFA is:

A1 → A3 | bA2 | ε

A2 → aA2 | aA3 | bA3

A3 → aA1

• The derivation of string aababa is:

A1 ⇒ A3 ⇒ aA1 ⇒ aA3 ⇒ aaA1 ⇒ aabA2

⇒ aabaA2 ⇒ aababA3 ⇒ aababaA1 ⇒ aababaε = aababa

Draw parsing tree (in the classroom):

1

2. Use the grammar rules given in Question 2.1 (page 154), give parsing trees for thefollowing strings

(1) (a+ a)× (a× a)(2) ((a)× a) + a

Solutions:

• The grammar is:

E → E + T |T

T → T × F |F

F → (E) | a

(1) the left-most derivation of string (a+ a)× (a× a):

E ⇒ T ⇒ T × F ⇒ F × F ⇒ (E)× F ⇒ (E + T )× F ⇒ (T + F )× F

⇒ (F + F )× F ⇒ (a+ F )× F ⇒ (a+ a)× F ⇒ (a+ a)× (E)⇒ (a+ a)× (T )

⇒ (a+ a)× (T × F )⇒ (a+ a)× (F × F )⇒ (a+ a)× (a× F )⇒ (a+ a)× (a× a)

Draw the parsing tree (in the classroom);

(2) the left-most derivation of string ((a)× a) + a:

E ⇒ E + T ⇒ T + T ⇒ F + T ⇒ (E) + T ⇒ (T ) + T ⇒ (T × F ) + T

⇒ (F × F ) + T ⇒ ((E)× F ) + T ⇒ ((T )× F ) + T ⇒ ((F )× F ) + T

⇒ ((a)× F ) + T ⇒ ((a)× a) + T ⇒ ((a)× a) + F ⇒ ((a)× a) + a

Draw the parsing tree (in the classroom).

3. Question 2.4 (c) and (e) on page 155.

Solutions:

(c) L = {w |w starts and ends with the same symbol }:

S → 0M0 | 1M1 | ε

A→ 0M | 1M | ε

(e) L = {w |w = wR, that is, w is a palindrome}:

S → 0S0 | 1S1 |M

M → ε | 0 | 1

4. Convert the following CFG to an equivalent CFG in Chomsky normal form, given theprocess given in Theorem 2.9.

S → ε, S → 0S1S, S → 1S0S

Solutions: (20 points in total)

Follow the 4 steps, i.e.,

(1) adding new start variable S0:

S0 → S, S → ε, S → 0S1S, S → 1S0S

(2). removing ε rules (S → ε):

S0 → S, S0 → ε, S0 → S, S → 01S, S → 0S1, S → 01, S → 0S1S,

S → 10S, S → 1S0, S → 10, S → 1S0S

(3) removing unit rules (S0 → S):

S0 → ε, S0 → 01S, S0 → 0S1, S0 → 01, S0 → 0S1S,

S0 → 10S, S0 → 1S0, S0 → 10, S0 → 1S0S

S → 01S, S → 0S1, S → 01, S → 0S1S,

S → 10S, S → 1S0, S → 10, S → 1S0S

(4) patching to proper forms, introducing new rules

Z → 0, W → 1, Ws →WS, Sw → SW, Zs → ZS, Sz → SZ,

Sws → SWs, Szs → SZs

patching to get additional rules:

S0 → ZWs, S0 → ZSw, S0 → ZW, S0 → ZSws, S0 →WZs, S0 →WSz, S0 →WZ, S0 →WSzs

S → ZWs, S → ZSw, S → ZW, S → ZSws, S →WZs, S →WSz, S →WZ, S →WSzs

and old rules:S0 → ε

5. Question 2.14 on page 156.

Solutions:

• The grammar isA→ BAB |B | ε

B → 00 | ε

• add the new start variable S0:

S0 → A

• remove ε-rules (B → ε):

S0 → A, A→ AB |BA |A |BAB |B | ε, B → 00

remove ε-rules (A→ ε):

S0 → A | ε, A→ AB |BA |BB |BAB |B, B → 00

• remove unit rules (A→ B):

S0 → A | ε, A→ AB |BA |BAB |BB | 00, B → 00

remove unit rules (S0 → A):

S0 → ε |AB |BA |BAB |BB | 00

A→ AB |BA |BAB |BB| 00, B → 00

• patching, introducing new rules:

Z → 0, Ab → AB

thenS0 → ε |AB |BA |BAb |BB |ZZ

A→ AB |BA |BAb |BB|ZZ, B → 00

6. Question 2.5 on page 155 (for question 2.4 (b) only).

Solutions: