Theory of Automata Solved Assignments Semester Spring 2010 Assignment 1
Question No.1
Marks: 4
a) Give regular expressions of the following languages over Σ={0,1}: 1. All strings having no pair of consecutive zeros. 2. All strings having exactly two 1’s or three 1’s not more than it. b) Show that the Regular expression ^ + 0(0+1)*+(0+1)*00(0+1)* is equivalent to ((0*1)*01*)* Solution: a) 1. RE for the language having all strings having no pair of consecutive zeros.
(1 + 01)*(0 + 1*) Other possibilities might be (1 + 01)*(0 + λ) + 1* or (1 + 01)* + (1 + 01)*0 + 1*
2. RE for the language language having all strings having exactly two 1’s or three three 1’s not more than it.
0*10*10* + 0*10*10*10*
b) After generating the string from both regular expressions :
^ + 0(0+1)*+(0+1)*00(0+1)*= ^, 0, 00, 01,001,010,100,011,000,0011,1100…… 01,001,010,100,011,000,0011,1100……… … ((0*1)*01*)* = ^,0,00,01,10,010,100,011,101,110,000, ^,0,00,01,10,010,100,011,101,110,000,…………….. ……………..
We can see that first RE cannot generate the string ‘10’ while this s tring can be generated by second RE. So they are not expressing the same language. Hence it is concluded that given Regular expressions are NOT equivalent.
Question No. 2
Marks: 4
={-,0,1,2,3,4,5,6,7,8,9}, a) Give recursive definition for the language ODD, of strings defined over ∑={-,0,1,2,3,4,5,6,7,8,9}, b) Give recursive definition for the language of palindromes having odd length
Solution:
a) Step1: 1 is in ODD Step2: if x is in ODD, then x+2 and x-2 are also in ODD Step3: No Strings except those constructed above, are allowed to be in ODD
b) Step1: a and b are in PALINDROME. Step2: if x is in PALINDROM then axa and bxb are also in PALINDROME Step3: No Strings except those constructed above, are allowed to be in PALINDROME
Question No. 3
Three Finite Automata (FAs) have been given below:
FA1
Marks: 6
FA2
FA3
Match the following Regular Expressions (RE’s) with corresponding Finite Automata (FA’s) given above. Also describe their languages (in English).
Regular Expression a(aa)*(^+a)b+b aa*b(aa*b)*
Finite Automaton (FA)
Language Description
(a+b)*a(a+b)*b(a+b)*
Solution:
Regular Expression
Finite Automaton (FA)
Language Description
a(aa)*(^+a)b+b
FA3
All strings in which “b” occurs only one time and that is only at the end of the string.
aa*b(aa*b)*
FA1
All strings which begins with “a” and ends at “b” and not containing the substring “bb”.
(a+b)*a(a+b)*b(a+b)*
FA2
All strings having substring “ab”.
Question No. 4
Marks: 6
Build an FA over Σ={a, b} that accepts only those words that do not contain the substring “ba”. Also make transition table for that FA.
Solution: FA
Transition table corresponding to above FA
New States Old States
After Reading a
After Reading b
q0-
q3
q1
q1+
q2
q1
q2
q2
q3
q1
q2 q3+
Assignment 2
Question No.1
Find the Regular Expression corresponding to the given FA. Show all steps. [Hint: FA to GTG a GTG to RE.]
Solution:
Step1:
Mark: 5
Step2:
Eliminating loop from q1
Step3:
Eliminating state q1
Step4:
To connect the initial state to the final state by single transition edge, the above TG can be reduced to the following
Hence the Required RE is
((0+1)0*11*0 )* . (0+1)0*11*
Question No.2
Mark: 5
A student walks into a classroom and sees o n the blackboard a diagram of a TG wi th two states that accepts only the string Λ. The student reverses the direction of exactly one edge, leaving all other edges and all labels and all final states (+’s) and start state (-) the same. The new TG accepts the language a*. What was the original TG?
Solution:
Original TG
accepts only Λ
New TG accepts the language a*
Question No. 3
Determine RE corresponding to the following TG. Show all steps. [Hint: TG to GTG a GTG to RE.]
Solution:
Step 1:
By eliminating states 7,8,10,11
Mark: 10
Step 2:
By eliminating states 6
Step 3:
By eliminating states 9 and 12
Step 4:
By combining loops at state 3
Step 5:
By eliminating states 2
Step 6:
By eliminating states 4
Step 7:
By eliminating states 3
Hence the Required RE is ba(abba+aabb)*(b+Λ)ab ba(abba+aabb)*(b+Λ)ab
Assignment 3
Question No 1:
Marks: 12
Let r1= (a+b)*ab(a+b)*
r2= (a*ba)*a*b
(The RE given in assignment was incorrect so it has been corrected now
According to the requirement of the question you had to work with FAs so mistake in RE will not effect your solution)
FA1 corresponding to r1
FA2 corresponding to r2
Using the FAs corresponding to r 1 and r2, Build an FA corresponding to
a) r1+ r2 (Union of r1 and r2). b) r1r2 (Concatenation of r1 and r2). c) r1* (Closure of r1).
Note : You
have to specify transition table and transition diagram for each part(a)(b)(c)
Solution: a)
Transition table corresponding to required FA
Old States
New states after reading a
z1 − ≡ z2 ≡
( x1 ,
( x2 ,
y1 ) y1 )
b
( x2 ,
y1 ) ≡ z2
( x1 ,
y2 ) ≡ z3
( x2 ,
y1 ) ≡ z2
( x3 ,
y2 ) ≡ z4
z3 + ≡
( x1 ,
y2 )
( x2 ,
y1 ) ≡ z2
( x1 ,
y3 ) ≡ z5
z4 + ≡
( x3 ,
y2 )
( x3 ,
y1 ) ≡ z7
( x3 ,
y3 ) ≡ z6
( x2 ,
y3 ) ≡ z8
( x1 ,
y3 ) ≡ z5
z5 ≡
( x1 ,
y3 )
z6 + ≡
( x3 ,
y3 )
( x3 ,
y3 ) ≡ z6
( x3 ,
y3 ) ≡ z6
z7 + ≡
( x3 ,
y1 )
( x3 ,
y1 ) ≡ z7
( x3 ,
y2 ) ≡ z4
( x2 ,
y3 ) ≡ z8
( x3 ,
y3 ) ≡ z6
z8 ≡
( x2 ,
y3 )
FA corresponding to r1+ r2
b)
Transition table corresponding to required FA
Old States
New states after reading a
b
z1 − ≡ x1
x2 ≡ z 2
x1 ≡ z1
z2 ≡ x2
x2 ≡ z 2
( x3 ,
y1 ) ≡ z3
z3 ≡
( x3 ,
y1 )
( x3 ,
y1 ) ≡ z3
( x3 ,
y1 , y2 ) = z4
z4 + ≡
( x3 ,
y1 , y2 )
( x3 ,
y1 ) = z3
( x3 ,
y1 , y2 , y3 ) ≡ z5
z5 + ≡
( x3 ,
y1 , y2 , y3 )
( x3 ,
y1 , y3 ) ≡ z6
( x3 ,
y1 , y2 , y3 ) ≡ z5
( x3 ,
y1 , y3 ) ≡ z6
( x3 ,
y1 , y2 , y3 ) ≡ z5
z6 ≡
( x3 ,
y1 , y3 )
FA corresponding to r1r2
c)
Transition table corresponding to required FA
Old States
New states after reading a
b
x2 ≡ z3
x1 ≡ z 2
Non-final z2 ≡ x1
x2 ≡ z3
x1 ≡ z 2
z3 ≡ x2
x2 ≡ z3
( x3 , x1 ) ≡
z4
Final z1 ± ≡
x1
z4 + ≡
( x3 , x1 )
( x3 , x1 , x2 ) ≡
z5
( x3 , x1 ) =
z4
z5 + ≡
( x3 , x1 , x2 )
( x3 , x1 , x2 ) ≡
z5
( x3 , x1 ) =
z4
FA corresponding to r1*
Question No 2:
Convert the following NFA into FA.
Marks: 4
Solution:
Transition table corresponding to required FA
Old States
New states after reading a
b Φ
z1 ± =
0
(1, 2) ≡
z2 + ≡
(1, (1, 2)
(1, 2,3) 2, 3) ≡
z3
(1, (1, 3) ≡ z4
z3 + ≡
(1, 2,3) 2, 3)
(1, 2,3) 2, 3) ≡
z3
(1, (1, 3) ≡ z4
z4 + ≡
(1, (1, 3)
(1, 2,3) 2, 3) ≡
z3
z2
Φ
FA corresponding to given NFA
Question No 3:
Build an NFA equivalent to the FA * where
Marks: 4
FA
Solution:
NFA equivalent to the FA * where
Assignment 4
Question No 1:
Marks: 5
Let L be the language of strings s trings defined over Σ= {0, 1} having odd no. of 1’s.Now Find Distinguishable and Indistinguishable strings by taking z as “1” belonging to Σ*.
Solution:
x
y
Distinguishable/Indistinguishable
11000
10000
Distinguishable
00000
11111
Distinguishable
10010
00110
Indistinguishable
10011
10000
Indistinguishable
00111
10111
Distinguishable
Question No 2: Marks:10
Convert the following Moore machine to an equivalent Mealy machine:
a) Construct transition table for Mealy machine. b) Run the string “babaaba” on both machines and find the output. o utput.
Solution:
Mealy machine corresponding to Moore machine
a) Transition table for Mealy machine. input States
q0
a
b
Output
Next State
Output
Next State
1
q2
0
q1
q1
1
q2
0
q3
q2
0
q3
1
q2
q3
0
q3
0
q3
b) Run the string “babaaba” on both machines and find the output. o utput.
Input States
q0
Mealy Moore
1
b
a
b
a
a
b
a
q1
q2
q2
q3
q3
q3
q3
0
1
1
0
0
0
0
0
1
1
0
0
0
0
Question No 3: Marks:5
Using De-Morgan’s law determine determine an FA corresponding corresponding to FA1∩FA2
FA1:
FA2:
Solution:
FA1c:
FA2c:
Transition table for for FA1c U FA2c
New States After Reading Old states
a
b
z1 ± ≡ (x0,y0)
z2 ≡ (x1,y0)
z3 ≡ (x1,y1)
z2 + ≡ (x1,y0)
z1 ≡ (x0,y0)
z4 ≡ (x0,y1)
z3+ ≡ (x1,y1)
z5 ≡ (x0,y2)
z4 ≡ (x0,y1)
z4+ ≡ (x0, y1)
z6 ≡ (x1,y2)
z3 ≡ (x1,y1)
z5 ≡ (x0,y2)
z6 ≡ (x1,y2)
z6 ≡ (x1,y2)
z6+ ≡ (x1,y2)
z5 ≡ (x0,y2)
z5 ≡ (x0,y2)
FA corresponding to FA1c U FA2c
FA corresponding to ( FA1c U FA2c)c = FA1 ∩ FA2
Assignment 5
Question No 1:
Marks: 5
Find the prefixes of Q in R where Q and R are languages give below:
Q={ ababa , aabbbbbbb,baabba , babbaaaaaa }
R={ aababa , bbaabba , bbaababa, aaaabbbbbbb, abbabbaaaaaa}
Solution:
Pref(Q in R)={a,b,aa,ab,bba, }
Question No. 2
By marking the states, decide whether the following FA accepts any word or not?
Note:
Show all steps
Marks: 5
Solution:
Marking and processing state q0
Processing state q1
Processing state q3
Processing state q2
Since no new state is marked and the final state q4 is unmarked therefore, the given FA does not accept any word.
Question No. 3
Marks: 5
Describe the language (in English) generated by the following CFG:
S → XY X → aX | bX | Λ Y → baa | bab | bba | bbb
Solution:
The given CFG generates the language L = {w ∈ {a, b}*: length(w ) ≥ 3 and third letter of w , from right is b.
Question No. 4
Marks: 5
Construct the CFG that generates the language L = {w ∈ {a, b}*: length(w ) ≥ 2 and w begins with b and ends in aa or bb}.
Solution:
S → bXY|bb X → aX | bX | Λ Y → aa | bb
Assignment 6
Question:
Mark:5+5+5+5
Consider the following Context Free Grammar (CFG)
S → aY | Ybb | Y X→Λ|a Y → aXY | bb | XXa
a) b) c) d)
Is the CFG ambiguous, if Yes then justify your answer. Draw a Total Language Tree (TLT) (T LT) for the given CFG. Convert the CFG into CNF. Build the PDA corresponding to the CFG (in CNF) of part (c).
Solution :
a) The CFG is said to be ambiguous if there exists atleast one word of it’s language that can be generated by the different production trees.
The word abb can be generated by the following three d ifferent trees
S
⇒
aY
S
⇒
Ybb
S
⇒
Y
S
⇒
abb
S
⇒
XXabb
S
⇒
aXY
S
⇒
Xabb
S
⇒
aY
S
⇒
abb
S
⇒
abb
Hence it has been concluded that the given CFG is ambiguous.
b) Total Language Tree
We can see that Y is repeating it self so Tree is growing in recursive manner.
c) Converting CFG into CNF
Remove the null production X →Λ.The new CFG, without null productions, is:
S → aY | Ybb | Y X →a
Y →aXY | aY | bb | XXa | Xa | a
The only unit production is S → Y, where the Y can be replaced with all Y’s non-unit productions (i.e. all of them).
The new CFG, without wi thout unit productions, is:
S → aY | Ybb | aXY | aY | bb | XXa | Xa | a X →a Y →aXY | aY | bb | XXa | Xa | a
Introduce the non-terminal A →a B →b
The new CFG, consisting of only solid non-terminals or one terminal is:
S → AY | YBB | AXY | AY | BB | XXA | XA | a X→a Y → AXY | AY | BB | XXA | XA | a A→a B→b
Shorten the strings of non-terminals non-terminals to length 2.Create new, intermediate non-terminals to accomplish this.
C → BB D → XY E → XA
The new CFG, in CNF is:
S → AY | YC | AD | AY | BB | XE | XA | a X→a Y → AD | AY | BB | XE | XA | a A→a B→b C → BB D → XY E → XA
d) PDA corresponding to the CFG
