The document discusses various concepts related to finite automata. It begins by defining a finite automaton as a mathematical model of a system with discrete inputs and outputs that can be in a finite number of states. A finite automaton consists of a finite set of states and transitions between states based on input symbols. The document then discusses formal languages, the functions of a head pointer and finite control, the two main types of finite automata (DFA and NFA), ways to represent automata, definitions of languages and transitions, regular expressions and languages, two-way finite automata, epsilon closure, equivalence of NFAs and DFAs, Moore and Mealy machines, and applications of finite automata such as lexical analysis.

1. UNIT –I
1) What is meant by finite automata?
Finite automaton is a mathematical model of a system with discrete inputs and outputs.
The system can be in any one of the finite number of states. The state of the system
summaries the information concerning past inputs that is needed to determine the behavior of
the systems on subsequent inputs. A finite automaton consists of a finite set of states and set
of transitions from state to state that occur on input symbols chosen from an alphabet ∑.
2) What is a formal language?
Language is a set of valid strings from some alphabet. The set may be empty, finite or
infinite. L(M) is the language defined by machine M and L(G) is the language defined by
context free grammar. The two notations for specifying formal languages are:
 Grammar or regular expression(Generative approach)
 Automaton( Recognition approach)
3) What are the functions of head pointer and finite control?
Head pointer:
The head examines only one square at a time and can move one square either to the left
or to the right but we restrict the movement of the reading head pointer only to the right-side.
Finite control:
The finite control contains the routines that instruct the reading head pointer to move
from one state to the next state by recognizing each symbol or alphabet.
4) What are the two main types of finite automata?
There are two types of finite automata,
i. Deterministic Finite Automata (DFA).
ii. Non deterministic Finite Automata (NFA).
DFA
For each and every state and for each and every input symbol there exist atmost one
transition. The transition mapping for DFA is Q×∑→Q.
NFA
For each and every state and for each and every input symbol there exist more than one
transition. The transition mapping for NFA is Q × (∑ U { }) → .

2. 5) What are the two ways of representing an automaton?
The two ways of representation of finite automata are ,
i. Transition Diagram.
ii. Transition Table.
Transition Diagram:
The transition diagram consists of finite set of states, symbols, initial state and
final state. It is a directed graph which shows the transition from one state to another.
Transition Table:
The transition table is the tabular representation of the transition function “delta
(∂)” with the rows denoting states and columns denoting input symbol.
6) Define a language.
An alphabet is a finite set of symbols. A language is a set of strings of symbols from
someone alphabet.
e.g.
If ∑ = {0,1},then ∑* }.
7) What is meant by transition?
Transition is the process of moving from one state to another state on reading an input
symbol. It can be represented as ∂ (q , a).
e.g.:
a
8) What is regular expression and regular language?
The language accepted by finite automata is described by a simple expression called as
regular expression. The language accepted by regular expression is called as regular
language.
9) What are two- way finite automata?
The deterministic finite automata as a control unit that reads a tape moving one square
right at each moved. We needed non-determinism to the model which allowed many “copies”
of the control unit to exist and scan the tape simultaneously .Next we added €- transitions,
which allowed change of state without reading the input symbol or moving the tape head.
Q0
Q1

3. Next interesting extension is to allow the tape head with the ability to move left as well as
right, such a finite automaton is called two way finite automata.
10) What is meant by -closure?
The “ ” is a character used to indicate the null string (i.e) the string which is used simply
for transition from one state to the other without any input. The NFA with € transition
function can be extended to ∂’ as ∂’=Q×∑*→
The -closure is represented as
11) State the equivalence theorem of NFA and DFA.
Let L be a language accepted by non-deterministic finite automata then there exist a
deterministic finite automata that accepts the same language “L”. It can also be said that the
language accepted by NFA is equal to the language accepted by DFA.
(i.e.) L(M)=L(M’)
Where,
M is the NFA and M’ is the DFA.
12) Define the language accepted by NFA with epsilon moves.
The language accepted by NFA with epsilon transition is represented as,
M= (Q,∑,∂,q0,F)
Where,
M is the NFA with moves.
Q is the finite set of non-empty states.
∑ is the finite set of non-empty symbols.
∂ is the finite set of transitions.
q0 is the initial state.
F is the final state.
13) What are the difference between NFA and DFA?
DFA NFA

4. i. For each and every state and
for each and every input
symbol there exist atmost one
transition.
ii. The transition mapping for
DFA is Q×∑→Q.
iii. The language accepted by
DFA is denoted as L(M).
iv. Epsilon transition is not
possible.
i. For each and every state and for
each and every input symbol
there exist more than one
transition.
ii. The transition mapping for NFA
is Q × (∑ U { }) → .
iii. The language accepted by NFA is
denoted by L(M’).
iv. Epsilon transition is possible.
14) Define regular expression theorem.
Let ‘r’ be the regular expression then there exists a NFA with -transition that accepts
L(r)
(i.e.) L(M)=L(r).
Where the regular expression “r” belonging to a language “L” can be accepted by
automata with .
15) Write down the rules for defining regular expression.
The rules that define the regular expression over alphabet ∑ are as follows,
 ø is a regular expression that denotes the empty set .
 is the regular expression that denotes the set
 For each ‘a’ in ∑, then the regular expression of a is denoted as {a}.
 If r and s are regular expression denoting language R and S respectively then
(r+s), rs and (r*) are regular expression that denote the set RUS, RS, R*
respectively.
16) Write down the operations of the regular expression.
There are main operations that are closed under regular expression,
i. Union.
ii. Concatenation.
iii. Kleene Closure.

5. Union :
Let r1 and r2 be the regular expressions and the union of these two regular
expressions are denoted as r1 U r2 or r1+r2 the general form of representation of union
operation is given by,
M==(Q1UQ2 U{q0,f0},∑1U∑2,∂,q0,{f0}).
Concatenation:
Let r1 and r2 be the regular expressions and the union of these two regular
expressions are denoted as r1. r2 or r1r2 the general form of representation of
concatenation operation is given by,
M==(Q1UQ2 ,∑1U∑2,∂,q0,{f0})
Kleene Closure :
Let r1 be the regular expressions and the union of these two regular expressions are
denoted as r1* or the general form of representation of kleene closure is given by,
M==(Q1U{q0,f0},∑,∂,q0,{f0})
17) State some applications of regular expression.
Some of the basic applications of regular expression are ,
 Lexical Analysis.
 Pattern searching.
18) State Arden’s theorem.
According to Arden’s theorem if the expression is of the form R=Q+RP, then we can
write it as the form R=QP*.
Where, P and Q are the regular expression.
Arden’s theorem is mainly used for checking the equivalence of two regular expression
as well as in conversion of DFA to regular expression.
19) What are the two types of finite automata with output?
There are two types of finite automata,
 Mealy machine.
 Moore machine.
An automaton in which the output depends on the states of the states of the
machine is called Moore machine.
An automaton in which the output depends on the state as well as on the input at
any instant of time is called a Mealy machine.

6. 20) State the difference between Mealy and Moore machine.
MEALY MACHINE MOORE MACHINE
i. For each and every transition there
will be a output.
ii. An automaton in which the output
depends on the state as well as on the
input at any instant of time is called a
Mealy machine.
iii. The transition mapping is given by,
λ:Q×∑→Δ
i. For each and every state there will be
a output.
ii. An automaton in which the output
depends on the states of the states of
the machine is called Moore machine.
iii. The transition mapping is given by,
λ:Q→Δ
21)State some applications of finite automata.
There are varieties of software design problems that are simplified by automatic
conversion of regular expression notation to an efficient computer implementation to the
corresponding finite automaton. Two such applications are
 Lexical analyser.
 Text Editor.
22) Construct a DFA with for all the set of strings with {0, 1} that has even number of 0’s
and 1’s.
1
1
0 1 0 0
1
1
23)Construct DFA for set of all strings {0,1} with 011 as substring.
0 0 1
1
1 0 1/
q4 q3
q1
q0 q1 q2
q0
q3

7. 24)Draw the block diagram of a finite automata?
25) Construct a NFA with -moves when a regular expression is given: r =01/101?
Taking r =01,
r1=0 r2=1
0 1
r3=r1r2
0  1
Taking r=101,
r4=1 r5=0 r6=1
1 0 1
r7=r4r5r6
1  0  1
¢ a b a $
FC
q0 q1 q2 q3
q0 q1 q2 q3
q4 q5 q6 q7 q8 q9
q4 q5 q6 q7 q8 q9

8. Now r=01/101
 0  1 
 1  0  1 
26) Explain the six tuples in a mealy and a moore machines ?
The moore and a mealy machine is being represented by M. Both the machine consists of
6-tuples.
M = (Q, ,,,,q0)
Where,
Q- set of finite states.
- set of input alphabets.
- set of output alphabets.
- set of input transition.
- set of output transition.
q0- initial states.
27) Explain finite automata with -moves ?
The ‘’ is a character used to indicate the null string (i,e) the string which is used
simply for transition from one state to the other without any input . The NFA with -
moves can be shown below:
a  
a b

28) Explain the transition mapping regarding finite automata with -moves?
q10
q0 q1 q2 q3
q11
q4 q5 q6 q7 q8 q9
q0 q2
q1
q3

9. The transition mapping regarding finite automata with -moves is :
Q x ({})  2^Q.
29) Give the transition mapping for moore and mealy machine?
For moore machine the transition mapping is:
:Q x   .
For mealy machine the transition mapping is:
: Q  .
30) Define the rules for transition diagram?
Rules for drawing transition diagram are:
1. The transition diagram starts with a arrow pointer.
2. The initial state is being represented by a circle.
3. Each alphabet or symbol is being represented above the edge or arcs.
4. The final state is represented by concentric circle.
31) What is the transition mapping for NFA and DFA?
For NFA:
:Q x Q
For DFA:
:Q x Q’ or 2^Q
32) Give an example for NFA? a
a b
a,b a,b
33)Give an example for DFA?
a b b
b a
34)Why are switching circuits called as finite state systems?
A switching circuit consists of a finite number of gates, each of which can be in any one
of the two conditions 0 or 1. Although the voltages assume infinite set of values, the
q0 q1 q2
q0 q1 q2

10. electronic circuitry is designed so that the voltages corresponding to 0 or 1 are stable and all
others adjust to these values. Thus control unit of a computer is a finite stat system.
35)Give the examples/applications designed as finite state system.
Text editors and lexical analyzers are designed as finite state systems. A lexical analyzer
scans the symbols of a program to locate strings corresponding to identifiers, constants etc,
and it has to remember limited amount of information.
36)What are the applications of automata theory?
Some of the applications of automata theory are:
 In complier construction.
 In switching theory and design of digital circuits.
 To verify the correctness of a program.
 Design and analysis of complex software and hardware systems.
 To design finite state machines such as Moore and Mealy machine.
37)Define Finite Automaton(FA)
FA consists of a finite set of states and a set of transitions from state to state that occur on
input symbols chosen from an alphabet ∑. Finite automaton is denoted by a 5-tuple (Q, ∑, δ,
q0, F), where
Q is the finite set of states,
∑ is the finite input alphabet,
δ is the transition mapping function,
q0 is the initial state.
F is the set of final state.
UNIT -2
1. What is meant by pumping lemma for regular language/regular set?
Let L be a regular set. Then there is a constant n such that if Z is any word in L,
and |z|≥n. we may write Z=uvw in such a way that |uv| ≤ n, |v| ≥ 1and for all L ≥0 then
uviw is in L.
2. What are the application of pumping lemma for regular set?
The applications of pumping lemma for regular set are:
 The pumping lemma is a powerful tool providing and proving certain language is
regular or not.

11.  It is also useful in the development of algorithms to answer certain question
concerning finite automata, such as whether the language accepted by a given FA
is finite or infinite.
3. What is homomorphisms?
A string homomorphism is a function on strings that works by substituting a
particular string for each symbol.
Example: The function h given by h(0) = abb and h(1) = ba is a homomorphism
which replaces each 0 by abb and each 1 by ba. Thus h(1011) = baabbbaba.
4. What is inverse homomorphisms?
Homomorphisms can also applied in reverse and in this mode they also preserve
regular language. That is, suppose h is a homomorphism from some alphabet ∑ to strings
in another (possibly the same) alphabet T-1. Let L be a language over alphabet T. Then h-1
is the set of strings w in ∑* such that h(w) is in L.
5. Write the principal closure properties for regular language?
The principal closure properties for regular language are:
i. The union of two regular languages is regular.
ii. The intersection of two regular languages is regular.
iii. The complement of a regular language is regular.
iv. The difference of two regular languages is regular.
6. Write the closure properties of regular set under Boolean operations?
The first closure properties are the three Boolean operations: Union, Intersection,
and Complementation:
i. Let L and M be language over alphabet ∑. Then LUM is the language that
contains all strings that are in either or both of L and M.
ii. Let L and M be language over alphabet ∑. Then L∩M is the language that
contains all strings that are in both of L and M.
iii. Let L be a language over alphabet ∑. Then , the complement of L, is the
set of strings in ∑* that are not in L.
7. What is reversal?
The reversal of a string a1a2...anis the string written backwards, that is, anan-1…a1.
We use wR for the reversal of string w. thus, 0010R is 0100, and ЄR = Є. The reversal of
language L, written LR, is the language consisting of the reversals of all its strings. For
instances, if L ={001, 10, 111}, then LR ={100,01,111}.
8. What is meant by minimization of DFA?

12. For each DFA we can find an equivalent DFA that has a few states as any DFA
accepting the same language. Moreover, except for our ability to call the states by
whatever names we choose, this minimum-state DFA is unique for the language. The
algorithm is as follows:
a. First, eliminate any state that cannot be reached from the start state.
b. Then, partition the remaining states into blocks, so that all states in the same block
are equivalent, and no pair of states from different blocks are equivalent.
9. What is meant by minimizing the states of an NFA?
We all may think that the same state-partition technique that minimizes the states
of a DFA could also be used to find a minimum-state NFA equivalent to a given NFA or
DFA. We can do it, by a process of exhaustive enumeration, find an NFA with as few
states as possible accepting a given regular language, we cannot simply group the states
of some given NFA for the language.
10. Verify the language Lpal of palindromes of 0’s and 1’s is regular language or not?
To verify this we use pumping lemma. If Lpal is a regular language, let n be the
associated constant, and consider the palindrome w =0n10n. If Lpal is regular, then we can
break w into w = xyz, such that y consist of one or more 0’s from the first group. Thus,
xz, which would also have to be Lpal if Lpal were regular, would have fewer 0’s to the left
of the lone 1 than there are to the right of the 1. Therefore xz cannot be a palindrome. So
we have now contradicted the assumption that Lpal is a regular language.
11. What is meant by Context-Free Grammar(CFG)?
A CFG is a way of describing languages by recursive rules called productions. A
CFG consists of a set of variables, a set of terminal symbols, and a start variable, as well
as the productions. Each production consists of a head variable and a body consisting of a
string of zero or more variables and/or terminals.
12. Write down the components of CFG?
There are four components present in CFG, or just grammar. We shall represent a
CFG G by its four components, that is, G=(V,T,P,S), where V is the set of variables, T
the terminals, P the set of productions, and S the start symbol.
13. What is derivation?
Beginning with the start symbol, we derive terminal strings by repeatedly
replacing a variable by the body of some production with that variable in the head. The
language of the CFG is the set of terminal strings we can so drive; it is called a context-
free language.
14. What is meant by leftmost derivation and rightmost derivation?

13. At each step when the leftmost variable is replaced by one of its bodies then it is
called leftmost derivation. It is denoted by . At each step when the rightmost variable
is replaced by one of its bodies then it is called rightmost derivation. It is denoted by .
15. Write about sentential form?
Derivations from the start symbol produce strings tht have a special role. We call
these “sentential forms”. That is, if G=(V,T,P,S) is a CFG, then any string α in (V U T)*
such that S ==> α is a sentential form. If S α, then α is left-sentential form, and if
S α, then α is a right-sentential form.
16. What is meant by parse tree?
The parse tree is a tree that shows the essentials of a derivation. Interior nodes are
labeled by variables, and leaves are labeled by terminals or Є. For each internal node,
there must be a production such that the head of the production is the label of the node,
and the labels of its children, read from right to left, from the body of that production.
17. What are the conditions to construct the parse tree?
The condition to construct parse tree are:
a. Each interior node is labeled by a variable in V.
b. Each leaf is labeled by either a variable, a terminal, or Є. However, if the leaf is
labeled Є, then it must be the only child of its parent.
c. If an interior node is labeled A, and its children are labeled X1,X2,…,Xn
respectively, from the left, then A  X1,X2,…,Xn is a production in P. Note that
the only time one of the X’s can be Є is if that is the label of the only child, and
A Є is a production of G.
18. What are the applications of context free language?
Context free languages are used in:
a. Defining programming languages.
b. Formalizing the notation of parsing.
c. Translation of programming languages.
d. String processing applications.
19. What are the uses of context free grammars?
a. Construction of compliers.
b. Simplified the definition of programming languages.
c. Describes the arithmetic expressions with arbitrary nesting of balanced
parenthesis {(, )}.

14. d. Describes block structure in programming languages.
e. Model neural nets.
20. Define a context free grammar.
A context free grammar (CFG) is denoted as G = (V, T, P, S),
Where,
 V and T are finite set of variables and terminals respectively. V
and T are disjoint set.
 P is a finite set of productions each is of the form A∑ where A
is a variable and ∑ is the string of symbols from (V U T)*.
 S is the start symbol.
21. What is the language generated by CFG or G?
The language generated by G(L(G)) is {w in T* | S w}. That is a string is in
L(G) if:
i. The string consists solely of terminals.
ii. The string can be derived from S.
22. What is meant by ambiguous grammars?
For some CFG’s, it is possible to find a terminal string with more than one parse
tree, or equivalently, more than one leftmost derivation or more than one rightmost
derivation. Such a grammar is called ambiguous grammar.
23. What is meant by unambiguous?
For some CFG’s, it is possible to find a terminal string with at most one parse
tree, or equivalently, at most one leftmost derivation or at most one rightmost derivation.
Such a grammar is called ambiguous grammar.
24. State Chomsky normal form?
A context-free grammar “G” is in CNF if every production is of the form Aa
(Non-Terminal Terminal) or ABC (Non-Terminal Non-Terminal*Non-Terminal)
is in G. For example consider “G” whose production are SAB, Aa, Bb. The G is
in Chomsky normal form.
25. State Greibach normal form?
A context-free grammar “G” is in GNF if every production is of the form Aaα
and Aa where α Є VN* and aЄ∑. Non-TerminalTerminal. Any number of Non-
Terminal. For example: SaAB, Abc , Bb, Cc is in GNF.

15. 26. What is meant by parsers?
The CFG is an essential concept for the implementation of compilers and other
programming-language processors. Tools such as YACC take a CFG as input and
produce a parser, the component of a complier that deduces the structure of the program
being complied.
27. What is meant by equivalence of parse trees and derivation?
A terminal string is in the language of a grammar if and only if it is the yield of at
least one parse tree. Thus, the existence of leftmost derivation, rightmost derivation, and
parse trees are equivalent conditions that each define exactly the strings in the language
of a CFG.
28. What is meant by inherent ambiguity?
A context-free language L is said to be inherent ambiguous if all its grammars are
ambiguous. If even one grammar for L is unambiguous, then L is an unambiguous
language.
29. What is meant by eliminating ambiguity?
For many useful grammars, such as those that describe the structure of programs
in a typical programming language, it is possible to find an unambiguous grammar that
generates the same language. Unfortunately, the unambiguous grammar is frequently
more complex than the simplest ambiguous grammar for the language. There are also
some context-free languages, usually quite contrived, that are inherently ambiguous,
meaning that every grammar for that language is ambiguous.
30. What is yield of a parse tree?
If we look at the leaves of any parse tree and concatenate them from the left, we
get a string, called the yield of the tree, which is always a string that is derived from the
root variable. The fact that the yield is derived from the root will be proved shortly. Of
special importance are those parse tree such that:
a. The yield is a terminal string. That is, all leaves are labeled either with a terminal
or with ε.
b. The root is labeled by the start symbol.
31. What is meant by language of a grammar?
If G (V,T,P,S) is a CFG, the language of G, denoted L(G), is the set of terminal strings
that have derivations from the start symbol.
L(G)={w in T* | S w}
If a language L is the language of some context-free grammar, then L is said to be a CFL.

16. 32. What are the notation for CFG derivation?
There are a number of conventions in common use that help us remember the role of the
symbols we use when discussing CFG’s. Here are the conventions we shall use:
a. Lower-case letters like a, b are called terminal symbols.
b. Upper-case letters like A, B are variables.
c. Lower-case letter near the end of the alphabet are string of terminals.
d. Upper-case letters near the end of the alphabet are either terminals or variables.
e. Lower-case Greek letters are strings consisting of terminals and/or variables.
33. The following grammar generates the grammar of the language consisting of all
strings of even length:
S→AS|ε
A→ aa|ab|ba|bb
Give leftmost and rightmost derivations for the following strings:
aabbba
Solution:
Leftmost derivation:
S AS
 aaS
 aaAS
 aabbAS
 aabbbaS
 aabbba
Rightmost derivation:
SAS
AAS
AAAS
AAA
AAba
Abbba
aabbba

17. 34. Prove whether {0n | n is a power of 2} is regular set or not?
L={0n, n is a power of 2}
L={00,0000,00000000, …}
CASE 1:
Consider z= 00
Here u=0, w=0, V=ε
V must not have ε value.
CASE 2:
Consider z= 0000
Here u=0, w=0 and v=00
|uv|≤n
3 ≤ 4
|v|≥1
 uviw
= 0(00)i0
When i=1
=0(00)10
=0000
When i=2
=0(00)20
=000000
Thus the string 000000 is not present in the set so it is not a regular set.
35. Prove whether {0n1n | n ≥ 1} is regular set or not?
L={0n1n, n≥1}
L={01,0011,000111, …}
CASE 1:
Consider z= 01
Here u=0, w=1, V=ε
V must not have ε value.
CASE 2:

18. Consider z= 0011
Here u=0, w=1 and v=01
|uv|≤n
3 ≤ 4
|v|≥1
 uviw
= 0(01)i1
When i=1
=0(01)11
=0011
When i=2
=0(01)21
=001011
Thus the string 001011 is not present in the set so it is not a regular set.
36. Prove whether {0n10n | n ≥ 1} is regular set or not?
L={0n10n, n≥1}
L={010,00100,0001000, …}
CASE 1:
Consider z= 010
Here u=0, w=0, V=1
|uv|≤n
2 ≤ 3
|v|≥1
 uviw
= 0(1)i0
When i=1
=0(1)10
=010
When i=2
=0(1)20

19. =0110
Thus the string 0110 is not present in the set so it is not a regular set.
37. Prove whether {01n1| n ≥ 1} is regular set or not?
L={01n1, n≥1}
L={011,0111,01111, …}
CASE 1:
Consider z= 011
Here u=0, w=1, V=1
|uv|≤n
2 ≤ 3
|v|≥1
 uviw
= 0(1)i1
When i=1
=0(1)11
=011
When i=2
=0(1)21
=0111
Thus the string 0111 is present in the set so it is a regular set or regular language.
38. What is the language generated by the grammar G=(V,T,P,S) where P={S->aSb,
a. S->ab}?
S==> aSb
==>aaSbb
==>aaabbb
i.e the general form is anbn
Thus the language L(G)={ anbn | n>=1}. The language has strings with
equal number of a’s and b’s.

20. 39. What are the three ways to simplify a context free grammar?
a. By removing the useless symbols from the set of productions.
b. By eliminating the empty productions.
c. By eliminating the unit productions.
40. What are the properties of the CFL generated by a CFG?
 Each variable and each terminal of G appears in the derivation of some word in L.
 There are no productions of the form AB where A and B are variables.
41. Find the grammar for the language L={ a2nbc, where n>1 }
Let G=( {S,A,B}, {a,b,c}, P, {S} ) where P:
SAbc
AaaA
42. Find the language generated by :S0S1 | 0A | 0 | 1B | 1 ; A0A | 0; B1B | 1
The minimum string is S0 |1
S0S1 =>001
S0S1 =>011
S0S1 =>00S11 =>000S111 =>0000A111 =>00000111
Thus L= {0n 1m | m not equal to n, and n,m>=1}
43. Differentiate sentences Vs sentential forms
Sentences Sentential forms
A sentence is a string of terminal
symbols.
A sentential form is a string
containing a mix of variables and
terminal symbols or all variables.
This is an intermediate form in
doing a derivation.
44. What is BNF?
a. BNF stands for Backus-Naur Form.
i. Computer scientists describes the programming languages by a notation
called Backus-Naur Form. This is a context free grammar notation with
minor changes in format and some shorthand.
ii.

21. UNIT – III
1. What is Push down automata?
Push down automata is mainly used to recognize the CFL. It is named as Push
down automata because it makes use of stack as Push down list. Here stack acts as Cache
memory and Auxiliary memory.
2. Define Push down automata.
The Push down automata (M) are
M= {Q, Σ, δ, Γ, q0, z0, F)
where,
Q  Finite set of states.
Σ  Finite set of input symbols.
δ  Set of Transitions.
Γ  Finite set of stack symbols.
q0Initial state.
z0Top of the stack.
F  Set of Final state.
3. Explain the transition mapping of PDA.
The transition mapping for PDA can be given as
Q * (Σ U e) * Γ  Q* Γ*
where Γ is a finite set of stack symbols and Γ* consist of stack symbols with z0.
4. Why there is a need for stack in PDA?
In PDA stack acts as Cash memory and Auxiliary memory. Cash memory is used
to store the frequently accessing data and the data can be retrieved as fast as possible. If
there is no enough space in input tape then the data can be stored in external memory.
5. What are the additional features of PDA has when compared with NFA?
In PDA we use push down list called stack which is not used in NFA. Also we use
cache memory and auxiliary memory which is not present in NFA. We have 7 tuples in
PDA they are M= {Q, Σ, δ, Γ, q0, z0, F)
The additional 2 tuples which are present in PDA and not in NFA are Γ and Z0.
6. Write down the application of Pumping lemma for CFL?

22. The applications of pumping lemma for CFL are
 To check whether the given language is CFL or not.
 To check whether the given string is finite or infinite
7. Define Top-down parsing?
Top-down parsing is a methodology in which we construct the derivation tree or
parse tree starting from root node to the leaf node. Top-down parsing can be viewed as
an attempt to find a left most derivation for an input string which is left sentential forms.
8. Define Bottom-up parsing?
Bottom-up parsing is a methodology in which we construct the derivation tree or
parse tree starting from leaf node to the root node. In Bottom-up parsing we use reverse
of right most derivation which is right sentential forms.
9. Define handle and handle pruning?
When the right side of the production can be replaced by the left hand side of the
production then the left part is known as handle. The methodology by which we replace
is known as handle pruning. Handle pruning is also called as right sentential form. Right
sentential form example:
id+id*id
→E+id*i
→E+E*id
→E*id
→E*E
→E
It is the reverse of right derivation.
10. Explain the actions used in Bottom up parsing.
The actions used in Bottom up parsing are:
a) Shift
b) Reduce
c) Accept
d) Error
Shift: Shifting action is that we shift the symbol from input tape to the stack.
Reduce: If handle is present at the top of the stack then it is replace by the left hand
side of the derivation. This action is known as reduce.
Accept: When the $ symbol encounter then the input string is accepted.
Error: If $ symbol does not encounter then it leads to error.
11. State decision algorithm

23. Decision algorithm for CFL is mainly to check the string generated by a language
is empty, finite and infinite.
12. State the relationship between derivation and derivation tree.
Let G be the context free grammar indicated by G= {V, T, P, S} then S => α if
only if there is a derivation tree in G.
13. Write down the assumptions involved in decision algorithm.
The assumptions involved in decision algorithm are:
 Production must be in CNF.
 Production must not contain any ø production.
 Let S be the starting symbol then rank of S is r then we cannot produce a string of
length greater than 2r. i.e. l ≤ 2r.
 The other non-terminals must have rank lesser than r. i.e. the rank must be r-1,
r-2 etc.,
14. How to eliminate the left recursive patterns?
When the given production is of the format A→Aα/β then left recursive occurs.
To eliminate the left recursive we must introduce two new production which is of the
form A→βA’ and A’→αA’/ε.
For example:
Given production are E→E+T/T
This production is of the form A→Aα/β
where A= E, α= +T, β=T
Now the production will change to
E→TE’
E’→+TE’/ε.
15. State the “pumping lemma for context-free languages”.
Let L be a CFL. Then there exists a constant n such that if z is any string in L
such that |z| is at least n, then we can write z=uvwxy, subject to the following conditions:
1. |vwx|<=n. That is, the middle portion is not too long.
2. |vx|>=1 . Since v and x are the pieces to be “pumped”, this condition says that
at least one of the string we pump must not be empty.
3. For all i>=0, uviwxiy is in L. That is, the two strings v and x may be
“pumped” any number of times, including 0, and the resulting string will still
be a member of L.
16. What are the conditions to select a sub tree in Pumping lemma for CFL?
The conditions to select a sub tree in Pumping Lemma for CFL are

24. 1. The two vertices of the sub tree must be very close to each other at the bottom of
the tree.
2. The root of the sub tree must be very close to the root of the tree but it should not
be the root.
17. When the pumping lemma is generally used?
The pumping lemma for CFL is used to check whether the given language is CFL
or not and to determine whether the string generated by the CFL is finite or not.
18. Define ambiguous and unambiguous grammar.
An ambiguous grammar is a one which has more than one derivation tree whereas
unambiguous grammar refers to the grammar with exactly one derivation tree.
19. Define directed acyclic graph.
A directed acyclic graph is a directed graph which has no cycles in it.
Example:
This directed graph can be called as a directed acyclic graph since it has no cycles in it.
20. What is mean by left recursion?
Left recursion is the major drawback of the top down parsing. A production is
said to be left recursive if it is of the form A->αA/β. The left recursion can be eliminated
by producing two productions of the form A->βA’ and A’->αA’/ε.
21. State the closure properties of CFL.
The closure properties of CFL are
 CFL’s are closed under union, concatenation and Kleene closure.
 CFL’s are closed under homomorphism
 CFL’s are closed under substitutions
 If L is a CFL and R is a regular language then, L/R is a CFL.
 CFL’s are not closed under intersection
 CFL’s are closed under complementation.
A
B C
D

25. 22. Give an example for PDA.
STEP 1:
¢ a a b b $
F.C
Transition mapping : δ(q0,a, Z0) = (q1, aZ0)
STEP 2:
¢ a a b b $
Transition mapping : δ(q1,a, a) = (q2, aaZ0)
STEP 3:
¢ a a b b $
Transition mapping: δ(q2,b, a) = (q3, aZ0
a
Z0
F.C
a
a
Z0
F.C
a
a
Z0

26. STEP 4:
¢ a a b b $
Transition mapping: δ(q3,b, , Z0 ) =(q4,Z0)
23. Consider G whose productions are SaAS |a , ASbA| SS |ba ,Show that S =>
aabbaa and construct a derivation tree.
The derivation for aabbaa can be given as follows.
S  aAS
aSbAS
aabAS
 aabbaS
 aabbaa
The derivation tree for the above derivation is as follows :
F.C
a
Z0
a
S
A S
S b A a
a b a

27. 24. Explain the shit reduce parser for id1*id2 with actions if EE+E , EE*E, E(E),
Eid.
Input string Stack Diagram Action
id1*id2 $ Start
*id2 id1
id1
Shift
*id2 E
E
Reduce
id2 E*
*
E
Shift
$ E*id2
id2
*
E
Shift
$ E*E
E
*
E
Reduce
$ E
E
Reduce
$ E
E
Accept

28. 25. Whether id1 – id2 is accepted by the shift reduce parser? Justify.
Input string Stack Diagram Action
id1 - id2 $ Start
-id2 id1
id1
Shift
-id2 E
E
Reduce
id2 E-
-
E
Shift
$ E-id2
id2
-
E
Shift
$ E-E
E
-
E
Reduce
$ E-E
E
-
E
Error
26. In a production if A is the start symbol with rank zero then show that the length of
the string is finite.
Consider the production Aa,
The directed acyclic graph for above production can be given as,

29. By the assumption of Decision algorithm , a non terminal with rank r cannot generate a
string of length greater than 2r ,
Here r = 0,
l ≤ 2r
l ≤ 20
l ≤ 1
Hence it is proved with string of length 1 which is finite.
27. What is the Equivalence of PDA’s and CFG’s.
The languages defined by PDA’s are exactly the context-free languages. The
equivalence is given as
1. The context-free languages, i.e the languages defined by CFG’s.
2. The languages that are accepted by final state by some PDA.
3. The languages that are accepted by empty stack by some PDA.
are all same class. This can be given as
28. Consider the grammar with productions S AB, AB, Bb. whether the length
of the string generated by the grammar is finite or infinite. Justify.
The given productions are,
S AB
AB
Bb
The directed graph for the above production can be given as
A
Grammar PDA by
emptystack
PDA by
final state

30. The graph does not contain any cycles. Hence the length of the string produced by
the grammar is finite. This can also be proved by the assumption of Decision algorithm,
A non terminal with rank r cannot generate a string of length greater than 2r ,
Here r = 2 for the non terminal S,
l ≤ 2r
l ≤ 22
l ≤ 4 which is finte.
29. Compare NFA and PDA.
NFA PDA
The language accepted by NFA is the
regular language.
The language accepted by PDA is context
free language.
NFA has no memory. PDA is essentially an NFA with a
stack(memory).
It can store only limited amount of
information.
It stores unbounded limit of information
A language/string is accepted only by
reaching the final state.
It accepts a language either by empty stack
or by reaching a final state.
30. Specify the two types of moves in PDA?
The moves dependent on the input symbol(a) scanned is:
δ (q,a,Z) = {(p1,aZ1), (p2,aZ2),………… (pm,aZm)}
where q and p are states, a is in ∑, Z is a stack symbol and ∑i is in Γ*.
PDA is in state q, with input symbol a and Z the top symbol on state enter state pi
replace symbol Z by string ∑i.
The moves dependent on the input symbol is (ε-move)
δ (q,ε,Z) = {(p1,Z1), (p2,Z2),………… (pm,Zm)}
Is that PDA is in state q, independent of input symbol being scanned and
with Z the top symbol on the stack enter a state pi replace symbol Z by string ∑i
S
BA

31. 31. What are the different types of language acceptances by a PDA.
There are 4 types of language acceptances of a PDA. They are
 Acceptance by final state
 Acceptance by empty state
 From empty to final state
 From final state to empty state.
32. Define acceptance by final state.
Let P = {Q, Σ, δ, Γ, q0, z0, F} be a PDA. Then L(P), the language accepted by P
by final state, is
{ w | (q0,w,z0) ├*
p (q,ε,α)}
For some state q in F and any state string α. That is, starting in the initial ID with
w waiting on the input, P consumes w from the input and enters an accepting state. The
contents of the stack at that time is irrelevant.
33. Define acceptance by empty stack.
For each PDA P= {Q, Σ, δ, Γ, q0, z0, F}, we also define
N(P) = { w | (q0,w,z0) ├* (q,ε,ε)}
For any state q. That is, N(P) is the set of input w that P can consume and
at the same tme empty its stack.
34. Is it true that the language accepted by a PDA by empty stack and final states are
different languages.
No, because the languages accepted by PDA’s by final state are exactly the
language accepted by PDA’s empty stack.
35. When is a string accepted by a PDA?
The input string is accepted by the PDA if:
 The final state is reached.
 The stack is empty.
36. What is the significance of PDA?
Finite automata is used to model regular expression and cannot be used to
represent non regular languages. Thus to model a context free language, a push down
automata is used.
37. How we can represent the configuration of a PDA?
We can represent the configuration of PDA by triple (q, w, γ), where

32.  q is the state,
 w is the remaining input
 γ is the stack contents
38. Define Instantaneous description in PDA.
ID describe the configuration of a PDA at a given instant. ID is a triple such as (q,
w, γ), where
 q is the state,
 w is the remaining input
 γ is the stack contents
If P = {Q, Σ, δ, Γ, q0, z0, F} is a PDA we say that
(q, aw, Z) ├p (p, w, αβ) if δ(q, a, Z) contains (p, α)
Example: (q1, BG) is in δ (q,0,G) tells that (q1, 011,GGR) ├ (q1,11,BGGR).
39. What are the components of Pushdown Automata?
The PDA usually consists of four components:
 A control unit
 Read unit
 Input tape
 Memory unit
40. Compare NPDA and DPDA.
NPDA DPDA
NPDA is the standard PDA used in
automata theory.
The standard PDA in practical situation is
DPDA.
Every PDA is NPDA unless otherwise
specified.
The PDA is deterministic in the sense, that
at most one move is possible from any ID.
UNIT-4
1. What is the difference between the Turing machine and the Finite automata?
Finite automata Turing machine
Here the input tape is closed at both the
ends.
Here the input tape is infinite at one end
either at the left or at right usually at the
right side.

33. The reading pointer moves only one side. Here the read/write head pointer moves to
right or left or stay at a square after a
read/write.
2. What are the components of Turing machine?
The components of Turing machine are:
 finite control
 an input tape
 a tape head
3. How many tuples are in the Turing machine? What are they?
Formally a Turing machine(TM) is denoted
M = (Q,∑,┌,§,q0,B,F),
Where
Q is the finite set of states,
┌ is the finite set of allowable tape symbols,
B a symbol of┌ ,is the blank,
∑ a subset of┌ not including B, is the set if input symbols,
§ is the next move function, a mapping from Q×┌×{L,R},
q0 in Q is the start state,
F€Q is the set of final states.
4. What is computational function?
Let S *
and let f be a function f : S -> *
. Then we say T computes f or
f is computable if for every x S,
(q0, x) *(h, f(x)) and for every x * that is not in S, T does not halt on x.
5. What do you meant by “Acceptance of Turing Machine”?

34. In the Turing machines there are two halt states: "accept halt" and "reject halt". A
string is accepted by a Turing machine if given the string, the Turing machine eventually
goes into the accept halt state. A language is a phrase structure (type 0) language if and
only if it is Turing-acceptable in either sense and it has no effects on decidability.
6. What are the different types of Turing machine?
 Turing Machines with Two Dimensional Tapes
 Turing Machine with One Dimensional Tape
 Turing Machines with Multiple Tapes
 Turing Machines with Multiple Heads
 Turing Machines with Infinite Tape
 Nondeterministic Turing Machines
7. What is Turing machine with two dimensional tapes?
This is a kind of Turing machines that have one finite control, one read-write head
and one two dimensional tape. The tape has the top end and the left end but extends
indefinitely to the right and down. It is divided into rows of small squares.
8. What is Turing machine with one dimensional tape?
For any Turing machine of this type there is a Turing machine with a one
dimensional tape that is equally powerful, that is, the former can be simulated by the
latter.
One Dimensional Tape
v 1 v 2 3 h 4 5 6 v 7
8 9 10 h 11 . . . . . .
9. What is Turing machine with multiple tapes?
A multitape turing machine consists of a finite control with k tape heads and k
tapes; each tape is infinite in both directions. On a single move, depending on the state of
the finite control and the symbol scanned by each of the tape heads, the machine can:
1) Change state
2) Prints a new symbol on each of the cells scanned by its tape heads;
3) Move each of its tape heads, independently, one cell to the left , or right, or keep
it stationary.
Initially the input appears on the first tape, and the other tapes are blank.

35. 10. What is Turing machine with multiple heads?
A k-head turing machine has some fixed number, k, of heads. The heads are
numbered 1 through k, and a move of the TM depends on the state and on the symbol
scanned by each head. In one move , the heads may each move independent left, right, or
remain stationary.
11. What is Turing machine with infinite tape?
In case of a Turing machine the input tape is infinite at one end either at the left or
at the right usually the right is called a Turing machine with infinite tape.
12. What is Non-deterministic Turing machine?
A Nondeterministic turing machine is a device with a finite control and a single ,
one way infinite tape. For a given state and tape symbol scanned by the tape head, the
machine has a finite number of choices for the next move. Each choice consists of a new
state, a tape aymbol to print and a direction of head motion.
13. What is a recursive algorithm?
A recursive algorithm is an algorithm applied to an problem then it can be solved
or decidable. The problem that can be solved by applying an algorithm is called recursive
algorithm.
14. What is recursive enumerable language?
Recursive enumerable language is a set of language or a set of problems that can
be solved or decided.
15. What is meant by Off-line Turing machines?
An off-line Turing machine is a multitape turing machine whose input tape is read
only. The input is surrounded by end markers ,¢ on the left and $ on the right.The Turing
machine is not allowed to move the input tape head off the region between ¢ and $.The
off-line Turing machine copy its own input onto the extra tape, and it then simulates M as
if the extra tape were M’s input.
16. What is semi infinite tape?
In case of a Turing machine one side of the input tape is infinite hence it is called
as the semi infinite tape.
17. What is the difference between Semi infinite and two way infinite tapes?
Semi infinite tape Two way infinite tape
In case of semi infinite tape TM only one In case of two way infinite tape both the

36. side of the input tape is infinite. sides of the input tape is infinite.
The head pointer moves only to the infinite
side of the input tape.
Here the head pointer moves in both sides
infinitely.
18. What are the uses of checking of symbol?
Checking off symbols is a useful trick for visualizing how a TM recognizes
languages defined by repeated strings, such as
{ww|w in ∑*}, {wcy|w and y in ∑*,w≠y} or {wwR |w in ∑*}.
It is also useful when lengths of substrings must be compared , such as in the languages
{ai bi|i>=1} or{aibick|i≠j or j≠k}.
We extract an extra track on the tape that holds a blank or √. The √ appears when
the symbol below it has been considered by the TM in one of its comparisons.
19. Write about the storage in the Finite control?
The finite control can be used to hold a finite amount of information. The state is
written as a pair of elements,one exercising control and the other storing a symbol.It
should be emphasized that this arrangement is for conceptual purposes only. No
modification in the definition of the Turing Machine has been made.
20. Write about shifting over in the turing machine?
A Turing machine can make space on its tape by shifting all nonblank symbols a
finite number of cells to the right . The tape head makes an excursion to the right
,repeatedly storing the symbols read in its finite control and replacing them with symbols
read from cells to the left.The TM can return to the vacated cells and print symbols of its
choosing .If space is available ,it can push block of symbols left in a similar manner.
21. What is a turing machine?
Turing machine is a simple mathematical model of a computer. TM has unlimited
and unrestricted memory and is a much more accurate model of a general purpose
computer. The turing machine is a FA with a R/W Head. It has an infinite tape divided
into cells ,each cell holding one symbol.
22. Define Instantaneous description of TM.
The ID of a TM M is denoted as |--*1q2. Here q is the current state of M is in Q;
1 2 is the string in ∑* that is the contents of the tape up to the rightmost nonblank
symbol or the symbol to the left of the head, whichever is the rightmost.

37. 23. What are the applications of TM?
TM can be used as:
 Recognizers of languages.
 Computers of functions on non negative integers.
 Generating devices.
24. What is the basic difference between 2-way FA and TM?
Turing machine Two way finite automata
Turing machine can change symbols on its
tape.
FA cannot change
symbols on tape.
TM has a tape head that moves both left
and right side
FA doesn’t have such a tape head.
TM has infinite input tape Input tape is finite in FA
25. What is (a)total recursive function and (b)partial recursive function
If f(i1,i2,………ik) is defined for all i1,…..ik then we say f is a total recursive
function. They are similar to recursive languages as they are computed by TM that
always halt.
A function f(i1,…ik) computed by a Turing machine is called a partial recursive
function. They are similar to r.e languages as they are computed by TM that may or may
not halt on a given input.
26. Define a move in TM.
Let X1 X2…X i-1 q Xi…Xn be an ID.
The left move is: if _ (q, Xi )= (p, Y,L) ,if i>1 then
X1 X2…X i-1 q Xi…Xn |---- X1X2… X i-2 p X i-1 Y X i+1…Xn.
M
The right move is if _ (q, Xi )= (p, Y,R) ,if i>1 then
X1 X2…X i-1 q Xi…Xn |---- X1X2… X i-1Y p X i+1…Xn.
M
27. What is the language accepted by TM?
The language accepted by M is L(M) , is the set of words in ∑ * that cause M to
enter a final state when placed ,justified at the left on the tape of M, with M at qo and the
tape head of M at the leftmost cell. The language accepted by M is:

38. { w | w in ∑ * and q0w |--- α pβ for some p in F and α, β is in ∑ * }.
28. Give examples of total recursive functions.
All common arithmetic functions on integers such as multiplication , n!, [log2n]
and 22n are total recursive functions.
29. What are(a) recursively enumerable languages (b) recursive sets?
The languages that is accepted by TM is said to be recursively enumerable (r. e )
languages. Enumerable means that the strings in the language can be enumerated by the
TM. The class of r. e languages include CFL’s.
The recursive sets include languages accepted by at least one TM that halts on all
inputs.
30. What are the various representation of TM?
We can describe TM using:
 Instantaneous description.
 Transition table.
 Transition diagram.
31. What are the possibilities of a TM when processing an input string?
 TM can accept the string by entering accepting state.
 It can reject the string by entering non-accepting state.
 It can enter an infinite loop so that it never halts.
32. What are the techniques for Turing machine construction?
 Storage in finite control.
 Multiple tracks.
 Checking off symbols.
 Shifting over
 Subroutines.
33. Differentiate PDA and TM.
PDA TM
1. PDA uses a stack for storage. 1. TM uses a tape that is infinite .
2.The language accepted by PDA is CFL. 2. Tm recognizes recursively enumerable
languages.

39. 34. How can a TM used as a transducer?
A TM can be used as a transducer. The most obvious way to do this is to treat the
entire nonblank portion of the initial tape as input , and to treat the entire blank portion of
the tape when the machine halts as output. Or a TM defines a function y=f(x) for strings
x ,y _ _* if: q0X | --- qfY, where qf is the final state.
35. What is a multidimensional TM?
The device has a finite control , but the tape consists of a k-dimensional array of
cells infinite in all 2k directions, for some fixed k. Depending on the state and symbol
scanned , the device changes state , prints a new symbol and moves its apehead in one of
the 2k directions, either positively or negatively ,along one of the k-axes.
36. When a recursively enumerable language is said to be recursive ? Is it true that the
language accepted by a non-deterministic Turing machine is different from
recursively enumerable language?
A language L is recursively enumerable if there is a TM that accepts L and
recursive if here is a TM that recognizes L. Thus r.e language is Turing acceptable and
recursive language is Turing decidable languages.
No , the language accepted by non-deterministic Turing machine is same as
recursively enumerable language.
37. What is Church’s Hypothesis?
The notion of computable function can be identified with the class of partial recursive
functions is known as Church-hypothesis or Church-Turing thesis. The Turing machine is
equivalent in computing power to the digital computer.
UNIT-5
1. What is computational complexity?
A computational complexity is based on the amount of time, space, or other
resource needed to recognize a language on some universal computing device such a
Turing Machine.
2. What are the complexity of classes?
DSPACE(S(n)) = determinastic Space complexity(S(n))
NSPACE(S(n)) = nondeterminastic Space complexity(S(n))
DTIME(T(n)) = determinastic Timecomplexity(T(n))
NTIME(T(n)) = nondeterminastic Timecomplexity(T(n))

40. 3. Define Time Complexity.
For every input word of length n, M (Turing machine) makes at most T(n) moves
before halting, then M is said to be a T(n) time-bounded Turing machine, or of time
complexity T(n). The language recognized by M is said to be of time complexity T(n).
4. Define Space Complexity.
For every input word of length n, M (Turing machine) scans at most S(n) cells on
any storage tape, then M is said to be an S(n) space-bounded Turing machine or of space
complexity S(n). The language recognized by M is said to be of Space complexity S(n).
5. Define Growth Rate of Functions.
When we have two algorithm for the same problem, we may require a comparison
between the running time of these two algorithms. With this in mind, we study the
growth rate of functions defined on the set of natural numbers.
6. Briefly explain f(n)=O(g(n)) Or State the condition under which f(n)=O(g(n)).
Let f , g: N->R+
(R+ being the set of all positive real numbers.) We say that f(n)=O(g(n)) if there exists
positive integer C and No such that
F(n)<C g(n) for all n>No. In these case we say “f” is the order of g.
f(n)=O(g(n)) it expresses a relation between two functions f and g.
7. Compare the Growth Rate of Polynomial Function with Exponential Function.
n f(n)=n^2 g(n)=n^2 +3n+9 q(n)=2^n
1 1 13 2
5 25 49 32
1000 1000000 1003009 1.07*10^301
From the table, it is easy to say that the function q(n) grows at a very fast rate when
compared to f(n) or g(n). Hence Exponential function grows at very faster rate when n
value increase and polynomial function grows at lesser rate when n value increases.
8. Define Travelling Salesman Problem.
The Travelling Salesman Problem(TSP) is a problem which contain a list of cities
and their pairwise distances, the task is to find a shortest possible tour that visits each city
exactly once.

41. 9. Define Hamiltonaian path or Traceable path.
A Hamiltonian path or Traceable path defines in a directed graph G is a directed
path that goes through each node exactly once.
10. Define Hamiltonian Cycle or Hamiltonian Circuit and give an example.
A Hamiltonian cycle (or Hamiltonian circuit) is a cycle in an undirected graph
which visits each vertex exactly once and also retuns to the starting vertex.
Example:
 A complete graph with more than two vertices is Hamiltonain
 Every cycle graph is Hamiltonian.
11. Define NP-completeness and give an example.
If the problem may or may not exists a solution then it is called NP-Complete.
In other word, A language B in NP- complete if it satisfies two conditions:
 B is in NP, and
 Every A in NP is polynomial time reducible to B.
Examples: Boolean satisfiablity problem, Knapsack problem, Hamiltonian path problem.
12 Define NP-hardness.
If the problem as a complete set of solution then it is called NP-hard.
13 What is clique problem and give an example graph?
A clique in an undirected graph is a subgraph wherein every two nodes are
connected by an edge. A k-clique is a clique that contains K nodes.
A graph with a 5-clique
14 Define Horn-satisfiability.
A clause is Horn if it contains at most one positive literal. This clauses are able to
express implication of on variable from a set of other variables.
Example: If the set of all values are true then Y needs to be true .

42. 15 What is Bollean satisfiablity problem?
Bollean satisfiablity problem is a decision problem, whose instance is a Boolean
expression written using only AND, OR, NOT, variables, and paranthesies.
16 Define class NP ?
A language L is in the class NP (Nondeterministic polynomial) if there is a
nondeterministic TM M and a polynomial time complexity T(n) such that L=L(M) and
when M is given a input of length n, there are no sequences of more than T(n) moves of
M.
Example: The Travelling salesman problem
17 Define class P and give an example .
P is the class of languages that are decidable in polynomial time on a
deterministic single-tape Turing machine.
Example: Kruskal’s algorithm
A graph; its minimum-weight spanning tree is indicated by heavy lines.
18 What is Knapsack problem or Rucksack problem?
The Knapsack problem or Rucksack problem is a problem in combinatorial
optimization: Given a set of items, each with a weight and a value, determine the number
of each item to include in a collection so that the total weight is less than a given limit
and the total value is as large as possible.
19 What is bounded and unbounded Knapsack problem ?
The bounded Knapsack problem restricts the number of item to a maximum
integer value and the Unbounded problem contain nonnegative integers of
weights(W1,…,Wn and W), then the Knapsack problem solved in pseudo-polynomial
time called Unbounded problem.
20. What is Vertex cover problem?
A Vertex cover of a graph is a set of vertices such that each edge of the graph is
incident to at least one vertex of the set.
21. When we say a problem is decidable? Give an example of undecidable problem?
A problem whose language is recursive is said to be decidable. Otherwise the
problem is said to be undecidable. Decidable problems have an algorithm that takes as
input an instance of the problem and determines whether the answer to that instance is
“yes” or “no”.
(eg) of undecidable problems are (1)Halting problem of the TM.

43. 22. Give examples of decidable problems.
1. Given a DFSM M and string w, does M accept w?
2. Given a DFSM M is L(M) = _ ?
3. Given two DFSMs M1 and M2 is L(M1)= L(M2) ?
4. Given a regular expression r and a string w ,does r generate w?
5. Given a NFSM M and string w ,does M accept w?
23. Give examples of recursive languages?
i. The language L defined as L= { “M” ,”w” : M is a DFSM that accepts w} is
recursive.
ii. L defined as { “M1” U “M2” : DFSMs M1 and M2 and L(M1) =L(M2) } is
recursive.
24. Differentiate recursive and recursively enumerable languages.
Recursive languages Recursively enumerable languages
1. A language is said to be recursive if and
only if there exists a membership algorithm for
it.
1. A language is said to be r.e if there exists a
TM that accepts it.
2. A language L is recursive iff there is a TM
that decides L.
(Turing decidable languages). TMs that decide
languages are algorithms.
2. L is recursively enumerable iff there is a TM
that semi-decides L.
(Turing acceptable languages). TMs that semi-
decides languages are not algorithms.
25. What are UTMs or Universal Turing machines?
Universal TMs are TMs that can be programmed to solve any problem, that can
be solved by any Turing machine. A specific Universal Turing machine U is:
Input to U: The encoding “M “ of a Tm M and encoding “w” of a string w.
Behavior : U halts on input “M” “w” if and only if M halts on input w.
26. What is the crucial assumptions for encoding a TM?
There are no transitions from any of the halt states of any given TM . Apart from
the halt state , a given TM is total.
27. What properties of recursive enumerable seta are not decidable?
 Emptiness
 Finiteness
 Regularity

44.  Context-freedom.
28. Define L .When is a trivial property?
L is defined as the set { <M> | L(M) is in Φ. } is a trivial property if Φ is empty
or it consists of all r.e languages.
29. What is a universal language Lu?
The universal language consists of a set of binary strings in the form of pairs
(M,w) here M is TM encoded in binary and w is the binary input string. Lu = { < M,w> |
M accepts w }.
30. What is a Diagonalization language Ld?
The diagonalization language consists of all strings w such that the TM M whose
code is w doesnot accept when w is given as input.
31. What properties of r.e sets are recursively enumerable?
 L ≠Φ
 L contains at least 10 members.
 w is in L for some fixed w.
 L ∩ Lu ≠ Φ
32. What properties of r.e sets are not r.e?
 L = Φ
 L = £ *.
 L is recursive
 L is not recursive.
 L is singleton.
 L is a regular set.
 L - Lu ≠ Φ
33. How can a TM acts as a generating device?
In a multi-tape TM ,one tape acts as an output tape, on which a symbol, once
written can never be changed and whose tape head never moves left. On that output tape ,
M writes strings over some alphabet ∑ , separated by a marker symbol # , G(M) ( where
G(M) is the set w in ∑ * such that w is finally printed between a pair of #’s on the output
device ).

45. 34. What are the different types of grammars/languages?
 Unrestricted or Phase structure grammar.(Type 0 grammar).(for TMs)
 Context sensitive grammar or context dependent grammar (Type1)(for Linear
Bounded Automata )
 Context free grammar (Type 2) (for PDA)
 Regular grammar (Type 3) ( for Finite Automata).
This hierarchy is called as Chomsky Hierarchy.
35. State a single tape TM started on blank tape scans any cell four or more times is
decidable?
If the TM never scans any cell four or more times , then every crossing sequence
is of length at most three. There is a finite number of distinct crossing sequence of length
3 or less. Thus either TM stays within a fixed bounded number of tape cells or some
crossing sequence repeats.
36. Does the problem of “ Given a TM M ,does M make more than 50 moves on input B
“?
Given a TM M means given enough information to trace the processing of a fixed
string for a certain fixed number of moves. So the given problem is decidable.
37. Show that AMBIGUITY problem is un-decidable.
Consider the ambiguity problem for CFGs. Use the “yes-no” version of AMB. An
algorithm for FIND is used to solve AMB. FIND requires producing a word with two or
more parses if one exists and answers “no” otherwise. By the reduction of AMB to FIND
we conclude there is no algorithm for FIND and hence no algorithm for AMB.
38. State the halting problem of TMs.
The halting problem for TMs is:
Given any TM M and an input string w, does M halt on w? This problem is undecidable
as there is no algorithm to solve this problem.
39. Define PCP or Post Correspondence Problem.
An instance of PCP consists of two lists , A = w1,w2,….wk and B = x1,…..xk of
strings over some alphabet _ .This instance of PCP has a solution if there is any sequence
of integers i1,i2,..im with m >=1 such that wi1, wi2,…wim = xi1,xi2 ,…xim
The sequence i1 ,i2 ,…im is a solution to this instance of PCP.

46. 40. Define MPCP or Modified PCP.
The MPCP is : Given lists A and B of K strings from _ * ,say A = w1 ,w2, …wk
and B= x1, x2,…..xk does there exists a sequence of integers i1,i2,…ir such that
w1wi1wi2…..wir = x1xi1xi2…xir?
41. What is the difference between PCP and MPCP?
The difference between MPCP and PCP is that in the MPCP ,a solution is
required to start with the first string on each list.
42. What are the concepts used in UTMs?
 Stored program computers.
 Interpretive Implementation of Programming languages.
 Computability.