Theory of Digital Computation
Course material for undergraduate students on IT
Department of Computer Science
University of Veszprem
Veszprem, Hungary
2002
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Table of Contents
Table of Contents

Alphabets and Languages

Finite Automata

Context-free Languages

Turing Machines

Universal Turing Machines

Examples

Problems
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Alphabets and Languages
Alphabets and Languages

Alphabet is a finite set of symbols.

A string over an alphabet is a finite sequence of symbols
from the alphabet.

A string may have no symbols at all in this case it is called
the empty string and is denoted by e.

The length of a string w, w, is its length as a sequence.
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Alphabets and Languages

Two strings over the same alphabet can be combined to
form a third by the operation concatenation. The
concatenation of strings x and y, written xy or simply xy, is
the string x followed by the string y:
w(j) = x(j)
for j = 1,...,x
w(x+ j) = y(j)
for j = 1,...,y
If w = xy then w=x+y.
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Alphabets and Languages

A string v is a substring of w iff there are strings x and y
such that w = xvy.

The reversal of a string w, denoted by wR, is the string
“spelling backwards”:
If w= 0, then wR = w = e.
If w= n+1 (>0), then w = ua for some a, and
wR = auR.
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
Any set of string over an alphabet is called language.
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Alphabets and Languages

Closure or Kleene star is a language operation of a single
language L, denoted by L*.
L* is the set of all strings obtained by concatenating zero or
more strings from L.
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Alphabets and Languages

Definition
A string over alphabet ( , ) , Ø , , *} is a regular
expression over alphabet  if it satisfies the following
conditions.
(1) Ø and each element of  is a regular expression
(2) If  and  are regular expressions then so is ()
(3) If  and  are regular expressions then so is ()
(4) If  is a regular expression, then so is *.
(5) Nothing is a regular expression unless it follows from
(1) through (4).
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Alphabets and Languages

Every regular expression represents a language.

Formally, the relation between regular expressions and the
corresponding languages is established by function L,
where L() is the language represented by regular
expression .
Thus, L is a function from strings to languages.
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Alphabets and Languages

Function L is defined as follows.
(1)
L(Ø) = Ø and L() = a for each  = a  
(2)
If  and  are regular expressions, then
L( () ) = L()L().
(3)
If  and  are regular expressions, then
L( () ) = L()L().
(4)
If  is a regular expression then L(*) = L()*.
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Finite Automata
Finite Automata

Definition
Deterministic finite automaton (DFA) is a quintuple
M = (K, , , s, F)
where
K is the set of states (finite set),
 is an alphabet,
s  K is the initial state,
F is the set of final states (F  K), and
, the transition function, is a function from K to K.
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Finite Automata

Graphical representation of a DFA: state diagram

state p:
- initial state q:
- final state r:

transition (p, b) = q:
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Finite Automata

A configuration of a DFA is determined by the current state
and the unread part of input. In other words, a configuration
of a deterministic finite automaton
(K, , , s, F) is an element of K*.
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
If (q, w) and (q’, w’) are two configurations of M, then (q, w)
├─M (q’, w’) iff w = w’ for some symbol   , and
(q, ) = q’. In this case, we say that configuration (q, w)
yields configuration (q’, w’) in one step.
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Finite Automata

We denote the reflexive, transitive closure of ├─M by ├─M*;
(q, w) ├─M* (q’, w’) is read, configuration (q, w) yields
configuration (q’, w’).
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
A string w  * is said to be accepted by M iff there is a
state qF such that (s, w) ├─M* (q, e).
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
The language accepted by M, L(M), is the set of all strings
accepted by M.
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Finite Automata

Definition
Non-deterministic finite automaton (NFA) is a quintuple
M = (K, , , s, F)
where
K is the set of states (finite),
 is an alphabet,
s  K is the initial state,
F is the set of final states (F  K), and
, the transition relation, is a finite subset of K×*×K.
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Finite Automata

Graphical representation of a NFA: state diagram

state p:
- initial state q:
- final state r:

transition (p, ba,q)  :
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Finite Automata

A configuration of M is an element of K×*.

The relation ├─M (yields in one step) between
configurations is defined as follows: (q, w) ├─M (q’, w’) iff
there is a u  * such that w = uw’ and (q, u, q’)  .

├─M* is the reflexive, transitive closure of ├─M.


w  * is accepted by M iff there is a state q  F such that
(s, w)├─M*(q, e).
The language accepted by M, L(M), is the set of all strings
accepted by M.
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Finite Automata

Definition
Finite automata M1, and M2 are said to be equivalent iff
L(M1) = L(M2).

Theorem
For each non-deterministic finite automaton, there is an
equivalent deterministic finite automaton.
(No proof here)
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Finite Automata

Theorem
Languages accepted by finite automata are closed under
union.

Proof:
Let NFA M1= (K1, , 1, s1, F1) and M2= (K2, , 2, s2, F2).
NFA M = (K, , , s, F) will be given such that
L(M) = L(M1)  L(M2).
s is a new state not in K1K2,
K = K1K2s,
(K1 and K2 are disjoint.)
F = F1 F2, and
 = 12(s, e, s1), (s, e, s2).
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Finite Automata

Theorem
Languages accepted by finite automata are closed under
concatenation.

Proof:
L(M) = L(M1) ○ L(M2)
Construct a NFA M = (K, , , s, F), where
K = K1K2
(K1 and K2 are disjoint.)
s = s1
F = F2
 = 12(F1  {e}  {s2}).
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Finite Automata

Theorem
Languages accepted by finite automata are closed under
Kleen-star.

Proof:
L(M) = L(M1)*
Construct a NFA M = (K, , , s, F), where
K = K1s
s is a new state not in K1
F = F1s
 = 1(F  {e}  {s1})
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Finite Automata

Theorem
Languages accepted by finite automata are closed under
complementation.

Proof:
L(M) = L(M1)
Construct a DFA M = (K, , , s, F)
K = K1
s = s1
F = K1 - F1
 = 1
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Finite Automata

Theorem
Languages accepted by finite automata are closed under
intersection.

Proof:
L(M) = L(M1)  L(M2)
Construct a DFA M
L(M) = * - ((* - L(M1))  ( * - L(M2))).
Note:
L(M1)  L(M2) = L(M1)  L(M2)
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Finite Automata

Theorem
There is an algorithm for answering the following question:
Given a finite automaton M, is L(M) = * ?

Proof:
- Tracing the state diagram can answer the question
wheather L(M’) = .
- L(M) = * iff L(M) = .
- L(M’) = L(M).
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Finite Automata

Theorem
There is an algorithm for answering the following question:
Given two finite automata M1 and M2, is L(M1)  L(M2)?

Proof:
L(M1)  L(M2) iff
(*-L(M2))  L(M1) = .
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Finite Automata

Theorem
There is an algorithm for answering the following question:
Given two finite automata M1 and M2, is L(M1) = L(M2) ?

Proof:
L(M1) = L(M2) iff
(L(M1)  L(M2) and L(M2)  L(M1)).
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Context-free Languages
Context-free Languages

Definition
A context-free grammar (CFG) G is a quadruple
(V, , R, S), where
V is an alphabet,
 (the set of terminals) is a subset of V,
R (the set of rules) is a finite subset of (V-)V*, and
S (the start symbol) is an element of V-.

The members of V- are called non-terminals.
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Context-free Languages



For A  V- and u  V*, A G u if (A, u)  R.
For any strings u, v  V*, u G v if there are strings
x, y, v’ V* and A  V- such that u = xAy, v = xv’y
and A G v’.
Relation G* is the reflexive, transitive closure of G.
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
Language generated by G, is L(G) = {w  *: S G* w}; we
also say that G generates each string in L(G).
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Context-free Languages

Definition
A language L is a context-free language if it is equal to L(G)
for some context-free grammar G.

Definition
A context-free grammar G = (V, , R, S) is regular iff
R  (V-)  *((V-)  e).
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Context-free Languages

Theorem
Context-free languages are closed under union.

Proof:
Let G1 = (V1, 1, R1, S1) and G2 = (V2, 2, R2, S2)
context-free grammars, V1-1 and V2-2 disjoint.
CF grammar G = (V, , R, S) will be given such that
L(G) = L(G1)  L(G2)
V = V1  V2  {S}
 = 1  2
R = R1  R2  {S  S1, S  S2}
S = a new symbol not in V1  V2.
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Context-free Languages

Theorem
Context-free languages are closed under concatenation.

Proof:
L(G) = L(G1)  L(G2)
Construct CF grammar G = (V, , R, S)
V = V1  V2  {S}
(V1-1 and V2-2 are disjoint.)
 = 1  2
R = R1  R2  {S  S1S2}
S = a new symbol not in V1  V2.
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Context-free Languages

Theorem
Context-free languages are closed under Kleen-star.

Proof:
L(G) = L(G1)*
Construct CF grammar G = (V, , R, S)
V = V1
 = 1
R = R1  {S1  e, S1  S1S1}
S = S1
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Context-free Languages

Definition
A pushdown automaton (PDA) is a sixtuple
M = (K, , , , s, F), where
K is a finite set of states,
 is an alphabet (the input symbols),
 is an alphabet (the stack symbols),
s is the initial state (s  K),
F is the set of final states (F  K), and
 is the transition relation, is a finite subset of
(K**)(K*).
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Context-free Languages

Graphical representation of a NFA: state diagram

state p:
- initial state q:
- final state r:

transition (p, ba, cd, q, dc)  :
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Context-free Languages

Intuitively, if ((p, u, ), (q, ))  , then M, whenever it is in
state p with  at the top of the stack, may read u from the
input tape,  replaced by  on the top of the stack, and
enter state q. ((p, u, ), (q, )) is called a transition of M.

To “push” a symbol is to add it to the top of the stack, to
“pop” a symbol is to remove it form the top of the stack.

A configuration is defined to be an element of K**:
The first component is the state of the machine, the second
is the partition of the input yet to be read, and the third is the
content of the pushdown store.
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Context-free Languages




For every transition ((p, u, ),(q, ))  , and for every x 
* and   *, we define (p, ux, ) ├─M (q, x, )),
moreover, ├─M (yields in one step) holds only between
configurations that can be represented in this form for some
transition, some x, and some .
The reflexive, transitive closure of ├─M is denoted by ├─M*.
We say that M accepts a string w  * iff (s, w, e) ├─M*
(p, e, e) for some p  F.
Language accepted by M, L(M), is the set of all strings
accepted by M.
E
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Turing Machines
Turing Machines

Definition
A Turing machine is a quadruple (K, , , s) where
K: is a finite set of states, not containing the halt state h;
(h  K),
: is an alphabet, containing the blank symbol #, but not
containing the symbols L and R;
s: is the initial state;
: is a function from K to (K {h})( {L, R}).

Configuration of a Turing machine M = (K, , ,s) is an
element of (K  {h})* (*(-{#}){e}).
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Turing Machines

Let (q1, w1, a1, u1) and (q2, w2, a2, u2) be configurations of
Turing machine M. Then (q1, w1, a1, u1) ├─M (q2,w2, a2, u2)
iff for some b    {L, R}, (q1, a1) = (q2, b) and either
(1)
b  , w1 = w2, u1= u2, and a2 = b; or
(2)
b = L, w1 = w2a2, and either
(3)
(a)
u2 = a1u1, if (a1  # or u1  e), or
(b)
u2 = e, if a1 = # and u1= e; or
b = R, w2 = w1 a1 and either
(a)
u1 = a2u2, or
(b)
u1 = u2 = e and a2 = #.
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Turing Machines

For any Turing machine M, relation ├─M* is the reflexive,
transitive closure of relation ├─M; We say configuration C1
yields configuration C2 if C1├─*M C2.

A computation by M is a sequence of configurations C0, C1,
C2,..., Cn (n0) such that
C0 ├─M C1 ├─M C2├─M … ├─M Cn .
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Turing Machines

Definition
Let f be a function from 0* to 1*. Turing machine M = (K,
, , s) , where 0, 1  , computes f if w  0* and
f(w) = u, then
(s, #w, #, e) ├─M* (h, #u, #, e).

If such a Turing machine exists, the function is called
Turing computable.
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Turing Machines

Let 0 an alphabet not containing the blank symbol. Let Y
and N be two fixed symbols not in 0. Then, language L 
0* is Turing decidable iff function xL: 0*  {Y, N} is
Turing computable, where for each w  0*,
xL(w) =
{
Y if w  L
N if w  L
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Turing Machines



Let 0 be an alphabet not containing #. M accepts string w
 0* if M halts on input w.
Thus, M accepts language L  0* iff L = {w  0*: M
accepts w}.
Definition
A language is Turing acceptable if there is some Turing
machine that accepts it.
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Turing Machines

Symbol writing Turing machines
For alphabet  and symbol a  ,
let Turing machine a = (K, , , s) =
({p}, , {(p, b, h, a): for any b  }, p)

Head moving Turing machines
For alphabet ,
let Turing machine L be defined as L = (K, , , s) =
({p}, , {(p, b, h, L): for any b  }, p), and
let Turing machine R be defined as R = (K, , , s) =
({p}, , {(p, b, h, R): for any b  }, p).
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Turing Machines

Definition
A machine schema is a triplet (m, h, M0) where
m is a finite set of Turing machines with common alphabet
 and disjoint sets of states;
M0  m is the initial machine; and
h is a function from subset of m to m.
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Turing Machines

Let m = {M0, M1, …, Mm} (m≥0), where
Mi = (Ki, , i, si) for i = 0, 1, …, m.
Let q0, q1, …, qm be new states not in any of the Ki.
Then machine schema (m, h, M0) is defined asTuring
machine M, where M = (K, , , s)
K = K0  ... Km  {q0, q2, …, qm},
s = s0, and
 is defined as follows.
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Turing Machines
Definition of 
(a)
if q  Ki (0 ≤ i ≤ m), a , i(q, a) = (p, b), and p  h,
then (q, a) = i(q, a) = (p, b);
(b)
if q  Ki (0 ≤ i ≤ m), a  , and i(q, a) = (h, b) then
(q, a) = (qi, b);
(c)
if a   and h(Mi , a) (0 ≤ i ≤ m) is not defined, then
(q, a) = (h, a);
(d)
if a   and h(Mi , a) = Mj (0 ≤ i ≤ m) and
j(sj, a) = (p, b), then
(qi, a) =
E
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(p, b) if p  h and
(qj, b) if p = h.
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Turing Machines

Theorem
Every Turing decidable language is Turing acceptable.


Proof:
Theorem
If L Turing decidable language, then it’s complement L is
also Turing-decidable.

Proof:
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Turing Machines

A language is Turing-decidable if and only if both it and its
complement are Turing acceptable .

Proof:
(only if)
L is a Turing decidable  L is also Turing decidable.
L, L is Turing decidable  L, L is Turing acceptable.
(if)
2-tape machine: M1 accepts L, M2 accepts L.
Parallel simulation: Which halts?
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Universal Turing Machines

Universal Turing Machines
The only feature possessed by electronic computers that
missing from the capability of Turing machines is the
programmability.

Difficulties:

The alphabet of any Turing machine must be finite.

The set of states of any Turing machine must be finite.
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
Hints:

Assume that there are fixed countably infinite sets
K = {q1, q2, q3, ...}, and
 = {a1, a2, a3, ...}
such that for every Turing machine, the set of states is
a subset of K and the alphabet is a subset of .
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
Encode the alphabet and states of the Turing machine as

()
qi
Ii+1
h
I
L
I
R
ai
II
Ii+2
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
Encode the Turing machine M over the alphabet {c, I}.
p(M) = cS0cS11S12 ... S1lS21S22 ... Sk1 ... Sklc
S0 encodes the initial state, S0 = (s).
Spr encodes the values of the transition function,
Spr = cw1cw2cw3cw4, where
w1 = (qi p), w2 = (qj r), w3 = (q’), w4 = (b)
for each
(qi p, qj r) = (q’, b).
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
The symbols on the tape must also be in encoded form.
p(w) = c(b1)c(b2)c ... c(bn)c
for each
w = b1b2 ... bn.
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
Universal Turing machine U simulates Turing machine M
on a three tape Turing machine.
Tape 1: encoding of the tape of Turing machine M.
Tape 2: encoding of Turing machine M
(i.e., it is the “program”).
Tape 3: encoding of the current state of Turing
machine M during the simulation.
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Examples
Alphabets and Languages
Examples

Symbol
Every character is a symbol, e. g., a, b, and c, etc.

Alphabet
 = {a, b, c} is an alphabet.

String
cab, abaca, a, and the empty string e are strings over the
alphabet  = {a, b, c}.

Length of a string
cab= 3, abaca= 5, a= 1, and e=0.
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Examples

Alphabets and Languages
Concatenation of strings
Strings cab and abaca can be concatenated to form string
cababaca.

Formally:
cababaca = cababaca
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Examples

Alphabets and Languages
Substring
Strings e, a, b, c, ab, ba, ac, ca, aba, bac, aca, abac, baca,
and abaca are the substrings of the string abaca.

Reversal of a string
eR = e
aR = a
abR = ba
bacR = cab
abacaR = acaba
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Examples

Alphabets and Languages
Language
Set L = {e, a, cab, abaca} is a language over the alphabet
 = {a, b, c}.
Thus, language L is a subset of * = {e, a, b, c, aa, ab, ac,
ba, bb, bc, ca, cb, cc, aaa, aab, aab, aba, …}.
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Examples

Alphabets and Languages
Language operation Kleene star
Closure or Kleene star of language L = {a, cab, abaca} the
language L* =
{e, a, cab, abaca, aa, acab, aabaca, caba, cabcab,
cababaca, …, abacaaa, abacaacab, …} =
{e, a, cab, abaca, aa, acab, aabaca, caba, cabcab,
cababaca, …, abacaaa, abacaacab, …}
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Examples

Alphabets and Languages
Regular expressions
Expressions
Ø, a, b,
(ab), ((ab)a),
(ab), (a(ab)), ((ab)((ab)a)), ((ab)c),
((ab)a), ((ab)(ab)),
a*, (ab)*, (ab)*,
(a*b), (a*b), ((ab)*((ab*)a)), and ((ab)*c*)
are regular expressions over the alphabet  = {a, b, c}.
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Examples

Alphabets and Languages
Language represented by regular expression
L(((ab)c)*) = ?
L(a) = {a}; L(b) = {b}; L(c) = {c};
L((ab)) = L(a)  L(b) = {a}  {b} = {a, b};
L(((ab)c)) = L((ab))L(c) = {a, b} {c} = {ac, bc};
L(((ab)c)*) = L(((ab)c))* = {ac, bc}* =
{e, ac, bc, acac, acbc, bcac, bcbc, acacac, acacbc, acbcac,
…, acbcacbcac, acbcacbcbc, … }
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Examples

Finite Automata
Deterministic finite automaton
M1= (K, , , s, F) =
(K= {p, q},
 = {a, b},
 = {(p, a, p), (p, b, q), (q, a, q), (q, b, p)},
s = p,
F = {q})
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Examples

Finite Automata
Graphical representation of DFA
M1= (K= {p, q},
 = {a, b},
 = {(p, a, p), (p, b, q), (q, a, q), (q, b, p)},
s = p,
F = {q})
is the state diagram:
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Examples

Finite Automata
Configuration of DFA M1
is: (p, babaa)
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Examples

Finite Automata
Yields in one step
(p, babaa)
├─M
(q, abaa)
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Examples

Finite Automata
Yields
(p, babaa)
├─M*
(p, aa)
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Examples

Finite Automata
String accepted by DFA M1: abbba
(p, abbba)
├─M*
(q, e)
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Examples

Finite Automata
Language accepted by DFA
M1= (K= {p, q},
 = {a, b},
 = {(p, a, p), (p, b, q), (q, a, q), (q, b, p)},
s = p,
F = {q})
L(M1) = {w  {a, b}*: the number of b’s in w is odd}
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Examples

Finite Automata
Graphical representation of NFA
M2= (K= {p, q},
 = {a, b},
 = {(p, ab, q), (p, e, q), (q, ba, q)},
s = p,
F = {q})
is the state diagram:
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Examples

Finite Automata
Language accepted by NFA
M2= (K= {p, q},
 = {a, b},
 = {(p, ab, q), (p, e, q), (q, ba, q)},
s = p,
F = {q})
is L(M2) = L( ((ab)  e)(ba)*) ).
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Examples

Context-free Languages
For context-free grammar
G = (V = {a, b, A, B, S},
 = {a, b},
R = {(S, aSb), (S, e)},
S)
we can write

SaSb, Se

SaSbaaSbbaaaSbbbaaabbb

S * aaSbb * aaabbb, and S * aaabbb
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Examples

Context-free Languages
Language generated by CFG
G = (V = {a, b, A, B, S},
 = {a, b},
R = {(S, aSb), (S, e)},
S)
is L(G) = {anbn: n  0}.
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Examples

Context-free Languages
Graphical representation of PDA
M3= (K= {p, q},
 = {a, b},
 = {c},
 = {(p, a, e, p, c), (p, e, q), (q, b, c, q, e)},
s = p,
F = {q})
is the state diagram:
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Examples

Context-free Languages
Language accepted by PDA
M3= (K= {p, q},
 = {a, b},
 = {c},
 = {(p, a, e, p, c), (p, e, q), (q, b, c, q, e)},
s = p,
F = {q})
is L(M2) = {anbn: n  0}.
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Examples
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Turing Machines
Turing machine represented by the following machine
schema operates as follows: (s, #w, #, e) ├─M* (h, #u, #, e),
where w  {a, b}* and u is derived from w by replacing
each a in w to b and each b in w to a.
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Problems
Problems

Alphabets and Languages

Which strings are the elements of the following
languages?
- {w: for some u{a, b}{a, b}, w = uuRu}
- {w: ww = www}
- {w: for some u, wuw = uwu}
- {w: for some u, www = uu}
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Problems

Is it true or false?
- baa  {a}*{b}*{a}*{b}*
- {b}*{a}*{a}*{b}* = {a}*{b}*
- {a}*{b}*{c}*{d}* = 
- abcd  ({a} ({c}{d})* {b})*
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Problems

Write regular expression R with
- L(R) = {w{a, b}*: there is at most three a's in w}
- L(R) = {w{a, b}*: there is no ab substring in w}
- L(R) = {w{a, b}*: w has exactly one aaa substring}
- L(R) = {w{a, b}*: in w every a is preceded and
followed by a b}
- L(R) = {w{a, b, c}*: in w there is no aa, bb, cc
substring in w}
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Problems
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Finite Automata

Draw state diagram of DFA M with
- L(M) = {w{a, b}*: in w the number of a's is even and
the number of b's is odd}
- L(M) = {w{a}*: in w the number of a's is even, but not
divisible by 4}
- L(M) = {w{a}*: in w the number of a's is either even or
divisible by 3}
- L(M) = {w{a}*: in w the number of a's is divisible by 3}
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Problems

Draw state diagram of NFA M with L(M) = L(R) where
- R = (a*b*)*(ab)
- R = (aba)*(aab)*
- R = ( (a*b*)(ab) )*
- R = ((ab)(ab))*
- R = (ab*)(a*b)
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Problems
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Context-free Languages

Which words can be derived in at most four steps in G =
(V, , R, S) from S?
V = {a, b, A, B, S},
 = {a, b}, and
- R = {SA, SabA, SaB, ASa, Bb}
- R = {SA, SabA, SaB, Aa, BSb}
- R = {SABS, AaA, BbB, Se, Aa, Bb}
- R = {SaSB, SbSA, Sa, Sb, AaS, BbS}
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Problems

Give a CF grammar G with
- L(G) = {w{a, b}*: in w the number of b's is even}
- L(G) = {anbman: n, m > 0}
- L(G) = {ancbnc: n  0}
- L(G) = {bnan: n  0}  {a3nbn: n  0}
- L(G) = {ambncpdr: m+n = p+r}
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Problems
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Give a pushdown automaton M with
- L(M) = {w{a, b}*: in w the number of a's is larger than
the number of b's}
- L(M) = {w{a, b}*: w = wR}
- L(M) = {w{a, b}*: in w the number of b's is odd}
-  = {a, +, *, (, )}
L(M) = {syntactically correct arithmetic expressions
involving + and * over the variable a}
- L(M) = {ambncp: m = n or m = p or n = p}
- L(M) = {ambncndm: m, n > 0}
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Problems
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Turing machines

Give a Turing machine which decides language L
- L = {w{a, b}*: in w the number of a's is the double of
the number of b's}
- L = {w{a, b}*: in w the number of a's is larger than the
number of b's}
- L = {w{a, b}*: in w the number of b's is even}
- L = {w{a, b}*: w = wR}
- L = {anbncn: n > 0}
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