Theory of Digital Computation

1
Theory of Digital Computation

• Course material for undergraduate students on IT
• Department of Computer ScienceUniversity of
  VeszpremVeszprem, Hungary
• 2002

2
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3
Table of Contents
Table of Contents

• Alphabets and Languages
• Finite Automata
• Context-free Languages
• Turing Machines
• Universal Turing Machines
• Examples
• Problems

4
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.

5
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?.

6
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? n1 (gt0), then w ua for some a??, and
  wR auR.
• Any set of string over an alphabet is called
  language.

7
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.

8
Alphabets and Languages

• DefinitionA 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).

9
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.

10
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

• DefinitionDeterministic 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

13
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??.
• 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.

14
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).
• A string w ? ? is said to be 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

• DefinitionNon-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

• DefinitionFinite automata M1, and M2 are said to
  be equivalent iff L(M1) L(M2).
• TheoremFor each non-deterministic finite
  automaton, there is an equivalent deterministic
  finite automaton.
• (No proof here)

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Finite Automata

• TheoremLanguages 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

• TheoremLanguages 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

• TheoremLanguages 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

• TheoremLanguages 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

• TheoremLanguages 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

• TheoremThere is an algorithm for answering the
  following questionGiven 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

• TheoremThere is an algorithm for answering the
  following questionGiven 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

• TheoremThere is an algorithm for answering the
  following questionGiven 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

• DefinitionA 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  xvy and A ?G v.
• Relation ?G is the reflexive, transitive closure
  of ?G.
• 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

• DefinitionA language L is a context-free
  language if it is equal to L(G) for some
  context-free grammar G.
• DefinitionA context-free grammar G (V, ?, R,
  S) is regular iff R ? (V-?) ? ?((V-?) ? e).

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Context-free Languages

• TheoremContext-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
  thatL(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

• TheoremContext-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.

32
Context-free Languages

• TheoremContext-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

• DefinitionA 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) ? ?

35
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.

36
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.

37
Turing Machines
Turing Machines

• DefinitionA Turing machine is a quadruple (K, S,
  ?, s) where
• K is a finite set of states, not containing the
  halt state h (h ? K),
• S 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).

38
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
• (a) u2 a1u1, if (a1 ? or u1 ? e), or
• (b) u2 e, if a1 and u1 e or
• (3) b R, w2 w1 a1 and either
• (a) u1 a2u2, or
• (b) u1 u2 e and a2 .

39
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 .

40
Turing Machines

• DefinitionLet 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.

41
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

42
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.
• DefinitionA language is Turing acceptable if
  there is some Turing machine that accepts it.

43
Turing Machines

• Symbol writing Turing machines
• For alphabet ? and symbol a ? ?,
• let Turing machine a (K, S, ?, s) (p, S,
  (p, b, h, a) for any b ? ?, p)
• Head moving Turing machines
• For alphabet ?,
• let Turing machine L be defined as L (K, S, ?,
  s) (p, S, (p, b, h, L) for any b ? ?, p),
  and
• let Turing machine R be defined as R (K, S, ?,
  s) (p, S, (p, b, h, R) for any b ? ?, p).

44
Turing Machines

• DefinitionA machine schema is a triplet (m, h,
  M0) where
• m is a finite set of Turing machines with common
  alphabet S and disjoint sets of states
• M0 ? m is the initial machine and
• h is a function from subset of m?? to m.

45
Turing Machines

• Let m M0, M1, , Mm (m0), 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, ?, s)
• K K0 ? ... ?Km ? q0, q2, , qm,
• s s0, and
• ? is defined as follows.

46
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) (p, b) if p ? h and
• (qj, b) if p h.

47
Turing Machines

• TheoremEvery Turing decidable language is Turing
  acceptable.
• Proof
• TheoremIf L Turing decidable language, then its
  complement L is also Turing-decidable.
• Proof

48
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?

49
Universal Turing Machines
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.

50
Universal Turing Machines

• 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 ??.

51
Universal Turing Machines

• Encode the alphabet and states of the Turing
  machine as

? ?(?)
qi Ii1
h I
L I
R II
ai Ii2
52
Universal Turing Machines

• 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).

53
Universal Turing Machines

• 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.

54
Universal Turing Machines

• 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.

55
Examples
Examples
Alphabets and Languages

• 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.

56
Examples
Alphabets and Languages

• Concatenation of strings
• Strings cab and abaca can be concatenated to
  form string
• cababaca.
• Formally
• cab?abaca cababaca

57
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

58
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, .

59
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,

60
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),
• (ab), (a?b), ((ab)?((ab)a)), and ((a?b)?c)
• are regular expressions over the alphabet ?
  a, b, c.

61
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,

62
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)

63
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

64
Examples
Finite Automata

• Configuration of DFA M1
• is (p, babaa)

65
Examples
Finite Automata

• Yields in one step
• (p, babaa) -M (q, abaa)

66
Examples
Finite Automata

• Yields
• (p, babaa) -M (p, aa)

67
Examples
Finite Automata

• String accepted by DFA M1 abbba
• (p, abbba) -M (q, e)

68
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 bs in w is
  odd

69
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

70
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)) ).

71
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

72
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.

73
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

74
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.

75
Examples
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.

76
Problems
Problems

• Alphabets and Languages
• Which strings are the elements of the following
  languages?
• w for some u?a, ba, b, w uuRu
• w ww www
• w for some u, wuw uwu
• w for some u, www uu

77
Problems

• Is it true or false?
• baa ? abab
• ba?ab a?b
• ab?cd ?
• abcd ? (a (cd) b)

78
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

79
Problems

• 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

80
Problems

• Draw state diagram of NFA M with L(M) L(R)
  where
• R (ab)(a?b)
• R (ab?a)?(aa?b)
• R ( (a?b)(ab) )
• R ((a?b)(a?b))
• R (a?b)(a?b)

81
Problems

• 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

82
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 gt 0
• L(G) ancbnc n ? 0
• L(G) bnan n ? 0 ? a3nbn n ? 0
• L(G) ambncpdr mn pr

83
Problems

• 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 gt 0

84
Problems

• 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 gt 0