Automata, Grammars and Languages

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Automata, Grammars and Languages

• Discourse 02
• Preliminaries

2
Sets

• Set primitive notion of aggregatefrom which
  all of mathematics and logic can be constructed
• One small hierarchy of concepts in this course

derives
Grammar
real
sequence
string
function
rational
relation
integer
tuple
set
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Sets (contd)

• Predicate P(x) a statement about a variable x
  that is true or false when x is replaced by a
  particular object
• P(x) x is odd
• Main predicate for set-membership x ? A
• Some axioms of set theory
• Axiom of Extension a set is determined by its
  extension
• Axiom of Specification For every set A and
  predicate P(x) there is a set
  of all elements of A for which P is
  true.
• Ex x?Z x is positive and not prime
  4,6,8,9,10,12,

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Sets (contd)

• Operations and relations on sets
• subset
• proper subset
• union
• intersection
• complement
• difference
• size of set
• Special sets
• empty set
• natural numbers
• 
  integers
• Sets of sets
• Power set

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Logical Implication (Material implication)

• R ? S If you do not pay us 1M by midnight (R),
  we will shoot your ambassador (S)

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Logical Implication (contd)

• P ? Q If you pay us 1M by midnight (P), we
  will not shoot your ambassador (Q)

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Quantifiers

• ?
• ?
• Ex defining big Oh relationship between
  functions
• Ex continuity of a function at a point
  (epsilon-delta defn)
• Abe Lincolns quote canfool(p,t) can fool
  person p at time t

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Quantifiers (contd)

• Relationship between ? and ?
• Ex cannot fool all of the people all of the
  time
• Ex non-continuity at a point

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Sets Predicates (U universe)

• Sets
• P x P(x)
• A?B
• A?B
• A - B
• P ? ?
• P U
• U - P
• x? P
• P?Q
• PQ

• Logical Predicates
• P(x)
• A(x) ? B(x)
• A(x) ? B(x)
• A(x) ? ? B(x)
• (?x) P(x)
• (?x) P(x)
• ? P(x)
• P(x) true
• (?x) P(x) ? Q(x)
• (?x) P(x) ? Q(x)

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Tuples

• 3,7 unordered pair 2-tuple
• (3,7) ordered pair (3,7) ? 3,7,3 (7,3)
  ? 3,7,7
• Generalize to n-tuple (a1, a2, a3,, an)
• Defn. Cartesian Product
• A x B ? (a,b) a?A ? b?B
• Generalization A1x A2 x x An

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Binary Relations

• Defn a binary relation R from A to B is a
  subset of A ? B
• Defn a function f from A to B, written f A ?
  B, is a relation f ? A ? B that is
  single-valued, i.e.,
• Defn one-to-one (injection), onto (surjection),
  one-to-one correspondence (bijection)
• See Definition 4.12, p. 175, text. Also see
  below.

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Binary Relations 3 views
predicate
set
relation
(postfix)
(infix)
(prefix)
(a,b) ? R
aRb
R (a,b)
lt (3,10)
3 lt 10
(3,10) ? lt
isFatherof (Charles, Andrew)
(Charles,Andrew) ? isFatherof
Charles isFatherof Andrew
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Why Relations? Generalize Functions.

• Ex functions
• Ex division with remainder E ? N?N?N?N
• Ex circle C ? R?R
• Ex Relational Database
• Grammar derives relation

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Relational Calculus
S
R
?
?
?
a
?
?
c
?
?
b
?
?
?
?
B
C
A
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Relational Inverse

• R ? A ? B
• ________________
• lt
• ?
• FatherOf
• DivisorOf
• Hits

• R -1 ? B ? A
• _______________
• gt
• ?
• ChildOf
• MultipleOf
• Is hit by

R
R-1
A B
B A
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The Calculus
A
A

• Proposition. If

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Proving a Proposition about Relations

• Thm.
• Pf (a)
• Let Then
• and
• by definition of ?. So
• and so
  Since (c, a) was chosen arbitrarily,
• (b)
• Let Then
• and So
• implying
  Hence
• Since (c,a) was chosen arbitrarily, (b)
  follows. ?

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Relational Properties R ? A?B
Relational calculus predicate calculus
name
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Family Relationships
is Child of
Grandparent
Great n Grandparent
Parent (of child with offspring!)
Parent
Sibling
Sibling or self!
Sibling
Self (erasexual reproduction only )
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Family Relationships (contd)
i
Nephew, niece or child
c
h
Uncle, Aunt or
f
g
b
Child (w. offspring)
a
e
d
1st Cousin Once Removedor
1st Cousin Once Removed or
Parent or Grandparent
Ancestor
Transitive closure of P
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Binary Relations on A to itself (A)

• Ex

Theorem
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Properties of
Name Defn. Relational Calculus
R reflexive
R symmetric
R transitive
R an equivalence relation ? reflexive,
symmetric transitive
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A False Proof About Relations

• Theorem? Clearly any symmetric and transitive
  relation R must be reflexive.
• Pf? Assume that R is symmetric and
  transitive. Then
• By transitivity
• Since a was chosen arbitrarily, it follows
  that
• Whats wrong?

If (?a)(?b) aRb is true, then the argument is
correct!
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Binary R ? A?A Digraphs 0-1 Matrices
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Binary Relations, Digraphs, Matrices (contd)
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Binary Relations, Digraphs, Matrices (contd)
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Binary Relations, Digraphs, Matrices (contd)
a
b
d
c
Repeats!
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Binary Relations, Digraphs, Matrices (contd)
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Transitive Closure (finite graph)
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Transitive Closure ? Reachability

• Defn a reaches b in relation (digraph) R iff
• Prop a reaches b in R iff aRb
• Thm Let be a relation
  where Then
• Pf Longest possible path in G(R) that will
  not repeat an edge is of length n. This path
  will result in an edge in Rn.
• Ex may need to go all the way up to Rn

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Strings and Languages

• In this course, a language is simply a set of
  strings a programming language is much more
  complex
• alphabet ? - a finite set of symbols
• String (word) over ? - finite sequence of symbols
• - the empty or null string
• w - length of string w
• What is a string precisely? String w of length n
  is a function
• String ops
• concatenation
• powers

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Strings and Languages (contd)

• ?? w w is a string over ? (note
  ? ? ?? )
• Language L over ? a subset L ? ??
• Ex
• Ex ? ASCII codes ( blank \040 ?
  ? )

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Strings and Languages (contd)

• Language ops
• Set operators
• Concatenation
• Powers
• Ex
• Ex

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Strings and Languages (contd)

• Language ops (contd)
• Defn Kleene Closure (Star)
• Note
• Defn
• Ex
• Ex

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Strings and Languages (contd)

• Theorem
• Pf
• Ex

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Strings and Languages (contd)

• Ex
• Ex
• Ex

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Methods of Proof

• Construction exhibit the object guaranteed by
  the theorem.
• Ex Construction of a regular expression, given
  a FA.
• Contradiction To show P Assume ?P and derive
  a contraction or clear falsity (reduction ad
  absurdum)
• Ex our proof of undecidability of the halting
  problem assumed halt program existed and
  derived an absurd contradiction
• Induction to prove P(n) holds for all
  non-negative integers n

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Induction
?
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A Rule of Inference
base
step
conclusion
k ? 0
Prove P(0)
P(k)
k n
T
k ? k1
F
P(n)
Prove P(k) ?P(k1)

• halts ?n
• algorithm ? ?n P(n)

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Kinds of Induction

• Simple induction
• _________________
• Equivalent
• ______________________________
• Course-of-Values Induction
• ___________________

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Ex Induction Argument Balanced Parens

• Defn The strings having balanced parentheses
  over (,) are defined (inductively) by
• The empty string ? is balanced
• If w is balanced, so is (w)
• If w, x are balanced, so is wx
• Nothing else is balanced except by the above
  rules
• Remark a grammar for balanced strings is
• Examples
• Balanced
• Unbalanced

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Parentheses (contd)

• Defn (C) A string w over (,) has the prefix
  property C iff
• Note the prefix property can be checked in a L-R
  scan of the string using a counter (this is what
  calculators do)
• Thm A string w is balanced iff it has the
  prefix property.
• Comment this iff (?, a logical equivalence)
  means we have to prove 2 directions
• If a string is balanced, it has the prefix
  property (?) (Lemma 1 below)
• If a string has the prefix property, then it is
  balanced (?) (Lemma 2)

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Parentheses (contd)
( ( ( ) ( ) ) (
( ( ) ) ) )
( ) ) (
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Parentheses (contd)

• Thm A word w is balanced iff it has the prefix
  property.
• Lemma 1 w balanced ? w has prefix property C.
• Pf Induction on w
• Base w0 ? w? ? w satisfies C.
• Step Let wn. Assume
• (IH) all strings shorter than n that are
  balanced satisfy C.
• Let w be balanced. Two cases are possible
• Case wuv where u,v are balanced. By IH, u,
  v satisfy C. Then
• and so w satisfies C(a).
• Next consider a prefix s of wuv. If s is a
  prefix of u then
• because by IH,
  then w satisfies C(b) for
• this prefix.

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Parentheses (contd)

• If s ut where t is a prefix of v then
• by IH, and so
• and so w satisfies C(b) in this case.
• Case w(u) where u is balanced. By IH u
  satisfies C, and
• so clearly so does (u) ?

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Parentheses (contd)

• Lemma 2 w satisfies C ? w is balanced.
• Pf Induction on w
• Base w? is balanced by definition.
• Step Let wn gt0. Assume
• (IH) all strings shorter than n that
  satisfy C are balanced.
• Let w have prefix property C. Let x be the
  shortest prefix
• of w such that
  Such a prefix exits since
• w has this property.
• Case xw Then w (u) where u satisfies C.
  By IH, u is
• balanced, and so then so is w.
• Case xvw with v??. Now x satisfies C(a) by
  assumption
• and satisfies C(b) since w does. So by IH x
  is balanced.

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Parentheses (contd)

• We claim that v has property C. Since
• and then
• and so v satisfies C(a). Suppose there were
  a prefix y of v such that
  Then
• Which would violate the prefix property of
  w. Thus it must
• be that So v
  satisfies C(b).
• By IH, v is balanced.
• Since both x and v are balanced, wxv is
  balanced. ?