Transcript
Mathematical Preliminaries
Hw (p.13) 1, 4, 7, 8, 9, 13, 23, 26, 30, 32
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Mathematical Preliminaries
• Sets • Functions• Relations• Graphs• Proof Techniques
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A set is a collection of elements
SETS
}3,2,1{A},,,{ airplanebicyclebustrainB
We writeA1
Bship
1 is a member (or element) of set A
ship is not a member (or element) of set B
Membership of a given set
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Set Representations
C = { a, b, c, d, e, f, g, h, i, j, k }
C = { a, b, …, k }
S = { 2, 4, 6, … }
S = { j : j > 0, and j = 2k for some integer k > 0 }
S = { j : j is nonnegative and even }
finite set
infinite set
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A = { 1, 2, 3, 4, 5 }
Universal Set: all possible elements
1 2 34 5
AU
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8910
U = { 1 , … , 10 }
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Set OperationsA = { 1, 2, 3 } B = { 2, 3, 4, 5}
• Union A U B = { 1, 2, 3, 4, 5 } • Intersection
A B = { 2, 3 }• Difference
A - B = { 1 }B - A = { 4, 5 }
U
A B231 4
5
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Venn diagrams
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A
• ComplementUniversal set = {1, …, 7} A = { 1, 2, 3 } A = { 4, 5, 6, 7}
1 2 34
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7A
A = A
8
024
61
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7even
{ even integers } = { odd integers }
odd
Integers
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DeMorgan’s Laws
A U B = A BU
A B = A U BU
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Empty, Null Set:= { }
S U = S
S =
S - = S
- S =
U = Universal Set
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SubsetA = { 1, 2, 3} B = { 1, 2, 3, 4, 5 }
A B
U
Proper Subset: A B
UAB
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Disjoint SetsA = { 1, 2, 3 } B = { 5, 6}
A B = UA B
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Set Cardinality• For finite setsA = { 2, 5, 7 }
|A| = 3
(set size)
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PowersetsA powerset is a set of subsets
Powerset of S = the set of all the subsets of S
S = { a, b, c }
2S = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} }
Observation: | 2S | = 2|S| ( 8 = 23 )
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Cartesian ProductA = { 2, 4 } B = { 2, 3, 5 }
A X B = { (2, 2), (2, 3), (2, 5), ( 4, 2), (4, 3), (4, 5) }
A X B is an ordered set, i.e. A X B ≠ B X A
|A X B| = |A|·|B|
Generalizes to more than two sets
A X B X … X Z
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Relation from sets A to BRelation from sets A to B Ex., A = { 1, 2, …, 7 }, B={ 1, 2, …,
50}– {(x, y): xA, yB, and y= x2} – {(x, y): xA, yB, and x < y}Are relations from A to B.
A relation on set A Equivalence relation a partition on set A
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FUNCTIONSdomain
123
ab
c
range f : A -> B
A B
If A = domain then f is a total function otherwise f is a partial function
f(1) = a4
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In general, we mean this.
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GRAPHSA directed graph
• Nodes (Vertices) V = { a, b, c, d, e }• Edges (Ordered pairs) E = { (a,b), (b,c), (b,e),(c,a), (c,e), (d,c), (e,b), (e,d) }
node
edge
ab
c
d
e
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Labeled Graph
ab
c
d
e
1 35 6
262
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Walk
ab
c
d
e
Walk is a sequence of adjacent edges(e, d), (d, c), (c, a)(e, d), (d, c), (c, a) is a walk from e to a of length 3 (or denoted as e-d-c-ae-d-c-a )Length = # of edges
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Path
ab
c
d
e
PathPath is a walk where no edge is repeated
Simple path: no node is repeated
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Path
ab
c
d
e
PathPath is a walk where no edge is repeated
Simple path: no node is repeated(e, b), (b, e), (e, d), (d, c), (c, a) is a path from e to a but it is not a simple path.
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Cycle
ab
c
d
e
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Cycle: a walk from a node (base) to itself without repeated edgesSimple cycle: only the base node is repeated Loop: an edge from a node to itself
base
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Find All Simple Paths starting from c
ab
c
d
e
origin
The longest simple path has at most length 4.Since every vertex can only be visited at most once, and there are 4 other vertices.
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(c, a)(c, e)
Step 1
ab
c
d
e
originStarting from vertex c, list all outgoing edges as long as they do not lead to any vertex already used in the path. At this point, we have all paths of length one starting at c.
For all vertices a,e reached by c, we list all outgoing edges originating at a or e according the same way we did before.
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(c, a)(c, a), (a, b)(c, e)(c, e), (e, b)(c, e), (e, d)
Step 2
ab
c
d
e
origin
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Step 3
ab
c
d
e
origin(c, a)(c, a), (a, b)(c, a), (a, b), (b, e)(c, e)(c, e), (e, b)(c, e), (e, d)
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Step 4
ab
c
d
e
origin
(c, a)(c, a), (a, b)(c, a), (a, b), (b, e)(c, a), (a, b), (b, e), (e,d)(c, e)(c, e), (e, b)(c, e), (e, d)
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Treesroot
leaf
parent
child
are connected directed graphs without cycles such that there is a special vertex called “root” having exactly one path to every other vertices.
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root
leaf
Level 0
Level 1
Level 2
Level 3
Height 3
The level associated with each vertex is the number of edges in the path form the root to the vertex.The height of the tree is the largest level number of any vertex.
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Binary Trees
A binary tree is a tree in which no parent can have more than two children.
A binary tree is a tree in which no parent can have more than two children.(p.10) Example 1.5.
Prove that a binary tree of height n has at most 2n leaves.
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PROOF TECHNIQUES
• Proof by induction
• Proof by contradiction
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Induction
We have statements P1, P2, P3, …
If we know• for some m that P1, P2, …, Pm are true• for any k >= m that
P1, P2, …, Pk imply Pk+1
Then Every Pi is true
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Proof by Induction• Inductive basis
Find P1, P2, …, Pm which are true• Inductive hypothesis
Let’s assume P1, P2, …, Pk are true, for any k >= m
• Inductive stepShow that Pk+1 is true
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ExampleTheorem: A binary tree of height n has at most 2n leaves. (p.10)
Proof by induction: let L(i) be the maximum number of leaves of any subtree at height i
We want to show: L(n) ≦ 2n for n = 0, 1, 2,….
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We want to show: L(n) ≦ 2n for n = 0, 1, 2,….
• Inductive basis L(0) =1 ≦ 20 (the root node : height=0)
• Inductive hypothesisLet’s assume L(i) ≦ 2i for all i = 0, 1, …, k
• Induction stepwe need to show that L(k + 1) ≦ 2k+1 , k≧0
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Induction Step
From Inductive hypothesis:
height
k
k+1
Let’s assume L(i) ≦ 2i for all i = 0, 1, …, k
need to show that L(k + 1) ≦ 2k+1
0
…
L(k) ≦ 2k
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L(k) ≦ 2k
L(k+1) ≦ 2 · L(k) ≦ 2· 2k = 2k+1
Induction Stepheight
k
k+1
(we add at most two nodes for every leaf of level k)
…
need to show that L(k + 1) ≦ 2k+1
To get a binary tree of height k+1 from one of height k, we can create at most 2 leaves in place of each previous one
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RemarkRecursion is another thing
Example of recursive function:
f(n) = f(n-1) + f(n-2)
f(0) = 1, f(1) = 1
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Proof by Contradiction
We want to prove that a statement P is true
• we assume that P is false• then we arrive at an incorrect conclusion• therefore, statement P must be true
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ExampleTheorem: is not rational
Proof:Assume by contradiction that it is rational = n/m n, m are nonzero integers and without common factorsWe will show that this is impossible
2
2
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= n/m 2 m2 = n2
Therefore, n2 is even n is evenn = 2 k
2 m2 = 4k2 m2 = 2k2m is evenm = 2 p
Thus, m and n have a common factor 2
Contradiction!
2
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LanguagesLanguages
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A language is a set of strings
String: A sequence of letters
Examples: “cat”, “dog”, “house”, …
Defined over an alphabet: zcba ,...,,,
Non-empty and finite
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Alphabets and StringsAlphabets and Strings We will use small alphabets:
Strings
abbawbbbaaavabu
ba,
baaabbbaabababaabbaaba
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Empty StringEmpty String
A string with no letters
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String OperationsString Operations
m
n
bbbvaaaw
......
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21
bbbaaaabba
mn bbbaaawv ...... 2121
Concatenation
abbabbbaaa
w = ? w = ?
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Empty StringEmpty StringObservations:
abbaabbaabba
www
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12... aaaw nR
naaaw ...21 ababaaabbb
Reverse
bbbaaababa
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String LengthString Length
Length:
Examples:
naaaw ...21nw
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4
aaaabba
| | = ?
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Length of Length of ConcatenationConcatenation
Example:
vuuv
8538
5,3,
vuuvaababaabuv
vabaabvuaabu
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Example 1.8 Example 1.8 (p.17) |uv| = |u| + |v|
A recursive definition of the length of a string: | a | =1,
| wa | = | w | + 1
For all a, w is any string from
Fix the string u and consider all possible strings v ( the length of v can be 1, 2, …. ( 0 is trivial) )
The proof is done by induction on the length of v ( for any given u )
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Example 1.8 Example 1.8 (p.17) |uv| = |u| + |v|
For a string u, consider the length of uv, concatenation of u with a string v
Basis
Inductive Step
Induction Assumption
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SubstringSubstring Substring of string: a
subsequence of consecutive letters
String Substring
bbabbabbaab
abbababbababbababbab
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Prefix and SuffixPrefix and Suffix Prefixes Suffixes
abbab
abbababbaabbaba
babbabbbababbab uvw
prefixsuffix
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Another Another OperationOperation
Example: abbaabbaabba 2
?0 w
n
n wwww
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Another OperationAnother Operation
Definitions:
abbawwabba 11
0w
0abba
wwwwn ...
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The * OperationThe * Operation : the set of all possible strings from alphabet
Example: = { a, b} then = ?
*
,,,,,,,,,*
,aabaaabbbaabaaba
ba
*
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The + Operation : the set of all possible strings from alphabet except
,...,,,,,,,,*,
aabaaabbbaabaababa
}{*
,...,,,,,,, aabaaabbbaabaaba
Fall 2008 Automata 60
LanguagesLanguages*
,...,,,,,,,*,
aaabbbaabaababa
},:{
},,,,,{,,,,
nmevenisnba
aaaaaaabaababaabbaaabaaa
mn
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Note that:
}{}{
0}{
1}{
0
Sets
Set size
Set size
String length
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Another ExampleAnother Example An infinite language }0:{ nbaL nn
aaaaabbbbbaabbab
L Labb
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Operations on LanguagesOperations on Languages The usual set operations
Complement: ??
aaaaaabbbaaaaaba
ababbbaaaaabaaaaabbabaabbbaaaaaba
,,,,}{,,,
},,,{,,,
LL * ,,,,,,, aaabbabaabbaa
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When we talk about a language, we must know what ground does this language stands on …..
LanguagesLanguages
We should know the ALPHABETSALPHABETS that constitute the language
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Complement ExampleComplement Example The complement
of},{ baaL
Universal Set?
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ReverseReverse Definition:
Examples:
}:{ LwwL RR
ababbaabababaaabab R ,,,,
}0:{
}0:{
nabL
nbaLnnR
nn
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ReverseReverse
}:{
})(:{}:{
Luu
LwwLwwLR
RRRRR
RRR LLLL )()()( 2121
RRR LLwthusLLw )(...,,: 2121
Hw # 10 (a) Prove or Disprove:
RRR LLwthusLLw 2121 ....,,)(:
i.e., wR L w LR
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ConcatenatioConcatenationn
Definition:
Example:
2121 ,: LyLxxyLL
baaabababaaabbaaaab
aabbaaba,,,,,
,,,
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2121
,:)(
)}(:{)(
LvLuuv
LLwwLLR
RR
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ConcatenationConcatenation
RRRRR uvuvww )()(
RRR LLLL 1221 )( Hw 8. Prove
21
2121
..
,)(:
LvandLusomeforuvwei
LLwthenLLwR
RR
2121 ,: LyLxxyLL
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Another OperationAnother Operation Definition:
Special case:
bbbbbababbaaabbabaaabaaa
babababa,,,,,,,
,,,, 3
0
0
,, aaabbaa
L
)timesitselfeconcatenat(... nLLLLLn
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More ExamplesMore Examples}0:{ nbaL nn
}0,:{2 mnbabaL mmnn
2Laabbaaabbb
LL 1
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Star-Closure (Kleene *)Star-Closure (Kleene *) Definition:
Example:
210* LLLL
,,,,,,,,
,,,
*,
abbbbabbaaabbaaabbbbbbaabbaa
bbabba
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Positive ClosurePositive ClosureDefinition:
21 LLL
,,,,,,,,
,,,
abbbbabbaaabbaaabbbbbbaabbaa
bbabba
*LLIt is not necessary thatIf L then L+ L* - {}
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True or FalseTrue or False
L *L
*L *)(L
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True or FalseTrue or False
*L *)( L
How to prove your answer?
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Try Hw# 9 & 10(b) on p.28
What does wL2 mean?What does wL* mean?
210* LLLL
.0..,,, 1010
*
ksomeforuuuwtsLuuu
Lw
kk
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More ExamplesMore Examples Consider a language on = {a, b}
)}()(:{ wnwnwL ba
What is L2 ? L*?
e.g. 4)(&2)(, wnwnabbabbw ba
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GrammarsGrammars
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Another ExampleAnother Example Grammar:
Derivation of sentence :
SaSbS
abaSbS ab
aSbS S
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Grammar:
Derivation of sentence :
aabbaaSbbaSbS
aSbS S
aabb
SaSbS
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Other derivations for
aaabbbaaaSbbbaaSbbaSbS
aaaabbbbaaaaSbbbbaaaSbbbaaSbbaSbS
SaSbSGrammar:
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Language of the grammar
SaSbS
}0:{ nbaL nn
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More NotationMore Notation Grammar PSTVG ,,,
:V
:T
:S
:P
Set of variables
Set of terminal symbols
Start variable
Set of Production rules
p.21
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ExampleExample Grammar :
SaSbSG
PSTVG ,,,
}{SV },{ baT
},{ SaSbSP
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More NotationMore Notation Sentential Form: A sentence that contains variables
Example:aaabbbaaaSbbbaaSbbaSbS
sentenceSentential Forms
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We write:
Instead of:
aaabbbS*
aaabbbaaaSbbbaaSbbaSbS
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In general we write:
If:
nww*
1
nwwww 321
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By default: ww*
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ExampleExample
S
aSbSGrammar
aaabbbS
aabbS
abS
S
*
*
*
*
Derivations
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Another Grammar ExampleAnother Grammar Example
Grammar :
bbaS nn
AaAbAAbS
Derivations:G
From A aAb and A , we know
A, ab, aabb, aaabbb, … *
aabbbAbbSabbAbSbAbS
*
*
*
S
aSbSbSaSbS
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Language defined by a GrammarLanguage defined by a GrammarFor a grammar with start variable :
GS
}:*{)( wSTwGL
String of terminals
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|aAbA
ExampleExample For grammar :
AaAbAAbS
}0:{)( 1 nbaGL nn
Since bbaS nn
G
Pf: show L(G) {anbn+1} & L(G) {anbn+1} from A aAb we get AanAbn when it is applied n times. Together with A, we get Aanbn for n = 0, 1, 2, …. : w L(G), i.e. S* w
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A Convenient NotationA Convenient Notation
AaAbA
|aAbA
In general, we need to give a proof that a given language indeed generated by a certain grammar.
Back to last Example
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ExampleExample For grammar :
}0:{ 1 nbaLLet nn
AaAbAAbS
}0:{)( 1 nbaGL nn
GTo show
L(G) L
&
L(G) L
Need to show?)( GLw
?LwPf: show L(G) {anbn+1} & L(G) {anbn+1} from A aAb we get AanAbn when it is applied n times. Together with A, we get Aanbn for n = 0, 1, 2, …. : w L(G), i.e. S* w
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More Examples on GrammarsMore Examples on Grammars Find grammars for L on {a, b} and give brief
arguments to explain why they work.
L contains all strings with exactly one a
L contains all strings with at least one a
L3
}0:{ 2 nba nn
},0:{ 2 nmnba mn
}3:{ 32 nba nn
At Least: S BaB; BaB | bB |
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1. Problems on p.27 1. Problems on p.27 You should be able to do 2 ~ 17, 21 Hand in: 9, 10, 11c, 14ef, 15c, 17
2. Read 2. Read P. 37~ 41P. 37~ 41, and try to describe , and try to describe LL((MM)) in Fig. 2.6. in Fig. 2.6.
Homework for next week.
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