Transcript Set operations ( § 1.7)

Set Operations
CS/APMA 202, Spring 2005
Rosen, section 1.7
Aaron Bloomfield
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Sets of Colors
Monitor gamut
(M)
Printer
gamut
(P)
• Pick any 3 “primary” colors
• Triangle shows mixable
color range (gamut) – the
set of colors
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Set operations: Union 1
Monitor gamut
(M)
• A union of the sets contains
all the elements in EITHER
set
Printer
gamut
(P)
• Union symbol is
usually a U
• Example:
C=MUP
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Set operations: Union 2
AUB
U
A
B
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Set operations: Union 3
• Formal definition for the union of two sets:
A U B = { x | x  A or x  B }
• Further examples
– {1, 2, 3} U {3, 4, 5} = {1, 2, 3, 4, 5}
– {New York, Washington} U {3, 4} = {New York,
Washington, 3, 4}
– {1, 2} U  = {1, 2}
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Set operations: Union 4
• Properties of the union operation
–AU=A
–AUU=U
–AUA=A
–AUB=BUA
– A U (B U C) = (A U B) U C
Identity law
Domination law
Idempotent law
Commutative law
Associative law
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Set operations: Intersection 1
Monitor gamut
(M)
• An intersection of the sets
contains all the elements in
BOTH sets
Printer
gamut
(P)
• Intersection symbol
is a ∩
• Example:
C=M∩P
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Set operations: Intersection 2
A∩B
U
A
B
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Set operations: Intersection 3
• Formal definition for the intersection of two
sets: A ∩ B = { x | x  A and x  B }
• Further examples
– {1, 2, 3} ∩ {3, 4, 5} = {3}
– {New York, Washington} ∩ {3, 4} = 
• No elements in common
– {1, 2} ∩  = 
• Any set intersection with the empty set yields the
empty set
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Set operations: Intersection 4
• Properties of the intersection operation
–A∩U=A
–A∩=
–A∩A=A
–A∩B=B∩A
– A ∩ (B ∩ C) = (A ∩ B) ∩ C
Identity law
Domination law
Idempotent law
Commutative law
Associative law
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Disjoint sets 1
• Two sets are disjoint if the
have NO elements in
common
• Formally, two sets are
disjoint if their intersection
is the empty set
• Another example:
the set of the even
numbers and the
set of the odd
numbers
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Disjoint sets 2
U
A
B
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Disjoint sets 3
• Formal definition for disjoint sets: two sets
are disjoint if their intersection is the empty
set
• Further examples
– {1, 2, 3} and {3, 4, 5} are not disjoint
– {New York, Washington} and {3, 4} are disjoint
– {1, 2} and  are disjoint
• Their intersection is the empty set
–  and  are disjoint!
• Their intersection is the empty set
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Set operations: Difference 1
Monitor gamut
(M)
• A difference of two sets is
the elements in one set
that are NOT in the other
Printer
gamut
(P)
• Difference symbol is
a minus sign
• Example:
C=M-P
• Also visa-versa:
C=P-M
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Set operations: Difference 2
A-A
B
B
U
A
B
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Set operations: Difference 3
• Formal definition for the difference of two
sets:
A - B = { x | x_  A and x  B }
A - B = A ∩ B  Important!
• Further examples
– {1, 2, 3} - {3, 4, 5} = {1, 2}
– {New York, Washington} - {3, 4} = {New York,
Washington}
– {1, 2} -  = {1, 2}
• The difference of any set S with the empty set will
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be the set S
Set operations: Symmetric
Difference 1
Monitor gamut
(M)
Printer
gamut
(P)
• A symmetric difference of
the sets contains all the
elements in either set but
NOT both
• Symetric diff.
symbol is a 
• Example:
C=MP
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Set operations: Symmetric
Difference 2
• Formal definition for the symmetric difference of
two sets:
A  B = { x | (x  A or x  B) and x  A ∩ B}
A  B = (A U B) – (A ∩ B)  Important!
• Further examples
– {1, 2, 3}  {3, 4, 5} = {1, 2, 4, 5}
– {New York, Washington}  {3, 4} = {New York,
Washington, 3, 4}
– {1, 2}   = {1, 2}
• The symmetric difference of any set S with the empty set will
be the set S
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Complement sets 1
Monitor gamut
(M)
• A complement of a set is all
the elements that are NOT
in the set
Printer
gamut
(P)
• Difference symbol is
a bar above
_ _the set
name: P or M
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Complement sets 2
_
A
B
U
A
B
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Complement sets 3
• Formal definition for the complement of a
set: A = { x | x  A }
– Or U – A, where U is the universal set
• Further examples (assuming U = Z)
– {1, 2, 3} = { …, -2, -1, 0, 4, 5, 6, … }
– {New York, Washington} - {3, 4} = {New York,
Washington}
– {1, 2} -  = {1, 2}
• The difference of any set S with the empty set will
be the set S
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Complement sets 4
• Properties of complement sets
¯
¯
–A=A
– A U A¯ = U
– A ∩ A¯ = 
Complementation law
Complement law
Complement law
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Quick survey

a)
b)
c)
d)
I understand the various set operations
Very well
With some review, I’ll be good
Not really
Not at all
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A last bit of color…
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Set identities
• Set identities are basic laws on how set
operations work
– Many have already been introduced on previous
slides
• Just like logical equivalences!
–
–
–
–
Replace U with 
Replace ∩ with 
Replace  with F
Replace U with T
• Full list on Rosen, page 89
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Set identities: DeMorgan again
• These should look
very familiar…
A B  A B
A B  A B
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How to prove a set identity
• For example: A∩B=B-(B-A)
• Four methods:
– Use the basic set identities (Rosen, p. 89)
– Use membership tables
– Prove each set is a subset of each other
• This is like proving that two numbers are equal by
showing that each is less than or equal to the other
– Use set builder
equivalences
notation
and
logical
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What we are going to prove…
A∩B=B-(B-A)
A
B
B-(B-A)
A∩B
B-A
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Proof by using basic set identities
• Prove that A∩B=B-(B-A)
A  B  B-(B  A )
Definition of difference
 B  (B  A )
Definition of difference
 B  (B  A )
DeMorgan’s law
 B  (B  A)
Complementation law
 (B  B )  (B  A)
Distributive law
   (B  A)
Complement law
 (B  A)
Identity law
 A B
Commutative law
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What is a membership table
• Membership tables show all the combinations of
sets an element can belong to
– 1 means the element belongs, 0 means it does not
• Consider the following membership table:
A
1
1
0
0
B
1
0
1
0
AUB A∩B A-B
1
1
0
1
0
1
1
0
0
0
0
0
• The bottom
third
second
top row
row
row
row
is
is all
isisall
elements
allelements
elements
that
that
that
that
belong
belong
belong
belong
tototo
set
to
both
neither
B
setbut
sets
A not
but
set
A
and
not
set
A
orA
set
B
setBB
– Thus, these elements
elements are
arein
neither
inthe
theunion
union,
union
the union,
and
but
anddifference,
not
intersection,
thethe
intersection,
intersection
but but
not nor
not
the
or
the difference
intersection
difference
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Proof by membership tables
• The following membership table shows that
A∩B=B-(B-A)
A
1
1
0
0
B
1
0
1
0
A∩B
1
0
0
0
B-A
0
0
1
0
B-(B-A)
1
0
0
0
• Because the two indicated columns have the
same values, the two expressions are identical
• This is similar to Boolean logic!
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Proof by showing each set
is a subset of the other 1
• Assume that an element is a member of one of
the identities
– Then show it is a member of the other
• Repeat for the other identity
• We are trying to show:
– (xA∩B→ xB-(B-A))  (xB-(B-A)→ xA∩B)
– This is the biconditional!
• Not good for long proofs
• Basically, it’s an English run-through of the proof
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Proof by showing each set
is a subset of the other 2
• Assume that xB-(B-A)
– By definition of difference, we know that xB and xB-A
• Consider xB-A
– If xB-A, then (by definition of difference) xB and xA
– Since xB-A, then only one of the inverses has to be true
(DeMorgan’s law): xB or xA
• So we have that xB and (xB or xA)
– It cannot be the case where xB and xB
– Thus, xB and xA
– This is the definition of intersection
• Thus, if xB-(B-A) then xA∩B
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Proof by showing each set
is a subset of the other 3
• Assume that xA∩B
– By definition of intersection, xA and xB
• Thus, we know that xB-A
– B-A includes all the elements in B that are also not in A not
include any of the elements of A (by definition of difference)
• Consider B-(B-A)
– We know that xB-A
– We also know that if xA∩B then xB (by definition of
intersection)
– Thus, if xB and xB-A, we can restate that (using the definition
of difference) as xB-(B-A)
• Thus, if xA∩B then xB-(B-A)
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Proof by set builder notation
and logical equivalences 1
• First, translate both sides of the set
identity into set builder notation
• Then massage one side (or both) to make
it identical to the other
– Do this using logical equivalences
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Proof by set builder notation
and logical equivalences 2
B  ( B  A)
 {x | x  B  x  ( B  A)}
Original statement
 {x | x  B  ( x  ( B  A))}
Negating “element of”
 {x | x  B  ( x  B  x  A)}
Definition of difference
 {x | x  B  ( x  B  x  A)}
 {x | x  B  x  B  x  B  x  A}
DeMorgan’s Law
 {x | x  B  ( x  B)  x  B  x  A}
 {x | F  x  B  x  A}
 {x | x  B  x  A}
 A B
Definition of difference
Distributive Law
Negating “element of”
Negation Law
Identity Law
Definition of intersection
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Proof by set builder notation
and logical equivalences 3
• Why can’t you prove it the “other” way?
– I.e. massage A∩B to make it look like B-(B-A)
• You can, but it’s a bit annoying
– In this case, it’s not simplifying the statement
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Quick survey

a)
b)
c)
d)
I understand (more or less) the four ways of
proving a set identity
Very well
With some review, I’ll be good
Not really
Not at all
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Today’s demotivators
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Computer representation of sets 1
• Assume that U is finite (and reasonable!)
– Let U be the alphabet
• Each bit represents whether the element in U is in the set
• The vowels in the alphabet:
abcdefghijklmnopqrstuvwxyz
10001000100000100000100000
• The consonants in the alphabet:
abcdefghijklmnopqrstuvwxyz
01110111011111011111011111
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Computer representation of sets 2
• Consider the union of these two sets:
10001000100000100000100000
01110111011111011111011111
11111111111111111111111111
• Consider the intersection of these two sets:
10001000100000100000100000
01110111011111011111011111
00000000000000000000000000
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Rosen, section 1.7 question 14
•
a)
b)
c)
d)
Let A, B, and C be sets. Show that:
(AUB)  (AUBUC)
(A∩B∩C)  (A∩B)
(A-B)-C  A-C
(A-C) ∩ (C-B) = 
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Quick survey

a)
b)
c)
d)
I felt I understood the material in this slide set…
Very well
With some review, I’ll be good
Not really
Not at all
44
Quick survey

a)
b)
c)
d)
The pace of the lecture for this slide set was…
Fast
About right
A little slow
Too slow
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