1 Set Operations CS/APMA 202, Spring 2005 Rosen, section 1.7 Aaron Bloomfield

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Set Operations

CS/APMA 202, Spring 2005

Rosen, section 1.7

Aaron Bloomfield

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• Triangle shows mixable color range (gamut) – the set of colors

Sets of ColorsMonitor gamut (M)

Printer gamut

(P)

• Pick any 3 “primary” colors

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• A union of the sets contains all the elements in EITHER set

• Union symbol is usually a U

• Example: C = M U P

Monitor gamut (M)

Printer gamut

(P)

Set operations: Union 1

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U

A B

A U B

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• 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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• Properties of the union operation– A U = A Identity law– A U U = U Domination law– A U A = A Idempotent law– A U B = B U A Commutative law– A U (B U C) = (A U B) U C Associative law

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• An intersection of the sets contains all the elements in BOTH sets

• Intersection symbol is a ∩

• Example: C = M ∩ P

Monitor gamut (M)

Printer gamut

(P)

Set operations: Intersection 1

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U

BA

A ∩ B

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• 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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• Properties of the intersection operation– A ∩ U = A Identity law– A ∩ = Domination law– A ∩ A = A Idempotent law– A ∩ B = B ∩ A Commutative law– A ∩ (B ∩ C) = (A ∩ B) ∩ C 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• A difference of two sets is

the elements in one set that are NOT in the other

• Difference symbol is a minus sign

• Example: C = M - P

Monitor gamut (M)

Printer gamut

(P)

• Also visa-versa: C = P - M

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Set operations: Difference 2

U

A B

B - AA - B

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• 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 be the set S

Set operations: Difference 3

_

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• 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

Monitor gamut (M)

Printer gamut

(P)

Set operations: Symmetric Difference 1

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• 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

Set operations: Symmetric Difference 2

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• A complement of a set is all the elements that are NOT in the set

• Difference symbol is a bar above the set name: P or M

__

Monitor gamut (M)

Printer gamut

(P)

Complement sets 1

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Complement sets 2

U

A

A

B

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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• Properties of complement sets

– A = A Complementation law– A U A = U Complement law– A ∩ A = Complement law

Complement sets 4

¯

¯

¯

¯

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Quick surveyQuick survey

I understand the various set I understand the various set operationsoperations

a)a) Very wellVery well

b)b) With some review, I’ll be goodWith some review, I’ll be good

c)c) Not reallyNot really

d)d) Not at allNot at all

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A last bit of color…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

BABA

BABA

• These should lookvery familiar…

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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 notation and logical equivalences

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What we are going to prove…

A∩B=B-(B-A)

A B

A∩B B-AB-(B-A)

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Definition of difference


DeMorgan’s law

Complementation law

Distributive law

Complement law

Identity law

Commutative law

Proof by using basic set identities• Prove that A∩B=B-(B-A)

)AB-(BBA

)A(BB

)AB(B

A)B(B

A)(B)B(B

A)(BA)(BBA

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• The top row is all elements that belong to both sets A and B– Thus, these elements are in the union and intersection, but not

the difference

• The second row is all elements that belong to set A but not set B– Thus, these elements are in the union and difference, but not the

intersection

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 B A U B A ∩ B A - B1 1 1 1 01 0 1 0 10 1 1 0 00 0 0 0 0

• The third row is all elements that belong to set B but not set A– Thus, these elements are in the union, but not the intersection or

difference

• The bottom row is all elements that belong to neither set A or set B– Thus, these elements are neither the union, the intersection, nor

difference

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Proof by membership tables

• The following membership table shows that A∩B=B-(B-A)

• Because the two indicated columns have the same values, the two expressions are identical

• This is similar to Boolean logic!

A B A ∩ B B-A B-(B-A)1 1 1 0 11 0 0 0 00 1 0 1 00 0 0 0 0

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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:– (xA∩B→ xB-(B-A)) (xB-(B-A)→ xA∩B)– This is the biconditional!

• Not good for long proofs• Basically, it’s an English run-through of the proof

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• Assume that xB-(B-A)– By definition of difference, we know that xB and xB-A

• Consider xB-A– If xB-A, then (by definition of difference) xB and xA– Since xB-A, then only one of the inverses has to be true

(DeMorgan’s law): xB or xA

• So we have that xB and (xB or xA)– It cannot be the case where xB and xB– Thus, xB and xA– This is the definition of intersection

• Thus, if xB-(B-A) then xA∩B

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• Assume that xA∩B– By definition of intersection, xA and xB

• Thus, we know that xB-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 xB-A– We also know that if xA∩B then xB (by definition of

intersection)– Thus, if xB and xB-A, we can restate that (using the definition

of difference) as xB-(B-A)

• Thus, if xA∩B then xB-(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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Original statement


Negating “element of”


DeMorgan’s Law

Distributive Law

Negating “element of”

Negation Law

Identity Law

Definition of intersection

)}(|{ AxBxBxx

)( ABB )}(|{ ABxBxx ))}((|{ ABxBxx

}|{ AxBxx

)}(|{ AxBxBxx }|{ AxBxBxBxx })(|{ AxBxBxBxx

}|{ AxBxFx

BA

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• 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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I understand (more or less) the four I understand (more or less) the four ways of proving a set identityways of proving a set identity





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Today’s demotivatorsToday’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:abcdefghijklmnopqrstuvwxyz10001000100000100000100000

• The consonants in the alphabet:abcdefghijklmnopqrstuvwxyz01110111011111011111011111

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Computer representation of sets 2

• Consider the union of these two sets: 1000100010000010000010000001110111011111011111011111 11111111111111111111111111

• Consider the intersection of these two sets: 1000100010000010000010000001110111011111011111011111 00000000000000000000000000

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Rosen, section 1.7 question 14

• Let A, B, and C be sets. Show that:

a) (AUB) (AUBUC)

b) (A∩B∩C) (A∩B)

c) (A-B)-C A-C

d) (A-C) ∩ (C-B) =

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I felt I understood the material in this I felt I understood the material in this slide set…slide set…





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The pace of the lecture for this The pace of the lecture for this slide set was…slide set was…

a)a) FastFast

b)b) About rightAbout right

c)c) A little slowA little slow

d)d) Too slowToo slow

Documents

1 Set Operations CS/APMA 202, Spring 2005 Rosen, section 1.7 Aaron Bloomfield