1 lecture 2 (part 1) partial orders reading: epp chp 10.5

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Lecture 2 (part 1)

Partial OrdersReading: Epp Chp 10.5

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Outline1. Revision2. Definition of poset3. Examples of posets

a. In lifeb. Finite posetsc. Infinite posets

4. Notation5. Visualization Tool: Hasse Diagram6. Definitions

– maximal, greatest, minimal, least.– 2 Theorems

7. More Definitions– Comparable, chain, total-order, well-order

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1. RevisionConcrete World

Abstract World

a. ___ has been to ___ {John, Mary, Peter} {Tokyo, NY, HK} {(John,Tokyo), (John,NY), (Peter, NY)}

b. ___ is in ___ {Tokyo, NY} {Japan, USA} {(Tokyo,Japan),(NY,USA)}

c. ___ divides ___ {1,2,3,4} {10,11,12} {(1,10),(1,11),(1,12), (2,10), (2,12),(3,12), (4,12)}

d. ___ less than ___ {1,2,3} {1,2,3} {(1,2),(1,3),(2,3)}

___ R ___ A B R A B

Q: What can you do with relations?A: (1) Set Operations; (2) Complement; (3) Inverse; (4) Composition

Relation R from A to B

Q: What happens if A = B?

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1. RevisionConcrete World

a. ___ same age as ___ {John, Mary, Peter} {(John,John), (Mary,Mary) (Peter,Peter),(Mary,Peter), (Peter,Mary)}

b. ___ same # of elements as ___

{ {}, {1}, {2}, {3.4} } { ({},{}), ({1},{1}), ({2},{2}) ({3,4},{3,4})({1},{2}), ({2},{1})

c. ___ ___ { {}, {1}, {2}, {1,2} } { ({},{}), ({},{1}), ({},{2}), ({},{1,2}),({1},{1}), ({1},{1,2}), ({2},{2}), ({2},{1,2})

({1,2},{1,2}) }

d. ___ ___ {1,2,3} {(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)}

___ R ___ A R A2 Relation R on A

“Everyone is related to himself” Reflexive

“If x is related to y and y is related to z, then x is related to z.” Transitive

“If x is related to y, then y is related to x” Symmetric

“If x is related to y and y is related to x, then x = y.” Anti-Symmetric

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1. Revision Given a relation R on a set A,

– R is reflexive iffxA, x R x– R is symmetric iffx,yA, x R y y R x– R is anti-symmetric iffx,yA, x R y y R x x=y– R is transitive iffx,yA, x R y y R z x R z

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2. Definition Given a relation R on a set A,

– R is an equivalence relation iff R is reflexive, symmetric and transitive. ( Last Lecture)

– R is a partial order iff R is reflexive, anti-symmetric and transitive.(This Lecture)

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2. Definition Given a relation R on a set A,

– R is an partial order (or partially-ordered set; or poset) iff R is reflexive, anti-symmetric and transitive.

Q: How do I check whether a relation is an partial order?

A: Just check whether it is reflexive, anti-symmetric and transitive. Always go back to the definition.

Q: How do I check whether a relation is reflexive, symmetric and transitive?

A: Go back to the definitions of reflexive, symmetric and transitive.

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3.1 Examples (Partial Orders in life) PERT - Program Evaluation and Review

Technique. CPM - Critical Path Method Used to deal with the complexities of

scheduling individual activities needed to complete very large projects.

Let T be the set of all tasks. We define a relation R on T such that x R y iff x = y or task x must be done before task y.

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3.1 Examples (Partial Orders in life)


Task 1 7 hrs

Task 2 6 hrs

Task 3 3 hrs

Task 7 1 hrs

Task 5 3 hrs

Task 8 2 hrs

Task 9 5 hrs

Task 6 1 hrs

Task 4 6 hrs

Q: How long does it take to complete the entire project?

(7)

(13)

(10) (14)

(16)

(19)

(21)

(20)

(26)

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3.1 Examples (Partial Orders in life)


Task 1 7 hrs

Task 2 6 hrs

Task 3 3 hrs

Task 7 1 hrs

Task 5 3 hrs

Task 8 2 hrs

Task 9 5 hrs

Task 6 1 hrs

Task 4 6 hrs

Q: Critical Path?

(7)

(13)

(10) (14)

(16)

(19)

(21)

(20)

(26)

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3.2 Examples (Finite Partial Orders) Let A = {0,1,2,3,4} Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4),

(3,3), (4,4)} Is R a partial order?

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Q1: Is R reflexive? Reflexive : xA, x R x

(Always go back to the definition) Yes, R is reflexive.

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Q2: Is R anti-symmetric? Anti-symmetric :

x,yA, x R y y R x x=y(Again, the definition!)

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True Always false

LHS: False

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LHS: False

Vacuously/blankly/stupidly True

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LHS: False

Vacuously True

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LHS: False

Vacuously True

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Carry on checking… Yes, it’s anti-symmetric

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Q3: Is R transitive? Transitive :

x,y,zA, x R y y R z x R z(DEFINITION!!!)

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True True True

True

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True True True

True

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Carry on checking… Yes, R is transitive.

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3.3 Examples (Common Infinite Posets) Let R be a relation on Z, such that

x R y iff x y R is a partial order

Reflexive: xZ, x x Anti-symmetric:x,yZ, xy yx x=y Transitive: x,y,zZ, xy yz xz

We will abbreviate the description of this relation to R = (Z, )

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3.4 Examples (Common Infinite Posets) Let R be a relation on Z+, such that

x R y iff x | y R is a partial order

Reflexive: xZ+, x | x Anti-symmetric:x,yZ+, x|y y|x x=y Transitive: x,y,zZ+, x|y y|z x|z

We will abbreviate the description of this relation to R = (Z+, |)

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3.5 Examples (Common Infinite Posets) Let R be a relation on P(A), such that

X R Y iff X Y R is a partial order

Reflexive: XP(A), XX Anti-symmetric:X,YP(A), XY YX X=Y Transitive: X,Y,ZP(A), XY YZ XZ

We will abbreviate the description of this relation to R = (P(A), )

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4. Notation In general, if we describe a partial order

relation as:

Let R be a relation on A, such thatx R y iff x op y

…we will shorten the description toR = (A, op)

Of course, this can be done only when the relation can be described in terms of a simple operator. We will not be able to this if the relation is described by a complicated logical expression

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4. Notation In general, if we describe a partial order

relation as:

Let R be a relation on A, such thatx R y iff x op y

…we will shorten the description toR = (A, op)

Hence we have: 1. R = (Z, )

2. R = (Z+, | )

3. R = (P(A), )

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4. Notation There are times when we discuss partial

orders in general. In such cases we may write:

R = (A, )as a general partial order.

We choose the ‘’ symbol to represent a general ordering operator because it looks like ‘’.

This is done due to the fact that the ordering of the elements in the set convey the idea of one below the other (something like on Z).

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5. Visualisation Tool: Hasse Diagram Let A = {0,1,2,3,4} Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4),

(3,3), (4,4), (1,4)}

0

3 1

4

2

Let’s simplify the diagram1. Eliminate all reflexive loops.

0

3 1

4

2

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(3,3), (4,4), (1,4)}

0

3 1

4

2

Let’s simplify the diagram2. Eliminate all transitive arrows.

0

3 1

4

2

0

3 1

4

2

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(3,3), (4,4), (1,4)}

0

3 1

4

2

Let’s simplify the diagram3. (a) Draw all arrow heads pointing upwards, and (b) eliminate arrow heads.

0

3 1

4

2

0

3 1

4

2

0

3

1

4

2

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(3,3), (4,4), (1,4)}

0

3 1

4

2

The result is a Hasse Diagram.

0

3 1

4

2

0

3 1

4

2

0

3

1

4

2

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5. Visualisation Tool: Hasse Diagram

Let A = {0,1,2,3,4}. Let R = (A, Draw the Hasse Diagram.

0

4 2

1

3

1. Eliminate all reflexive loops.

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0

4 2

1

3

1. Eliminate all reflexive loops.2. Eliminate all transitive arrows.

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0

4 2

1

3

1. Eliminate all reflexive loops.2. Eliminate all transitive arrows.

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0

4 2

1

3

1. Eliminate all reflexive loops.2. Eliminate all transitive arrows.3. (a) Draw all arrow heads pointing upwards, and (b) eliminate arrow heads.

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0

4

2

1

3

1. Eliminate all reflexive loops.2. Eliminate all transitive arrows.3. (a) Draw all arrow heads pointing upwards, and (b) eliminate arrow heads.

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Let A = {1,2,3,…,10}. Let R = (A, | Draw the Hasse Diagram.

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2 3 5 7

10

4

8 96

You may draw the Hasse Diagram immediately if you are able to.

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Let A = {1,2,3}. Let R = (P(A), Draw the Hasse Diagram.

R = ({ {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} } , )

{}

{1} {3}{2}

{1,2} {1,3}

{1,2,3}

{2,3}

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To be continued

1 lecture 2 (part 1) partial orders reading: epp chp 10.5

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