Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Lecture 2 (part 1) Partial Orders Reading: Epp Chp 10.5.

Similar presentations


Presentation on theme: "1 Lecture 2 (part 1) Partial Orders Reading: Epp Chp 10.5."— Presentation transcript:

1 1 Lecture 2 (part 1) Partial Orders Reading: Epp Chp 10.5

2 2 Outline 1.Revision 2.Definition of poset 3.Examples of posets a.In life b.Finite posets c.Infinite posets 4.Notation 5.Visualization Tool: Hasse Diagram 6.Definitions –maximal, greatest, minimal, least. –2 Theorems 7.More Definitions –Comparable, chain, total-order, well-order

3 3 1. Revision Concrete 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),(1,3),(2,3)} ___ R ___ AB R  A  BR  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 ?

4 4 1. Revision Concrete 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  A2R  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

5 5 1. Revision n Given a relation R on a set A, –R is reflexive iff  x  A, x R x –R is symmetric iff  x,y  A, x R y  y R x –R is anti-symmetric iff  x,y  A, x R y  y R x  x=y –R is transitive iff  x,y  A, x R y  y R z  x R z

6 6 2. Definition n 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)

7 7 2. Definition n 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. n Q: How do I check whether a relation is an partial order? n A: Just check whether it is reflexive, anti-symmetric and transitive. Always go back to the definition. n Q: How do I check whether a relation is reflexive, symmetric and transitive? n A: Go back to the definitions of reflexive, symmetric and transitive.

8 8 3.1 Examples (Partial Orders in life) n PERT - Program Evaluation and Review Technique. n CPM - Critical Path Method n 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.

9 9 3.1 Examples (Partial Orders in life) 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. 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)

10 10 3.1 Examples (Partial Orders in life) 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. 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)

11 11 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order?

12 12 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q1: Is R reflexive? Reflexive :  x  A, x R x (Always go back to the definition) n Yes, R is reflexive.

13 13 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q2: Is R anti-symmetric? n Anti-symmetric :  x,y  A, x R y  y R x  x=y (Again, the definition!)

14 14 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q2: Is R anti-symmetric? n Anti-symmetric :  x,y  A, x R y  y R x  x=y (Again, the definition!) TrueAlways false LHS: False

15 15 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q2: Is R anti-symmetric? n Anti-symmetric :  x,y  A, x R y  y R x  x=y (Again, the definition!) LHS: False Vacuously/blankly/ stupidly True

16 16 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q2: Is R anti-symmetric? n Anti-symmetric :  x,y  A, x R y  y R x  x=y (Again, the definition!) LHS: False Vacuously True

17 17 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q2: Is R anti-symmetric? n Anti-symmetric :  x,y  A, x R y  y R x  x=y (Again, the definition!) LHS: False Vacuously True

18 18 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q2: Is R anti-symmetric? n Anti-symmetric :  x,y  A, x R y  y R x  x=y (Again, the definition!) n Carry on checking… n Yes, it’s anti-symmetric

19 19 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q3: Is R transitive? n Transitive :  x,y,z  A, x R y  y R z  x R z (DEFINITION!!!)

20 20 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q3: Is R transitive? n Transitive :  x,y,z  A, x R y  y R z  x R z (DEFINITION!!!) True

21 21 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q3: Is R transitive? n Transitive :  x,y,z  A, x R y  y R z  x R z (DEFINITION!!!) True

22 22 3.2 Examples (Finite Partial Orders) n Let A = {0,1,2,3,4} n Let R = {(0,0), (3,1), (1,1), (0,4), (2,2), (3,4), (3,3), (4,4)} n Is R a partial order? Q3: Is R transitive? n Transitive :  x,y,z  A, x R y  y R z  x R z (DEFINITION!!!) n Carry on checking… n Yes, R is transitive.

23 23 3.3 Examples (Common Infinite Posets) n Let R be a relation on Z, such that x R y iff x  y n R is a partial order Reflexive:  x  Z, x  x Anti-symmetric:  x,y  Z, x  y  y  x  x=y Transitive:  x,y,z  Z, x  y  y  z  x  z We will abbreviate the description of this relation to R = (Z,  )

24 24 3.4 Examples (Common Infinite Posets) n Let R be a relation on Z +, such that x R y iff x | y n R is a partial order Reflexive:  x  Z +, x | x Anti-symmetric:  x,y  Z +, x|y  y|x  x=y Transitive:  x,y,z  Z +, x|y  y|z  x|z We will abbreviate the description of this relation to R = (Z +, |  )

25 25 3.5 Examples (Common Infinite Posets) n Let R be a relation on P(A), such that X R Y iff X  Y n R is a partial order Reflexive :  X  P(A), X  X Anti-symmetric :  X,Y  P(A), X  Y  Y  X  X=Y Transitive :  X,Y,Z  P(A), X  Y  Y  Z  X  Z We will abbreviate the description of this relation to R = (P(A),  )

26 26 4. Notation n In general, if we describe a partial order relation as: Let R be a relation on A, such that x R y iff x op y …we will shorten the description to R = (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

27 27 4. Notation n In general, if we describe a partial order relation as: Let R be a relation on A, such that x R y iff x op y …we will shorten the description to R = (A, op) Hence we have: 1. R = (Z,  ) 2. R = (Z +, | ) 3. R = (P(A),  )

28 28 4. Notation n There are times when we discuss partial orders in general. In such cases we may write: R = (A, 9) as a general partial order. We choose the ‘9’ 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).

29 29 5. Visualisation Tool: Hasse Diagram n Let A = {0,1,2,3,4} n 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 diagram 1. Eliminate all reflexive loops. 0 3 1 4 2

30 30 5. Visualisation Tool: Hasse Diagram n Let A = {0,1,2,3,4} n 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 diagram 2. Eliminate all transitive arrows. 0 3 1 4 2 0 3 1 4 2

31 31 5. Visualisation Tool: Hasse Diagram n Let A = {0,1,2,3,4} n 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 diagram 3. (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

32 32 5. Visualisation Tool: Hasse Diagram n Let A = {0,1,2,3,4} n 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 The result is a Hasse Diagram. 0 3 1 4 2 0 3 1 4 2 0 3 1 4 2

33 33 5. Visualisation Tool: Hasse Diagram Let A = {0,1,2,3,4}. Let R = (A,   n Draw the Hasse Diagram. 0 4 2 1 3 1. Eliminate all reflexive loops.

34 34 5. Visualisation Tool: Hasse Diagram Let A = {0,1,2,3,4}. Let R = (A,   n Draw the Hasse Diagram. 0 4 2 1 3 1. Eliminate all reflexive loops. 2. Eliminate all transitive arrows.

35 35 5. Visualisation Tool: Hasse Diagram Let A = {0,1,2,3,4}. Let R = (A,   n Draw the Hasse Diagram. 0 4 2 1 3 1. Eliminate all reflexive loops. 2. Eliminate all transitive arrows.

36 36 5. Visualisation Tool: Hasse Diagram Let A = {0,1,2,3,4}. Let R = (A,   n Draw the Hasse Diagram. 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.

37 37 5. Visualisation Tool: Hasse Diagram Let A = {0,1,2,3,4}. Let R = (A,   n Draw the Hasse Diagram. 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.

38 38 5. Visualisation Tool: Hasse Diagram Let A = {1,2,3,…,10}. Let R = (A, |  n Draw the Hasse Diagram. 1 2357 10 4 8 9 6 You may draw the Hasse Diagram immediately if you are able to.

39 39 5. Visualisation Tool: Hasse Diagram Let A = {1,2,3}. Let R = (P(A),   n 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}

40 40 n To be continued


Download ppt "1 Lecture 2 (part 1) Partial Orders Reading: Epp Chp 10.5."

Similar presentations


Ads by Google