Partial Orders on the integers. In this case ( a , b ) R if a b . a a - - PowerPoint PPT Presentation

partial orders
SMART_READER_LITE
LIVE PREVIEW

Partial Orders on the integers. In this case ( a , b ) R if a b . a a - - PowerPoint PPT Presentation

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide01.html Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide02.html Partial Orders prev |


slide-1
SLIDE 1

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide01.html 1 of 1 09/11/2003 03:52 PM prev | slides | next

Partial Orders

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide02.html 1 of 1 09/11/2003 03:52 PM

Partial Orders

prev | slides | next

A relation R on a set S is a partial ordering or partial order if it is reflexive, antisymmetric and transitive. A set S together with a partial order R is called a partially ordered set or poset and is denoted (S,R). For example, the relation "less than or equal to" is a partial ordering

  • n the integers. In this case (a,b) R if a
  • b.

a

  • a so R is reflexive.

a

  • b implies that b
  • a only when a = b so R is

antisymmetric. if a

  • b and b
  • c then a
  • c so R is transitive.

So (Z,

  • ) is a poset.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide03.html 1 of 1 09/11/2003 03:52 PM

Partial Orders

prev | slides | next

Show that the divisibility relation is a partial ordering on Z+: (a, b) R if a|b Need to show that R is reflexive, antisymmetric, transitive.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide04.html 1 of 1 09/11/2003 03:52 PM

Partial Orders

prev | slides | next

It is important to note that while this partial ordering is on Z+, there are pairs of elements a and b from Z+ such that neither (a,b) nor (b,a) are in R, the divisibility relation. 3 does not divide 5, nor does 5 divide 3. The elements a and b of a poset (S,R) are called comparable if either (a,b) R or (b,a) R. When a and b are elements of S such that neither ordered pair of them is in R then a and b are called incomparable.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

slide-2
SLIDE 2

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide05.html 1 of 1 09/11/2003 03:52 PM

Partial Orders

prev | slides | next

If (S,R) is a poset and every two elements of S are comparable, S is called a totally ordered set and R is called a total order. The "less than or equal to" relation on Z is totally ordered. The "divides" relation on Z+ is not totally ordered. (S,R) is a well-ordered set if it is a poset such that R is a total order and such that every nonempty subset of S has a least element.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide06.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Lexicographic Order This is sometimes called the "dictionary ordering" because of it’s similarity to how words are arranged in a dictionary. Consider the following order: lash < lashing < lass < lasso < last < lasting The length of the word has little to do with the order. What system is actually used to determine the order? What system is used to determine the order of these ordered pairs? (2,3) < (3,2) < (3,3) < (3,4) < (4,1) < (4,9) < (5,2)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide07.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Lexicographic Order The lexicographic ordering on A× B for two posets (A,R1) and (B,R2) is defined (a1, b1) < (a2, b2) if a1 < a2 or (a1 = a2 and b1 < b2) where a1,a2 A and b1,b2 B.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide08.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Hasse Diagrams Consider the partial ordering on {1, 2, 3} R = {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3)} and it’s diagraph

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

slide-3
SLIDE 3

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide09.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Hasse Diagrams Since to be a partial ordering the relation R must be reflexive we know that there will be loops at each vertex. Thus we can omit them and understand them to be present.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide10.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Hasse Diagrams Since partial ordering relations are transitive, we can omit the "short cut" edges and, as before, understand them to be present.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide11.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Hasse Diagrams Finally, because partial order relations are antisymmetric, we can always draw the digraph so that all edges (except the "loop" edges which we’ve already omitted) point up. If we agree to always do this then we can change the arrows to lines; we are coding the direction in the orientation of the graph rather than in the arrows. Our graph now looks like This diagram is an example of a Hasse Diagram.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide12.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Hasse Diagrams Exercise: Construct the Hasse Diagram for ({2, 5, 8, 10, 20}, |). The relation here is the "divides" relation. (Solution is on the next slide...)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

slide-4
SLIDE 4

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide13.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Hasse Diagrams The Hasse diagram for ({2, 5, 8, 10, 20}, |): The maximal elements of this poset are the "tops"; in this case {8, 20}. The minimal elements of this poset are the "bottoms", in this case {2, 5}.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide14.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Hasse Diagrams If there is only one maximal element then it is the greatest

  • element. If there is more than one maximal element then there is no

greatest element. Similarily, if there is a unique minimal element then it is the least

  • element. If there is more than one minimal element then there is no

least element. Consider a subset A of a poset (S, R). The upper bounds of A are the elements that are "above" every element in A. This means that that every upper bound can be obtained by tracing up from each element in A. The lower bounds of A are defined in a similar way; they are "below" all elements in A.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Partial Orders http://localhost/~senning/courses/ma229/slides/partial-orders/slide15.html 1 of 1 09/11/2003 03:53 PM

Partial Orders

prev | slides | next

Hasse Diagrams

If A={a, b, c} then the upper bounds are {e, f, h, j} and the least upper bound is e. If A={d, e, f} the upper bounds are {f, h, j}, the least upper bound is f. The lower bounds are {a, b} and the greatest lower bound is b.

The least element is a and there is no greatest element.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15