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## PowerPoint Slideshow about 'Partial Orders Definitions A relation R on set A is a partial order if ...' - bernad

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Definitions

- A relation R on set A is a partial order if it is:
- reflexive
- antisymmetric
- transitive.
- A is called a partially ordered set or poset.
- [A;R] means A is partially ordered by R.
- Example: [ P({1,2}); ] (draw its digraph).

Definitions ...

- is the prototype of a partial order
- a, bA are comparable under if either ab or ba.
- Otherwise, they are said to be incomparable
- If a, bA are comparable, then [A; ] is totally ordered and A is a chain.

Examples

- [ Z; ],
- where Z is the set of integers
- has the usual meaning.
- Z is totally ordered.
- [ Z+; | ],
- where Z+ is a set of positive integers
- | is the (evenly) divides operator (e.g., 2 | 10 )

Hasse Diagrams

- A Hasse diagram is a directed graph (digraph), where
- self-loops are omitted
- arcs implied by transitivity are omitted.
- Let Dn denote the set of positive divisors of n.
- Draw Hasse diagram for [ D8; | ]
- Draw Hasse diagram for [ D6; | ]

Hasse Diagrams

- What n yields a poset [Dn; | ] whose Hasse diagram is a cube of dimension:
- 0
- 1
- 2
- 3

Examples

- Let S be a set. [ P(S); ] is a poset.
- Draw the Hasse diagram for:
- S =
- S = {1}
- S = {1, 2}
- S = {1, 2, 3}
- What do you think the Hasse diagram looks like for S = {1, 2, 3, 4}? For S = {1, . . ., n}?

Computer Science Example

- Consider this sequence of Java assignments:

a = b + d + c;

d = a*(b + d) + c;

e = (b + d)*c;

(Draw an operator graph for these statements.)

- Sequence these operations in a way that is compatible with their partial order.
- Where are the longest paths?

Topological Sorting

- Let G = (V,E) be a directed graph where
- v V represents a task;
- (u, v) E means that task u must be completed before v can be started.
- G cannot have cycles.
- Problem: Find a schedule for G that respects the partial order.

Composing Relations

- Let R be a relation from A to B.
- Let S be a relation from B to C.
- The composition of R & S, denoted RS, is

RS = { xRSy | zB, xRz zSy }.

- Example:
- R = { (1,2), (3,4), (2,4) }
- S = { (2,4), (2,3), (4,1) }
- RS = { (1,4), (1,3), (3,1), (2,1) }

The Transitive Closure

- Let R be a relation from S to S.
- RR is usually denoted R2.
- Ri+1 = RiR.
- The transitive closure, R+, of a binary relation R is:

Reflexive-Transitive Closure

- R0 = { (x, x) | x S }
- The reflexive-transitive closure, denoted R*, is: R* = R0 R+. That is,

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