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## Partial Orders

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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,**Example**R a b c d e R0 Self-loops not shown a b c d e R+ a b c d e R* Self-loops not shown a b c d e