Relations and Partial Orders, Slides of Discrete Mathematics

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Relations

Outline

• (^) Relation:
• (^) Properties
• (^) Combining relations
• (^) Representing relations
• (^) Closure of relations
  • (^) Reflexive closure, diagonal relation, Warshall’s Algorithm,
• (^) Equivalence relations:
  • (^) Equivalence class, partitions,
  • (^) Partial order (POSET)
  • (^) Hasse Diagram, POSET, Minimal, Maximal, Minimum, Maximum, Upper bound, lower bond, Least Upper bound, greatest lower bound, Lattices

Reflexive Closure

• (^) In general, the reflexive closure of a relation R

on A is R where ={ (a,a) | aA}

• (^)  is the diagonal relation on A
• (^) Question : How can we compute the diagonal

relation using

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Reflexive Closure

• (^) Example: Consider the relation R = {(1,1), (1,2), (2,1), (3,2)} on set {1,2,3}
  • (^) Is it reflexive?
  • (^) How can we produce a reflective relation containing R that is as small as possible?

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Reflexive Closure – Cont.

• (^) The diagonal relation on A is:  = {( a , a ) | a  A }.
• (^) The reflexive closure of R is then: R  .
• (^) Properties:
  • (^) R  ( R  );
  • (^) R   is reflexive;
  • (^)  S ( R  S  S is reflexive)  ( R  )  S.
• (^) In zero-one matrix notation: M R  M
• Turn on all the diagonal bits!

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Example

• (^) Consider R = {( a , b ) Z  Z | a < b }
• (^) The reflexive closure of relation R is: R   = {( a , b ) Z  Z | a < b }  {( a , a ) | a  Z } = {( a , b ) Z  Z | a  b }

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Symmetric Closure (optional)

• (^) Example: Consider

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

• (^) R is not symmetric; the pairs missing are: (2,1), (1,3).
• (^) If we add those, we obtain the new relation: {(1,1), (1,2), ( 1 , 3 ), ( 2 , 1 ), (2,2), (2,3), (3,1), (3,2)}. The new relation is symmetric.

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Symmetric Closure (optional)

• (^) When a relation R on a set A is not symmetric:
  • (^) How to minimally augment R (adding the minimum number of ordered pairs) to have a symmetric relation?
  • (^) The symmetric closure of R.

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Transitive Closures

• (^) Consider R = {(1,3), (1,4), (2,1), (3,2)}.
  • (^) R is not transitive;
  • (^) What are the missing terms?
    • (^) Few are: (1,2), (2,3), (2,4), (3,1).
  • (^) If we add those, we obtain the new relation: {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2)}.
  • (^) Is the above relation transitive?
  • (^) No, it is not. Missing terms such as (1,1), (2,2)… Transitive closure is more complicated to build.

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Transitive Closure

• (^) When a relation R on a set A is not transitive:
  • (^) How to minimally augment R (adding the minimum number of ordered pairs) to have a transitive relation?
  • (^) The transitive closure of R.

Warshall’s Algorithm: Example

• (^) Compute the transitive closure of
  • (^) The relation R={(1,1),(1,2),(1,4),(2,2),(2,3),(3,1), (3,4),(4,1),(4,4)}
  • (^) On the set A={1,2,3,4}

Warshall’s Algorithm

• (^) Let A={a,b,c,d}
• (^) R={(a,b),(b,a),(b,c),(c,d)}

Equivalence Relation

• (^) Consider the set of every person in the world
• (^) Now consider a R relation such that (a,b) R if a and b are siblings.
• (^) Clearly this relation is
  • (^) Reflexive
  • (^) Symmetric, and
  • (^) Transitive
• (^) Such as relation is called an equivalence relation
• (^) Definition : A relation on a set A is an equivalence relation if it is reflexive, symmetric, and transitive

Equivalence Relation

• (^) A={1,2,3}
• (^) R= {(1,1),(2,2),(3,3)}