Representing B has everything PartialOrders that A has and more: - - PowerPoint PPT Presentation

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rep-po.1 Albert R Meyer March 22, 2013

Mathematics for Computer Science

MIT 6.042J/18.062J

Representing Partial Orders

rep-po.2 Albert R Meyer March 22, 2013

means B has everything that A has and more:

proper subset relation

A ⊂ B B⊄A

proper subset relation

{1,2,3,5,10,15,30} {1,3,5,15} {1,2,5,10} {1,3} {1,2} {1} {1,5} partial order: properly divides

1 1

3

3 1 on {1,2,3,5,10,15,30}

Albert R Meyer March 22, 2013 rep-po.4 Albert R Meyer March 22, 2013 rep-po.3

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rep-po.5 Albert R Meyer March 22, 2013

same shape as ⊂ example

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proper subset

{1} {1,3,5,15} {1,2} {1,3} {1,5} {1,2,5,10} {1,2,3,5,10,15,30} partial order: properly divides

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1

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• n {1,2,3,5,10,15,30}

1

rep-po.8 Albert R Meyer March 22, 2013

same shape as ⊂ example

isomorphic

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¡5 ¡

313

Albert R Meyer March 19, 2012

All that matters are the connections: graphs with the same connections are isomorphic

Isomorphism

rep-po.9 rep-po.10 Albert R Meyer March 22, 2013

Isomorphism

two graphs are isomorphic when there is an edge-preserving matching

• f their vertices.

matching

bijection

Albert R Meyer March 19, 2012

Formal Def of Graph Isomorphism

G1 isomorphic to G2 iff ∃ bijection f:V1 V2 with

uv in E1 IFF f(u)f(v) in E2

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p.o. represented by ⊂ Theorem: Every strict

partial order is isomorphic to a collection of subsets partially ordered by ⊂.

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rep-po.13 Albert R Meyer March 22, 2013

proof: map element, a, to

the set of elements below it. a maps to

p.o. isomorphic to ⊂ b ∈ A | b Ra OR b = a

{ }

rep-po.14 Albert R Meyer March 22, 2013

proof: map element, a, to

the set of elements below it. a maps to

p.o. isomorphic to ⊂ b ∈ A | b Ra OR b = a

{ }

f(a) ::= R−1(a) ∪ a

{ }

subsets from divides

2 {1,2} 5 {1,5} 3 {1,3} 15 {1,3,5,15} 10 {1,2,5,10} 30 {1,2,3,5,10,15,30} 1 {1}

Albert R Meyer March 22, 2013 rep-po.15

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