Survey
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* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
A relation R on a set A is called a partial order if R is
Reflexive, Antisymmetric and Transitive.
Reflexive – has a cycle of 1 at every vertex.
Example: (1,1), (2,2)
Antisymmetric – contains no back arrows – all one way
paths and can contain cycles of 1.
Example: (1,1), (1,2), (2,3)
Transitive – if you have (a,b) and (b,c), you must have (a,c).
Example:
a,b b,c a,c
(1,2), (2,3), (1,3)
The set A together with the partial order R is called a
partially ordered set or poset. This is denoted (A,R) or
(A, ≤).
The relation of divisibility (aRb if and only if a divides
b) is a partial order on Z+ .
A = {1,2,3,4} x ≤ y iff x|y
B = {1,7,11,14} x ≤ y in the regular way
x
y
x
y
a,b
a,b
a,b
a,b
(1,7) ≤ (3,14)
(2,7) ≤ (3,14)
Because 1|3 and 7 ≤ 14.
Because 2 does not divide 3.
a and b of set A are said to be comparable if a ≤ b or b ≤
a. Every pair of elements does not need to be
comparable aRb iff a|b is a partial order on Z+ .
2 and 7 are not comparable since 2|7 and 7|2.
Partial means some elements may not be comparable.
Lexicographic (lex sick o graphic)
Means alphabetic order.
park < part
help < hind
Jump < mump
Hasse Diagrams
Hasse diagrams, for simplicity, remove one cycles,
remove transitive property aRc, remove arrows, and all
edges point upwards.
Jt does not mean one cycles, transitive property and
arrows don’t exist….they still exist.
A = {1,2,3,4,12}
a and b are ∈ of A
a ≤ b if and only if a|b
Draw the Hasse diagram of the poset(A, ≤).
(1,1),(1,2),(1,3),(1,4),(1,12),
(2,2),(2,4),(2,12),
(3,3),(3,12),
(4,4),(4,12),
(12,12)
Cross off one cycles and (a,c) of transitive property.
(1,1),(1,2),(1,3),(1,4),(1,12),
(2,2),(2,4),(2,12),
(3,3),(3,12),
(4,4),(4,12),
(12,12)
12
12
4
4
3
2
3
1
2
1
Finite linear ordered set
1 2 3
1
2
3
(1,1), (1,2), (1,3),
(2,2), (2,3)
(3,3)
3
3
2
2
1
1
The process of constructing a linear order is called
topological sorting. With topological sorting, you
need to be able to insert a relation in the linear order
without breaking an existing relation.
3
1 is related to 2
2
1 is also related to 3
1
3
4 could be placed after 1 if 4 is related to 1 and this
would not break the relations of 1 to 2 and 1 to 3,
meaning 4 is related to 2 and 3
2
4
1
Isomorphism
A = {1,2,3,6}
≅
f
Under divisibility, x ≤ y iff x|y
B = {{}, {x},{y}, {x,y}}
Under subset, (a ≤ b) iff a ≤ b
Hasse diagrams of A & B
1
2
3
6
≅
{}
{x}
{y}
{x,y}
{x,y}
6
3
2
1
{y}
{x}
{}
The function A B is one to one. The function f is an isomorphism.
Incomparable
a ≤ b and b ≤ a. There is no relation between a and b so
we cannot compare them.
b
a
c
d
A poset (A, R -1) is called a dual poset…..it is the
inverse.
f
b
a
e
d
e
d
f
a
b
c
c