Transcript Binary Relation: - PLU | Pacific Lutheran University

Binary Relation:
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A binary relation between sets A and B is a subset of the
Cartesian Product A x B. If A = B we say that the relation is a
relation on A.
Suppose A = {1, 3, 5, 7} and B = {2, 4, 6, 8}.
Further suppose R = { (1,2), (3,4), (5,6), (7,8)}. This is a subset
of A x B so is a binary relation between A and B
B
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Binary Relation Exercise:
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If R = { (a,b): a < b}. Show R on the diagram below.
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A
Binary Relations as Sets of Ordered Pairs:
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Because we mention one set before another in a Cartesian
Product, A x B, the element, (a,b), in any relation, R, over A and
B must have its first element from A and its second element
from B.
So we say that the elements of R form ordered pairs.
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Exercise:
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If X = {1, 2, 3, 4, 5, 6}, find R = { (x,y): x is a divisor of y}
R = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6) }
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Graph Representation of a Binary Relation:
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If A and B are two finite sets and R is a binary relation between
A and B we can represent this relation as a graph (set of
vertices and edges).
A = {1, 3, 5 ,7}, B = {2, 4, 6, 8}
R = {(a,b): a < b}
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We call this graph a directed graph.
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Example:
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A = {1, 2, 3, 4, 5, 6}. R is a relation on A defined by the following
directed graph.
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Exercise: Give a rule-based
definition of R
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R = {(a,b): a divides b}
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Matrix Representation of a Binary Relation:
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If A and B are finite sets and R is a binary relation between A
and B then create a matrix, M, with the following properties:
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the rows of the matrix are indexed by the elements of A
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the columns of the matrix are indexed by elements of B
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M(ai,bj) = 1 if (ai,bj) belongs to R; 0 otherwise
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Exercise: Give a rule-based definition of R
R = {(a,b): a + 1 < b}
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Notation:
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If R is a binary relation on a set, X, we write x R y whenever
(x,y) ε R.
Example: sister_of relationship on girls in a school can be expressed
as x is_sister_of y.
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RelationProperties:
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Suppose R is a relation on a set A.
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We say
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R is reflexive if a R a for all a ε A.
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R is symmetric if x R y ==> y R x for all x, y ε A.
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R is antisymmetric when (x R y and y R x ==> x == y) for all
x, y ε A.
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R is transitive when (x R y and y R z ==> x R z) for all
x, y, z ε A.
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Understanding Relations as Ordered Pairs:
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R is reflexive if (x,x) ε R for all x ε A.
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R is symmetric if when (x,y) ε R then (y,x) ε R for all x,y ε A.
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R is antisymmetric if when (x,y) ε R and x != y then (y,x) ε R.
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R is transitive if when (x,y) ε R and (y,z) ε R then (x,z) ε R
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Understanding Relations as digraphs.
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If R is a relation represented as a di-graph then
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R is reflexive if every node has a loop to itself attached.
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R is symmetric if every directed edge is directed in both
directions.
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R is antisymmetric if there is no bi-directional edge.
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If there is a directed edge from x to y and another from y to z
then there is a directed edge from x to z
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Understanding Relations as Matrices:
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Given a binary relation R on a finite set X.
Let M be the matrix whose rows and columns are indexed by
the elements of X.
R is reflexive if the elements on the leading diagonal are all
1(T).
R is symmetric if the matrix is symmetric about the main
diagonal.
R is antisymmetric if there are no symmetrical elements. Hence
if mij == 1 the mji != 1.
The text says transitivity is not readily apparent. We'll see!
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Exercise:
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Evaluate the following relations for the previously mentioned
properties:
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x divides y on the natural numbers
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x != y on the integers.
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x is the same age as y in the set of all people.
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a/b has the same value as c/d in the set of all rationals, Q.
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Equivalence Relation:
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A relation R on a set A that is reflexive, symmetric and transitive
is said to be an equivalence relation.
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Equivalence relations generalize the concept of equality.
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Exercise: Show = is an equivalence relation.
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Exercise: Show a/b has the same value as c/d in the set of all
rationals, Q is an equivalence relation.
Exercise: Show “has the same angles” is an equivalence
relation on the set of all triangles.
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Equivalence Relations and Partitions:
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The examples on the previous slide show there is a strong
relationship between equivalence relations and partitions.
A partition is a set of non-empty subsets A1, A2, ... of a set A
such that
– A1 U A2 U ... = A
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U
– Ai
Aj = O for every pair i, j where i != j.
We call each Ai a block of the partition.
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Example:
A
A2
A1
A4
A5
A3
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Equivalence Class:
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If R is an equivalence relation on a set E, then
Ex = { y ε E: y R x }
is called the equivalence class of x.
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An easy way to represent an equivalence class of an element x
is [x].
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Here it comes
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!
Theorem:
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Let R be an equivalence relation on a set A, then the
equivalence classes of R form a partition of A.
Proof: All equivalence classes are non-empty subsets of A.
This is because for every x ε A, xRx and so x ε Ex.
Second, if xRy, then Ex = Ey. Suppose xRz, then z ε Ex.
However zRx since R is symmetric and so zRy by transitivity
and so z ε Ey. Hence Ex Ey. Similarly Ey Ex. Hence Ex = Ey.
Now we must show that equivalence classes satisfy the first
property of a partition. Namely, that the union of all the
equivalence classes is the entire set. First Ex A since the
relation is defined over A.
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Proof (cont):
So
U
x Ex
A
Since every x in A belongs to an Ex (remember xRx) we can say
A
Ux Ex
Hence
A = xU Ex
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Proof (cont):
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We are left having to prove the second part of the definition of a
partition. Namely that any two equivalence classes that are not
equal are disjoint.
P: Two equivalence classes are not equal
Q: Two equivalence classes are disjoint
To prove: P --> Q
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The contrapositive is actually easier to prove. Namely that if two
equivalence classes are not disjoint, they must be equal.
~P: Two equivalence classes are equal
~Q: Two equivalence classes are not disjoint
To prove: ~Q --> ~P
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Proof (cont):
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Suppose
Ex Ey = O
So there is a z in both Ex and Ey. This means xRz and yRz.
However if yRz then zRy because R is symmetric.
Finally, since xRz and zRy we can use transitivity to say xRy.
From this we can conclude that Ex and Ey are equal.
Thus the equivalence classes of an equivalence relation satisfy
both properties of being blocks of a partition and so form a
partition. QED.
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Example:
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Let R be a relation on R satisfying xRy iff x – y is an integer.
Show R is an equivalence relation and find the equivalence
classes – [0], [½] and [sqrt(2)].
Reflexive: xRx iff x-x is an integer. Since x-x = 0 and this is an integer, R is reflexive
Symmetric: if xRy then x-y is an integer. Hence y-x is also an integer so yRx.
Hence the relation is symmetric.
Transitive: if xRy and yRz then x-y is an integer and y-z is an integer. Therefore
(x-y)+(y-z) is also an integer. But this is x-z and so xRz. Hence the relation is
transitive.
[0] = { x ε R: x-0 is an integer} = Z.
[½] = { ±0.5, ±1.5, ±2.5, ...}
[sqrt(2)] = {sqrt(2), sqrt(2)±1, sqrt(2)±2, ...}
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Partial Orders:
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A binary relation on a set A that is reflexive, anti-symmetric and
transitive is called a partial order and the set is called a partially
ordered set with respect to the relation or a poset for short.
Partial orders allow elements to “precede” one another but not
necessarily.
Examples:
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subsets of a set are ordered partially
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Partial Orders:
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Example:
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subsets of a set are ordered partially
{1,2,3}
{1,2}
{2,3}
{1,3}
Reflexive:
{1}
{2}
v
{3}
Anti-symmetric:
Transitive:
{}
Hasse Diagram
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Partial Orders:
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Example:
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is a divisor of in the set of natural numbers.
Reflexive:
n is a divisor of n?
Anti-symmetric:
n is a divisor of m and m a divisor of n implies n == m?
Transitive:
n is a divisor of m and m a divisor of p
implies n is a divisor of p?
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Posets:
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A set on which a partial order is defined is called a poset.
If R is a partial order on a set A and xRy we say that x is a
predecessor of y and y is a successor of x.
An element of A can have many predecessors but if xRy and
there is no z such that xRz and zRy we say x is an immediate
predecessor of y (written x ≺ y).
We use a Hasse Diagram to represent immediate
predecessors.
The vertices of a Hasse Diagram represent the elements of A
and if they are directly connected then the lower vertex is the
immediate predecessor of the upper vertex.
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Example:
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Let A = {1,2,3,6,12,18}. Let R be the relation “is a divisor of”.
R is reflexive (a is a divisor of a), anti-symmetric (a is a divisor
of b and b is a divisor of a implies a == b) and transitive (a is a
divisor of b and b is a divisor of c implies a is a divisor of c).
R is a partial order.
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aRb as long as there
is a path up the diagram
from a to b.
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Total Order:
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A total order is a partial order in which every two elements of
the set are related. Hence if a, b  A, either aRb or bRa.
Examples:
- <= on the real numbers.
- lexicographical ordering of a dictionary.
- {1,2,6,12} is a totally ordered subset of A using “is a divisor of” relation.
NOTE: In Computer Science we often need to sort elements. In order to
do this the comparison operator needs to be a total order.
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