let a and b be any sets a binary relation r from a to b is a subset of axb given an ordered pair (x,...

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Relations Among Sets Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This is denoted If x is not related to y by R, we say The term “binary” means that the relation is defined for two elements. Other types of relations, called n-ary relations, also exist. If you want, look them up. They’re very useful in CS

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Page 1: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Relations Among Sets Let A and B be any sets

• A binary relation R from A to B is a subset of AxB• Given an ordered pair (x, y), x is related to y by R

iff (x, y) is in R.• This is denoted • If x is not related to y by R, we say

The term “binary” means that the relation is defined for two elements. Other types of relations, called n-ary relations, also exist. If you want, look them up. They’re very useful in CS

Page 2: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

A Simple Relation Consider sets A={0,1,2} and

B={1,2,3}. AxB is the set of all possible ordered pairs from A and B• If element x from A is related to element y

from B, we say xRy, “x is related to y” If x<y, which elements in AxB are in set R? Answer: {(0,1), (0,2), (0,3), (1,2), (1,3), (2,3)} Which elements are not in R? Answer: {(1,1), (2,1), (2,2)}

Page 3: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

The Circle Relation For any (x, y) in the real number plane

(RxR), (x, y) are in relation C iff “x2 + y2 = 1”• This is the equation for a circle• This relation has infinitely many solutions

because it is a subset of the cross product of two infinitely large sets (two sets of all real numbers)

• To check the relation, test some x and y values (0,0) => 02 + 02 = 0 ≠ 1 (1,0) => 12 + 02 = 1 (2, 0) => 22 + 02 = 4 ≠ 1

Page 4: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Properties of Relations Given set A and relation S:

• S is reflexive iff for all x in A, (x, x) is in S Informal: Each element is related to itself

• S is symmetric iff for all x and y in A, if (x, y) is in S, then (y, x) is in S Informal: If any one element is related to any other element, then

the second element is related to the first• S is transitive iff for all x, y, and z in A, if (x, y) and (y, z) are

in S, then (x, z) is in S Informal: If any one element is related to a second, and that second

is related to a third, then the first is related to the third These definitions are each universal statements

• These properties can be proven true by proving that every element x, y, and z must satisfy the property

• Prove them false by finding a counter example

Page 5: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Examples S = {(0,0), (0,1), (0,3), (1,0), (1,1), (2,2), (3,0), (3,3)} T ={(0,1), (2,3)} W = {(0,0), (0,2), (0,3), (2,3)}

• Which properties do S, T, and W satisfy? S is reflexive: Each element is related to itself – (0,0)…

S is symmetric: For every ordered pair in S, there is also a reverse pair – The reverse of (x, x) is (x, x)!!

S is not transitive: There is point (1,0) and point (0,3) but there is no point (1,3) – (x, y), (y, z), but not (x, z)

T is not reflexive: Not every point (x, x) is in T T is not symmetric: Not every point (x, y) has a corresponding (y, x) point T is transitive: It is not obvious why. Notice that there are no points (x ,y) and (y, z) in

T, so there is never an opportunity for transitivity to be false. We say that T is vacuously transitive, meaning the property is true by default

W is not reflexive: 1 is between 0 and 2, so is in set A on which W is a relation. The point (1,1), among other points like (2,2), is not in W W is not symmetric: There is a (0,2) but not a (2,0) W is transitive: There is a (0,2), (2,3) and a (0,3) – these are the only points of the form

(x, y), (y, z), (x, z). There are no other (x, y), (y, z) points. It must be transitive These properties extend readily to infinite relations like y =

x

Page 6: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

The Transitive Closure of a Relation

We have a relation H that is not transitive, but we want it to be transitive• The transitive closure of H, denoted Ht, is the

set that contains H but also contains all the points necessary to make Ht a transitive relation

• Ht satisfies the following properties Ht is transitive H is a subset of or is equal to Ht

If S is any other transitive relation that contains H, then it also contains Ht

Page 7: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Examples A = {0,1,2,3} and R = {(0,1), (1,2),

(2,3)} What is the transitive closure of R?

• Every ordered pair in R is in Rt

• Every grouping of points (x, y) and (y, z) must yield an (x, z) point (0,1) + (1,2) → (0,2) (1,2) + (2,3) → (1,3) Since (0,2) is now in Rt, (0,2) + (2,3) → (0,3)

• Rt = {(0,1), (0,2), (0,3), (1,2), (1,3), (2,3)}

Page 8: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

To Define Functions… Consider the function F from a set A

to a set B that satisfies the following two properties• For every element x in A there is an element

y in B such that (x, y) is in F. • For all elements x in A and y, z in B

If (x, y) is in F and (x, z) is in F, then y = z• If F is a function from A to B, we say

y = F(x) iff (x, y) is in F

Page 9: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Functions and Relations on Finite Sets

Sets A = {2,4,6} and B = {1,3,5} are connected by two relations R = {(2,5), (4,1), (4,3), (6,5)} and S = “for all (x, y) in AxB, (x, y) is in S iff y = x + 1”• Relation R is not a function because the x value 4

yields to separate y values 1 and 3 In a function, according to property 2, an x value (input)

can yield only one unique y value (output)• Relation S is not a function because there is no

element y in B such that y = 6 + 1 = 7 Property 1 states that every element in A must yield an

element in B

Page 10: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Domain and Range If a function is defined from set X to set Y, we

say that the first set, X, is the domain of the function and the second set, Y, is the range• When you think of a function in terms of a machine,

you can visualize the domain as the set of possible inputs and the range as the set of possible outputs

• If X is the set of all real numbers, there is no guarantee that Y will also contain all real numbers, as is demonstrated with the function y = x2

• The process of using a function to transform its domain into range values is known as mapping and consists of feeding in the inputs and recording the outputs You will do a lot of relational mapping in computer science

Page 11: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Injective Vs Surjective Functions For a function to be injective, or one-to-one, every

element in its domain must yield a unique value• If F(x) = F(x[2]), then x = x[2]

F(2) can’t equal F(3). An example is the function F(x)=x, because F(x[2]) can only equal x[2], so if x[2] ≠ x, F(x[2]) won’t equal F(x)

For a function to be surjective, or onto, every element in the function’s domain must be matched with an element from its range• A = {1,2,3}, B = {a,b,c,d}

F(A) = {a,b,c} This function is not surjective because 1 yields a, 2 yields b, 3 yields c,

but nothing yields d, which is in the range of F(x) because it is defined over the sets A and B

• Formally, this is described as: Given sets X and Y, F(X) is onto Y iff for every element y in Y, there exists an element x in X such that F(x) = y

Page 12: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Examples Is y= F(x) = x2 for all real numbers x

injective/surjective?• F(x) is not injective

(-2)2 = 22 = 4, therefore F(-2) = F(2) but -2 ≠ 2.• F(x) is not surjective if Y (the range) is defined as all real

numbers The reason is straightforward: We have y = x2. However, there is no

negative real number y such that x2 = y. EX: (-1/2)2 = 1/4. All squared real numbers are positive, so none of the negative real values of y are matched with F(x) values

Y = G(x) = x3 is both injective and surjective• This is true because every real number raised to the third

power has a unique value. G(-x) ≠ G(x), and it can be easily shown that every real value of y can be yielded from a particular value of x

Page 13: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Real Number Functions and the Vertical Line Test

For a function to be defined over the real numbers, it must be true that for every real value of x, there is a real value y as the output

Also, if a relation is graphed on a RxR plane, it cannot be a function if a vertical line can be drawn through the graph and intersect two points

Both of these properties stem from properties 1 and 2• Notice that a circle is not a real-number

function because it does not satisfy either property

• The second relation, however, is a real number function because it will continue to infinity in both directions without folding back into itself

Page 14: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Inverse Functions An inverse function is a function that “undoes” another

function by mapping the range of a function back onto its domain• In order for F(x) to have a single, all-encompassing inverse function

F-1(y), F(x) must be both injective and surjective• An inverse function is defined as follows: F-

1(y) = x y = F(x)• If every range value, y, is produced by the function F(x), and no two

domain values, x, yield the same y value, then F-1(y) will yield every x value and no two y values will yield the same x value

• Most functions do not have a single, perfect inverse function. In these cases, the inverse relation is actually a system of inverse functions, each defined over a set domain and range To obtain this set algebraically, swap the locations of all domain and range

values and solve for the range values EX: y = x2 → x = y2 → ±(x1/2) = y. Notice the ±. This means that the inverse of y = x2 is actually a system of two

functions, one having all positive and another having all negative values As you would expect, these systems tend to be more complicated, and are a

subject of higher mathematics

Page 15: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Inverse Relations If S is a relation from A to B, then

its inverse S-1 is defined as follows• S-1 = {(y, x) in BxA such that (x, y) is in

S}• In other words, (y, x) is in S-1 iff (x, y) is

in S To obtain an inverse relation, switch the x

and y values for all of the ordered pairs• To take the inverse of a real number

graph, rotate the graph 90 degrees clockwise Use the horizontal line test to determine

whether a graph’s inverse is a function

Both relations arefunctions, but onlyrelation 2 has a single inverse function.To see this, you canperform the horizontal line test

Page 16: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Hashing Functions Now we get to some practical applications

• A hashing function is one which takes a variety of inputs (strings, large integers, binary data) and transforms those inputs into a distinct numerical output Such functions are used to establish methods for sorting items,

encrypting data, and improving the efficiency of storing data Say you need a quick way to sort a data base of students based

on social security numbers These numbers are very large, so it would be impractical to sort data

based on the numbers alone Instead, we will use the hashing function H(n) = n mod 7, where mod is

the modulo division operator defined by “n mod m = n-m*(n div m)”. Div is an operator which returns the integer component of division

This function yields a manageable way to sort data. However, sometimes the function will return the same value for different n values. This is known as a collision, and must be resolved using an algorithm

Page 17: Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This

Checksum Functions A checksum function is a special class of hash

functions that is designed to avoid and correct collision errors• Essentially, you use a bit of programming magic to

ensure that the hash is sustainably implemented• As a conceptual example, consider the social security

hash on the previous slide If all of the numbers you entered into the system yielded

the same hash value (yes this is possible), then the sort would be totally ineffective

To ensure redundancy, you would have to add other parameters to the sort, like last names, birthdays, or student IDs