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CS1022 Computer Programming & Principles

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CS1022Computer Programming & Principles

Lecture 5.1

Functions (1)

- Why functions?
- Inverse relations and composition of relations (background to functions)
- Functions
- Terminology
- Digraphs and graphs of functions
- Properties of functions

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- Functions capture transformations
- Input arguments used to compute output results

- Input/output is key to computing
- Advantages:
- No commitments to computers/architectures
- Abstract, clean, compact, focussed

- Functions represent computations
- Underpinned by -calculus (a kind of “interpreter”)
- Programming paradigm: functional languages

- What is x's age? Given a value for x, output the result: What is x's age?(Bill) => 34

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- Factorial: 5! = 5 4 3 2 1 = 120
- Formally: or

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- Given a binary relation R between sets A and B
- The inverse relation R–1 between A and B is
R–1 (b, a) : (a, b) R

- For instance, the inverse of relation is a parent of on the set of People is the relation is a child of
- Graphically, the inverse of a binary relation is obtained by reversing all arrows in the digraph of original relation

R (p,1), (p,2), (q,2)

R–1 (1, p), (2, p), (2, q)

R

R–1

1

1

2

2

p

p

q

q

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- Let R be a binary relation between sets A and B
- Let S be a binary relation between sets B and C
- The composition of R and S, represented as SR, is a binary relation between A and C given by
SR(a,c) : aA, cC, aR b, bS c for some bB

- The new relation relates elements of A and C using elements of B as “intermediaries”

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- If R is a relation “is a sister of” on a set of people
- If S is a relation “is the mother of” on the same set
- The compositionSR is
SR(a,c) : ais a sister of b, bis mother of c

SR(a,c) : ais a sister of themother of c

SR(a,c) : ais an aunt of c

- What about SS?
SS(a,c) : ais mother ofb, bis mother of c

SS(a,c) : ais mother ofthe mother of c

SS(a,c) : ais a grandmother of c

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- Let R and S be binary relations defined as
Then

aR 1 and 1 Sy implies (a,y) SR

aR 2 and 2 Sx implies (a,x) SR

aR 3 and 3 Sx implies (a,x) SR

bR 2 and 2 Sx implies (b,x) SR

R(a,1), (a,2), (a,3), (b,2)

2

1

2

3

S(1,y), (2,x), (3,x)

R

SR

S

1

x

x

y

y

3

a

a

b

b

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- Relations represented as a matrices for
- It easy to see relationship properties by inspecting the matrix, e.g. relationship is reflexive where it is 1 on the middle diagonal.
- Computers view relations as a two dimensional array

- How to compute compositions using matrices?
- Problem:
- Given the logical matrices of two relations R and S
- Compute the logical matrix of their composition SR

- Result is logical (or Boolean) matrix product

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- Let
Aa1,a2,,an Bb1,b2,,bm Cc1,c2,,cp

- Relations
R relates A and B S relates B and C

- The logical matrix M representing R is given by
- M(i, j) T (true) if (ai, bj) R
- M(i, j) F (false) if (ai, bj) R

- The logical matrix N representing S is given by
- N(i, j) T (true) if (bi, cj) S
- N(i, j) F (false) if (bi, cj) S

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- Logical matrix P represents SR
- If there is bk B with aiRbkandbkScj then
M(i, k) = T andN(k, j) T

Since ai (SR)cj then P(i, j) is also T

- Otherwise
M(i, k) = F orN(k, j) F, for all k,

and P(i, j) F

- If there is bk B with aiRbkandbkScj then
- Logical matrix P representing SR is given by
- P(i, j) T (true) if M(i, k) T andN(k, j) T, for some k
- P(i, j) F (false) otherwise

bk is the bridge between the relations that allows the product relation.

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A Boolean product of two matrices, where the matrices have T/F values (1/0) –

It is the conjunction (and) of the disjunctions (ors), where a, b, and c reference the matrix and xi,jis the row and column of x in the matrix:

etc...

etc...

Find a pair with a bridge between relations and mark 1/0 in result.

Rbilljillphilmary

bill1000

jill0100

phil1010

mary1010

Sbilljillphilmary

bill0110

jill0110

phil0011

mary1100

R S R S

c1,2 is the same row in R, but the next column in S.

- Alternatively,
P(i, j) (M(i, 1) andN(1, j))

or (M(i, 2) andN(2, j))

...

or (M(i, n) andN(n, j))

- We write
PMN

For this Boolean matrix product

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- Functions: special kind of (restricted) relation
- Function from A to Bis a binary relation in which
- Every element of A is associated with a uniquely specified element of B

- That is, for all a A, there is exactly one pair (a, b)
- In terms of digraphs, a function is a relation with precisely one arc leaving every element of A

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Are these relations functions? Justify.

- Relation x is a sibling of y on the set of People
- No, since a person x may have several siblings (or none)

- R = (x, x2) : x Z (defined on Z)
- Yes, since for any integer x, the value x2 is unique

- S = (x, y) : x y2 (defined on R)
- No, because (2, 2) and (2, 2) belong to S

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- Let f be a function from set A to set B
- For each x A there exists a unique y B,(x, y) f
- We write y f(x)
- We refer to f(x) as the image of x under f
- We also write f : A B and it reads:
- f is a function which transforms/maps each element of A to a uniquely determined element of set B

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- A is the domain of f and B is the co-domain of f
- The range of f is the set of images of all elements of A under f, and is denoted as f(A)
f(A) f(x) : x A

- Venn diagram:

A

B

f

f(A)

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- When f : A B and sets A, B are infinite, we cannot draw a digraph of the function
- We can plot the pairs to build a picture of f
- For instance, f : R R, f(x) x2 is as follows:

y

- Horizontal axis xis the domain R
- Vertical axis yis the co-domain R
- Curve consists of points (x, y) in the Cartesian product R R for which f(x) x2

(2, 4)

x

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- Let f : A B be a function
- We say f is injective (or one-to-one) function if
a1a2 ((a1, a2 A) and f(a1) f(a2)) a1 a2

- That is,
If output is the same then input must be the same

- An injective function never repeats values
- Different inputs give different outputs:
a1a2 ((a1, a2 A and a1 a2) f(a1) f(a2))

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- Let f : A B be a function
- f is surjective (or onto) if its range coincides with its co-domain
- For every bBthere is an a A with b f(a)
b a (bB and a A and b f(a))

- In other words:
- Each element of the co-domain is a value of the function

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- Let f : A B be a function
- f is bijective if it is both injective and surjective

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- Consider the functions below (as graphs)

- Not injective as f(a) f(b) 1
- Not surjective as 2 in codomain is not a value of anyf(x)

a

a

a

1

1

1

b

b

b

2

2

2

c

c

c

3

3

3

- Is injective as f(x) are all different
- Is surjective as range and codomain are equal
- Is bijective (injective and surjective)

- Is injective as f(x) are all different
- Not surjective as 2 in codomain is not a value of anyf(x)

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You should now know:

- Inverse relations and composition of relations
- Functions as a (special type) of relations
- Properties of functions and some terminology

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- R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd. 2002. (Chapter 5)
- Wikipedia’s entry
- Wikibooks entry

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