chapter 5. functions

Chapter 5. Functions

Weiqi Luo (骆伟祺 )School of Software

Sun Yat-Sen UniversityEmail ： [email protected] Office ： A309

mailto:[email protected]

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5.1. Functions 5.2. Functions for Computer Science 5.3. Growth of Functions 5.4. Permutation Functions

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Chapter five: Functions

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Function (mapping, transformation) Let A and B be nonempty sets. A function f from A to B,

which is denoted f: AB, is a relation from A to B such that for all a in Dom(f), f(a), the f-relative set of a, contains just one element of B.

5.1. Functions

3

A B

ab=f(a)

f a: an argument of f

f(a): the value of f for the argument a or the image of a under f

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Example 1 Let A={1, 2, 3, 4} and B={a, b, c, d}, and let

f={(1,a), (2,a), (3,d), (4,c)}

Then we have

f(1)=a, f(2)=a, f(3)=d, f(4)=c

since each set f(n) is a single value, f is a function.

Note ： a in B appears as the second element of two different ordered pairs in f. This does not conflict with the definition of a function.

5.1 Functions

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Example 2 Let A={1,2,3} and B={x, y, z}. Consider the relations

R={(1,x),(2,x)} and S={(1,x),(1,y),(2,z),(3,y)}

The relation S is not a function since S(1)={x, y}.

The relation R is a function, and Dom(R)={1,2}, and Ran(R)={x}

5.1 Functions

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Example 3 Let P be a computer program that accepts an integer as input

and produces an integer as output. Let A=B=Z. Then P determines a relation fp defined as follow:

(m, n) in fp means that n is the output produced by program P where the input is m.

It is clear that fp is a function, since any particular input corresponds to a unique output (we assume that the computer results are reproducible, that is, they are the same each time the program is run)

Note: Functions can be regarded as “input-output” relations

5.1 Functions

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Example 5 Labeled digraph is a digraph in which the vertices or the

edges (or both) are labeled with information from a set.

V: vertices , E: edges , L: labels

f: VL

g: EL

e.g.

5.1 Functions

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Boston

Worcester

Providence

New Haven

Hartford

4449

106

64

39

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Example 7 Let A=Z and B={0,1}. Let f: AB be found by

Example 8

Let A be an arbitrary nonempty set. The identity function on A, denoted by 1A, is defined by

1A(a)=a

5.1 Functions

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1 ( )

0

if a is evenf a

if a is odd

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Composition Let f: AB and g: BC are functions. Then the

composition of f and g, gof, is a relation from A to C.

5.1 Functions

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gof

A

f

ab=f(a)

B

C

g

c=f(b)= gof (a)

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Example 9 Let A=B=Z, and C be the set of even integers.

Let f: AB, and g: BC be defined by

f(a)= a +1 f(b)= 2b

Find gof

Solution: we have

(gof)(a)

= g(f(a))

= g(a+1)

=2(a+1)

5.1 Functions

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Let f be a function from A to B Everywhere defined : Dom(f) = A

Onto : Ran(f) = B

One to one : If f(a) = f(a’) then a=a’

Bijection :

Dom(f) one to one Ran(f)

One-to-one correspondence between A and B

Dom(f) one to one Ran(f)

& Dom(f)=A and Ran(f)=B

5.1 Functions

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Example 11 Let A=B=Z and let f: AB be defined by

f (a) = a+1 for all a in A

Everywhere defined? Yes, Dom(f) =A=Z

Onto? Yes, for all b in B, we have a=b-1

One to one? Yes, a+1 = a’+1, then we have a=a’

5.1 Functions

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Invertible A function f: AB is said to be invertible if its inverse

relation, f-1, is also a function.

Example 14

Let f = {(1,a), (2,a), (3,d), (4,c)}

Then f-1 = {(a,1), (a,2), (d,3), (c,4)}

We see that f-1 is not a function, since f-1 (a)={1,2}

5.1 Functions

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Theorem 1 Let f: AB be a function

(a) Then f-1 is a function from B to A if and only if f is one to one.

If f-1 is a function, then

(b) the function f-1 is also one to one

(c) f-1 is everywhere defined if and only if f is onto

(d) f-1 is onto if and only if f is everywhere defined

5.1 Functions

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Example 15

Let A=B=Z and let f: AB be defined by

　　　　　　 f(a) = a + 1 for all a in A

　 Since it is everywhere defined, one to one, and onto, f is a one-to-one correspondence between A and B.

　 Thus f is invertible, and f-1 is a one-to-one correspondence between B and A.

5.1 Functions

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Theorem 2Let f: AB be any function. Then

(a) 1B o f = f

(b) f o 1A=f

If f is a one-to-one correspondence between A and B, then

(c) f-1 o f = 1 A

(d) f o f-1 =1 B

5.1 Functions

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Theorem 3 (a) Let f :AB and g: BA be functions such that

g o f = 1A and f o g = 1 B.

Then f is a one-to-one correspondence between A and B, g is a one-to-one correspondence between B and A, and each is the inverse of the other.

(b) Let f : AB and g: BC be invertible. The g o f is invertible, and (g o f) -1 =f-1 o g-1

5.1 Functions

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Theorem 4 Let A and B be two finite sets with the same number of

elements, and let f: AB be an everywhere defined function

(a) If f is one to one, then f is onto

(b) If f is onto, then f is one to one

(Refer to Section 3.3. Pigeonhole Principle)

5.1 Functions

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Example 18

CKAJADPDGKJYEIDOT

JOHNHASALONGBEARD

5.1 Functions

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Homework

Ex. 2, Ex. 10, Ex. 20, Ex. 32, Ex. 36, Ex.41

5.1 Functions

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The characteristic function of A Let A be a subset of the universal set U={u1,u2,…un}. The

characteristic function of A is defined as a function from U to {0,1} by the following:

e.g. If A={4,7,9} and U={1,2,…,10} , then

fA(2)=0, fA(4)=1, fA(7)=1, fA(12) is undefined.

It is everywhere defined and onto, but is not one to one

5.2 Functions for Computer Science

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1 ( )

0 i

A ii

if u Af u

if u A

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Boolean function A function from a set A to B, where B ={True, False} is

called a Boolean function.

Note: the predicates in Section 2.1 are examples of Boolean functions.

For instance:

Let P(x): x is even and Q(y): y is odd. Then P and Q are Boolean functions from Z to B.

R(x, y): x is even or y is odd is a Boolean function of two variables from Z×Z to B


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Floor Function

the largest integer less than or equal to q

where q is rational numbers

Ceiling Function

the smallest integer greater than or equal to q

where q is rational numbers


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( )f q q

( )f q q

(1.5) 1.5 1, ( 3) 3 3, ( 2.7) 2.7 3f f f

(1.5) 1.5 2, ( 3) 3 3, ( 2.7) 2.7 2f f f

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Polynomial function

Exponential function

Logarithm function

Factorial function


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20 1 2( ) ... n

np x a a x a x a x

( ) 2zf z

( ) log ( )n nf x x

( ) !f n n

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Example 7 (a) l: A*Z as l(w) is the length of the string w

(b) B is a finite subset of U

pow(B) denotes the power set of B

Note: pow is a function from V, the power set of U, to the power set of V

(c) Let A=B=the set of all 2-by-2 martices with entries and let t(M)=MT, the transpose of M


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Hash Functions Suppose that 10,000 customer account records must be stored

and processed. The company’s computer is capable of searching a list of 100 items in an acceptable amount of time. We decide to create 101 liked lists for storage, because if the hashing function works well in randomly assigning records to lists, we would expect to see roughly 100 records per list. We define a hashing function from the set of set of seven-digit account numbers to the set {0,1,2,3,…,100} as follows

h(n) = n (mod 101) h(2011) = 92

the record with account number 2011 will be assigned to 92.


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Collision Since the function in Ex. 10 is not one-to-one, different

accounts may be assigned to the same list. In such a case, the collision occurred.


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List #0

List #1

List #92 :2011193 1001…

List #100

…

…

customer account records

2011

193

1001

…Around 100 per list

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Fuzzy set and Fuzzy logic

Refer to

http://en.wikipedia.org/wiki/Fuzzy_mathematics


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Homework

Ex. 4, Ex. 7, Ex. 8, Ex.18, Ex. 22, Ex. 24


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Let f and g be functions whose domains are subsets of Z+. We say that f is O(g), read f is big-Oh of g, if there exist constants c and k such that |f(n)|<=c |g(n)| for all n >=k.

If f is O(g), then f grows no faster than g does.

5.3 Growth of Functions

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| ( ) |lim ,

| ( ) |n

f nc c R

g n

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Example 2

The function f(n) = ½ n3 + ½ n2 is O(g) for g(n)=n3. To see this, consider

Choosing 1 for c and 1 for k, we have shown that |f(n)| <=c |g(n)| for all n>=1 and f is O(g).


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3 2 3 3 31 1 1 1, 1

2 2 2 2n n n n n if n

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Let f and g be functions whose domains are subsets of Z+. We say that f and g have the same order if f is O(g) and g is O(f).

f is O(g) : f grows no faster than g does

g is O(f) : g grows no faster than f does


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2 1 1 2

| ( ) |lim ,

| ( ) |n

f nc c c c R

g n

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Example 3

Let f(n) = 3n4-5n2 and g(n) = n4 be defined for positive integers n. Then f and g have the same order. First,

Let c=8 and k=1, then |f(n)| <=c |g(n)| for all n>=k. Thus f is O(g). Conversely,

Let c=1 and k=2, we have g is O(f).


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4 2 4 2 4 4 43 5 3 5 3 5 8 , 1n n n n n n n if n

4 4 4 4 23 2 3 5 , 2n n n n n if n

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Let f and g be functions whose domains are subsets of Z+. We say that f is lower order than g or that f grows more slowly than g if

f is O(g) but g is not O(f)


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| ( ) |lim 0

| ( ) |n

f n

g n

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Example 4

f(n) = n5, g(n) = n7

Clearly,

f(n) is O(g), since n5<=n7 , n>=1

Support g(n) is O(f), then there exists c and k such that

n7 <=cn5 , for n>=k

Choose N so that N>k and N2>c, then

N7 <=cN5 < N2 N5= N7

This is a contradiction.


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We define a relation Θ, big-theta, on functions whose domain are subsets of Z+ as f Θ g if and only if f and g have the same order.

Theorem 1: The relation Θ is an equivalence relation

Proof:

Reflexive: |f(n)| <= c |f(n)| , c=1 for all n>=1

Symmetric: by the definition of the same order

(g and f have the same order iff f is O(g) and g is O(f) )


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Transitive:

support f and g have the same order

|f(n)| <=c1|g(n)| for all n>=k1 (1)

|g(n)| <=c2|f(n)| for all n>=k2 (2)

support g and h have the same order

|g(n)| <=c3|h(n)| for all n>=k3 (3)

|h(n)| <=c4|g(n)| for all n>=k4 (4)

|f(n)| <= c1|g(n)| <=c1c3|h(n)| for all n>=max(k1,k3)

|h(n)| <=c4|g(n)| <=c2c4|f(n)| for all n>= max(k2,k4)

Thus, f and h has the same order


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Example 5 All functions that have the same order as g(n)=n3 are

said to have order Θ(n3).

The common orders in computer science applications are

Θ(1), Θ(lg(n)), Θ(n), Θ(nlg(n)), Θ(n2), Θ(n3), and Θ(2n)

Note:

Θ(1): the classes of constant functions

lg is the base 2 log function


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Example 6 Every logarthmic function f(n)=logb(n) has the same order as

g(n)=lg(n). There is a logarithmic change-of-base identity

in which loga(b) is a constant. Thus

and, conversely,

Therefore g is O(f) and f is O(g).


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log ( )log ( )

log ( )a

ba

xx

b

1| log ( ) | | lg( ) |," "

| lg( ) |b x n holdsb

| lg( ) | lg( ) | log ( ) |," "bn b n holds

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Rules for determining the Θ-Class of a Function

1. Θ(1) functions are constant and have zero growth, the slowest growth possible.

2. Θ(lg(n)) is lower than Θ(nk) if k>0

3. Θ(na) is lower than Θ(nb) iff b>a>0

4. Θ(an) is lower than Θ(bn) iff b>a>0

5. Θ(nk) is lower than Θ(an) for any power nk and any a>1

6. If r is a not zero, and Θ(rf)= Θ(f) for any function f

7. If h is a nonzero function and Θ(f) is lower than (or the same as) Θ(g), then Θ(fh) is lower than (or the same as Θ(gh) )

8. If Θ(f) is lower than Θ(g), then Θ(f+g) = Θ(g)


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Example 7

Determine the Θ-class of each of the following

(a)f(n) = 4n4-6n7+24n3

(b)g(n) = lg(n) -3n

(c)h(n) = 1.1n+n15

(a) Θ(f) = Θ(n7) (Rules 3,6 and 8)

(b) Θ(g)= Θ(n) (Rules 2,6 and 8)

(c) Θ(h)= Θ(1.1n) (Rules 5 and 8)


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Example 8 Using the rules for ordering Θ-classes, arrange the

following in order from lowest to highest

Θ(nlg(n)) Θ(100n2-n) Θ(n0.2) Θ(1,000,000)

Θ(1.3n) Θ(n+107)

Θ(1,000,000) , Θ(n0.2), Θ(n+107) , Θ(nlg(n)),

Θ(100n2-n) Θ(1.3n)

Θ(n1.2) ?


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Homework

Ex. 6, Ex. 10, Ex. 12, Ex. 25, Ex. 29


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Permutation A bijection from a set A to itself is called a

permutation of A.

For instance,

Let A=R and let f: AA be defined by f(a)=2a+1.

Since f is one to one and onto, it follows that f is a permutation of A.

5.4 Permutation Functions

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If A ={a1,a2,…an} is a finite set and p is a bijection on A, we list the elements of A and the corresponding function values p(a1), p(a2), …, p(an) in the following form:

P is a permutation of a finite set A= {a1,a2,…an} , then the sequence p(a1) , p(a2), …, p(an) is just a rearrangement of the elements of A.


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1 2

1 2

, , ...,

( ), ( ), ..., ( )n

n

a a ap

p a p a p a

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Example 2 Let A={1,2,3}. Then all the permutations of A are


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1 2 31

1 2 3A

1

1 2 3

1 3 2p

2

1 2 3

2 1 3p

3

1 2 3

2 3 1p

4

1 2 3

3 1 2p

5

1 2 3

3 2 1p

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Example 3 Using the permutations of Example 2, compute (a) p4

-1 (b) p3

o p2

(a) Viewing p4 as a function, we have

P4={(1,3), (2,1), (3,2)}

then P4-1 ={(3,1), (1,2), (2,3)}


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4

1 2 3

3 1 2p

14 3

1 2 3

2 3 1p p

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(b)


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3 2

1 2 3 1 2 3

2 3 1 2 1 3p p

1 2 3

3 2 1

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Theorem 1 If A={a1,a2,…,an} is a set containing n elements, then

there are n!=n× (n-1) × …2 × 1 permutations of A

p(a1) --- n ways

p(a2) --- n-1 ways (p(a1) is fixed)

…

p(an-1) --- 2 ways (p(a1),…p(an-2) are fixed)

p(an) --- 1 way (p(a1),…p(an-1) are fixed)

Totally, there are n! permutations of A.


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Cyclic Permutation Let b1, b2, …, br be distinct elements of the set A={a1,a2,…,an}. The

permutation p: AA defined by

p(b1)= b2

p(b2)= b3

…

p(br-1)= br

p(br) = b1

p(x)=x, if x in A, x is not in {b1,b2,…,br}

is called a cyclic permutation of length r, or simply a cycle of length r, denoted by (b1,b2,…,br)


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br-1

br

b1

b2

b3

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Example 4

Let A={1,2,3,4,5}.The cycle (1,3,5) denotes the permutation


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1 2 3 4 5

3 2 5 4 1

1

35

Note: (1, 3, 5) = (3, 5, 1) = ( 5,1, 3)

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Example 5

Let A={1,2,3,4,5,6}. Compute (4,1,3,5) o (5,6,3) and (5,6,3) o (4,1,3,5)

Solutions:


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1 2 3 4 5 6(4,1,3,5)

3 2 5 1 4 6

1 2 3 4 5 6(5,6,3)

1 2 5 4 6 3

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Example 5


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1 2 3 4 5 6(4,1,3,5) (5,6,3)

3 2 4 1 6 5

1 2 3 4 5 6(5,6,3) (4,1,3,5)

5 2 6 1 4 3

(4,1,3,5) (5,6,3) (5,6,3) (4,1,3,5)

Neither product is a cycle. (why?)

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Two cycles of a set A are said to be disjoint if no element of A appears in both cycles.

For instance,

Let A={1,2,3,4,5,6}.

Then the cycles (1,2,5) and (3,4,6) are disjoint,

whereas the cycles (1,2,5) and (2,4,6) are not.


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Theorem 2 A permutation of a finite set that is not the identity or a

cycle can be written as a product of disjoint cycles of length >=2

For instance:


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12345678

34652187p

(7,8) (2,4,5) (1,3,6)

(2, 4,5) (1,3,6) (7,8) ...

p

Note: the product is unique except for the order of the cycles

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Transposition A cycle of length 2 is called a transposition.

p(ai)=aj ; p(aj)=ai

Note:

p o p = 1A - the identity permutation of A

Every cycle can be written as a product of transpositions.

(b1, b2, …, br) = (b1,br) o (b1,br － 1)…(b1,b3) o (b1,b2)

(not unique! Why?)

The proof is shown as below


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Basic Step

r=2, then the cycle is (b1,b2) , which has the proper form

Induction Step

Use P(k) to show P(k+1)

(b1,b2,…,bk,bk+1) be cycle of length k+1

(b1,b2,…,bk,bk+1) = (b1,bk+1) o (b1,b2,…,bk) (why?)

Using P(k),

(b1,b2,…,bk) = (b1,bk) o (b1,bk-1) … (b1,b2)

Thus, by substitution

(b1,b2,…,bk, bk+1) = (b1, bk+1) o (b1,bk) o (b1,bk-1)… (b1,b2)


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Corollary 1

Every permutation of a finite set with at least two elements can be written as a product of transpositions


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Example 9 Write the following permutation as a product of

transposition


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12345678

34652187p

(7,8) (2,4,5) (1,3,6)p

(2, 4,5) (2,5) (2,4)

(1,3,6) (1,6) (1,3)

(7,8) (2,5) (2,4) (1,6) (1,3)p

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Every permutation on a set of two or more elements can be written as a product of transpositions in many ways

For instance


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(1,2,3) (1,3) (1,2)

(2,1) (2,3)

(1,3) (3,1) (1,3) (1,2) (3,2) (2,3)

Even Transpositions

Even Transpositions

Even Transpositions

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Theorem 3 If a permutation of a finite set can be written as a

product of an even number of transpositions, then it can never be written as a product of an odd number of transpositions, and conversely

Note: a permutation of a finite set is called even if it can be written as a product of an even number of transpositions, and it is called odd if it can be written as a product of an odd number of transpositions.


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Example 10 Is the permutation

even or odd?


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1234567

2457631p

(3,5,6) (1,2,4,7)p

(1,2,4,7) (1,7) (1,4) (1,2) (3,5,6) (3,6) (3,5)

(3,6) (3,5) (1,7) (1,4) (1,2)p Odd Transpositions

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From the definition of even and odd permutations, it follows (see Ex. 22-24) that

(a) the product of two even permutations is even

(b) the product of two odd permutations is even

(c) the product of an even and an odd permutation is odd


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Theorem 4 Let A={a1,a2,…,an} be a finite set with n elements, n>=2.

There are n!/2 even permutations and n!/2 odd permutations.

Proof:

An be the set of all even permutations of A

Bn be the set of all odd permutations of A.

We shall define a function f: An Bn, which we show is one to one and onto, and this will show that An and Bn have the same number of elements.


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Since n>=2, we can choose a particular transposition q0 of A. say that q0=(an-1,an). Define f: AnBn by

f(p) = q0 o p p in An

note: p is even permutation, then f(p) is odd and thus f(p) in Bn.

one to one : support p1 and p2 are in An and f(p1)=f(p2)

q0 o p1 = q0 o q2 q0 o (q0 o p1) = q0 o( q0 o q2 )

by associative property

q0 o (q0 o p1) = (q0 o q0 )o p1 = p1 (q0 oq0)=1A

q0 o (q0 o p2) = (q0 o q0 )o p2 = p2 (q0 oq0)=1A

Therefore, p1=p2


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Onto: let q in Bn, then q0oq in An, and

f(q0 o q) = q0 o (q0 o q) = (q0 o q0) o q = q

Therefore, f is an onto function.

Since f is one to one and onto, then An and Bn have the same number of elements. Note that An∩ Bn = ф and by Theorem 1 |An U Bn|= n!, we then have

n! = |An U Bn| = |An| + |Bn| - | An ∩ Bn |= 2 |An|

so

|An|=|Bn|= n!/2


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Homework

Ex. 2, Ex. 6, Ex.24, Ex. 26, Ex. 27, Ex. 37


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chapter 5. functions

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