1 Discrete Structures CNS2300 Text Discrete Mathematics and Its Applications Kenneth H. Rosen (5 th Edition) Chapter 2 The Fundamentals: Algorithms,

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Discrete Structures – CNS2300Text

Discrete Mathematics and Its Applications

Kenneth H. Rosen (5th Edition)

Chapter 2The Fundamentals: Algorithms,

the Integers, and Matrices

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Section 2.5

Integers and

Algorithms

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Lemma (A short theorem not used for much except proving another theorem)

Let a = bq + r, where a,b,q and r are integers. Then gcd(a,b) = gcd(b,r)

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Euclidean Algorithmprocedure gcd(a,b: positive integers)x := ay := bwhile y > 0begin r := x mod y x := y y := rend {gcd(a,b) is x}

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Example

18 = (1)(12) + 6

12 = (2)(6) + 0

gcd(12,18)

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Example

18 = (1)(12) + 6

12 = (2)(6) + 0

gcd(12,18)

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Example

gcd(123,277)

277 = (2)(123) + 31

123 = (3)(31) + 30

31 = (1)(30) + 130 = (30)(1) + 0

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Representation of IntegersLet b be a positive integer greater than 1. Then if n is a positive integer, it can be expressed uniquely in the form

n a b a b a b ak kk k

1 1 0

1 . . .where k is a nonnegative integer, a0,a1,…,ak are nonnegative integers less than b, and

Which means?????

a k 0

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You can change bases

351 = (101011111)2

351 = ( 537)8

351 = (15F)16

351 = (253)12

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PeopleThere are only 10 types of people.

Those who understand binary numbers and those who don’t.

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How?

351 = (29)(12) + 3

29 = (2)(12) + 5

2 = (0)(12) + 2

The remainders tell us the number in base 12

351 = 25312

Convert 351 to base 12

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Convert 351 to base 16 (hex)

351 = 21(16) + 15 (F in base 16)

21 = 1(16) +5

Therefore

351 = 15F16

1 = 0(16) +1

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How About the Other Way?

5378 =

5 8 3 8 7 82 1 0

5 64 3 8 7 1

320 + 24 + 7 =

351

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Special Relationships

101011111

15

Special Relationships

1010111117

16

Special Relationships

10101111173

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Special Relationships

101011111735

18

Special Relationships

101011111735

1010111112 = 5378

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Try it Again

101011111

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This time, group by 4 bits

101011111F

21

Special Relationships

101011111F5

22

Special Relationships

101011111F51

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Special Relationships

101011111F51

1010111112 = 15F16

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0000

0001

0010

0011

0101

0100

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

11617256

912819256

56421

25611

12823

2563

3225

25613

12827256

7642925615

12831

256

1817128

96419128

53221

128116423

1283

1625

128136427

12873229

128156431

128

1417649

3219645

16216411322364382564133227647

16296415323164

1217329

161932582132111623323425321316273278293215163132

1

111611831

1611451

1631871

1611291

16518111163141311671815116

2

128124328122528324728

3

138134338132538334738

4

144142344

5

154152354

6

164162364

7

174172374

001 010 011 100 101 110 111E

M

range1 1, )

16 8[ 1 1, )

8 4[ 1 1, )

4 2[ 1 ,1)

2[ 1,2)[ 2,4)[ 4,8)[

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Addition of Integersprocedure add(a,b: positive integers)c:= 0for j:= 0 to n-1begin d := (aj + bj + c)/2 sj := aj + bj + c - 2d c:= dendsn := c

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Multiplying Integersprocedure multiply(a,b: positive integers)for j:= 0 to n-1begin if bj = 1 then cj := a shifted j places

else cj := 0endp := 0for j := 0 to n - 1 p := p + cj

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finished