Chapter 6 Number Theory Benchaporn Jantarakongkul Faculty of

Chapter 6 Number Theory Benchaporn Jantarakongkul Faculty of Informatics, Burapha University 1

Chapter 6 Number Theory Benchaporn Jantarakongkul Faculty of Informatics, Burapha University 1

The Integers and Division • �� hash functions, cryptography, digital signatures Faculty of

The Integers and Division • �� hash functions, cryptography, digital signatures Faculty of Informatics, Burapha University 2

Divisors-Examples �� ? 1. 77 | 7 2. 7 | 77 3. 24 |

Divisors-Examples �� ? 1. 77 | 7 2. 7 | 77 3. 24 | 24 4. 0 | 24 5. 24 | 0 Faculty of Informatics, Burapha University 4

Facts re: the Divides Relation • Theorem: a, b, c Z: 1. a≠ 0

Facts re: the Divides Relation • Theorem: a, b, c Z: 1. a≠ 0 a|0 �� a|a 2. (a|b a|c) a | (b + c) 3. a|bc 4. (a|b b|c) a|c 5. [a|(b+c) a|b)] a|c �� : w w 17|0 17|34 17|170 17|204 17|340 6|12 12|144 6 | 144 Faculty of Informatics, Burapha University 6

Division �� d the divisor a the dividend q the quotient r the remainder

Division �� d the divisor a the dividend q the quotient r the remainder 24 + 3· 31 = 117 a = dq + r Faculty of Informatics, Burapha University 16

Division �� q the quotient d the divisor a the dividend 1 r the

Division �� q the quotient d the divisor a the dividend 1 r the remainder 1+ (4 -)· 3 = 11 -ขอ �� : �� a = dq + r Faculty of Informatics, Burapha University 17

GCD shortcut • �� prime factorizations �� , �� . �� GCD ��

GCD shortcut • �� prime factorizations �� , �� . �� GCD �� : • �� : – – – a=84=2· 2· 3· 7 = 22· 31· 71 b=96=2· 2· 2· 3 = 25· 31· 70 0 =University Faculty 2 of Informatics, Burapha gcd(84, 96) = 2· 31· 7 2· 2· 3 = 12. 20

Greatest Common Divisor Relatively Prime Q: 1. 2. 3. 4. �� . �� gcd

Greatest Common Divisor Relatively Prime Q: 1. 2. 3. 4. �� . �� gcd �� : gcd(11, 77) gcd(33, 77) gcd(24, 36) gcd(24, 25) Faculty of Informatics, Burapha University 22

Greatest Common Divisor Relatively Prime A: 1. 2. 3. 4. gcd(11, 77) = 11

Greatest Common Divisor Relatively Prime A: 1. 2. 3. 4. gcd(11, 77) = 11 gcd(33, 77) = 11 gcd(24, 36) = 12 gcd(24, 25) = 1 �� 24 �� 25 �� (relatively prime( �� : �� relatively prime �� Faculty of Informatics, Burapha University 23

Euclid’s Algorithm Example gcd(44, 32) = gcd(44 mod 32, 32) = gcd(12, 32) =

Euclid’s Algorithm Example gcd(44, 32) = gcd(44 mod 32, 32) = gcd(12, 32) = gcd(32 mod 12, 12) = gcd(8, 12) = gcd(12 mod 8, 8) = gcd(4, 8) = gcd(8 mod 4, 4) = gcd(0, 4) =4 Faculty of Informatics, Burapha University 25

Euclid’s Algorithm Pseudocode procedure gcd(a, b: positive integers) while b 0 begin r ≔

Euclid’s Algorithm Pseudocode procedure gcd(a, b: positive integers) while b 0 begin r ≔ a mod b; a ≔ b; b ≔ r; end return a Faculty of Informatics, Burapha University 26

Euclidean Algorithm. Example gcd(33, 77): Step r = a mod b a b 0

Euclidean Algorithm. Example gcd(33, 77): Step r = a mod b a b 0 - 33 77 Faculty of Informatics, Burapha University 27

Euclidean Algorithm. Example gcd(33, 77): Step r = a mod b a b 0

Euclidean Algorithm. Example gcd(33, 77): Step r = a mod b a b 0 - 33 77 1 33 mod 77 = 33 77 33 Faculty of Informatics, Burapha University 28

Euclidean Algorithm. Example gcd(33, 77): Step r = a mod b a b 0

Euclidean Algorithm. Example gcd(33, 77): Step r = a mod b a b 0 - 33 77 77 33 33 11 1 2 33 mod 77 = 33 77 mod 33 = 11 Faculty of Informatics, Burapha University 29

Euclidean Algorithm. Example gcd(33, 77): Step r = a mod b a b 0

Euclidean Algorithm. Example gcd(33, 77): Step r = a mod b a b 0 - 33 77 77 33 33 11 11 0 1 2 3 33 mod 77 = 33 77 mod 33 = 11 33 mod 11 =0 Faculty of Informatics, Burapha University 30

Euclidean Algorithm. Example gcd(244, 117): Step r = a mod b a b 0

Euclidean Algorithm. Example gcd(244, 117): Step r = a mod b a b 0 - 244 117 1 244 mod 117 = 10 117 10 2 3 4 117 mod 10 = 7 10 mod 7 = 3 7 mod 3 = 1 10 7 3 1 5 3 mod 1=0 1 0 �� 244 �� 117 �� relatively prime �� gcd(244, 117)=1 Faculty of Informatics, Burapha University 31

Exercise on Board �� gcd �� Euclidean algorithm • gcd(372, 164 ( • gcd(299,

Exercise on Board �� gcd �� Euclidean algorithm • gcd(372, 164 ( • gcd(299, 26) • gcd(414, 662) • gcd(1740, 1120) • gcd(1246, 132) Faculty of Informatics, Burapha University 32

Pairwise relatively prime • �� {a 1, a 2, …} �� (pai rwise relatively

Pairwise relatively prime • �� {a 1, a 2, …} �� (pai rwise relatively prime) �� (ai, aj), �� i j, �� relatively prime Q: �� pairwise relatively prime �� { 17 , 169 , 15 , 21 , 28 , 44 } Faculty of Informatics, Burapha University 33

Pairwise relatively prime A: �� pairwise relatively prime �� {44, 28, 21, 15, 169,

Pairwise relatively prime A: �� pairwise relatively prime �� {44, 28, 21, 15, 169, 17} : �� {17, 169, 28, 15} �� {17, 169, 44, 15} �� : 15, 17, �� 27 �� pairwise relatively prime �� ? • �� , �� gcd(15, 27) = 3 15, 17, �� 28 �� pairwise relatively prime �� ? • �� , �� gcd(15, 17) = 1, gcd(15, 28) = 1 �� gcd(17, 28) =1 Faculty of Informatics, Burapha University 34

�� (Least Common Multiple( Q: 1. 2. 3. 4. 5. �� . �� lcm

�� (Least Common Multiple( Q: 1. 2. 3. 4. 5. �� . �� lcm �� : lcm(10, 100) lcm(7, 5) lcm(9, 21) lcm(3, 7) lcm(4, 6) Faculty of Informatics, Burapha University 36

�� (Least Common Multiple( A: 1. 2. 3. 4. 5. lcm(10, 100) = 100

�� (Least Common Multiple( A: 1. 2. 3. 4. 5. lcm(10, 100) = 100 lcm(7, 5) = 35 lcm(9, 21) = 63 lcm(3, 7) = 21 lcm(4, 6) =12 �� : a = 60 =22 31 51 b = 54 = 21 33 50 lcm(a, b) = lcm(60, 54) = 22 33 51 = 4 27 5 = 540 Faculty of Informatics, Burapha University 37

GCD �� LCM a = 60 = 22 31 51 b = 54 =

GCD �� LCM a = 60 = 22 31 51 b = 54 = 21 33 50 gcd(a, b) = 21 3 1 5 0 =6 lcm(a, b) = 22 3 3 5 1 = 540 Theorem: a b = gcd(a, b) lcm(a, b) Faculty of Informatics, Burapha University 38

mod function �� 1. 113 mod 24: 2. -29 mod 7 Faculty of Informatics,

mod function �� 1. 113 mod 24: 2. -29 mod 7 Faculty of Informatics, Burapha University 40

Modular Congruence Q: 1. 2. 3. 4. �� ? 3 3 (mod 17) 3

Modular Congruence Q: 1. 2. 3. 4. �� ? 3 3 (mod 17) 3 -3 (mod 17) 172 177 (mod 5) -13 13 (mod 26) Faculty of Informatics, Burapha University 42

Spiral Visualization of mod ��≡ 0 (mod 5) modulo-5 20 15 10 ≡ 4

Spiral Visualization of mod ��≡ 0 (mod 5) modulo-5 20 15 10 ≡ 4 19 14 9 (mod 5) 4 8 13 18 ≡ 3 (mod 5) ≡ 1 (mod 5) 5 0 3 2 1 6 11 16 21 7 12 17 22 ≡ 2 (mod 5) Faculty of Informatics, Burapha University 44

Useful Congruence Theorems • Theorem: �� a, b Z, m Z+ �� : a

Useful Congruence Theorems • Theorem: �� a, b Z, m Z+ �� : a b (mod m) k Z a=b+km Proof. direction: �� a = b + km, �� (a-b ) = km �� m | (a-b) �� a b (mod m) direction: �� a b (mod m), �� m | (a-b ) �� k �� (a-b ) = km 45 Faculty of Informatics, Burapha University

Useful Congruence Theorems • Theorem: �� a, b, c, d Z, m, n Z+

Useful Congruence Theorems • Theorem: �� a, b, c, d Z, m, n Z+ 1. �� a b (mod m) �� c d (mod m), �� : ▪ a+c b+d (mod m), �� ▪ ac bd (mod m) 2. �� a b (mod m) �� b c (mod m), �� a c (mod m) 3. �� a b (mod m) �� an bn (mod m) Faculty of Informatics, Burapha University 46

Useful Congruence Theorems �� (1. ) a b (mod m) c d (mod m)

Useful Congruence Theorems �� (1. ) a b (mod m) c d (mod m) ac bd (mod m) Proof. �� k �� l �� a = b + mk �� c = d + ml �� ac = (b + mk)(d + ml ) =bd +m(kd+lb+mkl) �� (ac-bd ) = m(kd+lb+mkl) �� m �� ac bd (mod m) Faculty of Informatics, Burapha University 47

Modular arithmetic harder examples Q: �� 1. 3071001 mod 102 2. (-45 · 77)

Modular arithmetic harder examples Q: �� 1. 3071001 mod 102 2. (-45 · 77) mod 17 3. Faculty of Informatics, Burapha University 48

Modular arithmetic harder examples 1. �� 3071001 mod 102 �� a b (mod

Modular arithmetic harder examples 1. �� 3071001 mod 102 �� a b (mod m) �� an bn (mod m) : �� 307 1 (mod 102) �� 307 n 1 n (mod 102) : 3071001 (mod 102) 11001 (mod 102) �� , 3071001 mod 102 = 1 Faculty of Informatics, Burapha University 49

Modular arithmetic harder examples 2. �� (-45 · 77) mod 17 �� a

Modular arithmetic harder examples 2. �� (-45 · 77) mod 17 �� a b (mod m) �� c d (mod m), �� ac bd (mod m) �� -45 6 (mod 17) �� 77 9 (mod 17), �� -45· 77 6· 9 (mod 17): (-45· 77) (mod 17) (6· 9) (mod 17) 54 (mod 17) 3 (mod 17) of Informatics, Burapha University �� (-45· 77)Faculty mod 17 = 3 50

Modular arithmetic harder examples 3. �� 10 -1 (mod 11) �� 104

Modular arithmetic harder examples 3. �� 10 -1 (mod 11) �� 104 (-1)4 (mod 11), �� 104 1(mod 11) �� 105 (-1)5 (mod 11), �� 105 -1(mod 11)… : �� a b (mod m) �� c d (mod m), �� a+c b+d (mod m) Faculty of Informatics, Burapha University 51

Hashing Functions Example • �� m=111 �� 64212848 �� 37149212 �� 14 ��

Hashing Functions Example • �� m=111 �� 64212848 �� 37149212 �� 14 �� 65 �� h(64212848) = 64212848 mod 111 = 14 h(37149212) = 37149212 mod 111 = 65 �� 24666707 �� 65 �� Faculty of Informatics, Burapha University 53

Caesar’s Cipher �� f (a) = (a+3) mod 26 �� “YESTERDAY” 1. YESTERDAY 2.

Caesar’s Cipher �� f (a) = (a+3) mod 26 �� “YESTERDAY” 1. YESTERDAY 2. 24 0 3 17 4 19 18 4 24 3. 1 3 6 20 7 22 21 7 1 4. “BHVWHUGDB” Faculty of Informatics, Burapha University 57

Caesar’s Cipher • �� “WHQ” • A : 22 7 16 f -1(a) =

Caesar’s Cipher • �� “WHQ” • A : 22 7 16 f -1(a) = (a-3) mod 26 19 4 13 “TEN” Faculty of Informatics, Burapha University 58

Digital Signature Faculty of Informatics, Burapha University 60

Digital Signature Faculty of Informatics, Burapha University 60

Slides: 60

Download presentation