CSE 311 Lecture 11: Modular Arithmetic. Emina Torlak and Kevin Zatloukal

Transcription

1 CSE 311 Lecture 11: Modular Arithmetic Emina Torlak and Kevin Zatloukal 1

2 Topics Sets and set operations A quick wrap-up of Lecture 10. Modular arithmetic basics Arithmetic over a finite domain (a.k.a computer arithmetic). Modular arithmetic properties Congruence, addition, multiplication, proofs. Modular arithmetic and integer representations Unsigned, sign-magnitude, and two s complement representation. Applications of modular arithmetic Hashing, pseudo-random numbers, ciphers. 2

3 Sets and set operations A quick wrap-up of Lecture 10.

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5 Sets and set operations A set is a collection of elements. Write a B (or a B) to say that a is (or isn t) an element in the set B. The order of elements doesn t matter, and duplicates don t matter. A B A = B x. x A x B A B A B A B x. x A x B Sets and are equal if they have the same elements. is a subset of if every element of is also in. Sets can be built from predicates or using set operations. A B = {x : (x A) (x B)} A B = {x : (x A) (x B)} A B = {x : (x A) (x B)} A B = {x : (x A) (x B)} A = {x : x A} = {x : (x A)} 4

6 This is Boolean algebra again A B = {x : (x A) (x B)} A B = {x : (x A) (x B)} A = {x : x A} = {x : (x A)} Union is defined using. Intersection is defined using. Complement works like. This means that all equivalences from Boolean algebra translate directly into set theory, and you can use them in your proofs! 5

7 More sets Power set, Cartesian product, and Russell s paradox. 6

8 Power set A (A) = {B : B A} Power set of a set is the set of all subsets of. Examples Let Days = {M, W, F}. (Days) = ( ) = A 7

9 Power set A (A) = {B : B A} Power set of a set is the set of all subsets of. Examples Let Days = {M, W, F}. (Days) = {, {M}, {W}, {F}, {M, W}, {M, F}, {W, F}, {M, W, F}} ( ) = A 7

10 Power set A (A) = {B : B A} Power set of a set is the set of all subsets of. Examples Let Days = {M, W, F}. (Days) = {, {M}, {W}, {F}, {M, W}, {M, F}, {W, F}, {M, W, F}} ( ) = { } A 7

11 Cartesian product The Cartesian product of two sets is the set of all of their ordered pairs. A B = {(a, b) : a A b B} Examples R R Z Z is the real plane. is the set of all pairs of integers. 8

12 Cartesian product The Cartesian product of two sets is the set of all of their ordered pairs. A B = {(a, b) : a A b B} Examples R R is the real plane. Z Z is the set of all pairs of integers. If A = {1, 2}, B = {a, b, c}, then A B = 8

13 Cartesian product The Cartesian product of two sets is the set of all of their ordered pairs. A B = {(a, b) : a A b B} Examples R R is the real plane. Z Z is the set of all pairs of integers. If A = {1, 2}, B = {a, b, c}, then. A B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} 8

14 Cartesian product The Cartesian product of two sets is the set of all of their ordered pairs. A B = {(a, b) : a A b B} Examples R R is the real plane. Z Z is the set of all pairs of integers. If A = {1, 2}, B = {a, b, c}, then. A B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} A = 8

15 Cartesian product The Cartesian product of two sets is the set of all of their ordered pairs. A B = {(a, b) : a A b B} Examples R R is the real plane. Z Z is the set of all pairs of integers. If A = {1, 2}, B = {a, b, c}, then. A B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} A = {(a, b) : a A b } = {(a, b) : a A F} =. 8

16 Russell s paradox Let S S = {x : x x} be the set of all sets that don t contain themselves. 9

17 Russell s paradox Let S S = {x : x x} S be the set of all sets that don t contain themselves. The definition of is contradictory, hence the paradox. 9

18 Russell s paradox Let S S = {x : x x} S be the set of all sets that don t contain themselves. The definition of is contradictory, hence the paradox. Suppose that S S. Then, by definition of S, S S, which is a contradiction. 9

19 Russell s paradox Let S S = {x : x x} S be the set of all sets that don t contain themselves. The definition of is contradictory, hence the paradox. Suppose that S S. Then, by definition of S, S S, which is a contradiction. Suppose that S S. Then, by definition of S, S S, which is a contradiction too. 9

20 Russell s paradox Let S S = {x : x x} S be the set of all sets that don t contain themselves. The definition of is contradictory, hence the paradox. Suppose that S S. Then, by definition of S, S S, which is a contradiction. Suppose that S S. Then, by definition of S, S S, which is a contradiction too. To avoid the paradox Define with respect to a universe of discourse. S S = {x U : x x} S S S U With this definition, and there is no contradiction because. 9

21 Working with sets Representing sets as bitvectors and applications of bitvectors. 10

22 Representing sets as bitvectors Suppose that universe is. We can represent every set U {1, 2,, n} B U b 1 b 2 b n where b i = 1 if i B = 0 if i B b i as a vector of bits: This is called the characteristic vector of set. B 11

23 Representing sets as bitvectors Suppose that universe is. We can represent every set U {1, 2,, n} B U b 1 b 2 b n where b i = 1 if i B = 0 if i B b i as a vector of bits: This is called the characteristic vector of set. Given characteristic vectors for and, what is the vector for A B = A B = A B B 11

24 Representing sets as bitvectors Suppose that universe is. We can represent every set U {1, 2,, n} B U b 1 b 2 b n where b i = 1 if i B = 0 if i B b i as a vector of bits: This is called the characteristic vector of set. A A B = ( a 1 + b 1 ) ( a n + b n ) A B = Given characteristic vectors for and, what is the vector for B B 11

25 Representing sets as bitvectors Suppose that universe is. We can represent every set U {1, 2,, n} B U b 1 b 2 b n where b i = 1 if i B = 0 if i B b i as a vector of bits: This is called the characteristic vector of set. A A B = ( a 1 + b 1 ) ( a n + b n ) A B = ( a 1 b 1 ) ( a n b n ) Given characteristic vectors for and, what is the vector for B B 11

26 Unix/Linux file permissions $ ls -l drwxr-xr-x... Documents/ -rw-r--r--... file1 Permissions maintained as bitvectors. Letter means the bit is 1. - means the bit is zero. 12

27 Bitwise operations z = x y z = x & y z = x ^ y (x y) y = x Note that. 13

28 Private key cryptography Alice wants to communicate a message secretly to Bob, so that eavesdropper Eve who sees their conversation can t understand. Alice and Bob can get together ahead of time and privately share a secret key. K How can they communicate securely in this setting? m m Sender (Alice) plaintext message encrypt key cyphertext decrypt key plaintext message Receiver (Bob) Eavesdropper (Eve) 14

29 One-time pad Alice and Bob privately share a random -bitvector. Eve doesn t know. n K Later, Alice has -bit message to send to Bob. Alice computes C = m K. Alice sends C to Bob. Bob computes, which is. Eve can t figure out from unless she can guess. And that s very unlikely for large m m = C K (m K) K = m m C K n n K 15

30 Modular arithmetic basics Arithmetic over a finite domain (a.k.a computer arithmetic). 16

31 Modular arithmetic in action What does this Java program output? public class Seconds { final static int SEC_IN_YEAR = 364*24*60*60; public static void main(string args[]) { System.out.println("Number of seconds in years: " + (SEC_IN_YEAR*10100)); } } 17

32 Modular arithmetic in action What does this Java program output? public class Seconds { final static int SEC_IN_YEAR = 364*24*60*60; public static void main(string args[]) { System.out.println("Number of seconds in years: " + (SEC_IN_YEAR*10100)); } } $javac Test.java $java -Xmx128M -Xms16M Test Number of seconds in years: You ll recognize this as integer overflow. It happens because computers use modular arithmetic to operate on finite integer data types, such as int. 17

33 It all starts with divisibility Divisibility is the core concept behind modular arithmetic. a b a b a Z, b Z a b k Z. b = ka Definition: divides, written as. For,. 18

34 It all starts with divisibility Divisibility is the core concept behind modular arithmetic. a b a b a Z, b Z a b k Z. b = ka Definition: divides, written as. For,. Examples: which of the following are true and which are false?

35 It all starts with divisibility Divisibility is the core concept behind modular arithmetic. a b a b a Z, b Z a b k Z. b = ka Definition: divides, written as. For,. Examples: which of the following are true and which are false? 5 1 F 1 5 T

36 It all starts with divisibility Divisibility is the core concept behind modular arithmetic. a b a b a Z, b Z a b k Z. b = ka Definition: divides, written as. For,. Examples: which of the following are true and which are false? 5 1 F 1 5 T 25 5 F 5 25 T

37 It all starts with divisibility Divisibility is the core concept behind modular arithmetic. a b a b a Z, b Z a b k Z. b = ka Definition: divides, written as. For,. Examples: which of the following are true and which are false? 5 1 F 1 5 T 25 5 F 5 25 T 5 0 T 0 5 F

38 It all starts with divisibility Divisibility is the core concept behind modular arithmetic. a b a b a Z, b Z a b k Z. b = ka Definition: divides, written as. For,. Examples: which of the following are true and which are false? 5 1 F 1 5 T 25 5 F 5 25 T 5 0 T 0 5 F 2 3 F 3 2 F 18

39 Division theorem Division theorem For a Z, d Z with d > 0, there exist unique integers with such that. a = dq + r q, r 0 r < d a d q = a div d r = a mod d If we divide by, we get a unique quotient and nonnegative remainder. 19

40 Division theorem Division theorem For a Z, d Z with d > 0, there exist unique integers with such that. a = dq + r q, r 0 r < d a d q = a div d r = a mod d If we divide by, we get a unique quotient and nonnegative remainder. r 0 a < 0 mod is not % Note that even if, so. public class NotMod { public static void main(string args[]) { System.out.println("-5 mod 2 = 1."); System.out.println("-5 % 2 = " + (-5 % 2)) } } $javac NotMod.java $java -Xmx128M -Xms16M NotMod -5 mod 2 = % 2 = -1 19

41 Example: arithmetic mod 7 + a + 7 b = (a + b) mod a 7 b = (a b) mod

42 Modular arithmetic properties Congruence, addition, multiplication, proofs. 21

43 Congruence modulo a positive integer a b m a b (mod m) a, b, m Z m > 0 a b (mod m) m (a b) Definition: is congruent to modulo, written as For with, 22

44 Congruence modulo a positive integer a b m a b (mod m) a, b, m Z m > 0 a b (mod m) m (a b) Definition: is congruent to modulo, written as For with, Examples: what do these mean and when are they true? x 0 (mod 2) 1 19 (mod 5) y 2 (mod 7) 22

45 Congruence modulo a positive integer a b m a b (mod m) a, b, m Z m > 0 a b (mod m) m (a b) Definition: is congruent to modulo, written as For with, Examples: what do these mean and when are they true? x 0 (mod 2) True for every x 1 19 (mod 5) y 2 (mod 7) that is divisible by 2, i.e., even. 22

46 Congruence modulo a positive integer a b m a b (mod m) a, b, m Z m > 0 a b (mod m) m (a b) Definition: is congruent to modulo, written as For with, Examples: what do these mean and when are they true? x 0 (mod 2) True for every x 1 19 (mod 5) 1 19 = 20 y 2 (mod 7) that is divisible by 2, i.e., even. True because is divisible by 5. 22

47 Congruence modulo a positive integer a b m a b (mod m) a, b, m Z m > 0 a b (mod m) m (a b) Definition: is congruent to modulo, written as For with, Examples: what do these mean and when are they true? x 0 (mod 2) True for every x that is divisible by 2, i.e., even (mod 5) True because 1 19 = 20 is divisible by 5. y 2 (mod 7) y y = 2 + 7k k Z True for every of the form where. 22

48 Congruence and equality Congruence property Let a, b, m Z with m > 0. Then, if and only if. a b (mod m) a mod m = b mod m 23

49 Congruence and equality Congruence property Let a, b, m Z with m > 0. Then, if and only if. a b (mod m) Proof: Suppose that. a b (mod m) a mod m = b mod m a mod m = b mod m Suppose that. 23

50 Congruence and equality Congruence property Let a, b, m Z with m > 0. Then, if and only if. a b (mod m) Proof: Suppose that. a b (mod m) Then by definition of congruence. So for some by definition of divides. Therefore,. Taking both sides modulo, we get Suppose that. a mod m = b mod m m a b a b = km k Z a = b + km m a mod m = (b + km) mod m = b mod m. a mod m = b mod m 23

51 Congruence and equality Congruence property Let a, b, m Z with m > 0. Then, if and only if. a b (mod m) Proof: Suppose that. a b (mod m) Then by definition of congruence. So for some by definition of divides. Therefore,. Taking both sides modulo, we get Suppose that. a mod m = b mod m m a b a b = km k Z a = b + km m a mod m = (b + km) mod m = b mod m. a mod m = b mod m a = mq + (a mod m) b = ms + (b mod m) By the division theorem, and for some q, s Z. Then, a b = (mq + (a mod m)) (ms + (b mod m)) = m(q s) + (a mod m b mod m) = m(q s), since a mod m = b mod m. Therefore, and so. m (a b) a b (mod m) 23

52 mod m (mod m) The function vs the predicate mod m a Z a mod m {0, 1,, m 1} The function takes any and maps it to a remainder. In other words, remainder modulo mod m m places all integers that have the same into the same group (a.k.a. congruence class ). (mod m) a, b Z a b mod m The predicate compares and returns true if and only if and are in the same group according to the function. 24

53 Modular addition property Modular addition property Let m be a positive integer ( m Z with m > 0). If and, then. a b (mod m) c d (mod m) a + c b + d (mod m) 25

54 Modular addition property Modular addition property Let m be a positive integer ( m Z with m > 0). If and, then. a b (mod m) c d (mod m) a + c b + d (mod m) Proof: Suppose that a b (mod m) and c d (mod m). By the definition of congruence, there are k and j such that a b = km and c d = jm. Adding these equations together, we get (a + c) (b + d) = m(j + k). Reapplying the definition of congruence, we get that. (a + c) (b + d) (mod m) 25

55 Modular multiplication property Modular multiplication property Let m be a positive integer ( m Z with m > 0). If and, then. a b (mod m) c d (mod m) ac bd (mod m) 26

56 Modular multiplication property Modular multiplication property Let m be a positive integer ( m Z with m > 0). If and, then. a b (mod m) c d (mod m) ac bd (mod m) Proof: Suppose that a b (mod m) and c d (mod m). By the definition of congruence, there are k and j such that a b = km and c d = jm. So, a = km + b and c = jm + b. Multiplying these equations together, we get ac = (km + b)(jm + d) = kj m 2 + kmd + bjm + bd. Rearranging gives us ac bd = m(kjm + kd + bj). Reapplying the definition of congruence, we get that. ac bd (mod m) 26

57 Example: a proof using modular arithmetic n Z n 2 0 (mod 4) n 2 1 (mod 4) Let, and prove that or. 27

58 Example: a proof using modular arithmetic n Z n 2 0 (mod 4) n 2 1 (mod 4) Let, and prove that or. Let s look at a few examples: 0 2 = 0 0 (mod 4) 1 2 = 1 1 (mod 4) 2 2 = 4 0 (mod 4) 3 2 = 9 1 (mod 4) 4 2 = 16 0 (mod 4) 27

59 Example: a proof using modular arithmetic n Z n 2 0 (mod 4) n 2 1 (mod 4) Let, and prove that or. Let s look at a few examples: 0 2 = 0 0 (mod 4) 1 2 = 1 1 (mod 4) 2 2 = 4 0 (mod 4) 3 2 = 9 1 (mod 4) 4 2 = 16 0 (mod 4) It looks like n 0 (mod 2) n 2 0 (mod 4) n 1 (mod 2) n 2 1 (mod 4) 27

60 Example: a proof using modular arithmetic n Z n 2 0 (mod 4) n 2 1 (mod 4) Let, and prove that or. Let s look at a few examples: 0 2 = 0 0 (mod 4) 1 2 = 1 1 (mod 4) 2 2 = 4 0 (mod 4) 3 2 = 9 1 (mod 4) 4 2 = 16 0 (mod 4) It looks like n 0 (mod 2) n 2 0 (mod 4) n 1 (mod 2) n 2 1 (mod 4) Proof by cases: Case 1 ( is even): n n Case 2 ( is odd): 27

61 Example: a proof using modular arithmetic n Z n 2 0 (mod 4) n 2 1 (mod 4) Let, and prove that or. Let s look at a few examples: 0 2 = 0 0 (mod 4) 1 2 = 1 1 (mod 4) 2 2 = 4 0 (mod 4) 3 2 = 9 1 (mod 4) 4 2 = 16 0 (mod 4) It looks like n 0 (mod 2) n 2 0 (mod 4) n 1 (mod 2) n 2 1 (mod 4) Proof by cases: Case 1 ( is even): n n 0 (mod 2) Suppose. Then for some integer k. So n 2 = (2k) 2 = 4k 2. Therefore, by definition of congruence,. n Case 2 ( is odd): n 2 n = 2k 0 (mod 4) 27

62 Example: a proof using modular arithmetic n Z n 2 0 (mod 4) n 2 1 (mod 4) Let, and prove that or. Let s look at a few examples: 0 2 = 0 0 (mod 4) 1 2 = 1 1 (mod 4) 2 2 = 4 0 (mod 4) 3 2 = 9 1 (mod 4) 4 2 = 16 0 (mod 4) It looks like n 0 (mod 2) n 2 0 (mod 4) n 1 (mod 2) n 2 1 (mod 4) Proof by cases: Case 1 ( is even): n n 0 (mod 2) Suppose. Then for some integer k. So n 2 = (2k) 2 = 4k 2. Therefore, by definition of congruence,. n Case 2 ( is odd): n 2 n = 2k 0 (mod 4) n 1 (mod 2) n = 2k + 1 Suppose. Then for some integer k. So n 2 = (2k + 1 ) 2 = 4 k 2 + 4k + 1 = 4( + k) + 1. Therefore, by definition of congruence,. n 2 k 2 1 (mod 4) 27

63 Modular arithmetic and integer representations Unsigned, sign-magnitude, and two s complement representation. 28

64 Unsigned integer representation x n x = n 1 i=0 b i2 i b i {0, 1} bn 1 b 2 b 1 b 0 Represent integer as a sum of powers of 2: If where each, then the representation is. 29

65 Unsigned integer representation x n x = n 1 i=0 b i2 i b i {0, 1} bn 1 b 2 b 1 b 0 Represent integer as a sum of powers of 2: If where each, then the representation is. Examples: 99 = = n = 8 99 = = So for : 29

66 Unsigned integer representation x n x = n 1 i=0 b i2 i b i {0, 1} bn 1 b 2 b 1 b 0 Represent integer as a sum of powers of 2: If where each, then the representation is. Examples: 99 = = n = 8 99 = = So for : This works for unsigned integers. How do we represented signed integers? 29

67 Sign-magnitude integer representation 2 n 1 < x < 2 n 1 x n If, represent with bits as follows: Use the first bit as the sign (0 for positive and 1 for negative), and the remaining bits as the (unsigned) value. n 1 Examples: 99 = = n = 8 99 = = So for : 30

68 Sign-magnitude integer representation 2 n 1 < x < 2 n 1 x n If, represent with bits as follows: Use the first bit as the sign (0 for positive and 1 for negative), and the remaining bits as the (unsigned) value. n 1 Examples: 99 = = n = 8 99 = = = So for : The problem with this representation is that our standard arithmetic algorithms no longer work, e.g., adding the representation of -18 and 99 doesn t give the representation of

69 Two s complement integer representation x n Represent with bits as follows: If 0 x < 2 n 1, use the n-bit unsigned representation of x. If, use the -bit unsigned representation of. 2 n 1 x < 0 n x 2 n 31

70 Two s complement integer representation x n Represent with bits as follows: If 0 x < 2 n 1, use the n-bit unsigned representation of x. If, use the -bit unsigned representation of. 2 n 1 x < 0 n x 2 n Key property: Two s complement representation of any number so arithmetic works. y mod 2 n mod 2 n y is equivalent to 31

71 Two s complement integer representation x n Represent with bits as follows: If 0 x < 2 n 1, use the n-bit unsigned representation of x. If, use the -bit unsigned representation of. 2 n 1 x < 0 n x 2 n Key property: Two s complement representation of any number so arithmetic works. y mod 2 n mod 2 n y is equivalent to Examples: So for : 99 = = = = 238 = = n = 8 99 = = =

72 Computing the two s complement representation 2 n 1 x < 0 x n 2 n x For, is represented using the -bit unsigned representation of. Here is an easy way to compute this value: n n Compute the -bit unsigned representation of. Flip the bits of x to get the representation of 2 1 x. This works because the string of all 1 s represents 2 n 1. Add 1 to get. 2 n x x 32

73 Applications of modular arithmetic Hashing, pseudo-random numbers, ciphers. 33

74 Hashing Problem: We want to map a small number of data values from a large domain {0, 1,, M 1} into a small set of locations {0, 1,, n 1} to be able to quickly check if a value is present. Solution: Compute hash(x) = x mod p for a prime p close to n. Or, compute for a prime close to. hash(x) = ax + b mod p p n This approach depends on all of the bits of data the data. Helps avoid collisions due to similar values. But need to manage them if they occur. 34

75 Pseudo-random number generation Linear Congruential method x n+1 = (a x n + c) mod m, a, c, m x 0 Choose random and produce a long sequences of s. x n 35

76 Simple ciphers Ceasar or shi! cipher: Treat letters as numbers: A = 0, B = 1, f (p) = (p + k) mod 26 f 1 (p) = (p k) mod 26 More general version: f (p) = (ap + b) mod 26 f 1 (p) = ( a 1 (p b)) mod 26 36

77 Summary Sets can be represented efficiently using bitvectors. This representation is used heavily in the real world. With this representation, set operations reduce to fast bitwise operations. Modular arithmetic is arithmetic over a finite domain. Key notions are divisibility and congruency. mod m Modular arithmetic is the basis of computing. Used with two s complement representation to implement computer arithmetic. Also used in hashing, pseudo-random number generation, and cryptography. 37