CSE 311: Foundations of Computing. Lecture 10: Set Operations & Representation, Modular Arithmetic

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CSE 311: Foundations of Computing Lecture 10: Set Operations & Representation, Modular Arithmetic

Definitions A and B are equal if they have the same elements A = B x (x A x B) A is a subset of B if every element of A is also in B A B x (x A x B) Note:

Set Operations A B = { x x A x B } A B = { x x A x B } A \ B = { x x A x B } Union Intersection Set Difference A = {1, 2, 3} B = {3, 5, 6} C = {3, 4} QUESTIONS Using A, B, C and set operations, make [6] = A B C {3} = A B = A C {1,2} = A \ B = A \ C

More Set Operations A B = { x x A x B } ഥA = x x A = x (x A) (with respect to universe U) Symmetric Difference Complement A = {1, 2, 3} B = {1, 2, 4, 6} Universe: U = {1, 2, 3, 4, 5, 6} A B = {3, 4, 6} ഥA = {4,5,6}

It s Boolean algebra again Definition for based on A B = { x x A x B } Definition for based on A B = { x x A x B } Complement works like ഥA = x (x A)

De Morgan s Laws Proof technique: To show C = D show x C x D and x D x C

Distributive Laws A B C = A B A C A B C = A B A C A B A B C C

A Simple Set Proof Prove that for any sets A and B we have (A B) A Remember the definition of subset? X Y x (x X x Y)

A Simple Set Proof Prove that for any sets A and B we have (A B) A Remember the definition of subset? X Y x (x X x Y) Proof: Let A and B be arbitrary sets and x be an arbitrary element of A B. Then, by definition of A B, x A and x B. It follows that x A, as required.

Power Set Power Set of a set A = set of all subsets of A P A = { B B A } e.g., let Days={M,W,F} and consider all the possible sets of days in a week you could ask a question in class P(Days)=? P( )=?

Power Set Power Set of a set A = set of all subsets of A P A = { B B A } e.g., let Days={M,W,F} and consider all the possible sets of days in a week you could ask a question in class P(Days)= M, W, F, M, W, M, F, W, F, M, W, F, P( )={ }

Cartesian Product A B = { a, b : a A, b B } R R is the real plane. You ve seen ordered pairs before. These are just for arbitrary sets. 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} =

Representing Sets Using Bits Suppose universe U is {1,2,, n} Can represent set B U as a vector of bits: b 1 b 2 b n where b i = 1 when i B b i = 0 when i B Called the characteristic vector of set B Given characteristic vectors for A and B What is characteristic vector for A B? A B?

UNIX/Linux File Permissions ls l drwxr-xr-x... Documents/ -rw-r--r--... file1 Permissions maintained as bit vectors Letter means bit is 1 -- means bit is 0.

Bitwise Operations 01101101 Java: z=x y 00110111 01111111 00101010 Java: z=x&y 00001111 00001010 01101101 Java: z=x^y 00110111 01011010

A Useful Identity If x and y are bits: (x y) y =? What if x and y are bit-vectors?

Private Key Cryptography Alice wants to communicate message secretly to Bob so that eavesdropper Eve who hears their conversation cannot tell what Alice s message is. Alice and Bob can get together and privately share a secret key K ahead of time.

One-Time Pad Alice and Bob privately share random n-bit vector K Eve does not know K Later, Alice has n-bit message m to send to Bob Alice computes C = m K Alice sends C to Bob Bob computes m = C K which is (m K) K Eve cannot figure out m from C unless she can guess K

Russell s Paradox S = { x x x } Suppose that S S

Russell s Paradox S = { x x x } Suppose that S S. Then, by definition of S, S S, but that s a contradiction. Suppose that S S. Then, by definition of the set S, S S, but that s a contradiction, too. This is reminiscent of the truth value of the statement This statement is false.

Number Theory (and applications to computing) Branch of Mathematics with direct relevance to computing Many significant applications Cryptography Hashing Security Important tool set

Modular Arithmetic Arithmetic over a finite domain In computing, almost all computations are over a finite domain

I m ALIVE! public class Test { final static int SEC_IN_YEAR = 364*24*60*60*100; public static void main(string args[]) { System.out.println( I will be alive for at least + SEC_IN_YEAR * 101 + seconds. ); } }

I m ALIVE! public class Test { final static int SEC_IN_YEAR = 364*24*60*60*100; public static void main(string args[]) { System.out.println( I will be alive for at least + SEC_IN_YEAR * 101 + seconds. ); } } Prints : I will be alive for at least -186619904 seconds.

Divisibility Definition: a divides b For a Z, b Z with a 0: a b k Z (b = ka) Check Your Understanding. Which of the following are true? 5 1 25 5 5 0 3 2 1 5 5 25 0 5 2 3

Divisibility Definition: a divides b For a Z, b Z with a 0: a b k Z (b = ka) Check Your Understanding. Which of the following are true? 5 1 25 5 5 0 3 2 5 1 iff 1 = 5k 25 5 iff 5 = 25k 5 0 iff 0 = 5k 3 2 iff 2 = 3k 1 5 5 25 0 5 2 3 1 5 iff 5 = 1k 5 25 iff 25 = 5k 0 5 iff 5 = 0k 2 3 iff 3 = 2k

Division Theorem Division Theorem For a Z, d Z with d > 0 there exist unique integers q, r with 0 r < d such that a = dq + r. To put it another way, if we divide d into a, we get a unique quotient q = a div d and non-negative remainder r = a mod d Note: r 0 even if a < 0. Not quite the same as a%d.

Division Theorem Division Theorem For a Z, d Z with d > 0 there exist unique integers q, r with 0 r < d such that a = dq + r. To put it another way, if we divide d into a, we get a unique quotient q = a div d and non-negative remainder r = a mod d public class Test2 { public static void main(string args[]) { int a = -5; int d = 2; System.out.println(a % d); } } Note: r 0 even if a < 0. Not quite the same as a%d.

Arithmetic, mod 7 a + 7 b = (a + b) mod 7 a 7 b = (a b) mod 7 + 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5 X 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 1 3 5 3 0 3 6 2 5 1 4 4 0 4 1 5 2 6 3 5 0 5 3 1 6 4 2 6 0 6 5 4 3 2 1

Modular Arithmetic Check Your Understanding. What do each of these mean? When are they true? x 0 (mod 2) Definition: a is congruent to b modulo m For a, b, m Z with m > 0 a b mod m m (a b) -1 19 (mod 5) y 2 (mod 7)

Modular Arithmetic Check Your Understanding. What do each of these mean? When are they true? x 0 (mod 2) -1 19 (mod 5) y 2 (mod 7) Definition: a is congruent to b modulo m For a, b, m Z with m > 0 a b mod m m (a b) This statement is the same as saying x is even ; so, any x that is even (including negative even numbers) will work. This statement is true. 19 - (-1) = 20 which is divisible by 5 This statement is true for y in {..., -12, -5, 2, 9, 16,...}. In other words, all y of the form 2+7k for k an integer.

Modular Arithmetic: A Property Let a, b, m be integers with m > 0. Then, a b (mod m) if and only if a mod m = b mod m. Suppose that a b (mod m). Then, m (a b) by definition of congruence. So, a b = km for some integer k by definition of divides. Therefore, a = b+km. Taking both sides modulo m we get: a mod m=(b+km) mod m = b mod m. Suppose that a mod m = b mod m. By the division theorem, a = mq + (a mod m) and b = ms + (b mod m) for some integers q,s. Then, a b = (mq + (a mod m)) (mr + (b mod m)) = m(q r) + (a mod m b mod m) = m(q r) since a mod m = b mod m Therefore, m (a-b) and so a b (mod m).

Modular Arithmetic: A Property Let a, b, m be integers with m > 0. Then, a b (mod m) if and only if a mod m = b mod m. Suppose that a b (mod m). Then, m (a b) by definition of congruence. So, a b = km for some integer k by definition of divides. Therefore, a = b + km. Taking both sides modulo m we get: a mod m = (b + km) mod m = b mod m. mod m = b Suppose that a mod m = b mod m. By the division theorem, a = mq + (a mod m) and b = ms + (b mod m) for some integers q,s. 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, m (a b) and so a b (mod m).