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CSci 2011 Discrete Mathematics Lecture 15

CSci 2011 Discrete Mathematics Lecture 15. Yongdae Kim. Admin. Due dates and quiz Groupwork 7 is due on Oct 26 th . Homework 4 is due on Oct 28 th . Quiz 4: Nov 4 th . 1 page cheat sheet is allowed. E-mail: csci2011-help@cselabs.umn.edu Put [2011] in front. Check class web site

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CSci 2011 Discrete Mathematics Lecture 15

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  1. CSci 2011 Discrete MathematicsLecture 15 Yongdae Kim CSci 2011

  2. Admin • Due dates and quiz • Groupwork 7 is due on Oct 26th. • Homework 4 is due on Oct 28th. • Quiz 4: Nov 4th. • 1 page cheat sheet is allowed. • E-mail: csci2011-help@cselabs.umn.edu • Put [2011] in front. • Check class web site • Read syllabus, Use forum. CSci 2011

  3. Recap • Propositional operation summary • Check translation • Definition • Tautology, Contradiction, logical equicalence CSci 2011

  4. Recap CSci 2011

  5. Recap • Quantifiers • Universal quantifier: x P(x) • Negating quantifiers • ¬x P(x) = x ¬P(x) • ¬x P(x) = x ¬P(x) xy P(x, y) • Nested quantifiers • xy P(x, y): “For all x, there exists a y such that P(x,y)” • xy P(x,y): There exists an x such that for all y P(x,y) is true” • ¬x P(x) = x ¬P(x), ¬x P(x) = x ¬P(x) CSci 2011

  6. Recap CSci 2011

  7. Recap • p→q • Direct Proof: Assume p is true. Show that q is also true. • Indirect Proof: Assume ¬q is true. Show that p is true. • Proof by contradiction • Proving p: Assume p is not true. Find a contradiction. • Proving p→q • ¬(p→q)   (p  q)  p  ¬q • Assume p is tue and q is not true. Find a contradiction. • Proof by Cases: [(p1p2…pn)q]  [(p1q)(p2q)…(pnq)] • If and only if proof: pq (p→q)(q→p) • Existence Proof: Constructive vs. Non-constructive Proof • Uniqueness: 1) Show existence 2) find contradiction if not unique • Forward and backward reasoning • Counterexample CSci 2011

  8. What is a set? • A set is a unordered collection of “objects” • set-builder notation: D = {x | x is prime and x > 2} • a S if an elementa is an element of a set S. a S if not. • U is the universal set. • empty (or null) set  = { }, if a set has zero elements. • S  T if  x (x  S  x  T), S = T if S  T and if T  S. • S  T if S is a subset of T, and S is not equal to T. • The cardinality of a set, |A| is the number of elements in a set. • The power set of S, P(S), is the set of all the subsets of S. • If a set has n elements, then the power set will have 2n elements • A x B = { (a,b) | a A and b  B } • A U B = { x | x  A or x  B } • A ∩ B = { x | x  A and x  B } • two sets are disjoint if their intersection is the empty set • A - B = { x | x  A and x  B } • Ac = { x | x  A } CSci 2011

  9. Set identities CSci 2011

  10. Functions • A function takes an element from a set and maps it to a UNIQUE element in another set • Domain, co-domain, range • A function f is one-to-one (injection) if f(x) = f(y) implies x = y. • A function f is onto (surjection) if for all y  C, there exists x  D such that f(x)=y. • A function f is bijection if it is 1-1 and onto. • A function f is an identity function if f(x)=x. • Composition of functions: (f°g)(x) = f(g(x)). • f-1 is an inverse function of f if (f°f-1)(x)=(f-1°f)(x)=x. CSci 2011

  11. Recap • Useful functions • x = n if and only if n ≤ x < n+1 • x = n if and only if n-1 < x ≤ n • round(x) =  x+0.5  • n! = n * (n-1) * (n-2) * … * 2 * 1 • Sequence and Series • Arithmetic Progression: a, a+d, a+2d, …, a+nd, … • an = a + (n-1) d • Geometric Progression: a, ar, ar2, ar3, …, arn-1, … • an = arn-1 • 1 + 2 + 3 + … + n = n(n+1)/2 n 1 ar a + - n if r 1  j  ar r 1 = - j 0 ( n 1 ) a if r 1 = + = CSci 2011

  12. Recap • For finite and infinite sets, two sets A and B have the same cardinality if there is a one-to-one correspondence from A to B • Countably infinite: elements can be listed • Anything that has the same cardinality as the integers • Uncountably infinite: elements cannot be listed • An algorithm is “a finite set of precise instructions for performing a computation or for solving a problem” • Complexity • f(x)O(g(x)) if kR, cR, xR, xk  0f(x)|cg(x)|. • f(x)(g(x)) if kR, cR, xR, xk  f(x)|  c g(x)|. • f(x)(g(x)) if f(x)O(g(x)) and f(x)(g(x)). Equivalently, if kR, c1,c2R, xR, xk  c1 |g(x)|  |f(x)|  c2 |g(x)|. • Def: a | b (a (a0) divides b) if  c such that b = ac. • Theorem: Let a, b, c be integers. Then • a | b, a | c  a | (b + c). • a | b  a | bc  c  Z. • a | b, b | c  a | c. CSci 2011

  13. Division Algorithm • Theorem: Let a be an integer and d a positive integer. Then there exist unique q and r, with 0r<d, such that a = dq + r. • Definition • q = a div d, quotient • r = a mod d, remainder • 101 = 7  14 + 3 • -11 = 7  (-2) + 3 CSci 2011

  14. Modular Arithmetic • Def: a, b: integers, m: positive integer, a is congruent to b modulo m if m | (a - b). • a  b mod m • a  b (mod m) iff a mod m = b mod m. • 17  12 mod 5, 17 mod 5 = 2, 12 mod 5 = 2 • -3  17 mod 10, -3 mod 10 = 7, 17 mod 10 = 7 CSci 2011

  15. Modular Arithmetic • Theorem: Let m be a positive integer. a  b mod m iff  k such that a = b + km. CSci 2011

  16. More… • Theorem: Let m be a positive integer. If a  b mod m and c  d mod m, then • a + c  b + d mod m. • ac  bd mod m. CSci 2011

  17. More… • Prove or Disprove • If ac  bc (mod m), where a, b, c, m Z (with m  2), then a  b (mod m). • a, b, c, d, m Z, c, d > 0, m  2, if a  b (mod m) and c  d (mod m) then ac bd mod m. • a = 2, b = 5, c = 4, d = 1, m = 3 CSci 2011

  18. Caesar Cipher • Alphabet to number: a~0, b~1, … , z~25. • Encryption: EK(x) = x + K mod 26. • Decryption: • Caesar used K = 3. • If you don’t know K and you have a secret text, would you be able to find K? How? CSci 2011

  19. ch3.5 Yongdae Kim CSci 2011

  20. Showing a number is prime • Show that 113 is prime • Solution • The only prime factors less than 113 = 10.63 are 2, 3, 5, and 7 • Neither of these divide 113 evenly • Thus, by the fundamental theorem of arithmetic, 113 must be prime CSci 2011

  21. Primes are infinite • Theorem (Euclid): There are infinitely many prime numbers (Proof) Proof by contradiction Assume there are a finite number of primes List them as follows: p1, p2 …, pn. Consider the number q = p1p2 … pn + 1 Since we have only finite number of primes and q is not one of them, pi should divide q for some i. Obviously pi | p1p2 … pn. Recall that a | b, a | c  a | b + c. Therefore, pi | (q - p1p2 … pn) = 1. (*) CSci 2011

  22. The prime number theorem • The radio of the number of primes not exceeding x and x/ln(x) approaches 1 as x grows without bound • Rephrased: the number of prime numbers less than x is approximately x/ln(x) • When x = 2512, # of primes = 2512/512  2503 CSci 2011

  23. Greatest common divisor • The greatest common divisor of two integers a and b is the largest integer d such that d | a and d | b • Denoted by gcd(a,b) • Examples • gcd (24, 36) = 12 • gcd (17, 22) = 1 • gcd (100, 17) = 1 CSci 2011

  24. Relative primes • Two numbers are relatively prime if they don’t have any common factors (other than 1) • Rephrased: a and b are relatively prime if gcd (a,b) = 1 • gcd (25, 39) = 1, so 25 and 39 are relatively prime CSci 2011

  25. Pairwise relative prime • A set of integers a1, a2, … an are pairwise relatively prime if, for all pairs of numbers, they are relatively prime • Formally: The integers a1, a2, … an are pairwise relatively prime if gcd(ai, aj) = 1 whenever 1 ≤ i < j ≤ n. • Example: are 10, 17, and 21 pairwise relatively prime? • gcd(10,17) = 1, gcd (17, 21) = 1, and gcd (21, 10) = 1 • Thus, they are pairwise relatively prime • Example: are 10, 19, and 24 pairwise relatively prime? • Since gcd(10,24) ≠ 1, they are not CSci 2011

  26. More on gcd’s • Given two numbers a and b, rewrite them as: • Example: gcd (120, 500) • 120 = 23*3*5 = 23*31*51 • 500 = 22*53 = 22*30*53 • Then compute the gcd by the following formula: • Example: gcd(120,500) = 2min(3,2)3min(1,0)5min(1,3) = 223051 = 20 CSci 2011

  27. Least common multiple • The least common multiple of the positive integers a and b is the smallest positive integer that is divisible by both a and b. • Denoted by lcm (a, b) • Example: lcm(10, 25) = 50 • What is lcm (95256, 432)? • 95256 = 233572, 432=2433 • lcm (233572, 2433) = 2max(3,4)3max(5,3)7max(2,0) = 243572 = 190512 CSci 2011

  28. lcm and gcd theorem • Let a and b be positive integers. Then a*b = gcd(a,b) * lcm (a, b) • Example: gcd (10,25) = 5, lcm (10,25) = 50 • 10*25 = 5*50 • Example: gcd (95256, 432) = 216, lcm (95256, 432) = 190512 • 95256*432 = 216*190512 CSci 2011

  29. Example Proof • Prove or disprove that n2 - 79n + 1601 is prime, whenever n is a positive integer. (Disprove) When n = 1601, n2 - 79n + 1601 = 1601 (1601 - 79 + 1) • Prove or disprove that p_1p_2…p_n+1 is a prime, whenever n is a positive integer. 2*3*5*7*11*13+1 = 30031 = 59 * 509 CSci 2011

  30. ch 3.6 Yongdae Kim CSci 2011

  31. Representation of Integers • Positive integer n can be uniquely written as n = akbk+ ak-1bk-1+ … + a1b + a0 • k  N, 0 < ai < b • The base b expansion of n is denoted by (akak-1… a1a0)b. • b = 2 Binary representation • b = 16 Hexadecimal representation • (245)8 = 2 82 + 4 8 + 5 = 165 = (11110101)2 = (F5)16 CSci 2011

  32. Eucledean Algorithm • Let a = b q + r, where a, b, q, r be integers. Then gcd (a, b) = gcd (b, r). • Proof on book is not enough! CSci 2011

  33. Eucledian Algorithm • procedure gcd (a, b: positive integer) • x := a, y := b • while y  0 • r := x mod y • x := y • y := r • Return r CSci 2011

  34. Square-and-Multiply • 213mod 17 = 223+22+1mod 17 =(((22)2)2) ((22)2) 2 • INPUT: a  Zn, and k < n where k = Sti=0ki2i • OUTPUT: ak mod n. • Algorithm • Set b = 1. If k = 0 then return(b). • Set A = a. • If k0 = 1 then set b = a. • For i from 1 to t do the following: • Set A = A2 mod n. • If ki = 1 then set b = A b mod n. • Return(b). CSci 2011

  35. Square-and-Multiply • a = 2 • k = 13 • n = 17 • k = Sti=0ki2i • Set A = a. • If k0 = 1 then set b = a. • For i from 1 to t • Set A = A2 mod n. • If ki = 1, set b = Ab mod n. CSci 2011

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