1 / 35

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

jeb
Download Presentation

CSci 2011 Discrete Mathematics Lecture 15

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related