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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: [email protected] Put [2011] in front. Check class web site

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admin
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: [email protected]
    • Put [2011] in front.
  • Check class web site
    • Read syllabus, Use forum.

CSci 2011

recap
Recap
  • Propositional operation summary
  • Check translation
  • Definition
    • Tautology, Contradiction, logical equicalence

CSci 2011

recap1
Recap

CSci 2011

recap2
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

recap3
Recap

CSci 2011

recap4
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

what is a set
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

functions
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

recap5
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

recap6
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

division algorithm
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

modular arithmetic
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

modular arithmetic1
Modular Arithmetic
  • Theorem: Let m be a positive integer. a  b mod m iff  k such that a = b + km.

CSci 2011

slide16
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

slide17
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

caesar cipher
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

ch3 5

ch3.5

Yongdae Kim

CSci 2011

showing a number is prime
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

primes are infinite
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

the prime number theorem
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

greatest common divisor
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

relative primes
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

pairwise relative prime
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

more on gcd s
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

least common multiple
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

lcm and gcd theorem
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

example proof
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

ch 3 6

ch 3.6

Yongdae Kim

CSci 2011

representation of integers
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

eucledean algorithm
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

eucledian algorithm
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

square and multiply
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

square and multiply1
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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