1. Lecture 2: Math Review and Asymptotic Analysis

2. Common Math Functions Floors and Ceilings: x-1 < └ x ┘ < x < ┌ x ┐ < x+1. Modular Arithmetic: a mod n = a – └ a/n ┘ n. Factorials: n! = 1 if n = 0 n! = n * (n-1)! If n > 0. n! < n n

3. Exponentials a 0 = 1 a 1 = a a -1 = 1/a (a m ) n = a mn a m a n = a m+n 0 0 = 1..when convenient

4. Logarithm Rules lg(n) = log 2 (n)Binary logarithm ln(n) = log e (n)Natural logarithm lg k (n) = (lg(n))kExponentiation lglg(n) = lg(lg(n))Composition log b (mn) = log b (m) + log b (n) log b (m/n) = log b (m) – log b (n) log b (m n ) = n · log b (m) log b (x) = log d (x) / log d (b)Change of Base Rule ln(x) = log d (x) / log d ( e ) log(1) = 0, lg(1) = 0 log(10) = 1, lg(2) = 1 log(100) = 2, lg(4) = 2

5. Summation Formulas Arithmetic series: Geometric series: –Special case if x < 1: Harmonic series: Other:

6. Fibonacci Numbers F 0 = 0 F 1 = 1 F i = F i-1 + F i-2

7. Proofs A proof is a logical argument that, assuming certain axioms, some statement is necessarily true. A few common proof techniques: –Direct Proof –Proof by Induction –Proof by Contradiction –Proof by Contraposition –Proof by Exhaustion –Proof by Counterexample

8. Direct Proof A conclusion is established by logically combining earlier definitions, axioms and theorems

9. Direct Proof: Example Pythagorean Theorem: For a right triangle, with sides (legs) a and b, and hypotenuse c, c²=a²+b². Proof (courtesy Legendre): 1.ABC, CBX, and ACX are similar triangles. 2.Corresponding parts of similar triangles are proportional: a/x=c/a → a²=cx b/(c-x)=c/b → b²=c²-cx → c²=cx+b² 3.Substituting a² for cx, we find c²=a²+b². CA B X x c b a

10. Proof by Induction A base case is proven, and an induction rule used to prove an series (possibly infinite) of other cases.

11. Proof by Induction: Example Theorem: 1 + 2 + … + n = n(n+1) / 2 Proof (courtesty Gauss): 1.Base Case: If n = 0, 0 = 0(0+1) / 2. 2.Inductive Hypothesis: Assume the statement is true for n = m, i.e.,1 + 2 + … + m = m(m+1) / 2 3.Inductive Step: Show that n = m + 1 holds: 1 + 2 + … + m + m + 1 = (m+1)(m+1+1) / 2 Since we know the theorem holds for n=m, we subtract those terms from each side.. m + 1 = (m 2 + 3m + 2) / 2 - (m 2 + m) / 2 m + 1 = (2m + 2)/2

12. Proof by Contradiction One shows that if a statement were false, a logical contradiction occurs, and thus the statement must be true.

13. Proof by Contradiction: Example Theorem: There exists an infinite number of prime numbers. Proof (courtesy of Euclid): 1.Assume that there are a finite number of primes. 2.Then there is a largest prime, p. Consider the number q = (2x3x5x7x...xp)+1. q is one more than the product of all primes up to p. q > p. And, q is not divisible by any prime up to p. For example, it is not divisible by 7, as it is one more than a multiple of 7. 3.If q is not a prime and it is not divisible by any prime p. 4.That is a contradiction, as p was assumed to be the largest prime. So, there is no largest prime. In other words, there are infinitely many primes.

14. Proof by Contraposition In order to prove A→B, prove ¬B → ¬A.

15. Proof by Contraposition Theorem: For all natural numbers: If n 2 is even, then n is even as well. Proof: 1.Contraposition: If n is odd, then n 2 is odd. 2.Let n = 2q+1, where q is a natural number. Then n 2 = (2q+1) 2 = 4q 2 +4q+1 = 2(2q 2 +2q)+1. 3.Let p = 2q 2 +2q, then n 2 = 2p+1. Since 2p must be even, n 2 must be odd proving the contraposition.

16. Proof by Exhaustion The conclusion is established by dividing the problem into a finite number of exhaustive cases and proving each one separately.

17. Proof by Exhaustion: Example Theorem: Every cube number is either a multiple of 9 or is 1 more or 1 less than a multiple of 9. Proof: Each cube number is the cube of an integer n. n is either a multiple of 3, or is 1 more or 1 less than a multiple of 3. The following 3 cases are exhaustive: 1.Case 1: If n is a multiple of 3 then the cube of n is a multiple of 27, and so certainly a multiple of 9. 2.Case 2: If n is 1 more than a multiple of 3 then the cube of n is 1 more than a multiple of 9. 3.Case 3: If n is 1 less than a multiple of 3 then the cube of n is 1 less than a multiple of 9.

18. Proof by Counterexample Given an assertion of the form : For all x, P(x) is true, disprove it by showing that there is a c such that ¬P(c) is true.

19. Proof by Counterexample: Example Conjecture: The square root of every integer is irrational. Counterexample: Given the function f(x) = √n, let n = 4. Clearly, √4 = 2 and 2 is rational.

20. Loop Invariants Another proof technique for proving correctness for loops is to prove these properties hold for a loop invariant (we saw an example last time): Initialization: It is true prior to the first iteration. Maintenance: It is true before an iteration of the loop and remains true before the next iteration. Termination: When the loop terminates, the invariant yields a useful property which helps show the algorithm is correct.

21. Loop Invariants: Example Show that the following pseudocode correctly determines if x is in the array A: for i=0 to len(A): if (x == A[i]) return true return false Loop Invariant: At the start of each for loop, the subsequence A[0]..A[i-1] does not contain the variable x.

22. Loop Invariants: Example Initialization: The loop invariant is true at initialization because the subsequence A[0]..A[-1] is an empty subsequence and by definition x is not contained in the subsequence. Maintenance: As we consider the (k-1)st element of A, if it is in A we return true. If it is not we continue. Therefore if we consider the kth element the (k-1)st element must not have been equal to x and subsequently any earlier element. Termination: The loop terminates when i=len(A) or when x is found in the array. From the maintenance property we know that if i=len(A), then A[0]..A[len(A)-1] does not contain x. If i != len(A) then the loop terminated because x was found.

23. Danger: Proof by Example Conjecture: For arbitrary sets A and B, the sets A \ B, B \ A, and A ∩ B are pair wise disjoint. Proof: Let A = {1,3,5,7} and B = {2,4,6,8}. Then... NO! In general, an example is almost never a proof. * Not to be confused with proof by construction (which you are unlikely to use in this class).

24. Danger: Proof by Example Conjecture: The Fibonacci sequence, F(x), is always odd. Proof: Consider x=2; F(2)=1. Or consider x=10; F(10)=55, etc.. Clearly the Fibonacci sequence is always odd. NO! Even if the conjecture were true this is not a correct proof.

25. Growth of Functions Review 1Constant log nLogarithmic nLinear n log n n 2 Quadratic n 3 Cubic 2 n Exponential

26. Asymptotic Notation How do we describe asymptotic efficiency? O(g(n)): g(n) is an asymptotic upper bound o(g(n)): g(n) is an upper bound that is not asymptotically tight Ω(g(n)): g(n) is an asymptotic lower bound ω(g(n)): g(n) is a lower bound that is not asymptotically tight Θ(g(n)): g(n) is an asymptotically tight bound

27. O-Notation O(g(n)) is pronounced “Big-oh of g of n” or sometimes just “Oh of g of n” We use this notation when we can only find an upper bound on a function. O(g(n)) = { f(n): there exists some positive constants c and n 0 such that: 0 n 0 }.

28. O-Notation Example Prove: ½ n 2 – 3n = O(n 2 ) Proof: 1.We must choose c and n 0 s.t.: ½ n 2 – 3n n 0. 2.Dividing by n 2 yields: ½ – 3/n < c 3.This holds if we choose c > ½ and n 0 > 7.

29. o-Notation o(g(n)) is pronounced “Little-oh of g of n” O(g(n)) may or may not be tight. We use o(g(n)) to specify that it is not asymptotically tight. o(g(n)) = { f(n): for any positive constant c>0 there exists a positive constant n 0 >0 such that: 0 n 0 }.

30. o-Notation Example For example, 2n = o(n 2 ) because n 2 is not an asymptotically tight upper bound for 2n. It is, however, an upper bound. But, 2n 2 ≠ o(n 2 ) because n 2 would be an asymptotically tight upper bound.

31. o-Notation Example Prove: 2n = o(n 2 ) Proof: 1.Let n 0 = 3/c. 2.Substituting n 0 for n in 2n < cn 2 : 2(3/c) < c(3/c) 2 6/c < 9/c. 3.Clearly the o-Notation definition holds.

32. Ω-Notation Ω(g(n)) is pronounced “Big-omega of g of n” or sometimes just “Omega of g of n” We use this notation when we can only find a lower bound on a function. Ω(g(n)) = { f(n): there exists some positive constants c and n 0 such that: 0 n 0 }.

33. Ω-Notation Example Prove: ½ n 2 – 3n = Ω (n 2 ) Proof: 1.We must choose c and n 0 s.t.: cn 2 n 0. 2.Dividing by n 2 yields: c < ½ – 3/n 3.This holds if we choose c > 1/14 and n 0 > 7.

34. ω-Notation ω(g(n)) is pronounced “Little-omega of g of n” Ω(g(n)) may or may not be tight. We use ω(g(n)) to specify that it is not asymptotically tight. ω(g(n)) = { f(n): for any positive constant c>0 there exists a positive constant n 0 >0 such that: 0 n 0 }.

35. ω-Notation Example For example, n 2 /2 = ω(n) because n is not an asymptotically tight lower bound for n 2 /2. It is, however, a lower bound. But, n 2 /2 ≠ ω (n 2 ) because n 2 would be an asymptotically tight upper bound.

36. ω-Notation Example Prove: n 2 /2 = ω(n) Proof: 1.Let n 0 = 3c. 2.Substituting n 0 for n in cn < n 2 /2 c(3c) < (3c) 2 /2 6c 2 < 9c 2 3.Clearly the ω-Notation definition holds.

37. Θ - Notation Θ(g(n)) is pronounced “Theta of g of n” We use this notation when g(n) provides an upper and lower bound for a function. That is, it is asymptotically tight. This is a stronger statement. Θ(g(n)) = { f(n): there exists some positive constants c 1, c 2 and n 0 such that: 0 n 0 }.

38. Θ - Notation You may notice that f(n) = Θ(g(n)) implies f(n) = O(g(n)) and f(n) = Ω(g(n)). The converse is not true. f(n) = Θ(g(n)) sandwhiches f(n) between an upper and lower bound.

39. Θ -Notation Example Prove: ½ n 2 – 3n = Θ(n 2 ) Proof: 1.We must choose c and n 0 s.t.: c 1 n 2 n 0. 2.Dividing by n 2 yields: c 1 < ½ – 3/n < c 2 3.This holds if we choose c 1 > 1/14 and c 2 > ½ and n 0 > 7.

40. Θ – Notation Example Prove: ½ n 2 – 3n = Θ(n 2 ) Alternate Proof: 1.We already proved ½ n 2 – 3n = O(n 2 ) and that ½ n 2 – 3n = Ω(n 2 ), then clearly the n 2 bound on ½ n 2 – 3n is asymptotically tight and ½ n 2 – 3n = Θ(n 2 ).

41. Asymptotic Notation in Equations and Inequalities 2n 2 + 3n + 1 = 2n 2 + Θ(n) means 2n 2 + 3n + 1 = 2n 2 + f(n) where f(n) is a function in the set Θ(n). We use this to impart meaning and remove clutter: 2n 2 + 3n + 1 = 2n 2 + Θ(n) = Θ(n 2 )

42. Transitivity f(n) = O(g(n)) && g(n) = O(h(n)) → f(n) = O(h(n)). f(n) = o(g(n)) && g(n) = o(h(n)) → f(n) = o(h(n)). f(n) = Ω(g(n)) && g(n) = Ω(h(n)) → f(n) = Ω(h(n)). f(n) = ω(g(n)) && g(n) = ω(h(n)) → f(n) = ω(h(n)). f(n) = Θ(g(n)) && g(n) = Θ(h(n)) → f(n) = Θ(h(n)).

43. Reflexivity f(n) = O(f(n)), f(n) = Ω(f(n)), f(n) = Θ(f(n)) But NOT: f(n) = o(f(n)), f(n) = ω(f(n)).

44. Symmetry f(n) = Θ(g(n)) if and only if g(n) = Θ(f(n)). Does not hold for O, o, Ω, or ω but the following transpose symmetries do: f(n) = O(g(n)) if and only if g(n) = Ω(f(n)). f(n) = o(g(n)) if and only if g(n) = ω(f(n)).

45. An Analogy The comparison of functions roughly analogizes to the comparison of real numbers: f(n) = O(g(n))…a < b f(n) = o(g(n))…a < b f(n) = Ω(g(n))…a > b f(n) = ω(g(n))…a > b f(n) = Θ(g(n))…a = b

46. Insertion Sort.. again Last time we came up with this running time for insertion sort: T(n) = (c 4 +c 5 +c 6 )n 2 /2 + (c 1 +c 2 +c 3 +c 4 /2- c 5 /2-c 6 /2c 7 )n – (c 2 + c 3 + c 4 + c 7 ) With some hand waving I claimed that T(n) = Θ(n 2 ) Prove more formally this is true. Simplification: T(n) = dn 2 + en – f where d, e and f are constants.

47. A Warm Up Exercise Show for any real constants a and b > 0, (n + a) b = Θ(n b ). Recall: Θ(g(n)) = { f(n): there exists some positive constants c 1, c 2 and n 0 such that: 0 n 0 }.

48. A Warm Up Exercise Show for any real constants a and b > 0, (n + a) b = Θ(n b ). Proof: 1.c 1 n b < (n + a) b < c 2 n b 2.(n+a) b = n b (1+a) b 3.Let c 1 = 1 4.Let c 2 = (2+a) b 5.Let n 0 = 1.

49. Insertion Sort.. Again Prove: T(n) = dn 2 + en – f Recall: Θ(g(n)) = { f(n): there exists some positive constants c 1, c 2 and n 0 such that: 0 n 0 }.

50. Insertion Sort.. again Prove: T(n) = dn 2 + en – f = Θ(n 2 ) Proof: c 1 n 2 < dn 2 + en – f < c 2 n 2 Let n0 = f*max(1,1/d)*max(1,1/e) Note: min(d,e)n 2 < dn 2 + en – f Let c1 = min(d,e). Note: dn 2 + en – f < (d+e)n 2 Let c2 = (d+e).

51. More Examples Prove: lg(n!) = Θ(n lg n)

52. More Examples Prove: lg(n!) = O(n lg n) Proof: 1.n! < n n 2.lg(n n ) = n lg (n) 3.lg(n n ) > lg(n!) 4.n lg (n) > lg(n!) 5.lg(n!) = O(n lg n)