Asymptotic Notation Review of Functions Summations 28 Aug

Asymptotic Notation, Review of Functions & Summations 28 Aug 2003

Asymptotic Complexity w Running time of an algorithm as a function of input size n for large n. w Expressed using only the highest-order term in the expression for the exact running time. s Instead of exact running time, say (n 2). w Describes behavior of function in the limit. w Written using Asymptotic Notation. asymp - 1 Comp 122

Asymptotic Notation w , O, , o, w w Defined for functions over the natural numbers. s Ex: f(n) = (n 2). s Describes how f(n) grows in comparison to n 2. w Define a set of functions; in practice used to compare two function sizes. w The notations describe different rate-of-growth relations between the defining function and the defined set of functions. asymp - 2 Comp 122

-notation For function g(n), we define (g(n)), big-Theta of n, as the set: (g(n)) = {f(n) : positive constants c 1, c 2, and n 0, such that n n 0, we have 0 c 1 g(n) f(n) c 2 g(n) } Intuitively: Set of all functions that have the same rate of growth as g(n) is an asymptotically tight bound for f(n). asymp - 3 Comp 122

-notation For function g(n), we define (g(n)), big-Theta of n, as the set: (g(n)) = {f(n) : positive constants c 1, c 2, and n 0, such that n n 0, we have 0 c 1 g(n) f(n) c 2 g(n) } Technically, f(n) (g(n)). Older usage, f(n) = (g(n)). I’ll accept either… f(n) and g(n) are nonnegative, for large n. asymp - 4 Comp 122

Example (g(n)) = {f(n) : positive constants c 1, c 2, and n 0, such that n n 0, 0 c 1 g(n) f(n) c 2 g(n)} w 10 n 2 - 3 n = (n 2) w What constants for n 0, c 1, and c 2 will work? w Make c 1 a little smaller than the leading coefficient, and c 2 a little bigger. w To compare orders of growth, look at the leading term. w Exercise: Prove that n 2/2 -3 n= (n 2) asymp - 5 Comp 122

Example (g(n)) = {f(n) : positive constants c 1, c 2, and n 0, such that n n 0, 0 c 1 g(n) f(n) c 2 g(n)} w Is 3 n 3 (n 4) ? ? w How about 22 n (2 n)? ? asymp - 6 Comp 122

O-notation For function g(n), we define O(g(n)), big-O of n, as the set: O(g(n)) = {f(n) : positive constants c and n 0, such that n n 0, we have 0 f(n) cg(n) } Intuitively: Set of all functions whose rate of growth is the same as or lower than that of g(n) is an asymptotic upper bound for f(n) = (g(n)) f(n) = O(g(n)). (g(n)) O(g(n)). asymp - 7 Comp 122

Examples O(g(n)) = {f(n) : positive constants c and n 0, such that n n 0, we have 0 f(n) cg(n) } w Any linear function an + b is in O(n 2). How? w Show that 3 n 3=O(n 4) for appropriate c and n 0. asymp - 8 Comp 122

-notation For function g(n), we define (g(n)), big-Omega of n, as the set: (g(n)) = {f(n) : positive constants c and n 0, such that n n 0, we have 0 cg(n) f(n)} Intuitively: Set of all functions whose rate of growth is the same as or higher than that of g(n) is an asymptotic lower bound for f(n) = (g(n)). asymp - 9 Comp 122

Example (g(n)) = {f(n) : positive constants c and n 0, such that n n 0, we have 0 cg(n) f(n)} w n = (lg n). Choose c and n 0. asymp - 10 Comp 122

Relations Between , O, asymp - 11 Comp 122

Relations Between , , O Theorem : For any two functions g(n) and f(n), f(n) = (g(n)) iff f(n) = O(g(n)) and f(n) = (g(n)). w I. e. , (g(n)) = O(g(n)) Ç (g(n)) w In practice, asymptotically tight bounds are obtained from asymptotic upper and lower bounds. asymp - 12 Comp 122

Running Times w “Running time is O(f(n))” Worst case is O(f(n)) w O(f(n)) bound on the worst-case running time O(f(n)) bound on the running time of every input. w (f(n)) bound on the worst-case running time (f(n)) bound on the running time of every input. w “Running time is (f(n))” Best case is (f(n)) w Can still say “Worst-case running time is (f(n))” s Means worst-case running time is given by some unspecified function g(n) (f(n)). asymp - 13 Comp 122

Example w Insertion sort takes (n 2) in the worst case, so sorting (as a problem) is O(n 2). Why? w Any sort algorithm must look at each item, so sorting is (n). w In fact, using (e. g. ) merge sort, sorting is (n lg n) in the worst case. s Later, we will prove that we cannot hope that any comparison sort to do better in the worst case. asymp - 14 Comp 122

Asymptotic Notation in Equations w Can use asymptotic notation in equations to replace expressions containing lower-order terms. w For example, 4 n 3 + 3 n 2 + 2 n + 1 = 4 n 3 + 3 n 2 + (n) = 4 n 3 + (n 2) = (n 3). How to interpret? w In equations, (f(n)) always stands for an anonymous function g(n) (f(n)) s In the example above, (n 2) stands for 3 n 2 + 2 n + 1. asymp - 15 Comp 122

o-notation For a given function g(n), the set little-o: o(g(n)) = {f(n): c > 0, n 0 > 0 such that n n 0, we have 0 f(n) < cg(n)}. f(n) becomes insignificant relative to g(n) as n approaches infinity: lim [f(n) / g(n)] = 0 n g(n) is an upper bound for f(n) that is not asymptotically tight. Observe the difference in this definition from previous ones. Why? asymp - 16 Comp 122

w -notation For a given function g(n), the set little-omega: w(g(n)) = {f(n): c > 0, n 0 > 0 such that n n 0, we have 0 cg(n) < f(n)}. f(n) becomes arbitrarily large relative to g(n) as n approaches infinity: lim [f(n) / g(n)] = . n g(n) is a lower bound for f(n) that is not asymptotically tight. asymp - 17 Comp 122

Comparison of Functions f g a b f (n) = O(g(n)) a b f (n) = (g(n)) a = b f (n) = o(g(n)) a < b f (n) = w (g(n)) a > b asymp - 18 Comp 122

Limits w nlim [f(n) / g(n)] = 0 f(n) o(g(n)) w nlim [f(n) / g(n)] < f(n) O(g(n)) w 0 < nlim [f(n) / g(n)] < f(n) (g(n)) w 0 < nlim [f(n) / g(n)] f(n) (g(n)) w nlim [f(n) / g(n)] = f(n) w(g(n)) w nlim [f(n) / g(n)] undefined can’t say asymp - 19 Comp 122

Properties w Transitivity f(n) = (g(n)) & g(n) = (h(n)) f(n) = (h(n)) f(n) = O(g(n)) & g(n) = O(h(n)) f(n) = O(h(n)) f(n) = (g(n)) & g(n) = (h(n)) f(n) = (h(n)) f(n) = o (g(n)) & g(n) = o (h(n)) f(n) = o (h(n)) f(n) = w(g(n)) & g(n) = w(h(n)) f(n) = w(h(n)) w Reflexivity f(n) = (f(n)) f(n) = O(f(n)) f(n) = (f(n)) asymp - 20 Comp 122

Properties w Symmetry f(n) = (g(n)) iff g(n) = (f(n)) w Complementarity f(n) = O(g(n)) iff g(n) = (f(n)) f(n) = o(g(n)) iff g(n) = w((f(n)) asymp - 21 Comp 122

Common Functions 28 Aug 2003 Comp 122, Spring 2004

Monotonicity w f(n) is s s asymp - 23 monotonically increasing if m n f(m) f(n). monotonically decreasing if m n f(m) f(n). strictly increasing if m < n f(m) < f(n). strictly decreasing if m > n f(m) > f(n). Comp 122

Exponentials w Useful Identities: w Exponentials and polynomials asymp - 24 Comp 122

Logarithms x = logba is the exponent for a = bx. Natural log: ln a = logea Binary log: lg a = log 2 a lg 2 a = (lg a)2 lg lg a = lg (lg a) asymp - 25 Comp 122

Logarithms and exponentials – Bases w If the base of a logarithm is changed from one constant to another, the value is altered by a constant factor. s Ex: log 10 n * log 210 = log 2 n. s Base of logarithm is not an issue in asymptotic notation. w Exponentials with different bases differ by a exponential factor (not a constant factor). s Ex: 2 n = (2/3)n*3 n. asymp - 26 Comp 122

Polylogarithms w For a 0, b > 0, lim n ( lga n / nb ) = 0, so lga n = o(nb), and nb = w(lga n ) s Prove using L’Hopital’s rule repeatedly w lg(n!) = (n lg n) s Prove using Stirling’s approximation (in the text) for lg(n!). asymp - 27 Comp 122

Exercise Express functions in A in asymptotic notation using functions in B. A 5 n 2 + 100 n B 3 n 2 + 2 A (B) A (n 2), n 2 (B) A (B) log 3(n 2) log 2(n 3) A (B) logba = logca / logcb; A = 2 lgn / lg 3, B = 3 lgn, A/B =2/(3 lg 3) A (B) nlg 4 3 lg n alog b = blog a; B =3 lg n=nlg 3; A/B =nlg(4/3) as n A o (B) lg 2 n n 1/2 lim ( lga n / nb ) = 0 (here a = 2 and b = 1/2) A o (B) n asymp - 28 Comp 122

Summations – Review 28 Aug 2003 Comp 122, Spring 2004

Review on Summations w Why do we need summation formulas? For computing the running times of iterative constructs (loops). (CLRS – Appendix A) Example: Maximum Subvector Given an array A[1…n] of numeric values (can be positive, zero, and negative) determine the subvector A[i…j] (1 i j n) whose sum of elements is maximum over all subvectors. 1 asymp - 30 -2 2 Comp 122 2

Review on Summations Max. Subvector(A, n) maxsum ¬ 0; for i ¬ 1 to n do for j = i to n sum ¬ 0 for k ¬ i to j do sum += A[k] maxsum ¬ max(sum, maxsum) return maxsum n n j w. T(n) = 1 i=1 j=i k=i w. NOTE: This is not a simplified solution. What is the final answer? asymp - 31 Comp 122

Review on Summations w Constant Series: For integers a and b, a b, w Linear Series (Arithmetic Series): For n 0, w Quadratic Series: For n 0, asymp - 32 Comp 122

Review on Summations w Cubic Series: For n 0, w Geometric Series: For real x 1, For |x| < 1, asymp - 33 Comp 122

Review on Summations w Linear-Geometric Series: For n 0, real c 1, w Harmonic Series: nth harmonic number, n I+, asymp - 34 Comp 122

Review on Summations w Telescoping Series: w Differentiating Series: For |x| < 1, asymp - 35 Comp 122

Review on Summations w Approximation by integrals: s For monotonically increasing f(n) s For monotonically decreasing f(n) w How? asymp - 36 Comp 122

Review on Summations w nth harmonic number asymp - 37 Comp 122

Reading Assignment w Chapter 4 of CLRS. asymp - 38 Comp 122