CS 4407 Algorithms Lecture 2: Growth Functions

Transcription

1 CS 4407 Algorithms Lecture 2: Growth Functions Prof. Gregory Provan Department of Computer Science University College Cork 1

2 Lecture Outline Growth Functions Mathematical specification of growth functions Function specification Examples

3 Today s Learning Objectives Describe mathematical principles for specifying the growth of run-time of an algorithm Classify the growth functions Θ, O, Ω, o, ω Understand relationships among the growth functions

4 Analyzing Algorithms Assumptions generic one-processor, random access machine running time (others: memory, communication, etc) Worst Case Running Time: the longest time for any input of size n upper bound on the running time for any input in some cases like searching, this is close Average Case Behavior: the expected performance averaged over all possible inputs it is generally better than worst case behavior sometimes it s roughly as bad as worst case

5 A Simple Example INPUT: a sequence of n numbers OUTPUT: the smallest number among them 1. x T[1] 2. for i 2 to n do 3. if T[i] < x then x T[i] * Performance of this algorithm is a function of n.

6 Runtime Analysis Elementary operation: an operation whose execution time can be bounded above by a constant depending on implementation used. Assume all elementary operations can be executed at unit cost. This is not true, but they are within some constant factor of each other. We are mainly concerned about the order of growth.

7 Order of Growth For very large input size, it is the asymptotic rate of growth (or order of growth) that matters. We can ignore the lower-order terms, since they are relatively insignificant for very large n. We can also ignore leading term s constant coefficients, since they are not as important for the rate of growth in computational efficiency for very large n. Higher order functions of n are normally considered less efficient.

8 Comparisons of Algorithms Multiplication classical technique: O(nm) divide-and-conquer: O(nm ln1.5 ) ~ O(nm 0.59 ) For operands of size 1000, takes 40 & 15 seconds respectively on a Cyber 835. Sorting insertion sort: Θ(n 2 ) merge sort: Θ(n lg n) For 10 6 numbers, it took 5.56 hrs on a supercomputer using machine language and min on a PC using C/C++.

9 Why Order of Growth Matters Computer speeds double every two years, so why worry about algorithm speed? When speed doubles, what happens to the amount of work you can do?

10 Effect of Faster Machines H/W Speed Comp. of Alg. No. of items sorted 1 M * 2 M Gain * Million operations per second. Ο(n 2 ) O(n lgn) Higher gain with faster hardware for more efficient algorithm. Results are more dramatic for higher speeds: for the n lg n algorithm, doubling the speed almost doubles the number of items that can be sorted.

11 Classifying Functions Functions Double Exp Exp Exponential Polynomial Poly Logarithmic Logarithmic Constant n5 25n log n (log n) 5 n 5 2 5n 2

12 Ordering Functions Functions Cons stant Loga arithmic Poly Polyn nomial Expo onential Exp << << << << << << Logarithmic 5 << 5 log n << (log n) 5 << n 5 << 2 5n << 2 << Doub ble Exp 2 n5 25n

13 Classifying Functions Functions Double Exp Exp Exponential Polynomial Poly Logarithmic Constant Logarithmic nθ(1) 2θ(n) 2 θ(1) θ(log n) (log n) θ(1) n θ(1) 2 θ(n) 2

14 Asymptotic Notation Θ, O, Ω, o, ω Used to describe the running times of algorithms Instead of exact running time, say Θ(n 2 ) Defined for functions whose domain is the set of natural numbers, N Determine sets of functions, in practice used to compare two functions

15 Notation Theta f(n) = θ(g(n)) f(n) c g(n) BigOh f(n) = O(g(n)) f(n) c g(n) Omega f(n) = Ω(g(n)) f(n) c g(n) Little Oh f(n) = o(g(n)) f(n) << c g(n) Little Omega f(n) = ω(g(n)) f(n) >> c g(n)

16 Θ-notation For a given function g(n), we denote by Θ(g(n)) the set of functions Θ(g(n)) = {f(n): there exist positive constants c 1, c 2 and n 0 such that 0 c 1 g(n) f(n) c 2 g(n), for all n n 0 } We say g(n) is an asymptotically tight bound for f(n)

17 Example 3n 2 + 7n + 8= Θ(n 2 ) What constants for n 0, c 1, and c 2 will work? Make c 1 a little smaller than leading coefficient, and c 2 a little bigger To compare orders of growth, look at the leading term

18 Definition of Theta f(n) = θ(g(n)) c c n, n n, c g( n) f ( n) c g( n) 1, 2, f(n) is sandwiched between c 1 g(n) and c 2 g(n) for some sufficiently small c 1 (= ) for some sufficiently large c 2 (= 1000)

19 Definition of Theta f(n) = θ(g(n)) c c n, n n, c g( n) f ( n) c g( n) 1, 2, n 0 For all sufficiently large n For some definition of sufficiently large

20 Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) c c n, n n, c g( n) f ( n) c g( n) 1, 2,

21 Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) c c n, n n, c g( n) f ( n) c g( n) 1, 2, ?? c 1 n 2 3n 2 + 7n + 8 c 2 n 2

22 Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) c c n, n n, c g( n) f ( n) c g( n) 1, 2, n 2 3n 2 + 7n n 2

23 Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) c c n, n n, c g( n) f ( n) c g( n) 1, 2, False

24 Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) c c n, n n, c g( n) f ( n) c g( n) 1, 2, False

25 Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) c c n, n n, c g( n) f ( n) c g( n) 1, 2, True

26 Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) c c n, n n, c g( n) f ( n) c g( n) 1, 2, n 8 3 n 2 3n 2 + 7n n 2

27 O-notation For a given function g(n), we denote by O(g(n)) the set of functions O(g(n)) = {f(n): there exist positive constants c and n 0 such that 0 f(n) cg(n), for all n n 0 } We say g(n) is an asymptotic upper bound for f(n)

28 Ω-notation For a given function g(n), we denote by Ω(g(n)) the set of functions Ω(g(n)) = {f(n): there exist positive constants c and n 0 such that 0 cg(n) f(n) for all n n 0 } We say g(n) is an asymptotic lower bound for f(n)

29 Relations Between Θ, Ω, O For any two functions g(n) and f(n), f(n) = Θ(g(n)) if and only if f(n) = O(g(n)) and f(n) = Ω(g(n)). i.e., Θ(g(n)) = O(g(n)) Ω(g(n)).

30 Running Times The running time is O(f(n)) Worst case is O(f(n)) Running time is Ω(f(n)) Best case is Ω(f(n)) Can still say Worst case running time is Ω(f(n)) Means worst case running time is given by some unknown function g(n) Ω(f(n))

31 Example Insertion sort takes Θ(n 2 ) worst case time, so sorting (as a problem) is O(n 2 ) Any sort algorithm must look at each item, so sorting is Ω(n) In fact, using (e.g.) merge sort, sorting is Θ(n lg n) in the worst case

32 Asymptotic Notation in Equations Asymptotic notation is used to replace expressions containing lower-order terms. For example, 4n 3 + 3n 2 + 2n + 1 = 4n 3 + 3n 2 + Θ(n) = 4n 3 + Θ(n 2 ) = Θ(n 3 ) In equations, Θ(f(n)) always stands for an anonymous function g(n) Θ(f(n)). In the example above, Θ(n 2 ) stands for 3n 2 + 2n + 1.

33 o-notation For a given function g(n), we denote by o(g(n)) the set of functions: o(g(n)) = {f(n): for any positive constant c > 0, there exists a constant n 0 > 0 such that 0 f(n) < cg(n) for all n n 0 } f(n) becomes insignificant relative to g(n) as n approaches infinity: lim [f(n) / g(n)] = 0 n We say g(n) is an upper bound for f(n) that is not asymptotically tight.

34 O(*) versus o(*) O(g(n)) = {f(n): there exist positive constants c and n 0 such that 0 f(n) cg(n), for all n n 0 }. o(g(n)) = {f(n): for any positive constant c > 0, there exists a constant n 0 > 0 such that 0 f(n) < cg(n) for all n n 0 }. Thus o(f(n)) is a weakened O(f(n)). For example: n 2 = O(n 2 ) n 2 o(n 2 ) n 2 = O(n 3 ) n 2 = o(n 3 )

35 ω-notation For a given function g(n), we denote by ω(g(n)) the set of functions ω(g(n)) = {f(n): for any positive constant c > 0, there exists a constant n 0 > 0 such that 0 cg(n) < f(n) for all n n 0 } f(n) becomes arbitrarily large relative to g(n) as n approaches infinity: lim [f(n) / g(n)] = n We say g(n) is a lower bound for f(n) that is not asymptotically tight.

36 Limits lim [f(n) / g(n)] = 0 f(n) ο(g(n)) n lim [f(n) / g(n)] < f(n) Ο(g(n)) n 0 < lim [f(n) / g(n)] < f(n) Θ(g(n)) n 0 < lim [f(n) / g(n)] f(n) Ω(g(n)) n lim [f(n) / g(n)] = f(n) ω(g(n)) n lim [f(n) / g(n)] undefined can t say n

37 Lecture Summary Defined the mathematical principles for specifying the runtime of an algorithm Classified the growth functions Θ, O, Ω, o, ω Defined several relationships among the growth functions Theta f(n) = θ(g(n)) f(n) c g(n) BigOh f(n) = O(g(n)) f(n) c g(n) Omega f(n) = Ω(g(n)) f(n) c g(n) Little Oh f(n) = o(g(n)) f(n) << c g(n) Little Omega f(n) = ω(g(n)) f(n) >> c g(n)

38 Additional Useful Relations The following slides provide many useful relations for orders of growth These relations may prove helpful in homeworks and exams

39 Comparison of Functions f g a b f (n) = O(g(n)) a b f (n) = Ω(g(n)) a b f (n) = Θ(g(n)) a = b f (n) = o(g(n)) a < b f (n) = ω (g(n)) a > b

40 Properties Transitivity f(n) = Θ(g(n)) & g(n) = Θ(h(n)) f(n) = Θ(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) = 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)) Symmetry f(n) = Θ(g(n)) if and only if g(n) = Θ(f(n)) Transpose Symmetry 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))

41 Useful Facts For a 0, b > 0, lim n ( lg a n / n b ) = 0, so lg a n = o(n b ), and n b = ω(lg a n ) Prove using L Hopital s rule and induction lg(n!) = Θ(n lg n)

42 Examples A B 5n n 3n log 3 (n 2 ) log 2 (n 3 ) n lg4 3 lg n lg 2 n n 1/2

43 Examples A B 5n n 3n A Θ(B) A Θ(n 2 ), n 2 Θ(B) A Θ(B) log (n ) log 2 (n ) n lg4 3 lg n lg 2 n n 1/2

44 Examples A B 5n n 3n A Θ(B) A Θ(n 2 ), n 2 Θ(B) A Θ(B) log (n ) log 2 (n ) A Θ(B) log b a = log c a / log c b; A = 2lgn / lg3, B = 3lgn, A/B =2/(3lg3) n lg4 3 lg n lg 2 n n 1/2

45 Examples A B 5n n 3n A Θ(B) A Θ(n 2 ), n 2 Θ(B) A Θ(B) log (n ) log 2 (n ) A Θ(B) log b a = log c a / log c b; A = 2lgn / lg3, B = 3lgn, A/B =2/(3lg3) n lg4 3 lg n A ω(b) a log b = b log a ; B =3 lg n =n lg 3 ; A/B =n lg(4/3) as n lg 2 n n 1/2

46 A Examples B 5n n 3n A Θ(B) A Θ(n 2 ), n 2 Θ(B) A Θ(B) log 3 (n 2 ) log 2 (n 3 ) A Θ(B) log b a = log c a / log c b; A = 2lgn / lg3, B = 3lgn, A/B =2/(3lg3) n lg4 3 lg n A ω(b) a log b = b log a ; B =3 lg n =n lg 3 ; A/B =n lg(4/3) as n lg 2 n n 1/2 A ο (B) lim ( lg a n / n b ) = 0 (here a = 2 and b = 1/2) A ο (B) n

47 Review on Summation Why do we need summation formulas? We need them for computing the running times of some algorithms. (CLRS/Chapter 3) 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

48 Reviews on Summation MaxSubvector(a, n) maxsum 0; for i 1 to n for j = i to n sum 0; for k i to j sum += a[k] maxsum max(sum, maxsum); return maxsum; n n j T(n) = 1 i=1 j=i k=i NOTE: This is not a simplified solution. What is the final answer?

49 Reviews on Summation Constant Series: For n 0, Linear Series: For n 0, b 1 = b - a + 1 i=a n i = n = n(n+1) / 2 i=1

50 Reviews on Summation Quadratic Series: For n 0, n i 2 = n 2 = (2n 3 + 3n 2 + n) / 6 i=1 Linear-Geometric Series: For n 0, n ic i = c + 2c nc n = [(n-1)c (n+1) - nc n + c] / (c-1) 2 i=1