Algorithm efficiency can be measured in terms of: Time Space Other resources such as processors, network packets, etc.

Transcription

1 Algorithms Analysis Algorithm efficiency can be measured in terms of: Time Space Other resources such as processors, network packets, etc. Algorithms analysis tends to focus on time: Techniques for measuring all resources are basically the same Historically, time dominated 3.1

2 The efficiency of a program or system is affected by: Programming Language Physical characteristics of the computer Room temperature Amount of data to process (or actual parameters) Algorithms used In the real world all factors are considered, and any one can dominate in any particular situation. When developing algorithms in the abstract, most such factors are ignored because they are: Out of our control, as programmers and algorithm developers. Difficult to predict. For that reason we say they are arbitrary. Consequently, we are interested in algorithm efficiency, not program or system efficiency. 3.2

3 The efficiency of an algorithm is specified as a running time, sometimes also called complexity. In our abstract world, the input length or parameters have the most influence on running time of an algorithm. The running time of an algorithm measures: Total number of operations executed by an algorithm, or Total number of statements executed by an algorithm Running times can be based on: Worst case (most frequently used) Average case (most difficult) Best case 3.3

4 Big Oh Notation Note that it is tempting to think that O(f(n)) means approximately f(n). Although used in this manner frequently, this interpretation or use of big Oh is absolutely incorrect. Frequently, O(f(n)) is pronounced on the order of f(n) or order f(n), but keep in mind that this also does not mean approximately f(n). 3.4

5 Definitions: Let f(n) and g(n) be functions: f(n) g(n) n 3.5

6 f(n) is said to be less than g(n) if f(n) <= g(n) for all n. g(n) f(n) n For example, n 2 is less than n

7 To within a constant factor, f(n) is said to be less than g(n) if there exists a positive constant c such that f(n) <= cg(n) for all n. cg(n) f(n) g(n) n 3.7

8 Example: g(n) = 3n f(n) = 6n Note: 6n is not less than 3n However: To within a constant factor 6n is less than 3n Proof: Let c=9, then we see that: 6n <= 9(3n 2 + 2) = 27n

9 By the way, for these two functions it is also the case that: To within a constant factor 3n is less than 6n Proof: Let c=1, then we see that: 3n <= 1(6n 2 + 3) = 6n In fact 3n is actually less than 6n In other words, asymptotically, there ain t much difference in the growth of the two functions, as n goes to infinite. (always enjoy using that word in class) 3.9

10 Question: g(n) = n 2 f(n) = 2 n Is f(n), to within a constant factor, less than g(n)? In other words, is there a constant c such that: 2 n <= cn 2? No! Not even if we let c = 1,000,000, or larger! 2 n is significantly different, and bigger than n 2 more than just a constant factor. 3.10

11 Big O Definition Definition #1: f(n) is said to be O(g(n)) if there exists two positive constants c and n 0 such that f(n) <= cg(n) for all n >= n 0. cg(n) f(n) n n 0 Intuitively, we say that, as n goes to infinity, f(n) is, to within a constant factor, less than g(n). Stated another way, as n goes to infinity, f(n) is, to within a constant factor, bounded from above by g(n). 3.11

12 From the definition of big-o, it should now be clear that saying T(n) is O(g(n)) means that g(n) is an upper bound on T(n), and does not mean approximately. An analogy: Consider two integers x and y, where x <= y. Is y approximately equal to x? Of course not! Similarly for the definition of big-o. 3.12

13 Example: g(n) = n 2 f(n) = n cg(n) f(n) Claim: n is O(n 2 ) Proof: Let c=2 and n 0 = 1, then we see that: n 0 n <= 2n 2 for all n >= n 0 = 1. Note that in this case f(n) is not less than g(n), even to within a constant, but it is as n goes to infinite. 3.13

14 Give a polynomial, a simple procedure is to drop the lower-order terms, and the coefficient of the highest order term. f(n) = n f(n) is O(n 2 ) g(n) = n + 6 g(n) is O(n) h(n) = n 35 + n 16 + n 4 + 3n h(n) is O(n 35 ) r(n) = (1/2)n r(n) is O(n) s(n) = 4n 2 + 2n + 25 s(n) is O(n 2 ) 3.14

15 More generally: If: f(n) = a m-1 n m-1 + a m-2 n m a 1 n 1 + a 0 Then: f(n) is O(n m-1 ). Explanation: Let: c = a m-1 + a m-2 + a 1 + a 0 n 0 = 1 Then it will be the case that: f(n) <= cn m-1 for all n>= n

16 Why is f(n) <= cn m-1 for all n>=n 0 =1? cn m-1 = a m-1 n m-1 + a m-2 n m a 1 n m-1 + a 0 n m-1 f(n) = a m-1 n m-1 + a m-2 n m a 1 n 1 + a

17 A bit more formally: f(n) = a m-1 n m-1 + a m-2 n m a 1 n 1 + a 0 <= ( a m-1 n m-1 + a m-2 n m a 1 n 1 + a 0 n 0 ) <= ( a m-1 n m-1 + a m-2 n m a 1 n m-1 + a 0 n m-1 ) = ( a m-1 + a m-2 + a 1 + a 0 ) n m-1 = cn m

18 Another thing worth noting: If, for example, f(n) = 5n n 2 + 3n Then f(n) is O(n 3 ), but also O(n 4 ), O(n 5 ), etc. This may seem counter-intuitive, but it is a useful property of big-oh to be, i.e., an upper bound. More technically O(n 3 ) is said to be an asymptotically tight upper bound for f(n), while O(n 4 ), O(n 5 ), etc., are not. 3.18

19 Recall the definition of big-o: f(n) is said to be O(g(n)) if there exists two positive constants c and n 0 such that f(n) <= cg(n) for all n >= n 0. The authors give two slightly different definitions: We write f(n) = O(g(n)) if there exist constants c > 0 and n 0 > 0 such that 0 f(n) cg(n) for all n n 0. O(g(n)) = { f(n) : there exist constants c > 0 and n 0 > 0 such that 0 f(n) cg(n), for all n n 0 } What s the difference? 3.20

20 What s the significance of the difference? Fundamentally, not much, but there is a notational difference: 4n 2 + 2n + 25 is O(n 2 ) 4n 2 + 2n + 25 = O(n 2 ) 4n 2 + 2n + 25 O(n 2 ) Unless stated otherwise, you are free to choose, just so long is there is no misunderstanding that big-o is an upper-bound. 3.21

21 All of the notations are routinely abused: 4n 2 + O(n) T(n) = O(n 2 ) O(n 2 ) + O(n) = O(n 2 ) O(1) In most such cases, the big-o term represents an anonymous function. This is true for big-o, big-, and big- 3.22

22 Big Definition Definition: f(n) is said to be (g(n)) if there exists two positive constants c and n 0 such that cg(n) <= f(n) for all n >= n 0. f(n) cg(n) n 0 As n goes to infinity, f(n) is, to within a constant factor, bounded from below by g(n). 3.23

23 Example: n 2 1/2n cg(n) f(n) Claim: 1/2n is (n 2 ) Proof: Let c=1/2 and n 0 = 1, then we see that: n 0 1/2n 2 <= 1/2n for all n >= n 0 = 1. Note that in this case n 2 is not less than 1/2n 2 + 1, even to within a constant, but it is as n goes to infinite. 3.24

24 Give a polynomial, can we say that a simple procedure is to drop the lower-order terms, and the coefficient of the highest order term? f(n) = n f(n) is (n 2 ) g(n) = n + 6 g(n) is (n) h(n) = n 35 + n 16 + n 4 + 3n h(n) is (n 35 ) r(n) = (1/2)n r(n) is (n) s(n) = 4n 2 + 2n + 25 s(n) is (n 2 ) What s the proof? 3.25

25 Another thing worth noting: If, for example, f(n) = 5n n 2 + 3n Then f(n) is (n 3 ), but also (n 2 ), (n), etc. This may seem counter-intuitive, but this is useful property of big-, i.e., it is a lower bound. More technically (n 3 ) is said to be an asymptotically tight lower bound for f(n), while (n 2 ), (n), etc., are not. 3.26

26 Example: 6n is (3n 2 + 2) Proof: Let c=1 and n 0 = 1 then we see that: c(3n 2 + 2) <= 6n n <= 6n for all n >= n 0 =

27 Similarly: 3n is (6n 2 + 3) Proof: Let c=1/3 and n 0 = 1 then we see that: 1/3(6n 2 + 3) <= 3n n <= 3n for all n >= n 0 =

28 Lastly consider: Is 2 n = (n 2 )? Yes, if we let c = 1, then cn 2 <= 2 n for all n>=1. Is n 2 = (2 n )? It is not the case that c2 n <= n 2 for all n>=n 0, for any constants c and n 0. Not even if we let c = 1/1,000,000, or smaller! 3.29

29 Just like with big-oh, the authors give two slightly different definitions: We write f(n) = (g(n)) if there exist constants c > 0 and n 0 > 0 such that 0 cg(n) f(n) for all n n 0. (g(n)) = { f(n) : there exist constants c > 0 and n 0 > 0 such that 0 cg(n) f(n), for all n n 0 } But once again, the difference is largely cosmetic. 3.30

30 Big Definition Definition: f(n) is said to be (g(n)) if there exist positive constants c 1, c 2 and n 0 such that c 1 g(n) <= f(n) <= c 2 g(n) for all n >= n 0. If f(n) is (g(n)) then it is O(g(n)), and also (g(n)). 3.31

31 Give a polynomial, a simple procedure is to drop the lower-order terms, and the coefficient of the highest order term. f(n) = n f(n) is (n 2 ) g(n) = n + 6 g(n) is (n) h(n) = n 35 + n 16 + n 4 + 3n h(n) is (n 35 ) r(n) = (1/2)n r(n) is (n) s(n) = 4n 2 + 2n + 25 s(n) is (n 2 ) 3.32

32 Big Definition Just like with big-o and big-, the authors give two slightly different definitions: We write f(n) = (g(n)) if 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. (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.} But once again, the difference is largely cosmetic. 3.33

33 Big Definition One advantage of the set-theoretic definition of big- is that it can be defined simply as: (g(n)) = O(g(n)) (g(n)) 3.34

34 Little o Definition Definition: f(n) is said to be o(g(n)) if, for any positive constant c, there exists a positive constant n 0 such that f(n) <= cg(n) for all n >= n 0. Contrast the above with the definition of big-o: f(n) is said to be O(g(n)) if there exists two positive constants c and n 0 such that f(n) <= cg(n) for all n >= n 0. What s the difference? 3.35

35 Little o Definition Definition: f(n) is said to be o(g(n)) if, for any positive constant c, there exists a positive constant n 0 such that f(n) <= cg(n) for all n >= n 0. What s the significance of the difference? Suppose you wanted to make cg(n) smaller than f(n). How would you do it? Pick a really small value for c. There is still a point (n 0 ) where cg(n) over-takes f(n). 3.36

36 Little o Definition Definition: f(n) is said to be o(g(n)) if, for any positive constant c, there exists a positive constant n 0 such that f(n) <= cg(n) for all n >= n 0. This means that no matter what value for c you pick, and no matter now small it is, there is a point where cg(n) becomes greater than f(n). Therefore, g(n) is significantly greater than f(n); in particular, by more than just a constant factor. More technically, g(n) denotes an upper bound on f(n) that is not asymptotically tight. 3.37

37 Little ω Definition Definition: f(n) is said to be ω(g(n)) if, for any positive constant c, there exists a positive constant n 0 such that cg(n) <= f(n) for all n >= n 0. Contrast the above with the definition of big- : f(n) is said to be (g(n)) if there exists two positive constants c and n 0 such that cg(n) <= f(n) for all n >= n 0. What s the difference? 3.38

38 Little ω Definition Definition: f(n) is said to be ω(g(n)) if, for any positive constant c, there exists a positive constant n 0 such that cg(n) <= f(n) for all n >= n 0. What s the significance of the difference? Suppose you wanted to make cg(n) bigger than f(n). How would you do it? Pick a really big value for c. If f(n) is ω(g(n)) there is still a point (n 0 ) where f(n) over-takes cg(n). 3.39

39 Little ω Definition Definition: f(n) is said to be ω(g(n)) if, for any positive constant c, there exists a positive constant n 0 such that f(n) <= cg(n) for all n >= n 0. This means that no matter what value for c you pick, and no matter now big it is, there is a point where f(n) becomes greater than cg(n). Therefore, f(n) is significantly greater than g(n); in particular, by more than just a constant factor. More technically, g(n) denotes a lower bound on f(n) that is not asymptotically tight. 3.40

40 Fill in the Blanks Suppose f(n) is O(g(n)) Yes No Maybe f(n) is (g(n)) f(n) is (g(n)) f(n) is o(g(n)) f(n) is ω(g(n)) g(n) is O(f(n)) g(n) is (f(n)) g(n) is (f(n)) g(n) is o(f(n)) g(n) is ω(f(n)) X X X X X X X X X 3.41

41 Fill in the Blanks! Suppose f(n) is (g(n)) Yes No Maybe f(n) is O(g(n)) f(n) is (g(n)) f(n) is o(g(n)) f(n) is ω(g(n)) g(n) is O(f(n)) g(n) is (f(n)) g(n) is (f(n)) g(n) is o(f(n)) g(n) is ω(f(n)) 3.42

42 Fill in the Blanks! Suppose f(n) is (g(n)) Yes No Maybe f(n) is O(g(n)) f(n) is (g(n)) f(n) is o(g(n)) f(n) is ω(g(n)) g(n) is O(f(n)) g(n) is (f(n)) g(n) is (f(n)) g(n) is o(f(n)) g(n) is ω(f(n)) 3.43

43 Fill in the Blanks! Suppose f(n) is o(g(n)) Yes No Maybe f(n) is O(g(n)) f(n) is (g(n)) f(n) is (g(n)) f(n) is ω(g(n)) g(n) is O(f(n)) g(n) is (f(n)) g(n) is (f(n)) g(n) is o(f(n)) g(n) is ω(f(n)) 3.44

44 Fill in the Blanks! Suppose f(n) is ω(g(n)) Yes No Maybe f(n) is O(g(n)) f(n) is (g(n)) f(n) is (g(n)) f(n) is o(g(n)) g(n) is O(f(n)) g(n) is (f(n)) g(n) is (f(n)) g(n) is o(f(n) g(n) is ω(f(n)) 3.45

45 Warning! The following are all common misunderstandings of the notation. Big-O is a worst-case analysis Big- is an average case analysis Big- is is a best case analysis Worst case running time can be analyzed using any of the three, and so can average case or best case. 3.46

46 Warning! Example - Quicksort Worst case: O(n 2 ) (n 2 ) already sorted list (lower-bound example) Therefore, worst-case is (n 2 ) Average case: O(nlogn) (nlogn) Therefore, average-case is (nlogn) Sometimes there is no distinction between best, worst, and average cases. Input a list of n integers, and add them up (n) - best, worst, and average case. 3.47

47 Identities Transitivity: if f(n) is O(g(n)) and g(n) is O(h(n)) then f(n) is O(h(n)) if f(n) is (g(n)) and g(n) is (h(n)) then f(n) is (h(n)) if f(n) is (g(n)) and g(n) is (h(n)) then f(n) is (h(n)) if f(n) is o(g(n)) and g(n) is o(h(n)) then f(n) is o(h(n)) if f(n) is ω(g(n)) and g(n) is ω(h(n)) then f(n) is ω(h(n)) Reflexivity: f(n) is (f(n)) f(n) is O(g(n)) f(n) is (g(n)) Symmetry: f(n) is (g(n)) if and only if g(n) is (f(n)) Transpose Symmetry: f(n) is O(g(n)) if and only if g(n) is (f(n)) f(n) is o(g(n)) if and only if g(n) is ω(f(n)) 3.48

48 All the notations subsume constants: 4O(n 2 ), O(6n 2 ), and O(n 2 /5) are all just O(n 2 ) Similarly for (n 2 ), (n 2 ), o(n 2 ) and ω(n 2 ) Lower-order terms are subsumed by higher-order terms: O(n 2 ) + O(n) + O(1) is just O(n 2 ) o(n 2 ) + o(n) + o(1) is just o(1) Constant time is always expressed as O(1), never O(2), O(57), etc. 3.49