1. 1 CSE 326: Data Structures Program Analysis Lecture 3: Friday, Jan 8, 2003

2. 2 Outline Empirical analysis of algorithms Formal analysis of algorithms Reading assignment: sec. 2.4.3 (maximum subsequence)

3. 3 Determining the Complexity of an Algorithm Empirical measurements: –pro: discover if constant factors are significant –con: may be running on “wrong” inputs Formal analysis (proofs): –pro: no interference from implementation/hardware details –con: hides constants; may be hard In theory, theory is the same as practice, but in practice it is not

4. 4 Measuring Empirical Complexity: Linear vs. Binary Search Linear Search Binary Search Time to find one item: Time to find N items: Find a item in a sorted array of length N Binary search algorithm:

5. 5 int bfind(int x, int a[], int left, int right) { if (left+1 == right) return –1; m = (left + right) / 2; if (x == a[m]) return m; if (x < a[m]) return bfind(x, a, left, m); else return bfind(x, a, m, right); } int bfind(int x, int a[], int left, int right) { if (left+1 == right) return –1; m = (left + right) / 2; if (x == a[m]) return m; if (x < a[m]) return bfind(x, a, left, m); else return bfind(x, a, m, right); } for (i=0; i<n; i++) a[i] = i; for (i=0; i<n; i++) lfind(i,a,n); int lfind(int x, int a[], int n) { if (n==0) return –1; if (x == a[n-1]) return n-1; return lfind(x, a, n-1); } int lfind(int x, int a[], int n) { if (n==0) return –1; if (x == a[n-1]) return n-1; return lfind(x, a, n-1); } or bfind(i,a,-1,n)

6. 6 Graphical Analysis

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9. 9 slope  2 slope  1 Recall: we search n times Linear = O(n 2 ) Binary = O(n log n)

10. 10 Property of Log/Log Plots On a linear plot, a linear function is a straight line On a log/log plot, any polynomial function is a straight line! slope  y/  x = exponent vertical axis slope Proof: suppose y = cx k log(y) = log(cx k ) log(y) = log(c) + log(x k ) log(y) = log(c) + k log(x) horizontal axis

11. 11 slope  1 Why does O(n log n) look like a straight line?

12. 12 Empirical Complexity Large data sets may be required to gain an accurate empirical picture When running time is expected to be polynomial, use Log/log plots  slope = exponent When the running time is expected to be exponential, use log on the y axis When running time is expected to be log, then use long on the x axis Best: try all three, and see which one is linear

13. 13 Analyzing Code primitive operations consecutive statements function calls conditionals loops recursive functions

14. 14 Conditionals Conditional if C then S 1 else S 2 Suppose you are doing a O( ) analysis? Time(C) + Max(Time(S1),Time(S2)) or Time(C)+Time(S1)+Time(S2) Suppose you are doing a  ( ) analysis? Time(C) + Min(Time(S1),Time(S2)) or Time(C)

15. 15 Nested Loops for i = 1 to n do for j = 1 to n do sum = sum + 1

16. 16 Nested Dependent Loops for i = 1 to n do for j = i to n do sum = sum + 1

17. 17 Nested Dependent Loops for i = 1 to n do for j = i to n do sum = sum + 1 Compute it the hard way: Compute it the smart way: substitute n - i+1 with j

18. 18 Other Important Series Sum of squares: Sum of exponents: Geometric series: Novel series: –Reduce to known series, or prove inductively

19. 19 Linear Search Analysis Best case, tight analysis: Worst case, tight analysis: void lfind(int x, int a[], int n) { for (i=0; i<n; i++) if (a[i] == x) return i; return –1; } void lfind(int x, int a[], int n) { for (i=0; i<n; i++) if (a[i] == x) return i; return –1; }

20. 20 Iterated Linear Search Analysis Easy worst-case upper-bound: Worst-case tight analysis: for (i=0; i<n; i++) a[i] = i; for (i=0; i<n; i++) lfind(i,a,n);

21. 21 Analyzing Recursive Programs 1.Express the running time T(n) as a recursive equation 2.Solve the recursive equation For an upper-bound analysis, you can optionally simplify the equation to something larger For a lower-bound analysis, you can optionally simplify the equation to something smaller

22. 22 Binary Search What is the worst-case upper bound? int bfind(int x, int a[], int left, int right) { if (left+1 == right) return –1; m = (left + right) / 2; if (x == a[m]) return m; if (x < a[m]) return bfind(x, a, left, m); else return bfind(x, a, m, right); } int bfind(int x, int a[], int left, int right) { if (left+1 == right) return –1; m = (left + right) / 2; if (x == a[m]) return m; if (x < a[m]) return bfind(x, a, left, m); else return bfind(x, a, m, right); }

23. 23 Binary Search Introduce some constants… b = time needed for base case c = time needed to get ready to do a recursive call Size is n = right-left Running time is thus: int bfind(int x, int a[], int left, int right) { if (left+1 == right) return –1; m = (left + right) / 2; if (x == a[m]) return m; if (x < a[m]) return bfind(x, a, left, m); else return bfind(x, a, m, right); } int bfind(int x, int a[], int left, int right) { if (left+1 == right) return –1; m = (left + right) / 2; if (x == a[m]) return m; if (x < a[m]) return bfind(x, a, left, m); else return bfind(x, a, m, right); }

24. 24 Binary Search Analysis One sub-problem, half as large Equation: T(1)  b T(n)  T(n/2) + c for n>1 Solution: T(n)  T(n/2) + cwrite equation  T(n/4) + c + cexpand  T(n/8) + c + c + c  T(n/2 k ) + kcinductive leap  T(1) + c log n where k = log nselect value for k  b + c log n = O(log n)simplify

25. 25 Solving Recursive Equations by Telescoping Create a set of equations, take their sum

26. 26 Inductive Proof If you know the closed form solution, you can validate it by ordinary induction

27. 27 Amortized Analysis Stack Stack operations –push –pop –is_empty Stack property: if x is on the stack before y is pushed, then x will be popped after y is popped A BCDEFBCDEF E D C B A F What is biggest problem with an array implementation?

28. 28 Stretchy Stack Implementation int[] data; int maxsize; int top; Push(e){ if (top == maxsize){ temp = new int[2*maxsize]; for (i=0;i<maxsize;i++) temp[i]=data[i]; data = temp; maxsize = 2*maxsize; } data[++top] = e; } int pop() { return data[--top]; } int[] data; int maxsize; int top; Push(e){ if (top == maxsize){ temp = new int[2*maxsize]; for (i=0;i<maxsize;i++) temp[i]=data[i]; data = temp; maxsize = 2*maxsize; } data[++top] = e; } int pop() { return data[--top]; } Best case Push = O( ) Worst case Push = O( )

29. 29 Stretchy Stack Amortized Analysis Consider sequence of n push/pop operations Amortized time = (T 1 + T 2 +... + T n ) / n We compute this next n push(e 1 ) push(e 2 ) pop() push(e 3 ) push(e 4 ) pop()... push(e k ) push(e 1 ) push(e 2 ) pop() push(e 3 ) push(e 4 ) pop()... push(e k )  time = T 1  time = T n

30. 30 Stretchy Stack Amortized Analysis The length of the array increases like this: 1, 2, 4, 8,..., 2 k,..., n For each T i we have one of the following –T i = O(1) for pop( ), and for some push(e i ) –T i = O(2 k ) for some push(e i ) Hence

31. 31 Stretchy Stack Amortized Analysis Let’s compute this sum: And therefore: In an asymptotic sense, there is no overhead in using stretchy arrays rather than regular arrays!

32. 32 Geometric Series

33. 33 Stretchy Stack Amortized Analysis Careful ! We must be clever to get good amortized performance ! Consider “smart pop”: int pop(){ int e = data[--top]; if (top <= maxsize/2){ maxsize = maxsize/2; temp = new int[maxsize]; for (i=0;i<maxsize;i++) temp[i]=data[i]; data = temp;} return e; } int pop(){ int e = data[--top]; if (top <= maxsize/2){ maxsize = maxsize/2; temp = new int[maxsize]; for (i=0;i<maxsize;i++) temp[i]=data[i]; data = temp;} return e; }

34. 34 Stretchy Stack Amortized Analysis Take the sequence of 3n push/pop operations: push(e 1 ) push(e 2 )... push(e n ) pop() push(e n ) pop() push(e n ) pop()... push(e n ) pop() push(e 1 ) push(e 2 )... push(e n ) pop() push(e n ) pop() push(e n ) pop()... push(e n ) pop() n 2n Suppose n = 2 k +1 Hence amortized time is: T = (  (1) +... +  (1) +  (n) +...+  (n))/3n = (n  (1) + 2n  (n))/3n = 2/3  (n) Hence T =  (n) !!!