1. Chapter 2

2. Complexity Space Complexity Space Complexity Instruction Space Instruction Space Data Space Data Space Enviroment Space Enviroment Space Time Complexity Time Complexity Operation count Operation count Asymptotic Notation Asymptotic Notation

3. Instruction Space Operation a+b+b*c+(a+b-c)7(a+b)+4

4. Data Space Borland C++

5. Enviromental Space template T Rsum(T a[], int n) {// Return sum of numbers a[0:n - 1]. if (n > 0) return Rsum(a, n-1) + a[n-1]; return 0;} Recursive Program Rsum(a,n) Rsum(a,n-1) Rsum(a,n-2)... Rsum(a,1) Rsum(a,0)

6. Space Complexity Space Required Fixed PartVariable Part The Space Requirement S(P) = c + S P (instance characteristics)

7. Time Complexity Compilation Time Compilation Time Run Time Run Time n : instance characteristics

8. Example max element template int Max(T a[], int n) {// Locate the largest element in a[0:n-1]. int pos = 0; for (int i = 1; i < n; i++) if (a[pos] < a[i]) pos = i; return pos; } Total n-1 comparisons

9. Example Polynomial Evaluation template T PolyEval(T coeff[], int n, const T& x) {// Evaluate the degree n polynomial with // coefficients coeff[0:n] at the point x. T y = 1, value = coeff[0]; for (int i = 1; i <= n; i++) { // add in next term y *= x; value += y * coeff[i]; } return value; } Number of addition : n Number of Multiplication : 2n

10. Example (2) Polynomial Evaluation Number of addition : n Number of Multiplication : n Horner’s rule template T Horner(T coeff[], int n, const T& x) {// Evaluate the degree n polynomial with // coefficients coeff[0:n] at the point x. T value = coeff[n]; for (int i = 1; i <= n; i++) value = value * x + coeff[n - i]; return value; }

11. Example Rank The rank of an element in a sequence is the number of smaller elements in the sequence plus the number of equal elements that appear to its left Ex: a[4 3 9 3 7] : rank r[2 0 4 1 3] template void Rank(T a[], int n, int r[]) {// Rank the n elements a[0:n-1]. for (int i = 0; i < n; i++) r[i] = 0; //initialize // compare all element pairs for (int i = 1; i < n; i++) for (int j = 0; j < i; j++) if (a[j] <= a[i]) r[i]++; else r[j]++; } The total number of comparison : 1 + 2 + …+ n-1 = (n-1)*n/2

12. Example Rank Sort template void Rearrange(T a[], int n, int r[]) {// Rearrange the elements of a into sorted order // using an additional array u. T *u = new T [n+1]; // move to correct place in u for (int i = 0; i < n; i++) u[r[i]] = a[i]; // move back to a for (int i = 0; i < n; i++) a[i] = u[i]; delete [] u; } The number of elements moved : 2n

13. Example Selection Sort The number of comparison = 1 + 2 + 3 +… + n-1 = (n-1)*n/2 The number of elements moved : 3 (n-1) template void SelectionSort(T a[], int n) {// Sort the n elements a[0:n-1]. for (int size = n; size > 1; size--) { int j = Max(a, size); Swap(a[j], a[size - 1]); }

14. Example Buble Sort The number of comparison = 1 + 2 + 3 +… + n-1 = (n-1)*n/2 The number of elements moved : 3 (n-1) (worsed case) template void Bubble(T a[], int n) {// Bubble largest element in a[0:n-1] to right. for (int i = 0; i < n - 1; i++) if (a[i] > a[i+1]) Swap(a[i], a[i + 1]); } template void BubbleSort(T a[], int n) {// Sort a[0:n - 1] using bubble sort. for (int i = n; i > 1; i--) Bubble(a, i); }

15. What is worst case what is best case Sequential search Sequential search Best case : 1 Worst case : n On average :

16. Insertion into Sorted Array template void Insert(T a[], int & n, const T& x) {// Insert x into the sorted array a[0:n-1]. // Assume a is of size > n. int i; for (i = n-1; i >= 0 && x < a[i]; i--) a[i+1] = a[i]; a[i+1] = x; n++; // one element added to a } On average :