ITEC2620 Introduction to Data Structures

Transcription

1 ITEC2620 Introduction to Data Structures Lecture 3a Complexity Estimation Best, Worst, and Avg Cases I Why do people buy lottery tickets? Win millions! Case? Best Best, Worst, and Avg Cases II Why should you invest your retirement savings in the stock market? Best long-term investment (compared to cash and bonds) Case? Average Best, Worst, and Avg Cases III Why should you not fly a kite in a thunderstorm? Get killed! Case? Worst 1

2 Other Cases I Buying lottery tickets Average case Waste of money Other Cases II Investing in the stock market Worst case Market crash! Other Cases III Flying a kite in a thunderstorm Best case Get famous! (lie ) Which case should we use? Average Case? Most often Best Case? When reasonable expectation exists Worst Case? When guarantees are required 2

3 Best, Worst, and Average Cases in Algorithm Analysis I Best, Worst, and Average cases exist when there is conditional execution while loops Branches with different cost alternatives If it runs the same way every time, there is only one case Best, Worst, and Average Cases in Algorithm Analysis II Best case Minimal execution Worst case Maximal execution Average case Weighted average of all possible cases Only (best+worst)/2 when they have equal probability Complexity Estimation I The most important thing about algorithm analysis is to determine big- Oh the approximate complexity Allows us to estimate program run times Do we need all the math? Not really Complexity Estimation II Two key questions How many times? What does it cost? 3

4 Complexity Estimation III How many times? is the summation What does it cost? is what s inside the summation more accurate questions What does it cost each time? What does it cost on average? Complexity Estimation IV Notes How many times? implies multiplication What does it cost? can be another question pair Example I for (int i = 0; i < n; i++) if (Math.random() > 0.5) if (i%2 == 0) O(logn) else O(1) else for (int j = 0; j < i; j++) for (int k = 0; k < i; k++) O(1) Example II Is there a best, worst, and average case? Yes Branches with different cost alternatives 4

5 Example III Best case Math.random() always > 0.5 How many times? n What does it cost (on average)? (1+logn) / 2 Overall? n(1+logn)/2 O(nlogn) Example IV Average Case Best and Worst case have probability (average them) Example V Worst case Math.random() never > 0.5 How many times? n What does it cost (on average)? Example VI Go straight to big-oh notation on each line for (int i = 0; i < n; i++) // O(n) for (int j = 0; j < i; j++) // O(n) for (int k = 0; k < i; k++) // O(n) O(1) 5

6 Example VII Worst case is O(n) * O(n) * O(n) * O(1) O(n 3 ) First Code Example I Matrix multiplication A nn * B nn = C nn Average case is ( O(nlogn) + O(n 3 ) ) /2 O(n 3 ) A B C First Code Example II for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { c[i][j] = 0; for (int k = 0; k < n; k++) c[i][j] += a[k][j] * b[j][k]; } First Code Example III for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { c[i][j] = 0; for (int k = 0; k < n; k++) c[i][j] += a[k][j] * b[j][k]; } //O(n) //O(n) //O(1) //O(n) //O(1) 6

7 First Code Example IV How many times (does the first for loop happen)? n What does it cost (each time through the loop? Another loop The stuff inside the loop First Code Example V O(n) * O(n) * { O(1) + [ O(n) * O(1) ] } First loop Second loop Third loop First Code Example VI O(n) * O(n) * { O(1) + [ O(n) * O(1) ] } Cost of third loop = O(n) * O(1) = O(n) O(n) * O(n) * { O(1) + O(n) } Inside second loop = O(1) + O(n) = O(n) O(n) * O(n) * O(n) O(n 3 ) Second Code Example I for (int i = 0; i < a.length; i++) int min = a[i]; for (int j = i; j < a.length; j++) if (a[j] < a[i]) min = a[j]; minlocation = j; swap(i, minlocation) 7

8 Second Code Example II for (int i = 0; i < a.length; i++) // O(n) int min = a[i]; // O(1) for (int j = i; j < a.length; j++) // O(n) if (a[j] < a[i]) // O(1) min = a[j]; // O(1) minlocation = j; // O(1) swap(i, minlocation) // O(1) Second Code Example III O(n) * { O(1) + [ O(n) * O(1) ] + O(1) } Cost of second loop = O(n) * O(1) = O(n) O(n) * { O(1) + O(n) + O(1) } Inside loop = O(1) + O(n) + O(1) = O(n) O(n) * O(n) O(n 2 ) Second Code Example IV Pseudocode loop through all elements (have to put n elements into place) for loop loop through all remaining elements and find smallest initialization, for loop, and branch swap smallest element into correct place single method Second Code Example V What does it cost? n times through the outer loop n elements to sort What does it cost (on average)? Look at n/2 elements swap smallest element into correct place O(n) * {O(n) + O(1) } = O(n 2 ) 8

9 Third Code Example I a is an array with n elements selectionsort(a); for (int i = 0; i < n; i++) binarysearch(i,a); Third Code Example II selectionsort(a); // O(n 2 ) for (int i = 0; i < n; i++) // O(n) binarysearch(i,a); // O(logn) Third Code Example III O(n 2 ) + { O(n) * O(logn) } loop Happens once Readings and Assignments Suggested Readings from Shaffer (third edition) 3.2, Exercise 3.12 O(n 2 ) + O(n logn) O(n 2 ) 9