1. Computer Science 112 Fundamentals of Programming II Searching, Sorting, and Complexity Analysis

2. Things to Desire in a Program Correctness Robustness Maintainability Efficiency

3. Measuring Efficiency Empirical - use clock to get actual running times for different inputs Problems: –Different machines have different running times –Some running times are so long that it is impractical to check them

4. Measuring Efficiency Analytical - use pencil and paper to determine the abstract amount of work that a program does for different inputs Advantages: –Machine independent –Can predict running times of programs that are impractical to run

5. Complexity Analysis Pick an instruction that will run most often in the code Determine the number of times this instruction will be executed as a function of the size of the input data Focus on abstract units of work, not actual running time

6. Example: Search for the Minimum def ourMin(lyst): minSoFar = lyst[0] for item in lyst: minSoFar = min(minSoFar, item) return minSoFar We focus on the assignment ( = ) inside the loop and ignore the other instructions. for a list of length 1, one assignment for a list of length 2, 2 assignments. for a list of length n, n assignments

7. Big-O Notation Big-O notation expresses the amount of work a program does as a function of the size of the input O(N) stands for order of magnitude N, or order of N for short Search for the minimum is O(N), where N is the size of the input (a list, a number, etc.)

8. Common Orders of Magnitude ConstantO(k) LogarithmicO(log 2 n) LinearO(n) QuadraticO(n 2 ) ExponentialO(k n )

9. Graphs of O(n) and O(n 2 )

10. Common Orders of Magnitude n O(log 2 n) O(n) O(n 2 )O(2 n ) 2 1 2 4 4 4 2 4 16 64 8 3 8 64 256 16 4 16 256 65536 32 5 32 1024 4294967296 64 6 64 4096 19 digits 128 7 128 16384 yikes! 256 8 256 65536 512 9 512 262144 1024 10 1024 1048576

11. Suppose an algorithm requires exactly 3N + 3 steps As N gets very large, the difference between N and N + K becomes negligible (where K is a constant) As N gets very large, the difference between N and N / K or N * K also becomes negligible Use the highest degree term in a polynomial and drop the others (N 2 – N)/2  N 2 Approximations

12. Example Approximations n O(n) O(n) + 2 O(n 2 ) O(n 2 ) + n 2 2 4 4 6 4 4 6 16 20 8 8 10 64 72 16 16 18 256 272 32 32 34 1024 1056 64 64 66 4096 5050 128 128 130 16384 16512 256 256 258 65536 65792 512 512 514 262144 262656 1024 1024 1026 1048576 1049600

13. def ourIn(target, lyst): for item in lyst: if item == target: return True # Found target return False # Target not there Example: Sequential Search Which instruction do we pick? How fast is its rate of growth as a function of n? Is there a worst case and a best case? An average case?

14. Improving Search Assume data are in ascending order Goto midpoint and look there Otherwise, repeat the search to left or to right of midpoint 34415663728995 0 1 2 3 4 5 6 89 target 3 midpoint leftright

15. Improving Search Assume data are in ascending order Goto midpoint and look there Otherwise, repeat the search to left or to right of midpoint 34415663728995 0 1 2 3 4 5 6 89 target 5 midpoint rightleft

16. Example: Binary Search def ourIn(target, sortedLyst): left = 0 right = len(sortedLyst) - 1 while left <= right: midpoint = (left + right) // 2 if target == sortedLyst[midpoint]: return True elif target < sortedLyst[midpoint]: right = midpoint - 1 else: left = midpoint + 1 return False

17. Analysis while left <= right: midpoint = (left + right) // 2 if target == sortedLyst[midpoint]: return True elif target < sortedLyst[midpoint]: right = midpoint - 1 else: left = midpoint + 1 How many times will == be executed in the worst case?

18. Sorting a List 34415663728995 0 1 2 3 4 5 6 89566372413495 0 1 2 3 4 5 6 sort

19. Selection Sort For each position i in the list –Select the smallest element from i to n - 1 –Swap it with the ith one

20. Trace 89566372413495 0 1 2 3 4 5 6 smallest i Step 1: find the smallest element 34566372418995 0 1 2 3 4 5 6 i Step 2: swap with first element 34566372418995 0 1 2 3 4 5 6 i Step 3: advance i and goto step 1

21. for each i from 0 to n - 1 minIndex = minInRange(lyst, i, n) if minIndex != i swap(lyst, i, minIndex) Design of Selection Sort minInRange returns the index of the smallest element swap exchanges the elements at the specified positions

22. def selectionSort(lyst): n = len(lyst) for i in range(n): minIndex = minInRange(lyst, i, n) if minIndex != i: swap(lyst, i, minIndex) Implementation

23. def minInRange(lyst, i, n): minValue = lyst[i] minIndex = i for j in range(i, n): if lyst[j] < minValue: minValue = lyst[j] minIndex = j return minIndex def selectionSort(lyst): n = len(lyst) for i in range(n): minIndex = minInRange(lyst, i, n) if minIndex != i: swap(lyst, i, minIndex)

24. Implementation def swap(lyst, i, j): lyst[i], lyst[j] = lyst[j], lyst[i] def minInRange(lyst, i, n): minValue = lyst[i] minIndex = i for j in range(i, n): if lyst[j] < minValue: minValue = lyst[j] minIndex = j return minIndex def selectionSort(lyst): n = len(lyst) for i in range(n): minIndex = minInRange(lyst, i, n) if minIndex != i: swap(lyst, i, minIndex)

25. Analysis of Selection Sort The main loop runs approximately n times Thus, the function minInRange runs n times Within the function minInRange, a loop runs n - i times

26. Analysis of Selection Sort Overall, the number of comparisons performed in function minInRange is n - 1 + n - 2 + n - 3 +.. + 1 = (n 2 – n) / 2  n 2

27. For Thursday Finding Faster Algorithms