1. CSC 172 DATA STRUCTURES

2. THEORETICAL BOUND  Many good sorting algorithms run in O(nlogn) time.  Can we do better?  Can we reason about algorithms not yet invented?

3. THEORETICAL BOUND  We can think of any comparison based sorting algorithm as a decision tree

4. THEORETICAL BOUND  A binary tree of depth d has at most 2^d leaves.  A binay tree with L leaves must have depth at lease log(L)  There are n! Arrangements of n items  A tree with n! Leaves must have depth at least log(n!)  log(n!) is at least Ω (n log n)

5. THEORETICAL BOUND log(n!) = log(n(n-1)(n-2)...(2)(1)) = log(n) + log(n-1) +.... + log(2) + log(1) >= log(n) + log(n-1) +.... + log(n/2) >= (n/2) log(n/2) >=(n/2) log(n) – (n/2) = Ω (n log n)

6. EXAMPLE A post office routes mail letters are sorted into separate bags for different geographical areas, each of these bags is itself sorted into batches for smaller sub-regions, and so on until they are delivered. Radix Sort

7. RADIX SORT  Radix sort considers the structure of the keys  Assume keys are represented in base M  Sorting is done by comparing bits in the same position

8. RADIX EXCHANGE SORT  Examine bits from left to right  Sort array with the respect to the leftmost bit 1XXX 0XXX 1XXX

9. RADIX EXCHANGE SORT  Scanning Partition REPEAT ; scan top down to find a “1” scan bottom up to find a “0” exchange UNTIL: scans cross 1XXX 0XXX 1XXX 0XXX 1XXX 0XXX 1XXX

10. RADIX EXCHANGE SORT  Examine bits from left to right  Sort array with the respect to the leftmost bit 1XXX 0XXX 1XXX 0XXX 1XXX

11. RADIX EXCHANGE SORT  Examine bits from left to right  Sort array with the respect to the leftmost bit 1XXX 0XXX 1XXX 0XXX 1XXX TIME: O(b N)

12. DIVIDE AND CONQUER Mergesort Quicksort Maximum Subsequence Fast Fourier Transform

13. DIVIDE AND CONQUER Divide: Solve a subproblem by recursively calling on a subset of the data Conquer: The solution to the larger problem is formed from the solutions to the sub problems.

14. Example  One dimensional pattern recognition  Input: a vector x of n floating point numbers  Output: the maximum sum found in any contiguous subvector of the input.  X[2..6] or 187 84-23-939758-532659-4131

15. Obvious solution //check all pairs int sum; int maxsofar = 0; for (int i = 0; i<x.length;i++) for (int j = i; j<x.length;j++){ sum = 0; for (int k = i;k<=j;k++) sum += x[k]; maxsofar = max(sum,maxsofar); }

16. A better solution // check all pairs int sum; int maxsofar = 0; for (int i = 0; i<x.length;i++) { sum = 0; for (int j = i; j<x.length;j++){ sum += x[k]; // the sum of x[i..j] maxsofar = max(sum,maxsofar); }

17. Divide & Conquer  To solve a problem of size n, recursively solve two sub-problems of size of size n/2, and combine their solutions to yield a solution to the complete problem.

18. D&C for LCS ab x mama mbmb m a, m b or: mcmc

19. Recursive D&D LCS public int LCS(int[] x){ return LCS(x,0,x.length-1); } public int LCS(int[] x, int low, high ){ // the hard part

20. Recursive D&C LCS public int LCS(int[] x, int low, int high){ if (low>high) return 0; if (low == high) return max(0,x[low]); int mid = (low + high) /2; return max(LCS(x,low,mid), LCS(x,mid+1,high)); }// still need to do “m c ”

21. How to find m c ?  Note that mc consists of two parts  The part starting at the boundary and reaching up  The part ending at the boundary and reaching down  The sum of these is m c mcmc m clower m cup

22. public int LCS(int[]x,int low,int high){ if (low>high) return 0; if (low == high) return max(0,x[low]); int mid = (low + high) /2; int umax = findUmax(x,mid+1,upper); int lmax = findLmax(x,low,middle); return max(LCS(x,lower,middle), LCS(middle+1,upper), lmax + umax); }

23. findLmax int findLmax(int[]x,int low,int mid){ double lmax = 0, sum = 0; for (int j = mid;j>=low;j--){ sum+=x[j]; lmax = max(lmax,sum); } return lmax; } // Run Time? In terms of middle-lower?

24. findUmax int findLmax(int[]x, int mid1,int high){ int umax = 0, sum = 0; for (int j = midd1;j<=high;j++){ sum+=x[j]; umax = max(lmax,sum); } return umax; } // Run Time? In terms of high-mid1?

25. Runtime of Div&Conq Please : Read p 449 & 450 of Weiss

26. RUNTIME T(N) = 2T(N/2) + O(N) is O(N log N)

27. In General

28. by telescoping because