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MCS680: Foundations Of Computer Science

int MSTWeight(int graph[][], int size) { int i,j; int weight = 0; for(i=0; i<size; i++) for(j=0; j<size; j++) weight+= graph[i][j]; return weight; }. 1 1. O (1) O (1). O (n). O (n). n. n. Running Time = 2 O (1) + O (n 2 ) = O (n 2 ).

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MCS680: Foundations Of Computer Science

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  1. int MSTWeight(int graph[][], int size) { int i,j; int weight = 0; for(i=0; i<size; i++) for(j=0; j<size; j++) weight+= graph[i][j]; return weight; } 1 1 O(1) O(1) O(n) O(n) n n Running Time = 2O(1) + O(n2) = O(n2) MCS680:Foundations Of Computer Science Iteration Induction Recursion Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  2. Introduction • Iteration • Computers have the ability to execute a task repeatedly • Can perform repetitive operations fast • Many common operations can be implemented by using iteration • Searching, sorting, copying, ... • Recursion • A concept is defined in terms of itself • Recurrence relation • Example: • “A list either is empty or it is an element followed by a list” • Recursion is supported in programming languages by allowing a function to call itself. • Example Factorial: • F(x) = x * F(x-1) given F(1) = 1 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  3. Introduction • Recursion Example: Factorial • Induction is used in mathematical proofs • Used to show that a statement is true • Technique “Proof by Induction” • Basis Step • Induction Hypothesis • Inductive Step int Factorial(int n) { if (n < 0) return -1; if ((n == 0) || (n == 1)) return 1; else return n * Factorial(n-1); } Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  4. Iteration • Iteration supported by looping constructs in programming languages • for loop, while loop, do...while loop • Used in many algorithms • Basis for automating computed solutions • Example: Sorting with Selection Sort • Many sorting algorithms exist • Varing complexity and performance • Insertion, merge, quick, heap, ... • Selection Sort • Goal: Sort an array of integers into a non-decreasing order • Partition the array into a sorted and a non-sorted parts sorted unsorted n-1 i 0 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  5. Selection Sort • Selection Sort • Iterating step • Find the smallest element in the unsorted portion of the array (in range [i ... n-1]) • Exchange the smallest element with the first element in the unsorted portion of the array • Element at index i • Advancement step • The array is now sorted from [0 ... i] • Advance i so that it is the first element in the unsorted portion of the array • Advance i by one position • Termination step • Stop advancing i when it gets to the last position in the array sorted unsorted n-1 i 0 movement of i Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  6. Selection Sort • Implementation of the selection sort in ‘C’ void SelectionSort(int A[], int n) { int i,j,small,temp; for(i=0; i<(n-2); i++) { //Get initial smallest element small = i; //Find actual smallest element for(j=i+1; j<n; j++) if(A[j] < A[small]) small = j; //Put smallest element in the first //location of the unsorted portion of //the array temp = A[small]; A[small] = A[i]; A[i] = temp; } } Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  7. Selection Sort - Example Run Example: Sort (10, 3, 9, 4, 1, 7) 10 3 9 4 1 7 10 3 9 4 1 7 i j i j small 1 3 9 4 10 7 1 3 9 4 10 7 i j i j small 1 3 9 4 10 7 1 3 9 4 10 7 i j i j small 1 3 4 9 10 7 1 3 4 9 10 7 i j i j small 1 3 4 7 10 9 1 3 4 7 10 9 i j i j small 1 3 4 7 9 10 i j Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  8. Induction • Inductive Proofs • Mathematical induction is a technique for proving that a statement S(n) is true for all integers above some lower limit • Generally for integers > 0 • Example • Prove: • Inductive proofs are performed using a three step process • Basis • Show S(n) holds for a base case • Induction Hypothesis • Assume for some fixed n, the hypothesis holds • Inductive Step • Show that S(n) implies S(n+1) Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  9. Induction Example • Prove by induction on n: Basis n=1: Induction Hypothesis -- Assume the following is true: Induction Step -- Show the following: Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  10. Induction Example Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  11. Induction Example • Prove by induction on n: Basis n=1: Induction Hypothesis -- Assume the following is true: Induction Step -- Show the following: Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  12. Induction Example Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  13. Why Does Induction Work? • Suppose that we want to show that S(n) is true for a particular value of n • Assume for the lowest value of n that S(n) holds (this can easily be shown in the basis case) • If we can show that S(n) implies S(n+1) by the inductive step then we can show that • S(0) implies S(1) • S(1) implies S(2) • S(2) implies S(3) • eventually we will reach: • S(n-2) implies S(n-1) • S(n-1) implies S(n) • Thus by repetitively applying the inductive step we can show that S(n) is true for any n so long as the basis case holds and we can show that S(n) implies S(n+1) • S(n) implies S(n+1) is shown in the inductive step Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  14. Induction Example -Error Correcting Codes • Background • Computers operate using binary data • Information is transmitted between components using an analog representation of the data • The analog data must be converted back to a digital representation before it can be used for computation • Example: The ASCII code for ‘C’ is 67 • This is 1000011 in binary • We want to send this character over a computer network Sending Computer C DAC Amp Noise Receiving Computer A DAC Amp Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  15. Error Correcting Codes -Parity • Because transmission is susceptible to error additional information must be added to the message to verify its correctness • Goal: Design an encoding scheme where all codes differ by two or more bits • Consider ‘A’ and ‘C’ • A: 1000001 C: 1000011 • An error in the 6th or 7th bit will make the two characters indistinguishable • Solution: Add a parity bit to ensure that all codes differ by two or more bits • Even Parity: The sum of the on bits is even • Odd Parity: The sum of the on bits is odd Standard Code Even Parity Odd Parity ‘A’ 1000001 01000001 11000001 ‘C’ 1000011 11000011 01000011 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  16. Error Correcting Codes-Parity Signaling With Even Parity Sending Computer C DAC Amp Noise Receiving Computer xx DAC Amp Signaling With Odd Parity Sending Computer C DAC Amp Noise Receiving Computer xx DAC Amp Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  17. Error Correcting Codes-Induction • Show that if C is a set of bit strings of length n, then C contains at most 2n-1 strings. Assume that n = len(C) • Basis Step • Assume n=2 (Definition kind of funny for n=0 and n=1) • C = {(00),(1,1)} because C is error correcting, |C| = 2 • All strings differ by two or more bits • 2(2-1) = 21 =2 • Basis holds because for n=2, |C| = 2 and2(n-1) =2 • Induction Hypothesis • Assume for n>2 that C contains at most 2(n-1) strings • Induction Step • Show that for strings of length (n+1) that C contains at most 2n strings Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  18. Error Correcting Codes-Induction • Induction Step • By the induction hypothesis |C| = 2n-1 • 50% prefixed by a ‘0’, 50% prefixed by a ‘1’ • Split C into C0 and C1 based on the first bit: • |C0| = |C1| = (2n-1 /2) • Because C0 & C1 are error correcting, so are D0 and D1 • Adding a zero bit in front of D0 or a one bit in front of D1 still ensures that the resultant strings differ in two or more bits • If this was false C0 and C1 would not be error correcting D0 C0 ‘0’ C ‘1’ D1 C1 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  19. Error Correcting Codes-Induction • Now let n= length(D0) or length(D1) • Note: length(D0) = length(D1) • There are 2n-1 strings in D0 and D1 • D0 and D1 are error correcting (shown before) by the inductive hypothesis • Now, prefix all D0 strings by a zero bit (=C0) • Also, prefix all D1 strings by a one bit (=C1) • length(C0) = length(C1) = n+1 • |C0| = |C1| = 2n-1 • Because C0 and C1 were derived from D0 and D1 by adding a single bit D0 C0 ‘0’ C ‘1’ D1 C1 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  20. Error Correcting Codes-Induction • C = C0  C1 • |C| = |C0| + |C1| • |C| = 2n-1 + 2n-1 = 2n • length(C) = length(C0) = length(C1) = n+1 • Thus we have shown that given C of length n+1 that the total number of error correcting strings in C is |C| = 2n • This verifies the induction hypothesis • Thus we have proved that given C, which is a set of bit strings, contains at most 2n-1 strings. D0 C0 ‘0’ C ‘1’ D1 C1 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  21. Complete Induction • In ordinary induction we have: • A basis case • An induction hypothesis • An induction step that shows that S(n) implies S(n+1) • Thus because S(n) implies S(n+1) we may continuously reapply this definition to work up from the base case to an S(n) • Complete Induction • Same as ordinary induction for the base case and induction hypothesis • The induction step differs because we may use all values of i from the base case up until S(i) to show S(i+1) • Ordinary induction only uses S(i) to show S(i+1) Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  22. Recursive Definitions • Define a class of related objects in terms of the objects themselves • Recursive definitions must have • A basis case in which one or more simple objects are defined • An inductive or recursive step in which larger objects are defined in terms of the smaller ones in the collection • Example: Factorial • Definition: n! = 1*2*3* ... * (n-1) * n • Basis case: • Factorial(n) not defined for n < 0 • Factorial(n) = 1 for n  {0,1} • Recursive/Inductive Step: • Factorial(n) = n * Factorial(n-1) for n > 1 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  23. Induction on a Recursive Definition • Show that Factorial(n) = n * Factorial(n-1) for all n > 0 • Basis Case: Factorial(1) • Factorial(1) = 1 * Factorial(0) = 1*1 = 1 = 1! • Thus the basis case holds • Inductive Hypothesis • Assume Factorial(n) = n * Factorial(n-1) = n! for all n > 1 • Induction Step • Show that Factorial(n+1) = (n+1) * Factorial((n+1)-1) for all n>1 • By the inductive hypothesis we know that Factorial((n+1)-1) = Factorial(n) = n! = 1*2* ... * (n-1) * (n) • Thus Factorial(n+1) = (n+1) * Factorial(n) which is (n+1) * n! (from the previous step) • This can be rewritten as n! * (n+1) which equals 1*2* ... * (n-1) * n * (n+1) which is (n+1)! by the mathematical definition of the factorial operator Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  24. Recursive Procedures • A recursive procedure (or function) is one that is called from within its own body • Direct: f() calls f() from within its body • Indirect: f() calls f1() which in turn calls f2() which calls f() • This can be generalized for any number of indirect calls so long as the original function gets called from within its body • Recursive procedures are easy to code • Can directly map recursive definition into source code • Must have a basis case and an inductive part • Basis case resolves to a direct answer based on the base case of the recursive definition • Used an an exiting condition • Inductive part makes one or more call to the recursive procedure • Generally the procedure parameters are changed Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  25. Recursive Procedures-Factorial() Function • Recall the recursive definition of the factorial operation • Factorial(0) = Factorial(1) = 1 • Factorial(n) = n * Factorial(n-1) • Develop the code directly from the factorial recursive definition: • Notice how the above program fragment directly maps to the factorial recursive definition int Factorial(int n) { if (n < 0) return -1; if ((n == 0) || (n == 1)) return 1; else return n * Factorial(n-1); } Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  26. return 5 * Factorial(5-1) return 4 * Factorial(4-1) return 3 * Factorial(3-1) return 2 * Factorial(2-1) return 1 return 1* 2 return 2*3 return 6*4 return 24*5 Recursive Procedures-Factorial() Function Analyze a call to Factorial(5): • With recursive procedures the system stack is used to “remember” values needed to evaluate the recursive function once the basis case is reached • Stack must be restored to its original state • Efficiency issues: Stack space is limited and procedure calls are (relatively) slow Procedure Calls Stack <empty> 5 4, 5 3, 4, 5 2, 3, 4, 5 3, 4, 5 4, 5 5 <empty> Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  27. Divide and Conquer Techniques • Divide the problem into subproblems • Solve the subproblems directly or by futher dividing into more subproblems • Ideal for implementation with recursive procedures • Requirements • The subproblems must be simpler then the original problem • The inductive definition • After a finite number of subdivisions we must encounter a subproblem that can be directly solved • The basis case • If the above requirements are not met then the recursive procedure may continue to subdivide without ever reaching a condition that can be directly solved Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  28. Transformation of Selection Sort Into a Recursive Procedure Recall the original Selection Sort Algorithm: • Selection Sort - Recursive • Selection step • Find the smallest element in the unsorted portion of the array (in range [i ... n-1]) • Exchange the smallest element with the first element in the unsorted portion of the array • Element at index i • Advancement/Recursive step • The array is now sorted from [0 ... i] • Resort the array using the selection sort routine for [i+1...n] • Termination step • Stop resorting the array when i gets to the last position in the array sorted unsorted n-1 i 0 movement of i Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  29. Transformation of Selection Sort Into a Recursive Procedure Now the Recursive Selection Sort Algorithm: • Selection Sort • Iterating step • Find the smallest element in the unsorted portion of the array (in range [i ... n-1]) • Exchange the smallest element with the first element in the unsorted portion of the array • Element at index i • Advancement step • The array is now sorted from [0 ... i] • Advance i so that it is the first element in the unsorted portion of the array • Advance i by one position • Termination step • Stop advancing i when it gets to the last position in the array sorted unsorted n-1 i 0 movement of i Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  30. Transformation of Selection Sort Into a Recursive Procedure void SelectionSort(int A[], int idx, int n) { int j,small,temp; //BASIS CASE: If idx=(n-1)then array sorted if (idx < (n - 2)) { //Get initial smallest element small = idx; //Find actual smallest element for(j=idx+1; j<n; j++) if(A[j] < A[small]) small = j; //Smallest element put in the first //location in the unsorted array side temp = A[small]; A[small] = A[idx]; A[idx] = temp; //Recursively sort the rest of the array SelectionSort(A, idx+1, n); } } Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  31. Comparison: Iterative vs Recursive Selection Sort • Selection Sort • Initialization operations • Comparisons to find minimum element • Copy operation to save minimum element • Copies to exchange minimum element with the first element in the unsorted portion of the array • Iterative Algorithm • Initialization operationssmall = i • This operation executes (n-1) times in the outer loop • Comparisons to find minimum elementfor(j=i+1; j<n; j++) if(A[j] < A[small]) small = j; • This operation executes (n-i) times in the inner loop and (n-1) times in the outer loop Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  32. Comparison: Iterative vs Recursive Selection Sort • Iterative Algorithm (con’t) • Comparisons to find minimum element can be bounded by the following sum: • (n-i) is the worse case on the number of comparisons that are performed in the inner loop • There are (n-1) copies of j into small (worst case) • And the inner loop is executed (n-1) times • Thus the number of comparisons can be bounded by [n(n-1)]/2: Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  33. Comparison: Iterative vs Recursive Selection Sort • Iterative Algorithm (con’t) • Copies to exchange minimum element with the first element in the unsorted portion of the arraytemp = A[small]; A[small] = A[i]; A[i] = temp; • The above 3 copy statements are executed (n-1) times by the outer loop • Total performance of the iterative algorithm: Selection Sort Performance (Iterative)= (n-1) copies for initialization + [n(n-1)]/2 comparisons + (n-1) copies to save smallest element + 3(n-1) copies for the exchange = [n(n-1)]/2 comparisons + 5(n-1) copies Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  34. Comparison: Iterative vs Recursive Selection Sort • Consider the following lines of code in the recursive version of SelectionSort():void SelectionSort(int A[], int id, int n){ ... if(idx < (n-2)) { ... SelectionSort(A,idx+1,n); }} • Notice that the SelectionSort() routine is called (n-1) times • Initialization operationssmall = i • This operation executes (n-1) times due to (n-1) calls to SelectionSort() Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  35. Comparison: Iterative vs Recursive Selection Sort • Recursive Algorithm (con’t) • Comparisons to find minimum element can be bounded by the following sum: • (n-i) is the worse case on the number of comparisons that are performed in the inner loop • And the inner loop is executed (n-1) times due to (n-1) calls to SelectionSort() • Also there are (n-1) copies of j into small (worst case) • Thus the number of comparisons can be bounded by [n(n-1)]/2 • Shown before in the iterative algorithm Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  36. Comparison: Iterative vs Recursive Selection Sort • Recursive Algorithm (con’t) • Copies to exchange minimum element with the first element in the unsorted portion of the arraytemp = A[small]; A[small] = A[i]; A[i] = temp; • The above 3 copy statements are executed (n-1) times due to (n-1) calls to SelectionSort() • Total performance of the recursive algorithm: Selection Sort Performance (Recursive)= (n-1) copies for initialization + [n(n-1)]/2 comparisons + (n-1) copies to save smallest element + 3(n-1) copies for the exchange = [n(n-1)]/2 comparisons + 5(n-1) copies Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  37. Summary: Iterative vs Recursive Selection Sort • The iterative and recursive version of the selection sort have the same execution complexity: • [n(n-1)]/2 comparisons + 5(n-1) copies • If a comparison and a copy take about the same amount of execution time, execution complexity becomes: • [n(n-1)]/2 + 5(n-1) operations Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  38. Merge Sort: A “Real” Recursive Sorting Algorithm • A divide and conquer technique • List A = (a1, a2, ... , an) is sorted by • Divide the list into two lists that are half the size of the original list • Sort each of the half sized lists separately • Merge the two sorted sub-lists to produce a sorted list, Asorted • Merge sort grows much slower then the selection sort for larger values of n • Merging • Produce a sorted list from two sorted lists Sorted List 1 Merge Sorted List Sorted List 2 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  39. Merge Sort • Merge Algorithm • Examine the element at the front of both lists • Pick the smallest from one of the lists • Ties are broken arbitrarily • Add the smallest element (chosen above) to the sorted list • Remove the chosen element from the front of the list • Continue until both lists are empty • The resultant output list will be sorted • The algorithm assumes that the two input lists have been pre-sorted • Merge algorithm will not produce a sorted list unless the input lists are sorted! Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  40. Merging With Linked Lists • The merge routine will use the following data structure to represent a merge list:typedef struct _mList { int element; struct _mList next; }*mList; • The merge algorithm can then be recursively defined for two lists (given lst1 and lst2): • Basis cases: • lst1 is NULL, add lst2 to the merge list • lst2 is NULL, add lst1 to the merge list • if neither lst1 nor lst2 is null then add the smallest of element in the front of lst1 or lst2 to the merge list • Merge steps • if lst1->element < lst2->element • remove the first element from the lst1 list and add it to the merge list Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  41. Merging With Linked Lists • Merge Steps (con’t) • if lst2->element < lst1->element • remove the first element from the lst2 list and add it to the merge list • Termination step: Stop merging when both lst1 and lst2 are empty mList Merge(mList lst1, mList lst2) { mList *ml; if (lst1 == NULL) ml = lst2; else if (lst2 == NULL) ml = lst1; else if (lst1->element <= lst2->element) { //Ans: lst1->element + merge of remainder lst1->next = Merge(lst1->next, lst2); ml = lst1; } else { //Ans: lst2->element + merge of remainder lst2->next = Merge(lst1, lst2->next); ml = lst2; } return ml; //Return the merged list } Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  42. Splitting Linked Lists • Merge sort must be able to split a list into two equal parts • The lists may differ by one in length if the input list to be split is odd in length • Options • Split at the midpoint • Split taking every other element 7 1 3 5 7 1 3 5 SplitMidpoint() SplitEveryOther() 7 1 3 5 7 3 1 5 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  43. Split Routine (Every Other) • A recursive routine to split a list • Input list is modified with every other element removed • A new list is created based on the removed elements from the input list • The new list is returned from the Split() function mList Split(mList orig) { mList secondCell; if (orig == NULL) return NULL; else if (orig->next == NULL) return NULL; else { secondCell = orig->next; orig->next = secondCell->next; secondCell->next = Split(secondCell->next); return secondCell; } } Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  44. Merge Sort • Now we are able to split a list and merge a sorted list • Notice that a list of length 1 is sorted • MergeSort() algorithm • Recursively split the input lists until the list lengths are 1 • Lists of length 1 are sorted • Result is a tree • Recursively merge the sorted lists • Build the resultant sorted list by recursively merging the lists in the tree that was created in the previous step Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  45. MergeSort Algorithm Input List 742897721 Splitting 742897721 Direction of Execution 72971 4872 791 27 47 82 2 4 8 71 7 7 2 9 7 1 Merging 12247789 12779 2478 Direction of Execution 179 27 47 28 2 4 8 17 7 7 2 9 7 1 Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  46. MergeSort Algorithm • Recursively split list • Continue until list sizes are 1 • Recursively merge lists • Continue until entire list is sorted • Resultant list is sorted void MergeSort(mList list) { mList SecondList; if (list != NULL) { if(list->next != NULL) { SecondList = Split(list); MergeSort(list); MergeSort(SecondList); list = Merge(list,SecondList); } } } Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  47. Rough Analysis-MergeSort() • Merge() - 2(n-1) • (n-1) recursive calls • 2 assignments each call • Split() - 3(n-1) • (n-1) recursive calls • 3 assignments each call • MergeSort() - [10n - 6]lg n • 2 lg n recursive calls • MergeSort() is called twice from within its body • Each call is with a list of length (n/2) • Each recursive call contains • A call to Split() • A call to Merge() • 2 assignments Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  48. Rough Analysis-MergeSort() • Merge Sort Performance (con’t) Merge Sort Performance = (2 calls to MergeSort() * lg n internal recursive calls) * (time to Merge() + time to Split + 2 assignments) = (2 lg n) * [2(n-1) + 3(n-1) + 2] = (2 lg n) * [2n - 2 + 3n -3 + 2] = (2 lg n) * [5n - 3] = (10n - 6) lg n Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  49. Merge Sort Performance Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

  50. Merge vs Selection Sort • Selection Sort - Quadratic Growth • Merge Sort - Logarithmic Growth Brian Mitchell (bmitchel@mcs.drexel.edu) - Drexel University MCS680-FCS

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