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Sorting Algorithms 2

Sorting Algorithms 2. Quicksort. General Quicksort Algorithm: Select an element from the array to be the pivot Rearrange the elements of the array into a left and right subarray All values in the left subarray are < pivot All values in the right subarray are > pivot

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Sorting Algorithms 2

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  1. Sorting Algorithms2

  2. Quicksort General Quicksort Algorithm: • Select an element from the array to be the pivot • Rearrange the elements of the array into a left and right subarray • All values in the left subarray are < pivot • All values in the right subarray are > pivot • Independently sort the subarrays • No merging required, as left and right are independent problems [ Parallelism?!? ]

  3. Quicksort void quicksort(int* arrayOfInts, int first, int last) { int pivot; if (first < last) { pivot = partition(arrayOfInts, first, last); quicksort(arrayOfInts,first,pivot-1); quicksort(arrayOfInts,pivot+1,last); } }

  4. Quicksort int partition(int* arrayOfInts, int first, int last) { int temp; int p = first; // set pivot = first index for (int k = first+1; k <= last; k++) // for every other indx { if (arrayOfInts[k] <= arrayOfInts[first]) // if data is smaller { p = p + 1; // update final pivot location swap(arrayOfInts[k], arrayOfInts[p]); } } swap(arrayOfInts[p], arrayOfInts[first]); return p; }

  5. 9 9 9 18 5 5 5 18 3 3 3 18 20 20 20 3 5 9 18 20 Partition Step Through partition(cards, 0, 4) P = 0 K = 1 P = 1 K = 3 cards[1] < cards[0] ? No cards[3] < cards[0]? Yes P = 2 P = 0 K = 2 temp = cards[3] cards[2] < cards[0] ? Yes cards[3] = cards[2] P = 1 cards[2] = cards[3] temp = cards[2] P = 2 K = 4 cards[2] = cards[1] cards[4] < cards[0]? No cards[1] = temp temp = cards[2], cards[2] = cards[first] cards[first] = temp, return p = 2;

  6. Complexity of Quicksort • Worst case is O(n2) • What does worst case correspond to? • Already sorted or near sorted • Partitioning leaves heavily unbalanced subarrays • On average is O(n log2n), and it is average a lot of the time.

  7. Complexity of Quicksort Recurrence Relation: [Average Case] 2 sub problems ½ size (if good pivot) Partition is O(n) a = 2 b = 2 k = 1 2 = 21 Master Theorem: O(nlog2n)

  8. Complexity of Quicksort Recurrence Relation: [Worst Case] • Partition separates into (n-1) and (1) • Can’t use master theorem: b (subproblem size) changes n-1/n n-2/n-1 n-3/n-2 • Note that sum of partition work: n + (n-1) + (n-2) + (n-3) … Sum(1,N) = N(N+1)/2 = O(N2)

  9. Complexity of Quicksort • Requires stack space to implement recursion • Worst case: O(n) stack space • If pivot breaks into 1 element and n-1 element subarrays • Average case: O(log n) • Pivot splits evenly

  10. MergeSort • General Mergesort Algorithm: • Recursively split subarrays in half • Merge sorted subarrays • Splitting is first in recursive call, so continues until have one item subarrays • One item subarrays are by definition sorted • Merge recombines subarrays so result is sorted • 1+1 item subarrays => 2 item subarrays • 2+2 item subarrays => 4 item subarrays • Use fact that subarrays are sorted to simplify merge algorithm

  11. MergeSort void mergesort(int* array, int* tempArray, int low, int high, int size) { if (low < high) { int middle = (low + high) / 2; mergesort(array,tempArray,low,middle, size); mergesort(array,tempArray,middle+1, high, size); merge(array,tempArray,low,middle,high, size); } }

  12. MergeSort void merge(int* array, int* tempArray, int low, int middle, int high, int size) { int i, j, k; for (i = low; i <= high; i++) { tempArray[i] = array[i]; } // copy into temp array i = low; j = middle+1; k = low; while ((i <= middle) && (j <= high)) { // merge if (tempArray[i] <= tempArray[j]) // if lhs item is smaller array[k++] = tempArray[i++]; // put in final array, increment else // final array position, lhs index array[k++] = tempArray[j++]; // else put rhs item in final array } // increment final array position // rhs index while (i <= middle) // one of the two will run out array[k++] = tempArray[i++]; // copy the rest of the data } // only need to copy if in lhs array // rhs array already in right place

  13. MergeSort Example 20 3 18 9 5 Recursively Split 20 3 18 9 5

  14. MergeSort Example 20 3 18 9 5 Recursively Split 9 20 3 18 5

  15. MergeSort Example 20 3 18 9 5 Merge

  16. 3 k Merge Sort Example 2 cards Not very interesting Think of as swap 20 3 3 20 Temp Array Array Temp[i] < Temp[j] Yes 3 20 18 j i

  17. Array 3 18 20 k MergeSort Example Temp Array Array Temp[i] < Temp[j] No 3 20 18 3 18 j k i Update J, K by 1 => Hit Limit of Internal While Loop, as J > High Now Copy until I > Middle

  18. 5 9 3 5 9 18 20 i=3,j=5 i=1,j=3 i=1,j=4 i=1,j=5 i=2,j=5 MergeSort Example 2 Card Swap 9 5 5 9 3 18 20 Final after merging above sets i=0,j=3

  19. Complexity of MergeSort Recurrence relation: 2 subproblems ½ size Merging is O(n) for any subproblem Always moving forwards in the array a = 2 b = 2 k = 1 2 = 21 Master Theorem: O(n log2n) Always O(n log2n) in both average and worst case Doesn’t rely on quality of pivot choice

  20. Space Complexity of Mergesort • Need an additional O(n) temporary array • Number of recursive calls: • Always O(log2n)

  21. Tradeoffs • When it is more useful to: • Just search • Quicksort or Mergesort and search • Assume Z searches Search on random data: Z * O(n) Fast Sort and binary search: O(nlog2n) + Z *log2n

  22. Tradeoffs Z * n <= nlog2n + Zlog2n Z(n - log2n) <= n log2n Z <= (n log2n) / (n-log2n) Z <= (n log2n) / n [Approximation] Z <= log2n [Approximation] Where as before, had to do N searches to make up for cost of sorting, now only do log2N 1,000,000 items = 19 searches, instead of 1,000,000

  23. How Fast? • Without specific details of what sorting, O(n log2n) is the maximum speed sort possible. • Only available operations: Compare, Swap • Proof: Decision Tree – describes how sort operates • Every vertex represents a comparison, every branch a result • Moving down tree – Tracing a possible run through the algorithm

  24. How Fast? K1 <= K2 [1,2,3] Yes No K2 <= K3 K1 <= K3 [2,1,3] [1,2,3] Yes No Yes No [2,3,1] [1,2,3] [2,1,3] stop K2 <= K3 stop K1 <= K3 [1,3,2] Yes Yes No No stop stop stop stop [1,3,2] [3,1,2] [2,3,1] [3,2,1]

  25. How Fast? • There are n! possible “stop” nodes – effectively all permutations of the n numbers in the array. • Thus any decision tree representing a sorting algorithm must have n! leaves • The height of a this type of tree (a binary tree) is correlated with number of leaves: • Height k = 2^(k-1) leaves • Must be at least log2n! + 1 height

  26. How Fast? • Path from top to bottom of tree – trace of a run of the algorithm • Need to prove that (log2n!) is lower bounded by (n log2n) n! = (n)(n-1)(n-2)(n-3) … (3)(2)(1) > (n)(n-1)(n-2)(n-3) … ceil(n/2) // doing fewer multiplies > ceil(n/2) (ciel(n/2)) // doing multiplies of bigger things > approximately (n/2)(n/2) log 2 n! > log 2 (n/2)(n/2) log 2 n! > (n/2) log 2 (n/2) //exponentiation in logs = multiplication out front log 2 n! > (n/2)(log2n – log2 2) // division in logs = subtraction log 2 n! > (n/2)(log2n – 1) log 2 n! > (n/2)(log2n) – (n/2) log 2 n! > (1/2) [nlog2n – n] log 2 n! ~ O(n log2n)

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