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Description of the general sorting problem

Description of the general sorting problem. Given array A[1... n ] with n elements that can be compared directly (i.e. for each i and j either A[ i ] < A[ j ], or A[ i ] = A[ j ], or A[ i ] > A[ j ]), rearrange elements of A in such a way that: A[1]  A[1]  A[2]  …  A[ n ].

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Description of the general sorting problem

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  1. Description of the general sorting problem Given array A[1...n] with n elements that can be compared directly (i.e. for each i and j either A[i] < A[j], or A[i] = A[j], or A[i] > A[j]), rearrange elements of A in such a way that: A[1] A[1] A[2]  …A[n]. That is, in the general case we only assume that some given set X is totally ordered and there is an algorithm that for any two elements x,y from X returns one of three possible answers: x < y, x = y, x > y.

  2. To understand algorithms it is sufficient to assume that elements of array A are just integers. In practice elements of array A usually will be more complex and will have to be sorted according to some particular property (Key), e.g. (in C++) they may be defined as classes class Item { int Key;  };

  3. If elements of A are complex (use large amount of memory) it may be more efficient to construct new array B of keys, sort it, and then rearrange items in initial array according to the order of keys. Actually elements of B must contain also the information about the initial position of keys, i.e. they should be pairs (Key,Position).

  4. Insertion Sort procedureInsertionSort(array A[1…n]): for ifrom2to n do j i x A[i] while j 1and A[j1] > x do A[j] A[j1] j j  1 A[j] x

  5. Insertion Sort - Example

  6. Insertion Sort - Complexity procedureInsertionSort(array A[1...n]): for ifrom2to ndo j i x A[i] while j 1and A[j1] > x do A[j] A[j1] j j  1 A[j] x n1 0..i

  7. Insertion Sort - Complexity Time (number of operations) – worst case: T(n) = (c1+ c2) + (c1+ 2c2) + … + (c1+ (n  1) c2) T(n) =nc1+ (1/2 n (n  1)) c2  T(n) = (n2) Space (memory) – worst case: M(n) = n +const

  8. Insertion Sort - Complexity Average complexity? For space – obviously the same: M(n) = n +const For time – it can be shown that (n2) operations are required also in average case.

  9. Insertion Sort - Complexity (empirical) [Adapted from M.Lamont]

  10. Bubble Sort (version 1) procedureBubbleSort(array A[1...n]): for ifrom1to n do for jfrom2to ndo if A[j] > A[j1] then A[j] A[j1]

  11. Bubble Sort - Example

  12. Bubble Sort - Complexity procedureBubbleSort(array A[1...n]): for ifrom1to n do for jfrom2to n do if A[j] > A[j1] then A[j] A[j1] n n1

  13. Bubble Sort Complexity Worst case: T(n) = n(n–1) const = (n2) M(n) = n +const Average complexities are the same as in worst case

  14. Bubble Sort Complexity (empirical) [Adapted from M.Lamont]

  15. Bubble Sort (version 2) procedureBubbleSort(array A[0…n]): stop = false while stop =falsedo stop =true for jfrom2to n do if A[j] > A[j1] then stop = false A[j] A[j1]

  16. Simple sorting algorithms - empirical complexity [Adapted from M.Lamont]

  17. Merge Sort - Idea [Adapted from H.Lang]

  18. Merge Sort procedureMergeSort(array A[l...r], B[l...r]): if l < rthen m l + (r l)/2 MergeSort(A[l…m], B[l…m]) MergeSort(A[m+1...r], B[m+1...r]) Merge(A[l...r], B[l...m], B[m+1...r])

  19. Merge Sort procedure Merge(array A[l...r], B[l...m], C[m+1...r]) apos  l; bpos  l; cpos  m+ 1 whilebpos  mandcpos  rdo if B[bpos] < C[cpos] then A[apos]  B[bpos]; bpos  bpos+ 1; apos  apos+ 1 else A[apos]  C[cpos]; cpos  cpos+ 1; apos  apos+ 1 whilebpos  mdo A[apos]  B[bpos]; bpos  bpos+ 1; apos apos+ 1 whilecpos  rdo A[apos]  C[cbos]; cpos  cpos+ 1; apos  apos+ 1

  20. Merge Sort - Example

  21. Merge Sort - Complexity Worst case: Merge procedure: Tm(n) = n const = (n) MergeSortprocedure: T(n) = 2 T(n/2) + const + Tm(n) = 2 T(n/2) + (n) = (n log n)

  22. Merge Sort - Complexity Worst case: MergeSortprocedure: T(n) = 2 T(n/2) + const + Tm(n) = 2 T(n/2) + (n) T(n) = (n log n) M(n) = 2 n + const(can be impoved up to 3/2 n + const) Average complexities are the same as in worst case

  23. Quick Sort Sort - Idea [Adapted from H.Lang]

  24. Quick Sort (not very efficient version) procedureQuickSort(array A[1...n]): if n > 1 then xA[1] (or iRandom(n);x A[i]) bpos1; cpos 1 for ifrom 1to ndo if x > A[i] then B[bpos] = A[i]; bpos  bpos+ 1 if x A[i] then C[cpos] = A[i]; cpos  cpos+ 1 QuickSort(B[1...bpos]); QuickSort(C[1...cpos]) Append(x,A[1...n], B[1...bpos], C[1...cpos])

  25. Quick Sort (not very efficient version) procedure Append(x,A[1...n], B[1...bpos], C[1...cpos]) for ifrom 1 to bposdo A[i]  B[i] A[bpos] = x for ifrom 1 to cposdo A[bpos+i+1]  C[i]

  26. Quick Sort - Example [Adapted from T.Niemann]

  27. Quick Sort - Example 2 4 8 2 2 8 4

  28. Quick Sort - Complexity Worst case: Append procedure: Ta(n) = n const = (n) QuickSortprocedure: T(n) = T(bpos) + T(cpos) + (n)+ Ta(n) = = T(bpos) + T(cpos) + (n)

  29. Quick Sort - Complexity QuickSortprocedure: T(n) = T(bpos) + T(cpos) + (n) Case 1: bpos = cpos T(n) = 2 T(n/2) + (n) Case 2: bpos= 1, cpos = n– 1 T(n) = T(1) + T(n– 1) + (n) = (n log n) = (n2)

  30. Quick Sort - Complexity It can be shown that the worst situation is in Case 2 (or with bpos and cpos reversed), thus QuickSort worst case time complexity is T(n) = (n2)

  31. Quick Sort - Complexity Good news: • Average time complexity for QuickSort is (n log n) • This (inefficient) version uses 2n+ const of space, but that can be improved to n + const • On average QuickSort is practically the fastest sorting algorithm

  32. Quick Sort - Complexity [Adapted from K.Åhlander]

  33. Quick Sort - Complexity [Adapted from K.Åhlander]

  34. Quick Sort - Complexity [Adapted from K.Åhlander]

  35. Quick Sort - Complexity [Adapted from K.Åhlander]

  36. Quick Sort - Complexity [Adapted from K.Åhlander]

  37. Quick Sort (more efficient version) procedure QuickSort(array A[l...r]): if l < rthen i l; j r+ 1; x A[l] (or select x randomly) whilei < jdo i i+ 1 while i  rand A[i] < xdo i i+ 1 j j– 1 while j  l and A[j] > xdo j j– 1 A[i]  A[j] A[i]  A[j]; A[j]  A[l] QuickSort(A[l...j–1]) QuickSort(A[j+1...r])

  38. Quick Sort (more efficient version) i j After selecting of x move pointers i and j from start and array correspondingly and swap all pairs of elements for which A[i] > x and A[j] < x

  39. Quick Sort (more efficient version) - example Positions: 1 2 345678910 Keys: 9111171318 41214 5 > > < after 1while: 91 5171318 4121411 > < < < after 2while: 91 5 4131817121411 < >< < after 3while: 91 513 41817121411 2 extra swaps: 41 5 9131817121411 sort sub-arrays:1 4 5 9111213141718

  40. Complexity comparison

  41. Shell Sort - Idea Donald Shell 1959 h1= 1 < h2< h3 < h4 < h5 <   an increasing sequence of integers For given n select some hk < n. Repeatedly for inc = hk, hk-1,, h1= 1 and for sfrom 1 to inc InsertionSort the sequences A[s], A[s + inc],A[s + 2 inc], A[s + 3 inc], 

  42. Shell Sort - Idea [Adapted from H.Lang]

  43. Shell Sort procedure ShellSort(array A[1…n]): inc InitialInc(n) whileinc 1 do for ifrominctondo j i x  A[i] while j  incand A[j–inc] > xdo A[j]  A[j–inc] j  j – inc A[j]  x inc NextInc(inc,n)

  44. Shell Sort - Example Positions: 12345678910 Keys: 9 111171318 41214 5 sort for inc = 5: 9111171318412145 9111145184121713 sort for inc = 3: 9111145184121713 4111951713121814 sort forinc= 1: 4111 9 5 17 13121814 1 4 5 91112 13141718

  45. Shell Sort - Good increment sequences? • sequence 1, 2, 4, 8,  (hi= 2i-1) is not particularly good • the “best” increment sequence is not known • sequence 1, 4, 13, 40, 121,  (hi = (3i–1)/2 ) is one (of several known) that gives good results in practice

  46. Shell Sort - Complexity • in general the finding of time complexity for a given increment sequence is very difficult • for sequence 1, 2, 4, 8, we have T(n) = (n2) • it is known that for sequence 1, 4, 13, 40, 121,  T(n) = O(n1.5), but in practice it tends to perform comparably with n log n time algorithms • there exists sequences for which it has been proven that T(n) = O(n (log n)2)

  47. Shell Sort - Complexity • it is known that for any sequence the worst case complexity T(n) =  (n (log n/log logn)2) • it is not known whether average complexity T(n) = O(n log n) can be obtained

  48. Heap Sort - procedure "Heapify" [Adapted from T.Giang]

  49. Heap Sort - procedure "Heapify" [Adapted from T.Giang]

  50. Heap Sort - procedure "Heapify" procedure Heapify(array A[i...j]): if RC(j)  iand A[RC(j)]  A[LC(j)] and A[RC(j)] < A[j] then A[j]  A[RC(j)] Heapify(A[i…RC(j)]) else if LC(j)  i and A[LC(j)] < A[j] then A[j]  A[LC(j)] Heapify(A[i…LC(j)]) Here LC(j)= 2j–n–1 (left child) and RC(j)= 2j–n–2 (right child)

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