Sorting

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# Sorting - PowerPoint PPT Presentation

Sorting. Introduction. Assumptions Sorting an array of integers Entire sort can be done in main memory Straightforward algorithms are O(N 2 ) More complex algorithms are O(NlogN). Insertion Sort. Idea: Start at position 1 and move each element to the left until it is in the correct place

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## PowerPoint Slideshow about 'Sorting' - trula

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### Sorting

Introduction
• Assumptions
• Sorting an array of integers
• Entire sort can be done in main memory
• Straightforward algorithms are O(N2)
• More complex algorithms are O(NlogN)
Insertion Sort
• Idea: Start at position 1 and move each element to the left until it is in the correct place
• At iteration i, the leftmost i elements are in sorted order

for i = 1; i < a.size(); i++

tmp = a[i]

for j = i; j > 0 && tmp < a[j-1]; j--

a[j] = a[j-1]

a[j] = tmp

Insertion Sort – Analysis
• Worst case: Each element is moved all the way to the left
• Example input?
• Running time in this case?
Insertion Sort – Analysis
• Running time in this case?
• inner loop test executes p+1 times for each p

N

Σi = 2 + 3 + 4 + … + N = θ(N2)

i=2

Sorting – Lower Bound
• Inversion – an ordered pair (i, j) where i < j but a[i] > a[j]
• Number of swaps required by insertion sort
• Running time of insertion sort O(I+N)
• Calculate average running time of insertion sort by computing average number of inversions
Sorting – Lower Bound
• Theorem 7.1: The average number of inversions in an array of N distinct elements is N(N-1)/4
• Proof: Total number of inversions in a list L and its reverse Lr is N(N-1)/2. Average list has half this amount, N(N-1)/4.
• Insertion sort is O(N2) on average
Sorting – Lower Bound
• Theorem 7.2: Any algorithm that sorts by exchanging adjacent elements requires Ω(N2) time on average.
• Proof: Number of inversions is N(N-1)/4. Each swap removes only 1 inversion, so Ω(N2) swaps are required.
Shellsort
• Idea: Use increment sequence h1, h2, …ht. After phase using increment hk, a[i] <= a[i+ hk]

for each hk

for i = hk to a.size()

tmp = a[i]

for j = i j >= hk && tmp < a[j- hk] j-= hk

a[j] = a[j- hk ]

a[j] = tmp

Shellsort
• Conceptually, separate array into hk columns and sort each column
• Example:

81 94 11 96 12 35 17 95 28 58 41 75 15

Shellsort: Analysis
• Increment sequence determines running time
• Shell’s increments
• ht = floor(N/2) and hk=floor(hk+1/2)
• θ(N2)
• Hibbard’s increments
• 1, 3, 7, …, 2k-1
• θ(N3/2)
Shellsort – Analysis
• Theorem 7.3: The worst-case running time of Shellsort, using Shell’s increments, is θ(N2).
• Proof – part 1: There exists some input that takes Ω(N2). Choose N to be a power of 2. All increments except last will be even. Give as input array with N/2 largest numbers in even positions and N/2 smallest in odd positions.
Shellsort – Analysis

Example: 1 9 2 10 3 11 4 12 5 13 6 14 7 15 8 16

• At last pass, ith smallest number will be in position 2i-1 requiring i-1 moves. Total number of moves:

N/2

Σ i-1 = Ω(N2)

i=1

Shellsort – Analysis
• Proof – part 2: A pass with increment hk performs hk insertion sorts of lists of N/hk elements. Total cost of a pass is O(hk(N/ hk)2) = O(N2/ hk). Summing over all passes gives a total bound of

O(sum from 1-t(N2/hi))

= O(N2 (sum from 1-t1/hi)

= O(N2)

Heapsort
• Strategy
• Apply buildHeap function
• Perform N deleteMin operations
• Running time is O(NlogN)
• Problem: Simple implementation uses two arrays – N extra space
Heapsort
• Solution: After deletion, size of heap shrinks by 1, so simply insert into empty slot
• Problem: Results in array sorted in decreasing order – use max heap instead
• Example: 97 53 59 26 41 58 31
Mergesort
• Recursively sort elements 1-N/2 and N/2-N and merge the result
• This is a classic divide-and-conquer algorithm.
• Uses temporary array – extra space
Mergesort – Algorithm
• Base case
• List of 1 element, return
• Otherwise
• Mergesort left half
• Mergesort right half
• Merge
Mergesort – Algorithm
• Merge

leftPos = leftStart; rightPos = rightStart; tmpPos = tmpStart;

while(leftPos <=leftEnd && rightPos <= rightEnd)

if(a[leftPos] <= a[rightPos])

tmpArray[tmpPos++] = a[leftPos++]

else

tmpArray[tmpPos++] = a[rightPos++]

//copy rest of left half

//copy rest of right half

//copy tmpArray back

• Example 24 13 26 1 2 27 38 15
Mergesort – Analysis
• T(1) = 1
• T(N) = 2T(N/2) + N – divide by N
• T(N)/N = T(N/2)/N/2 + 1 – substitute N/2
• T(N/2)/N/2 = T(N/4)/N/4 + 1
• T(N/4)/N/4 = T(N/8)/N/8 + 1 …
• T(2)/2 = T(1)/1 + 1 – telescoping
• T(N)/N = T(1)/1 + logN – mult by N
• T(N) = N + NlogN = O(NlogN)
• Can also repeatedly substitute recurrence on right-hand side
Quicksort
• Base case: Return if number of elements is 0 or 1
• Pick pivot v in S.
• Partition into two subgroups. First subgroup is < v and last subgroup is > v.
• Recursively quicksort first subgroup and last subgroup
Quicksort
• Picking the pivot
• Choose first element
• Choose randomly
• Random number generation can be expensive
• Median-of-three
• Choose median of left, right, and center
• Good choice!
Quicksort
• Partitioning
• Move pivot to end
• i = start; j = end;
• i++ and j-- until a[i] > pivot and a[j] < pivot and i < j
• swap a[i] and a[j]
• swap a[last] and a[i] //move pivot
• Example 13 81 92 43 65 31 57 26 75 0
Quicksort – Analysis
• Quicksort relation
• T(N) = T(i) + T(N-i-1) + cN
• Worst-case – pivot always smallest element
• T(N) = T(N-1) + cN
• T(N-1) = T(N-2) + c(N-1)
• T(N-2) = T(N-3) + c(N-2)
• T(2) = T(1) + c(2) – add previous equations
• T(N) = T(1) + csum([2-N] of i) = O(N2)
Quicksort – Analysis
• Best-case – pivot always middle element
• T(N) = 2T(N/2) + cN
• Can use same analysis as mergesort
• = O(NlogN)
• Average-case (pg 278)
• = O(NlogN)
Sorting – Lower Bound
• A sorting algorithm that uses comparisons requires ceil(log(N!)) in worst case and log(N!) on average.
Decision Tree

a<b<c

a<c<b

b<a<c

b<c<a

c<a<b

c<b<a

b<a

a<b

a<b<c

a<c<b

c<a<b

b<a<c

b<c<a

c<b<a

b<c

a<c

c<b

c<a

a<b<c

a<c<b

c<a<b

b<a<c

b<c<a

c<b<a

a<c

b<c

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c<b

a<b<c

a<c<b

b<a<c

b<c<a

Sorting – Lower Bound
• Let T be a binary tree of depth d. The T has at most 2d leaves.
• A binary tree with L leaves must have depth at least ceil(log L).
• Any sorting algorithm that uses only comparisons between elements requires at least ceil(log(N!)) comparisons in the worst case.
• N! leaves because N! permutations of N elements
Sorting – Lower Bound
• Any sorting algorithm that uses only comparisons between elements requires Ω(NlogN) comparisons
• log(N!) comparisons are required
• log(N!) = log(N(N-1)(N-2)…(2)(1))
• = logN+log(N-1)+log(N-2)+…+log1
• >= logN+log(N-1)+log(N-2)+…+logN/2
• >= N/2log(N/2) >= N/2log(N)-N/2
• = Ω(NlogN)
Bucketsort