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 – Example
34 8 64 51 32 21
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
Σ i = 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.
Shell sort
• 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
Shell sort
• Conceptually, separate array into hk
columns and sort each column
• Example:
81 94 11 96 12 35 17 95 28 58 41 75 15
Shell sort: 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)
Shell sort – Analysis
• Theorem 7.3: The worst-case running time
of Shell sort, 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.
Shell sort – 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
Shell sort – 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-t 1/hi)
= O(N2)
Heap sort
• Strategy
– Apply buildHeap function
– Perform N deleteMin operations
• Running time is O(NlogN)
• Problem: Simple implementation uses two
arrays – N extra space
Heap sort
• 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
Merge sort
• Recursively sort elements 1-N/2 and N/2N and merge the result
• This is a classic divide-and-conquer
algorithm.
• Uses temporary array – extra space
Merge sort – Algorithm
• Base case
– List of 1 element, return
• Otherwise
– Merge sort left half
– Merge sort right half
– Merge
Merge sort – 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
Merge sort – 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
• Bad for presorted data
– 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
a<b
a<c
a<b<c
a<c<b
c<a<b
a<b<c
a<c<b
c<a
c<a<b
b<c
c<b
a<b<c
a<c<b
b<a
b<c
b<a<c
b<c<a
c<b<a
b<a<c
b<c<a
a<c
b<a<c
c<b
c<b<a
c<a
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)
Bucket sort
• Sorting algorithm that runs in O(N) time,
how? What is special about this
algorithm?
for i = 1 to n
increment count[Ai]