# Sorting by yaoyufang

VIEWS: 13 PAGES: 25

• pg 1
```									     Sorting – Part II

CS 367 – Introduction to Data Structures
Better Sorting
• The problem with all previous examples is
the O(n2) performance
– this may be acceptable for small data sets,
but not large ones
• Theoretically, O(n log n) is possible
– see proof in Section 9.2 of the book
Heap Sort
• Major problem with selection sort
– it has to search entire back end of array on
every search for next smallest item
– what if we could make this search faster?
• A heap always keeps the largest element
at the top
– it only takes O(log n) to remove the top
– O(log n) is much better than O(n) search time
of selection sort
Heap Sort
• Basic procedure
– build a heap
– swap the root with the last element
– rebuild the heap excluding the last element
• the last element is where it is supposed to be
– repeat until only one item left in the heap
Heap Sort - Conceptually
Z
0   1   2   3   4   5   6
X               M       queue                           Z

T       N       J       L

X
0   1   2   3   4   5   6
T               M       queue                       X   Z

L       N       J

T
0   1   2   3   4   5   6
N               M       queue                   T   X   Z

L       J
Heap Sort - Implementation
0   1      2    3   4   5       6
Z   X     M T       N   J   L

swap 0 and 6, rebuild

0   1      2    3   4   5       6
X   T     M L       N   J   Z                      0    1   2   3   4   5    6
J    L   M N     T   X    Z
swap 0 and 5, rebuild
swap 0 and 1, done
0   1      2    3   4   5       6
0    1   2   3   4   5    6
T   N     M L       J   X   Z
L    J   M N     T   X    Z
swap 0 and 4, rebuild
swap 0 and 2, rebuild
0   1      2    3   4   5       6                  0    1 2 3 4 5 6
N   L     M J       T   X   Z                      M L      J   N   T   X    Z
swap 0 and 3, rebuild
Building the Heap
• The heap will be build within the array
– no extra data structures will be needed
• Basic idea
– start at the last non-terminal node
– restore heap for tree rooted at this node
• simply swap this node with it’s largest child if the
child is larger
– repeat this process for all non-terminal nodes
Building the Heap
0   1    2   3    4    5   6
M    J    Z T      X    L N
compare Z with its children
(no move made)
0   1    2   3    4    5   6
M J       Z T      X    L N
compare J with its children
(swap it with X)
0   1    2   3    4    5   6
M X       Z T      J    L N

compare M with its children
(swap it with Z and then N)
0    1 2 3 4 5 6
Z    X    N T      J    L M        Valid Heap
Building the Heap
• Code to re-build the heap
void moveDown(Object[ ] data, int first, int last) {
int child = 2 * first + 1;
while(child <= last) {
if((child < last) && ((child + 1) <= last)) {
if(data[child] < data[child + 1]) { child++; }
if(data[first] < data[child]) {
swap(first, child);
first = child;
child = 2 * child + 1;
}
else { break; }
}
}
Heap Sort
• Code to build the heap and sort it
void heapSort(Object[ ] data) {
// build the heap out of the data
for(int i=data.length / 2; i >= 0; i--)
moveDown(data, i, data.length – 1);

// now sort it
for(int i = data.length – 1; i < 0; i--) {
swap(0, i);
moveDown(data, 0, i – 1);
}
}
Heap Sort
• Time to build the heap in worst case
– O(n)
– proof can be found in Section 6.9.2 of book
• Number of swaps to perform
– always (n – 1)
• Performance to rebuild the heap
– O(n log n)
• Overall performance
– O(n) + (n-1) + O(n log n) = O(n log n)
Quicksort
• Basic procedure
– divide the initial array into two parts
• all of the elements in the left side must be smaller than all of
the elements in the right side
– sort the two arrays separately and put them back
together
• we now have a completely sorted array
– however, before sorting the two arrays, divided them
each into two more arrays
• we now have a total of 4 arrays
• smallest elements in far left and largest in far right
– repeat this process until only 1 element arrays remain
• put them all together and the overall array is sorted
Quicksort
0       1       2       3   4       5       6
M       J       Z X         T       L N
break into two parts

0       1       2       3           0       1       2
M           J       L N                 Z       X       T

break into four parts

0       1           0       1               0       1               0
J       L           M       N               X       T           Z

break into 7 parts

0       0               0           0           0           0               0
J       L               M           N           T       X                   Z
Quicksort - Implementing
•    Steps
1. move the largest value to the highest spot
–   this prevents some array overflow problems
2. pick an upper bound for the left sub-array
–   pick the value in the center of the array
–   move this to first element so it doesn’t get moved
3.   move all elements less than this to left side
4.   move all elements greater to the right side
5.   bound will now be in its final position
6.   repeat with the two new arrays
–   from 0 to index(bound) – 1
–   from index(bound) + 1 to array.length - 1
Quicksort - Implementing
void quickSort(Object[ ] data) {
if(data.length < 2) { return; }
int max = 0;

// find the highest value and put it in top spot
for(int i=1; i<data.length; i++)
if(data[i] > data[max]) { max = i; }
swap(max, data.length – 1);

// start the real algorithm
quickSort(data, 0, data.length – 2);
}
Quicksort - Implementing
void quickSort(Object[ ] data, int first, int last) {
int lower = first + 1, upper = last;
swap(first, (first + last) / 2); // find the bound
Comparable bound = data[first];
while(lower <= upper) { // divides the array in half
while(data[lower] < bound) { lower++; } // lowers that are right
while(data[upper] > bound) { upper--; } // uppers that are right
if(lower < upper) { swap(lower++, upper--); }
else { lower++; } // arrays are already split
}
swap(upper, first); // puts bound in its final location
if(first < upper – 1) { quickSort(data, first, upper – 1); }
if(upper + 1 < last) { quickSort(data, upper + 1, last); }
}
Quicksort Performance
• Worst case
– consider selecting the smallest (or largest)
number as the bound
– then all of the numbers end up on one “side”
– consider the sorting the following array
• [5 3 2 1 4 6 8]
• 1 will be the first bound and end up in its proper
location
• however, there will still be n – 1 elements to sort
• this will happen on each iteration
– the result is an O(n2) algorithm
Quicksort Performance
• So what’s the average case?
– the answer is O(n log n)
• In practice, quicksort is usually the best
sorting algorithm
– the closer the bound is to the median, the
better it is
– beware, for arrays under 30 elements,
insertion sort is more efficient
• can you think how quicksort and insertion sort
could be combined?
Mergesort
• One of the first ever sorting algorithms
used on a computer
• It works on a principle similar to quicksort
– each array is broken into two parts and then
sorted separately
– this partition and sort method continues until
only single element arrays exist
– then all of the arrays are put back together to
form a sorted array
Mergesort
• Big difference from quicksort is that the
arrays are always broken into equal
partitions
– or in the case of an odd sized array, as close
as possible to even
• There is no bound selected
• To put the arrays back together, simply
select the smallest element from either
array and make it next
Mergesort
0       1           2       3       4       5           6
Z       M J R                       X       T V
break into 7 parts

0       0               0               0               0           0               0
Z       M               J           R               X           T                   V

0       1                   0       1                   0       1
M       Z               J           R                   T       X

0       1       2           3                               0           1   2
J       M R Z                                               T           V   X

0       1           2       3       4       5           6
J       M R T                       V       X Z
Merging
• The most sophisticated part of mergesort
is recombining (or merging) two separate
arrays
• Just go through each array selecting the
smallest remaining element from each
array
– add it to the new array
Merging
• Pseudo-code
merge(array, first, last) {
mid = (first + last) / 2;
i1= 0;
i2 = first;
i3 = mid + 1;
while( // both left and right sub-arrays contain elements ) {
if(array[i2] < array[i3]) { tmp[i1++] = array[i2++]; }
else { tmp[i1++ = array[i3++]; }
}
// load into temp array remaining elements of array
// copy elements in temp back into array
}
Mergesort
• Once the merge code is done, the code for
mergesort is easy
• Psuedo-code
mergeSort(data, first, last) {
if(first < last) {
mid = (first + last) / 2;
mergeSort(data, first, mid);
mergeSort(data, mid + 1, last);
merge(data, first, last);
}
}
Mergesort Performance
• Mergesort produces a lot of copying in
memory
• It also requires extra storage space for the
temporary array
– this can be prohibitive for very large data sets

```
To top