I know we can optimize quicksort by leveraging tail recursion by removing more than 1 recursion calls and reducing it to once single recursion call:-
void quickSort(int arr[], int low, int high)
{
if (low < high)
{
int pi = partition(arr, low, high);
quickSort(arr, low, pi - 1);
quickSort(arr, pi + 1, high);
}
}
void quickSort(int arr[], int low, int high)
{
start:
if (low < high)
{
int pi = partition(arr, low, high);
quickSort(arr, low, pi - 1);
low = pi+1;
high = high;
goto start;
}
}
But can we optimize randomized quicksort with tail recursion?
Tail stack recursion focuses on optimizing recursive calls. The only difference between randomized quicksort and normal quicksort is the partition function which selects a random pivot in case of randomized quicksort. Note that this partition function is non-recursive. As the recursive part of both randomized quicksort and normal quicksort is same, same optimization can be done in both cases. So, yes.
Related
This question already has answers here:
Sorting 64-bit structs using AVX?
(2 answers)
Fast merge of sorted subsets of 4K floating-point numbers in L1/L2
(6 answers)
Closed 8 months ago.
Do you know a way to use a sorting algorithm that uses vectors intrinsics efficiently ?
I have to use the capability of loading, storing 4 floats at one operation and also other vectors operations.
I found this code for "Quick Sort".
Can you help me understand how to implement it with SIMD ?
int partition(float *arr, int low, int high)
{
float pivot;
int i, j;
// pivot (Element to be placed at right position)
pivot = arr[high];
i = (low - 1); // Index of smaller element and indicates the
// right position of pivot found so far
for (j = low; j <= high - 1; j++) {
// If current element is smaller than the pivot
if (arr[j] < pivot) {
i++; // increment index of smaller element
swap(&arr[i], &arr[j]);
}
}
swap(&arr[i + 1], &arr[high]);
return (i + 1);
}
/* low –> Starting index, high –> Ending index */
void quickSort(float *arr, int low, int high)
{
int pi;
if (low < high) {
/* pi is partitioning index, arr[pi] is now at right place */
pi = partition(arr, low, high);
quickSort(arr, low, pi-1); // Before pi
quickSort(arr, pi + 1, high); // After pi
}
}
What is the best way to sort a dictionary with 1Gbyte size(255 char for each word) with 2G of RAM?
I have already tried quicksort and didn't get the acceptable result.
This the quicksort code:
#include <iostream>
#include <fstream>
#include <cstring>
#define MAXL 4000000
using namespace std;
void swap(char *&ch1,char *&ch2)
{
char *temp = ch1;
ch1 = ch2;
ch2 = temp;
}
int partition (char **arr, int low, int high)
{
string pivot = arr[high]; // pivot
int i = (low - 1); // Index of smaller element
for (int j = low; j <= high- 1; j++)
{
// If current element is smaller than or
// equal to pivot
if (arr[j] <= pivot)
{
i++; // increment index of smaller element
swap(arr[i], arr[j]);
}
}
swap(arr[i + 1], arr[high]);
return (i + 1);
}
void quickSort(char **arr, int low, int high)
{
if (low < high)
{
int pi = partition(arr, low, high);
// Separately sort elements before
// partition and after partition
quickSort(arr, low, pi - 1);
quickSort(arr, pi + 1, high);
}
}
int main()
{
fstream file("input.txt",ios::in|ios::out|ios::app);
fstream o("output.txt",ios::out);
char **arr = new char*[MAXL];
for(int i=0;i<MAXL;i++)
arr[i] = new char[255];
long long i=0;
while(file)
{
//words are sepearated by spcae
file.getline(arr[i],256,' ');
i++;
}
file.close();
quickSort(arr, 0, i-2);
for(long long j=0;j<i-1;j++)
{
o << arr[j] << "\n";
}
}
It takes more than 10 minutes to sort the mentioned list but it shouldn't take more than 20 seconds.
(MAXL is the number of words in the 1G file and input words are stored in a text file)
If you can't fit it all in memory, a file-based merge sort will work well.
In-place algorithms are your solution. Find more here:
As another example, many sorting algorithms rearrange arrays into sorted order in-place, including bubble sort, comb sort, selection sort, insertion sort, heapsort, and Shell sort. These algorithms require only a few pointers, so their space complexity is O(log n).
What signals the program to say, "Ok the first recursive quickSort call is done; proceed to the second recursive call"?
int partition (int arr[], int low, int high)
{
int pivot = arr[high]; // pivot
int i = (low - 1); // Index of smaller element
for (int j = low; j <= high- 1; j++)
{
if (arr[j] <= pivot)
{
i++; // increment index of smaller element
swap(&arr[i], &arr[j]);
}
}
swap(&arr[i + 1], &arr[high]);
return (i + 1);
}
void quickSort(int arr[], int low, int high)
{
if (low < high)
{
int pi = partition(arr, low, high);
quickSort(arr, low, pi - 1);
quickSort(arr, pi + 1, high);
}
}
Your actual question roots to the Recursion Stack.
Let's first understand Recursion, which basically constitutes a method that keeps calling itself on increasingly smaller cases and repeats the same non recursive procedure each time until it reaches base case, at which is stops.
In the case of QuickSort, the base case of the recursion are lists of size zero or one, which never need to be sorted. If this is not the case, the array not meant to be sorted. That's why we call the QuickSort method again, twice, on arrays of smaller sizes.
We recurse on the side of the array containing all the elements from A[0] to A[i - 2], and the side of array containing the elements A[i] to A[A.length - 1].
Why do we leave out A[i - 1]? Simple - It's already in its correct place.
i implemented the Quick Sort Algorithm which given pseudo code in Introduction to Algorithms (Cormen, 3rd Edition) 7.1
When i tried algorithm with small sized arrays, result is true. But when i tried with N=50000 and array is already sorted like this;
N = {1, 2, 3, ..., 50000};
It gives StackOverflowError. I think it's happening because the function recurse itself 50000 times.
QuickSort(A, 0, 49999) => QuickSort(A, 0, 49998) => QuickSort(A, 0, 49997)... so go on.
Can i solve this problem? Or should i use different pivot position?
Here is my code;
public void sort(int[] arr){ QuickSort(arr, 0, arr.length - 1); }
private void QuickSort(int[] A, int left, int right){
if(left < right){
int index = Partition(A, left, right);
QuickSort(A, left, index - 1);
QuickSort(A, index + 1, right);
}
}
private int Partition(int[] A, int left, int right){
int pivot = A[right];
int wall = left-1;
for(int i=left; i<right; i++){
if(A[i] <= pivot){
Swap(A, ++wall, i);
}
}
Swap(A, wall + 1, right);
return wall + 1;
}
private void Swap(int[] A, int x, int y){
int keeper = A[x];
A[x] = A[y];
A[y] = keeper;
}
Yes, this pivot scheme is not right choice for sorted array. It causes very unbalanced partition, leads to O(N^2) complexity and very deep recursion level, as you noticed.
There are some approaches to improve this behavior.
For example, you can use random index for pivot like pivotIdx = start + rand() % (end-start+1);, or choose median-of-three method (median of the first, last and middle elements in index range).
P.S. An option to avoid stack overflow - call recursion for shorter segment at first, then for longer one.
https://en.wikipedia.org/wiki/Quicksort#Choice_of_pivot
What I am is doing is using a quicksort algorithm, so that my pivot element(which will always be the first element of the array gets positioned to its appropriate position in the sorted array and I am calling this method again until I do not position the element at a given rank. Is there a better solution?
Here is my code:
public static int arbitrary(int a[],int x,int y,int rank)//x and y are first and last indecies of the array
{
int j=y,temp;
if(x<y)
{
for(int i=y;i>x;i--)
{
if(a[i]>a[x])
{
temp=a[i];
a[i]=a[j];
a[j]=temp;
j--;
}
}
temp=a[x];
a[x]=a[j];
a[j]=temp;
//System.out.println("j is "+j);
if(j==rank)
return a[j];
else if(rank<j)
return arbitrary(a,x,j-1,rank);
else
return arbitrary(a,j+1,y,rank);
}
else
return 0;
}
The algorithm you have implemented is called Quickselect.
Just select a random pivot and to get rid of the worst case with O(n²) time complexity.
The expected runtime is now about 3.4n + o(n).
Quickselect is probably the best tradeoff between performance and simplicity.
An even more advanced pivot selection strategy results in 1.5n + o(n) expected time
(Floyd-Rivest Algorithm).
Fun Fact: With deterministic algorithms you can't go better than 2n. BFPRT for example needs about 2.95n to select the median.
Best way to find Rank element by using the QuickSort method:
In QuickSort, at every iteration you are able to get fixed one pivot element.
When the RankElement == PivotIndex, and break the condition and return the value.
public class FindRank {
public void find(int[] arr, int low, int high, int k) {
if (low < high) {
int pivot = partition(arr, low, high, k);
find(arr, low, pivot - 1, k);
find(arr, pivot + 1, high, k);
}
}
public int partition(int[] arr, int low, int high, int k) {
int pivotIndex = high;
while (low < high) {
while (low < high && arr[low] <= arr[pivotIndex]) {
low++;
}
while (low > high && arr[high] >= arr[pivotIndex]) {
high--;
}
if (low < high) {
swap(arr, low, high);
}
}
swap(arr, pivotIndex, high);
if (pivotIndex == k) {
System.out.println("Array Value:" + arr[k] + " index:" + k);
return k;
}
return high;
}
private void swap(int[] arr, int low, int high) {
int temp = arr[low];
arr[low] = arr[high];
arr[high] = temp;
}
}