Preparata’s algorithm is a recursive Divide and Conquer Algorithm where the rank of each input key is computed and the keys are outputted according to their ranks.
m[i, j] := M[i, j] for 1 <= i, j <= n in parallel;
for r : = 1 to logn do
{
Step 1. In parallel set q[i, j, k] := m[i, j] + m[j, k] for
1 <= i, j, k <= n.
Step 2. In parallel set m[i, j] := min
{
q[i, l, j], q[i, 2, j], ..., q[i, n, j]
}
for
1 <= i, j <= n.
}
Put M(i)(i):=0 for all i and M(i)(j):=m[i, j] for i≠j
|
In the above procedure O(N3) global memory is used by using the variables m[i, j] for 1 ≤ i, j ≤ N and q[i, j, k] for 1 ≤ i, j, k ≤ N.
- Initializing m[ ] that takes O(N2) time.
- Step 1 of the above algorithm takes O(1) time while using 3 processors.
- In Step 2, N2 number of different m[i, j]’s are computed.
The computation of a single m[z, j] involves computing a minimum of N numbers and hence can be completed in O(1) time using N2 CRCW PRAM processors. In fact, this minimum can also be computed in O(1) time using n(1 + e) processors for any fixed e > 0.
Step 2 can be completed in O(1) time using n(3 + e) common CRCW PRAM processors. Thus, the for loop runs in O(log N) time. The final computation of M also can be done in O(1) time using N2 processors.
The correctness of the above algorithm can be proven by induction on R. It can be shown that the value of m[i, j] at the end of the rth iteration of the for loop is min, and the minimum is taken over all the sequences of elements of {1, 2, …, N} such that k < 2R. The above algorithm can be specialized to solve several problems including the transitive closure, connected components, minimum spanning tree, and so on.
Let k1, k2, …, kn be the input sequence. Preparata’s algorithm partitions the input into log N parts K1, K2, …, if log N; where there are N/log N keys in each part.
If k is any key in the input, its rank in the input is computed as follows. First, the rank Ri of k is computed for each i, 1< i < log N. Then, the total rank of k is computed as
The details of Preparata’s algorithm are given below.
Consider T(N) to be the run time of Preparata’s algorithm using N*log N processors. Clearly, step 1 takes T(N/log N) time and steps 2 and 3 together take O(log(log N)) time. Thus, there is
T(n) = T(N/log N) + O(log(log N))
which can be solved by repeated substitution to get T(N) = O(log N). Also, the number of processors used in each step is N*log N.
Preparata’s Sorting Algorithm
Below are the steps for the Preparata’s Sorting Algorithm:
- If N is a small constant, sort the keys using any algorithm and quit.
- Partition the given N keys into log N parts, with N/(log N) keys in each part.
- Sort each part recursively and separately in parallel, assigning N processors to each part. Let S1, S2, …, Slog N be the sorted sequences.
- Merge Si with Sj for 1< i, j < log N in parallel. This can be done by allocating N/(log N) processors to each pair(i, j). That is, using N*log N processors, this step can be accomplished in O(log(log N)) time with the algorithm. As a by-product of this merging step, the rank is computed of each key in each one of the Si‘s(1 < i < log N).
- Allocate log N processors to compute the rank of each key in the original input. This is done in parallel for all the keys by adding the log N ranks computed (for each key) in step 2. This can be done in O(log(log N)) time using the prefix computation algorithm.
- Finally, the keys are written in the order of their ranks.
Below is the implementation of the above approach:
// C++ program to implement Preparata's // a time-optimal parallel algorithm #include <iostream> #include <vector> using namespace std;
// Class for Point class Point {
public :
int x, y;
// Constructor
Point( int x, int y) {
this ->x = x;
this ->y = y;
}
}; // Finds the left most index of the given // set of points class LeftIndex {
public :
int leftMostIndex(Point points[], int n) {
int min = 0;
// Finding the point on plane
// with minimum x value
for ( int i = 1; i < n; i++) {
// Update the minimum index
if (points[i].x < points[min].x)
min = i;
else if (points[i].x == points[min].x) {
if (points[i].y > points[min].y)
min = i;
}
}
// Return the minimum index
return min;
}
// Perform the parallel process in the
// preparata's algorithm according to
// the value of p, q, r
int parallel(Point p, Point q, Point r) {
// For the three-dimensional
int val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y);
// If val is equals to 0, then
// the points are collinear
if (val == 0)
return 0;
else if (val > 0) // Clockwise
return 1;
else // Anti-clockwise
return 2;
}
// Perform the parallel process in the
// preparata's algorithm
void preparata(Point points[], int n) {
// There must be at least 3 points
if (n < 3)
return ;
// Find the left most point
int l = leftMostIndex(points, n);
vector<Point> pre;
// Start from left most point, keep
// moving counterclockwise until reach
// the start point again.
int p = l, q;
do {
// Add current point to result
pre.push_back(points[p]);
// Search for a point 'q' such that
// orientation(p, q, r) is counterclockwise
// for all points 'r'. The idea is to keep
// track of last visited most counterclock-
// wise point in q. If any point 'i' is more
// counterclock-wise than q, then update q.
q = (p + 1) % n;
for ( int i = 0; i < n; i++) {
// If i is more counterclockwise than
// current q, then update q
if (parallel(points[p], points[i], points[q]) == 2)
q = i;
}
// Now q is the most counterclockwise with
// respect to p. Set p as q for next iteration,
// so that q is added to result 'pre'
p = q;
} while (p != l);
// Print Result
for ( int i = 0; i < pre.size(); i++)
cout << "(" << pre[i].x << ", " << pre[i].y << ")" << endl;
}
}; // Driver Code int main() {
Point points[] = {Point(0, 3), Point(2, 2), Point(1, 1), Point(2, 1), Point(3, 0), Point(0, 0), Point(3, 3)};
int n = sizeof (points) / sizeof (points[0]);
LeftIndex obj;
obj.preparata(points, n);
return 0;
} //this code is contributed by bhardwajji |
// Java program to implement Preparata's // a time-optimal parallel algorithm import java.util.*;
// Class for Point class Point
{ int x, y;
// Constructor
Point( int x, int y)
{
this .x = x;
this .y = y;
}
} // Finds the left most index of the given // set of points class LeftIndex
{ int leftMostIndex(Point points[], int n)
{
// Initialize the minimum index
int min = 0 ;
// Finding the point on plane
// with minimum x value
for ( int i = 1 ; i < n; i++)
{
// Update the minimum index
if (points[i].x < points[min].x)
min = i;
else if (points[i].x == points[min].x)
{
if (points[i].y > points[min].y)
min = i;
}
}
// Return the minimum index
return min;
}
// Perform the parallel process in the
// preparata's algorithm according to
// the value of p, q, r
int parallel(Point p, Point q, Point r)
{
// For the three-dimensional
int val = (q.y - p.y) * (r.x - q.x) -
(q.x - p.x) * (r.y - q.y);
// If val is equals to 0, then
// the points are collinear
if (val == 0 )
return 0 ;
else if (val > 0 ) // Clockwise
return 1 ;
else // Anti-clockwise
return 2 ;
}
// Perform the parallel process in the
// preparata's algorithm
void preparata(Point points[], int n)
{
// There must be at least 3 points
if (n < 3 )
return ;
// Find the left most point
int l = leftMostIndex(points, n);
List<Point> pre = new ArrayList<Point>();
// Start from left most point, keep
// moving counterclockwise until reach
// the start point again.
int p = l, q;
do
{
// Add current point to result
pre.add(points[p]);
// Search for a point 'q' such that
// orientation(p, q, r) is counterclockwise
// for all points 'r'. The idea is to keep
// track of last visited most counterclock-
// wise point in q. If any point 'i' is more
// counterclock-wise than q, then update q.
q = (p + 1 ) % n;
for ( int i = 0 ; i < n; i++)
{
// If i is more counterclockwise than
// current q, then update q
if (parallel(points[p], points[i],
points[q]) == 2 )
q = i;
}
// Now q is the most counterclockwise with
// respect to p. Set p as q for next iteration,
// so that q is added to result 'pre'
p = q;
} while (p != l);
// Print Result
for ( int i = 0 ; i < pre.size(); i++)
System.out.println( "(" + pre.get(i).x +
", " + pre.get(i).y + ")" );
}
// Driver Code public static void main(String[] args)
{ Point points[] = new Point[ 7 ];
points[ 0 ] = new Point( 0 , 3 );
points[ 1 ] = new Point( 2 , 2 );
points[ 2 ] = new Point( 1 , 1 );
points[ 3 ] = new Point( 2 , 1 );
points[ 4 ] = new Point( 3 , 0 );
points[ 5 ] = new Point( 0 , 0 );
points[ 6 ] = new Point( 3 , 3 );
LeftIndex obj = new LeftIndex();
obj.preparata(points, 7 );
} } |
# Python program to implement Preparata's # a time-optimal parallel algorithm class func:
def __init__( self , x, y):
self .x = x
self .y = y
# Function to find the left index of # the given set of points def Left_index(points):
# Finding the point on plane
minn = 0
# Traverse the given points
for i in range ( 1 , len (points)):
# Update the value of minn
if points[i].x < points[minn].x:
minn = i
elif points[i].x = = points[minn].x:
if points[i].y > points[minn].y:
minn = i
# Return the value of min
return minn
# Function to perform the parallel # process in the preparata's algorithm # according to the value of p, q, r def parallel(p, q, r):
# For the three-dimensional
val = (q.y - p.y) * (r.x - q.x) - \
(q.x - p.x) * (r.y - q.y)
if val = = 0 :
return 0
elif val > 0 :
return 1
else :
return 2
# Function to perform the parallel # process in the preparata's algorithm def preparata(points, n):
# There must be at least 3 points
if n < 3 :
return
# Find the leftmost point
l = Left_index(points)
pre = []
p = l
q = 0
while ( True ):
# Add current point to result
pre.append(p)
q = (p + 1 ) % n
for i in range (n):
# If i is more counterclockwise
# than current q, then update q
if (parallel(points[p],
points[i], points[q]) = = 2 ):
q = i
p = q
# While it doesn't come to first point
if (p = = l):
break
# Print Result
for each in pre:
print (points[each].x, points[each].y)
# Driver Code algo = []
algo.append(func( 0 , 3 ))
algo.append(func( 2 , 2 ))
algo.append(func( 1 , 1 ))
algo.append(func( 2 , 1 ))
algo.append(func( 3 , 0 ))
algo.append(func( 0 , 0 ))
algo.append(func( 3 , 3 ))
# Function Call preparata(algo, len (algo))
|
using System;
using System.Collections.Generic;
// Class for Point public class Point
{ public int x, y;
// Constructor
public Point( int x, int y)
{
this .x = x;
this .y = y;
}
} // Finds the left most index of the given // set of points public class LeftIndex
{ public int leftMostIndex(Point[] points, int n)
{
int min = 0;
// Finding the point on plane
// with minimum x value
for ( int i = 1; i < n; i++)
{
// Update the minimum index
if (points[i].x < points[min].x)
min = i;
else if (points[i].x == points[min].x)
{
if (points[i].y > points[min].y)
min = i;
}
}
// Return the minimum index
return min;
}
// Perform the parallel process in the
// preparata's algorithm according to
// the value of p, q, r
public int parallel(Point p, Point q, Point r)
{
// For the three-dimensional
int val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y);
// If val is equals to 0, then
// the points are collinear
if (val == 0)
return 0;
else if (val > 0) // Clockwise
return 1;
else // Anti-clockwise
return 2;
}
// Perform the parallel process in the
// preparata's algorithm
public void preparata(Point[] points, int n)
{
// There must be at least 3 points
if (n < 3)
return ;
// Find the left most point
int l = leftMostIndex(points, n);
List<Point> pre = new List<Point>();
// Start from left most point, keep
// moving counterclockwise until reach
// the start point again.
int p = l, q;
do
{
// Add current point to result
pre.Add(points[p]);
// Search for a point 'q' such that
// orientation(p, q, r) is counterclockwise
// for all points 'r'. The idea is to keep
// track of last visited most counterclock-
// wise point in q. If any point 'i' is more
// counterclock-wise than q, then update q.
q = (p + 1) % n;
for ( int i = 0; i < n; i++)
{
// If i is more counterclockwise than
// current q, then update q
if (parallel(points[p], points[i], points[q]) == 2)
q = i;
}
// Now q is the most counterclockwise with
// respect to p. Set p as q for next iteration,
// so that q is added to result 'pre'
p = q;
} while (p != l);
// Print Result
for ( int i = 0; i < pre.Count; i++)
Console.WriteLine( "(" + pre[i].x + ", " + pre[i].y + ")" );
}
} // Driver class Program
{ static void Main( string [] args)
{
Point[] points = { new Point(0, 3), new Point(2, 2), new Point(1, 1), new Point(2, 1), new Point(3, 0), new Point(0, 0), new Point(3, 3) };
int n = points.Length;
LeftIndex obj = new LeftIndex();
obj.preparata(points, n);
Console.ReadLine();
}
} |
//JS equivalent //Function to find the left index of //the given set of points function Left_index(points) {
// Finding the point on plane
let minn = 0;
// Traverse the given points
for (let i = 1; i < points.length; i++) {
// Update the value of minn
if (points[i].x < points[minn].x) {
minn = i;
}
else if (points[i].x == points[minn].x) {
if (points[i].y > points[minn].y) {
minn = i;
}
}
}
// Return the value of min
return minn;
} // Function to perform the parallel // process in the preparata's algorithm // according to the value of p, q, r function parallel(p, q, r) {
// For the three-dimensional
let val = (q.y - p.y) * (r.x - q.x) -
(q.x - p.x) * (r.y - q.y);
if (val == 0) {
return 0;
}
else if (val > 0) {
return 1;
}
else {
return 2;
}
} // Function to perform the parallel // process in the preparata's algorithm function preparata(points, n) {
// There must be at least 3 points
if (n < 3) {
return ;
}
// Find the leftmost point
let l = Left_index(points);
let pre = [];
let p = l;
let q = 0;
while ( true ) {
// Add current point to result
pre.push(p);
q = (p + 1) % n;
for (let i = 0; i < n; i++) {
// If i is more counterclockwise
// than current q, then update q
if (parallel(points[p],
points[i], points[q]) == 2) {
q = i;
}
}
p = q;
// While it doesn't come to first point
if (p == l) {
break ;
}
}
// Print Result
for (let each of pre) {
console.log(points[each].x, points[each].y);
}
} // Driver Code let algo = []; algo.push({x: 0, y: 3}); algo.push({x: 2, y: 2}); algo.push({x: 1, y: 1}); algo.push({x: 2, y: 1}); algo.push({x: 3, y: 0}); algo.push({x: 0, y: 0}); algo.push({x: 3, y: 3}); // Function Call preparata(algo, algo.length); |
Output
0 3 0 0 3 0 3 3