Prefix sum (multi-processor-edition)
A. Second Edition
This is programming assignment of comp346. And in this edition I changed the implementation to follow professor's
What is "prefix sum"? Try to imagine that we have n numbers and I want to know the sum of first 1,
the sum of first 2, the sum of first 3... the sum of first n. How to calculate?
กก
A kind of repetition of previous assignment. Basically you need only calculate the sum AFTER your previous
neighbor finishes it. So, you simply declare n threads and each them just sum up its own value with its previous
neighbor's value.
Maybe I am still not quite clear about the algorithms of programming assignment2 because I am still not sure about if
the way PA2 is better than what we did in PA1. See below, is it correct about PA2? That is each column will calculate
its "sum" with "previous-hop-column" (sum+=a[level][j-hop]). You can see the total sum operation is FIVE.
col0 col1 col2 col3
level 0 a0 a1 a2 a3
level 1 a0 a01=a0+a1 a12=a1+a2 a23=a2+a3
level 2 a0 a01 a012=a0+a12 a0123=a01+a23
E.Further improvement
F.File listing
1. Cyclic.java
file name: Cyclic.java
public class Cyclic {
static int P = 8; // number of workers
static int logP = 3;
static int Pplus1 = 9; // 0-indexed array
static int logPplus1 = 4; // lengths
static Daemon daemon = new Daemon ();
static Semaphore[][] future = new Semaphore[logP][P];
static int[][] a = new int[logP][P];
static int col;
//static ...
public static void main (String[] args) {
Worker w[] = new Worker[P];
for (int k=0; k<logP; k++)
{
future[k]=new Semaphore[P];
//for( int j=0; j<e.length; j++ )
for (int j=0; j<P; j++)
{
future[k][j] = new Semaphore(0);
}
}
for( int j=0; j<P; j++ ) {
//test
//System.out.println("test signal here");
future[0][3].Signal();
col = daemon.column (j);
w[col] = new Worker (col);
w[col].start ();
}
for( int j=0; j<P; j++ ) { // wait for workers
try { w[j].join (); } // to terminate
catch (InterruptedException exc) { };
}
for( int j=1; j<Pplus1; j++ )
System.out.println ("Sum " + j + " = "+ a[logP-1][j-1]);
System.out.println ("System terminates normally.");
}
}
class Daemon {
private static int[] w = new int[] {3, 6, 5, 7, 4, 2, 1, 0};
private static int[] d = new int[] {0, 7, 4, 5, 4, 4, 9, 4, 1};
public int column (int j) {
return w[j];
}
public void interrupt (int j) {
if ( d[j] > 4 )
Thread.yield ();
}
}
class Semaphore {
private int value;
Semaphore (int value1) {
value = value1;
}
public synchronized void Wait () {
while( value <= 0 ) {
try { wait (); }
catch (InterruptedException e) { };
}
value--;
}
public synchronized void Signal () {
++value;
notify ();
}
}
class Worker extends Thread {
private int j;
private int sum;
private int hop = 1;
Worker (int col) {
sum = col+1;
j = col;
}
public void run () {
System.out.println ("Worker " + (j) + " begins execution.");
yield ();
for (int level=0; level<Cyclic.logP; level++) {
// if (j <= (Cyclic.P-hop))
{
System.out.println ("Worker " + j + " defines a[" + (level) + ","
+ j +"].");
Cyclic.a[level][j] = sum;
//System.out.println("a["+level+"]["+j+"]="+Cyclic.a[level][j]);
//System.out.println("exactly here before signal "+j);
Cyclic.future[level][j].Signal();
//System.out.println("exactly after signal of "+j);
// ...
}
if (j >= hop)
{
// ...
//test
//System.out.println("before wait of "+j);
Cyclic.future[level][j-hop].Wait();
System.out.println ("Worker " + j + " uses a[" + (level) + ","
+ (j-hop) +"].");
sum += Cyclic.a[level][j-hop];
System.out.println("worker "+(j)+" has sum of "+ sum);
}
//test
//System.out.println("before interrupt");
Cyclic.daemon.interrupt (j);
//test
//System.out.println("after interrupt");
hop = 2*hop;
}
Cyclic.a[Cyclic.logP-1][j]=sum;
System.out.println("Worker " + j + " terminates.");
}
}
The running result is something like following:
[qingz_hu@alpha PA2] % java Cyclic
Worker 3 begins execution.
Worker 6 begins execution.
Worker 5 begins execution.
Worker 7 begins execution.
Worker 4 begins execution.
Worker 2 begins execution.
Worker 1 begins execution.
Worker 0 begins execution.
Worker 3 defines a[0,3].
Worker 6 defines a[0,6].
Worker 5 defines a[0,5].
Worker 7 defines a[0,7].
Worker 7 uses a[0,6].
worker 7 has sum of 15
Worker 7 defines a[1,7].
Worker 4 defines a[0,4].
Worker 4 uses a[0,3].
worker 4 has sum of 9
Worker 4 defines a[1,4].
Worker 2 defines a[0,2].
Worker 1 defines a[0,1].
Worker 0 defines a[0,0].
Worker 0 defines a[1,0].
Worker 0 defines a[2,0].
Worker 0 terminates.
Worker 6 uses a[0,5].
worker 6 has sum of 13
Worker 5 uses a[0,4].
worker 5 has sum of 11
Worker 5 defines a[1,5].
Worker 3 uses a[0,2].
worker 3 has sum of 7
Worker 2 uses a[0,1].
worker 2 has sum of 5
Worker 2 defines a[1,2].
Worker 2 uses a[1,0].
worker 2 has sum of 6
Worker 2 defines a[2,2].
Worker 2 terminates.
Worker 1 uses a[0,0].
worker 1 has sum of 3
Worker 6 defines a[1,6].
Worker 6 uses a[1,4].
worker 6 has sum of 22
Worker 7 uses a[1,5].
worker 7 has sum of 26
Worker 7 defines a[2,7].
Worker 3 defines a[1,3].
Worker 4 uses a[1,2].
worker 4 has sum of 14
Worker 4 defines a[2,4].
Worker 4 uses a[2,0].
worker 4 has sum of 15
Worker 4 terminates.
Worker 1 defines a[1,1].
Worker 6 defines a[2,6].
Worker 6 uses a[2,2].
worker 6 has sum of 28
Worker 5 uses a[1,3].
worker 5 has sum of 18
Worker 5 defines a[2,5].
Worker 3 uses a[1,1].
worker 3 has sum of 10
Worker 1 defines a[2,1].
Worker 6 terminates.
Worker 3 defines a[2,3].
Worker 5 uses a[2,1].
worker 5 has sum of 21
Worker 5 terminates.
Worker 1 terminates.
Worker 7 uses a[2,3].
worker 7 has sum of 36
Worker 7 terminates.
Worker 3 terminates.
Sum 1 = 1
Sum 2 = 3
Sum 3 = 6
Sum 4 = 10
Sum 5 = 15
Sum 6 = 21
Sum 7 = 28
Sum 8 = 36
System terminates normally.