need an algorithm to find possible combinations

You may be able to adapt this code.

/*

The CombinationGenerator Java class systematically generates all combinations of n

elements, taken r at a time. The algorithm is described by Kenneth H. Rosen, Discrete

Mathematics and Its Applications, 2nd edition (NY: McGraw-Hill, 1991), pp. 284-286.

The class is very easy to use. Suppose that you wish to generate all possible

three-letter combinations of the letters "a", "b", "c", "d", "e", "f", "g".

Put the letters into an array. Keep calling the combination generator's getNext ()

method until there are no more combinations left. The getNext () method returns

an array of integers, which tell you the order in which to arrange your original

array of letters. Here is a snippet of code which illustrates how to use the

CombinationGenerator class.

String[] elements = {"a", "b", "c", "d", "e", "f", "g"};

int[] indices;

CombinationGenerator x = new CombinationGenerator (elements.length, 3);

StringBuffer combination;

while (x.hasMore ()) {

combination = new StringBuffer ();

indices = x.getNext ();

for (int i = 0; i < indices.length; i++) {

combination.append (elements[indices[i]]);

}

System.out.println (combination.toString ());

}

Another example of the usage of the CombinationGenerator is shown below in connection with the Zen Archery problem.

Source Code

The source code is free for you to use in whatever way you wish.

//--

// Systematically generate combinations.

//--

*/

import java.math.BigInteger;

public class CombinationGenerator {

private int[] a;

private int n; // all combinations of n elements

private int r; // taken r at a time

private BigInteger numLeft;

private BigInteger total;

//

// Constructor

//

public CombinationGenerator (int n, int r) {

if (r > n) {

throw new IllegalArgumentException ();

}

if (n < 1) {

throw new IllegalArgumentException ();

}

this.n = n;

this.r = r;

a = new int[r];

BigInteger nFact = getFactorial (n);

BigInteger rFact = getFactorial (r);

BigInteger nminusrFact = getFactorial (n - r);

total = nFact.divide (rFact.multiply (nminusrFact));

reset ();

}

//

// Reset

//

public void reset () {

for (int i = 0; i < a.length; i++) {

a[i] = i;

}

numLeft = new BigInteger (total.toString ());

}

//

// Return number of combinations not yet generated

//

public BigInteger getNumLeft () {

return numLeft;

}

//--

// Are there more combinations?

//--

public boolean hasMore () {

return numLeft.compareTo (BigInteger.ZERO) == 1;

}

//

// Return total number of combinations

//

public BigInteger getTotal () {

return total;

}

//

// Compute factorial

//

private static BigInteger getFactorial (int n) {

BigInteger fact = BigInteger.ONE;

for (int i = n; i > 1; i--) {

fact = fact.multiply (new BigInteger (Integer.toString (i)));

}

return fact;

}

//--

// Generate next combination (algorithm from Rosen p. 286)

//--

public int[] getNext () {

if (numLeft.equals (total)) {

numLeft = numLeft.subtract (BigInteger.ONE);

return a;

}

int i = r - 1;

while (a[i] == n - r + i) {

i--;

}

a[i] = a[i] + 1;

for (int j = i + 1; j < r; j++) {

a[j] = a[i] + j - i;

}

numLeft = numLeft.subtract (BigInteger.ONE);

return a;

}

}

ArtVandelaya at 2007-7-16 1:45:32 >

Yes you can also take this also as a Set operations.

Lets us say i have A,B,C,D as sets .

I want to find out elements in And b alone

that is (A and B) MINUS (A AND B AND C ) MINUS(A AND B AND D)

WHERE (A AND B AND C ) IS (A AND B AND ) MINUS ((A AND B AND C AND D))

(A AND B AND D) IS

(A AND B AND D) MINUS (A AND B AND C AND D)

DP_devAa at 2007-7-16 1:45:32 >