Minimum Set to satisfies the condition & formulae - Algorithm

Question

I need a solution for this problem. Please help..
Problem:</
Input:_
Let N = 100
{x y}
{1 2} - A
{2 3} - B
{1 2} - C
{1 5} - D
Condition:_
I can pick any set from the above, like {A, B} or {A, A, A, A....} or {A, B, D} or {A, A, B, B, C, C, C, ...} etc...
Formulae:_
example:
for {A, B}, the formulae is --> (Ax + Bx)^2 + (Ay + By)^2 = N^2
for {A, B, D}, the formulae is --> (Ax + Bx + Dx)^2 + (Ay + By + Dy)^2 = N^2
i.e) (sum of all selected element's x value) ^ 2 + (sum of all selected element's y value) ^ 2 = N ^ 2
Output: _
The minimal number of elements that can be used to achieve the condition.
example, if both {A, B} & {A, C, D} satisfies the formulae then my solution is {A, B} which has minimal elements in the set. so the output will be 2. if no solution, then return -1.
Hope I have explained it clearly. Please help...

Answer 1

First find the Pythagorean triples corresponding to a*a + b*b = N*N. This will give integer values of a and b which your x and y values should total to. This simplifies the next stage, as finding a set of values k1*x1+k2*x2+... + kn*xn = a reduces the search space somewhat - your last term will always be ( a - (k1*x1..kn-1*xn-1) ) / xn.
Order your values so xi > xi+1. This means you can stop easily when xi == 0, and it's typically bit faster to try removing the biggest values first.
You can also use dynamic programming to avoid some repeated calculations, but even without it's fairly fast, even for 20 or so different vectors in the basis.
    static class FindMinimimSet implements TripleProcessor {
        int bestTotal = Integer.MAX_VALUE;
        final int[][] basis;
        final int[] counts;
        final HashMap<Long,Integer> memo = new HashMap<Long,Integer>();
        // basis should be sorted by x then y with maximum values first,
        // eg { {4,3}, {4,2}, {0,4}, {0,2} }
        FindMinimimSet ( int[][] basis ) {
            this.basis = basis;
            this.counts = new int[basis.length];
        public void triple ( int a, int b, int c ) {
            bestTotal = findMinimumSet( a, b, bestTotal );
            if ( a != b )
                bestTotal = findMinimumSet( b, a, bestTotal );
        int findMinimumSet ( int a, int b, int bestTotal ) {
            return findMinimumSet ( a, b, 0, new int[basis.length], 0, bestTotal );
        int findMinimumSet ( int a, int b, int i, int[] counts, int total, int bestTotal ) {
            long key = (a * 104729L + b) * 104729L + i;
            Integer val = memo.get(key);
            if (val == null) {
                // TODO: record actual counts too
                val = reallyFindMinimumSet(a,b,i,counts,total,bestTotal);
                memo.put(key,val);
            return val;
        int reallyFindMinimumSet ( int a, int b, int i, int[] counts, int total, int bestTotal ) {
            if ( ( a == 0 ) && ( b == 0 ) ) {
                // todo - save the counts somewhere rather than just outputting them
                System.out.println( total + " -> " + Arrays.toString( counts ) );
                return total;
            if ( i >= basis.length )
                return Integer.MAX_VALUE / 2;
            int[] vector = basis;
if ( i < basis.length - 1 ) {
final int maxCount = Math.min ( bestTotal - total, Math.min ( ( vector[0] > 0 ) ? a / vector[0] : b, ( vector[1] > 0 ) ? ( b / vector[1] ) : a ) );
// if the next value of x is zero, and this value isn't, count must be exact
if ( ( vector[0] != 0 ) && ( basis[i+1][0] == 0 ) ) {
if ( vector[0] * maxCount != a )
return bestTotal;
} else {               
for ( int count = maxCount; count >= 0; --count ) {
counts[i] = count;
bestTotal = findMinimumSet( a - vector[0] * count, b - vector[1] * count, i+1, counts, total + count, bestTotal );
return bestTotal;
// only check the max count for last in vector/last with non-zero x
final int count = Math.min( ( vector[0] > 0 ) ? a / vector[0] : b, ( vector[1] > 0 ) ? ( b / vector[1] ) : a );
if ( total + count < bestTotal ) {
counts[i] = count;
bestTotal = findMinimumSet( a - vector[0] * count, b - vector[1] * count, i+1, counts, total + count, bestTotal );
return bestTotal;
The code to find Pythagorean triples, given the longest side:public interface TripleProcessor {
void triple ( int a, int b, int c ) ;
public static void findTriples ( int c, TripleProcessor out ) {
final int c2 = c * c;
final HashMap<Integer,Integer> squares = new HashMap<Integer,Integer>(c2);
for ( int i = 1; i < c; ++i )
squares.put(i*i, i);
for ( int a = 1; a < c; ++a ) {
final Integer b = squares.get(c2 - a*a);
if ( b != null )
if ( a <= b )
out.triple(a,b,c);
else
break;