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Iterating Over All Subsets of a Set | Reply
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
This is quite similar to my original cookbook article. I've added some more examples of subsets and how to map them to binary numbers. I think these help in understanding. This article is missing some example problems.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
Two last paragraphs can actually be replaced with a link to recipe http://forums.topcoder.com/?module=Thread&threadID=671561&start=0 . And yes, this needs two examples of problems solved using this. Otherwise looks good.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
Now let's solve VectorMatching. In this problem we are given n points and we are asked to find a vector matching that has the minimal length. The number of points is small (n=20) which allows us to iterate over all possible vector matchings. A vector matching consists of two sets of points A and B, such that the head of the i-th vector is the i-th point in A and its tail is the i-th point in B. A and B should have the same number of elements, namely n/2. Each point can be either in set A or in set B, so we have 2^n possible states. However, we only need to consider those states where the size of A is n/2. This reduces the problem to Choose(20,10) = 185756 states.

Suppose the point (x1,y1) is in A and the point (x2,y2) is in B. The vector represented by these two points is (x2-x1,y2-y1). Hence if a point is in A then it contributes negatively to the vector sum; otherwise it is in B and it contributes positively. So now we can determine the vector sum (which is a vector) as we iterate through the points in A and B; its x-component will be sumX and its y-component will be sumY:
public class VectorMatching
{
  public double minimumLength(int[] x, int[] y)
  {
    double result = Double.MAX_VALUE;
    int n = x.length;
    for (int i = 0; i < (1 << n); i++)
    {
      //size of A and B must be n/2
      if (2*Integer.bitCount(i) != n) continue;		
 
      long sumX = 0;
      long sumY = 0;
      for (int k = 0; k < n; k++)
      {
	//point (x[k],y[k]) is in B
	if ((i&(1<<k)) > 0)
	{
          sumX += x[k];
          sumY += y[k];
        }
	//point (x[k],y[k]) is in A
	else
	{
          sumX -= x[k];
          sumY -= y[k];
        }
      }
 
      double length=Math.sqrt(sumX*sumX + sumY*sumY);  //vector length
      result = Math.min(result,length);  //record the minimum length
    }
    return result;
  }
}
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
I think it should be mentioned that BouncingBalls can in fact be solved more easily without iterating over all subsets. (By changing the order of summation by summing over "i" inside the "k" and "m" loops, it is easy to show that it is enough to add 0.25 for each pair of balls which are close enough.)
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
// iterate over all the subsets with no more than m elements
for (int i = 0; i < (1<<n); i=Integer.bitCount(i) >< m ? i+1 : (i|(i-1))+1)
{
...
}

why i can't modify it as below?

for (int i = 0; i < (1<<n) && Integer.bitCount(i)><=m; i++)
{
...
}
Re: Iterating Over All Subsets of a Set (response to post by savon_cn) | Reply
First of all your code tests all 2^n elements, which makes it much slower. Secondly your code will terminate as soon as it reaches a number with more than m bits set to 1. This means that it will not find all subsets with no more than m elements, only some of them.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
Another useful bit twiddling hack is gospers hack, which calculates the next greater number, that has the same number of bits set in binary representation. This can be used in a loop, to run through those subsets with k elements.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
You said "Another trick is to iterate over all the subsets of a subset in O(3^n) time", can you explain a bit more how O(3^n) was obtained? Thanks.
Re: Iterating Over All Subsets of a Set (response to post by hacker007) | Reply
There are exactly 3^n pairs (A,B) where B is a set of elements of {1, ..., n} and A is a subset of B. To see this, notice that each pair of sets like this can be associated to a n-digit ternary number (with leading zeros, possibly): Put a 0 in the i-th position if i is in neither A nor B, an 1 if i is in B but not A, and a 2 if i is in both sets. Since this is clearly a bijection, we've proved the result.

If you take O(1) time per subset while iterating, you then have a O(3^n) algorithm for iterating over all pairs as above.
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
How will the method to go over all subsets of a set change if the numbers are not unique. For example if we need to find subsets of {1,1,2,3,4,4,4} such that no set is repeated i.e. We don't need to have {1} twice. But {1,1} will be a valid set.

What is the best way to do it?
Re: Iterating Over All Subsets of a Set (response to post by puneetm) | Reply
I don't know how to do that. I guess you could keep a HashSet of all subsets that you have seen (you need a good hash function). If you meet a subset that you have already seen then skip it.
Re: Iterating Over All Subsets of a Set (response to post by puneetm) | Reply
You can change to another representation of the set: (element,count). The set is then represented as {(1,2),(2,1),(3,1),(4,3)}. For each element, loop through each possible count. In this case, there are 3*2*2*4 subsets to loop through (including the empty set).
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
the cookbook example is good and its helping me. Can u site some problem of this type as i am learning how to program.

Thank in advance
Re: Iterating Over All Subsets of a Set (response to post by dimkadimon) | Reply
can you please explain what this line is doing " if ((i&(1<<k)) > 0) " ?
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