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 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 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< 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 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 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.
 Iterating Over All Subsets of a Set | Reply Problem:Iterate over all the subsets of a set.Solution:A subset of a set is a set that contains all or some of the elements of that set. For example if the set is {3, 5, 7} then some of its subsets are {} - the empty set, {3, 7}, {5, 7}, {5} and {3, 5, 7} - the original set. Often we need to find a subset with a particular property. If the original set is small, ie. has a small number of elements then it is sufficient to generate and test all its subsets. If the original set has n elements then it will contain 2^n subsets. Assuming that the computation required for each subset is small, we can usually do this for n up to 20.Let us represent a subset with a unique binary number i. This way there will be a total of 2^n such unique numbers. The k-th element of the set can be either included or not included in the subset. We can associate the inclusion of the k-th element with a value of 1 at the k-th bit of i, similarly if the k-th bit is 0 then the k-th element is not included. For example, here is how we can associate a unique binary number from 0 to 7 with every subset of {3, 5, 7}:`Binary Number Subset000 {}001 {7}010 {5}011 {5, 7}100 {3}101 {3, 7}110 {3, 5}111 {3, 5, 7}`Now all that remains is to iterate over all the subsets. Since these are just binary numbers, it suffices to iterate from 0 to 2^n-1 inclusive. In the following Java example we iterate over all the subsets of the set {"appple", "banana", "carrot"}. After generating a subset we can perform any action. Here for simplicity we print it. Notice that the 0-th subset is the empty subset, as it contains no elements.```String[] set={"apple", "banana", "carrot"}; int n=set.length;   // iterate over all the subsets for (int i=0; i < (1< 0) subset.add(set[k]); } // perform an action over the subset, here just print it System.out.print("Subset "+i+":"); for (int k=0; k 0; i2 = (i2-1) & i) { // generate the subset induced by i2 ... } } ```Warning: i=0 (empty subset) is a special case and must be treated separately.Finally, there are many other operations that can be performed on sets: union, intersection, subtraction, element addition, element deletion. All of these can be done with simple and extremely fast bitwise operations. If A and B are two sets whose bit-masks are a and b respectively then we have the following:Union of A and B: a|bIntersection of A and B: a&bA "minus" B: a&(~b). This is a set of elements in A that are not in B.Add k-th element to set A: a |= 1 << kDelete k-th element of set A: a &= ~(1 << k)For an excellent description of bitwise operations take a look at bmerry’s recipe.Now let's solve some problems that use these techniques. Let's solve BouncingBalls. For this problem, the number of balls (n<=12) is sufficiently small that we can generate all possible starting states. Each ball can be rolled either left or right, so we have 2^n possible starting states. Let dir[i] be the initial direction given to the i-th ball, 1 for right and -1 for left. A collision will occur between balls k and m if ball k is to the left of ball m (x[k]<=2*T). For each possible starting state, we count the number of possible bounces for every pair of balls k and m. To compute the expected number of bounces we divide this value by the total number of starting states (1<```public class BouncingBalls { public double expectedBounces(int[] x, int T) { int n=x.length; int bounces=0;   //iterate over all possible starting states for (int i=0; i<(1< 0) dir[k]=+1; //roll ball k right else dir[k]=-1; //roll ball k left } for (int k=0; k
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