Пример #1
0
 /**
  * Fill the list recursively with collideNum numbers that sum to <i>sum</i>. Assuming items
  * already in the array p sum to <i>usedSum</i>. The numbers must be > min. It stops if it finds
  * <i>limit</i> patterns and the probability of those found patterns become very small comparing
  * to the other patterns found (probability <1/freq)
  *
  * @param patterns
  * @param collideNum
  * @param sum
  * @param usedSum sum of items in array <i>p</i>
  * @param min: is used to make p a sorted array
  * @param level what is the current level of recursion (number of items in array <i>p</i>)
  * @param p keeps track of already decided items in the pattern, it will be a sorted array
  * @param oldDistribution
  * @param limit the number of patterns to be found
  * @param freqI the frequency of the sum items in the sketch
  * @return
  */
 private boolean getPatternRecursive(
     List<Map<Integer, Integer>> patterns,
     int collideNum,
     int sum,
     int usedSum,
     int min,
     int level,
     int[] p,
     Distribution oldDistribution,
     double limit,
     int freqI) {
   if (min > sum - usedSum) { // repetitive
     return false;
   }
   if (level == collideNum - 1) { // create the pattern with whatever remained out of sum
     p[level] = sum - usedSum;
     if (oldDistribution.getFreq(p[level]) == 0) {
       return false;
     }
     Map<Integer, Integer> pattern = getPattern(p);
     patterns.add(pattern);
     double prob = getProb(oldDistribution, pattern);
     if (sumProb == 0) {
       meanProb = prob * (level + 1);
     } else {
       meanProb = alpha * meanProb + (1 - alpha) * prob * (level + 1);
     }
     sumProb += prob;
   } else {
     for (int j = min; j <= sum - usedSum - (collideNum - level - 1); j++) {
       p[level] = j;
       if (oldDistribution.getFreq(p[level]) == 0) {
         continue;
       }
       // if I have found enough patterns and the probability of found items are very small
       if (patterns.size() > limit && meanProb / sumProb < 1.0 / freqI) {
         return true;
       }
       getPatternRecursive(
           patterns,
           collideNum,
           sum,
           usedSum + j,
           Math.max(min, j),
           level + 1,
           p,
           oldDistribution,
           limit,
           freqI);
     }
   }
   return false;
 }
Пример #2
0
 /**
  * Returns te probability of each of the key items in the pattern. It leverages Poisson
  * distribution. See the paper for its description
  *
  * @param distribution
  * @param pattern
  * @return
  */
 private double getProb(Distribution distribution, Map<Integer, Integer> pattern) {
   return pattern
       .entrySet()
       .stream()
       .mapToDouble(
           e -> {
             double l = distribution.getFreq(e.getKey()) / capwidth;
             return l == 0
                 ? 0
                 : (new PoissonDistribution(l).probability(e.getValue()) / FastMath.exp(-l));
           })
       .reduce((a, b) -> (a * b))
       .getAsDouble();
 }