Fast O(n) Integer Sorting Algorithm!

Update: This post was reposted @ dzone with lots of upvotes – Thanks!

Yesterday I learned that there is an O(n) integer sort algorithm (I should have read this before in a basic algorithm book :-/).

Now I wondered: is this necessary in real applications? E.g. somewhere in Java? Today I have taken the counting sort and I can argue: yes, you should use integer sort especially for large arrays!

And when in detail should you apply the fast integer sort? Apply it if

  • you have positive integer values to sort. The requirement ‘positive’ and ‘integer’ is necessary for the listed O(n) algorithm, but not if you implement your own possible better solution.
  • you have a limited interval for the integer values (preferable min and max=M should be known before you sort)
    E.g. if you know the maximum integer number in your array will be M=10^7 then you should use the integer sort if the array length n is roughly greater than M/2500 = 40000. This linear equation should hold true (for some values ;-)), because quick sort is nearly independent of M and the time-offset for integer sort increases nearly linear with M as you can see in the graph

Now take a look at the graph where y=time in seconds for 10 runs and x=array length:

Conclusion

I would apply this sorting algorithm only for n>10^7 where the difference between quicksort and integer sort could lay in the range of seconds. The memory consumption was not measured but should be ~twice times higher for the fast integer sort.

Java Sourcecode

//class LinearSort
public static void main(String[] args) {

 // init jvm
 new LinearSort().start(1000, 10000, 10000);
 new LinearSort().start(1000, 10000, 10000);

 // run performance comparison
 for (int maxInteger = 1000; maxInteger < 100000000; maxInteger *= 3) {
  for (int arrLength = 1000; arrLength < 100000000; arrLength *= 3) {
   System.gc();
   new LinearSort().start(arrLength, maxInteger, 10);
  }
 }
}

 private Random rand = new Random();

 // stop watch for integer sort with *unknown* range. marked as Lin in the plot
 private SimpleTimer linearStopWatch = new SimpleTimer();</pre>

 // stop watch for integer sort with known range. marked as Lin' in the plot
 private SimpleTimer linearKnownStopWatch = new SimpleTimer();

 private SimpleTimer qSortStopWatch = new SimpleTimer();

 private void start(int arrLength, int maxInteger, int times) {
 for (int count = 0; count < times; count++) {
   int[] list1 = new int[arrLength];
   for (int i = 0; i < arrLength; i++) {
     // do only allow positive integers until the specified 'max'-value
     list1[i] = Math.abs(rand.nextInt(maxInteger));
   }
   linearStopWatch.start();
   LinearSort.sort(list1);
   linearStopWatch.pause();

   int[] list2 = Arrays.copyOf(list1, arrLength);
   qSortStopWatch.start();
   Arrays.sort(list2);
   qSortStopWatch.pause();

   list2 = Arrays.copyOf(list1, arrLength);
   linearKnownStopWatch.start();
   LinearSort.sort(list2, 0, maxInteger);
   linearKnownStopWatch.pause();
 }

 System.out.println(maxInteger + ";" + arrLength + ";" + linearStopWatch
 + ";" + linearKnownStopWatch
 + ";" + qSortStopWatch); // + ";" + qSortListStopWatch);
}

 static int[] sort(int[] array, int min, int max) {
   //the range is useful to minmize the memory usage
   //countIntegers holds the number of each integer
   int[] countIntegers = new int[max - min + 1];

   for (int i = 0; i < array.length; i++) {
     countIntegers[array[i] - min]++;
   }

   int insertPosition = 0;
   //fill array in sorted order
   for (int i = min; i <= max; i++) {
     for (int j = 0; j < countIntegers[i - min]; j++) {
       array[insertPosition++] = i;
     }
   }
   return array;
 }

 static int[] sort(int[] array) {
   int min, max = min = array[0];
   //determine the max and min in the array
   for (int i = 1; i < array.length; i++) {
     if (array[i] < min)
       min = array[i];

     if (array[i] > max)
       max = array[i];
   }
   return sort(array, min, max);
 }

 //class SimpleTimer

 private long lastStart = -1;
 private long time;

 public void start() {
   if (lastStart != -1)
     throw new IllegalStateException("Call stop before!");

   lastStart = System.currentTimeMillis();
 }

 public void pause() {
   if (lastStart < 0)
     throw new IllegalStateException("Call start before!");

   time = time + (System.currentTimeMillis() - lastStart);
   lastStart = -1;
 }

 public String toString() {
   return time / 1000f + "";
 }

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2 thoughts on “Fast O(n) Integer Sorting Algorithm!

  1. Pingback: Сортировка за O(N)-время | О программировании

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