Help needed urgently on a problem..plzzz

Question

hi..this is a linear congruential generator. I have to implement it and i need the execution time for the program.
for your understanding i'm providing an example below.
Xn=(( a* xn-1 )+b) mod m
If X0=7 ; a = 7 ; b =7 ; m=10
Then
X0 = 7
X1 =((7 * 7) + 7))mod 10 = 6
X2 = ((6*7)+7))mod 10 = 9
X3 = ((9*7)+7) mod 10 = 0
X4 = ((0*7)+7) mod 10 = 7
Now since the cycle is being repeated i.e 7 appears again�so the period is 4 coz there are 4 diff nos. i.e 7,6,9,0�..
help required urgently....your help will be appreciated...thankyou..

Answer 1

Hi,
I wrote the code so that it catches any cycle (not only the "big" one).
Otherwise it will enter infinite loop...
The time complexity is O(N*logN): it can do at most N iterations (here N is your 'm'), and in each iteration there can be O(log N) comparisons (since I maintain TreeSet).
Interesting issue: is it possible to supply such (x0, a, b, m) tuple such that all possible values from 0 to m-1 will be output? I think no :)
Here is the program:
package recurr;
import java.util.TreeSet;
import java.util.Comparator;
public class Recurrences {
     private static long x0, a, b, m;
     private static TreeSet theSet;
     public static void main(String[] args)
          long l0, l1, l2, l3;
          try {
               x0 = Long.parseLong(args[0]);
               a = Long.parseLong(args[1]);
               b = Long.parseLong(args[2]);
               m = Long.parseLong(args[3]);
          } catch(NumberFormatException nfe) {
               nfe.printStackTrace();
          System.out.println("X[0]: " + x0 + "\n");
          long curr = x0;
          boolean cut = false;
          int i;
          // initialize the set
          theSet = new TreeSet(new LongComparator());
          // we can get at most m distinct values (from 0 to m-1) through recurrences
          for(i=1; i <= m; ++i) {
               // iterate until we find duplicate
               theSet.add(new Long(curr));
               curr = recurrence(curr);
               if(theSet.contains(new Long(curr))) {
                    cut = true;
                    break;
               System.out.println("X[" + i + "]: " + curr + "\n");
          if(cut) {
               System.out.println("Cycle found: the next will come " + curr + "\n");
          } else {
               System.out.println("No cycle found!");
          System.out.println("----------------------------------");
          System.out.println("Totally " + (i-1) + " iterations");
     private static long recurrence(long previous)
          return (a*previous + b)%m;
     static class LongComparator implements Comparator
          public int compare(Object o1, Object o2)
               if(((Long)o1).longValue() < ((Long)o2).longValue()) {
                    return -1;
               } else if(((Long)o1).longValue() > ((Long)o2).longValue()) {
                    return 1;
               } else return 0;
}