Project Euler Question 14 (Collatz Problem)

Question

The following iterative sequence is defined for the set of positive integers:

n ->n/2 (n is even) n ->3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 40 20 10 5 16 8 4 2 1 It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

I tried coding a solution to this in C using the bruteforce method. However, it seems that my program stalls when trying to calculate 113383. Please advise :)

#include <stdio.h>
#define LIMIT 1000000

int iteration(int value)
{
 if(value%2==0)
  return (value/2);
 else
  return (3*value+1);
}

int count_iterations(int value)
{
 int count=1;
 //printf("%d\n", value);
 while(value!=1)
 {
  value=iteration(value);
  //printf("%d\n", value);
  count++;
 }
 return count;
}

int main()
{
 int iteration_count=0, max=0;
 int i,count;


 for (i=1; i<LIMIT; i++)
 {
  printf("Current iteration : %d\n", i);
  iteration_count=count_iterations(i);
  if (iteration_count>max)
   {
   max=iteration_count;
   count=i;
   }

 }

 //iteration_count=count_iterations(113383); 
 printf("Count = %d\ni = %d\n",max,count);

}

Answer 1

The reason you're stalling is because you pass through a number greater than 2^31-1 (aka INT_MAX); try using unsigned long long instead of int.

I recently blogged about this; note that in C the naive iterative method is more than fast enough. For dynamic languages you may need to optimize by memoizing in order to obey the one minute rule (but this is not the case here).

---

Oops I did it again (this time examining further possible optimizations using C++).

Answer 2

Notice that your brute force solution often computes the same subproblems over and over again. For example, if you start with 10, you get 5 16 8 4 2 1; but if you start with 20, you get 20 10 5 16 8 4 2 1. If you cache the value at 10 once it's computed, and then won't have to compute it all over again.

(This is known as dynamic programming.)