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 → 1It 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.
Running time: 6100ms (6.1 sec)
Assessment: I had to read this problem several times before I really grokked it, and then I thought it was pretty nifty. I’m sure there’s a faster way of working out the answer than my brute force solution.
import java.util.*; public class Problem014 { public static void main(String[] args) { long begin = System.currentTimeMillis(); LinkedList<Long> list = new LinkedList<Long>(); long length = 0; int res = 0; for(int j = 10; j < 1000000; j++) { long i = j; while (i != 1) { if (i % 2 == 0) { i /= 2; list.add(i); } else { i = (3 * i) + 1; list.add(i); } } if(list.size() > length) { length = list.size(); res = j; } list.clear(); } long end = System.currentTimeMillis(); System.out.println(res+" with chain size: "+ length); System.out.println(end-begin + "ms"); } }
Very nice solution .. Thanks :)Â
This is pretty much identical to my solution, with the only difference being that I used ArrayList.
Got it in 28 milliseconds on my T6600, in C++ o/
package CoreJavaAssignments;
public class Problem14 {
public static void main(String[] args) {
long begin = System.currentTimeMillis();
int maxLength = 1;
int number = 0;
for (int n = 1; n maxLength) {
maxLength = length;
number = n;
}
}
long end = System.currentTimeMillis();
System.out.println(“Number:” + number + “Length:” + maxLength);
System.out.println(end – begin + “ms”);
}
private static int findTheChain(int n) {
long localNumber = n;
int count = 0;
while (localNumber != 1) {
if (localNumber % 2 == 0) {
localNumber = localNumber / 2;
count++;
} else {
localNumber = (3 * localNumber) + 1;
count++;
}
}
return count;
}
}
No need to use any data structure :)
public class EulerProject14
{
public static void main(String[] args)
{
int kingnum = 0;
int bignum = 0;
int count =0;
for( int i=10; i=kingnum)
{
kingnum = count;
bignum = i;
}
}
System.out.println(bignum);
}
}
mine minutes to run how can i make it faster
Could you delete the solution please otherwise my students going to copy it you can re post it after august.
Kind Regards
No.
Cheaters will cheat. You can’t scour the internet to have all Project Euler solutions removed. It’s idiotic on its face.
If you’re worried about it, maybe you should come up with your own problem sets? Instead of copying others’ work?