Project Euler – Problem # 14 – Solved with Go

Problem:

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.

package main

import "fmt"

func main() {
	largest := 0
	longest_sequence := 0
	for n := 2; n < 1000000; n++ {

		var (
			Number int64 = 0
		)
		Number = int64(n)
		counter := 1
		for Number > 1 {
			if Number%2 == 0 {
				Number = Number / 2
			} else {
				Number = Number*3 + 1
			}
			counter += 1
		}

		if counter > longest_sequence {
			longest_sequence = counter
			largest = n
		}

	}
	fmt.Println("Answer = ", largest)
	fmt.Println("longest sequence = ", longest_sequence)

}
Advertisements

Project Euler – Problem # 14 – Solved with Java

Problem:

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.


public class Problem_14 {
	public static void main(String args[]){
		long start = System.currentTimeMillis();
		
		int largest = 0;
		int longest_sequence = 0;
		for(int n = 2; n < 1000001; n++){
			long Number = n;
			int counter = 1;
			while(Number > 1){
				if(Number % 2 == 0){
					Number = Number / 2;
				} else{
					Number = (Number * 3) + 1;
				}
				counter += 1;
			}
			
			if(counter > longest_sequence){
				longest_sequence = counter;
				largest = n;
			}
		}
	
		System.out.format("Answer = %d", largest);
		System.out.println();
		System.out.format("longest_sequence = %d ", longest_sequence);
		System.out.println();
		long stop = System.currentTimeMillis();
		System.out.println("Run time = " + (stop - start) + " milliseconds");
	}

}

Project Euler – Problem # 14 – Solved with Python

Problem:

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.

**********

Help from this blog for the idea of a Cache of known numbers.

# Python version = 2.7.2
# Platform = win32

import time

def main():
    """Main Program"""
    start_time = time.clock()
    largest = 0
    longest_sequence = 0
    Cache = dict()
    for n in xrange(2, 1000001):
        Number = n
        counter = 0
        while Number > 1:
            try:
                counter += Cache[Number] - 1
                break
            except KeyError:
                pass
            
            if Number % 2 == 0:
                Number = Number / 2
            else:
                Number = Number * 3 + 1
            counter += 1

        counter += 1
        Cache[n] = counter

        if counter > longest_sequence:
            longest_sequence = counter
            largest = n

    print "Answer = %s" % largest
    print "longest sequence = %s" % longest_sequence
    
    run_time = time.clock() - start_time
    print "Run time = ", run_time
        
if __name__ == '__main__':
    main()