Collatz (3n+1) sequence in java
problem
You wish to write some code to generate the well known Collatz sequence. More specific, given a number n, calculate and show the numbers from n concluding to 1. Recall the Collatz sequence always diverges to 1.
SOLUTION
![{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}}\\[4px]3n+1&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec22031bdc2a1ab2e4effe47ae75a836e7dea459)
In plain English, if the number is even then divide it by 2 otherwise multiply it by 3 and add 1. Repeat until the result is 1 (which always does).
From the moment we have a repetition in our problem, it means that we need to use a type of loop. In this case a while loop will do the trick. Let’s see the code:
public void generate(int number){
System.out.println(number);
int result = collatz(number);
while(result != 1){
System.out.println(result);
result = collatz(result);
}
System.out.println(result);
}
And the collatz() method is as follows:
public int collatz(int number){
if (number % 2 == 0) {
return number / 2;
}else{
return number * 3 + 1;
}
}
running
Putting it all together with a main() method:
package com.programmerabroad;
public class CollatzExample {
public void generate(int number){
System.out.println(number);
int result = collatz(number);
while(result != 1){
System.out.println(result);
result = collatz(result);
}
System.out.println(result);
}
public int collatz(int number){
if (number % 2 == 0) {
return number/ 2;
}else{
return number * 3 + 1;
}
}
public static void main(String[] args) {
new CollatzExample().generate(10);
}
}
The code above shows the sequence starting from the number 10
output
Correct! The sequence prints 10, 5, 16, 8, 4, 2, 1 and it ends with 1.
conclusion
In this post we saw how to implement the Collatz sequence in Java. In order to do that we used a while loop and a bit of logic.
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Ellion
Professional IT consultant, writer, programmer enthusiast interested in all sorts of coding.
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