Unsolved Challange : Collatz Conjecture

Question

Problem Definition:

> Pick a random number.
> If it is even, devide by 2.
> If it is odd, multiply by 3 and add 1.
> Repeat the same procedure with new number.
> You will end up with 1.

It looks pretty simple right...!! But it isn't.

No one is able to prove this problem till date, but statement is valid and verified true upto 10^ 14 range. (as my knowledge)

But You can try here two things

1. You can try to prove this problem.
2. You can show highest number upto which this is valid with your code.

Answer 1

https://code.sololearn.com/caKB2SsH6GeV/?ref=app

Answer 2

My attempt in C++

https://code.sololearn.com/cYF3SNPxVYPT/

Answer 3

With this code within the Time Limit of SoloLearn I can check how many steps is required for each number from 1 to 155:
https://code.sololearn.com/c6pp146olucn/?ref=app

Answer 4

Max Poppe That's the beauty of maths and this problem..😀😀

It is believed in maths that proof by examples is not a proof.

but this phenomenon is true for all possible computation we have still no proof..😀😀😀😊

Answer 5

https://code.sololearn.com/cx3RcCWuiAhx/?ref=app

Answer 6

https://code.sololearn.com/cPfc378E3hEk/?ref=app
👆Simple code...

https://code.sololearn.com/c9uIkB0fFqPl/?ref=app
👆By creating a class and methods...

Please upvote, if you like it!

Answer 7

Max Poppe It's fine bro. We are just having coding fun here.. 😊😊✌✌

Answer 8

https://code.sololearn.com/com3ZCOZGil9/?ref=app
i ran this code on my computer and scanned every number until 1 million, even al 50 digit number, so you can probably run it until infinity (theoretically speaking)

Answer 9

Zoetic_Zeel I know right? even top mathematicians have failed to prove or disprove that it always can get to 0. But i’m not going to try to prove or disprove this, it’s just too hard and I don’t have so much time.