[Akane]

Done!

This doesn't count for anything since I added it after the challenge started though

Code:

i"1>""_2%;_3*1+;2/?p"h

Could you explain how the while loops work?

[Shebang]

Could you explain how the while loops work?

Sure!

So, in this instance, h is the character that triggers the while-loop. What it does is pop off the top three items from the stack:

• A string, which is the body of the while loop.
  
• A string, which is the condition the while loop executes on
  
• A value, which is passed to the condition and body.
  

In this case:

• _2%;_3*1+;2/?p is the body of the loop.
  
• 1> is the condition
  
• i is the value (coming from user input in this case).
  

So, what this code does is:

• Take user input.
  
• While this user input is greater than 1:
  • Check if it's divisible by 2.
    
  • If so, divide it by 2.
    
  • If not, multiply by 3 and add 1.
    
  • Print out the result.
    

That's basically it! It's essentially a direct conversion of the Python code.

[lux]

Sure!

So, in this instance, h is the character that triggers the while-loop. What it does is pop off the top three items from the stack:

• A string, which is the body of the while loop.
  
• A string, which is the condition the while loop executes on
  
• A value, which is passed to the condition and body.
  

In this case:

• _2%;_3*1+;2/?p is the body of the loop.
  
• 1> is the condition
  
• i is the value (coming from user input in this case).
  

So, what this code does is:

• Take user input.
  
• While this user input is greater than 1:
  • Check if it's divisible by 2.
    
  • If so, divide it by 2.
    
  • If not, multiply by 3 and add 1.
    
  • Print out the result.
    

That's basically it! It's essentially a direct conversion of the Python code.

Exact conversation of the JS code, too.

[Equinox]

I know this isn't exactly the smallest, and is quite large compared to some other entries, but I am entering since this is sort of a unique entry. I used the Wolfram language, which breaks off from both the Python and NodeJS entries.

Also 69 byte master race!

Code:

n=10;While[1!=n,Print[n];n=Part[{n/2,(n*3)+1},Mod[n,2]+1]]; Print[n];

[lux]

I know this isn't exactly the smallest, and is quite large compared to some other entries, but I am entering since this is sort of a unique entry. I used the Wolfram language, which breaks off from both the Python and NodeJS entries.

Also 69 byte master race!

Code:

n=10;While[1!=n,Print[n];n=Part[{n/2,(n*3)+1},Mod[n,2]+1]]; Print[n]

[img]-snip-[/img]

from the input

I don't see you inputting anything.

[Equinox]

I don't see you inputting anything.

;w; Completely missed that part

(I don't know how to get input from Wolfram, and honestly don't care to learn, so I guess that means I'm disqualified. meh)

[lux]

;w;

(I don't know how to get input from Wolfram, and honestly don't care to learn, so I guess that means I'm disqualified. meh)

It should still be considered, just a penalty applied. (Such as +10 bytes to score).

Go fast Haskell))) https://www.haskell.org/

[Equinox]

It should still be considered, just a penalty applied. (Such as +10 bytes to score).

Go fast Haskell))) https://www.haskell.org/

I guess technically it would be input, since the Wolfram programming cloud itself is in an interactive-mode type of input, though still not the input wanted in this challenge.

[Eclipse]

I know this isn't exactly the smallest, and is quite large compared to some other entries, but I am entering since this is sort of a unique entry. I used the Wolfram language, which breaks off from both the Python and NodeJS entries.

Also 69 byte master race!

Code:

n=10;While[1!=n,Print[n];n=Part[{n/2,(n*3)+1},Mod[n,2]+1]]; Print[n];

It should still be considered, just a penalty applied. (Such as +10 bytes to score).

Go fast Haskell))) https://www.haskell.org/

I agree with Lux, but your program isn't as small as Shebang's so you wouldn't win anyway, and I'd hate to add points to 69, so I'll accept your solution as-is.

[OversouL]

Well, firstly, that code doesn't work right It's printing 10 5 1 for 20, when it should be printing 10 5 16 8 4 2 1. Once you fix that I'll show you a couple ways of shortening it

Oh, I divided the odd number by 3. XD

Code:

#include <iostream> int main(){int n;std::cin >> n;while(n > 1){n%2==0?n/=2:n=n*3+1;std::cout<<n<<" ";}return 0;}

I wonder why n*=3+1 doesn't work, like n=n*3+1 ??? Anyway, it's 114 bytes now.

Code:

#include <iostream> int main(){int n;std::cin>>n;while(n>1){n%2==0?n/=2:n=n*3+1;std::cout<<n<<" ";}return 0;}

110 bytes after removing some spaces. :3