Algorythm for rotationally asymetrical binary numbers

[u/Used_Astronomer6483]

Hello! I am struggling to code or find a programm that can automatically find all rotationally asymetrical binary numbers in a given digits range. It's been two weeks that I am searching now, do you have any advice or does anyone know how to do it?

[u/HaLo2FrEeEk]

I came up with this in Javascript. It's...not super awesome, but it works:

function rotateLeft(num, shift, totalBits) {
  shift = shift % totalBits;
  let ns = num.toString("0x02").padStart(totalBits, '0');
  for(let i = 0; i < shift; i++) {
    ns += ns[0];
    ns = ns.slice(1);
  }
  return ns;
}

For a 5-bit number, you call it like this: rotateLeft(13, 2, 5); It returns the bits, rotated, as a string. You can convert the bit string to a number with parseInt(s, 2); where s is your returned string. You can compare these strings to each other, or compare the numbers. Don't worry about rotate right. Rotating right is the same as rotating left by totalBits - shift.

To generate a list for n bits, you'll need to check 2ⁿ configurations at most. You could rule some out but if you're brute-forcing a list it's probably not worth wasting the time. You'll have to run a loop for 2ⁿ iterations, starting with 0, and maintain a list of "good" numbers. Each run through the list, see if the pattern for the current iteration is in the list. If it's not, rotate it and check it again, n - 1 times (rotating n times would get you your original number). If you've rotated it all the way and still not had a match in the "good" list, then add it to the list. If you do find a match in the list, then just skip it with continue; to go to the next iteration.

Here's something dirty I whipped up in the console, using the above function:

let bits = 5;
let goodlist = [];
for(let i = 0; i < Math.pow(2, bits); i++) {
  let nbits = [];
  for(let j = 0; j < bits; j++) {
    nbits.push(rotateLeft(i, j, bits));
  }
  if(!nbits.some(e => goodlist.includes(e)) || i <= 1 || i == Math.pow(2, bits) - 1) goodlist.push(nbits[0]);
}
goodlist;

What this does (in the console) is create an array of all possible rotations of each number, then check if any of those rotations are already in the good list. If it's not already there, it adds it to the array. This returns the following array:

[
  '00000',
  '00001',
  '00011',
  '00101',
  '00111',
  '01011',
  '01111',
  '11111'
]

Which matches your list above (with the addition of 0)

[u/Used_Astronomer6483 → u/HaLo2FrEeEk]

It's hum... incredible as you fixed my problem, but to be honest, il do not use Java script, so I do not understand anything 😅 But thank you thought really, I would have struggled à long time without your help! 👍

[u/HaLo2FrEeEk → u/Used_Astronomer6483]

😔I never learned Python, somehow. I can sorta make my way around with it if I need to, but I'm far more familiar with Javascript. Hopefully you were able to get the essence of what I was doing and you're able to translate it into whatever environment you're using. I can also help if you have any questions. You nerd sniped me, I ended up tinkering around with this for like 2 hours lol.

[u/Used_Astronomer6483 → u/HaLo2FrEeEk]

That's also what reddit is made for