# LMD Async Tutoring

**URL:** https://discuss.criticalfallibilism.com/t/lmd-async-tutoring/1933
**Category:** Tutoring
**Created:** [April 23, 2024, 12:05am UTC](https://discuss.criticalfallibilism.com/t/lmd-async-tutoring/1933 "2024-04-23T00:05:39Z")
**Posts on this page:** 1
**Showing post:** 439

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### Author: ![LMD](https://discuss.criticalfallibilism.com/user_avatar/discuss.criticalfallibilism.com/lmd/32/3213_2.png) [@LMD](https://discuss.criticalfallibilism.com/u/LMD)
#### Post date: [July 24, 2024, 6:30am UTC](https://discuss.criticalfallibilism.com/t/lmd-async-tutoring/1933/439 "2024-07-24T06:30:15Z")

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> [@Eternity Async Tutoring](https://discuss.criticalfallibilism.com/t/eternity-async-tutoring/1932/437):
>
> You can find another one if you want. You could also do xor(x,y).

I got xor above: (x + y) mod 2.

Okay. Another remainder one. I can try.

So 1 mod x checks if x is greater than 1. In this configuration it functions as a ‘greater than’ comparator that outputs 0 or 1.

In my solution for xor, x mod 2 functions as an even number checker, checking if x is even.

or(x, y): (x + y + xy) mod 2

1, 1: (1 + 1 + (1 \* 1)) mod 2 = 1  
1, 0: (1 + 0 + (1 \* 0)) mod 2 = 1  
0, 1: (0 + 1 + (0 \* 1)) mod 2 = 1  
0, 0: (0 + 0 + (0 \* 0)) mod 2 = 0

This is like the remainder xor expression, but it makes the case where both inputs are 1 sum to an odd number by adding xy (which equals 1 only when the inputs are both 1)

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