rps.md

What is there to improve about a game like rock paper scissors?

Rock paper scissors was one of the first commands for Kllua and by far the most complex one for a long time as it was far above my skill level when I started. Though this mainly related to the asyncio things going on. However even with my limited knowledge something bothered me... the way the winner would be determined.

The original way

This is what I did at the start to determine who won:

async def rpsf(choice1, choice2):

    if choice1.lower() == 'rock' and choice2.lower() == 'scissors':
        return 1
    if choice1.lower() == 'rock' and choice2.lower() == 'rock':
        return 2
    if choice1.lower() == 'rock' and choice2.lower() == 'paper':
        return 3
    if choice1.lower() == 'paper' and choice2.lower() == 'rock':
        return 1
    if choice1.lower() == 'paper' and choice2.lower() == 'paper':
        return 2
    if choice1.lower() == 'paper' and choice2.lower() == 'scissors':
        return 3
    if choice1.lower() == 'scissors' and choice2.lower() == 'paper':
        return 1
    if choice1.lower() == 'scissors' and choice2.lower() == 'scissors':
        return 2
    if choice1.lower() == 'scissors' and choice2.lower() == 'rock':
        return 3

Better yet not great

Admitably this was horrible. So once I had gained some more experience I decided to rewrite it. This is what I came up with:

async def rpsf(choice1, choice2):
    if choice1.lower() == choice2.lower():
        return 2
    if choice1.lower() == 'rock' and choice2.lower() == 'scissors':
        return 1
    if choice1.lower() == 'paper' and choice2.lower() == 'rock':
        return 1
    if choice1.lower() == 'scissors' and choice2.lower() == 'paper':
        return 1

    return 3

Now this is as simple and small as you can get the code for this and most people would stop there. However I have never liked the idea of static if checks. I wanted something dynamic, smart and cool. So I turned to many people's nightmare: maths.

The math way

At the time I had the luck of having a very smart maths teacher who had studied maths at Oxford. After failing to come up for a foruma I eventually decided to ask him for help. My query was the following:

The problem is as follows: Assuming I assign each option in rock paper scissors a number, I am trying to find a formula which can determine if, when given two of those numbers, number 1 wins or not.

It took him less than a day to get back to. He suggested the following:

let rock = -1
let paper = 0
let scissors = 1

The outputs would be :

-1 Player 1 wins

0 Draw

1 Player 2 wins

And finally, the formula:

$$\ f(p, q) = sin(\pi/12*(q - p)*((q - p)^2 + 5))) $$ p would be the numerical representation of Player 1's choice, and q would be the numerical representation of Player 2's choice.

Let's take an example: Player 1 chooses rock, so p = -1. Player 2 chooses paper, so q = 0.

$$\ f(-1, 0) = sin(\pi/12*(0 - (-1))*((0 - (-1))^2 + 5))) = sin(\pi/12*(1)*(1 + 5)) = sin({\pi/12}*6) = sin(\pi/2) = 1 $$

So the function returns 1, which means that Player 2 wins which is... correct.

Implementing it

import math

def result(p: int, q: int) -> int:
    return int(math.sin(math.pi/12*(q-p)*((q-p)**2+5)))

That's it. One line. It's not the most readable code, probably also not the most efficient but it's very smart.

I have geniunely no idea how exactly this maths works but I do know that this proves that maths is not only cool but also magic.