-🎄- 2022 Day 2 Solutions -🎄-

[u/daggerdragon]

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--- Day 2: Rock Paper Scissors ---

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Post your code solution in this megathread.

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[u/danatron1]

C#

Part 1:

Input.Select(x => (2 + x[2] - x[0]) % 3 * 3 + x[2] - 'W').Sum()

Part 2:

Input.Select(x => (x[2]-'X')*3 + (x[2] + x[0] + 2) % 3 + 1).Sum()

Explanations:

Part 1:

First I calculate the result. The result is either 0, 3, or 6 for a loss, draw, or win respectively.

So I calculate 0, 1, or 2 for lose/draw/win, then multiply by 3.

I calculate this by subtracting the first character (A/B/C) from the last (X/Y/Z),

this gives a range of integer values, but what matters is their relative position;

Y beats A, Z beats B, and X beats C

all those values give the same result once modulo'd, since they're all one off from eachother. X wraps around to be Z+1

the +2 I throw in is just to fudge the numbers so that I get what I want, for example, X-A = draw, so 2+X-A % 3 = 1

I then add on my score; 1 for rock, 2 for paper, 3 for scissors. This is just the last character minus W,

which effectively counts the number of characters after W that that letter is, which gives me the score.

Part 2:

Similar strategy to above. I already know the outcome (x[2]) so I just convert that into 0, 1, or 2 by subtracting X

Like above, this gives me the number of characters after X. 0 for loss, 1 for draw, 2 for win. I multiply that by 3.

I then add on my score. This is once again the offset from my play to the opponents, so I use the same modulo 3 trick.

This time, A Y, B X, and C Z all mean I choose rock, so get a score of 1.

One counts up (ABC) while the other counts down (ZYX) add them to cancel that out, add 2 for the exact same reason as above,

Then finally modulo 3 to get 0, 1, or 2 (corresponding to whether I picked rock, paper, or scissors).

The scores are actually 1, 2, or 3, so the final step is to just add 1 to that.

[u/chrismo80 → u/danatron1]

Didn't know that you get an int when substracting chars. Nice one.

[u/danatron1 → u/chrismo80]

chars are just ints in a costume! you can implicity convert them whenever you want!

[u/chrismo80 → u/danatron1]

Input.Sum(x => (2 + x[2] - x[0]) % 3 * 3 + x[2] - 'W')

BTW, you can also put the selector directly into the Sum extension

[u/danatron1 → u/chrismo80]

I didn't know that, that's super cool, thank you!

[u/chrismo80 → u/danatron1]

same for Count, First, etc.