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Robert Mion
Robert Mion

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Rock Paper Scissors

#adventofcode #algorithms #javascript #programming

Advent of Code 2022 Day 2

Part 1

  1. It was bound to happen eventually
  2. Learning my ABCs and 123s
  3. Going from A to Z, so to speak
  4. My algorithm in JavaScript

It was bound to happen eventually

"Want to learn programming?" they say.
"Why not build a simple game?" they say.
"How about Rock Paper Scissors?" they say.

AoC has avoided it seven years. Until now.

How exciting!

Learning my ABCs and 123s

  • A and X and 1 for Rock
  • B and Y and 2 for Paper
  • C and Z and 3 for Scissors

Opponent's point of view:

  • A beats Z
  • C beats Y
  • B beats X

Player's point of view:

  • X beats C
  • Y beats A
  • Z beats B

And the formula for a round's score:

  • Amount above for what player chose, plus...
  • 0 for player loss
  • 3 for tie
  • 6 for player win

Going from A to Z, so to speak

I need to turn this:

A Y B X C Z
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Into this:

8 1 6
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In order to get 15.

Starting with the first round:

A Y
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I could use a few dictionaries to map winners and amounts:

const elf = { A: 'Z', C: 'Y', B: 'X' } const player = { X: 'C', Y: 'A', Z: 'B' } const score = { X: 1, Y: 2, Z: 3 }
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Three conditions:

if (player[col2] == col1) { // player wins  } else if (elf[col1] == col2) { // elf wins } else { // 'twas a draw }
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Would this work on the first round?

A Y player[Y] == 'A' // true player wins: 6 Y = 2 6 + 2 = 8 Success!
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Great! Second round?

B X player[X] == 'C' // false elf[B] == 'X' // true player loses: 0 X = 1 1 + 0 = 1 Success!
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Great! Third round?

C Z player[Z] == 'C' // false elf[C] == 'Z' // false draw: 3 Z = 3 3 + 3 = 6 Success!
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It seems this algorithm and data structure will help me solve Part 1!

My algorithm in JavaScript 

const elf = { A: 'Z', C: 'Y', B: 'X' } const player = { X: 'C', Y: 'A', Z: 'B' } const score = { X: 1, Y: 2, Z: 3 } return input .split('\n') .reduce((total, round) => { let [col1, col2] = round.split(' ') if (player[col2] == col1) { return total += score[col2] + 6 } else if (elf[col1] == col2) { return total += score[col2] } else { return total += score[col2] + 3 } }, 0)
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Part 2

  1. Just what I need
  2. My algorithm in JavaScript

Just what I need

An expectedly clever switch of the rules!

  • X needs me to lose
  • Y needs me to draw
  • Z needs me to win

How will I decide what I play?

I'll reference example round one again:

A Y
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Y needs me to draw (pick the same thing as my opponent).

What draw's to an A? X

How might I represent that - and losing and winning - as a data structure?

{ A: { X: 'Z', Y: 'X', Z: 'Y' }, B: { X: 'X', Y: 'Y', Z: 'Z' }, C: { X: 'Y', Y: 'Z', Z: 'X' }, }
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Here I've mapped the first column to the second column to the player's necessary choice.

Trying round one again:

A Y A: Y: X Y is a draw = 3 X = 1 3 + 1 = 4 Success!
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Trying round two:

B X B: X: X X is a loss = 0 X = 1 0 + 1 = 1 Success!
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C Z C: Z: X Z is a win = 6 X = 1 6 + 1 = 7 Success!
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It seems this algorithm and data structure will help me solve Part 2!

My algorithm in JavaScript 

const move = { A: { X: 'Z', Y: 'X', Z: 'Y' }, B: { X: 'X', Y: 'Y', Z: 'Z' }, C: { X: 'Y', Y: 'Z', Z: 'X' }, } const score = { X: 1, Y: 2, Z: 3 } const outcome = { X: 0, Y: 3, Z: 6 } return input .split('\n') .reduce((total, round) => { let [col1, col2] = round.split(' ') return total += score[move[col1][col2]] + outcome[col2] }, 0)
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I did it!!

  • I solved both parts!
  • I worked my way toward solutions by writing what I observed and thought, correcting myself along the way!
  • I was surprised that my code for Part 2 is shorter than my code for Part 1! I didn't need all the conditions!

This was my funnest time playing Rock Paper Scissors!

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