Advent Of Code - Day 22

I finished Day 22 at 7pm on December 29th. This wasn't due to 7 days of rigorous coding, failing, rethinking, mind you. I didn't do any work on it Christmas Eve, Christmas Day, the 26th, 27th or 28th. Long stories there. On the 23rd and 24th I took a break from 22 because it was clear I was not going about it the right way, and did days 23 and 24. I don't know how much coding I actually put into it. The leaderboard filled up at about 4am on the 22nd, I wouldn't have been on it if I did all of the coding at one time starting at midnight, I know that much. It might have been 8 hours for me.

There were a few little things that made this puzzle very fun but also a pain! Little subtle word clues that I initially just glossed over, turned out to lead to flaws in the logic that prevented me from solving this puzzle.

The Puzzle

The puzzle returns to day 21, where Little Henry Case is playing the same role playing game, but instead with swords and armor, you're now a magician. You have 5 spells at your disposal, each with varying effects. You are tasked with defeating the boss, who's hit points and damage are the puzzle's input. Mine happened to be 58 HP and 9 DMG. You must find the "path" to defeat the boss using as little mana as possible.

The Spells

These are the spells at your disposal:

  1. Magic Missile - Costs 53. Instant damage of 4.
  2. Drain - Costs 73. Instant damage of 2. Instant heal of 2.
  3. Shield - Costs 113. Initializes an effect that lasts 6 turns and adds 7 defense to you over those 6 turns.
  4. Poison - Costs 173. Lasts 6 turns. At the start of each turn, deals 3 damage
  5. Recharge - Costs 229. Lasts 5 turns. At the start of each turn, adds 101 mana to player.

Rules

The subtle wording comes in the rules and how the spells are interpreted.

  • On each of your turns, you must select one of your spells to cast. If you cannot afford to cast any spell, you lose.
  • Effects apply at the start of both the player's turns and the boss' turns.
  • Effects are created with a timer
  • At the start of each turn, after they apply any effect they have, their timer is decreased by one
  • You cannot cast a spell that would start an effect which is already active
  • However, effects can be started on the same turn they end.
  • Do not include mana recharge effects as "spending" negative mana

The Code!

Breakdown

  • Instant is a property on Effect because Instant effects are processed at the same time as reoccurring effects, and the rule states that you can't start an effect which is already active. However, if you have poison active, which causes Damage, you should be able to cast the Magic Missile effect, which also causes Damage, because it's an instant effect.
  • OneTime is a property on Effect to make Shield work correctly. Shield occurs immediately, and remains, but its effects aren't processed every turn. Just once it's cast, and removed when the timer counts down.
  • I use random generation to get the next spell. It will pick one based on how much mana is left on the player, and if that effect is not already active, and also checking the Instant flag on Effect, because we want to be able to cast Magic Missile even though Poison may be active.
  • I was a little inconsistent in how the functions work. For instance, ApplyEffects applies the Boss and Player hit points and mana directly (use pointers), while Attack just returns the damage to apply. I'm ok with this.
  • The Story struct is fun to read through, when it works, and you're verifying everything, and you say to yourself, "This might actually work!" :)

I had a lot of fun on these puzzles. I think I want to post one for Day 25 too, so check back soon for that one!

Advent of Code - Day 15

I've been going through the Advent of Code, and it's been pretty straight forward day to day. I got to day 15, and it took me a little bit to think of how to get an array of 4 digits that also adds up to 100. After some thought, I then gave into the easy, brute force solution. There was a previous puzzle for "incrementing" a password and I thought I could solve this one the same way.

In other words, I'd start with [0,0,0,0] and always increment the right-most digit (array with index = len(array)-1) and when it hit 100, it would increment the digit at position len(array)-2 and reset the current one back to 0. This would take F@#%inG FOREVER. But it would loop through all possible combinations.  Before calculating the "cookie recipe" based on the array, I would first check if they added up to 100, if not I would just increment until it did add up to 100. One quick brilliant optimization I made was to start with the initial array of [0,0,0,99] :)  (I kid on the brilliance of it... it saved 99 iterations in what could be a hundred million).

So then I thought again about it. There were two other previous puzzles where you could brute force the solution by first getting all possible permutations of the elements, and looping over all permutations, applying the logic, and sorting them to tell you the highest / lowest combined value.  I thought, this seems similar in a way to that, moreso than it did to the password incrementing.

So you'd start with inserting 100 into a blank array, then recursively call a method that would give you permutations of the remaining value, in this case would be 0. So in the end it would return [100,0,0,0]

If I inserted 50 into the array first, it would then give me permutations of the remaining 50.

So, [50,25,25,0], [50,25,0,25], [50,26,0,24] etc, all the way to [50,0,0,50], [50,50,0,0], [50,0,50,0].  Totally exhaustive but also, 100% correct and I never have to check if the sum of the array adds up to 100, because it's guaranteed to! :)

Here's that method in Go

func Perms(total, num int) [][]int {
    ret := [][]int{}
 
    if num == 2 {
        for i := 0; i < total/2 + 1; i++ {
            ret = append(ret, []int{ total-i, i })
            if i != total - i {
                ret = append(ret, []int{ i, total-i })
            }
        }
    } else {
        for i := 0; i <= total; i++ {
            perms := Perms(total-i, num-1)
            for _, p := range perms {
                q := append([]int{ i }, p...)
                ret = append(ret, q)
            }
        }
    }
    return ret
}

perms := Perms(5, 3)
        fmt.Println(perms)

And the output

[[0 5 0] [0 0 5] [0 4 1] [0 1 4] [0 3 2] [0 2 3] [1 4 0] [1 0 4] [1 3 1] [1 1 3]
[1 2 2] [2 3 0] [2 0 3] [2 2 1] [2 1 2] [3 2 0] [3 0 2] [3 1 1] [4 1 0] [4 0 1]
[5 0 0]]

YES!!! I thought about that one for a while. I knew there had to be a way to do it :)

More importantly, for the Perms(100,4) input in the puzzle, it returns only 176,851 possible permutations, and my original brute force method could have looped 100,000,000 (one hundred million) times!!  Quite the improvement.