# AoC 2023 Day 7: Pokinator

## Source: Day 7: Camel Cards

Full solution for today (spoilers!)

## Part 1

Simulate a limited poker game with no suits and break otherwise tied hands lexicographically (AAAA2 beats AKAAA) because the the hands are both four of a kind, the first cards are both A, but the second A beats the K. It doesn’t matter that the first hand’s off card was a 2

Order all hands then calculate the sum of the ordering of hands (1 for best etc) times the bet for each.

### Types and Parsing

#[derive(Debug, PartialEq, Eq, PartialOrd, Ord, Copy, Clone, Hash)]
pub enum Card {
Two,
Three,
Four,
Five,
Six,
Seven,
Eight,
Nine,
Ten,
Jack,
Queen,
King,
Ace,
}

impl From<char> for Card {
fn from(c: char) -> Self {
match c {
'2' => Card::Two,
'3' => Card::Three,
'4' => Card::Four,
'5' => Card::Five,
'6' => Card::Six,
'7' => Card::Seven,
'8' => Card::Eight,
'9' => Card::Nine,
'T' => Card::Ten,
'J' => Card::Jack,
'Q' => Card::Queen,
'K' => Card::King,
'A' => Card::Ace,
_ => panic!("Invalid card: {}", c),
}
}
}

#[derive(Debug, PartialEq, Eq, Copy, Clone)]
pub struct Hand {
pub cards: [Card; 5],
pub bid: u64,
}

fn hand(s: &str) -> IResult<&str, Hand> {
let (s, cards) = count(anychar, 5)(s)?;
let cards = cards.iter().map(|c| Card::from(*c)).collect::<Vec<_>>();
let (s, _) = space1(s)?;
let (s, bid) = complete::u64(s)?;

Ok((
s,
Hand {
cards: cards.try_into().unwrap(),
bid,
},
))
}

pub fn hands(s: &str) -> IResult<&str, Vec<Hand>> {
let (s, hands) = separated_list1(newline, hand)(s)?;
Ok((s, hands))
}


I probably overdid this. One nice thing that we get for free though is #[derive(PartialOrd, Ord)] on Card. This will later to the lexicographic ordering for free.

### Solving the Problem

Okay, we have almost everything we need, but we still want to be able to order Hand as well as Card. We can’t just do the trivial derive this time, unless we implement a enum HandType with Ord (which we certainly could do). But instead, let’s implement Ord (and PartialOrd) on Hand:

impl Hand {
fn counts(&self) -> Vec<usize> {
let mut counts: HashMap<Card, usize> =
self.cards
.into_iter()
.fold(HashMap::new(), |mut counts, c| {
*counts.entry(c).or_default() += 1;
counts
});

let mut counts = counts.values().cloned().collect::<Vec<_>>();
counts.sort();
counts.reverse();
counts
}
}

impl Ord for Hand {
fn cmp(&self, other: &Self) -> Ordering {
let self_counts = self.counts();
let other_counts = other.counts();

// Counts are sorted in descending order, so we can compare them directly
// IE five of a kind is <5>, four of a is <4, 1>, full house is <3, 2>, three of a is <3, 1, 1> etc
// If two counts are the same, compare the cards lexicographically (using Ord on Card)
if self_counts == other_counts {
self.cards.cmp(&other.cards)
} else {
self_counts.cmp(&other_counts)
}
}

fn max(self, other: Self) -> Self {...}
fn min(self, other: Self) -> Self {...}
fn clamp(self, _min: Self, _max: Self) -> Self {...}
}

impl PartialOrd for Hand {
fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
Some(self.cmp(other))
}
}


Okay, there’s a bit to unfold (heh) there. What I want to do is produce a list of counts of cards, ordered by the number in the set descending. So AKAAA should be <4, 1> etc. To do that, we can use a fold to build a HashMap of Card to count.

Then, get just the counts (we don’t actually care what kind of card it is for this ordering) and collect them into a Vec. This is nice because Vec is already Ord (if it’s elements are) and it’s lexicographic (I seem to be saying that word a lot).

So we just have to compare self_counts and other_counts. If they’re the same, compare self.cards and other.cards.

One bit of weirdness is that we have to implement both Ord and PartialOrd. If you do a custom implementation of one, it (for some reason) can’t derive the other automatically from it? Or at least cargo clippy didn’t want me to let it. I did just leave max, min, and clamp as unimplemented! though. We don’t need them for this.

And … that’s it!


fn main() -> Result<()> {
let stdin = io::stdin();
let (s, mut hands) = parse::hands(&input).unwrap();
assert_eq!(s.trim(), "");

hands.sort();

let result = hands
.iter()
.enumerate()
.map(|(i, h)| (i + 1) * h.bid as usize)
.sum::<usize>();

println!("{result}");
Ok(())
}


Parse the hands, sort them (why we did all the Ord things), and apply the scoring.

## Part 2

Treat J as Jokers rather than as Jacks. For the purposes of hand type (four of a kind etc), treat jokers as whatever card would score the hand highest. For the purposes of tie-breaking, treat it as the lowest value.

Well that’s neat. Especially because they go opposite ways. So to start off, we’ll expand the types and parsing with Jokers:

#[derive(Debug, PartialEq, Eq, PartialOrd, Ord, Copy, Clone, Hash)]
pub enum Card {
Joker,
Two,
// ...
}

impl From<char> for Card {
fn from(c: char) -> Self {
match c {
'*' => Card::Joker,
// ...
}
}
}


By adding Joker first to Card, it will automatically be sorted as the lowest value. By parsing * as a Joker rather than J, this functionality works for both part 1 and part 2 (we will have to rewrite the input though, we’ll come back to that).

And then we have to add a bit of code to counts:

impl Hand {
fn counts(&self) -> Vec<usize> {
let mut counts: HashMap<Card, usize> =
self.cards
.into_iter()
.fold(HashMap::new(), |mut counts, c| {
*counts.entry(c).or_default() += 1;
counts
});

// Special case for part 2, if we have any jokers assign them to the otherwise largest group
// For 5 jokers, treat as 5 Aces (but this won't actually matter)
if counts.contains_key(&Card::Joker) {
let best_type = counts.iter().fold(Card::Ace, |best_type, (&k, &v)| {
// Update if non-joker with more than current
// If there's nothing but jokers, replace with Aces
if k != Card::Joker && v > *(counts.get(&best_type).unwrap_or(&0)) {
k
} else {
best_type
}
});

*counts.entry(best_type).or_default() += counts[&Card::Joker];
counts.remove(&Card::Joker);
}

let mut counts = counts.values().cloned().collect::<Vec<_>>();
counts.sort();
counts.reverse();
counts
}
}


There are a few interesting edge cases to deal with there:

• No Jokers (we just skip the block)
• All Jokers (treat them all as Aces, although as noted this won’t matter since 5 of a kind is 5 of a kind)
• 1-4 jokers: go through all the types and find the one with the most entries; add the Jokers to this and remove them (so we don’t end up with more than 5 cards)

You can go through all the types to determine this for yourself, but this works because of how the Ord on Hand works. The most common count is always the one that will be most improved by adding Jokers to it.

It’s not the prettiest code (.unwrap_or(&0) remains a weird necessity), but it’s functional. Nothing changes in the Ord implementation, and there’s just one extra line in the solution:

let input = input.replace('J', "*");


And that’s it. We have part 2.

## Performance

$just time 7 1 hyperfine 'just run 7 1' Benchmark 1: just run 7 1 Time (mean ± σ): 85.3 ms ± 5.9 ms [User: 33.4 ms, System: 11.2 ms] Range (min … max): 80.8 ms … 105.2 ms 27 runs$ just time 7 2

hyperfine 'just run 7 2'
Benchmark 1: just run 7 2
Time (mean ± σ):      87.6 ms ±   5.2 ms    [User: 34.5 ms, System: 12.1 ms]
Range (min … max):    82.5 ms … 107.4 ms    27 runs


Plenty quick. Nothing much more to do here.