Project Euler #54: Poker hands

This article is part of a series where I'll be diving head first into the Project Euler puzzles. I want to document the challenge of solving such a puzzle and how I got to the answer. I want to prefix this by stating that I can't cheat for any of these challenges; with that I mean I can't look up any other implementations online. After the implementation, I will validate the answer by using this document or a similar sheet.

In this article I'll be solving: Project Euler #54.


I am explained the game of poker, which is a highly needed because I’ve only played a single game in my life. There’s text file containing poker hands. The format is: each line contains 10 cards, where the first 5 cards belong to Player 1 and the other 5 cards belong to Player 2. The puzzle wants to know how many times Player 1 wins the game over Player 2.

Calculating the value of a hand

The first step is calculating the value of “a hand”. According to the game of poker, there are 10 possible outcomes. Let’s start with that, and we’ll deal with the edge cases later. The first step is to take a hand and turn them into individual cards:

fn hand_to_cards(hand: &str) -> Vec<&str> {
    let mut cards = vec![];
    let mut p = 0;
    let max = hand.len();

    while p >= 0 {
        let card = &hand[p..=p + 1];
        p += 3;

        if p > max {

fn value_hand(hand: &str) -> u16 {
    let cards = hand_to_cards(hand);
    println!("{:?}", cards);

fn test_value_hand() {
    assert_eq!(value_hand("5H 5C 6S 7S KD"), 1)

In Rust, we can do something nice with the struct keyword like this:

struct PokerHand<'a>(Vec<&'a str>);

Meaning, I can use PokerHand as the place to store the individual cards and implement methods on the PokerHand struct, to see what kind of value it has. My idea currently is to start from the highest possible value (a royal flush) and go down to straight flush etc.

After an hour of coding, I have all the hands in place. It’s a little too much code to share nicely in a code block, so I hope the link to said commit would be fine.

The value_hand method can now be written like such:

fn value_hand(hand: &str) -> u16 {
    let cards = hand_to_cards(hand);

fn test_value_hand() {
    assert_eq!(value_hand("5H 5C 6S 7S KD"), 1)

The next step is to check how many games Player 1 just wins and print out the ties:

let mut wins = 0;

for game in &games {
    let v1 = value_hand(&game[0..14]);
    let v2 = value_hand(&game[15..]);

    if v1 > v2 {
        wins += 1
    } else if v1 == v2 {
        println!("{:?}", game);

It seems like Player 1 beats Player 2 261 times, but there are a lot of ties. The goal is to have no more ties. By far the most ties are with “high card” hands. However, before solving them one by one, let’s figure out a method to get rid of the ties.

hand current values
high card 0 5 (7), … 12 (A)**
one pair 1 0 (2), 1 (3), … 12 (A)
two pairs 2 Multiply pair values? So the highest is 11 (K) * 12 (A) = 132
three of a kind 3 0 (2), 1 (3), … 12 (A)
straight 4 There are 7 possible straights, meaning 0 till 6
flush 5 0*
full house 6 Multiply three of a kind values, with pair values, the max being 12 (A) * 11 (K) = 132
four of a kind 7 0 (2), 1 (3), … 12 (A)
straight flush 8 There are 7 possible straights, meaning 0 till 6*
royal flush 9 0*

* I don’t think a suite is higher than another suite.

** With 5 cards per hand, the lowest theoretical value is a single 7.

The lowest value hand would be 2S 3D 4C 5S 7D. This hand would be valued as a “high card” 7, I imagine. Turning the 7 value card into a 6 makes it a straight. Turning the 7 into any of the other values (2, 3, 4, 5) makes “one pair”.

The puzzle explains a lovely edge case, though:

Player 1 Player 2
4D 6S 9H QH QC 3D 6D 7H QD QS
Pair of Queens Pair of Queens
Highest card Nine Highest card Seven

In this case, the highest cards are compared. If those also tie, the next highest cards are compared, etc. I need an iterator on PokerHand, giving the highest card to the lowest card by calling .next(). Calling it once should give me the highest value, calling it another time should give me the next highest value, etc.

The first step here is sorting the cards before they go into the PokerHand struct. While implementing the sorting, I spotted a pretty gnarly bug in my is_straight() method, which I had to fix first. After some more fiddling and after reading some very extensive blog posts on how to implement an iterator properly, I got an iterator:

struct PokerHandIter<'a> {
    inner: &'a PokerHand<'a>,
    index: usize

impl Iterator for PokerHandIter<'_> {
    type Item = usize;

    fn next(&mut self) -> Option<usize> {
        if self.index >= self.inner.0.len() {
        } else {
            let card = self.inner.0[self.index];
            let value = CARDS
                .position(|&r| r == card.chars().nth(0).unwrap())

            self.index += 1;
            Some(CARDS.len() - value)

After writing the correct loop for it, I get an answer to problem 54. According to my current code, 491 games are awarded to Player 1. Upon checking the answer, which is 376, it seems that I’m awarding a little too many games to Player 1. Looking at my code, I do spot that I probably need to break when Player 2 holds a higher card than Player 1, which drops the amount of wins to 359. This means that there are probably cases with one pair and a single high card, where the single high card is higher than the value of the one pair.

One hour further

I managed to resolve it. Upon investigating all the hands, it seemed like the only duplicate ranks were for “one pair” and “high hand”. Not once was there a case where Player 1 and Player 2 had a “one pair”, followed by the same high hand.In most cases the resolve was like this:

Voila, solved!


This puzzle is not really that complicated, it’s just a lot of work. After wasting some more hours, I improved quite a bit to the existing code. I figured that not once in the p054_poker.txt a game exists with a similar “double pair” and a similar high hand; so there’s always an obvious winner. This means the whole iterator idea can be tossed out of the window. Because of this, I calculate the rank() as a vector. The first value contains the old rank; meaning a value from 0-9, where 0 is “high hand” and 9 means “royal flush”. The next value is only useful for high hand and any of the pairs to indicate the height of said pair. The full version can be seen in the GitHub link below.

The full solution is available on GitHub.