Problem 54: Poker Hands

In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:

High Card: Highest value card.
One Pair: Two cards of the same value.
Two Pairs: Two different pairs.
Three of a Kind: Three cards of the same value.
Straight: All cards are consecutive values.
Flush: All cards of the same suit.
Full House: Three of a kind and a pair.
Four of a Kind: Four cards of the same value.
Straight Flush: All cards are consecutive values of same suit.
Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.

The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.

If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives (see example 1 below). But if two ranks tie, for example, both players have a pair of queens, then highest cards in each hand are compared (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.

Consider the following five hands dealt to two players:

Hand	Player 1	Player 2	Winner
1	5H 5C 6S 7S KD Pair of Fives	2C 3S 8S 8D TD Pair of Eights	Player 2
2	5D 8C 9S JS AC Highest card Ace	2C 5C 7D 8S QH Highest card Queen	Player 1
3	2D 9C AS AH AC Three Aces	3D 6D 7D TD QD Flush with Diamonds	Player 2
4	4D 6S 9H QH QC Pair of Queens Highest card Nine	3D 6D 7H QD QS Pair of Queens Highest card Seven	Player 1
5	2H 2D 4C 4D 4S Full House With Three Fours	3C 3D 3S 9S 9D Full House with Three Threes	Player 1

The file, poker.txt, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards. You can assume that all hands are valid (no invalid characters or repeated cards), each player's hand is in no specific order, and in each hand there is a clear winner.

How many hands does Player 1 win?

Solution

I accomplish this by assigning each hand a six-digit base 13 integer representing its value, then seeing which of the hand values in each matchup is greater. Let the value of each card be the number of other card ranks it beats; so a 2 is worth 0, a 3 is worth 1, etc., a King is worth 11, and an Ace is worth 12. Hand values are assigned as follows:

High Card: \((0, c_4, c_3, c_2, c_1, c_0)_{13}\), where \(c_0 \lt c_1 \lt c_2 \lt c_3 \lt c_4\) are the card values.
Pair: \((1, 0, p, c_2, c_1, c_0)_{13}\), where \(p\) is the value of the pair and \(c_0 \lt c_1 \lt c_2\) are the values of the other cards.
Two Pairs: \((2, 0, 0, p_1, p_0, c)_{13}\), where \(p_0 \lt p_1\) are the values of the pairs and \(c\) is the value of the remaining card.
Three of a Kind: \((3, 0, 0, t, c_1, c_0)_{13}\), where \(t\) is the value of the triple and \(c_0 \lt c_1\) are the values of the remaining cards.
Straight: \((4, 0, 0, 0, 0, c_4)_{13}\), where \(c_4\) is the highest card in the straight (we don't need to compare any others, because the fact that the hand is a straight tells us their ranks).
Flush: \((5, c_4, c_3, c_2, c_1, c_0)_{13}\), where \(c_0 \lt c_1 \lt c_2 \lt c_3 \lt c_4\) are the card values.
Full House: \((6, 0, 0, 0, t, p)_{13}\), where \(t\) is the rank of the triple and \(p\) is the rank of the pair.
Four of a Kind: \((7, 0, 0, 0, q, c)_{13}\), where \(q\) is the rank of the quadruple and \(c\) is the rank of the remaining card.
Straight Flush / Royal Flush: \((8, 0, 0, 0, 0, c_4)_{13}\), where \(c_4\) is the highest card in the straight. (A royal flush is just an ace-high straight flush.)

In real poker, a straight can also be formed from a 5, 4, 3, 2, and Ace, with the 5 counting as the highest card; but this problem as written excludes ace-low straights, so my code does too.

Python Code

import urllib.request
def hand_value(hand_str):
	vals = sorted("23456789TJQKA".index(i) for i in hand_str[::3])
	unique_ranks = len(set(vals))
	if unique_ranks == 5:  # straight, flush, straight flush, or high card
		is_straight = vals[0] + 4 == vals[4]
		is_flush = len(set(hand_str[1::3])) == 1
		if is_straight and is_flush:
			return 8 * 13**5 + vals[4]
		elif is_straight:
			return 4 * 13**5 + vals[4]
		elif is_flush:
			return 5 * 13**5 + sum(vals[i] * 13**i for i in range(5))
		else:
			return sum(vals[i] * 13**i for i in range(5))
	elif unique_ranks == 4:  # one pair
		if vals[0] == vals[1]:
			p, c0, c1, c2 = vals[0], vals[2], vals[3], vals[4]
		elif vals[1] == vals[2]:
			p, c0, c1, c2 = vals[1], vals[0], vals[3], vals[4]
		elif vals[2] == vals[3]:
			p, c0, c1, c2 = vals[2], vals[0], vals[1], vals[4]
		else:
			p, c0, c1, c2 = vals[3], vals[0], vals[1], vals[2]
		return 1 * 13**5 + p * 13**3 + c2 * 13**2 + c1 * 13 + c0
	elif unique_ranks == 3:  # two pairs or three of a kind
		if vals[0] == vals[2] or vals[1] == vals[3] or vals[2] == vals[4]:
			if vals[0] == vals[2]:
				t, c0, c1 = vals[0], vals[3], vals[4]
			elif vals[1] == vals[3]:
				t, c0, c1 = vals[1], vals[0], vals[4]
			else:
				t, c0, c1 = vals[2], vals[0], vals[1]
			return 3 * 13**5 + t * 13**2 + c1 * 13 + c0
		else:
			if vals[0] != vals[1]:
				p0, p1, c = vals[1], vals[3], vals[0]
			elif vals[2] != vals[3]:
				p0, p1, c = vals[0], vals[3], vals[2]
			else:
				p0, p1, c = vals[0], vals[2], vals[4]
			return 2 * 13**5 + p1 * 13**2 + p0 * 13 + c
	else:  # four of a kind or full house
		if vals[0] == vals[3] or vals[1] == vals[4]:
			if vals[0] == vals[3]:
				q, c = vals[0], vals[4]
			else:
				q, c = vals[1], vals[0]
			return 7 * 13**5 + q * 13 + c
		else:
			if vals[0] == vals[2]:
				t, p = vals[0], vals[3]
			else:
				t, p = vals[2], vals[0]
			return 6 * 13**5 + t * 13 + p

def p54():
	url = "https://projecteuler.net/resources/documents/0054_poker.txt"
	f = urllib.request.urlopen(url)
	res = 0
	for line_raw in f:
		line = line_raw.decode('utf-8')
		hand1_str = line[0:14]
		hand2_str = line[15:29]
		if hand_value(hand1_str) > hand_value(hand2_str):
			res += 1
	return res

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