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Generating all 5 card poker hands

由 余生颓废 提交于 2019-12-03 18:24:19

问题


This problem sounds simple at first glance, but turns out to be a lot more complicated than it seems. It's got me stumped for the moment.

There are 52c5 = 2,598,960 ways to choose 5 cards from a 52 card deck. However, since suits are interchangeable in poker, many of these are equivalent - the hand 2H 2C 3H 3S 4D is equivalent to 2D 2S 3D 3C 4H - simply swap the suits around. According to wikipedia, there are 134,459 distinct 5 card hands once you account for possible suit recolorings.

The question is, how do we efficiently generate all these possible hands? I don't want to generate all hands, then eliminate duplicates, as I want to apply the problem to larger numbers of cards, and the number of hands to evaluate fast spirals out of control. My current attempts have centered around either generating depth-first, and keeping track of the currently generated cards to determine what suits and ranks are valid for the next card, or breadth-first, generating all possible next cards, then removing duplicates by converting each hand to a 'canonical' version by recoloring. Here's my attempt at a breadth-first solution, in Python:

# A card is represented by an integer. The low 2 bits represent the suit, while
# the remainder represent the rank.
suits = 'CDHS'
ranks = '23456789TJQKA'

def make_canonical(hand):
  suit_map = [None] * 4
  next_suit = 0
  for i in range(len(hand)):
    suit = hand[i] & 3
    if suit_map[suit] is None:
      suit_map[suit] = next_suit
      next_suit += 1
    hand[i] = hand[i] & ~3 | suit_map[suit]
  return hand

def expand_hand(hand, min_card):
  used_map = 0
  for card in hand:
    used_map |= 1 << card

  hands = set()
  for card in range(min_card, 52):
    if (1 << card) & used_map:
      continue
    new_hand = list(hand)
    new_hand.append(card)
    make_canonical(new_hand)
    hands.add(tuple(new_hand))
  return hands

def expand_hands(hands, num_cards):
  for i in range(num_cards):
    new_hands = set()
    for j, hand in enumerate(hands):
      min_card = hand[-1] + 1 if i > 0 else 0
      new_hands.update(expand_hand(hand, min_card))
    hands = new_hands
  return hands

Unfortunately, this generates too many hands:

>>> len(expand_hands(set([()]), 5))
160537

Can anyone suggest a better way to generate just the distinct hands, or point out where I've gone wrong in my attempt?


回答1:


Your overall approach is sound. I'm pretty sure the problem lies with your make_canonical function. You can try printing out the hands with num_cards set to 3 or 4 and look for equivalencies that you've missed.

I found one, but there may be more:

# The inputs are equivalent and should return the same value
print make_canonical([8, 12 | 1]) # returns [8, 13]
print make_canonical([12, 8 | 1]) # returns [12, 9]

For reference, below is my solution (developed prior to looking at your solution). I used a depth-first search instead of a breadth-first search. Also, instead of writing a function to transform a hand to canonical form, I wrote a function to check if a hand is canonical. If it's not canonical, I skip it. I defined rank = card % 13 and suit = card / 13. None of those differences are important.

import collections

def canonical(cards):
    """
    Rules for a canonical hand:
    1. The cards are in sorted order

    2. The i-th suit must have at least many cards as all later suits.  If a
       suit isn't present, it counts as having 0 cards.

    3. If two suits have the same number of cards, the ranks in the first suit
       must be lower or equal lexicographically (e.g., [1, 3] <= [2, 4]).

    4. Must be a valid hand (no duplicate cards)
    """

    if sorted(cards) != cards:
        return False
    by_suits = collections.defaultdict(list)
    for suit in range(0, 52, 13):
        by_suits[suit] = [card%13 for card in cards if suit <= card < suit+13]
        if len(set(by_suits[suit])) != len(by_suits[suit]):
            return False
    for suit in range(13, 52, 13):
        suit1 = by_suits[suit-13]
        suit2 = by_suits[suit]
        if not suit2: continue
        if len(suit1) < len(suit2):
            return False
        if len(suit1) == len(suit2) and suit1 > suit2:
            return False
    return True

def deal_cards(permutations, n, cards):
    if len(cards) == n:
        permutations.append(list(cards))
        return
    start = 0
    if cards:
        start = max(cards) + 1
    for card in range(start, 52):
        cards.append(card)
        if canonical(cards):
            deal_cards(permutations, n, cards)
        del cards[-1]

def generate_permutations(n):
    permutations = []
    deal_cards(permutations, n, [])
    return permutations

for cards in generate_permutations(5):
    print cards

It generates the correct number of permutations:

Cashew:~/$ python2.6 /tmp/cards.py | wc
134459



回答2:


Here's a Python solution that makes use of numpy and generates the canonical deals as well as their multiplicity. I use Python's itertools module to create all 24 possible permutations of 4 suits and then to iterate over all 2,598,960 possible 5-card deals. Each deal is permuted and converted to a canonical id in just 5 lines. It's quite fast as the loop only goes through 10 iterations to cover all deals and is only needed to manage the memory requirements. All the heavy lifting is done efficiently in numpy except for the use of itertools.combinations. It's a shame this is not supportedly directly in numpy.

import numpy as np
import itertools

# all 24 permutations of 4 items
s4 = np.fromiter(itertools.permutations(range(4)), dtype='i,i,i,i').view('i').reshape(-1,4)

c_52_5 = 2598960 # = binomial(52,5) : the number of 5-card deals in ascending card-value order
block_n = c_52_5/10
def all5CardDeals():
    '''iterate over all possible 5-card deals in 10 blocks of 259896 deals each'''
    combos = itertools.combinations(range(52),5)
    for i in range(0, c_52_5, block_n):
        yield np.fromiter(combos, dtype='i,i,i,i,i', count=block_n).view('i').reshape(-1,5)

canon_id = np.empty(c_52_5, dtype='i')
# process all possible deals block-wise.
for i, block in enumerate(all5CardDeals()):
    rank, suit = block/4, block%4     # extract the rank and suit of each card
    d = rank[None,...]*4 + s4[:,suit] # generate all 24 permutations of the suits
    d.sort(2)                         # re-sort into ascending card-value order
    # convert each deal into a unique integer id
    deal_id = d[...,0]+52*(d[...,1]+52*(d[...,2]+52*(d[...,3]+52*d[...,4])))
    # arbitrarily select the smallest such id as the canonical one 
    canon_id[i*block_n:(i+1)*block_n] = deal_id.min(0)
# find the unique canonical deal ids and the index into this list for each enumerated hand
unique_id, indices = np.unique(canon_id, return_inverse=True)
print len(unique_id) # = 134459
multiplicity = np.bincount(indices)
print multiplicity.sum() # = 2598960 = c_52_5



回答3:


Your problem sounded interesting, so i simple tried to implements it by just looping over all possible hands in a sorted way. I've not looked at your code in details, but it seems my implementation is quite different from yours. Guess what count of hands my script found: 160537

  • My hands are always sorted, e.g. 2 3 4 4 D
  • If there are 2 equal cards, the color is also sorted (colors are just called 0,1,2,3)
  • the first card has always color 0, the second color 0 or 1
  • A card can only have the color of an previous card or the next bigger number, e.g. if card 1+2 have color 0, card three can only have the colors 0 or 1

Are you sure, the number on wikipedia is correct?

count = 0
for a1 in range(13):
    c1 = 0
    for a2 in range(a1, 13):
        for c2 in range(2):
            if a1==a2 and c1==c2:
                continue
            nc3 = 2 if c1==c2 else 3
            for a3 in range(a2, 13):
                for c3 in range(nc3):
                    if (a1==a3 and c1>=c3) or (a2==a3 and c2>=c3):
                        continue
                    nc4 = nc3+1 if c3==nc3-1 else nc3
                    for a4 in range(a3, 13):
                        for c4 in range(nc4):
                            if (a1==a4 and c1>=c4) or (a2==a4 and c2>=c4) or (a3==a4 and c3>=c4):
                                continue
                            nc5 = nc4+1 if (c4==nc4-1 and nc4!=4) else nc4
                            for a5 in range(a4, 13):
                                for c5 in range(nc5):
                                    if (a1==a5 and c1>=c5) or (a2>=a5 and c2>=c5) or (a3==a5 and c3>=c5) or (a4==a5 and c4>=c5):
                                        continue
                                    #print([(a1,c1),(a2,c2),(a3,c3),(a4,c4),(a5,c5)])
                                    count += 1
print("result: ",count)



回答4:


I'm not a poker player, so the details of hand precedence are beyond me. But it seems like the problem is that you are traversing the space of "sets of 5 cards" by generating sets from the deck, when you should be traversing the space of "distinct poker hands".

The space of distinct hands will require a new grammar. The important thing is to capture exactly the information that is relevant to hand precedence. For example, there are only 4 hands that are royal flushes, so those hands can be described as the symbol "RF" plus a suit designator, like "RFC" for royal flush in clubs. A 10-high heart flush could be "FLH10" (not sure if there are other precedence characteristics of flushes, but I think that's all you need to know). A hand that is "2C 2S AH 10C 5D" would be a longer expression, something like "PR2 A 10 5" if I undestand your initial problem statement.

Once you have defined the grammar of distinct hands, you can express it as regular expressions and that will tell you how to generate the entire space of distinct hands. Sounds like fun!




回答5:


You could simply give all hands a canonical ordering of values (A to K), then assign abstract suit letters according to their order of first appearance in that order.

Example: JH 4C QD 9C 3D would convert to 3a 4b 9b Jc Qa.

Generation should work best as dynamic programming:

  • start with a set of a single hand that is empty,
  • make a new set:
    • for each hand in the old set, generate each possible hand by adding one of the remaining cards
    • canonicalize all new hands
    • remove duplicates



回答6:


Initial input:

H 0 0 0 0 0 0 0 0 0 0 0 0 0
C 1 0 0 0 0 0 0 0 0 0 0 0 0
D 1 0 0 0 0 0 0 0 0 0 0 0 0
S 1 1 0 0 0 0 0 0 0 0 0 0 0
+ A 2 3 4 5 6 7 8 9 T J Q K

Step 1: for each rank greater than or equal the highest rank used, set all suits in that rank to 0. you can get away with only checking higher cards because lower combinations will be checked by the lower starting points.

H 0 0 0 0 0 0 0 0 0 0 0 0 0
C 1 0 0 0 0 0 0 0 0 0 0 0 0
D 1 0 0 0 0 0 0 0 0 0 0 0 0
S 1 0 0 0 0 0 0 0 0 0 0 0 0
+ A 2 3 4 5 6 7 8 9 T J Q K

Step 2: Collapse to distinct rows

0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0 0 0
A 2 3 4 5 6 7 8 9 T J Q K

Step 3: Climb back up determining first suit that match each distinct row, and choose the suits which match the distinct rows (identified by a *)

H 0 * 0 0 0 0 0 0 0 0 0 0 0
C 1 0 0 0 0 0 0 0 0 0 0 0 0
D 1 * 0 0 0 0 0 0 0 0 0 0 0
S 1 1 0 0 0 0 0 0 0 0 0 0 0
+ A 2 3 4 5 6 7 8 9 T J Q K

Now showing the repeat for rank 3

H 0 0 0 0 0 0 0 0 0 0 0 0 0
C 1 0 0 0 0 0 0 0 0 0 0 0 0
D 1 0 0 0 0 0 0 0 0 0 0 0 0
S 1 1 0 0 0 0 0 0 0 0 0 0 0
+ A 2 3 4 5 6 7 8 9 T J Q K

0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 0 0 0 0 0
A 2 3 4 5 6 7 8 9 T J Q K

H 0 0 * 0 0 0 0 0 0 0 0 0 0
C 1 0 0 0 0 0 0 0 0 0 0 0 0
D 1 0 * 0 0 0 0 0 0 0 0 0 0
S 1 1 * 0 0 0 0 0 0 0 0 0 0
+ A 2 3 4 5 6 7 8 9 T J Q K

Step 4: Once there are 5 cells set to 1, increment the total possible suit abstracted hands count by 1 and recurse up.

The total number of suit abstracted hands possible is 134,459. This is the code I wrote to test it out:

using System;
using System.Collections.Generic;
using System.Linq;

namespace ConsoleApplication20
{
    struct Card
    {
        public int Suit { get; set; }
        public int Rank { get; set; }
    }
    
    class Program
    {
        static int ranks = 13;
        static int suits = 4;
        static int cardsInHand = 5;

        static void Main(string[] args)
        {
            List<Card> cards = new List<Card>();
            //cards.Add(new Card() { Rank = 0, Suit = 0 });
            int numHands = GenerateAllHands(cards);
    
            Console.WriteLine(numHands);
            Console.ReadLine();
        }
  
        static int GenerateAllHands(List<Card> cards)
        {
            if (cards.Count == cardsInHand) return 1;
    
            List<Card> possibleNextCards = GetPossibleNextCards(cards);
    
            int numSubHands = 0;
    
            foreach (Card card in possibleNextCards)
            {
                List<Card> possibleNextHand = cards.ToList(); // copy list
                possibleNextHand.Add(card);
                numSubHands += GenerateAllHands(possibleNextHand);
            }
    
            return numSubHands;
        }
    
        static List<Card> GetPossibleNextCards(List<Card> hand)
        {
            int maxRank = hand.Max(x => x.Rank);
            
            List<Card> result = new List<Card>();
    
            // only use ranks >= max
            for (int rank = maxRank; rank < ranks; rank++)
            {
                List<int> suits = GetPossibleSuitsForRank(hand, rank);
                var possibleNextCards = suits.Select(x => new Card { Rank = rank, Suit = x });
                result.AddRange(possibleNextCards);
            }
    
            return result;
        }
    
        static List<int> GetPossibleSuitsForRank(List<Card> hand, int rank)
        {
            int maxSuit = hand.Max(x => x.Suit);
    
            // select number of ranks of different suits
            int[][] card = GetArray(hand, rank);
    
            for (int i = 0; i < suits; i++)
            {
                card[i][rank] = 0;
            }
    
            int[][] handRep = GetArray(hand, rank);
    
            // get distinct rank sets, then find which ranks they correspond to
            IEnumerable<int[]> distincts = card.Distinct(new IntArrayComparer());
    
            List<int> possibleSuits = new List<int>();
    
            foreach (int[] row in distincts)
            {
                for (int i = 0; i < suits; i++)
                {
                    if (IntArrayComparer.Compare(row, handRep[i]))
                    {
                        possibleSuits.Add(i);
                        break;
                    }
                }
            }
    
            return possibleSuits;
        }
    
        class IntArrayComparer : IEqualityComparer<int[]>
        {
            #region IEqualityComparer<int[]> Members
    
            public static bool Compare(int[] x, int[] y)
            {
                for (int i = 0; i < x.Length; i++)
                {
                    if (x[i] != y[i]) return false;
                }
    
                return true;
            }
    
            public bool Equals(int[] x, int[] y)
            {
                return Compare(x, y);
            }
    
            public int GetHashCode(int[] obj)
            {
                return 0;
            }

            #endregion
        }

        static int[][] GetArray(List<Card> hand, int rank)
        {
            int[][] cards = new int[suits][];
            for (int i = 0; i < suits; i++)
            {
                cards[i] = new int[ranks];
            }

            foreach (Card card in hand)
            {
                cards[card.Suit][card.Rank] = 1;
            }
    
            return cards;
        }
    }
}

Hopefully it is broken up enough to be easily understandable.




回答7:


Here is a simple and straightforward algorithm for reducing hands to a canonical one based on suit permutatoins.

  1. convert hand to four bitsets, one per suit representing cards of the suit
  2. sort the bitsets
  3. convert bitsets back into hand

This is what the algorithm looks like in C++, with some implied Suit and CardSet classes. Note that the return statement converts the hand by concatenating the bitstrings.

CardSet CardSet::canonize () const
{
  int smasks[Suit::NUM_SUIT];
  int i=0;
  for (Suit s=Suit::begin(); s<Suit::end(); ++s)
    smasks[i++] = this->suitMask (s);

  sort (smasks, smasks+Suit::NUM_SUIT);

  return CardSet(
    static_cast<uint64_t>(smasks[3])                        |
    static_cast<uint64_t>(smasks[2]) << Rank::NUM_RANK      |
    static_cast<uint64_t>(smasks[1]) << Rank::NUM_RANK*2    |
    static_cast<uint64_t>(smasks[0]) << Rank::NUM_RANK*3);
}



回答8:


Generating equivalence classes for 5 card hands is not an easy task. When I need this I usually use the http://www.vpgenius.com/ webpage. At http://www.vpgenius.com/video-poker/games/ you can choose which variety of poker game you need, and in the "Programming tab" you have an section on "Unique Suit Patterns". So just copying that and loading into program might be easier than trying to generate your own.




回答9:


Take a look here:

http://specialk-coding.blogspot.com/

http://code.google.com/p/specialkpokereval/

These regard a 5-card hand (and a 7-card hand) as an integer, the sum the individual cards, which is independent of the suit. Exactly what you need.

This is part of a scheme for quickly ranking 7- and 5-card hands, written in Objective-C and Java.




回答10:


If you are just interested in hands that result in different hand rankings, there are actually only 7462 distinct hand classes that have to be considered (see Wikipedia).

By creating a table with an example for each class and their accompanying multiplicity you can check all relevant hands weighted with their probability quite fast. That is, assuming that no cards are known and therefore fixed beforehand already.




回答11:


Chances are you really want to generate the number of distinct hands, in the sense of non-equivalent. In that case, according to the wikipedia article there are 7462 possible hands. Here is a python snippet that will enumerate them all.

The logic is simple: there is one hand for each 5-set of ranks; in addition, if all the ranks are distinct, another, different kind of hand can be formed by making all the suits match.

count = 0 

for i in range(0,13):
    for j in range (i,13):
        for k in range(j,13):
            for l in range(k,13):
                for m in range(l,13):
                    d = len(set([i,j,k,l,m])) # number of distinct ranks
                    if d == 1: continue    # reject nonsensical 5-of-a-kind
                    count += 1
                    # if all the ranks are distinct then 
                    # count another hand with all suits equal
                    if d == 5: count += 1

print count   # 7462


来源：https://stackoverflow.com/questions/3829457/generating-all-5-card-poker-hands

标签
python
algorithm
permutation
combinatorics
poker
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