生成所有5张牌扑克牌手。

[英]Generating all 5 card poker hands


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.

乍一看,这个问题听起来很简单,但结果却比看上去要复杂得多。这让我暂时难住了。

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.

52c5 = 2,598,960种方法从52张牌牌中选择5张牌。不过,由于扑克中的套装是可互换的,所以其中的许多都是等价的——手2H 2C 3H 3S 4D相当于2D 2S 3D 3C 4H——简单地交换一下套装。根据维基百科,有134,459个不同的5张牌的手,一旦你解释了可能的衣服重新着色。

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:

问题是,我们如何有效地产生所有这些可能的手?我不想生成所有的手,然后消除重复,因为我想把这个问题应用到更多的卡片上,并且用手的数量来评估快速旋转失去控制。我目前的尝试都集中在生成深度优先级,并跟踪当前生成的卡片,以确定下一张卡片或宽度优先级的适合和级别是有效的,然后生成所有可能的下一种卡片,然后通过重新着色将每只手转换为“规范”版本,从而删除重复。这是我的尝试,在Python中:

# 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:

不幸的是,这产生了太多的手:

>>> 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?

有没有人能提出一种更好的方法来产生不同的手,或者指出我的尝试中哪里出了问题?

12 个解决方案

#1


18  

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.

你的整体方法是正确的。我很确定问题在于你的make_canonical函数。您可以尝试将num_cards设置为3或4,并查找您已经错过的对等项。

I found one, but there may be more:

我找到了一个,但可能还有更多:

# 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.

作为参考,下面是我的解决方案(在查看解决方案之前开发)。我使用了深度优先搜索而不是宽度优先搜索。另外,我编写了一个函数来检查手是否规范,而不是编写一个函数来将手转换成规范形式。如果不是canonical,我就跳过它。我定义了rank = card % 13, suit = card / 13。这些差异都不重要。

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:

它产生正确的排列数:

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

#2


3  

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.

这里有一个Python解决方案,它使用numpy并生成规范的协议以及它们的多样性。我使用Python的itertools模块创建了所有24种可能的4套组合,然后迭代所有25598960个可能的5张牌。每一项交易都是经过调整的,并在5行中转换成一个规范的id。它非常快,因为循环只经过10次迭代来覆盖所有的交易,并且只需要管理内存需求。除了使用itertools.组合之外,所有繁重的工作都在numpy中高效地完成。遗憾的是,这并不是直接在numpy中进行的。

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

#3


2  

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

您的问题听起来很有趣,所以我简单地尝试着用排序的方法来实现它。我没有详细地看你的代码,但是看起来我的实现和你的很不一样。猜猜看,我的脚本发现了什么?

  • My hands are always sorted, e.g. 2 3 4 4 D
  • 我的手总是排好序,例如:2 3 4 4 D。
  • If there are 2 equal cards, the color is also sorted (colors are just called 0,1,2,3)
  • 如果有两张相等的卡片,颜色也会被分类(颜色是0,1,2,3)
  • the first card has always color 0, the second color 0 or 1
  • 第一张牌的颜色总是0,第二个颜色是0或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
  • 一张卡片只能有前一张牌的颜色或下一个较大的数字,例如,如果卡片1+2有颜色0,卡片3只能有颜色0或1。

Are you sure, the number on wikipedia is correct?

你确定,维基百科上的数字是正确的吗?

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)

#4


1  

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".

我不是扑克玩家,所以手的优先权比我更重要。但问题似乎在于,当你要穿越“独特的扑克手”的空间时,你正在从甲板上生成“5张牌”的空间。

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.

不同的手的空间需要一个新的语法。重要的是准确地捕获与手动优先级相关的信息。举个例子,只有4只手是皇家的冲锋,所以这些手可以被描述为“RF”的符号,再加上一套衣服的名称,就像“RFC”(RFC)一样,可以在俱乐部里使用。10个高的心脏冲水可能是“FLH10”(不确定是否有其他优先级特征,但我认为这就是你所需要知道的)。一个“2C 2S AH 10C 5D”的手将会是一个更长的表达,比如“PR2 A 10 5”,如果我坚持你的初始问题的话。

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!

一旦你定义了不同的手的语法,你可以将它表达为正则表达式,它会告诉你如何生成完全不同的手的空间。听起来很有趣!

#5


1  

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.

你可以简单地给所有的人一个标准的值(a到K)的顺序,然后根据他们第一次出现的顺序来分配抽象的适合的字母。

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

示例:JH 4C QD 9C 3D将转换为3a 4b9b Jc Qa。

Generation should work best as dynamic programming:

在动态规划中,生成应该是最好的:

  • start with a set of a single hand that is empty,
  • 从一个空的单手开始,
  • make a new set:
    • for each hand in the old set, generate each possible hand by adding one of the remaining cards
    • 对于旧集合中的每只手,通过添加剩下的一张牌来生成每个可能的手。
    • canonicalize all new hands
    • 规范化所有新的手
    • remove duplicates
    • 删除重复的
  • 制作一个新的集合:在旧的集合中,每只手都可以通过添加一个将所有新手删除重复的牌的剩余的卡片来生成每一个可能的手。

#6


1  

Initial input:

初始输入:

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.

第一步:对于每个级别大于或等于最高等级的人,将所有的套装设置为0。你可以只检查更高的牌,因为较低的组合会被较低的起点检查。

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

步骤2:崩溃到不同的行。

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 *)

第3步:爬回确定匹配每一行的第一套衣服,并选择匹配不同行的套装(由*标识)

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

现在显示3级的重复。

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.

步骤4:一旦有5个单元格设置为1,就可以将所有可能的suit抽象为1并递归。

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

可抽调的衣服总数是134,459。这是我写的测试代码:

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.

希望它被分解的足够容易理解。

#7


1  

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

这是一个简单和直接的算法,以减少手到一个标准的基于适合的permutatoins。

  1. convert hand to four bitsets, one per suit representing cards of the suit
  2. 将手转换为4个位组,每一套代表套装的牌。
  3. sort the bitsets
  4. 对bitsets排序
  5. convert bitsets back into hand
  6. 将bitset转换回手动。

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.

这就是算法在c++中的样子,有一些默认的Suit和CardSet类。注意,return语句通过连接位串来转换手。

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);
}

#8


0  

Look at Pokersource. The problem gets even worse when you're considering completing hands given some cards already drawn.

看看Pokersource。当你考虑完成一些已经画好的牌时,问题就更严重了。

The guy behind PokerStove did a great job in this direction, but the source is disclosed.

poker炉子后面的人在这个方向上做得很好,但是消息来源被披露了。

#9


0  

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.

为5个卡片手生成等价类不是一件容易的事情。当我需要时,我通常使用http://www.vpgenius.com/网页。在http://www.vpgenius.com/videopoker/games/你可以选择你需要的扑克游戏种类,在“编程标签”中你有一节“独特的西装模式”。因此,复制并加载到程序中可能比创建自己的程序更容易。

#10


0  

Take a look here:

看一看:

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

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

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

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.

这些都是一个5张牌的手(和7张牌的手)作为一个整数,和单独的卡片,独立于诉讼。正是你需要的。

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

这是在Objective-C和Java中快速排序7- 5卡手的计划的一部分。

#11


0  

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).

如果你只是感兴趣的手,结果在不同的手排名,实际上只有7462个不同的手类必须被考虑(参见维基百科)。

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.

通过为每一个类创建一个表及其伴随的多样性,您可以快速地检查所有相关的手。也就是说,假设没有已知的卡片,因此事先已经确定了。

#12


0  

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.

很可能你真的想要产生不同的手的数量,在非等价的意义上。在这种情况下,根据维基百科的文章,有7462个可能的手。下面是将枚举它们的python代码片段。

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.

逻辑很简单:每一组5组的队伍都有一只手;此外,如果所有的等级都是不同的,另一种不同的手可以通过使所有的衣服匹配而形成。

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

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