確定德州撲克手贏家的算法

Question

好吧，我正在爲我的高級項目製作德州撲克AI。我已經創建了gui和投注/交易程序，但是我已經達到了需要確定誰贏了牌局的部分，但我不知道處理此問題的最佳方法。我正在使用python btw。 ATM我有2個列表，一個用於7個播放卡，另一個用於7個電腦卡。目前，所有卡都以結構形式存儲在列表中，其格式爲{'Number'：XX，'Suit'：x}，其中數字爲2-14，套數爲1-4。我要這樣做的方式是爲每種手型制定一個功能，從最高處開始。例如。 self.CheckRoyal（playerCards），並手動瀏覽列表並評估是否實現了皇家沖洗。必須有更好的，數字化的方式來做到這一點。

Answer 1

http://www.codingthewheel.com/archives/poker-hand-evaluator-roundup

你會得到最好的算法是7個在外觀尺寸100 MB查找表（如果我沒記錯）

Answer 2

Monte Carlo？這是我看到的第一個建議here。這是另一個高級項目。簡單而緩慢，但否則你可能會看到一些複雜的組合，我不會假裝知道很多。

Answer 3

ralu的帖子中使用的方法是迄今爲止我見過的最好的選擇。我在自己的項目中使用了這種方法，而且速度非常快。

懸崖：

做一些預處理，以產生一個表，包含用於每個不同的撲克手一個值。確保桌子是按手工排序的。

每個卡片值都有相應的素數值。該表格由手中每張牌值的乘積索引。因此，要找到手AAAAK的價值，你計算黃金乘法，並以此作爲索引表：

int prime = getPrime(hand); // Calculates A.getPrime()...*K.getPrime(); 
int value = table[prime]; 

（很抱歉的Java語法）。

這樣，AAAAK與KAAAA是同一隻手，並且您不需要5個暗表。

請注意，您需要通過所有最好的5張牌組合，以及可以選擇的7張牌來找到最大的值，這是手牌的真實價值。

您使用不同的表格進行沖洗。

表格變得很健壯，因爲這個實現有很多浪費的單元格。爲了解決這個問題，您可以在預處理過程中創建一個映射，它將較大的素數值映射爲整數值，並將其用作源代碼。

Answer 4

一個現成的德州撲克7和5卡評估器的例子可以找到here和進一步解釋here。這可能會幫助你提高性能。所有的反饋意見歡迎在其中找到的電子郵件地址。

Answer 5

import itertools 
from collections import Counter 

# gets the most common element from a list 
def Most_Common(lst): 
    data = Counter(lst) 
    return data.most_common(1)[0] 



# gets card value from a hand. converts A to 14, is_seq function will convert the 14 to a 1 when necessary to evaluate A 2 3 4 5 straights 
def convert_tonums(h, nums = {'T':10, 'J':11, 'Q':12, 'K':13, "A": 14}): 
    for x in xrange(len(h)): 

     if (h[x][0]) in nums.keys(): 

      h[x] = str(nums[h[x][0]]) + h[x][1] 

    return h 


# is royal flush 
# if a hand is a straight and a flush and the lowest value is a 10 then it is a royal flush 
def is_royal(h): 
    nh = convert_tonums(h) 
    if is_seq(h): 
     if is_flush(h): 
      nn = [int(x[:-1]) for x in nh] 
      if min(nn) == 10: 
       return True 

    else: 
     return False 


# converts hand to number valeus and then evaluates if they are sequential AKA a straight 
def is_seq(h): 
    ace = False 
    r = h[:] 

    h = [x[:-1] for x in convert_tonums(h)] 


    h = [int(x) for x in h] 
    h = list(sorted(h)) 
    ref = True 
    for x in xrange(0,len(h)-1): 
     if not h[x]+1 == h[x+1]: 
      ref = False 
      break 

    if ref: 
     return True, r 

    aces = [i for i in h if str(i) == "14"] 
    if len(aces) == 1: 
     for x in xrange(len(h)): 
      if str(h[x]) == "14": 
       h[x] = 1 

    h = list(sorted(h)) 
    for x in xrange(0,len(h)-1): 
     if not h[x]+1 == h[x+1]: 

      return False 
    return True, r 

# call set() on the suite values of the hand and if it is 1 then they are all the same suit 
def is_flush(h): 
    suits = [x[-1] for x in h] 
    if len(set(suits)) == 1: 
     return True, h 
    else: 
     return False 


# if the most common element occurs 4 times then it is a four of a kind 
def is_fourofakind(h): 
    h = [a[:-1] for a in h] 
    i = Most_Common(h) 
    if i[1] == 4: 
     return True, i[0] 
    else: 
     return False 


# if the most common element occurs 3 times then it is a three of a kind 
def is_threeofakind(h): 
    h = [a[:-1] for a in h] 
    i = Most_Common(h) 
    if i[1] == 3: 
     return True, i[0] 
    else: 
     return False 


# if the first 2 most common elements have counts of 3 and 2, then it is a full house 
def is_fullhouse(h): 
    h = [a[:-1] for a in h] 
    data = Counter(h) 
    a, b = data.most_common(1)[0], data.most_common(2)[-1] 
    if str(a[1]) == '3' and str(b[1]) == '2': 
     return True, (a, b) 
    return False 

# if the first 2 most common elements have counts of 2 and 2 then it is a two pair 
def is_twopair(h): 
    h = [a[:-1] for a in h] 
    data = Counter(h) 
    a, b = data.most_common(1)[0], data.most_common(2)[-1] 
    if str(a[1]) == '2' and str(b[1]) == '2': 
     return True, (a[0], b[0]) 
    return False 


#if the first most common element is 2 then it is a pair 
# DISCLAIMER: this will return true if the hand is a two pair, but this should not be a conflict because is_twopair is always evaluated and returned first 
def is_pair(h): 
    h = [a[:-1] for a in h] 
    data = Counter(h) 
    a = data.most_common(1)[0] 

    if str(a[1]) == '2': 
     return True, (a[0]) 
    else: 
     return False 

#get the high card 
def get_high(h): 
    return list(sorted([int(x[:-1]) for x in convert_tonums(h)], reverse =True))[0] 

# FOR HIGH CARD or ties, this function compares two hands by ordering the hands from highest to lowest and comparing each card and returning when one is higher then the other 
def compare(xs, ys): 
    xs, ys = list(sorted(xs, reverse =True)), list(sorted(ys, reverse = True)) 

    for i, c in enumerate(xs): 
    if ys[i] > c: 
     return 'RIGHT' 
    elif ys[i] < c: 
     return 'LEFT' 

    return "TIE" 


# categorized a hand based on previous functions 
def evaluate_hand(h): 

    if is_royal(h): 
     return "ROYAL FLUSH", h, 10 
    elif is_seq(h) and is_flush(h) : 
     return "STRAIGHT FLUSH", h, 9 
    elif is_fourofakind(h): 
     _, fourofakind = is_fourofakind(h) 
     return "FOUR OF A KIND", fourofakind, 8 
    elif is_fullhouse(h): 
     return "FULL HOUSE", h, 7 
    elif is_flush(h): 
     _, flush = is_flush(h) 
     return "FLUSH", h, 6 
    elif is_seq(h): 
     _, seq = is_seq(h) 
     return "STRAIGHT", h, 5 
    elif is_threeofakind(h): 
     _, threeofakind = is_threeofakind(h) 
     return "THREE OF A KIND", threeofakind, 4 
    elif is_twopair(h): 
     _, two_pair = is_twopair(h) 
     return "TWO PAIR", two_pair, 3 
    elif is_pair(h): 
     _, pair = is_pair(h) 
     return "PAIR", pair, 2 
    else: 
     return "HIGH CARD", h, 1 



#this monster function evaluates two hands and also deals with ties and edge cases 
# this probably should be broken up into separate functions but aint no body got time for that 
def compare_hands(h1,h2): 
    one, two = evaluate_hand(h1), evaluate_hand(h2) 
    if one[0] == two[0]: 

     if one[0] =="STRAIGHT FLUSH": 

      sett1, sett2 = convert_tonums(h1), convert_tonums(h2) 
      sett1, sett2 = [int(x[:-1]) for x in sett1], [int(x[:-1]) for x in sett2] 
      com = compare(sett1, sett2) 

      if com == "TIE": 
       return "none", one[1], two[1] 
      elif com == "RIGHT": 
       return "right", two[0], two[1] 
      else: 
       return "left", one[0], one[1] 

     elif one[0] == "TWO PAIR": 

      leftover1, leftover2 = is_twopair(h1), is_twopair(h2) 
      twm1, twm2 = max([int(x) for x in list(leftover1[1])]), max([int(x) for x in list(leftover2[1])]) 
      if twm1 > twm2: 
       return "left", one[0], one[1] 
      elif twm1 < twm2: 
       return "right", two[0], two[1] 


      if compare(list(leftover1[1]), list(leftover2[1])) == "TIE": 
       l1 = [x[:-1] for x in h1 if x[:-1] not in leftover1[1]] 
       l2 = [x[:-1] for x in h2 if x[:-1] not in leftover2[1]] 
       if int(l1[0]) == int(l2[0]): 
        return "none", one[1], two[1] 
       elif int(l1[0]) > int(l2[0]): 
        return "left", one[0], one[1] 
       else: 
        return "right", two[0], two[1] 
      elif compare(list(leftover1[1]), list(leftover2[1])) == "RIGHT": 
       return "right", two[0], two[1] 
      elif compare(list(leftover1[1]), list(leftover2[1])) == "LEFT": 
       return "left", one[0], one[1] 


     elif one[0] == "PAIR": 
      sh1, sh2 = int(is_pair(h1)[1]), int(is_pair(h2)[1]) 
      if sh1 == sh2: 

       c1 = [int(x[:-1]) for x in convert_tonums(h1) if not int(sh1) == int(x[:-1])] 
       c2 = [int(x[:-1]) for x in convert_tonums(h2) if not int(sh1) == int(x[:-1])] 
       if compare(c1, c2) == "TIE": 
        return "none", one[1], two[1] 
       elif compare(c1, c2) == "RIGHT": 
        return "right", two[0], two[1] 
       else: 
        return "left", one[0], one[1] 




      elif h1 > h2: 
       return "right", two[0], two[1] 
      else: 
       return "left", one[0], one[1] 

     elif one[0] == 'FULL HOUSE': 

      fh1, fh2 = int(is_fullhouse(h1)[1][0][0]), int(is_fullhouse(h2)[1][0][0]) 
      if fh1 > fh2: 
       return "left", one[0], one[1] 
      else: 
       return "right", two[0], two[1] 
     elif one[0] == "HIGH CARD": 
      sett1, sett2 = convert_tonums(h1), convert_tonums(h2) 
      sett1, sett2 = [int(x[:-1]) for x in sett1], [int(x[:-1]) for x in sett2] 
      com = compare(sett1, sett2) 
      if com == "TIE": 
       return "none", one[1], two[1] 
      elif com == "RIGHT": 
       return "right", two[0], two[1] 
      else: 
       return "left", one[0], one[1] 



     elif len(one[1]) < 5: 
      if max(one[1]) == max(two[1]): 
       return "none", one[1], two[1] 
      elif max(one[1]) > max(two[1]): 
       return "left", one[0], one[1] 
      else: 
       return "right", two[0], two[1] 
     else: 
      n_one, n_two = convert_tonums(one[1]), convert_tonums(two[1]) 
      n_one, n_two = [int(x[:-1]) for x in n_one], [int(x[:-1]) for x in n_two] 

      if max(n_one) == max(n_two): 
       return "none", one[1], two[1] 
      elif max(n_one) > max(n_two): 
       return "left", one[0], one[1] 
      else: 
       return "right", two[0], two[1] 
    elif one[2] > two[2]: 
     return "left", one[0], one[1] 
    else: 
     return "right", two[0], two[1] 



''' 
a = ['QD', 'KD', '9D', 'JD', 'TD'] 
b = ['JS', '8S', 'KS', 'AS', 'QS'] 
print compare_hands(a,b) 
'''