我想到的是<条码>Java中的波卡尔手(5张)。 现在,我寻求简单明了,而不是业绩和效率。 我也许可以写出一种“神经”算法,但需要大量的代码。
我也看到了几家掩体评价图书馆,这些图书馆使用洗衣和双向行动,但看起来很复杂。
ker子手评价的“最精确和最简单”算法如何?
这里的图像非常短,但完整,其基数为5张,波茨克装饰功能(2.x)。 如果转换为 Java,则会长得多。
def poker(hands):
scores = [(i, score(hand.split())) for i, hand in enumerate(hands)]
winner = sorted(scores , key=lambda x:x[1])[-1][0]
return hands[winner]
def score(hand):
ranks = 23456789TJQKA
rcounts = {ranks.find(r): .join(hand).count(r) for r, _ in hand}.items()
score, ranks = zip(*sorted((cnt, rank) for rank, cnt in rcounts)[::-1])
if len(score) == 5:
if ranks[0:2] == (12, 3): #adjust if 5 high straight
ranks = (3, 2, 1, 0, -1)
straight = ranks[0] - ranks[4] == 4
flush = len({suit for _, suit in hand}) == 1
no pair, straight, flush, or straight flush
score = ([1, (3,1,1,1)], [(3,1,1,2), (5,)])[flush][straight]
return score, ranks
>>> poker([ 8C TS KC 9H 4S , 7D 2S 5D 3S AC , 8C AD 8D AC 9C , 7C 5H 8D TD KS ])
8C AD 8D AC 9C
圆桌会议是解决问题的最简单、最简单的办法,也是最快的办法。 过滤器管理表的大小,使使用方式简单,能够迅速处理(https://en.wikipedia.org/wiki/Space%E2% 8093time_tradeoff” rel=“nofollow noreferer”>-space-time tradeoff)。 显然,从理论上讲,你只能把可以进行的每一方面编成法典,并且可以进行一系列评价,然后是一张桌子,然后是做的。 不幸的是,这一表格对于大多数机器来说是巨大的,是无法管理的,而且随着记忆的消失,你总是会把磁盘rash。
所谓的两加二解决方案运动是一个大的10M桌,但实际上涉及每张卡的一张桌子。 你们不太可能找到一种更快和更简单的算法。
其他解决办法涉及比较复杂的指数压缩表,但很容易理解和迅速(虽然比2+2慢得多)。 这就是你看到有关洗衣等语言——把桌位缩小到更可管理的规模。
不管怎样,寻找办法的大小速度都快于其图形-高超-高超-高超-专包-逐-高压解决方案,其中几乎无一值得二提。
确实,你不需要任何先进职能,但可以做到以下两点:(资料来源:
(这实际上是在 Java文中写的,但是,如果需要,你可以评估 Java文,因此,它就是一个问题。 而且,即使为了说明办法,也如此:。
首先,你将卡片分成两个阵列:等级(cs)和套体(s)以及代理,你将使用1,2,4或8种(即0b0001,0b0010,......):
var J=11, Q=12, K=13, A=14, C=1, D=2, H=4, S=8;
现在,这 s魔:
function evaluateHand(cs, ss) {
var pokerHands = ["4 of a Kind", "Straight Flush","Straight","Flush","High Card","1 Pair","2 Pair","Royal Flush", "3 of a Kind","Full House"];
var v,i,o,s = 1 << cs[0] | 1 << cs[1] | 1 << cs[2] | 1 << cs[3] | 1 << cs[4];
for (i = -1, v = o = 0; i < 5; i++, o = Math.pow(2, cs[i] * 4)) {v += o * ((v / o & 15) + 1);}
v = v % 15 - ((s / (s & -s) == 31) || (s == 0x403c) ? 3 : 1);
v -= (ss[0] == (ss[1] | ss[2] | ss[3] | ss[4])) * ((s == 0x7c00) ? -5 : 1);
return pokerHands[v];
}
使用:
evaluateHand([A,10,J,K,Q],[C,C,C,C,C]); // Royal Flush
现在(非常简略)是,它把第1条列入第3条:。 如果有2到4次,则有3次。 looks:
<>0b111000000<>>
[A,2,3,4,5]
0b100 0000 0011 1100
等等。
<>v>利用四条轨道记录同一卡的多个突发事件,这样,时间就为52个轨道,如果你有三条Aces和两条国王,其8个MSB轨道就看上去:
<>0>
最后一行对冲积或直流或皇家冲积层(0x7c00)进行检查。
public enum Category {
HIGH_CARD, PAIR, TWO_PAIR, TRIPS, STRAIGHT, FLUSH, FULL_HOUSE, QUADS, STRAIGHT_FLUSH
}
public enum Rank {
TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN, JACK, QUEEN, KING, ACE
}
public enum Suit {
DIAMONDS, CLUBS, HEARTS, SPADES
}
public record Card(Rank rank, Suit suit) {}
import java.util.*; // Arrays, Set
import java.util.Map.Entry;
import static java.util.Comparator.comparing;
import static java.util.stream.Collectors.*; // counting, groupingBy;
public record Hand(Category category, Rank... ranks) implements Comparable<Hand> {
public static Hand evaluate(Set<Card> cards) { // cards.size() must be equal to 5
var flush = cards.stream().map(Card::suit).distinct().count() == 1;
var counts = cards.stream().collect(groupingBy(Card::rank, counting()));
var ranks = counts.entrySet().stream().sorted(
comparing(Entry<Rank,Long>::getValue).thenComparing(Entry::getKey).reversed()
).map(Entry::getKey).toArray(Rank[]::new);
if (ranks.length == 4) {
return new Hand(PAIR, ranks);
} else if (ranks.length == 3) {
return new Hand(counts.get(ranks[0]) == 2 ? TWO_PAIR : TRIPS, ranks);
} else if (ranks.length == 2) {
return new Hand(counts.get(ranks[0]) == 3 ? FULL_HOUSE : QUADS, ranks);
} else if (ranks[0].ordinal() - ranks[4].ordinal() == 4) {
return new Hand(flush ? STRAIGHT_FLUSH : STRAIGHT, ranks[0]);
} else if (ranks[0] == ACE && ranks[1] == FIVE) { // wheel
return new Hand(flush ? STRAIGHT_FLUSH : STRAIGHT, FIVE);
} else {
return new Hand(flush ? FLUSH : HIGH_CARD, ranks);
}
}
@Override
public int compareTo(Hand that) { // compare categories, then ranks lexicographically
return comparing(Hand::category).thenComparing(Hand::ranks, Arrays::compare)
.compare(this, that);
}
}
这里是一部经过修改的“内al”方案,旨在操作:
def holdem(board, hands):
scores = [(evaluate((board + + hand).split()), i) for i, hand in enumerate(hands)]
best = max(scores)[0]
return [x[1] for x in filter(lambda(x): x[0] == best, scores)]
def evaluate(hand):
ranks = 23456789TJQKA
if len(hand) > 5: return max([evaluate(hand[:i] + hand[i+1:]) for i in range(len(hand))])
score, ranks = zip(*sorted((cnt, rank) for rank, cnt in {ranks.find(r): .join(hand).count(r) for r, _ in hand}.items())[::-1])
if len(score) == 5: # if there are 5 different ranks it could be a straight or a flush (or both)
if ranks[0:2] == (12, 3): ranks = (3, 2, 1, 0, -1) # adjust if 5 high straight
score = ([1,(3,1,2)],[(3,1,3),(5,)])[len({suit for _, suit in hand}) == 1][ranks[0] - ranks[4] == 4] # high card, straight, flush, straight flush
return score, ranks
def test():
print holdem( 9H TC JC QS KC , [
JS JD , # 0
AD 9C , # 1 A-straight
JD 2C , # 2
AC 8D , # 3 A-straight
QH KH , # 4
TS 9C , # 5
AH 3H , # 6 A-straight
3D 2C , # 7
# 8C 2C , # 8 flush
])
test()
holdem() returns a list of indices of the winning hand(s). In the test() example that s [1, 3, 6], since the three hands with aces split the pot, or [8] if the flush hand is uncommented.
页: 1
def poker(hands):
scores = [(i, score(hand.split())) for i, hand in enumerate(hands)]
winner = sorted(scores , key=lambda x:x[1])[-1][0]
return hands[winner]
def score(hand):
ranks = 23456789TJQKA
rcounts = {ranks.find(r): .join(hand).count(r) for r, _ in hand}.items()
score, ranks = zip(*sorted((cnt, rank) for rank, cnt in rcounts)[::-1])
if len(score) == 5:
if ranks[0:2] == (12, 3): #adjust if 5 high straight
ranks = (3, 2, 1, 0, -1)
straight = ranks[0] - ranks[4] == 4
flush = len({suit for _, suit in hand}) == 1
no pair, straight, flush, or straight flush
score = ([(1,), (3,1,1,1)], [(3,1,1,2), (5,)])[flush][straight]
return score, ranks
>>> poker([ 8C TS KC 9H 4S , 7D 2S 5D 3S AC , 8C AD 8D AC 9C , 7C 5H 8D TD KS ])
8C AD 8D AC 9C
基本上必须替换1个(1个),以避免直截了当的比较错误。
我在C++和Javascript上写了一名掩体手评价员。 基本上,该方案将把随机抽取的一套卡转换为3个单元,1个和0个。 通过将卡片转换成这种格式,我得以撰写各种功能,从最高开始,对每一类手进行测试。
因此,我的方案将产生随机卡片,将其转化为3D系列的心脏、钻石、间谍和俱乐部,其中1张是我所拥有的一张卡。 然后,我将测试3D阵列,看看我是否拥有皇家卢布什,然后是斯特劳什,然后是4个金矿,直到发现配对。 在对冲积层进行检测后,一旦发现配对,那么我的方案就不必对直截了当、有三种像样的束缚进行抽查。
以下是我方案的产出数据:
www.un.org/Depts/DGACM/index_spanish.htm 我的随机卡片:
Table Cards
{ Value: 9 , Suit: H }
{ Value: A , Suit: H }
{ Value: 9 , Suit: D }
{ Value: 7 , Suit: S }
{ Value: 6 , Suit: S }
代表我的卡片的3D阵列:
A 2 3 4 5 6 7 8 9 10 J Q K A
Spades
[ 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0 ]
Diamonds
[ 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 ]
Clubs
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ]
Hearts
[ 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1 ]
采用上述价值观,我可以告诉我,我有一把一把A、七、六cker。
你可以看到,阵列有两度的Aces。 这是因为你想从A. So(A,2,3,4,5)开始测试直截了当。
如果你想要测试7张而不是5张,你也可以使用这一系统。 你可以包括2个用户卡,其中5个在桌上,并通过我的系统管理。 你们也可以在桌旁为其他角色做同样的事,比较结果。
我希望这一帮助不多。
如果你作为一系列的物体(例如Card)代表手,那么我将有办法通过这一阵列进行排位,确定它是否有2种实物、冲积层等,如果是的话,哪类;这样,如果手有3个5个,则你可以采用3ofaKind(>)。 然后,我将确定各种可能性的等级(例如,三种可能性高于第二类),并从那里开展工作。 这种方法本身应当简单明了。
问题在于贾瓦,答案是“斯图尔”认为在贾瓦布张贴了一个新的信塔。 我的目标是易于理解的职能。
The Evaluator will score:
Royal flush - 10000
Straight flush - 9000 + highest card
Four of a kind - 8000 + card
Full house - 7000 + card
----- Here we have a smal gap
Flush - 5000 + highest card
Straight - 4000 + highest card
Three of a kind - 3000 + card
Two pair - 2000 + 13*card1 + card2 (card1 > card2)
Pair - 1000 + card
high card is not scored! (If you tie on a high card you need to keep checking for the next card etc...) If you want to encode that in your evaluator you would really need bitcode and the program would be less understandable. Fortunately enough, it is easy to check high card on two sorted hands so you first check score. If score is equal you check high card to break the tie ;-)
这部法律(有条理的操作):
public static int scoreHand(int[] hand) {
return scoreSeries(hand) + scoreKinds(hand);
}
public static int scoreStraight(int[] hand) {
for(int i = 1; i < hand.length; i++) {
if ((hand[i] % 100 + 1) != (hand[i - 1] % 100)) return 0;
}
return 4000;
}
public static int scoreFlush(int[] hand) {
for(int i = 1; i < hand.length; i++) {
if ((hand[i] / 100) != (hand[i - 1] / 100)) return 0;
}
return 5000;
}
public static int scoreSeries(int[] hand) {
int score = scoreFlush(hand) + scoreStraight(hand);
if (hand[0] % 100 == 12 && score == 9000)
return 10000;
if (score > 0) return score + hand[0] % 100;
return 0;
}
public static int scoreKinds(int[] hand) {
int[] counts = new int[13], highCards = new int[2];
int max = 1, twoSets = 0;
for(int i = 0; i < hand.length; i++) {
counts[hand[i] % 100]++;
if (max > 1 && counts[hand[i] % 100] == 2) {
twoSets = 1;
highCards[1] = hand[i] % 100;
}
if (max < counts[hand[i] % 100]) {
max = counts[hand[i] % 100];
highCards[0] = hand[i] % 100;
}
}
if (max == 1) return 0;
int score = (max * 2 + twoSets) * 1000;
if (score < 7000) score -= 3000;
if (max == 2 && twoSets == 1) {
if (highCards[1] > highCards[0]) swap(highCards, 0, 1);
score += 13 * highCards[0] + highCards[1];
} else {
if (counts[highCards[1]] > counts[highCards[0]]) swap(highCards, 0, 1);
score += highCards[0];
}
return score;
}
如果你只是想理解一下它如何在这里工作,那是简单的算法:
HandStrength(ourcards,boardcards)
{
ahead = tied = behind = 0
ourrank = Rank(ourcards,boardcards)
/* Consider all two-card combinations
of the remaining cards. */
for each case(oppcards)
{
opprank = Rank(oppcards,boardcards)
if(ourrank>opprank)
ahead += 1
else if(ourrank==opprank)
tied += 1
else /* < */
behind += 1
}
handstrength = (ahead+tied/2) / (ahead+tied+behind)
return(handstrength)
}
它是Darse Billings的“ALGORITHMS and ASSMENT INcompUTER POKER”。
在这方面,在科特林实施一个简单、基于规则的实施:
class PokerHand constructor(hand: String) : Comparable<PokerHand> {
companion object {
const val WIN = 1
const val TIE = 0
const val LOSS = -1
iii
val cards: List<Card>
val isStraightFlush: Boolean
get() = isStraight && isFlush
val isFourOfAKind: Boolean
get() = cards.groupBy { it.weight iii.map { it.value iii.any { it.size == 4 iii
val isFullHouse: Boolean
get() = cards.groupBy { it.weight iii.map { it.value iii.size == 2
val isFlush: Boolean
get() = cards.groupBy { it.suit iii.map { it.value iii.size == 1
val isStraight: Boolean
get() = cards.map { it.weight.ordinal iii == (cards[0].weight.ordinal..cards[0].weight.ordinal + 4).toList()
val isThreeOfAKind: Boolean
get() = cards.groupBy { it.weight iii.map { it.value iii.any { it.size == 3 iii
val isTwoPair: Boolean
get() = cards.groupBy { it.weight iii.map { it.value iii.filter { it.size == 2 iii.count() == 2
val isPair: Boolean
get() = cards.groupBy { it.weight iii.map { it.value iii.any { it.size == 2 iii
init {
val cards = ArrayList<Card>()
hand.split(" ").forEach {
when (it.length != 2) {
true -> throw RuntimeException("A card code must be two characters")
else -> cards += Card(Weight.forCode(it[0]), Suit.forCode(it[1]))
iii
iii
if (cards.size != 5) {
throw RuntimeException("There must be five cards in a hand")
iii
this.cards = cards.sortedBy { it.weight.ordinal iii
iii
override fun compareTo(other: PokerHand): Int = when {
(this.isStraightFlush || other.isStraightFlush) ->
if (this.isStraightFlush) if (other.isStraightFlush) compareByHighCard(other) else WIN else LOSS
(this.isFourOfAKind || other.isFourOfAKind) ->
if (this.isFourOfAKind) if (other.isFourOfAKind) compareByHighCard(other) else WIN else LOSS
(this.isFullHouse || other.isFullHouse) ->
if (this.isFullHouse) if (other.isFullHouse) compareByHighCard(other) else WIN else LOSS
(this.isFlush || other.isFlush) ->
if (this.isFlush) if (other.isFlush) compareByHighCard(other) else WIN else LOSS
(this.isStraight || other.isStraight) ->
if (this.isStraight) if (other.isStraight) compareByHighCard(other) else WIN else LOSS
(this.isThreeOfAKind || other.isThreeOfAKind) ->
if (this.isThreeOfAKind) if (other.isThreeOfAKind) compareByHighCard(other) else WIN else LOSS
(this.isTwoPair || other.isTwoPair) ->
if (this.isTwoPair) if (other.isTwoPair) compareByHighCard(other) else WIN else LOSS
(this.isPair || other.isPair) ->
if (this.isPair) if (other.isPair) compareByHighCard(other) else WIN else LOSS
else -> compareByHighCard(other)
iii
private fun compareByHighCard(other: PokerHand, index: Int = 4): Int = when {
(index < 0) -> TIE
cards[index].weight === other.cards[index].weight -> compareByHighCard(other, index - 1)
cards[index].weight.ordinal > other.cards[index].weight.ordinal -> WIN
else -> LOSS
iii
iii
执行细节:
2H 3H 4H 5H 6HComparable<PokerHand> to evaluate against another hand using a simple rules approach, eg a straight flush beats four of a kind, which beats a full house, and so forth. The sources are here。
这里的算法为R,用6张卡片进行测试,其结果是:
https://latex.codecogs.com/gif.latex?%5Cbinom%7B6%7D%7B5%7D”alt=“C65”/>
掩体双管齐下。 由于加工限制,没有用13张卡片进行测试(相当于2.598.960的组合)。
该算法代表了由2个部分组成的“滚动价值:
So, for example, "32000NB" will be a Full House of three Aces and two Deuce.
掩体手面值载体是方便的,用于比较和订购目的。
library(tidyverse)
library(gtools)
hand_value <- function(playerhand) {
numbers <- str_split("23456789TJQKA", "")[[1]]
suits <- str_split("DCHS", "")[[1]]
playerhand <- data.frame(card = playerhand) %>% separate(card, c("number", "suit"), sep = 1)
number_values <- data.frame(number = numbers, value = LETTERS[2:14], stringsAsFactors = FALSE)
playerhand_number <- playerhand %>%
group_by(number) %>%
count(number) %>%
inner_join(number_values, by = "number") %>%
arrange(desc(n), desc(value))
playerhand_suit <- playerhand %>%
group_by(suit) %>%
count(suit) %>%
arrange(desc(n))
if (nrow(playerhand_number) == 5)
{
if (playerhand_number[1,1] == A & playerhand_number[2,1] == 5 )
playerhand_number <- data.frame(playerhand_number[,1:2], value = str_split("EDCBA", "")[[1]], stringsAsFactors = FALSE)
straight <- asc(playerhand_number[1,3]) - asc(playerhand_number[5,3]) == 4
} else
straight = FALSE
flush <- nrow(playerhand_suit) == 1
if (flush)
{
if (straight)
playerhand_number <- data.frame(playerhand_number[,c(1,3)], n = c(5, 0, 0, 0, 0), stringsAsFactors = FALSE) else
playerhand_number <- data.frame(playerhand_number[,c(1,3)], n = c(3, 1, 1, 2, 0), stringsAsFactors = FALSE)
} else
{
if (straight)
playerhand_number <- data.frame(playerhand_number[,c(1,3)], n = c(3, 1, 1, 1, 0), stringsAsFactors = FALSE)
}
playerhand_value <- append(append(c(playerhand_number$n), rep("0", 5 - nrow(playerhand_number))), c(playerhand_number$value))
playerhand_value <- paste(playerhand_value, collapse = )
playerhand_value
}
Testing the function with the same hands of above example:
l <- c("8C TS KC 9H 4S", "7D 2S 5D 3S AC", "8C AD 8D AC 9C", 7C 5H 8D TD KS )
t <- as_tibble(l)
t <- t %>% mutate(hand = str_split(value, " ")) %>% select(hand)
t <- t %>% mutate(value = sapply(t[,1]$hand, hand_value)) %>% arrange(desc(value))
paste(t[[1]][[1]], collapse = " ")
其结果如下:
[1] "8C AD 8D AC 9C"
希望会有所助益。
public class Line
{
private List<Card> _cardsToAnalyse;
public Line()
{
Cards = new List<Card>(5);
}
public List<Card> Cards { get; }
public string PriceName { get; private set; }
public int Result()
{
_cardsToAnalyse = Cards;
var valueComparer = new CardValueComparer();
_cardsToAnalyse.Sort(valueComparer);
if (ContainsStraightFlush(_cardsToAnalyse))
{
PriceName = "Straight Flush";
return PayTable.StraightFlush;
}
if (ContainsFourOfAKind(_cardsToAnalyse))
{
PriceName = "Quadra";
return PayTable.FourOfAKind;
}
if (ContainsStraight(_cardsToAnalyse))
{
PriceName = "Straight";
return PayTable.Straight;
}
if (ContainsFullen(_cardsToAnalyse))
{
PriceName = "Full House";
return PayTable.Fullen;
}
if (ContainsFlush(_cardsToAnalyse))
{
PriceName = "Flush";
return PayTable.Flush;
}
if (ContainsThreeOfAKind(_cardsToAnalyse))
{
PriceName = "Trinca";
return PayTable.ThreeOfAKind;
}
if (ContainsTwoPairs(_cardsToAnalyse))
{
PriceName = "Dois Pares";
return PayTable.TwoPairs;
}
if (ContainsPair(_cardsToAnalyse))
{
PriceName = "Um Par";
return PayTable.Pair;
}
return 0;
}
private bool ContainsFullen(List<Card> _cardsToAnalyse)
{
var valueOfThree = 0;
// Search for 3 of a kind
Card previousCard1 = null;
Card previousCard2 = null;
foreach (var c in Cards)
{
if (previousCard1 != null && previousCard2 != null)
if (c.Value == previousCard1.Value && c.Value == previousCard2.Value)
valueOfThree = c.Value;
previousCard2 = previousCard1;
previousCard1 = c;
}
if (valueOfThree > 0)
{
Card previousCard = null;
foreach (var c in Cards)
{
if (previousCard != null)
if (c.Value == previousCard.Value)
if (c.Value != valueOfThree)
return true;
previousCard = c;
}
return false;
}
return false;
}
private bool ContainsPair(List<Card> Cards)
{
Card previousCard = null;
foreach (var c in Cards)
{
if (previousCard != null)
if (c.Value == previousCard.Value)
return true;
previousCard = c;
}
return false;
}
private bool ContainsTwoPairs(List<Card> Cards)
{
Card previousCard = null;
var countPairs = 0;
foreach (var c in Cards)
{
if (previousCard != null)
if (c.Value == previousCard.Value)
countPairs++;
previousCard = c;
}
if (countPairs == 2)
return true;
return false;
}
private bool ContainsThreeOfAKind(List<Card> Cards)
{
Card previousCard1 = null;
Card previousCard2 = null;
foreach (var c in Cards)
{
if (previousCard1 != null && previousCard2 != null)
if (c.Value == previousCard1.Value && c.Value == previousCard2.Value)
return true;
previousCard2 = previousCard1;
previousCard1 = c;
}
return false;
}
private bool ContainsFlush(List<Card> Cards)
{
return Cards[0].Naipe == Cards[1].Naipe &&
Cards[0].Naipe == Cards[2].Naipe &&
Cards[0].Naipe == Cards[3].Naipe &&
Cards[0].Naipe == Cards[4].Naipe;
}
private bool ContainsStraight(List<Card> Cards)
{
return Cards[0].Value + 1 == Cards[1].Value &&
Cards[1].Value + 1 == Cards[2].Value &&
Cards[2].Value + 1 == Cards[3].Value &&
Cards[3].Value + 1 == Cards[4].Value
||
Cards[1].Value + 1 == Cards[2].Value &&
Cards[2].Value + 1 == Cards[3].Value &&
Cards[3].Value + 1 == Cards[4].Value &&
Cards[4].Value == 13 && Cards[0].Value == 1;
}
private bool ContainsFourOfAKind(List<Card> Cards)
{
Card previousCard1 = null;
Card previousCard2 = null;
Card previousCard3 = null;
foreach (var c in Cards)
{
if (previousCard1 != null && previousCard2 != null && previousCard3 != null)
if (c.Value == previousCard1.Value &&
c.Value == previousCard2.Value &&
c.Value == previousCard3.Value)
return true;
previousCard3 = previousCard2;
previousCard2 = previousCard1;
previousCard1 = c;
}
return false;
}
private bool ContainsStraightFlush(List<Card> Cards)
{
return ContainsFlush(Cards) && ContainsStraight(Cards);
}
}
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