是否有比这更简单的方法来计算扑克中的直线?
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25-09-2019 - |
题
我有一种用于计算玩家的手是否在德克萨斯州Hold'em中握住的算法。它可以正常工作,但是我想知道是否有一种更简单的方法可以不涉及数组/字符串转换等。
这是我拥有的简化版本。假设玩家正在处理一只52个元素的卡值阵列:
var rawHand = [1,0,0,0,0,0,0,0,0,0,0,0,0, //clubs
0,0,0,0,0,0,0,0,0,0,0,0,0, //diamonds
0,1,1,0,1,0,0,0,0,0,0,0,0, //hearts
0,0,0,1,0,0,0,0,1,0,0,0,0];//spades
1表示该值插槽中的卡。上面的手有一个2俱乐部,没有钻石,3次心脏,4个心脏和6次心脏,一个5跨度和10跨度。现在,我看着它找到一个笔直。
var suits = []; //array to hold representations of each suit
for (var i=0; i<4; i++) {
var index = i*13;
// commenting this line as I removed the rest of its use to simplifyy example
//var hasAce = (rawHand[i+13]);
//get a "suited" slice of the rawHand, convert it to a string representation
//of a binary number, then parse the result as an integer and assign it to
//an element of the "suits" array
suits[i] = parseInt(rawHand.slice(index,index+13).join(""),2);
}
// OR the suits
var result = suits[0] | suits[1] | suits[2] | suits[3];
// Store the result in a string for later iteration to determine
// whether straight exists and return the top value of that straight
// if it exists; we will need to determine if there is an ace in the hand
// for purposes of reporting a "low ace" straight (i.e., a "wheel"),
// but that is left out in this example
var resultString = result.toString(2);
//Show the result for the purposes of this example
alert("Result: " + resultString);
这里的诀窍是或各种套件,因此只有一个2到ACE表示。我认为必须有一种更简单的方法来做到这一点吗?
解决方案
您的代码所做的几乎所有工作都是类型转换。如果您刚刚以刻度格式存储的手开始(需要> 32位类型),则可以做类似的事情:
var mask = 2^13 - 1; // this will zero out all but the low 13 bits
var suits = (rawHand | rawHand>>13 | rawHand>>26 | rawHand>>39) & mask;
使用一线循环的等效物将是:
var suits = [];
for(var i=0; i < 13; i++) {
suits[i] = rawHand[i] || rawHand[i+13] || rawHand[i+26] || rawHand[i+39];
}
这更短,更容易理解。
与使用位或操作员节省的时间相比,从位表现转换和从比特表示需要更多的代码和CPU时间。
其他提示
好吧,直道必须包括5或10,因此,如果没有一个或另一个,您可以首先扔掉手:
if (rawHand[3] || rawHand[16] || rawHand[29] || rawHand[42] ||
rawHand[8] || rawHand[21] || rawHand[34] || rawHand[47]) {
// do some more checks
} else {
// not a straight
}
您可以将整数值用作卡值的比特菲尔德,ACE获得两个斑点低和高。然后,您将比较末端与十个可能的直线进行比较。
或使用循环并检查五个连续的数字 - 有效地是相同的。
这个问题使我感兴趣。我最终走了。并编写一个网页来计算任何手。它可能不是最有效的,但确实有效。我只用javascript(没有jQuery)做到这一点。这里有一个 演示 http://jsbin.com/izuto4/2/
以下是代码:
<html>
<head>
<script>
// var myrawHand = [1,0,0,0,0,0,0,0,0,0,0,0,0, //clubs
// 0,0,0,0,0,0,0,0,0,0,0,0,0, //diamonds
// 0,1,1,0,1,0,0,0,0,0,0,0,0, //hearts
// 0,0,0,1,0,0,0,0,0,0,0,0,0];//spades
function getCardsInHand(rawHand) {
var cardsInHand = new Array();
var counter = 0;
for (var i = 0; i < rawHand.length; i ++) {
if (rawHand[i]) {
cardsInHand[counter] = i;
counter ++;
}
}
return cardsInHand;
}
function cardsfiltered(rawHand) {
var cards = getCardsInHand(rawHand)
var cardsfiltered = new Array();
for (var j = 0; j < cards.length; j ++){
cardsfiltered[j] = cards[j] - (parseInt(cards[j] / 13) * 13);
}
cardsfiltered.sort();
return {cards : cards, cardsfiltered : cardsfiltered};
}
function whatIsMyHand(rawHand) {
var cardObject = cardsfiltered(rawHand);
if (((cardObject.cards[0] == 0 && cardObject.cards[1] == 9)
|| (cardObject.cards[0] == 13 && cardObject.cards[1] == 22)
|| (cardObject.cards[0] == 26 && cardObject.cards[1] == 35)
|| (cardObject.cards[0] == 39 && cardObject.cards[1] == 48))
&& cardObject.cards[4] == cardObject.cards[3] + 1 &&
cardObject.cards[3] == cardObject.cards[2] + 1 &&
cardObject.cards[2] == cardObject.cards[1] + 1) {
return "Royal Flush";
}
else if (cardObject.cards[4] == cardObject.cards[3] + 1 &&
cardObject.cards[3] == cardObject.cards[2] + 1 &&
cardObject.cards[2] == cardObject.cards[1] + 1 &&
cardObject.cards[1] == cardObject.cards[0] + 1) {
return "Straight Flush";
}
else if ((cardObject.cardsfiltered[1] == cardObject.cardsfiltered[2]
&& cardObject.cardsfiltered[2] == cardObject.cardsfiltered[3])
&& (cardObject.cardsfiltered[0] == cardObject.cardsfiltered[1]
|| cardObject.cardsfiltered[3] == cardObject.cardsfiltered[4])) {
return "Four of a Kind";
}
else if ((cardObject.cardsfiltered[0] == cardObject.cardsfiltered[1]
&& cardObject.cardsfiltered[1] == cardObject.cardsfiltered[2]
&& cardObject.cardsfiltered[3] == cardObject.cardsfiltered[4])
|| (cardObject.cardsfiltered[0] == cardObject.cardsfiltered[1]
&& cardObject.cardsfiltered[2] == cardObject.cardsfiltered[3]
&& cardObject.cardsfiltered[3] == cardObject.cardsfiltered[4])) {
return "Full House";
}
else if (parseInt(cardObject.cards[0] / 13) == parseInt(cardObject.cards[1] / 13)
&& parseInt(cardObject.cards[0] / 13) == parseInt(cardObject.cards[2] / 13)
&& parseInt(cardObject.cards[0] / 13) == parseInt(cardObject.cards[3] / 13)
&& parseInt(cardObject.cards[0] / 13) == parseInt(cardObject.cards[4] / 13)) {
return "Flush";
}
else if ((cardObject.cardsfiltered[4] == cardObject.cardsfiltered[3] + 1
&& cardObject.cardsfiltered[3] == cardObject.cardsfiltered[2] + 1
&& cardObject.cardsfiltered[2] == cardObject.cardsfiltered[1] + 1
&& cardObject.cardsfiltered[1] == cardObject.cardsfiltered[0] + 1)
|| (cardObject.cardsfiltered[0] == 0
&& cardObject.cardsfiltered[1] == 10
&& cardObject.cardsfiltered[2] == 11
&& cardObject.cardsfiltered[3] == 12
&& cardObject.cardsfiltered[4] == 9)) {
return "Straight";
}
else if ((cardObject.cardsfiltered[0] == cardObject.cardsfiltered[1]
&& cardObject.cardsfiltered[1] == cardObject.cardsfiltered[2])
|| (cardObject.cardsfiltered[1] == cardObject.cardsfiltered[2]
&& cardObject.cardsfiltered[2] == cardObject.cardsfiltered[3])
|| (cardObject.cardsfiltered[2] == cardObject.cardsfiltered[3]
&& cardObject.cardsfiltered[3] == cardObject.cardsfiltered[4])) {
return "Three of a Kind";
}
else if ((cardObject.cardsfiltered[0] == cardObject.cardsfiltered[1]
&& (cardObject.cardsfiltered[2] == cardObject.cardsfiltered[3]
|| cardObject.cardsfiltered[3] == cardObject.cardsfiltered[4]))
|| (cardObject.cardsfiltered[1] == cardObject.cardsfiltered[2]
&& cardObject.cardsfiltered[3] == cardObject.cardsfiltered[4])) {
return "Two Pair"
}
else if (cardObject.cardsfiltered[0] == cardObject.cardsfiltered[1]
|| cardObject.cardsfiltered[1] == cardObject.cardsfiltered[2]
|| cardObject.cardsfiltered[2] == cardObject.cardsfiltered[3]
|| cardObject.cardsfiltered[3] == cardObject.cardsfiltered[4]) {
return "Pair";
}
else {
return "High Card";
}
}
var CardCheckCount = 0;
function MaxCardCheck(element) {
if (element.checked) {
if (CardCheckCount < 5) {
CardCheckCount++;
return true;
}
}
else {
CardCheckCount--;
return true;
}
element.checked = !element.checked;
alert("You can only pick 5 cards.");
return false;
}
function calculateHand() {
var checkboxes = document.getElementsByTagName("input");
var myrawHand = new Array();
for (var i = 0, element; element = checkboxes[i]; i++) {
myrawHand[parseInt(element.name)] = element.checked ? element.value : 0;
}
alert(whatIsMyHand(myrawHand));
}
</script>
</head>
<body>
<table>
<thead>
<tr>
<td> A</td>
<td> 2</td>
<td> 3</td>
<td> 4</td>
<td> 5</td>
<td> 6</td>
<td> 7</td>
<td> 8</td>
<td> 9</td>
<td>10</td>
<td> J</td>
<td> Q</td>
<td> K</td>
<td> </td>
</tr>
</thead>
<tbody>
<tr>
<td><input name="0" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="1" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="2" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="3" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="4" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="5" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="6" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="7" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="8" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="9" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="10" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="11" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="12" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td>Clubs</td>
</tr>
<tr>
<td><input name="13" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="14" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="15" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="16" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="17" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="18" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="19" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="20" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="21" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="22" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="23" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="24" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="25" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td>Diamonds</td>
</tr>
<tr>
<td><input name="26" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="27" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="28" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="29" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="30" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="31" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="32" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="33" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="34" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="35" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="36" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="37" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="38" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td>Hearts</td>
</tr>
<tr>
<td><input name="39" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="40" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="41" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="42" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="43" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="44" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="45" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="46" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="47" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="48" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="49" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="50" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td><input name="51" type="checkbox" value="1" onclick="MaxCardCheck(this);"/></td>
<td>Spades</td>
</tr>
</tbody>
</table>
<button onclick="calculateHand()">Calculate Hand</button>
</body>
</html>
不,那很简单。不久前,我研究了扑克手评估,我认为最快的方法使用了像您这样的方法。查看第一个结果 这个网站. 。它使用位操作来计算手。
编辑:首先,我的意思是“ Pokersource poker-eval评估者”。
为什么不按等级对卡进行排序,然后检查每个等级是否比上一个级别多。 “等级”是一个长度为5的数组,具有ace = 1,2 = 2,... j = 11,q = 12,k = 13。我认为这比此处介绍的其他方法更简单。
function isStraight( ranks) {
ranks.sort();
return (ranks[0] + 1 == ranks[1] || (ranks[0] == 1 && ranks[4] == 13)) &&
(ranks[1] + 1 == ranks[2]) &&
(ranks[2] + 1 == ranks[3]) &&
(ranks[3] + 1 == ranks[4]);
}
您可以使用Specialk的7卡和5张评估器发布 这里 并要求它排名。如果返回的等级在5854和5863(包括)或7453和7462(都包含在内)之间,那么您的手分别为5或7张卡,是或包含直的。
请注意,黑桃的王牌用0表示,王牌的王牌是1,到由51代表的两个俱乐部。该算法将“添加”您的卡片,并在小桌子中查找排名9MB RAM。它也会做更多的事情,但是...