扑克: Holdem包straight1功能？

Question

假设x是一只全息牌(棋盘+孔牌)，其中J:A = 11:14，A可能= 1。西装不重要。你只是在检查直线。

x <- c(2,5,6,7,8,9,14)

这是来自holdem包的straight1函数。除了函数末尾的for循环之外，我什么都能理解。有谁能解释一下这部分是怎么工作的吗？我迷路了。

function(x){
a1 = sort(unique(x))
if (length(a1)<4.5) return(0)
a3 = 0
n = length(a1)
if(a1[n] == 14) a1 = c(1,a1) ## count ace as both 1 and 14
a2 = length(a1)
for(j in c(5:a2)){ ## j will be the potential highest card of straight
if( sum(15^c(1:5) * a1[(j-4):j]) == sum(15^c(1:5) * ((a1[j]-4):a1[j]))) a3 = a1[j]
}
a3
}   ## end of straight1

Answer 1

我认为这个问题需要澄清一下：

Qstraight <-function(x){
     a1 = sort(unique(x))
     if (length(a1)<4.5) return(0)
     a3 = 0
     n = length(a1)
     if(a1[n] == 14) a1 = c(1,a1) ## count ace as both 1 and 14
     a2 = length(a1)
     for(j in c(5:a2)){ 
         ## j will be the potential highest card of straight
         if( sum(15^c(1:5) * a1[(j-4):j]) == 
             sum(15^c(1:5) * ((a1[j]-4):a1[j]))) a3 = a1[j]
                        }
    a3
    }   ## end

所以结果是..。

 Qstraight(x)
#[1] 9  # i.e a "nine-high straight
x2 <- c(2,5,6,7,8,10,14)
 Qstraight(x2)
#[1] 0   # i.e not a straight at all.

我可能会对唯一值进行排序，然后取$values == 1的最大长度(diff(唯一(排序(X)。或者不那么依赖于模数运算。