编写孪生素数函数

Question

我们知道孪生素数是一对素数(x，y)，因此y=x+ 2。我想要创建一个函数Thatit()，它可以列出n下面的所有孪生素数。如果知道函数Eratosthenes()列出了n的素数(即介于2到n之间)，我如何修改我的代码以得到这样的函数？

    Thatit <- function(n){
      twin <- c()
      twin2 <- Eratosthenes(n)
    for(i in twin2){
      if((i+2) in twin2){
        twin <- c(twin, "(", (i, i+2), ")")
      } else next()
    } return(twin)
}

Answer 1

我们可以避免for循环，并有：

TwinPrimesLong = function(n) {

if (n < 5) stop("Input value of n should be at least 5.")

primesLessThanN = Eratosthenes(n)

#Combine differenced sets with original prime set
twinPrimeSet = cbind( (primesLessThanN-2) ,primesLessThanN)

#Keep only those rows where differenced numbers are also primes i.e. belong to prime set
twinPrimeSet = data.frame(matrix(twinPrimeSet[(primesLessThanN-2) %in% primesLessThanN,],ncol=2))

colnames(twinPrimeSet)=c("TwinPrime1","TwinPrime2")

cat("The twin primes less than n = ",n,"in row form are :\n")

cat(paste0("(",twinPrimeSet[,1],",",twinPrimeSet[,2],")",collapse=","),"\n")

cat("\n\nThe twin primes less than n = ",n,"in table form are :\n")

print(twinPrimeSet)

#If you wish to return,uncomment below
# return(twinPrimeSet)

}

更新：

使用diff，它可以进一步简化为

(primesLessThanN[diff(primesLessThanN)==2],primesLessThanN[diff(primesLessThanN)==2]+2)

TwinPrimesShort = function(n) {

if (n < 5) stop("Input value of n should be at least 5.")

primesLessThanN = Eratosthenes(n)


#Use diff function to compute difference with previous value and check if == 2
TwinPrime1 = primesLessThanN[diff(primesLessThanN)==2]
TwinPrime2 = TwinPrime1 + 2

#Keep only those rows where differenced numbers are also primes i.e. belong to prime set 
twinPrimeSet = data.frame(TwinPrime1=TwinPrime1,TwinPrime2=TwinPrime2)

cat("The twin primes less than n = ",n,"in row form are :\n")

cat(paste0("(",twinPrimeSet[,1],",",twinPrimeSet[,2],")",collapse=","),"\n")

cat("\n\nThe twin primes less than n = ",n,"in table form are :\n")

print(twinPrimeSet)

}

输出：

TwinPrimesLong(100)
#The twin primes less than n =  100 in row form are :
#(3,5),(5,7),(11,13),(17,19),(29,31),(41,43),(59,61),(71,73) 
#
#
#The twin primes less than n =  100 in table form are :
#  TwinPrime1 TwinPrime2
#1          3          5
#2          5          7
#3         11         13
#4         17         19
#5         29         31
#6         41         43
#7         59         61
#8         71         73

Answer 2

除对(3,5)外，所有孪生素数对某些k的形式为6k±1，因此，要找到小于n的孪生素数，首先计算筛分素数小于sqrt(n)，初始化长度n/4的位数组，使所有值都为真，然后再筛两次。在第一次筛选中，当k= 0，1，2时，位数组的指数代表6k+1.。。因此，对于每个筛分质数，找出最小的6k+1，这是从这一点开始的筛分素和筛子的倍数。在第二次筛选中，当k= 0，1，2时，位数组的索引为6k+1.。。因此，对于每个筛分质数，找出最小的6k+1，这是从这一点开始的筛分素和筛子的倍数。在这两个筛选中幸存的位数组位置表示孪生素数；您可以从索引中计算出两个孪生素数。这个算法非常快，它只是Eratosthenes的筛子。我在我的博客讨论了该算法。