Problem: Find the sum of all the primes below two million.

Commentary: If you’re not careful, this problem can really get you in R. In poking around the internet, there are a few solutions to this problem, but all the ones I’ve tested are slow. Some, even from people who are better at R than me, are VERY slow. That linked one in particular took 6 minutes and 8.728000 seconds to complete. My solutions runs in under 2 seconds.

So we’ll be using that old prime algorithm I’ve already discussed to death. Other than a few tricks in there that I never see get implemented, there’s a crucial trick to this solution. Since we only want the sum, let’s not be slow about finding the primes we want. I only have to find all primes below , where . It takes basically no time to find them. From there, I start evaluating my sum, and like a good little R user, I vectorize by getting the possible list of possible primes above . From there, I check if any of my known primes divide into those possible primes. This is actually much faster than generating primes and storing them.** **

R Code:

ElapsedTime <- system.time({
##########################
n <- 2000000
 
### Function to get primes below specified n; option to get only those below sqrt(n)
PrimesBelow <- function(n, below.sqrt=FALSE){
  if (below.sqrt == TRUE){
    m <- ceiling(sqrt(n))
  } else {
    m <- n
  }
 
  primes <- c(2,3)
  i <- 3
  while (i < m-1){
    flag <- 0
    i <- i+2
    if (i%%6 == 3){
      flag <- 1
    }
    if (flag == 0){
      s <- sqrt(i)+1
      possible.primes <- primes[primes<s]
      for (prime in possible.primes){
        if ((i%%prime == 0)){
          flag <- 1
          break
        }
      }
      if (flag == 0){
        primes <- c(primes, i)
      }
    }
  }
  primes
}
###
 
primes <- PrimesBelow(n, below.sqrt=TRUE)
 
biglist <- ceiling(sqrt(n)):n
biglist <- biglist[-which(biglist%%2 == 0)]
biglist <- biglist[-which(biglist%%6 == 3)]
 
for (prime in primes){
  biglist <- biglist[which(biglist%%prime != 0)]
}
 
answer <- sum(primes)+sum(as.numeric(biglist))
##########################
})[3]
ElapsedMins <- floor(ElapsedTime/60)
ElapsedSecs <- (ElapsedTime-ElapsedMins*60)
cat(sprintf("\nThe answer is:  %f\nTotal elapsed time:  %d minutes and %f seconds\n", answer, ElapsedMins, ElapsedSecs))

Output: The answer is: 142913828922.000000 Total elapsed time: 0 minutes and 1.915000 seconds