Month: July 2008

Problem #8 on the Project Euler problem list requires finding the greatest product of five consecutive digits in a large string of integers.

For this problem, I copied and pasted the large number into a file, and read it in using scan(). As there were line breaks in the copied and pasted numbers, I used paste() with the collapse argument to restore it to a single line. From there, it was a matter of taking 5-character substrings, converting them to numbers, and using prod() to calculate the product of each group of 5 numbers.

gprod <- function() {
 # Read number as string
 snum <- scan(file="number.dat", what=character(0),
  strip.white=TRUE, quiet=TRUE)
 # Concatenate lines into a single line
 snum <- paste(snum, collapse="", sep="")
 mp <- 0
 for (i in 1:(nchar(snum)-4)) {
  s <- substr(snum, i, i+4)
  m <- prod(as.numeric(strsplit(s,"")[[1]]))
  if (m > mp) {
   mp <- m
}
 }
 mp
}

Problem 7 on the Project Euler site is the following challenge:

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10001st prime number?

The routine below returns a list of primes, each successive prime being calculated by testing none of the currently known primes divide the current integer counter value evenly. If not, the current integer value is added to the list of known primes.

nprime <- function(n) {
        if (n == 1) {
                return (2)
        }
        psofar <- c()
        psofar[1] <- 2
        j <- 2
        p <- 1
        i <- 2
        while (p < n)  {
                if ( all((i %% psofar) > 0) ) {
                        psofar[j] <- i
                        j <- j + 1
                        p <- p + 1
                }
                i <- i + 1
        }
        psofar
}

To return the 10001^st prime, we invoke it like so:

>tail(psofar(10001),1)

Problem 6 in the Project Euler list asks us to:

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

The literal translation of this is:

sumdiff <- function(n) {
    sum(c(1:n))^2-sum(c(1:n)^2)
}

However, this is another problem (like Problem #1) that has a closed-form solution, and so can run in constant time.

The closed form expression for the sum of the squares is

\[ sum_{i=1}^{n}i^2 = \frac{1}{6}n(2n+1)(n+1) \]

The closed form expression for the square of the sum is

\[ \left(\sum_{i=1}^{n}i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2 \]

So if we subtract and simplify, we end up with the solution

\[ \frac{1}{12}n(3n^3+2n^2-3n-2) \]

The solution is shown below:

sumdiff2 <- function(n) {
    #(1/4)*(n^2)*(1+n)^2 - (1/6)*n*(1+n)*(1+2*n)
    (1/12)*(n)*(-2-3*n+2*n^2+3*n^3)
}

We can calculate the answer as shown:

> sumdiff2(100)
[1] 25164150


Needless to say, closed form solutions are preferable where possible.