Category: Coding

Problem 4 in on Project Euler’s list is the following:

Find the largest palindrome made from the product of two 3-digit numbers.

We can attack this as follows: firstly, we know that for an n-digit number, the minimum n-digit number is \(10^{n-1}\) and the maximum is \(9\sum_{i=0}^{n-1}10^{i}\). So we need the cartesian product of all numbers in this range, which we can then test for the palindrome property, and finally filter for the maximum element.

Testing for the palindrome property requires that we check each digit in the number for equality on either side. We do this first by calculating the number of digits in the product (using \(\lfloor\log_{10}(x)\rfloor\) and then just comparing the digits by using the modulus and shifting. Note a couple of points in the R code below:

• %/% is integer division;
• Note the vectorization at work in statements like (p %/% 10^x): this will divide p by successive powers of 10, where the powers are defined in the x vector;
• Note the use of the logical all() function to test whether all palindromic contenders are equal.

pal < - function(N) {
 palindromes <- c()
 min_no <- 1*10^(N-1)
 max_no <- sum(9*10^(seq(0,N-1)))
 n <- 1
 for (i in seq(max_no,min_no,-1)) {
  for (j in seq(i,min_no,-1)) {
   p <- i * j
   digits <- floor(log10(p))
   x <- seq(0, digits %/% 2)
    if (all( ((p %/% 10^x) %% 10)==((p %/% 10^(digits-x)) %% 10) )) {
     palindromes[n] <- p
     n <- n + 1
    }
  }
 }
 max(palindromes)
}

The third problem in Project Euler is an interesting one:

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

In order to compute prime factors, we can use a sieving method. The simplest and oldest sieving algorithm is the sieve of Erastothenes, which we can implement in R as follows:

sieve <- function(n) {
    if (n <= 1) {
        return(numeric(0))
    }
    if (n == 2) {
        return(c(2))
    }

    i <- 1
    while (length(a) > 0) {
        b[i] <- a[1]
        i <- i + 1
        a <- a[!(a %% a[1] == 0)]
    }
    b
}

So now we need to compute the largest prime factor of 600851475143. An upper bound for the factorisation is \(p \leq \lfloor\sqrt{x}\rfloor\), so we can halt our search at this point (alternatively, we may be able to begin our search here and work downwards). Let’s do it the obvious (and inefficient) way, starting at 1 and working up to \(\lfloor\sqrt{x}\rfloor\), building up a list of prime factors along the way:

pfactor <- function(x) {
    factors <- c()
    j <- 1
    primes <- sieve( floor(sqrt(x)) )
    for (prime in primes) {
        if (x %% prime == 0) {
            factors[j] <- prime
            j <- j + 1
    }
    }
    factors
}

Now to compute the maximum factor, we just take the maximum of this list:

max( pfactor(600851475143) )