Euler Problem 13 asks to add one hundred numbers with fifty digits. This seems like a simple problem where it not that most computers are not designed to deal with numbers with a lot of integers. For example:

When asking R to compute this value we get 1.844674e+19, losing most of the digits and limiting the accuracy of the results. Computers solve this problem using Arbitrary-precision Arithmetic. There are many software libraries that can process long integers without loosing accuracy. Euler Problem 13 requires this type of approach.

## Euler Problem 13 Definition

Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.

## Solution

The easy way to solve this problem is to use the gmp package for working with very large integers. This package uses a special number types such as Big Rational and Big Integer. The number of digits in these number types is only limited by the size of the memory.

library(gmp) numbers <- readLines("Euler/p013_numbers.txt") digits <- sum(as.bigz(numbers)) answer <- substr(as.character(digits),1,10)

# Using Base-R

To find the solution to this problem using only base R, I wrote a function to add numbers using strings instead of integers. The function adds leading zeros to the smallest number to make them both the same length. The function then proceeds to add numbers in the same way we were taught in primary school. This function can in principle be used for several other Euler Problems using large integers.

# Add numbers with many digits big.add <- function(a, b) { # Add leading zeros to smallest numer if (nchar(a) < nchar(b)) a <- paste0(paste(rep(0, nchar(b) - nchar(a)), collapse = ""), a) if (nchar(a) > nchar(b)) b <- paste0(paste(rep(0, nchar(a) - nchar(b)), collapse = ""), b) solution <- vector() remainder <- 0 for (i in nchar(b):1) { p <- as.numeric(substr(a, i, i)) q <- as.numeric(substr(b, i, i)) r <- p + q + remainder if (r >= 10 & i!=1) { solution <- c(solution, r %% 10) remainder <- (r - (r %% 10))/10 } else { solution <- c(solution, r) remainder <- 0 } } return(paste(rev(solution), collapse = "")) }

With this function, the problem is easy to solve. The second part of the code runs this function over the one hundred numbers provided on the Euler Problem page and calculates the answer.

numbers <- readLines("Euler/p013_numbers.txt") for (i in numbers) { answer <- big.add(answer, i) } answer <- substr(answer, 1, 10)

## Multiplying Big Numbers

You can expand this function to multiply a very large number with a smaller number using the Reduce function. This function adds the number a to itself, using the *big.add* function. The outcome of the addition is used in the next iteration until it has been repeated *b* times. The number b in this function needs to be a ‘low’ number because it uses a vector of the length b.

big.mult <- function(a, b) { Reduce(big.add, rep(a, as.numeric(b))) }

View the code on GitHub.

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So give already: what’s the result of microbenchmark(gmp_method, big.add) ?

Hi Carl,

The base-r method is obviously much slower than the gmp library. I wrote this function because I want to solve Euler problems without using libraries.

By the way, you are mis-using gmp. In fact, “as.numeric(numbers[1])” loses the fidelity you desire. You need to convert directly: sum(as.bigz(numbers)) to get the actual values. as.bigz does accept char strings. Once you do that, the full result is identical to your minifunction.

Thanks for the tip Carl. I have updated the code in the post.

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