Largest palindrome product - euler project

Question

I was trying to solve project Euler problem 4 which is:

A palindromic number reads the same both ways. The largest palindrome made from the product of tw

Answer 1

You are also assuming that the first palindrome you'll find will be the largest. The first palindrome you'll find is 580085 which isn't the right answer.

You are also assuming that the first palindrome you'll find is the product of two 3 digit numbers. You should also use two different numbers instead of multiplying 999 with 999 and iterating down to 100 * 100.

Answer 2

Since no one did in R. This a solution that gives the answer to the problem.

Project Euler Question 4

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers. R doesn't have a built-in function to check palindromes so I created one although 'rev' can be used :). Also, the code is not optimized for speed on purpose primarily to increase readability.

       reverse <- function(x){
        Args:
       x : object whose elements are to be reversed, can be of type 
       'character' or 'vector' of length = 1
       Returns:
          x : The object's elements are reversed e.g boy becomes yob and 23 
       becomes 32
       Error Handling:
      if (is.vector(x) == TRUE & length(x) > 1){
        stop("Object whose length > 1 cannot be used with reverse(x) 
        consider vector.reverse(x)")
          }
         Function Execution
         if (is.character(x) == TRUE){
           v <- unlist(strsplit(x, ''))
             N <- length(v)
           rev.v <- v
          for (i in 0:(N - 1)){
         rev.v[i + 1] <- v[N - i]
        }
         rev.v <- paste0(rev.v, collapse = '')
            return(rev.v)
         } else {
            x <- as.character(x)
       v <- unlist(strsplit(x, ''))
           rev.v <- v
            N <- length(v)
 for (i in 0:(N - 1)){
   rev.v[i + 1] <- v[N - i]
 }
rev.v <- paste0(rev.v, collapse = '')
rev.v <- as.numeric(rev.v)
  return(rev.v)
   }
 }

    the function vector.reverse() has been deleted to reduce the length of 
    this code

    is.palindrome <- function(x){
  Args:
     x : vector whose elements will be tested for palindromicity, can be of 
 length >= 1
  Returns:
    TRUE : if an element in x or x is palindromic
    FALSE: if an element in x or x is not palindromic

   Function Execution:
  if (is.vector(x) == TRUE & length(x) > 1){
    x.prime <- vector(length = length(x))
    for (i in 1:length(x)){
      x.prime [i] <- reverse(x [i])
 }
   return(x.prime == x)
 } else {
 ifelse(reverse(x) == x, return(TRUE), return(FALSE))
     }
   }

  palindromes between 10000 and 999*999
 Palin <- (100*100):(999*999)
 Palin <- Palin [is.palindrome(Palin) == 1]
 i.s <- vector('numeric', length = length(Palin))
 j.s <- vector('numeric', length = length(Palin))

   Factoring each of the palindromes
   for (i in 100:999){
     for (j in 100:999){
       if (sum(i * j == Palin) == 1){
         i.s[i-99] <- i
         j.s[i-99] <- j

      }
     }
   }
 product <- i.s * j.s
 which(i.s * j.s == max(product)) -> Ans
 paste(i.s[Ans[1]], "and", j.s[Ans[1]], "give the largest two 3-digit 
  palindrome")



ANSWER
  993 * 913 = 906609

ENJOY!

Answer 3

Another simple solution written in C#

private static void Main(string[] args)
        {
            var maxi = 0;
            var maxj = 0;
            var maxProd = 0;
            for (var i = 999; i > 100; i--)
            for (var j = 999; j > 100; j--)
            {
                var product = i * j;
                if (IsPalindrome(product))
                    if (product > maxProd)
                    {
                        maxi = i;
                        maxj = j;
                        maxProd = product;
                    }
            }
            Console.WriteLine(
                "The highest Palindrome number made from the product of two 3-digit numbers is {0}*{1}={2}", maxi, maxj,
                maxProd);
            Console.ReadKey();
        }

        public static bool IsPalindrome(int number)
        {
            var numberString = number.ToString();
            var reverseString = string.Empty;
            for (var i = numberString.Length - 1; i >= 0; --i)
                reverseString += numberString[i];
            return numberString == reverseString;
        }