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66 changes: 66 additions & 0 deletions src/bin/4.rs
Original file line number Diff line number Diff line change
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// A palindromic number reads the same both ways. The largest palindrome made
// from the product of two 2-digit numbers is 9009 = 91 × 99.

/// Checks if `num` is a palindrome.
/// Increasingly compares beginning and ending digits
fn is_palindrome(num: u64) -> bool {
let mut number = num;
let mut leading_div = 1;
while number > 10 {
number /= 10;
leading_div *= 10;
}
number = num;
while number > 10 {
let ending = number % 10;
let leading = number / leading_div;
if ending != leading {
return false;
}
number = (number % leading_div) / 10;
leading_div /= 100;
}
true
}

/// Finds the largest palindrome made from the product of two n-`digit` numbers.
/// This is an extremely naive approach bruteforce method.
/// It finds all possible palindromes and then pick the largest one.
///
/// TODO: Find a more efficient method.
fn problem_4_sol_1(digits: u32) -> Result<u64, &'static str> {
let a = 10u64.pow(digits - 1)..10u64.pow(digits);
let mut max_palindrome = 0;
for a_value in a {
for b_value in (10u64.pow(digits - 1)..10u64.pow(digits)).rev() {
let candidate = a_value * b_value;
if is_palindrome(candidate) && candidate > max_palindrome {
max_palindrome = candidate;
}
}
}
if max_palindrome == 0 {
Err("No palindrome product found")
} else {
Ok(max_palindrome)
}
}

fn main() {
let digits = 3;
println!("Largest palindromic product for 2 numbers with {} digits {}",
digits, problem_4_sol_1(digits).ok().unwrap());
}

#[test]
fn test_is_palindrome() {
assert_eq!(true, is_palindrome(9009));
assert_eq!(true, is_palindrome(1234554321));
assert_eq!(true, is_palindrome(1));
assert_eq!(false, is_palindrome(12));
}

#[test]
fn test_largest_palindrome() {
assert_eq!(9009, problem_4_sol_1(2).ok().unwrap());
}