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magic_5_gon_ring.rs
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magic_5_gon_ring.rs
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use crate::utils;
pub fn solve() -> i64 {
// Assign an index to each of the 10 nodes of the 5-gon ring. Write the
// indices of each triplet of nodes one after the other. In particular, if
// the indices of one triplet of nodes are 0, 1 and 2, those of the next
// node (moving clockwise) will be 3, 2 and 4. Effectively, this maps a
// permutation to a potential solution.
const INDICES: [usize; 15] = [0, 1, 2, 3, 2, 4, 5, 4, 6, 7, 6, 8, 9, 8, 1];
// These will be written into the nodes. Starting with this order ensures
// that we find the answer early.
let mut numbers = [7, 1, 2, 3, 4, 5, 6, 8, 9, 10];
let result = 'numbers: loop {
// Skipping the first permutation is safe because it cannot be the
// answer. If any permutation is a solution, it must begin with at most
// 6 (because the first external node must be the smallest of all
// external nodes) which the first call to this function will ensure.
utils::prev_permutation(&mut numbers);
let sum = numbers.iter().take(3).sum();
for i in (0..INDICES.len()).step_by(3).skip(1) {
// The first external node must be the smallest of all external
// nodes.
if numbers[INDICES[i]] < numbers[0] {
continue 'numbers;
}
// The sum of the nodes in each triplet must be the same.
if numbers[INDICES[i]] + numbers[INDICES[i + 1]] + numbers[INDICES[i + 2]] != sum {
continue 'numbers;
}
}
let solution = INDICES.iter().fold(0, |solution, &index| {
let multiplier = if numbers[index] == 10 { 100 } else { 10 };
solution * multiplier + numbers[index]
});
if solution < 10i64.pow(16) {
// Technically, the first solution we find need not be the largest
// one, because the solution is not a true permutation. (It is an
// expansion, since a 10-size permutation was expanded to 15
// elements.) Luckily, though, it is the answer!
break solution;
}
};
assert_eq!(result, 6531031914842725);
result
}