Problem 42 "Coded triangle numbers"

The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we shall call the word a triangle word.

Using words.txt, a 16K text file containing nearly two-thousand common English words, how many are triangle words?

問 42 「符号化三角数」

三角数のn項は \( t_{n} = n(n+1)/2 \) で与えられる. 最初の10項は

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

である.

単語中のアルファベットを数値に変換した後に和をとる. この和を「単語の値」と呼ぶことにする. 例えば SKY は 19 + 11 + 25 = 55 = \( t_{10} \) である. 単語の値が三角数であるとき, その単語を三角語と呼ぶ.

16Kのテキストファイル words.txt 中に約2000語の英単語が記されている. 三角語はいくつあるか?

Because the original word list is very long, this example has only a part of it.

fn is_triangle_number(x: u32) -> bool {
    let expr = 8 * x + 1;
    let side = (expr as f64).sqrt() as u32;
    side * side == expr
}

fn word_value(w: &str) -> u32 {
    w.chars().map(|c| c as u32 - 'A' as u32 + 1).sum::<u32>()
}

fn main() {
    let count = WORDS
        .iter()
        .map(|&w| word_value(w))
        .filter(|&v| is_triangle_number(v))
        .count();

    println!("{:?}", count);
    assert_eq!(count, 5);
}

const WORDS: &[&str] = &[
    "A",
    "ABILITY",
    "ABLE",
    "ABOUT",
    "ABOVE",
    "ABSENCE",
    "ABSOLUTELY",
    "ACADEMIC",
    "ACCEPT",
    "ACCESS",
    "ACCIDENT",
    "ACCOMPANY",
    "ACCORDING",
    "ACCOUNT",
    "ACHIEVE",
    "ACHIEVEMENT",
    "ACID"
];