計算機科学のブログ

Building Abstractions with Data - Hierarchical Data and the Closure Property - Sequences as Conventional Interfaces - Sequence Operations - accumulate, tree, leaves

Structure and Interpretation of Computer Programs: JavaScript Edition(Harold Abelson(著)、Gerald Jay Sussman(著)、Julie Sussman(著)、The MIT Press)のChapter 2(Building Abstractions with Data)、2.2(Hierarchical Data and the Closure Property)、2.2.3(Sequences as Conventional Interfaces)、Sequence Operations、Exercise 2.35の解答を求めてみる。

コード

function stringfy(x) {
    if (is_null(x)) {
        return "null";
    }
    if (is_pair(x)) {
        return "[" + stringfy(head(x)) + ", " + stringfy(tail(x)) + "]";
    }
    return x.toString();
}
function pair(x, y) {
    return [x, y];
}
function head(z) {
    return z[0];
}
function tail(z) {
    return z[1];
}
function is_pair(x) {
    return Array.isArray(x);
}
function is_null(x) {
    return x === null;
}
function list(...args) {
    return args.length === 0 ?
        null :
        pair(args[0], list(...args.slice(1)));
}
function display(x) {
    return console.log(stringfy(x));
}
function accumulate(op, initial, sequence) {
    return is_null(sequence) ?
        initial :
        op(
            head(sequence),
            accumulate(op, initial, tail(sequence))
        );
}
function map(f, sequence) {
    return accumulate(
        (x, y) => pair(f(x), y),
        null,
        sequence
    );
}
function count_leaves(tree) {
    return accumulate(
        (x, y) => x + y,
        0,
        map(
            (x) =>
                is_null(x) ? 0 :
                    is_pair(x) ?
                        count_leaves(x) :
                        1,
            tree
        )
    );
}

const tree = pair(list(1, 2), list(3, 4));
const tree1 = list(tree, tree);
display(tree);
display(count_leaves(tree));
display(tree1);
display(count_leaves(tree1));

入出力結果(Terminal, Zsh)

% node answer2.35.js
[[1, [2, null]], [3, [4, null]]]
4
[[[1, [2, null]], [3, [4, null]]], [[[1, [2, null]], [3, [4, null]]], null]]
8
%