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
%