Eloquent JavaScript #07# Project: A Robot
注释即笔记：
const roads = [ "Alice's House-Bob's House", "Alice's House-Cabin", "Alice's House-Post Office", "Bob's House-Town Hall", "Daria's House-Ernie's House", "Daria's House-Town Hall", "Ernie's House-Grete's House", "Grete's House-Farm", "Grete's House-Shop", "Marketplace-Farm", "Marketplace-Post Office", "Marketplace-Shop", "Marketplace-Town Hall", "Shop-Town Hall" ]; function buildGraph(edges) { // graph的存储格式类似于 // graph.place_a: [place_b, place_c] // 表示由a能够直接到达b或者c let graph = Object.create(null); function addEdge(from, to) { // 首先判断该起点有没有被添加 if(graph[from] == null) { graph[from] = [to]; } else { graph[from].push(to); } } // 'Alice's House-Bob's House'.split("-") // → ['Alice's House', 'Bob's House'] for(let [from, to] of edges.map(r => r.split("-"))) { addEdge(from, to); addEdge(to, from); } return graph; } const roadGraph = buildGraph(roads); /** * 不要条件反射地把每个概念都搞成对象， * 这样的程序通常是难以理解和维护的。 * 相反，用最小的数值集合来描述村庄&机器人 * 状态就可以了。 */ class VillageState { constructor(place, parcels) { this.place = place; this.parcels = parcels; } move(destination) { // move表示机器人的一次移动。 // 首先检查是否存在一条从当前位置this.place到目的地destination的路 // 如果不存在就说明不是合法的移动，返回旧的VillageState。 // 如果存在，就更新机器人的当前位置，也就是更新VillageState.place // 为destination，当然同时需要更新包裹的状态：1-还没被机器人拿到 // 的包裹不去管它，2-拿到的包裹则更新当前位置place（目的地address不改变） // 3-最后过滤掉已经送到的包裹（目的地就在本地） // PS. 整个move方法实际上是重造了一个VillageState，并没改变旧的 // VillageState对象 if(!roadGraph[this.place].includes(destination)) { return this; } else { let parcels = this.parcels.map(p => { if(p.place != this.place) return p; return { place: destination, address: p.address }; }).filter(p => p.place != p.address); return new VillageState(destination, parcels); } } } /** * 可以用Object.freeze冻结对象 * 所有对该对象属性的写入操作会被忽略 * 这需要计算机做一些额外工作 * let object = Object.freeze({value: 5}); object.value = 10; console.log(object.value); // → 5 */ /** * But the most important limit * on what kind of systems we can build * is how much we can understand. * Anything that makes your code easier * to understand makes it possible * to build a more ambitious system. * 写程序最重要的限制是我们能够理解多少 */ // robot实际上是一个函数接口， // 该函数输入state和memory // 输出决策action用于实际移动 // 这样写就能够动态更改策略了。 function runRobot(state, robot, memory) { for(let turn = 0;; turn++) { if(state.parcels.length == 0) { console.log(`Done in ${turn} turns`); break; } let action = robot(state, memory); state = state.move(action.direction); memory = action.memory; console.log(`Moved to ${action.direction}`); } } function randomPick(array) { let choice = Math.floor(Math.random() * array.length); return array[choice]; } // 最愚蠢但有效的策略--随机决定下一个方向。 // 虽然这里只有一个参数，但是js允许传 // 入更多（或者更少）的参数。在这里，传入的 // memeory被忽略了 function randomRobot(state) { return { direction: randomPick(roadGraph[state.place]) }; } // 初始化 VillageState.random = function(parcelCount = 5) { let parcels = []; for(let i = 0; i < parcelCount; i++) { let address = randomPick(Object.keys(roadGraph)); let place; do { place = randomPick(Object.keys(roadGraph)); // 包裹的起点和终点不可以是同一个地方 } while (place == address); parcels.push({ place, address }); } return new VillageState("Post Office", parcels); }; // runRobot(VillageState.random(), randomRobot); // 版本一，步数不稳定 // runRobotAnimation(VillageState.random(), randomRobot); // 作者写的动画版本，相当直观酷炫。。 // 第二个策略：事先指定一条可以通过所有地点的路线 // 走两遍就可以确保投递所有邮件 const mailRoute = [ "Alice's House", "Cabin", "Alice's House", "Bob's House", "Town Hall", "Daria's House", "Ernie's House", "Grete's House", "Shop", "Grete's House", "Farm", "Marketplace", "Post Office" ]; // [a,b,c].slice(1) // → [b,c] // [a,b,c].slice(1, 2) // → [b] // 包括start不包括end function routeRobot(state, memory) { if(memory.length == 0) { memory = mailRoute; } // memory类似于队列 // 等价: return {direction: memory.shift(), memory: memory} return { direction: memory[0], memory: memory.slice(1) }; } // runRobot(VillageState.random(), routeRobot, []); // 版本二，最多26步 /** * The problem of finding a route * through a graph is a typical search problem. * We can tell whether a given solution (a route) * is a valid solution, but we can’t directly compute * the solution the way we could for 2 + 2. * Instead, we have to keep creating potential solutions * until we find one that works. */ // 返回一点到另一点的最短路线，参考：C++ 电路布线/最短路径问题 function findRoute(graph, from, to) { let work = [{at: from, route: []}]; // 其实也是个队列 for(let i = 0; i < work.length; i++) { let {at, route} = work[i]; // 原来还可以这样赋值。。 for(let place of graph[at]) { // 搜索四周围，如果找到了目的地就直接+1返回。 if(place == to) return route.concat(place); if(!work.some(w => w.at == place)) { // 判断点是否已经入队 work.push({at: place, route: route.concat(place)}); } } } } function goalOrientedRobot({place, parcels}, route) { // 首先判断当前制定的路线走完没有， // 走完就重新制定下一条路线 // 逐个包裹处理（当然也有可能顺 // 路完成其它包裹的fetch和投递） if(route.length == 0) { let parcel = parcels[0]; if(parcel.place != place) { // 制定取包裹路线 route = findRoute(roadGraph, place, parcel.place); } else { // 制定投递路线 route = findRoute(roadGraph, place, parcel.address); } } return {direction: route[0], memory: route.slice(1)}; } runRobot(VillageState.random(), goalOrientedRobot, []); // 版本三，平均十来步的样子
Exercises
function testRobot(state, robot, memory) { for(let turn = 0;; turn++) { if(state.parcels.length == 0) { return turn; break; } let action = robot(state, memory); state = state.move(action.direction); memory = action.memory; } } function compareRobots(robot1, memory1, robot2, memory2) { let tasks = []; for (let i = 0; i != 100; ++i) { tasks.push(VillageState.random()); } let total1 = 0, total2 = 0; for (let task of tasks) { total1 += testRobot(task, robot1, memory1); total2 += testRobot(task, robot2, memory2); } console.log(`average turns: robot1 ${total1 / 100}, robot2 ${total2 / 100}`); } compareRobots(routeRobot, [], goalOrientedRobot, []); // → average turns: robot1 18.07, robot2 15.03
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没做任何优化的穷举，而且还用递归。。。6个以上包裹浏览器应该会直接崩溃掉。。
/** max表示一个包裹要被处理的次数 arr为长度为包裹数量的全0数组 arr[i]表示第i个包裹被处理的次数 当arr为[max, max, max, ...]时 表示所有包裹已经被处理完 返回的sequences包含处理包裹的 所有顺序集合 */ function makeSequences(max, arr) { const sequences = []; const fillArrWith = (max, arr, start, sequence) => { // 填充起始点 arr[start] += 1; sequence.push(start); // 判断是否已经填充满 let sum = 0; for (let x of arr) { sum += x; }; if (sum == max * arr.length) { sequences.push(sequence); return; } // 寻找下一个填充点 for (let i = 0; i != arr.length; ++i) { if (arr[i] < max) fillArrWith(max, arr.slice(), i, sequence.slice()); } }; for (let i = 0; i != arr.length; ++i) { fillArrWith(max, arr.slice(), i, []); } return sequences; } /** 把生成的序列转化为具体业务相关的表达。 得到route并不是实际的route 而是可能不相邻的地点 routes包含所有能够完成任务的路线 */ function sequencesToRoutes(sequences, {place, parcels}) { const routes = []; const flag = parcels.map(() => 0); // 用于之后拷贝用 // 逐个序列处理 for (let sequence of sequences) { let route = [place]; // 添加起点 let localFlag = flag.slice(); // 标记包裹的状态 for (let num of sequence) { if (localFlag[num] == 0) { // 第一次处理包裹num：到place取包裹 localFlag[num]++; // 包裹num已取,这样可以保证某包裹的place一定优先于该包裹的address入队 if (route[route.length - 1] != parcels[num].place) { // 避免出现两个连续重复place route.push(parcels[num].place); } } else { // 第二次处理包裹num： 送包裹，去该包裹的目的地 if (route[route.length-1] != parcels[num].address) { route.push(parcels[num].address); } } } routes.push(route); } return routes; } /** * 计算单个路线需要的最短步数 * turnsMap用于保存已经计算的两点值，避免重复计算 */ function turnsOfRoute(route, turnsMap=new Map()) { let totalTurns = 0; for (let i = 0; i != route.length - 1; ++i) { // 两点、两点处理。 let routeKey = route[i].concat(route[i + 1]); let turns = turnsMap.get(routeKey); if (turns != undefined) { totalTurns += turns; } else { turns = findRoute(roadGraph, route[i], route[i + 1]).length; // 计算 a到b 的最小步数 ↑ // 保存计算结果 ↓ turnsMap.set(routeKey, turns); routeKey = route[i + 1].concat(route[i]); // a到b和b到a的最短路是一样的。 turnsMap.set(routeKey, turns); totalTurns += turns; } } return totalTurns; } /** * 寻找最短路线 */ function shortestRoute(routes) { let min = Infinity; let tempRoute; let turnsMap = new Map(); // 用于保存已经计算的两点值，避免重复计算 for (let route of routes) { let turns = turnsOfRoute(route, turnsMap); if (turns < min) { min = turns; tempRoute = route; // 保存最短路线 } } // 将最短路线转化为相邻的可以实际移动的地点序列 let result = []; for (let i = 0; i != tempRoute.length - 1; ++i) { // 仍然是两点、两点处理 let midRoute = findRoute(roadGraph, tempRoute[i], tempRoute[i + 1]); if (result[result.length - 1] != midRoute[0]) { // 避免出现两个连续重复place result = result.concat(midRoute); } else { result = result.concat(midRoute.shift()); } } return result; } function simpleRobot({place, parcels}, route) { if(route.length == 0) { let sequences = makeSequences(2, parcels.map(() => 0)); // 转化成一种比较抽象的东西。。。 let routes = sequencesToRoutes(sequences, {place, parcels}); route = shortestRoute(routes); } return {direction: route[0], memory: route.slice(1)}; } //runRobot(VillageState.random(), simpleRobot, []); // 版本四，穷举。。 平均 10.64 //runRobotAnimation(VillageState.random(), simpleRobot, []);
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结果没错，实现上有点跑偏。。（英语阅读能力不足造成的）
class PGroup { add(x) { let result = Object.create(PGroup.prototype); if (this.container == undefined) { this.container = []; } if (!this.container.includes(x)) { result.container = this.container.concat(x); } else { result.container = this.container; } return result; } delete(x) { let result = Object.create(PGroup.prototype); if (this.container != undefined) { result.container = this.container.filter(a => !(a === x)); } return result; } has(x) { if (this.container != undefined) { return this.container.includes(x); } return false; } } PGroup.empty = Object.create(PGroup.prototype); let a = PGroup.empty.add("a"); let ab = a.add("b"); let b = ab.delete("a"); console.log(b.has("b")); // → true console.log(a.has("b")); // → false console.log(b.has("a")); // → false
