初中级Lower intermediate在线实验室Online lab可打印Printable

猴子共和国：给猴子装上记忆

Monkey Republic: Give Your Monkey a Memory

把行为写成策略卡，逐轮预测，再用一次误动作和一次换对手来检查程序。

Write a strategy card, predict every round, then test one mistaken action and a change of opponent.

沿用香蕉糖或纸香蕉，这次照规则卡行动：“对方上一轮怎么对我，我这一轮就怎么对它。”先写到能执行，再问：记的是谁的上一轮？

Keep the candy or paper bananas; this time follow a rule card: “Treat the other monkey as it treated me last round.” Make that executable, then ask: whose last round did we remember?

开始之前Before you start

写清第一步、看哪段历史和下一步的规则；准确回放五轮，并区分当前对手的记忆与全局上一招。

Specify the first action, remembered history and next action; trace five rounds and distinguish per-opponent memory from the last action globally.

先做猴子初级收益表；会读“如果／否则”。不要求先学感知机或 Python。

Complete the monkey payoff lesson and read if/otherwise instructions. No perceptron or Python prerequisite.

一步一步试试看Try it step by step

先猜（5分钟）：三张规则卡：一直 S、一直 G、Copycat“第一轮 S，以后复制对方上一轮实际动作”。两边必须根据旧历史同时决定。Predict (5 min): use always S, always G and Copycat: S first, then the opponent’s previous executed action. Both choose from the old history simultaneously.
动手（10分钟）：Copycat 分别对一直 S、一直 G，各五轮。记录实际动作、每轮收益和总分。换对手前重置历史，不拿一场总分宣告普遍最优。Play (10 min): run Copycat against always S and always G for five rounds each. Record actions and scores. Reset history between matches; one score does not prove universal superiority.
观察（10分钟）：两个 Copycat 对战，第 2 轮只把右猴的动作翻转一次。先填意图与实际动作，再让后续程序读取实际动作，观察回应链。Observe (10 min): play two Copycats and flip only the right action in round two. Record intended versus executed actions; later choices read executed history.
挑战（10分钟）：依次遇到友好 A、抢夺 B、友好 A。比较“全局上一招”和“给每个对手单独记忆”，说明程序按哪条信息决定。Challenge (10 min): meet friendly A, grabbing B, then A again. Compare global-last-action with per-opponent memory and explain which information each rule reads.

一起算一个例子Work through one example

Win-Win 下 Copycat 对一直 G：第一轮 S/G 得 −1/5，后四轮 G/G 各 0/0，五轮总计 −1/5。对一直 S 则每轮 S/S，共 15/15。

Under Win-Win, Copycat versus always G starts S/G at −1/5, then four G/G rounds at 0/0, totaling −1/5. Against always S it gets five S/S rounds, totaling 15/15.

先想一想，再看答案Think first, then check

两个 Copycat 只在第 2 轮翻转右猴动作，第 3 轮各会出什么？B 抢过以后，A 的个人历史是否也应该变成 G？

After only the right Copycat is flipped in round two, what does each do in round three? Does B’s grab change A’s personal history?

查看参考答案Show the expected answer

第三轮为 G/S，之后第 4 轮 S/G、第 5 轮 G/S。完整五轮 SS、SG、GS、SG、GS，各得 11；无翻转时各 15。A→B→A：全局记忆动作 S,S,G；按对手记忆 S,S,S，B 的记录不覆盖 A。总分全局 7、按对手 5，这不证明记忆正确就一定得分更多。原信任游戏同样 Copycat 对 G 五轮为 4/9，这里为 −1/5，源于不同收益表。

Round three is G/S, followed by S/G then G/S. The full SS,SG,GS,SG,GS trace totals 11 each; no flip gives 15 each. For A→B→A, global memory produces S,S,G; per-opponent memory S,S,S. B does not overwrite A. Their totals are seven and five: correct memory scope need not maximize score. The original trust game gives 4/9 against G, versus −1/5 here, because the payoffs differ.

这个实验说明了什么What this experiment shows

这是 Duel 的明确策略卡，不是神经网络或自动学会的策略。Society 的同名角色使用参数引擎，不能直接套这张逐轮答案；随机失误率也不等于指定某一轮的一次翻转。

These are explicit Duel rules, not neural networks or automatically learned strategies. Society’s similarly named characters use a dial engine; this exact trace does not transfer. A random mistake rate is not a scheduled single flip.