Can you solve Dongle's Difficult Dilemma? - Dennis E. Shasha

翻译人员: Ruishi (Jerry) Zhang 校对人员: Helen Chang

According to legend, when this planet was young and molten,

传说中，当这个星球 还年轻并炽热的时候

three galactic terraformers shaped it into a paradise.

三个来自星系之外的异种 把这里变成了人间福地

When their work was done, they sought out new worlds,

他们改造完星球后 开始了寻找新的星球的旅程

but left the source of their power behind:

但是他们把他们的力量源泉 留在了星球里

three powerful golden hexagons,

三个有着强大力量的黄金六角宝石

hidden within dungeons full of traps and monsters.

这三个宝石被放在 充满陷阱和怪物的地下城里

If one person were to bring all three hexagons together,

如果有人集齐了这三个宝石

they could reinvent the world however they saw fit.

他就可以依意愿来重新改造这个世界

That was thousands of years ago.

这件事发生在上千年前

Today, you’ve learned of Gordon, an evil wizard dead set on collecting the hexagons

今天，你得知一个叫戈登的邪恶巫师 要集齐这三个宝石

and enslaving the world to his will.

并奴役这个世界

So you set off on a quest to get them first,

所以你决定抢在戈登之前 拿到这三个宝石

adventuring through fire, ice, and sand.

你经过了火焰、寒冰，和狂沙的试炼

Yet each time, you find that someone else got there first.

但是每次都有另一个人 在你之前拿到宝石

Not Gordon, but a merchant named Dongle.

这个人不是戈登 他是一个叫当构的商人

At the end of the third dungeon. you find a note inviting you to Dongle’s castle.

在第三个地下城里 你找到了来自当构城堡的邀请函

You show up with a wallet bursting with the 99 gems

你带着在冒险时得到的99个宝石

you’ve collected in your travels,

到达了当构的城堡

arriving just moments before Gordon, who also has 99 gems.

戈登在你之后也到了 而且他也有99个宝石

Dongle has not only collected the golden hexagons,

当构不止拿到了仅有的三个六边金宝石

but he’s used them to create 5 silver hexagons,

他还用金宝石创造了五个六边银宝石

just as powerful as their golden counterparts.

并且银宝石和金宝石有着相同的力量

Why did Dongle do all this?

为什么当构要这么做呢？

Because there’s one thing he loves above all else: auctions.

因为当构最喜欢拍卖

You and the evil wizard will compete to win the hexagons,

你和邪恶的巫师要去竞拍这些六边宝石

starting with the three golden ones,

从三个金宝石开始

making one bid for each item as it comes up.

一个物品能出价一次

The winners of ties will alternate, starting with you.

如果出价相同，胜者会轮换

从你开始

Whoever first collects a trio of either golden or silver hexagons

第一个拿到三个相同颜色宝石的人 （不论金色还是银色）

can use their power to recreate the world.

可以用它们的力量重塑这个世界

You’ve already bid 24 gems on the first,

当你第一轮已经出了24个宝石的时候

when you realize that your rival has a dastardly advantage:

你发现你的对手用了卑鄙的手段

a mirror that lets him see what you’re bidding.

一个可以让他看到你出价的镜子

He bids zero, and you win the first hexagon outright.

他的出价是0个宝石 你赢得了你一个宝石

What’s your strategy to win a matching trio of hexagons before your rival?

你要怎么在戈登之前 集齐三个六边宝石呢？

Pause here to figure it out for yourself.

如果想自己先算出来，在这里暂停

Answer in 3

三

Answer in 2

二

Answer in 1

一

Dongle’s dangled a difficult dilemma.

当构出了一个让人进退两难的难题

Do you spend big to try to win the golden hexagons outright?

你要花大钱试图直接赢得 全部的金六边宝石吗？

Save as much as possible for silver? Or something in-between?

还是攒钱去买银六边宝石？ 或者在这两者之间？

Gordon can use his magic mirror and 99 gems

戈登可以用他的神奇镜子和99个宝石

to make sure that no matter what you bid on the second gold,

去确保不论你对第二个 金六边宝石的出价是多少

he can bid one more and block you.

他都能比你多出一个宝石，胜过你

So the real question is— how can you force Gordon to spend enough on the golds

那么问题来了——如何让戈登 在金六边宝石上花上够多的钱

to guarantee that you’ll win on the silvers?

从而确保你能赢下银色的六边宝石呢？

Here’s a hint.

给你点提示

Let’s say at the start of the silver auctions

如果在开始竞拍银六边宝石的时候

you had a one gem advantage, such as 9 to 8.

你比戈登多一个宝石 比如你有九个，他有八个

You need to win three auctions,

你需要赢下三次竞拍

so could you divide your gems into three groups of three and win?

那么你能把你的宝石 分成三组并取得胜利吗？

For simplicity, let’s assume a set of rules that’s worse for you,

为了让情况变得更简单 我们假设有一个对你很不利的规则

where Gordon wins every tie.

那就是每次出价一样都算是戈登获胜

If you bid 3 each time, the best he could do

如果你每次出价三个宝石 戈登最多能赢下两个宝石

is win two silver hexagons, and have two gems left—

并且只剩两个宝石

which you’ll beat with three bids of 3.

那你就可以用你的三个宝石打败他

Any one-gem advantage where your starting total is divisible by three

一样的逻辑 只要你的宝石数可以除以三

will lead to victory by the same logic.

你就肯定能赢得最终的胜利

So knowing that, how can you force Gordon's hand in the gold auctions

知道了这个后，如何让戈登 把更多钱花在金六边宝石上

so you go into silver with an advantage?

以让你在银六边宝石的竞拍上 赢得优势呢？

Let’s first imagine that Gordon lets you win the second gold auction

试想戈登出了和你一样数量的宝石

by betting some amount X with a tie.

让你赢了第二个金宝石

You could then bid everything you have left on the third gold hexagon,

接着不管你在第三个金宝石上出价多少

and he'd have to match you to block.

他必须比你出的多 从而阻止你获得三个金宝石

So if you could bid 51 on the third gold,

所以如果你在第三个金宝石上 出价51个宝石

you'd go into silver with a 51 to 48 advantage, which you know you can win.

你将会以51比48的优势 参加银宝石竞拍，并且可以赢

Solving for X reveals that in order to have 51 on round three,

经过计算，如果你想在 第三个金宝石竞拍时有51个宝石

you should bid at most 24 on round two.

你第二轮最多出24个宝石

But what about the other possibility,

但是如果戈登在第二轮

where Gordon wins the second gold against your bid of 24—

出价超过24个宝石

would this strategy still work?

这个计策还有用吗？

The least he could bid to win the second gold is 25,

如果戈登想赢第二轮 他最少出25个宝石

making the total 75 to Gordon’s 74.

这样的话你就有75个宝石 戈登有74个

No one would then bid on round three,

紧接着没有人赢得第三轮

since you’ve each blocked the other from getting three golds.

因为你和戈登互相牵制对方

After that, you could bid 25 every time to win three silvers.

然后，你就可以每轮出价25 并赢下三个银宝石

The bidding war was close,

竞拍结束了

but your ingenuity kept you one link ahead in the chain of inference,

但是你的聪明才智使你比戈登更快一步

and the silver tri-source is yours.

你得到了三个银六边宝石

Now... what will you do with it?

那么现在，你要用它们做什么呢？