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

譯者: Lilian Chiu 審譯者: Amanda Zhu

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

根據傳說，

當這顆行星在很年輕的熔岩狀態時，

three galactic terraformers shaped it into a paradise.

三位銀河地貌形塑師 將它塑造成天堂。

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

他們完工之後，

就去尋找新世界了，

but left the source of their power behind:

而留下了他們的力量來源：

three powerful golden hexagons,

三個強大的六角黃金，

hidden within dungeons full of traps and monsters.

藏在滿是陷阱和怪物的地牢中。

If one person were to bring all three hexagons together,

如果有人能把三個六角黃金 通通集合在一起，

they could reinvent the world however they saw fit.

就可以照自己的意思來重造世界。

That was thousands of years ago.

這是數千年前的事。

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

現在，你聽說邪惡的巫師高登

下定決定要找到這些六角黃金

and enslaving the world to his will.

讓世界臣服在他腳下。

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

於是你展開旅程，要搶先找到，

adventuring through fire, ice, and sand.

冒險穿過烈火、冰天雪地、沙地。

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

但每一次你都發現有人捷足先登。

Not Gordon, but a merchant named Dongle.

這個人不是高登， 而是名為丹戈的商人。

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

在第三個地牢的盡頭， 你找到一張字條，

邀請你去丹戈的城堡。

You show up with a wallet bursting with the 99 gems

你依邀請前往時，

皮夾裡裝滿了在旅途上 收集來的九十九顆寶石。

you’ve collected in your travels,

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

高登隨後來到，

同樣帶著九十九顆寶石。

Dongle has not only collected the golden hexagons,

丹戈不僅收集到了所有六角黃金，

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

且已經用它們創造出了 五個六角白銀，

just as powerful as their golden counterparts.

和六角黃金一樣強大。

Why did Dongle do all this?

丹戈為什麼要這麼做？

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

因為他對某件事的愛勝過一切：

拍賣。

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

你要和邪惡巫師競爭以贏得六角金銀，

starting with the three golden ones,

先從三個六角黃金開始，

making one bid for each item as it comes up.

每個放出來時，可出價一次。

The winners of ties will alternate, starting with you.

平手第一次算你贏，之後輪流。

Whoever first collects a trio of either golden or silver hexagons

先收集到三個六角黃金 或三個六角白銀的人

can use their power to recreate the world.

就可以用它們的力量去重建世界。

You’ve already bid 24 gems on the first,

競標第一件物品時， 你已經用掉了二十四個寶石，

when you realize that your rival has a dastardly advantage:

此時你才發現你的對手 有個很卑鄙的優勢：

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

他有一面鏡子可以看到你的出價。

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

他出價零個寶石， 讓你贏得第一個六角黃金。

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

你要用什麼策略

才能比對手更早取得 三個黃金或三個白銀？

Pause here to figure it out for yourself.

［自行解題，請在此暫停 答案公佈倒數：3］

Answer in 3

Answer in 2

［2］

Answer in 1

［1］

Dongle’s dangled a difficult dilemma.

丹戈製造了一個兩難的局面。

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

你會直接出高價取得六角黃金嗎？

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

還是盡可能保留給六角白銀？

或者折衷？

Gordon can use his magic mirror and 99 gems

高登可以用他的魔法鏡子 以及九十九個寶石，

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

不論你對第二個六角黃金出價多少，

he can bid one more and block you.

他都能多出價一個寶石來阻止你。

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

所以真正的問題是——

你要如何迫使高登 在六角黃金上花費夠多寶石，

to guarantee that you’ll win on the silvers?

以確保你能贏得三個六角白銀？

Here’s a hint.

提示如下。

Let’s say at the start of the silver auctions

假設在六角白銀拍賣開始時，

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

你有多一顆寶石的優勢，

比如九比八。

You need to win three auctions,

你得贏三場拍賣，

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

所以，你可以把你的寶石 三個三個分為一組並獲勝？

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

為了簡化問題，

咱們就假設你們用的是 一組對你不利的規則，

where Gordon wins every tie.

也就是每次平手都是高登贏。

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

如果你每次出價三顆寶石，

對他來說最好的狀況 就是贏得兩個六角白銀

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

並剩下兩顆寶石——

which you’ll beat with three bids of 3.

你後面三次都出價三顆寶石就贏了。

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

只要開始時你的寶石 總數比對方多一個

且可以被三整除，

will lead to victory by the same logic.

都能依同樣的邏輯讓你獲勝。

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

知道這一點之後，

你要如何迫使高登依你的意思 投入六角黃金的拍賣，

so you go into silver with an advantage?

讓你在六角白銀的拍賣上能佔優勢？

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

先想像一下，如果高登 跟你一樣出價 x 顆寶石，形成平手，

by betting some amount X with a tie.

讓你贏了第二件六角黃金。

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

你就可以把你剩下的一切 都押在第三個六角黃金上，

and he'd have to match you to block.

他得出同價才能阻止你。

So if you could bid 51 on the third gold,

所以，如果你能出價五十一顆 寶石買第三個六角黃金，

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

進入六角白銀的拍賣時， 你就有五十一比四十八的優勢，

你知道這樣你就贏定了。

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

未知數 X 的解法：

為了在第三輪時擁有五十一顆寶石，

you should bid at most 24 on round two.

在第二輪你最高只能 出價二十四顆寶石。

But what about the other possibility,

但另一種可能性呢？

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

如果你出價二十四顆寶石，高登出價 更高取得了第二個六角黃金，

would this strategy still work?

這項策略還行得通嗎？

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

要取得第二個六角黃金， 他最少要出二十五顆寶石，

making the total 75 to Gordon’s 74.

這樣你共有七十五顆寶石， 高登只有七十四顆。

No one would then bid on round three,

第三輪就沒有人會出價，

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

因為你們都已經成功阻止對方 取得三個六角黃金了。

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

接下來，你可以 每次出價二十五顆寶石，

贏得三個六角白銀。

The bidding war was close,

拍賣戰爭即將開始，但你的足智多謀

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

讓你能在推論鏈上領先一環，

and the silver tri-source is yours.

三個六角白銀的力量來源 就歸你所有了。

Now... what will you do with it?

那現在……你要拿它來做什麼？