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According to legend, when this planet was young and molten, three galactic terraformers shaped it into a paradise.
传说中，当这个星球还年轻并炽热的时候，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, but left the source of their power behind:
他们改造完星球后，开始了寻找新的星球的旅程，但是他们把他们的力量源泉留在了星球里：
three powerful golden hexagons, hidden within dungeons full of traps and monsters.
三个有着强大力量的黄金六角宝石，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, they could reinvent the world however they saw fit. That was thousands of years ago.
如果有人集齐了这三个宝石，他就可以依意愿来重新改造这个世界。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 and enslaving the world to his will.
今天，你得知一个叫戈登的邪恶巫师要集齐这三个宝石，并奴役这个世界。
So you set off on a quest to get them first, adventuring through fire, ice, and sand.
所以你决定抢在戈登之前拿到这三个宝石，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. At the end of the third dungeon, you find a note inviting you to Dongle's castle.
这个人不是戈登，他是一个叫当构的商人。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 you've collected in your travels, arriving just moments before Gordon, who also has 99 gems.
你带着在冒险时得到的99个宝石，到达了当构的城堡，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, but he's used them to create 5 silver hexagons, just as powerful as their golden counterparts.
当构不止拿到了仅有的三个六边金宝石，他还用金宝石创造了五个六边银宝石，并且银宝石和金宝石有着相同的力量。
Why did Dongle do all this? Because there's one thing he loves above all else: auctions.
为什么当构要这么做呢？因为当构最喜欢拍卖。
You and the evil wizard will compete to win the hexagons, starting with the three golden ones, making one bid for each item as it comes up.
你和邪恶的巫师要去竞拍这些六边宝石，从三个金宝石开始，一个物品能出价一次。
The winners of ties will alternate, starting with you. Whoever first collects a trio of either golden or silver hexagons can use their power to recreate the world.
如果出价相同，胜者会轮换。从你开始，第一个拿到三个相同颜色宝石的人，（不论金色还是银色），可以用它们的力量重塑这个世界。
You've already bid 24 gems on the first, when you realize that your rival has a dastardly advantage: a mirror that lets him see what you're bidding.
当你第一轮已经出了24个宝石的时候，你发现你的对手用了卑鄙的手段：一个可以让他看到你出价的镜子。
He bids zero, and you win the first hexagon outright. What's your strategy to win a matching trio of hexagons before your rival?
他的出价是0个宝石，你赢得了你一个宝石。你要怎么在戈登之前集齐三个六边宝石呢？
Dongle's dangled a difficult dilemma. Do you spend big to try to win the golden hexagons outright? Save as much as possible for silver? Or something in-between?
当构出了一个让人进退两难的难题，你要花大钱试图直接赢得全部的金六边宝石吗？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 to make sure that no matter what you bid on the second gold, he can bid one more and block you.
戈登可以用他的神奇镜子和99个宝石，去确保不论你对第二个金六边宝石的出价是多少，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 to guarantee that you'll win on the silvers?
那么问题来了--如何让戈登在金六边宝石上花上够多的钱，从而确保你能赢下银色的六边宝石呢？
Here's a hint. Let's say at the start of the silver auctions you had a one gem advantage, such as 9 to 8.
给你点提示。如果在开始竞拍银六边宝石的时候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, so could you divide your gems into three groups of three and win?
你需要赢下三次竞拍，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, where Gordon wins every tie.
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，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 is win two silver hexagons, and have two gems left -- which you'll beat with three bids of 3.
如果你每次出价三个宝石，戈登最多能赢下两个宝石，并且只剩两个宝石，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 will lead to victory by the same logic.
一样的逻辑，只要你的宝石数可以除以三，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 so you go into silver with an advantage?
知道了这个后，如何让戈登把更多钱花在金六边宝石上，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 by betting some amount X with a tie.
试想戈登出了和你一样数量的宝石，让你赢了第二个金宝石。
You could then bid everything you have left on the third gold hexagon, and he'd have to match you to block.
接着不管你在第三个金宝石上出价多少，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, you'd go into silver with a 51 to 48 advantage, which you know you can win.
所以如果你在第三个金宝石上出价51个宝石，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, you should bid at most 24 on round two.
经过计算，如果你想在第三个金宝石竞拍时有51个宝石，你第二轮最多出24个宝石。
But what about the other possibility, where Gordon wins the second gold against your bid of 24 -- would this strategy still work?
但是如果戈登在第二轮出价超过24个宝石，这个计策还有用吗？
The least he could bid to win the second gold is 25, making the total 75 to Gordon's 74.
如果戈登想赢第二轮，他最少出25个宝石，这样的话你就有75个宝石，戈登有74个。
No one would then bid on round three, since you've each blocked the other from getting three golds.
紧接着没有人赢得第三轮，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, but your ingenuity kept you one link ahead in the chain of inference, and the silver tri-source is yours. Now... what will you do with it?
竞拍结束了，但是你的聪明才智使你比戈登更快一步，你得到了三个银六边宝石。那么现在，你要用它们做什么呢？
