一道計數謎題最出乎意料的答案：英語精讀講義

Sometimes math and physics conspire in ways that just feel too good to be true. 慣用語

有時候，數學和物理會以一種美好得令人難以置信的方式不謀而合。

Let's play a strange sort of mathematical croquet. 文化背景

讓我們來玩一場奇怪的數學槌球遊戲。

We're going to have two sliding blocks and a wall.

我們有兩個滑動方塊和一面牆。

The first block starts by coming in at some velocity from the right

第一個方塊一開始以一定的速度從右邊滑過來，

while the second one starts out stationary.

而第二個方塊最初是靜止的。

Being overly idealistic physicists, 文法

作為過度理想主義的物理學家，

let's assume that there's no friction and all of the collisions are perfectly elastic, 用詞

假設沒有摩擦力，並且所有的碰撞都是完全彈性的，

which means no energy is lost.

這意味著沒有能量損失。

The astute among you might complain that such collisions would make no sound, 語氣意圖

你們中敏銳的人可能會抱怨，這種碰撞不會發出聲音，

but your goal here is gonna be to count how many collisions take place.

但你在這裡的目標是計算發生了多少次碰撞。

So, in slight conflict with that assumption, 片語搭配

所以，稍微違背一下那個假設，

I want to leave a little clack sound to better draw your attention to that count. 片語搭配

我想保留一點「喀噠」聲，以便更好地引起你對計數的注意。

The simplest case is when both blocks have the same mass.

最簡單的情況是兩個方塊質量相同。

The first block hits the second,

第一個方塊撞擊第二個方塊，

transferring all of its momentum, 用詞

轉移了它所有的動量，

then the second one bounces off the wall 片語動詞

然後第二個方塊從牆上反彈，

and then transfers all of its momentum back to the first,

接著把所有的動量傳回給第一個方塊，

which then sails off towards infinity.

第一個方塊隨後向著無窮遠處滑去。

Three total clacks.

總共三下「喀噠」聲。

What about if the first block was 100 times the mass of the second one?

如果第一個方塊的質量是第二個的 100 倍會怎樣？

I promise I will explain to you all the relevant physics in due course. 片語搭配

我保證我會在適當的時候向你解釋所有相關的物理知識。

It's not entirely obvious how you would predict the dynamics here, 語氣意圖

在這裡，你該如何預測動態並不完全明顯，

but in the spirit of getting to the punchline, 講解

但本著直奔主題的精神，

let's watch what happens.

讓我們看看會發生什麼。

The second one will keep bouncing back and forth between the wall and the first block 100 times its mass, 片語搭配

第二個方塊會在牆壁和質量是其 100 倍的第一個方塊之間不斷來回反彈，

like a satisfying game of Breakout, 文化背景

就像一場令人滿足的《打磚塊》遊戲一樣，

slowly and discreetly redirecting that first block's momentum to point in the opposite direction.

慢慢地、悄無聲息地改變第一個方塊的動量，使其指向相反的方向。

In total,

總共，

there will be 31 collisions before each block is sliding off towards infinity,

在每個方塊向無窮遠處滑去之前會有 31 次碰撞，

never to be touched again. 文法

它們將再也不會發生接觸。

What if the first block was 10,000 times the mass of the second one?

如果第一個方塊的質量是第二個的 10,000 倍呢？

In that case,

在那種情況下，

there would be quite a few more clacks,

會多出相當多的「喀噠」聲，

all happening very rapidly at one point.

全都在某一時刻非常迅速地發生。

Adding up in all to 313 total collisions. 片語動詞

總共發生 313 次碰撞。

Well, actually, hang on. 語氣意圖

嗯，實際上，等一下。

Wait for it. 語氣意圖

等等看。

Wait for it.

再等等看。

Okay, 314 clacks.

好了，314 次「喀噠」聲。

If the first block was 1 million times the mass of the other,

如果第一個方塊的質量是另一個的 100 萬倍，

then again,

那麼同樣，

with all of our crazy idealistic conditions,

在我們所有瘋狂的理想條件下，

almost all of the clacks happened in one big burst,

幾乎所有的「喀噠」聲都會在一次大爆發中發生，

this time resulting in a total of 3,141 collisions.

這次總共會導致 3,141 次碰撞。

Perhaps you see the pattern here,

也許你看到了這裡的規律，

though it's forgivable if you don't since it defies all expectation. 講解

儘管沒看出來也情有可原，因為它完全出乎意料。

When the mass of that first block is some power of 100 times the mass of the second, 用詞

當第一個方塊的質量是第二個方塊的 100 的某次方倍時，

the total number of collisions have the same digits as pi. 文法

碰撞的總數與圓周率的數字相同。

This absolutely blew my mind when it was first shared with me. 慣用語

當別人第一次告訴我這件事時，我簡直驚呆了。

Credit to the viewer Henry Cavill for introducing me to this fact, 表達脈絡

感謝觀眾 Henry Cavill 向我介紹了這個事實，

which was originally discovered by the mathematician Gregory Galperin in 1995 and published in 2003.

它最初由數學家 Gregory Galperin 在 1995 年發現，並於 2003 年發表。

Part of what I love about this is that if ever there were Olympic games for algorithms that compute pi, 講解

我之所以喜歡它，部分原因在於，如果有一天要為計算圓周率的演算法舉辦奧運會，

this one would have to win medals both for being the most elegant and for being the most comically inefficient. 文法

那麼這個演算法一定會因為最優雅和最滑稽的低效而贏得獎牌。

I mean,

我的意思是，

think about the actual algorithm here.

想想這裡實際的演算法。

Step one,

第一步，

implement a physics engine.

實作一個物理引擎。

Step two,

第二步，

choose the number of digits d of pi that you'd like to compute.

選擇你想計算的圓周率的位數 d。

Step three,

第三步，

set the mass of one of the blocks to be 100 to the power d minus one, 用詞

將其中一個方塊的質量設定為 100 的 d 減一次方，

then send it traveling on a frictionless surface towards a block of mass one. 文法

然後讓它在無摩擦的表面上向質量為 1 的方塊移動。

And step four,

第四步，

count all of the collisions.

計算所有的碰撞。

So, for example,

所以，舉個例子，

to calculate only 20 digits of pi,

只計算圓周率的 20 位數字，

which fits so cleanly on the screen,

這個數字非常俐落地適合在螢幕上顯示，

one block would have to have 100 billion billion billion billion times the mass of the other, 文法

一個方塊的質量必須是另一個的 1000 億億億億倍，

which if that small block was 1 kg,

如果那個小方塊是 1 公斤，

means the big one has a mass about 10 times that of the supermassive black hole at the center of the Milky Way. 文法

意味著大方塊的質量大約是銀河系中心超大質量黑洞的 10 倍。

That means you would need to count about 31 billion billion collisions.

這意味著你需要計算大約 310 億億次碰撞。

And at one point in this virtual process,

在這個虛擬過程中的某一個時刻，

the frequency of clacks would be around 100 billion billion billion billion clacks per second.

「喀噠」聲的頻率將達到每秒約 1000 億億億億次。

So, let's just say that you would need very good numerical precision to get this working accurately, 語氣意圖

所以，可以說你需要非常好的數值精度才能讓它準確運行，

and it would take a very long time for the algorithm to complete.

並且該演算法需要很長時間才能完成。

I'll emphasize again that this process is way over-idealized, 用詞

我再次強調，這個過程過度理想化了，

quickly departing from anything that could possibly happen in real physics.

完全脫離了現實物理中可能發生的一切。

But of course,

但當然，

you all know this is not interesting because of its potential as an actual pi computing algorithm

你們都知道，這並不有趣，並非因為它作為實際計算圓周率的演算法的潛力，

or as a pragmatic physics demonstration.

或者作為一種實用的物理演示。

It's mind-boggling because why on earth would pi show up here? 講解

令人難以置信的是，為什麼圓周率會出現在這裡？

And it's in such a weird way, too.

而且出現的方式如此奇怪。

Its decimal digits are counting something,

它的十進位數字正在計算某種東西，

but usually pi shows up when its precise value is describing something continuous.

但通常圓周率出現在其精確值描述某些連續事物的時候。

I will show you why this is true.

我會向你展示為什麼這是真的。

Where there is pi, there is a hidden circle. 表達脈絡

哪裡有圓周率，哪裡就有一個隱藏的圓。

And in this case, that hidden circle comes from the conservation of energy. 用詞

在這種情況下，那個隱藏的圓來自於能量守恆。

In fact, you're going to see two separate methods,

事實上，你將看到兩種不同的方法，

which are each as stunning and surprising as the fact itself. 文法

每一種都和事實本身一樣令人驚嘆和驚訝。

Delaying gratification, though, 片語搭配

儘管要延遲滿足，

I will make you wait until the next video to see what's going on. 文法

我會讓你等到下一個影片再看看是怎麼回事。

In the meantime,

與此同時，

I highly encourage you to take a stab at it yourself and be social about it. 慣用語

我強烈建議你自己嘗試一下，並與他人交流。

It's a hard puzzle,

這是一個很難的謎題，

so it never hurts to recruit some other smart minds to the task. 片語搭配

所以招募一些其他聰明的大腦來完成這項任務也無妨。