一个计数谜题最出人意料的答案：英语精读讲义

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

有时候，数学和物理会以一种美好得令人难以置信的方式不谋而合。

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

让我们来玩一场奇怪的数学槌球游戏。

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

我们有两个滑动方块和一面墙。

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

第一个方块一开始以一定的速度从右边滑过来，

while the second one starts out stationary.

而第二个方块最初是静止的。

Being overly idealistic physicists, 語法

作为过度理想主义的物理学家，

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

假设没有摩擦力，并且所有的碰撞都是完全弹性的，

which means no energy is lost.

这意味着没有能量损失。

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

你们中敏锐的人可能会抱怨，这种碰撞不会发出声音，

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

但你在这里的目标是计算发生了多少次碰撞。

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

所以，稍微违背一下那个假设，

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

我想保留一点“咔哒”声，以便更好地引起你对计数的注意。

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

最简单的情况是两个方块质量相同。

The first block hits the second,

第一个方块撞击第二个方块，

transferring all of its momentum, 用詞

转移了它所有的动量，

then the second one bounces off the wall 短語動詞

然后第二个方块从墙上反弹，

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

接着把所有的动量传回给第一个方块，

which then sails off towards infinity.

第一个方块随后向着无穷远处滑去。

Three total clacks.

总共三下“咔哒”声。

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

如果第一个方块的质量是第二个的 100 倍会怎样？

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

我保证我会在适当的时候向你解释所有相关的物理知识。

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

在这里，你该如何预测动态并不完全明显，

but in the spirit of getting to the punchline, 講解

但本着直奔主题的精神，

let's watch what happens.

让我们看看会发生什么。

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

第二个方块会在墙壁和质量是其 100 倍的第一个方块之间不断来回反弹，

like a satisfying game of Breakout, 文化背景

就像一场令人满足的《打砖块》游戏一样，

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

慢慢地、悄无声息地改变第一个方块的动量，使其指向相反的方向。

In total,

总共，

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

在每个方块向无穷远处滑去之前会有 31 次碰撞，

never to be touched again. 語法

它们将再也不会发生接触。

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

如果第一个方块的质量是第二个的 10,000 倍呢？

In that case,

在那种情况下，

there would be quite a few more clacks,

会多出相当多的“咔哒”声，

all happening very rapidly at one point.

全都在某一时刻非常迅速地发生。

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

总共发生 313 次碰撞。

Well, actually, hang on. 語氣意圖

嗯，实际上，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“咔哒”声。

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

如果第一个方块的质量是另一个的 100 万倍，

then again,

那么同样，

with all of our crazy idealistic conditions,

在我们所有疯狂的理想条件下，

almost all of the clacks happened in one big burst,

几乎所有的“咔哒”声都会在一次大爆发中发生，

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

这次总共会导致 3,141 次碰撞。

Perhaps you see the pattern here,

也许你看出了这里的规律，

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

尽管没看出来也情有可原，因为它完全出乎意料。

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

当第一个方块的质量是第二个方块的 100 的某次方倍时，

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

碰撞的总数与圆周率的数字相同。

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

当别人第一次告诉我这件事时，我简直惊呆了。

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

感谢观众 Henry Cavill 向我介绍了这个事实，

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

它最初由数学家 Gregory Galperin 在 1995 年发现，并于 2003 年发表。

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

我之所以喜欢它，部分原因在于，如果有一天要为计算圆周率的算法举办奥运会，

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

那么这个算法一定会因为最优雅和最滑稽的低效而赢得奖牌。

I mean,

我的意思是，

think about the actual algorithm here.

想想这里实际的算法。

Step one,

第一步，

implement a physics engine.

实现一个物理引擎。

Step two,

第二步，

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

选择你想计算的圆周率的位数 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

将其中一个方块的质量设置为 100 的 d 减一次方，

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

然后让它在无摩擦的表面上向质量为 1 的方块移动。

And step four,

第四步，

count all of the collisions.

计算所有的碰撞。

So, for example,

所以，举个例子，

to calculate only 20 digits of pi,

只计算圆周率的 20 位数字，

which fits so cleanly on the screen,

这个数字非常利落地适合在屏幕上显示，

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

一个方块的质量必须是另一个的 1000 亿亿亿亿倍，

which if that small block was 1 kg,

如果那个小方块是 1 千克，

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

意味着大方块的质量大约是银河系中心超大质量黑洞的 10 倍。

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

这意味着你需要计算大约 310 亿亿次碰撞。

And at one point in this virtual process,

在这个虚拟过程中的某一个时刻，

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

“咔哒”声的频率将达到每秒约 1000 亿亿亿亿次。

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

所以，可以说你需要非常好的数值精度才能让它准确运行，

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

并且该算法需要很长时间才能完成。

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

我再次强调，这个过程过度理想化了，

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

完全脱离了现实物理中可能发生的一切。

But of course,

但当然，

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

你们都知道，这并不有趣，并非因为它作为实际计算圆周率的算法的潜力，

or as a pragmatic physics demonstration.

或者作为一种实用的物理演示。

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

令人难以置信的是，为什么圆周率会出现在这里？

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

而且出现的方式如此奇怪。

Its decimal digits are counting something,

它的十进制数字正在计算某种东西，

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

但通常圆周率出现在其精确值描述某些连续事物的时候。

I will show you why this is true.

我会向你展示为什么这是真的。

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

哪里有圆周率，哪里就有一个隐藏的圆。

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

在这种情况下，那个隐藏的圆来自于能量守恒。

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

事实上，你将看到两种不同的方法，

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

每一种都和事实本身一样令人惊叹和惊讶。

Delaying gratification, though, 短語搭配

尽管要延迟满足，

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

我会让你等到下一个视频再看看是怎么回事。

In the meantime,

与此同时，

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

我强烈建议你自己尝试一下，并与他人交流。

It's a hard puzzle,

这是一个很难的谜题，

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

所以招募一些其他聪明的大脑来完成这项任务也无妨。