行星游行

Planetary Parade

专题: Probability / 概率
难度: L8
来源: Jane Street

题目详情

本月初我们正在享受行星巡游在地球上，但我的外星朋友来自遥远的皮克诺星球（很好地建模为完美的球体）并没有留下太深刻的印象。他们的单星系统除了 Pyrknot 之外还有六颗行星每晚这些行星独立出现的混乱轨道皮克诺斯天空中均匀随机的位置（行星不是白天可见）。如果在某个特定时刻存在某个地方在 Pyrknot 上，所有六颗行星都是可见的，然后从我朋友的他们在表面上的位置也有 $\alpha$ 概率看到所有行星。

我的朋友正在考虑建造一座塔来提高他们的机会看到这些行星游行。假设塔楼允许观看从地球上至少一个位置可以看到的每一个天体 Pyrknot 的表面距底部的距离r 塔。在 r 与半径 R 相比较小的情况下 Pyrknot，从行星上观看行星游行的新概率塔顶的线性近似为* $\alpha$ * + $\beta$ ·(r/R)。查找准确地说， $\alpha$ 和 $\beta$ 。

We are enjoying a planetary parade during the beginning of this month here on Earth, but my alien friend from the distant planet Pyrknot (well modeled as a perfect sphere) isn’t too impressed. Their single-star system has six planets other than Pyrknot with such chaotic orbits that each night these planets independently appear in uniformly random locations in the Pyrknothian sky (the planets aren’t visible in daylight). If at a given moment there exists somewhere on Pyrknot that all six planets are visible, then from my friend’s position on the surface they have an α probability of also seeing all of the planets.

My friend is considering building a tower to improve their chances of seeing these planetary parades. Assume the tower allows the viewing of every celestial body that could be seen from at least one location on the surface of Pyrknot less than a distance r from the base of the tower. In the limit of r being small compared to the radius R of Pyrknot, the new probability of seeing the planetary parade from the top of the tower is linearly approximated by α + β·(r/R). Find α and β in exact terms.

解析

这个月我们必须计算一些棘手的概率，涉及皮克诺特恒星系统中的行星可观测性。我们被告知皮克诺特的六颗“共行星”在天空，并且在白天不可见（所以我们还需要处理恒星作为第七天体，不能在天空中存在一段时间行星游行发生）。因此，为了找到 $\alpha$ ，我们可以推理以下方式：

从 Pyrknot 表面的每一点，人们都可以看到恰好一半的面积。天空的整个球体，因此每个天体都定义了一个大圆将皮克诺特分成可以看到身体的一半不能的一半。现在，这些大圆圈切碎了表面 Pyrknot 转换为具有多个顶点 V、边 E 和的图满足欧拉特征公式的面（或区域）F 对于 V - E + F = 2 的球体。我们知道

所有大圆对都在两个顶点相交，因此 V = 2 * (7 * 6 / 2) = 42。
七个大圆中的每一个都从这些顶点分为 12 条边，因此 E = 7 * 12 = 84。
因此 F = 2 - V + E = 2 - 42 + 84 = 44。

所以你最终会得到 44 个不同的区域，每个区域都会看到不同的行星的子集忽略了这样一个事实：当它们的恒星处于可见，没有行星可见。这些区域中的一个或零个可以在特定时刻看到行星游行，因此有条件区域看到一个，您有 1/44 位于与该区域相同的区域内某人（这些区域在对称性方面的预期面积都相等）。

现在，如果你建造一座可以看到任何天体的塔从表面上 r 距离处可见 Pyrknot，首先你要向外推进边界区域r，因此提高你的概率与该区域的周长。但也不完全是，因为 1/7 该区域的边界（预期）是由白天/夜晚造成的 Pyrknot 的边界，如果你的塔靠近这个边界，你可以实际上毁了在你的塔顶观看行星游行在白天。哎呀！

该区域的预期周长仅为

(1/44) * [所有区域的总周长] = 1/44 * [14 * 2 $\pi$ R] = (7/11) $\pi$ R。

现在该周长的 6/7 扩展了距离 r 以及该周长的 1/7 周长收缩距离r。首先订购绝对面积差为

(6/11) $\pi$ R⋅r - (1/11) $\pi$ R⋅r = (5/11) $\pi$ R⋅r。

作为 Pyrknot 整个面积的百分比 4 $\pi$ R^{2}，此差异为

[(5/11) $\pi$ R⋅r] / [4 $\pi$ R^{2}] = (5/44)(r/R)

因此 $\alpha = \frac{1}{44}$ 和 $\beta = \frac{5}{44}$ 。

恭喜求解者能够推理出这个棘手的天文设置！

Original Explanation

This month we had to compute some tricky probabilities involving planetary observability in the Pyrknot star system. We are told that the six “coplanets” to Pyrknot have indepedent uniform positions in the sky, and are not visible in daylight (so we also need to treat the star as a seventh celestial body that can’t be in the sky for a planetary parade to occur). So to find α we can reason in the following way:

From each point on Pyrknot’s surface one can see exactly half of the full sphere of the sky, and so each of the celestial bodies defines a great circle dividing Pyrknot into the half that can see that body and the half that cannot. Now, these great circles chop up the surface of Pyrknot into a graph with a number of vertices V, edges E, and faces (or regions) F, that satisfy the Euler characteristic formula for a sphere of V - E + F = 2. We know

All pairs of great circles intersect each other at two vertices, so V = 2 * (7 * 6 / 2) = 42.
Each of the seven great circles is divided into 12 edges from these vertices, so E = 7 * 12 = 84.
Therefore F = 2 - V + E = 2 - 42 + 84 = 44.

So you end up with 44 different regions that each would see a distinct subset of the planets ignoring the fact that when their star is visible no planets can be seen. Either one or zero of those regions can see a planetary parade at a given moment and so conditional on one region seeing one, you are 1/44 to be inside the same region as that someone (the regions are all equal in expected area by symmetry).

Now, if you build a tower that allows you to see any celestial body visible from a point up to r distance away on the surface of Pyrknot, to first order you are pushing outward the borders of the region by r and so improve your probability proportional to the length of the perimeter of the region. But not quite, because 1/7 of the boundary of the region (in expectation) is caused by the day/night boundary on Pyrknot, and if your tower is near this boundary you could actually ruin seeing a planetary parade by the top of your tower being in daylight. Egads!

The expected perimeter of the region is just

(1/44) * [total perimeter of all regions] = 1/44 * [14 * 2πR] = (7/11)πR.

Now 6/7 of that perimeter expands by a distance r and 1/7 of that perimeter contracts by a distance r. To first order the absolute difference in area is

(6/11)πR⋅r - (1/11)πR⋅r = (5/11)πR⋅r.

As a percentage of the full area of Pyrknot, 4πR², this difference is

[(5/11)πR⋅r] / [4πR²] = (5/44)(r/R)

And thus α = 1/44 and β = 5/44.

Congrats to the solvers able to reason out this tricky astronomical setup!