在一個 1*1 的正方形隨機選2個點，如何求這兩個點的期望歐幾里得距離？

02-02

這個好像是Square Picking問題中最簡單的一個了...

因為竟然可以直接積分而不用幾何轉換...

一個比較傳統的做法是這個樣子的:Square Line Picking

這類問題算起來非常煩,極端的繁瑣

不過這個特例其實有比較黑科技的做法...

Concerning $int_0^1 cdots int_0^1 {(x_1^2 + cdots + x_k^2 )} ^{{1 / 2}} dx_1 cdots ,dx_k $ and a Taylor Series Method (1976年的書)研究了對高維超立方體的一般形式...然後得到了一個結論...

這個結論二維情況就是:

$[egin{gathered} int_0^1 {int_0^1 {int_0^1 {int_0^1 {sqrt {{{left( {{x_1} - {x_2}} ight)}^2} + {{left( {{y_1} - {y_2}} ight)}^2}} } } } } d{y_2}d{y_1}d{x_2}d{x_1} hfill \ = int_0^infty {left( {frac{{{ ext{erf}}(x)}}{{3x}}{e^{ - {x^2}}} + frac{4}{{15sqrt pi }}{e^{ - 2{x^2}}} + frac{4}{{15sqrt pi }}{e^{ - {x^2}}}} ight)} {mkern 1mu} dx hfill \ = frac{1}{{15}}left( {sqrt 2 + 2 + 5{{sinh }^{ - 1}}(1)} ight) hfill \ end{gathered} ]$

三維情況就是:

$[egin{gathered} int_0^1 {int_0^1 {int_0^1 {int_0^1 {int_0^1 {int_0^1 {sqrt {{{left( {{x_1} - {x_2}} ight)}^2} + {{left( {{y_1} - {y_2}} ight)}^2} + {{left( {{z_1} - {z_2}} ight)}^2}} } } } } } } d{z_2}d{z_1}d{y_2}d{y_1}d{x_2}d{x_1} hfill \ = int_0^infty {left( { - frac{1}{5}2sqrt pi {e^{ - {x^2}}}{ ext{erf}}{{(x)}^2} + frac{{4{e^{ - 2{x^2}}}{ ext{erf}}(x)}}{{5x}} + frac{{{e^{ - {x^2}}}{ ext{erf}}(x)}}{{5x}} - frac{{12{e^{ - 3{x^2}}}}}{{35sqrt pi }} + frac{{68{e^{ - 2{x^2}}}}}{{105sqrt pi }} + frac{{8{e^{ - {x^2}}}}}{{105sqrt pi }}} ight)} {mkern 1mu} dx hfill \ = frac{1}{{105}}left( {17sqrt 2 - 6sqrt 3 - 7pi + 4 + 21{{sinh }^{ - 1}}(1) + 84{{sinh }^{ - 1}}left( {frac{1}{{sqrt 2 }}} ight)} ight) hfill \ = frac{1}{{105}}left( {17sqrt 2 - 6sqrt 3 - 7pi + 4 - 42log (2) + 21log left( {sqrt 2 + 1} ight) + 84log left( {sqrt 3 + 1} ight)} ight) hfill \ end{gathered} ]$ 同時給出了一個n維中不錯的估計上界:

$[l(n) < frac{{sqrt n }}{3}sqrt {sqrt {1 - frac{3}{{5n}}} + frac{1}{2}} ]$

$frac{2+sqrt{2}+5ln(1+sqrt{2})}{15} approx 0.521$

具體請參考：https://www.youtube.com/watch?v=i4VqXRRXi68 (VERY HARD Puzzle: What Is The Distance Between Two Random Points In A Square?)