Which is Bigger - The Number of Atoms in Universe or Complexity of Go? 围棋的走法和宇宙原子总量谁更多?

in #busy6 years ago (edited)

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The board of Go Game has 19x19 lines. Players place stone on the line intersections. Therefore the complexity of all state space is pow(3, 361) because each intersection could have three 3 possibilities: black, white or nothing.

Researchers have estimated the number of Atoms in universe is somewhat between 10^78 to 10^82 [Ref] let's assume it is 10^80.

So, which is bigger?

Of course, with the help of computers, these two numbers can be calculated first and then compared. But we human don't (and can't) calculate the actual values of both. Instead, we do a little math here.

Let's denote N is the number of atoms in universe i.e. 3^361 and M, the all state space complexity of Go i.e. 10^80. We know which is bigger by comparing the value of M/N with one.

Let's take lg(M/N), we know that equals lg(M) - lg(N) so

    lg(3^361) - lg(10^80)
= 361*lg(3) - 80*lg(10)
= 172.24 - 80
= 92.24 

Therefore, M/N = 10^92.24 that means M is a lot larger than N! The Go State complexity is way more huge than the number of atoms in universe. Even the computers cann't bruteforce all the possibilities!

Reposted to my blog: https://helloacm.com/which-is-bigger-the-number-of-atoms-in-universe-or-complexity-of-go/

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围棋的棋盘是19乘于19条线,棋手在线交叉的位置上放棋子。每个地方可以有三种状态:白的、红的、或者为空。所以不考虑棋子有效性,围棋所有棋盘的数目是 3的361次方。

科学家们估计了宇宙中原子的数目大约在 10的78次方到10的82次方间 [Ref] 我们就姑且认为是 10的80次方吧。

问题来,哪一个更多呢?

当然,计算机可以先分别算出两个值的大小直接粗暴的比大小。不过,我们却不能。相反,我们可以简单的推算一下。

假定围棋状态数为M,宇宙原子 数为N,那么我们只需要比较M/N和1的大小即可。

取LOG 10为底 lg(M/N),转换一下变成 lg(M) - lg(N)

    lg(3^361) - lg(10^80)
= 361*lg(3) - 80*lg(10)
= 172.24 - 80
= 92.24 

所以, M/N = 10^92.24 也就是说 M 比N 要大得多。围棋的状态数远远比宇宙的原子数多,怪不得计算机穷举是不可能的!

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Sort:  

是“3的361次方”,还是“3的361序”组合排列之和?

333*......3 // 361个3

“序”,数论的概念

不懂。

红色表示涨,绿色表示跌。😢

英文的MN定義寫反了...

好眼力, fixed!

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