未來報告 3

天眼救未來(Next)| The power is in the principles, 5

Principles, aka differential equations, let you see part of the future. 

— Me@2013-07-06 06:05:21 PM

2013.07.08 Monday (c) All rights reserved by ACHK

Multinomial coefficient 2.4

二項式係數 4.4 | Binomial coefficient 4.4

這段改編自 2010 年 7 月 20 日的對話。

假設有 10 個友人,要乘坐計程車去郊遊。總共有兩輛計程車。第一輛車的載客量是 4 人,而第二輛的載客量是 6 人。換而言之,那 10 人要分成兩組乘車。那樣,總共有多少個分配方法呢?

換而言之,從 10 人中抽 4 人出來,組成第一隊樂隊,總共有多少個抽法呢?

在這個情況下,次序很明顯不重要。試想想,假設你從那 10 人中,抽了「ABCE」4 人出來。無論抽的先後次序是「ABCE」,還是「ACBE」,他們所組成樂隊都會「一樣」。兩個情況所組成的音樂組合,你都會視之為「同一隊」樂隊。

但是,如果問題改為:

從 10 人中抽 4 人出來,去參加一個音樂比賽,而沒有其他參賽者的話,總共有多少個可能的比賽排名結果呢?

那樣,被抽了出來的那 4 個人中,不同的人拿冠軍,為之不同的排名,不同的結果。所以,次序需要考慮。運算方面,詳細的版本是:

首先,考慮有「冠、亞、季、殿」軍 4 個空格:

(_)(_)(_)(_)

因為冠軍寶座有 10 個可能的奪得者,所以,第一格是 10:

(10)(_)(_)(_)

其中 1 人奪得冠軍後,亞軍還有 9 個可能的領獎人士:

(10)(9)(_)(_)

如此類推的話,我們就可以推斷到,總共有 5040 個可能的比賽結果:

(10)(9)(8)(7)

= 5040

精簡的版本則是:

題目明確地問,有多少個可能的比賽排名。所以,題目所問的,就相當於:

從 10 人中抽 4 人出來,而次序重要的話,總共有多少個抽法呢?

那是 permutation(排列)。答案明顯是 10_P_4,即是「10 排 4」,等於 5040。

10_P_4 =

10!
——-
(10-4)!

結論是,總共有 5040 個可能的比賽排名。

— Me@2013.07.08

2013.07.08 Monday (c) All rights reserved by ACHK

Looper, 5.2

Paradox 5.3 | Meta-time 4.3 | Cumulative concept of time, 13.3 | Two dimensional time 4.2 | 二次元時間 4.2

In the movie Looper, Young Joe (in the year 2044) influences Old Joe (in the year 2074) in the sense that Young Joe’s every action affects the state of Old Joe, because Old Joe is Young Joe’s future self.

For example, after Young Joe had hurt his own arm, the corresponding wound also appeared on Old Joe’s arm, even though Old Joe had already time-travelled back to the year 2044. 

All of Young Joe’s actions are the causes of Old Joe’s state. Young Joe is in the past of Old Joe.

Old Joe (2074-Joe) = [ …, Young Joe (2044-Joe), … ]

B = [ …, A, … ]

However, Old Joe (2074-Joe) had time-travelled back to the year 2044, meeting the Young Joe.

So, some of Old Joe’s actions would affect Young Joe’s decisions on his own actions. In this sense, Old Joe also influences Young Joe indirectly. Some of Old Joe’s actions are the causes of Young Joe’s state. Part of Old Joe is also in the past of Young Joe.

Young Joe (2044-Joe) = [ …, Old Joe (2074-Joe), … ]

A = [ …, B, … ]

However, it is logically impossible to have both

B is in the past of A 

and

A is in the past of B

just as it is logically almost impossible to have both

D is a part of C

and

C is a part of D

If you insist that it is the case, the only possibility is that

C = D

In this analogy, neither C nor D is really a “part” of another. In the time travel case, neither A nor B is really in the past of another. In other words, A (Young Joe) and B (Old Joe) have no time relationship. Neither’s actions are the causes of the state of another.

The real causes of Young Joe or Old Joe’s states are actually not within the movie story’s timeline. The real causes are the decisions of the author of the story.

— Me@2013-07-03 6:19 PM

2013.07.08 Monday (c) All rights reserved by ACHK