現代数学の系譜工学物理雑談古典ガロア理論も読む46

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48: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2017/11/12(日) 09:02:54.43 ID:cTg/FCp5

>>47 つづき

スレ45 https://rio2016.5ch.net/test/read.cgi/math/1508931882/472
472 自分返信：現代数学の系譜 工学物理雑談 古典ガロア理論も読む[sage] 投稿日：2017/11/06(月) 00:05:26.40 ID:1Au30FRy [6/13]

The strategy is as follows: Let 〜 be the equivalence relation on functions from Ｒ to Ｒ defined by f 〜 g iff for all but finitely many y, f(y) = g(y). Using the axiom of choice, pick a representative from each equivalence class.

In Step 2, choose x with uniform probability from [ 0,1 ].
When, in step 3, Bob reveals {(x0, f(x0)) ｜ x0 ≠ x }, you know what equivalence class f is in, because you know its values at all but one point. Let g be the representative of that equivalence class that you picked ahead of time. Now, in step 4, guess that f(x) is equal to g(x).

What is the probability of success of this strategy?
Well, whatever f that Bob picks, the representative g of its equivalence class will differ from it in only finitely many places.
You will win the game if, in Step 2, you pick any number besides one of those finitely many numbers.
Thus, you win with probability 1 no matter what function Bob selects.
(引用終り)

つづく

http://rio2016.5ch.net/test/read.cgi/math/1510442940/48

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