で、むしろ時枝記事に近いのは、君が>>295（>>304）で紹介した下記の方が、時枝に近いだろう ここでは、任意の関数f(x)の任意の貴方の選ぶ１点（”You pick an x ∈ Ｒ”）を、” whatever f Bob picked, you will win the game with probability 1!”、”it’s arbitrary: it doesn’t have to be continuous or anything”の条件で当てられるとあるよ
Ｎ⊂Ｒだから、”You pick an n ∈ Ｎ”とすれば、時枝記事の場合を含むことになろう で、時枝記事のように、どこの箱が当たるか分らず、また確率99/100に対して、これは自分で選んだｘであり、”with probability 1!”だから、こちらの解法がよほど優れている
https://xorshammer.com/2008/08/23/set-theory-and-weather-prediction/ SET THEORY AND WEATHER PREDICTION XOR’S HAMMER Some things in mathematical logic that I find interesting WRITTEN BY MKOCONNOR Blog at WordPress.com. AUGUST 23, 2008 (抜粋) Here’s a puzzle: You and Bob are going to play a game which has the following steps.
1)Bob thinks of some function f： Ｒ → Ｒ (it’s arbitrary: it doesn’t have to be continuous or anything). 2)You pick an x ∈ Ｒ. 3)Bob reveals to you the table of values {(x0, f(x0))｜ x0 ≠ x } of his function on every input except the one you specified 4)You guess the value f(x) of Bob’s secret function on the number x that you picked in step 2.
You win if you guess right, you lose if you guess wrong. What’s the best strategy you have? This initially seems completely hopeless: the values of f on inputs x0 ≠ x have nothing to do with the value of f on input x, so how could you do any better then just making a wild guess? In fact, it turns out that if you, say, choose x in Step 2 with uniform probability from [ 0,1 ], the axiom of choice implies that you have a strategy such that, whatever f Bob picked, you will win the game with probability 1! つづく 0052現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2017/11/30(木) 22:28:15.89ID:IqNIthYM>>51 つづき
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) 0053現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2017/11/30(木) 22:29:07.23ID:IqNIthYM>>52 つづき
先に私の見解を書いておくが、ピエロくんの紹介してくれた >>312 PDF が参考になるね（＾＾ The Mathematics of Coordinated Inference: A Study of Generalized Hat Problems (Developments in Mathematics) 2013 edition by Hardin, Christopher S., Taylor, Alan D.
P9 ”In Chapter 7 we start to move further away from the hat problem metaphor and think instead of trying to predict a function's value at a point based on knowing (something about) its values on nearby points. The most natural setting for this is a topological space and if we wanted to only consider continuous colorings, then the limit operator would serve as a unique optimal predictor. But we want to consider arbitrary colorings. Thus we have each point in a topological space representing an agent and if f and g are two colorings, then f ≡a g if f and g agree on some deleted neighborhood of the point a. It turns out that an optimal predictor in this case is wrong only on a set that is "scattered" (a concept with origins going back to Cantor). Moreover, this predictor again turns out to be essentially unique, and this is the main result in Chapter 8.”
１）下記、XOR’S HAMMERのYou and Bobのpuzzleを、任意関数の数当て解法としよう。 記 (>>471より) https://xorshammer.com/2008/08/23/set-theory-and-weather-prediction/ SET THEORY AND WEATHER PREDICTION XOR’S HAMMER Some things in mathematical logic that I find interesting WRITTEN BY MKOCONNOR Blog at WordPress.com. AUGUST 23, 2008 (抜粋) Here’s a puzzle: You and Bob are going to play a game which has the following steps.
２．任意関数の数当て解法は、射程として、可算無限個数列の数当て解法を含んでいるんだ。それを示そう １）XOR’S HAMMERの任意関数の数当て解法は、”In Step 2, choose x with uniform probability from [ 0,1 ].”で、”Thus, you win with probability 1 no matter what function Bob selects.”なのだから ２）やり方は、>>483に書いたように、時枝の可算無限個との対応は、1/1,1/2,1/3,・・・1/n,・・・とすれば、全て[0,1]内の実数と対応がつく ３）数列 s = (s1，s2，s3 ，・・・,sn，・・・)から、 f(1)=s1，f(1/2)=s2，f(1/3)=s3 ，・・・,f(1/n)=sn，・・・となる関数f(x)を作れば良い。 関数はなんでも良いので、簡単に例えばf(1/2)とf(1/3)とを直線で結ぶ これで、時枝の可算無限個を、関数に埋め込めたので、XOR’S HAMMERの任意関数の数当て解法が適用できる ３）”you”は、好きな”1/n”を選べば、XOR’S HAMMERの任意関数の数当て解法で、当たる確率１だ
つづく
注）ここ、「“with uniform probability from [ 0,1 ].”を除いて、もとの問題設定通り、任意にｘを選べるとすれば、」とするのが正確だったね。 “with uniform probability from [ 0,1 ].”だと、任意にｘを選べないから。（＾＾ 0056現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2017/11/30(木) 22:32:09.79ID:IqNIthYM sage 0057現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2017/11/30(木) 22:32:29.83ID:IqNIthYM sage 0058現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2017/11/30(木) 22:32:35.74ID:IqNIthYM>>55 つづき
３．さて、XOR’S HAMMERの任意関数の数当て解法が、関数論の数理に反していることは明白だ ”Bob thinks of some function f： Ｒ → Ｒ (it’s arbitrary: it doesn’t have to be continuous or anything).”（>>471より） なのだから、解析関数でもなく、まして、連続でもない関数の値f(a)は、a以外の点の関数値が分かったところで、関数値f(a)は決まらない だから、XOR’S HAMMERの任意関数の数当て解法は、数理ではなくパズルであって、「選択公理と同値類を使えば、こんな奇妙は結論がもっともらしく見える」というところが面白いのだ
