真實宣告的XYZ選定 - 線上
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By Freda
at 2018-10-28T22:52
at 2018-10-28T22:52
Table of Contents
※ 引述《evildark (莫裝逼, 裝逼遭雷劈)》之銘言:
: ※ 引述《kuoyipong (petohtalrayn)》之銘言:
: : 真實宣告
: : 充能次數是XYZ三數的總和,XYZ隨機分配
: : 實際上這個分配的方式我覺得比想像中平均,不知道有沒有人也跟我一樣的感覺
: : 我原本的想像中,XYZ的分配長這樣:
: : 在總長度為S的繩子上隨機找兩個點切割,每個點被選定的機率相同
: : 三段的長度就分別為X, Y, Z
: 根據NGA的程式碼源內容
: 假設充能N次
: 每一次都是X.Y.Z三選1
: 選N次
nga來源在哪
我看到的是有呼叫一個函數決定xyz,但那篇沒講那個函數裡面怎麼決定的
至於分布形式我後來隨便算了一下,不管是每個三元組均勻分布還是xyz常態分布,其實
出來的結果都差不了多少
所以我就懶得去檢驗了 Kappa
--
: ※ 引述《kuoyipong (petohtalrayn)》之銘言:
: : 真實宣告
: : 充能次數是XYZ三數的總和,XYZ隨機分配
: : 實際上這個分配的方式我覺得比想像中平均,不知道有沒有人也跟我一樣的感覺
: : 我原本的想像中,XYZ的分配長這樣:
: : 在總長度為S的繩子上隨機找兩個點切割,每個點被選定的機率相同
: : 三段的長度就分別為X, Y, Z
: 根據NGA的程式碼源內容
: 假設充能N次
: 每一次都是X.Y.Z三選1
: 選N次
nga來源在哪
我看到的是有呼叫一個函數決定xyz,但那篇沒講那個函數裡面怎麼決定的
至於分布形式我後來隨便算了一下,不管是每個三元組均勻分布還是xyz常態分布,其實
出來的結果都差不了多少
所以我就懶得去檢驗了 Kappa
--
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