翻訳と辞書 Words near each other ・ Baby, We're Really in Love ・ Baby, What a Big Surprise ・ Baby, What About You ・ Baby, You Don't Wanna Know ・ Baby, You Knock Me Out ・ Baby, You're a Rich Man ・ Baby, You're Something ・ Baby-Baby-Baby ・ Baby-Brousse ・ Baby-Face Mouse ・ Baby-faced Assassin ・ Baby-led potty training ・ Baby-led weaning ・ Baby-Sitters Little Sister ・ Baby-Sitting Is a Dangerous Job ・ Baby-step giant-step ・ Baby-Towarzystwo ・ Baby...You're Bummin' My Life out in a Supreme Fashion ・ Baby.com.br ・ Babyak ・ Babyak dialect ・ Babybel ・ Babybird ・ Babybjörn ・ Babybug ・ Babycakes ・ BabyCenter ・ Babych ・ BabyChakra ・ Babycham Dictionary Lists mini英和辞書 mini和英辞書 Webster 1913 Latin-English FOLDOC Wikipedia English ウィキペディア

翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Baby-step giant-step ： ウィキペディア英語版 Baby-step giant-step In group theory, a branch of mathematics, the baby-step giant-step is a meet-in-the-middle algorithm computing the discrete logarithm. The discrete log problem is of fundamental importance to the area of public key cryptography. Many of the most commonly used cryptography systems are based on the assumption that the discrete log is extremely difficult to compute; the more difficult it is, the more security it provides a data transfer. One way to increase the difficulty of the discrete log problem is to base the cryptosystem on a larger group. ==Theory== The algorithm is based on a space-time tradeoff. It is a fairly simple modification of trial multiplication, the naive method of finding discrete logarithms. Given a cyclic group $G$ of order $n$, a generator $\backslash alpha$ of the group and a group element $\backslash beta$, the problem is to find an integer $x$ such that : $\backslash alpha^x\; =\; \backslash beta\backslash ,.$ The baby-step giant-step algorithm is based on rewriting $x$ as $x\; =\; im\; +\; j$, with $m\; =\; \backslash left\backslash lceil\; \backslash sqrt\; \backslash right\backslash rceil$ and and . Therefore, we have: :$\backslash beta(\backslash alpha^)^i=\backslash alpha^j\backslash ,.$ The algorithm precomputes $\backslash alpha^j$ for several values of $j$. Then it fixes an $m$ and tries values of $i$ in the left-hand side of the congruence above, in the manner of trial multiplication. It tests to see if the congruence is satisfied for any value of $j$, using the precomputed values of $\backslash alpha^j$. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Baby-step giant-step」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.