Why RSA? A Pedagogical Comment

IF 1.5 Q3 EDUCATION & EDUCATIONAL RESEARCH Journal of Information Technology Education-Innovations in Practice Pub Date : 2017-01-01 DOI:10.12988/JITE.2017.61248
P. Olague, O. Kosheleva, V. Kreinovich
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Abstract

One of the most widely used cryptographic algorithms is the RSA algorithm in which a message m encoded as the remainder c of m modulo n, where n and e are given numbers – forming a public code. A similar transformation c mod n, for an appropriate secret code d, enables us to reconstruct the original message. In this paper, we provide a pedagogical explanation for this algorithm. 1 RSA Algorithm: A Pedagogical Puzzle RSA algorithm: a brief reminder. In many computer transaction, the communicated message is encoded, to avoid eavesdropping. This happens, e.g., every time a credit card information is passed over to some website. In most of such cases, a special RSA algorithm is used to encode the message m; see, e.g., [1]. In this algorithm, two specially selected and publicly available numbers n and e are used to encode the message. The encoded message c has the form of the remainder c = m mod n. The number n is usually at least 100 decimal digits long, and the number e is similarly large. For such large numbers, it is not feasible to compute m simply as m · . . . ·m, by starting with m and e− 1 times multiplying the result by m. Instead, the following much faster algorithm is performed. First, the number e is represented in the binary form, as the sum of powers of two: e = 21 + 22 + . . . + 2p for some k1 > k2 > . . . > kp. For example, 1110 is represented as 10112 = 2 3 + 2 + 2 = 8 + 2 + 1.
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为什么RSA ?教学评论
最广泛使用的加密算法之一是RSA算法,其中消息m编码为m模n的余数c,其中n和e是给定的数字-形成公共代码。对于适当的密码d,进行类似的变换c mod n,使我们能够重建原始消息。在本文中,我们对该算法提供了一个教学解释。RSA算法:一个简短的提示。在许多计算机交易中,通信信息都是经过编码的,以避免被窃听。例如,每次将信用卡信息传递到某个网站时都会发生这种情况。在大多数情况下,使用特殊的RSA算法对消息m进行编码;看,例如,b[1]。在该算法中,使用两个特别选择且公开可用的数字n和e对消息进行编码。编码后的消息c具有余数c = m mod n的形式。数字n通常至少有100位十进制数字长,数字e也同样大。对于如此大的数字,将m简单地计算为m·…是不可行的。·m,从m和e - 1开始乘以m。相反,执行以下更快的算法。首先,数字e以二进制的形式表示,即2的幂的和:e = 21 + 22 +…+ 2p对于一些k1 > k2 >…> kp。例如,1110表示为10112 = 2 3 + 2 + 2 = 8 + 2 + 1。
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来源期刊
CiteScore
1.90
自引率
33.30%
发文量
5
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