网络安全:数论初步.ppt

  1. 1、本文档共54页,可阅读全部内容。
  2. 2、有哪些信誉好的足球投注网站(book118)网站文档一经付费(服务费),不意味着购买了该文档的版权,仅供个人/单位学习、研究之用,不得用于商业用途,未经授权,严禁复制、发行、汇编、翻译或者网络传播等,侵权必究。
  3. 3、本站所有内容均由合作方或网友上传,本站不对文档的完整性、权威性及其观点立场正确性做任何保证或承诺!文档内容仅供研究参考,付费前请自行鉴别。如您付费,意味着您自己接受本站规则且自行承担风险,本站不退款、不进行额外附加服务;查看《如何避免下载的几个坑》。如果您已付费下载过本站文档,您可以点击 这里二次下载
  4. 4、如文档侵犯商业秘密、侵犯著作权、侵犯人身权等,请点击“版权申诉”(推荐),也可以打举报电话:400-050-0827(电话支持时间:9:00-18:30)。
查看更多
* The actual RSA encryption and decryption computations are each simply a single exponentiation mod (n). Note that the message must be smaller than the modulus. The “magic” is in the choice of the exponents which makes the system work. * Can show that RSA works as a direct consequence of Euler’s Theorem, so that raising a number to power e then d (or vica versa) results in the original number! * Here walk through example RSA key generation using “trivial” sized numbers. Selecting primes requires the use of a primality test. Finding d as inverse of e mod ?(n) requires use of Euclid’s Inverse algorithm (see Ch4) * Then show that the encryption and decryption operations are simple exponentiations mod 187. Rather than having to laborious repeatedly multiply, can use the square and multiply algorithm with modulo reductions to implement all exponentiations quickly and efficiently (see next). * To perform the modular exponentiations, you can use the “Square and Multiply Algorithm”, a fast, efficient algorithm for doing exponentiation. The idea is to repeatedly square the base, and multiply in the ones that are needed to compute the result, as found by examining the binary representation of the exponent. * State here one version of the “Square and Multiply Algorithm”, from Stallings Figure 9.7. * * To speed up the operation of the RSA algorithm using the public key, can choose to use a small value of e (but not too small, since its then vulnerable to attack). Must then ensure any p or q chosen are relatively prime to the fixed e (and reject and find another if not), for system to work. * To speed up the operation of the RSA algorithm using the private key, can use the Chinese Remainder Theorem (CRT) to compute mod p q separately, and then combine results to get the desired answer. This is approx 4 times faster than calculating “C^d mod n” directly. Note that only the owner of the private key details (who knows the values of p q) can do this, but of course that’s exactly w

文档评论(0)

shaoye348 + 关注
实名认证
内容提供者

该用户很懒,什么也没介绍

1亿VIP精品文档

相关文档