By Song Y. Yan
RSA is a public-key cryptographic method, and is the main well-known and widely-used cryptographic process in cutting-edge electronic global. Cryptanalytic assaults on RSA, a certified booklet, covers just about all significant recognized cryptanalytic assaults and defenses of the RSA cryptographic approach and its versions.
Since RSA relies seriously on computational complexity thought and quantity thought, historical past details on complexity idea and quantity thought is gifted first. this can be via an account of the RSA cryptographic method and its variants.
Cryptanalytic assaults on RSAis designed for a certified viewers composed of practitioners and researchers in undefined. This ebook is usually appropriate as a reference or secondary textual content booklet for complicated point scholars in desktop science.
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Additional resources for Cryptanalytic Attacks on RSA
In around 200BC, the Chinese mathematician Sun Tsu proposed a problem in his classic three-volume mathematics book Mathematical Manual: find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 5, and a remainder of 2 when divided by 7. In modern algebraic language, it is to find the smallest positive integer satisfying the following systems of congruences: x ≡ 2 (mod 3), x ≡ 3 (mod 5), x ≡ 2 (mod 7). The general form of the problem is as follows. 11 (The Chinese Remainder Theorem (CRT)).
As commented by Prof R. P. Brent of Australian National University (see ScienceWise@ANU – Science and Engineering at ANU, Vol 2, No 6, 2005, Page 4): One of the hardest things to prove is the difficulty of a problem. A problem is usually considered to be hard if no one can solve it despite a lot of people trying over a long time. But that’s not the same thing as proving it is difficult. It just says that no-one has been clever or lucky enough to come up with a solution. Thus, it is not easy to prove that a conjectured hard problem such as the IFP problem is indeed a hard problem.
Find x and y in gcd(1281, 243) = 1281x+243y. We perform the computation for gcd(a, b) at the left-hand side and in the same time, the calculation for (x, y) at the right-hand side as follows. 1281 = 243 = 66 = 45 = 21 = 243 · 5 + 66 ⇐⇒ 66 = 1281 − 243 · 5 66 · 3 + 45 ⇐⇒ 45 = 243 − 66 · 3 = 243 − 3(1281 − 243 · 5) = 16 · 243 − 3 · 1281 45 · 1 + 21 ⇐⇒ 21 = 66 − 45 · 1 = 1281 − 243 · 5 − 16 · 243 + 3 · 1281 = 4 · 1281 − 21 · 243 21 · 2 + 3 ⇐⇒ 3 = 45 − 21 · 2 = 16 · 243 − 3 · 1281 − 2(4 · 1281 − 21 · 243) = −11 · 1281 + 58 · 243 ↑ ↑ ↑ ↑ x a y b 3·7+0 The above process is well suited for computer programming, but of course, if the calculation of (x, y) are by hand, the following (bottom-up) process is convenient.