By Gilbert Baumslag

Cryptography has turn into crucial as financial institution transactions, bank card infor-mation, contracts, and delicate scientific details are despatched via inse-cure channels. This ebook is worried with the mathematical, specifically algebraic, features of cryptography. It grew out of many classes offered through the authors over the last 20 years at quite a few universities and covers a variety of issues in mathematical cryptography. it really is essentially geared in the direction of graduate scholars and complicated undergraduates in arithmetic and laptop technology, yet can also be of curiosity to researchers within the area.

Besides the classical tools of symmetric and personal key encryption, the ebook treats the math of cryptographic protocols and several other precise issues such as

- Group-Based Cryptography
- Gröbner foundation tools in Cryptography
- Lattice-Based Cryptography

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Additional info for A Course in Mathematical Cryptography

Sample text

To accomplish this, some form of padding is used, that is adding meaningless letters to the end of the message. There are different padding schemes and we refer to [Buc] for a discussion of these. Very simple block ciphers are obtained by using a fixed permutation or a fixed substitution. They are called permutation ciphers (or transpositions) and substitution ciphers, respectively. Such cryptosystems were used historically, but as we explained in Chapter 1, they are not secure. Therefore, to increase security, modern block ciphers are created by iterating the basic operations of substitution and permutation.

9 The following message has been encrypted with the Vernam one-time pad. The ciphertext is: PITTW What is the plaintext?

Mt and get possible values for n1 , . . , nt of the keyword. In 1920, a second test, the Friedmann test was further developed. We assume that N = 26 and the alphabet is {A, B, C, . . } but the method is entirely general. Let (an , . . , an ) be a sequence of length n of letters from the alphabet. Let n1 be the number of A???? s that occur, n2 the number of B???? s and so on. Then: n (n −1) (a) There are (n21 ) = 1 21 pairs of A’s in the sequence, (n22 ) pairs of B’s and so on. (b) There are altogether (n21 ) + ⋅ ⋅ ⋅ + (n226 ) pairs of equal letters.

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