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CRYPTOGRAPHY

CRYPTOGRAPHY. AND THE MATH MAJOR Dr. Mihai Caragiu. Mathematics Department Ohio Northern University. Cryptography : the art or science of transforming an intelligible message into one that is unintelligible, and then retransforming that message back to its original form….

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CRYPTOGRAPHY

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  1. CRYPTOGRAPHY AND THE MATH MAJOR Dr. Mihai Caragiu Mathematics Department Ohio Northern University

  2. Cryptography: the art or science of transforming an intelligible message into one that is unintelligible, and then retransforming that message back to its original form…

  3. Mathematics plays a crucial role in cryptography!

  4. 2000 years ago Julius Caesar used a simple substitution cipher (replace each letter of message by a letter a fixed distance – k – away) Say, for example, k = 3. Then SCIENCE transforms into: VFLHQFH This is not a secure cryptosystem! Still, there is some mathematics hidden here which is indeed very useful for the design of more reliable cryptosystems…

  5. MODULAR ARITHMETIC First let us associate numbers from 0 through 25 to the twenty six letters of the English alphabet: A  0 B  1 C  2 D  3 E  4 … X 23 Y 24 Z 25 Now, let us learn how to compute “modulo 26”. This means computing within a “universe” in which the only available numbers are those from 0 through 25: U = {0, 1, 2, …, 25}

  6. U = {0, 1, 2, …, 25} What about the other numbers? 26, 27, … Well, 26 is 0 in disguise! 27 is 1 in disguise! … 531 is 11 in disguise! – 17 is 9 in disguise! To “see” the “real face” of an integer modulo 26, divide it by 26 and take the remainder. What about 2001? 2001 = 26 · 76 + 25 Technically we denote this by 2001 (mod 26) = 25 Therefore 2002 will be simply… 0 (modulo 26) !

  7. How to add mod 26, then? Well, add as usual, then take the remainder! 17 + 15 = 6 22 + 18 = 14 … How to multiply? Multiply as usual, then take the remainder! 15 · 17 = 21 11 · 5 = 3 … Caesar’s cipher in modular arithmetic: X X + 3 (mod 26) Decryption: X X – 3 (mod 26)

  8. A(0) F(5) B(1) M(12) C(2) T(19) D(3) A(0) E(4) H(7) F(5) Q(14) G(6) V(21) H(7) C(2) I(8) J(9) J(9) Q(16) K(10) X(23) L(11) E(4) M(12) L(11) N(13) S(18) O(14) Z(25) P(15) G(6) Q(16) N(13) R(17) U(20) S(18) B(1) T(19) I(8) U(20) P(15) V(21) W(22) W(22) D(3) X(23) K(10) Y(24) R(17) Z(25) Y(24) VARIATIONS OF THE CAESAR’S CIPHER AFFINE SUBSTITUTIONS X a · X + b ( mod 26 )  a,b are elements of U, and a is relatively prime to 26 EXAMPLE: a = 7, b = 5 gives the following letter-by- letter encryption : X 7·X + 5 ( mod 26 )

  9. INVERTING THE AFFINE CIPHER X 7·X + 5 ( mod 26 ) (encryption formula) THE “INVERSE TRANSFORMATION” X  15·X + 3 (mod 26) (decryption formula) EXAMPLE Say, by using the encryption formula Alice encrypts “11” into 7·11 + 5 = 4 ( mod 26 ) and sends “4” over to Bob… Bob gets the “4” and wants to decrypt it by using the decryption formula. He computes: 4·15 + 3 = 63 = 52 + 11 = 11 (mod 26) and thus he recovers the “11”.

  10. UNFORTUNATELY, letter-by-letter encryption is easy to break (for example, by using a frequency analysis) EXAMPLE: Assume a smart eavesdropper Q suspects that Alice and Bob use an encryption of the type described above, that is, X  a·X+b (mod 26). But Q does not know the values of a and b. Well, Q keeps listening, and after a few moments realizes that the letter that has the highest frequency in the (otherwise unintelligible) cyphertext that Alice is sending over is H (7). Moreover, Q realizes that the letter coming next in the order of frequency is I (8). At this moment Q quickly opens a linguistics book and finds out that the letters having the two highest frequencies in English are E (4) (highest) and T (19) (second highest frequency). Finding a and b is not difficult: indeed, the encryption of 4 must be 7 and the encryption of 19 must be 8: a·4+b =7 (mod 26) a·19+b =8 (mod 26) This is a system of two equations with two unknowns (in modular arithmetic though), which is not difficult to solve.

  11. a·4+b =7 (mod 26) a·19+b =8 (mod 26) Substract the first equation out of the second to get 15·a = 1 (mod 26) from where a follows to be 7 [just check: 15·7=105 = 26·4+1=1 (mod 26); as a math major you will find out efficient ways of solving such equations of degree one in modular arithmetic). Once we know a=7, replace this value back into one of the two equations and you will find b=7 – a·4= 7 –7·4 = –21 = 5 (mod 26).

  12. TOPICS CRUCIAL TO CRYPTOGRAPHY A MATH MAJOR WILL GET TO KNOW: BBasic modular arithmetic. v PPrime numbers and factoring large integers. v AAlgorithms in number theory. v AAlgebra of matrices and polynomials.

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