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

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Presentation on theme: "CRYPTOGRAPHY AND THE MATH MAJOR Dr. Mihai Caragiu Mathematics Department Ohio Northern University."— Presentation transcript:

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 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 ) 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) 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) D(3) X(23)K(10) Y(24)R(17) Z(25)Y(24)

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.  PPrime numbers and factoring large integers.  AAlgorithms in number theory.  AAlgebra of matrices and polynomials.


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