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ENGG2013 Unit 10 n  n determinant and an application to cryptography

ENGG2013 Unit 10 n  n determinant and an application to cryptography. Feb, 2011. Yesterday – A formula for matrix inverse using cofactors. cofactors. Usually called the adjoint of A. Suppose that det A is nonzero.

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ENGG2013 Unit 10 n  n determinant and an application to cryptography

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  1. ENGG2013 Unit 10n  n determinant andan application to cryptography Feb, 2011.

  2. Yesterday – A formula for matrix inverse using cofactors cofactors Usually called the adjoint of A Suppose that det A is nonzero. Three steps in computing above formula1. for i,j = 1,2,3, replace each aij by cofactor Cij2. Take the transpose of the resulting matrix.3. divide by the determinant of A. ENGG2013

  3. Outline • nxn determinant • Caesar Cipher • Modulo arithmetic • Hill Cipher ENGG2013

  4. DETERMINANT IN GENERAL ENGG2013

  5. A pattern • Arrange the products so that the first subscripts are in ascending order. • All possible orderings of the second subscripts appear once and only once. ENGG2013

  6. Transposition • A transposition is an exchange of two objects in a list of objects. Examples: A B C D 2 1 4 5 3 A C B D 1 2 4 5 3 “Transposition” is anothermathematical term, and isnot the same as matrix tranpose. ENGG2013

  7. Another pattern • The sign of each term is closely related to the number of transpositions required to obtain the second subscripts, starting from (1,2) for the 2x2 case or (1,2,3) for the 3x3 case. ENGG2013

  8. The sign • Let p(1), p(2), …, p(n) be an order of 1,2,…,n. • For example p(1)=3, p(2) = 2, p(3)=1 is an ordering of 1, 2, 3. • Starting from (1,2,…,n), if we need an odd no. of transpositions to get ( p(1), p(2), …, p(n) ), we define the sign of (p(1), p(2),…,p(n)) be –1. • Otherwise, if we need an even no. of transpositions to get ( p(1), p(2), …, p(n) ), we define the sign of (p(1), p(2),…,p(n)) be +1. ENGG2013

  9. Definition of nn determinant 1 • The summation is over all n! possible orderings p = ( p(1), p(2), …, p(n) ) of 1,2,…,n. • There are n! terms. • sgn(p) is either +1 or –1, usually called the signature or signum of p. http://en.wikipedia.org/wiki/Determinant ENGG2013

  10. Properties of determinant • Determinant of nn identity matrix equals 1. • Exchange two rows (or columns)  multiply determinant by –1. • Multiply a row (or a column) by a constant k  multiply the determinant by k. • Add a constant multiple of a row (column) to another row (column)  no change • Additive property as in the 33 and 22 case. ENGG2013

  11. Cofactor and the adjoint formula for matrix inverse • Cofactors are defined in a similar way as in the 3x3 case. • The cofactor of the (i,j)-entry of a matrix A, denoted by Cij, is defined as (–1)i+j Aij, where A is the determinant of the sub-matrix obtained by removing the i-th row and the j-th column. • We have similar expansion along a row or a column (also called the Laplace expansion) as in the 3x3 case. • The adjoint formula: transpose nxn identity A adjoint of A The formula in this form holds when det A = 0 also ENGG2013

  12. CAESAR CIPHER ENGG2013

  13. Caesar and his army ATTACK Soldier carrying themessage “ATTACK” Message may be interceptedby enemy ENGG2013

  14. Caesar cipher http://en.wikipedia.org/wiki/Caesar_cipher ATTACK Soldier carrying theencrypted message“DWWDFN” The encrypted messagelooks random and meaningless ENGG2013

  15. Private key encryption Key The value of “key” is keptsecret Encryptionfunction Plain text Ciphertext Decryptionfunction Plain text Ciphertext key ENGG2013

  16. Mathematical description Caesar cipher is not secureenough, because the numberof keys is too small. Key =3 Shift to the rightby 3 ATTACK DWWDFN Shift to the leftby 3 ATTACK DWWDFN Key = 3 ENGG2013

  17. MODULO ARITHMETIC ENGG2013

  18. Mod 12 • Clock arithmetic 6+8= 2 mod 12 12 1 5+12 = 5 mod 12 11 10 2 9 3 4 8 7 5 6 ENGG2013

  19. Mod 7 • Week arithmetic 1+9 = 3 mod 7 2+3 = 5 mod 7 0 1 2 3 4 5 6 6 ENGG2013

  20. Mod 60 • 天干地支 arithmetic http://www.hko.gov.hk/gts/time/stemsandbranchesc.htm Year of rabbit ENGG2013

  21. Mod n – formal definition • n is a fixed positive integer • Definition: a mod nis the remainder of a after division by n. • Example: 25 = 1 mod 12. • Addition and multiplication: If the sum or product of two integers is larger than or equal to n, divide by n and take the remainder. • Example: 2+10 = 0 mod 12. • Example: 25 = 3 mod 12. ENGG2013

  22. More examples • 10 mod 7 = 3 • 4+5 mod 7 = 2 • 6+7 mod 7 = 6 • 27 mod 7 = 0 ENGG2013

  23. Mod 26 Fix a one-to-one correspondence between the English alphabetsand the integers mod 26. Caesar’s cipher: shifting a letter to the right by 3is the same as adding 3 in mod 26 arithmetic. ENGG2013

  24. Examples of mod 26 calculations • 3+19 = ? mod 26 • 13+20 = ? mod 26 • 34 = ? Mod 26 • 134 = ? Mod 26 ENGG2013

  25. Peculiar phenomena in modulo arithmetic • Non-zero times non-zero may be zero • 49 = 0 mod 12 • 22 = 0 mod 4 • Multiplicative inverse may not exist • Cannot find an integer x such that 4x = 1 mod 12. 4-1 does not exist mod 12. ENGG2013

  26. No fraction in modulo arithmetic • In mod 12, don’t write 1/3 or 3-1 because it does not exist. • But 5-1 is well-defined mod 12, because we can solve 5x=1 mod 12.Indeed, we have 55 = 1 mod 12. Therefore 5-1 = 5 mod 12. Fraction Fact from number theory: multiplicative inverse of x mod n existsif and only the gcd of x and n is 1. ENGG2013

  27. HILL CIPHER ENGG2013

  28. Hill cipher http://en.wikipedia.org/wiki/Hill_cipher • Invented by L. S. Hill in 1929. • Inputs : String of English letters, A,B,…,Z. An nn matrix K, with entries drawn from 0,1,…,25.(The matrix K serves as the secret key. ) • Divide the input string into blocks of size n. • Identify A=0, B=1, C=2, …, Z=25. • Encryption: Multiply each block by K and then reduce mod 26. • Decryption: multiply each block by the inverse of K, and reduce mod 26. ENGG2013

  29. Note • The decryption must be the inverse function of the encryption function. • It is required that K-1K = In mod 26. • Provided that det(K) has a multiplicative inverse mod 26, i.e., if det(K) and n has no common factor, the inverse of K can be computed by the adjoint formula for matrix inverse. • Inverse of an integer mod 26 can be obtained by trial and error. ENGG2013

  30. Example • Plain text: “LOVE”, Secret Key: • “LO”  • “VE”  • 2, 3, 16, 5 are transformed to cipher text “CDQF” ENGG2013

  31. How to decode? • Given “CDQF”, and the encryption matrix • How do we decrypt? • We need to compute the inverse of • Remind that all arithmetic are mod 26. There is no fraction and care should be taken in computing multiplicative inverse mod 26. ENGG2013

  32. Determinant • The determinant of equals 20(7)-3(15),which is 17 mod 26. • Find the multiplicative inverse of 17 mod 26, i.e., find integer x such that 17x = 1 mod 26. • Just try all 26 possibilities for x: 171 = 17 mod 26 172= 8 mod 26 173 = 25 mod 26 174 = 16 mod 26 175 = 7 mod 26 176 = 24 mod 26 177 = 15 mod 26 178 = 6 mod 26 179= 23 mod 26 1710 = 14 mod 26 1711 = 5 mod 26 1712 = 22 mod 26 1713 = 13 mod 26 1714 = 4 mod 26 1715 = 21 mod 26 1716= 12 mod 26 1717 = 3 mod 26 1718 = 20 mod 26 1719 = 11 mod 26 1720 = 2 mod 26 1721 = 19 mod 26 1722 = 10 mod 26 1723= 1 mod 26 1724 = 18 mod 26 1725 = 9 mod 26 170 = 0 mod 26 ENGG2013

  33. Computing the inverse mod 26 • From 1723= 1 mod 26, we know that the multiplicative inverse of 17 mod 26 is 23. • Using the formula for 2  2 matrix inverse we get Replace (17)-1 mod 26 by 23 ENGG2013

  34. Decryption • Given the ciphertext “CDQF”, we decrypt by multiplying by • From the table in p.23, 11, 14, 21, 4 is “LOVE”. ENGG2013

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