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

2. Yesterday – A formula for matrix inverse using cofactors kshumENGG20132 Suppose that det A is nonzero. Three steps in computing above formula 1. for i,j = 1,2,3, replace each a ij by cofactor C ij 2. Take the transpose of the resulting matrix. 3. divide by the determinant of A. Usually called the adjoint of A cofactors

3. Outline nxn determinant Caesar Cipher Modulo arithmetic Hill Cipher kshumENGG20133

4. DETERMINANT IN GENERAL kshumENGG20134

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. kshumENGG20135

6. Transposition A transposition is an exchange of two objects in a list of objects. kshumENGG20136 A B C D A C B D Examples: 2 1 4 5 3 1 2 4 5 3 “Transposition” is another mathematical term, and is not the same as matrix tranpose.

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. kshumENGG20137

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. kshumENGG20138

9. Definition of n  n determinant 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. kshumENGG20139 http://en.wikipedia.org/wiki/Determinant 11

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. kshumENGG201310

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 C ij, is defined as (–1) i+j A ij, 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: kshumENGG201311 nxn identity Aadjoint of A The formula in this form holds when det A = 0 also transpose

12. CAESAR CIPHER kshumENGG201312

13. Caesar and his army kshumENGG201313 ATTACK Soldier carrying the message “ATTACK” Message may be intercepted by enemy

14. Caesar cipher kshumENGG201314 http://en.wikipedia.org/wiki/Caesar_cipher ATTACK Soldier carrying the encrypted message “DWWDFN” The encrypted message looks random and meaningless

15. Private key encryption kshumENGG201315 Plain text Encryption function Ciphertext Plain text Decryption function Ciphertext Key key The value of “key” is kept secret

16. Mathematical description kshumENGG201316 ATTACK Shift to the right by 3 DWWDFN ATTACK Shift to the left by 3 DWWDFN Key =3 Caesar cipher is not secure enough, because the number of keys is too small.

17. MODULO ARITHMETIC kshumENGG201317

18. Mod 12 Clock arithmetic kshumENGG201318 12 1 2 93 6 4 57 8 10 11 6+8= 2 mod 12 5+12 = 5 mod 12

19. Mod 7 Week arithmetic kshumENGG201319 6 1+9 = 3 mod 7 2+3 = 5 mod 7 SunMonTueWedThrFriSat 1 2345678 9101112131415 16171819202122 23242526272829 3031 0123456

20. Mod 60 天干地支 arithmetic kshumENGG201320 http://www.hko.gov.hk/gts/time/stemsandbranchesc.htm 123456789101112 甲子甲子 乙丑乙丑 丙寅丙寅 丁卯丁卯 戊辰戊辰 己巳己巳 庚午庚午 辛未辛未 壬申壬申 癸酉癸酉 甲戌甲戌 乙亥乙亥 131415161718192021222324 丙子丙子 丁丑丁丑 戊寅戊寅 己卯己卯 庚辰庚辰 辛巳辛巳 壬午壬午 癸未癸未 甲申甲申 乙酉乙酉 丙戌丙戌 丁亥丁亥 252627282930313233343536 戊子戊子 己丑己丑 庚寅庚寅 辛卯辛卯 壬辰壬辰 癸巳癸巳 甲午甲午 乙未乙未 丙申丙申 丁酉丁酉 戊戌戊戌 己亥己亥 373839404142434445464748 庚子庚子 辛丑辛丑 壬寅壬寅 癸卯癸卯 甲辰甲辰 乙巳乙巳 丙午丙午 丁未丁未 戊申戊申 己酉己酉 庚戌庚戌 辛亥辛亥 495051525354555657585960 壬子壬子 癸丑癸丑 甲寅甲寅 乙卯乙卯 丙辰丙辰 丁巳丁巳 戊午戊午 己未己未 庚申庚申 辛酉辛酉 壬戌壬戌 癸亥癸亥 Year of rabbit

21. Mod n – formal definition n is a fixed positive integer Definition: a mod n is 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. kshumENGG201321

22. More examples 10 mod 7 = 3 4+5 mod 7 = 2 6+7 mod 7 = 6 2  7 mod 7 = 0 kshumENGG201322

23. Mod 26 ABCDEFGHIJKLM 0123456789101112 kshumENGG201323 NOPQRSTUVWXYZ 13141516171819202122232425 Fix a one-to-one correspondence between the English alphabets and the integers mod 26. Caesar’s cipher: shifting a letter to the right by 3 is the same as adding 3 in mod 26 arithmetic.

24. Examples of mod 26 calculations 3+19 = ? mod 26 13+20 = ? mod 26 3  4 = ? Mod 26 13  4 = ? Mod 26 kshumENGG201324 ABCDEFGHIJKLM 0123456789101112 NOPQRSTUVWXYZ 13141516171819202122232425

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. kshumENGG201325

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. kshumENGG201326 Fraction Fact from number theory: multiplicative inverse of x mod n exists if and only the gcd of x and n is 1.

27. HILL CIPHER kshumENGG201327

28. 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. kshumENGG201328 http://en.wikipedia.org/wiki/Hill_cipher

29. Note The decryption must be the inverse function of the encryption function. – It is required that K -1 K = I n 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. kshumENGG201329

30. Example Plain text: “LOVE”, Secret Key: “LO”  “VE”  2, 3, 16, 5 are transformed to cipher text “CDQF” kshumENGG201330 ABCDEFGHIJKLM 0123456789101112 NOPQRSTUVWXYZ 13141516171819202122232425

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. kshumENGG201331

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: kshumENGG201332 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

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 kshumENGG201333 Replace (17) -1 mod 26 by 23

34. Decryption Given the ciphertext “CDQF”, we decrypt by multiplying by From the table in p.23, 11, 14, 21, 4 is “LOVE”. kshumENGG201334