engg2013 unit 10 n n determinant and an application to cryptography feb, 2011
Post on 20-Dec-2015
221 views
TRANSCRIPT
ENGG2013 Unit 10n n determinant and
an application to cryptographyFeb, 2011.
Yesterday – A formula for matrix inverse using cofactors
kshum ENGG2013 2
Suppose that det A is nonzero.
Three steps in computing above formula1. for i,j = 1,2,3, replace each aij by cofactor Cij
2. Take the transpose of the resulting matrix.3. divide by the determinant of A.
Usually called the adjoint of A
cofactors
Outline
• nxn determinant• Caesar Cipher• Modulo arithmetic• Hill Cipher
kshum ENGG2013 3
DETERMINANT IN GENERAL
kshum ENGG2013 4
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.
kshum ENGG2013 5
Transposition
• A transposition is an exchange of two objects in a list of objects.
kshum ENGG2013 6
A B C D
A C B D
Examples:
2 1 4 5 3
1 2 4 5 3
“Transposition” is anothermathematical term, and isnot the same as matrix tranpose.
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.
kshum ENGG2013 7
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.
kshum ENGG2013 8
Definition of nn 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.
kshum ENGG2013 9
http://en.wikipedia.org/wiki/Determinant
1
Properties of determinant
• Determinant of nn 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 33 and 22 case.
kshum ENGG2013 10
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:
kshum ENGG2013 11
nxn identityA adjoint of A
The formula in this form holds when det A = 0 also
transpose
CAESAR CIPHER
kshum ENGG2013 12
Caesar and his army
kshum ENGG2013 13
ATTACK
Soldier carrying themessage “ATTACK”
Message may be interceptedby enemy
Caesar cipher
kshum ENGG2013 14
http://en.wikipedia.org/wiki/Caesar_cipher
ATTACK
Soldier carrying theencrypted message“DWWDFN”
The encrypted messagelooks random and meaningless
Private key encryption
kshum ENGG2013 15
Plain text Encryptionfunction Ciphertext
Plain text Decryptionfunction Ciphertext
Key
key
The value of “key” is keptsecret
Mathematical description
kshum ENGG2013 16
ATTACK Shift to the rightby 3 DWWDFN
ATTACKShift to the left
by 3 DWWDFN
Key =3
Key = 3
Caesar cipher is not secureenough, because the numberof keys is too small.
MODULO ARITHMETIC
kshum ENGG2013 17
Mod 12
• Clock arithmetic
kshum ENGG2013 18
121
2
9 3
6
4
57
8
10
11
6+8= 2 mod 12
5+12 = 5 mod 12
Mod 7
• Week arithmetic
kshum ENGG2013 19
6
1+9 = 3 mod 7
2+3 = 5 mod 7
Sun Mon Tue Wed Thr Fri Sat
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31
0 1 2 3 4 5 6
Mod 60• 天干地支 arithmetic
kshum ENGG2013 20
http
://w
ww
.hko
.gov
.hk/
gts/
time/
stem
san
dbra
nche
sc.h
tm
1 2 3 4 5 6 7 8 9 10 11 12甲子
乙丑
丙寅
丁卯
戊辰
己巳
庚午
辛未
壬申
癸酉
甲戌
乙亥
13 14 15 16 17 18 19 20 21 22 23 24丙子
丁丑
戊寅
己卯
庚辰
辛巳
壬午
癸未
甲申
乙酉
丙戌
丁亥
25 26 27 28 29 30 31 32 33 34 35 36戊子
己丑
庚寅
辛卯
壬辰
癸巳
甲午
乙未
丙申
丁酉
戊戌
己亥
37 38 39 40 41 42 43 44 45 46 47 48庚子
辛丑
壬寅
癸卯
甲辰
乙巳
丙午
丁未
戊申
己酉
庚戌
辛亥
49 50 51 52 53 54 55 56 57 58 59 60壬子
癸丑
甲寅
乙卯
丙辰
丁巳
戊午
己未
庚申
辛酉
壬戌
癸亥
Year of rabbit
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: 25 = 3 mod 12.
kshum ENGG2013 21
More examples
• 10 mod 7 = 3• 4+5 mod 7 = 2• 6+7 mod 7 = 6• 27 mod 7 = 0
kshum ENGG2013 22
Mod 26
A B C D E F G H I J K L M
0 1 2 3 4 5 6 7 8 9 10 11 12
kshum ENGG2013 23
N O P Q R S T U V W X Y Z
13 14 15 16 17 18 19 20 21 22 23 24 25
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.
Examples of mod 26 calculations
• 3+19 = ? mod 26• 13+20 = ? mod 26• 34 = ? Mod 26• 134 = ? Mod 26
kshum ENGG2013 24
A B C D E F G H I J K L M
0 1 2 3 4 5 6 7 8 9 10 11 12
N O P Q R S T U V W X Y Z
13 14 15 16 17 18 19 20 21 22 23 24 25
Peculiar phenomena in modulo arithmetic
• Non-zero times non-zero may be zero– 49 = 0 mod 12– 22 = 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.
kshum ENGG2013 25
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 55 = 1 mod 12. Therefore 5-1 = 5 mod 12.
kshum ENGG2013 26
FractionFact from number theory: multiplicative inverse of x mod n existsif and only the gcd of x and n is 1.
HILL CIPHER
kshum ENGG2013 27
Hill cipher• Invented by L. S. Hill in 1929.• Inputs : String of English letters, A,B,…,Z.
An nn 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.
kshum ENGG2013 28
http://en.wikipedia.org/wiki/Hill_cipher
Note
• The decryption must be the inverse function of the encryption function.– It is required that K-1 K = 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.
kshum ENGG2013 29
Example• Plain text: “LOVE”, Secret Key:• “LO”
• “VE” • 2, 3, 16, 5 are transformed to cipher text “CDQF”
kshum ENGG2013 30
A B C D E F G H I J K L M
0 1 2 3 4 5 6 7 8 9 10 11 12
N O P Q R S T U V W X Y Z
13 14 15 16 17 18 19 20 21 22 23 24 25
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.
kshum ENGG2013 31
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:
kshum ENGG2013 32
171 = 17 mod 26172= 8 mod 26173 = 25 mod 26174 = 16 mod 26175 = 7 mod 26176 = 24 mod 26177 = 15 mod 26
178 = 6 mod 26179= 23 mod 261710 = 14 mod 261711 = 5 mod 261712 = 22 mod 261713 = 13 mod 261714 = 4 mod 26
1715 = 21 mod 261716= 12 mod 261717 = 3 mod 261718 = 20 mod 261719 = 11 mod 261720 = 2 mod 261721 = 19 mod 26
1722 = 10 mod 261723= 1 mod 261724 = 18 mod 261725 = 9 mod 26170 = 0 mod 26
Computing the inverse mod 26
• From 1723= 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
kshum ENGG2013 33
Replace (17)-1 mod 26 by 23
Decryption
• Given the ciphertext “CDQF”, we decrypt by multiplying by
• From the table in p.23, 11, 14, 21, 4 is “LOVE”. kshum ENGG2013 34