“‘To be an effective Information Warrior, individuals need superior computer skills, as well as an in-depth understanding of information technology architectures, … protocols and processes. ’ Michael Erbschloe author of “Information Warfare: How to Survive Cyber Attacks” 1
AGENDA n n n MATHEMATICAL BACKGROUND n Revision (3 -7); ORDER of a mod n (8, 9); Primitive Root g of n (10, 11); Index of a (12 -14); a: quadratic residue mod p and Legendre Symbol (15 to 32); Square and non-square elements of Zp (33 to 36); dlogg, p(b) : Discrete Logarithm of b for base g (mod p) (37 to 39); Diffie-Hellman Key Exchange (40 to 50); El. Gamal’s PK System (51 to 57); Digital Signature Systems (58 to 62); Elliptic Curve Cryptosystem (ECC) (63 to 76); Identity Based Encryption (IBE) (77 to 93): ISO/IEC 11770 -3 Key Agreement Scheme, Shamir’s Method, (Cocks’s quadratic residues IBE scheme and Pairing-based methods: left out for self-study)
Modular Arithmetic: systematized by Carl Friedrich Gauss in his book Disquisitiones Arithmeticae, published in 1801 Reference: http: //programmingpraxis. com/2009/07/07/modular-arithmetic/ as of Dec 06, 2009 n n Exponentiation: repeated modular multiplication Square root: that number which, when multiplied by itself, equals the target number normal arithmetic: √ 4 = +2 or -2. n modular arithmetic: √ 18 mod 31 = 7 or 24. Since (24 + 7) mod 31 = 0, 7 and 24 may be considered to be ‘negative’ of each other. n n Consider x 2 (mod 13): x n 1 2 3 4 5 6 7 8 9 10 11 12 x 2 n 0 0 1 4 9 3 12 10 10 12 3 9 4 1 x 2 (mod 13) acquires the values of 0, 1, 3, 4, 9, 10, 12 ONLY. x 2 (mod 13) is NEVER equal to 2, 5, 6, 7, 8, 11 3
Modular Arithmetic: Square root of some numbers may not exist. n There is no x such that x 2 mod 13 = 7 the square root of 7 mod 13 does not exist; n the only numbers that have square roots modulo 13 are 1, 3, 4, 9, 10, and 12, or, equivalently, ± 1, ± 3, and ± 4. Another restriction: the modular square root is only defined if the modulus is an odd prime. Example: COMPOSITE MODULUS: For x 2 (mod 15): Please see the next slide: n 4
Composite modulus: No square root exists Example: COMPOSITE MODULUS: For x 2 (mod 15): x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 x 2 0 1 4 9 1 10 6 4 4 6 10 1 9 4 1 n n As x is varied from 0 to 14, 2 n x (mod 15) acquires the values of 0, 1, 4, 6, 9, 10 ONLY. 2 n x (mod 15) is NEVER equal to 2, 3, 5, 7, 8, 11, 12, 13, 14. 4 has two sets of conjugate square roots: ± 2 and ± 7 non-unique solution Hence the modular square root of 4 is said not to exist when the modulus is composite. 5
Revision Slide-1 Logarithms Logxy=a => y=xa 0 n Logx 1=0 (x = 1) 1 n Logxx=1 (x = x) n Logx(y. z)=Logx(y)+Logx(z) n Logx(yr)=r. Logx(y) n 6
Revision Slide-2 Euler’s theorem: Generalization of Fermat’s theorem: If a and n are relatively prime, (n) a = 1 mod n where (n) = Euler’s Totient Function = number of positive integers less than n and relatively prime to n 7
Order of a mod n Given: a and n are relatively prime. Let am=1 mod n. The smallest positive value of m for which the above equation is satisfied is called the ORDER of a mod n. n Examples: Order of a mod 17: (Please see the next slide. ) 44=1 mod 17 Order of 4 mod 17 = 4. Similarly 316=1 mod 17; 516=1 mod 17; 28=1 mod 17; 88=1 mod 17 Order of 3 mod 17 = 16; Order of 5 mod 17 = 16 Order of 2 mod 17 = 8; Order of 8 mod 17 = 8 8
Example: p = 17 …. 1 i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 i 1 1 1 1 2 i 2 4 8 16 15 13 9 1 3 i 3 9 10 13 5 15 11 16 14 8 7 4 1 4 i 4 16 13 1 5 i 5 8 7 i 7 8 i 8 6 4 16 13 1 1 1 12 2 6 4 16 13 1 3 15 7 1 8 5 1 16 9 4 15 12 13 7 9 14 1 16 1 13 14 2 10 16 12 9 11 4 15 3 4 12 16 10 2 14 13 6 13 2 16 9 . 11 9 4 15 1 8 13 2 . 11 i 11 2 5 4 10 8 3 16 6 16 1 . 16 i 16 1 9
m Example: a modulo 19 For a=2, 3, 10, 13, 14 or 15: Choose any one of the above 6 values for a. n As m is varied from 1 to 18, m n a (modulo 19) generates the entire set of non-zero integers from 1 to 18. (The example of a=15 is given below. ) i 15 i n 1 2 15 16 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 12 9 2 11 13 5 18 4 3 7 10 17 8 6 14 1 For each of the ( above) 6 values of a 18 n a =1 mod 19. Hence Order of a mod 19 for the above values of a is 18. 10
Primitive root: n n Definition: If, for some integer value of ‘a’, the ‘order of a mod n’ is equal to Φ(n), the integer value of ‘a’ is called the ‘Primitive Root of n’. n Primitive roots of a prime number p will be denoted by g. Property: For a primitive root and for every value of 0<m≤Φ(n), am generates a distinct number (mod n) and every such number is co-prime with n. An integer may - or may not have – a primitive root. Integer of type pα, 2 pα , where p: an odd prime number; α: a positive integer, have one or more primitive roots. 11
Examples of primitive roots n 2 3 4 5 6 7 9 10 Primitive 1 2 3 2, 3 5 3, 5 2, 5 3, 7 roots of n n 11 Primitive 2, 6, roots of n 7, 8 13 14 17 18 19 22 2, 6, 3, 3, 5, . . 2, 7, 11 … … … 7, . . gs(n): The smallest primitive root of an integer n Reference: http: //mathworld. wolfram. com/Primitive. Root. html as of Dec 06, 2009
Index of a number a Let modulus: n Primitive root of n: g An integer, co-prime to n: a If gx = a mod n, then x = v(a) is called the Index of a. Examples: modulus = 11, primitive root = 6, i 1 2 3 4 5 6 7 8 9 10 3 7 9 10 5 8 4 2 1 6 i 6 For a = 5, 66 = 5 mod 11; Therefore v(5) = 6; For b = 7, 63 = 7 mod 11; Therefore v(7) = 3.
Similarities between Log and Index Given: a mod n = g(v(a)) mod n (v(b)) b mod n = g mod n n n Log(a. b) = Log a + Log b axb mod n = g(v(a) + v(b)) mod n v(axb) = v(a) + v(b) Example: 5 x 7 mod 11 = 6(6 + 3) mod 11 b n Log(a ) = b Log a b (b. v(a)) n a mod n = g mod n b v(a ) = b. v(a) Examples: 57 mod 11 = 6(7 x 6) mod 11 =3 Similarly ba mod n = g(a. v(b)) mod n Example: 75 mod 11 = 6(5 x 3) mod 11 =10
References: 1. For the smallest primitive roots for the first few integers: http: //mathworld. wolfram. com/Primitive. Root. html as of Dec 06, 2009 2. For a list of first 1000 prime numbers: http: //primes. utm. edu/lists/small/1000. txt as of Dec 06, 2009 3. Primes by primitive roots: http: //www. research. att. com/~njas/sequences/Sindx_Pri. html as of Dec 06, 2009 4. G. A. Miller, “ Methods to Determine the Primitive Roots of a Number”, http: //www. jstor. org/view/00029327/di 994161/99 p 0203 o/0? frame=noframe&user. ID=89 cf 8 ca 8@uwindsor. ca/01 c 0 a 8346600501 ceadb 5&dpi=3&config=jstor as of Dec 1, 2007 http: //www. jstor. org/stable/2370177? &Search=yes&term=Number&term=Methods&term=Roots&term=Determ ine&term=Primitive&list=hide&search. Uri=%2 Faction%2 Fdo. Advanced. Search%3 Fq 0%3 DMethods%2 Bto%2 BDetermine%2 Bthe%2 BPrimitive%2 BRoots%2 Bof%2 Ba%2 BNumber%26 f 0%3 Dall%26 c 0%3 DAND%26 q 1 %3 D%26 f 1%3 Dall%26 c 1%3 DAND%26 q 2%3 D%26 f 2%3 Dall%26 c 2%3 DAND%26 q 3%3 D%26 f 3%3 Dall%2 6 wc%3 Don%26 Search%3 DSearch%26 sd%3 D%26 ed%3 D%26 la%3 D%26 jo%3 D&item=11&ttl=7332&retu rn. Article. Service=show. Article as of Dec 06, 2009 15
2 Solution for x = a mod p 2 PROBLEM: Given values of ‘a’ and ‘p’: x = a mod p where p: odd prime and a: an integer To solve for x: There are three possibilities: (i) No solution: ‘a’ is said to be a “quadratic non-residue mod p”. (ii) One solution if a = 0 mod p (iii) Two solutions ‘a’ is said to be a “quadratic residue mod p”. Reference: Henri Cohen, ”A Course in Computational Algebraic Number Theory”, Springer 1996, pp 27 16
Example: Existence of a solution Consider modulus = 11. Squares: 1, 3, 4, 5, 9 Non-squares: 2, 6, 7, 8, 10 2 For non-squares, a solution for x = a mod p does not exist. 2 Thus there is no value of x, which satisfies x = 6 mod 11. x 1 2 3 4 5 6 7 8 9 10 x 2 mod 11 1 4 9 5 3 3 5 9 4 1
Definition: Legendre-Jacobi-Kronecker Symbol Legendre Symbol (a/p): (i) (a/p) = -1 if a is quadratic non-residue mod p (ii) (a/p) = 0 if a = 0 (iii) (a/p) = 1 if a is quadratic residue mod p. 2 The number of solutions of x = a mod p will be (1 + (a/p)). 18
Solutions, if a is a quadratic residue mod p If (a/p) = 1. there exists an x such that 2 x = a mod p n An easy solution for half of the primes, which obey p = 3 mod 4: (p+1)/4 x = a mod p n For half of the remaining primes, which obey p = 5 mod 8, there are two possibilities: n n (p-1)/4 (p+3)/8 a = +1 The solution is x = a mod p. (p-1)/4 (p-5)/8 a = -1 The solution is x = 2 a. (4 a) mod p. For the remaining primes, which obey p = 1 mod 8, it is difficult to come to similar solutions. n 19
Example: 2 Solutions for x: x = a mod p For p =11: It obeys p = 3 mod 4. n Hence if (a/p) = 1, its solutions can be found by using (p+1)/4 x = a mod p For p =11, Given: a 1 3 4 5 9 n To Find: x 1 5 9 4 3
Algorithm for finding out the value of (a/p) (slides 22 -34) 21
Algorithm for evaluating Kronecker(a/b) where a, b ε Z Step 1: If b = 0, output = 0 if lal≠ 1 = 1 if lal= 1 END n Step 2: (for removing 2’s from b) Set v = 0 While b is even { set v (v + 1) b (b/2)} If v is even, set k 1. Otherwise k (-1) (a**2 – 1)/8 If b < 0, set b (-b), AND if in addition a < 0, set k (-k). n 22
Algorithm for evaluating Kronecker(a/b) where a, b ε Z contd. 2 Step 3 (for reducing size once) Note: At this stage b is odd and b > 0. Set a a mod b n Step 4: If a = 0, output = 0 if b > 1 = k if b = 1 END n Step 5 (for removing powers of 2) Set v = 0 While a is even { set v (v + 1) a (a/2)} If v is odd, set k (-1) (b**2 – 1)/8. k n 23
Algorithm for evaluating Kronecker(a/b) where a, b ε Z contd. 3 n Step 6: Subtract and apply reciprocity. Note: At this stage a and b are odd. Set r (b – a). (a-1). (b-1)/4 If r > 0, set k = (-1) . k b a r; Else set a (-r). Go to Step 4. 24
Legendre (a/b), where a, b ε Z and b is an odd prime Step 1: not required. n Step 2: (required only for initializing k) K is set to 1. n Step 3 (for reducing size once) Note: At this stage b is odd and b > 0. Set a a mod b n Step 4: If a = 0, output = 0 if b > 1 = k if b = 1 END n 25
Legendre (a/b), where a, b ε Z and b is an odd prime …. 2 Step 5 (for removing powers of 2 from a) Set v = 0 While a is even { set v (v + 1) a (a/2)} If v is odd, set k (-1) (b**2 – 1)/8. k n 26
Legendre (a/b), where a, b ε Z and b is an odd prime …. 3 n n n n Step 6: Subtract and apply reciprocity. Note: At this stage a and b are odd. Set r (b – a). (a-1). (b-1)/4 If r > 0, set k = (-1) . k b a r; Else set a (-r). Go to Step 4. 27
Example 1 for Legendre Symbol For modulus p = 11, we found n Squares: 1, 3, 4, 5, 9 n Non-squares: 2, 6, 7, 8, 10 (i) By using the algorithm (of the last three slides), it can be seen that for each of the square values, (a/p) = 1 (ii) By using the algorithm (of the last three slides), it can be seen that for each of the nonsquare values, (a/p) = -1. Note: Try the algorithm for one of the square values and one of the non-square values and confirm the above two statements.
Example 2 for Legendre Symbol 25 mod 11 Iteration 1: n Step 2: k = 1 n Step 3: a = 25 mod 11 = 3 n Step 4: a ≠ 0 n Step 5: v =0; Since v ≠ odd, no change in the value of k. n Step 6: r = 11 - 3 = 8 k = (-1) (a-1). (b-1)/4. k = (-1) 2. (10)/4. k = -1 b = 3 a = 8. n 29
Example 2 for Legendre Symbol … 2 Iteration 2: (begins at step 4) n Step 4: a ≠ 0 n Step 5: v =0; { v = 1, a = 4}; {v = 2, a = 2}; {v = 3, a = 1} Since v is odd, k = (-1) (b**2 – 1)/8. k = 1 n Step 6: r = 3 - 1 = 2 k = (-1) (a-1). (b-1)/4. k = (-1) 0. (2)/4. k = 1 b = 1 a = 2. 30
Example 2 for Legendre Symbol … 3 Iteration 3: (begins at step 4) n Step 4: a ≠ 0 n Step 5: v =0; { v = 1, a = 1}; Since v is odd, k = (-1) (b**2 – 1)/8. k = 1 n Step 6: r = 1 - 1 = 0; a = 0 Iteration 4: (begins at step 4) a =0 Since b = 1, output = k = 1 By slide 18, a solution exists. By slide 19, a solution for the primes, which obey p = 3 mod 4: (p+1)/4 x = a mod p. 2 Since 11 = 3 mod 4, the solution for x = 25 mod 11 is: x = 253 mod 11 = 33 mod 11 =5 31
Example 3 for Legendre Symbol 17 mod 11 Iteration 1: n Step 2: k = 1 n Step 3: a = 17 mod 11 = 6 n Step 4: a ≠ 0 n Step 5: v =0; { v = 1, a = 3} Since v is odd, k = (-1) (b**2 – 1)/8. k = - 1 n Step 6: r = 11 - 3 = 8 k = (-1) (a-1). (b-1)/4. k = (-1) 2. (10)/4. k = 1 b = 3, a = 8. n 32
Example 3 for Legendre Symbol … 2 Iteration 2: (begins at step 4) n Step 4: a ≠ 0 n Step 5: v =0; { v = 1, a = 4}; {v = 2, a = 2}; {v = 3, a = 1} Since v is odd, k = (-1) (b**2 – 1)/8. k = -1 n Step 6: r = 3 - 1 = 2 k = (-1) (a-1). (b-1)/4. k = (-1) 0. (2)/4. k = -1 b = 1 a = 2. 33
Example 3 for Legendre Symbol … 3 Iteration 3: (begins at step 4) n Step 4: a ≠ 0 n Step 5: v =0; { v = 1, a = 1}; Since v is odd, k = (-1) (b**2 – 1)/8. k = -1 n Step 6: r = 1 - 1 = 0; a = 0 Iteration 4: (begins at step 4) a =0 Since b = 1, output = k = -1. 2 By slide 18, no solution exists for x = 17 mod 11 34
Square and Non-square Elements (next 4 slides) 35
Example: p = 17 …. 1 i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 i 1 1 1 1 2 i 2 4 8 16 15 13 9 1 3 i 3 9 10 13 5 15 11 16 14 8 7 4 1 4 i 4 16 13 1 5 i 5 8 7 i 7 8 i 8 6 4 16 13 1 1 1 12 2 6 4 16 13 1 3 15 7 1 8 5 1 16 9 4 15 12 13 7 9 14 1 16 1 13 14 2 10 16 12 9 11 4 15 3 4 12 16 10 2 14 13 6 13 2 16 9 . 11 9 4 15 1 8 13 2 . 11 i 11 2 5 4 10 8 3 16 6 16 1 . 16 i 16 1 36
Example: p = 17 …. 2 Squares Elements of Zp = {1, 2 3, …(p-1)} can be either Squares (as) or Non-squares (an). n Squares: 1, 2, 4, 8, 9, 13, 15, 16 1 = 162 mod 17; 2 = 62 mod 17 = 112 mod 17 4 = 152 mod 17; 8 = 52 mod 17 = 122 mod 17 9 = 142 mod 17; 13 = 82 mod 17 = 92 mod 17 15 = 72 mod 17 = 102 mod 17; 16 =132 mod 17 For all i, asi mod p = a square element only. A square element cannot be a primitive root. n Non-squares: 3, 5, 6, 7, 10, 11, 12, 14 No. of as elements = No. of an elements = (p-1)/2 37
Example: p = 17 …. 3 Sub-groups Testing whether or not an element is square: an efficient algorithm called Legendre Symbol n Examples of groups, formed by n a=3, 5 : primitive roots; n n n Example: for p = 17, primitive roots: 3, 5, 7, 11 Finding primitive roots of a large prime number: computationally tough a= 2, 8 : two blocks of q = (p-1)/2 each a= 4 : four blocks of (p-1)/4 each 38
Example: p = 17 …. 4 Sub-groups n Depending upon the generator elements, size of Sub-groups of Zp: n Full group: (p-1) members, if the generator element is a primitive root n Size of sub-groups: (p-1)/m n Sub-group of size 1: g =1 n Sub-group of size 2: Members are 1 and (p -1) Example: Use a of 1, 16, 4, 2 or 8, 3 or 5 to get groups of size 1, 2, 4, 8 and 16 respectively. (See slide 36. ) 39
Logarithmic for Modular Arithmetic Consider a prime number ‘p’ and its primitive root g. (There is at least one primitive root for every Zp. ) For any integer b, we can find the exponent ‘i’ such that b=gi(mod p). Both g and i are members of Zp i. e. 0≤ i ≤ (p-1) i: Discrete Logarithm of b for base g (mod p) : dlogg, p(b) n 40
Discrete Logarithm Theorems n n dloga, p(1) = 0 dloga, p(a) = 1 n dloga, p(bc)= (dloga, pb + dloga, pc) mod Φ(p) n dloga, p(yr)= [r. dloga, p(y)] mod Φ(p) Compare: Logx 1=0 (x 0= 1) Logxx=1 (x 1= x) Logx(y. z)=Logx(y)+Logx(z) Logx(yr)=r. Logx(y) 41
Calculation of Discrete Logarithm Consider p: a prime number. Its primitive root : generator element=g. y = gx mod p n Given x, y can be calculated easily using CLRS algorithm. (as studied in RSA PK method) n For large prime numbers – Given y, for calculation of x: no method with a complexity lower than that for factorizing prime numbers exists. This is known as the Discrete Logarithm Problem (DLP). n 42
Diffie-Hellman Key Exchange (agreement) ……. . 1 n Diffie-Hellman Key Exchange: based on DLP n Alice selects a prime p and generator g of Gallois Field Zp n select a random number a < p, a n computes y =g mod p and n sends y , p and g to Bob n selects a random number b< p, b n computes z =g mod p and n sends z to Alice Reference: Whitfield Diffie and Martin E. Hellman, ”New Directions in Cryptography”, IEEE Transactions on Information Theory, IT 22(6): 644 -654, November 1976 43
Diffie-Hellman Key Exchange (agreement)……………. . 2 then Alice computes k = za mod p (= gab mod p) b ab n And Bob computes k = y mod p (=g mod p). n n Therefore Alice and Bob are able to get the same key securely without meeting together by sending messages on an insecure line. A Hacker knows p, g, y and z. But without knowing a or b, k cannot be determined. a = dlogg, p(y) and b = dlogg, p(x) cannot be found, since discrete log is difficult to evaluate for large numbers. 44
Diffie-Hellman Key Exchange Example … 1 n n n n Choose p = 11. Primitive roots of 11 are 2, 6, 7, 8 Alice and Bob choose g = 2 for p =11 for key exchange. She chooses a private key of a = 5. 25 mod 11 = 10. Alice sends y = 10 to Bob chooses a private key of b = 7. 27 mod 11 = 7. Bob sends z = 7 to Alice. 45
Diffie-Hellman Key Exchange Example … 2 SECRET KEY n He calculates the secret key k = 107 mod 11 = 10 n Alice calculates the secret key k = 75 mod 11 = 10 EVE: Knows about p = 11 and g = 2 Can sniff y =10 and z = 7. But does not know about the private keys. Reference: Example 5. 2 from Man Young Rhee, “Internet Security: 46 Cryptographic principles. algorithms and protocols”. Wiley 2003
Diffie-Hellman Key Exchange: To Find the private keys To find the private keys: For a: Solve the equation 2 a mod 11 = 10. i. e. a = dlog 2, 11(10) For b: Solve the equation 2 b mod 11 = 7 i. e. b = dlog 2, 11(7) Calculation of discrete logarithms for large prime numbers is very hard. 47
Diffie-Hellman Key Exchange: The Protocol Every user should publish her/his public key (p, g and y) in a directory. Then all users, whose keys are in the directory, can communicate with one another securely by calculating the secret key. Question: How authentic will the directory be? n Authenticate using the Diffie-Hellman key: If Alice and Bob recognize each other’s voice, voice samples may be encrypted by using the secret key and exchanged to confirm that there is no MITM. Problem: Will work till voice synthesis technology is able n to reproduce the exactly similar voice samples. 48
Diffie-Hellman Key Exchange Man-in-the-Middle attack… 1 n n n Alice sends y =ga mod p to Bob Eve intercepts it and sends w =gc mod p to Bob (believing that the message is from Alice) b n responds with z =g mod p; and b cb n creates the key k 1 = w = g Eve intercepts Bob’s message and c bc n is able to create the key k 1 = z = g d n sends v =g mod p to Alice. d ad n is able to create the key k 2= y = g Alice receives v and creates the key k 2= va = gda 49
Diffie-Hellman Key Exchange Man-in-the-Middle attack… 2 All future communication: n n Alice sends messages to Bob encrypted with k 2 Eve n n intercept the message and decrypts it using the key k 2 encrypts it/modified message using the key k 1 Sends the encrypted message to Bob receives the message and is able to decrypt it by using the key k 1 -- similar scenario for the messages from Bob to Alice n Thus Alice and Bob can be under the mistaken impression that they are talking to each other. 50 n
MITM attack and smaller Sub-groups For a prime number p, Zp = {1, 2 3, …(p-1)}, a primitive root g can generate all the members. During a MITM attack, Eve may send a non-primitive Root as g, leading to a small sub-group of Zp. This may compromise the security. n If g is a non-square: y = ga mod p n n n is a square if a is even and it is non-square if a is odd. Thus Eve can check y and find out the last bit of a ( ie whether a is even or odd) Use only squares? 51
Safe Prime n n If p = 2 q +1, where p and q are both prime numbers, p is called a safe prime. Choose a group n n n with modulo p, where p = 2 q + 1; which has q elements; for which g is a square. (Use Legendre Symbol function to verify. ) 52
Safe Primes: How to choose g for such a group? n g should be a square; n n n Since it is a square, it cannot contain all the 2 q elements. The number of elements must be a factor of (p-1). However since p-1 = 2 q, it can have only subgroups of 1, 2 and q. Choose a random number r in the range 2 ……………. (p-2). n Select g = r 2, except that it should not be either 1 or (p-1) n 53
El. Gamal’s PK System - keys El. Gamal proposed two systems for use in PK system and for encryption of plaintext messages. PK System: n Choose a prime number p and two random numbers g and d such that n n n g is the primitive root modulo p. 1 ≤ d ≤ (p-2) Calculate e =gd mod p Private key: d Public key: e, g and p 54
El. Gamal’s PK System - Security Example: Choose p =11, g = 6 and d = 8 e = 68 mod 11 = 4 Private key = 8; Public key: 4, 6 and 11 SECURITY: To find d from public key, one has to solve the equation 6 d mod 11 = 4 or d = dlog 6, 11(4). This is the Discrete Logarithm Problem. It is computationally infeasible for large values of p. 55
El. Gamal Encryption of plaintext message 0 ≤ m ≤ p-1 Bob wants to send a message securely to Alice. He knows Alice’s public key: e, g and p. Encryption Process by Bob: n Choose a random number k <p; k is to be kept secret by Bob k n Message Key: K = e mod p n The Cipher consists of two numbers: (C 1, C 2) k n C 1= g mod p C 2 = K. m mod p K masks the message by using the public key of Alice. Bob sends the masked message C 2 along with C 1 helps Alice calculate the mask K for decryption. 56 Inverse of K helps calculation of m. n
El. Gamal’s PK System - Encryption Example Given: Alice’s public key: e = 4, g = 6 and p = 11 Bob chooses a random number k = 7. Bob wants to send the message m = 5 to Alice. m El. Gamal Encrypter Public key (C 1, C 2) e, g, p Message Key: K = ek mod p = 47 mod 11 = 5 C 1 = gk mod p = 67 mod 11 = 8; C 2 = K. m mod p = 5 x 5 mod 11 = 3 Bob sends the Cipher (8, 3) to Alice. 57
El. Gamal’s PK System – Decryption Alice receives (C 1 and C 2). She has her private key d. To decrypt: K = ek mod p = gdk mod p = C 1 d mod p C 2 = K. m mod p or m = K-1. C 2 mod p (C 1, C 2) El. Gamal Decrypter m d 58
El. Gamal’s PK System – Comments Alice keeps d as a secret. n Bob keeps k as his secret. n Bob can compute the mask K = ek mod p. Bob does not know d. But he knows e, where e =gd mod p. Therefore K = gdk mod p n Bob sends C 2 along with C 1 where C 1= gk mod p. n Alice can compute K, without knowing k, since K = n C 1 d mod p. 59
El. Gamal’s PK System – Decryption Example Given: n Cipher = (C 1, C 2) = (8, 3) n Alice’s Private Key = d = 8 To Find: m K = C 1 d mod p = 88 mod 11 = 5 K-1. 5 mod 11 = 1; K-1 = 9 m = K-1. C 2 mod p = 9. 3 mod 11 = 5 Reference: Example 5. 8 from Man Young Rhee, “Internet Security: Cryptographic principles. algorithms and protocols”. Wiley 2003 60
Digital Signature n Association with the entity, which signs it: n n n The receiver can associate with the signing entity. The signer cannot repudiate it. Association with the message: n The message, which is authenticated, cannot be changed. 61
Attacks on RSA Systems …… 1 n Low Exponent Attack: e is sometimes chosen to be small ( eg 3) to make encryption faster. Coppersmith Theorem: In a modulo n polynomial f(x) of degree e, one can use an algorithm of complexity log n to find the roots if one of the roots is smaller than n 1/e. On applying theorem to c = me mod n, for e=3, if only two-third of the bits in m are known, the algorithm, can find all the bits. Recommendation: e may not be smaller than 216 + 1 = 65537. 62
Attacks on RSA Systems …… 2 Broadcast Attacks: If the same message is sent to many recipients with the same public key. Example: e = 3: n a 1 = m 3 mod n 1 a 2 = m 3 mod n 2 a 3 = m 3 mod n 3 CRT can be used to find A = m 3 mod n 1. n 2. n 3 m can then be found by using ordinary arithmetic. 63
Attacks on RSA Systems …… 3 Short Pad Attack: Bob wants to send a message m to Alice He pads it with x and encrypts m ll x to get C 1. The message is intercepted and dropped by Eve. Alice tells Bob that she has not received the message. Bob again pads m with y and encrypts m ll y to get C 2. The message is intercepted by Eve. If x and y are small, Coppersmith proved that Eve can find m. n Use Optimal Asymmetric Encryption Padding (OAEP) with G: a function for converting k bits to m bits, and, H: a function for converting m bits to k bits Reference: 1. M. Bellare, P. Rogaway. Optimal Asymmetric Encryption -- How to encrypt with RSA. Extended abstract in Advances in Cryptology - Eurocrypt '94 Proceedings, Lecture Notes in Computer Science Vol. 950, A. De Santis ed, Springer-Verlag, 1995 http: //cseweb. ucsd. edu/users/mihir/papers/oae. pdf as on 6 th Dec 2009 2. http: //en. wikipedia. org/wiki/Optimal_Asymmetric_Encryption_Padding as on 6 th Dec 64 2009
Comparison n How secure is RSA and Diffie-Hellman or El. Gamal? n n n RSA: based on factorization Diffie-Hellman and El. Gamal: based on DLP Have proved: n n Factoring a large prime is equivalent to solving DLP problem. Exist algorithms with a sub-exponential but super-polynomial complexity 65
Elliptic Curve Cryptosystem (ECC) n For ECC, the sub-exponential algorithm of breaking it has not been found. n n So ECC is more secure than RSA or El. Gamal Or to say, using much smaller key size can achieve the same security as RSA or El. Gamal with a larger key size, so more efficient. 66
Elliptic curve group over real number n n n y 2 = x 3 + ax + b, where x, y, a and b are real numbers. All (x, y) points, satisfying above equation, along with infinite point O and addition operation, form a group Suppose P=(x, y) then define –P=(x, -y). 67
![Definition of a Group [A 1] closure under addition: [A 2] Associativity of addition:](https://present5.com/presentation/d8b7aafd89cd9040afced32c70351d1a/image-68.jpg "Definition of a Group [A 1] closure under addition: [A 2] Associativity of addition:")
Definition of a Group [A 1] closure under addition: [A 2] Associativity of addition: [A 3] Additive identity: Group Abelian Group [A 4] Additive inverse: [A 5] Commutativity of addition: 68
Elliptic curve example 69
Addition operation (A Geometric Approach) n If P and Q are distinct, and if P -Q, define P+Q as follows: n n Draw a line through P and Q, then the line will intersect with the curve, the intersected point is denoted as –R, and define P+Q=R. Define P + (-P) = O If P=(x, 0), then P+P = O , (in fact, a vertical line) Otherwise, draw a tangent line through P, the intersected point is defined as –R, then P+P =2 P =R. 70
Definition of P+Q = R 71
Definition of P+(-P) 72
Definition of P+P (where y!=0) 73
Definition of P+P (where y=0) 74
n n Adding distinct points P and Q When P = (x. P, y. P) and Q = (x. Q, y. Q) and P Q, P -Q, P + Q = R where s = (y. P - y. Q) / (x. P - x. Q) x. R = s 2 - x. P - x. Q and y. R = -y. P + s(x. P - x. R) Note that s is the slope of the line through P and Q. Doubling the point P When y. P is not 0, 2 P = R where s = (3 x. P 2 + a) / (2 y. P ) x. R = s 2 - 2 x. P and y. R = -y. P + s(x. P - x. R) P + (-P) =O, If P = (x. P, y. P) and y. P =0, then P + P = 2 P = O. 75
Elliptic Curve Groups over Zp n n Zp = {0, 1, …, p-1} y 2 mod p = (x 3 + ax + b) mod p n n Where a and b are in Zp, and x, y are also in Zp. Addition with modular p. Example p=23, Zp=Z 23 , y 2 = x 3 + x Points lying on y 2 = x 3 + x: (0, 0) (1, 5) (1, 18) (9, 5) (9, 18) (11, 10) (11, 13) (13, 5) (13, 18) (15, 3) (15, 20) (16, 8) (16, 15) (17, 10) (17, 13) (18, 10) (18, 13) (19, 1) (19, 22) (20, 4) (20, 19) (21, 6) (21, 17) Point (1, 5): X=1 RHS=2, y 2 = 2 mod 23 y = 5 Point (21, 6): X=21 RHS=x 3+x=213+21=(15 + 21)mod 23 = 13, y 2 = 13 mod 23 y = 6 76
y 2 mod 23 = (x 3 + x) mod 23 77
Points on Elliptic curve along with addition operation form a group. n Given a point P (P (x, 0)), consider 2 P=P+P, 3 P=2 P+P, …. , n. P=(n-1)P+P, … n Given any n, it is easy to compute R=n. P. n However given R, it is very difficult to find n, such that n. P=R. This is called The Elliptic Curve Discrete Logarithm Problem (ECDLP). n 78
Many cryptosystems can be formed based on Elliptic Curve n Example: Diffie-Hellman key exchange n n Given elliptic curve E and a point P (public) Alice selects an a, computes A=a. P, sends A to Bob selects a b, computes B=b. P, sends B to Bob Then Alice can compute the key K=a. B=ab. P, similarly, Bob computes the key K=b. A=ab. P 79
“It is tough to make predictions, especially about the future. ” -- Yogi Berra 80
X. 509 v 3 1. Distinguished Name: Root CA: single point of failure 2. Validity period 3. Public Key Example: National CA/Univ of Windsor/CS/End User like Chris Smith 2075 Policy of CA Access Control through the certificate Certificate revocation lists (CRLs) Cross-certification is the black hole of PKI 81
CRL Problems n n n Not issued frequently enough to be effective against an attacker Expensive to distribute Vulnerable to simple DOS attacks Attacker can prevent revocation by blocking CRL delivery If a user caches a CRL, he may deal with an outdated CRL. 82
CRL Problems …… 2 Back-dated CRL can appear at any point in the future Destroys the entire concept of nonrepudiation n Revoking self-signed certificates is hairy when a Cert revokes itself, Applications may n – Accept the CRL as valid and revoke the certificate – Reject the CRL as invalid since it was signed with a revoked certificate – Crash to provide timely revocation exacerbates the problem Example: 10 M clients download a 1 MB CRL issued once a minute ~150 GB/s traffic n 83
Online Certificate Status Protocol, OCSP Reply is created on the spot in response to the request Ephemeral pseudo-CRL avoids CRL validity period n Problems: n n n Requires a signing operation for every query CAs charge fees to issue a certificate (Most expensive collection of bits in the world) Revocation checks may also cost. 84
Identity-based PK Systems 85
Differences between Identity-based System and a standard PK system n Different Methods of Constructing a key n Distributing a key n Authenticating a key n Using a key n Reference: 1. Liqun Chen, ”Identity-based Cryptography”, HP Laboratories, 2006, http: //www. sti. uniurb. it/events/fosad 06/papers/Chenfosad 06. pdf 2. A. Shamir. Identity-based cryptosystems and signature schemes. In Advances in Cryptology - Crypto '84, Springer-Verlag LNCS 196, 47 -53, 1984. 86
Public Key Infrastructure (PKI) System n n n Sender (Alice) requests the CA for the public key of the Receiver (Bob). Through an authenticated channel, CA sends the public key (of Bob) certificate, signed by the private key of CA. Alice decrypts the certificate using the public key of CA. Alice encrypts her message using the public key of Bob. Alice sends the message to Bob through Internet Bob gets his private key from CA through an authenticated channel and decrypts the message. 87
Identity Based Encryption (IBE) n n n Alice uses the identity of Bob to create his public key. Alice encrypts her message using the public key of Bob. Alice sends the message to Bob through Internet Bob gets his private key from the Master Key Generator by supplying to it his identity. Bob decrypts the message by using his private key. 88
Key Generator in IBE Private Key Identity Private Key Generator Master Key 89
IBE Schemes n n Shamir’s paper 1984 Three IBE schemes in 2001 n n n Sakai, Ohgishi and Kasahara Boneh and Franklin Cocks Sakai and Kasahara in 2003. . 90
Identity n n n E-mail address Photo Phone number Postal address Role-based access based upon the role of a person in his organization 91
Shamir’s Method: IB Private key for Bob Identity may be the digest of any data string associated with Bob: Thus ID = H(bob@uwindsor. ca) Let the Master private and public keys be (d, n) and (e, n) respectively. Private key = SID = IDd mod n n n For signing a message: Choose r: a random number Compute t = re mod n Find f = H(t, m) where m = message 92
Shamir’s Method: Verification of Signatures s = SID. rf mod n where SID = IDd mod n Output Signatures: (s, t) and f is the signed message. Verification of Signatures Compute LHS = se Compute RHS = ID. t. H(t, m) mod n, where f = H(t, m) and t = re mod n n If LHS = RHS, the signature is acceptable. PROOF: LHS = se = IDd. e. rf. e mod n =ID. rf. e mod n RHS = ID. re. f mod n 93
ISO/IEC 11770 -3 Key Agreement Scheme n n n Developed by Guillou and Quisquater, based on Shamir’s scheme IDA and IDB : identities of Alice and Bob respectively. Master Key Generator: n private key: (d, n) public key: (e, n) e n Two elements g and h such that g = h mod n Master Key Generator: creates private keys for Alice and Bob as follows: SA = (1/IDA)d mod n SB = (1/IDB)d mod n 94 94
ISO/IEC 14888 -2 Signature Scheme Key Exchange Alice selects a random number a and computes t. A = SA. ha mod n and sends it to Bob. n Bob selects a random number b and computes t. B = SB. hb mod n and sends it to Alice. n Both Alice and Bob are able to compute the common key KAB as follows: KAB =((t. B)e. IDB)a = gab and KAB =((t. A)e. IDA)b = gab n The common symmetric key can be used by Alice and Bob to exchange messages. 95
Cock’s IBE Scheme 96
Cocks’s quadratic residues IBE scheme n based on the hardness of the quadratic residues problem, i. e. 2 n y : x = y mod n n n = pq where p and q are two large primes, like in RSA n does not use pairing Reference: C. Cocks. An identity-based encryption scheme based on quadratic residues. In Proceedings of Cryptography and Coding, LNCS 2260, pp. 360 -363, Springer-Verlag, 2001 97
Cocks’s quadratic residues IBE scheme . . . 2 n n is quite fast encrypts a message bit by bit, and it requires log n bits of ciphertext per bit of plaintext Reference: C. Cocks. An identity-based encryption scheme based on quadratic residues. In Proceedings of Cryptography and Coding, LNCS 2260, pp. 360 -363, Springer-Verlag, 2001. 98
Pairings in IBE n pairings, which have been used in identity-based cryptography: the Weil pairing and the Tate pairing and their variants. References: 1. P. Barreto, H. Kim, B. Lynn, and M. Scott, Efficient algorithms for pairing-based cryptosystems, Proceedings of CRYPTO 2002, LNCS 2442, pages 354– 369, Springer-Verlag, 2002. 2. D. Boneh and M. Franklin. Identity based encryption from the Weil pairing. In Advances in Cryptology - Crypto 2001, Springer. Verlag LNCS 2139, 213 -229, 2001.
THANKS 100
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