Download William Stallings, Cryptography and Network Security 3/e

Cryptography and Network
Security
M. Sakalli
Source: Third Edition
by William Stallings
And
Modified Lecture Slides of Lawrie Brown
Sun-Tsu’s Chinese Remainder Thr
 States that when the moduli of a system of linear congruencies are
pairwise prime, there is a unique solution of the system modulo, the
product of the moduli.
 ax = b (mod m).
The 1st Century CE (Common Era, ~400 AD), the Chinese mathematician Sun
Tsu Suan-Ching asking the following problem:
“There are certain things whose number is unknown. When divided by 3, the
remainder is 2; when divided by 5, the remainder is 3; and when divided
by 7, the remainder is 2. What will be the number of things?”
Discrete Math Kenneth H Rosen, page 186.
 x = 2 mod(3)
 x = 3 mod(5)
 x = 2 mod(7)
Let m1, m2, …, mn be (pairwise) relatively prime numbers. Then the system:
 x = a1 mod (m1) = a2 mod (m2) = …. = an mod (mn)
 Has a unique solution modulo
 M = m 1 m2 … m n .
The CRT says that only one number of x mod(3x5x7) satisfies all eqns.
x = 23 (mod 105),. x = 23 = 7*3 + 2 = 2 (mod 3),
x = 23 = 4*5 + 3 = 3 (mod 5), x = 23 = 3*7 + 2 = 2 (mod 7)
How to construct the solution in mod(M)
23 mod(105) = 23 + 105 n  { …, -292, -187, -82,
23, 128, 233, 338, …}
Therefore, all these congruent numbers are
solutions of Sun-Tsu’s three equations.
•
•
i.
ii.
i.
i.
M = (πk=1:n mi) = m1m2 … mn all mk’s have to be
pairwise relatively prime.
For each equation of x= ak mod(mk) calculate
Mk =M/mk; all mi except for mk.
yk inverse of Mk from Mk yk=1 mod(mk)=(Mkmod(mk)) yk
x = 2 mod (3) (5*7) y1 = 1 mod(3)  y1 =2.
x = 3 mod (5) (3*7) y2 = 1 mod(5)  y2 =1
x = 2 mod (7) (3*5) y3 = 1 mod(7)  y3 =1
x = Σ(aiMiyi) = (a1 M1 y1 + a2 M2 y2 + a3 M3 y3)
= 233 = 23 mod(105)
Why does this work? (without going into detail)
Suppose the solution x and “mod” it by m1:
M1y1 is equal to a1, since M1y1 = 1 mod (m1).
M2y2, M3y3 , …, every other term is zero mod(m1), since MK is a multiple of m1.
x = a1M1y1 + a2M2y2 + … + anMnyn.
= 2 (5*7) 2 + 3 (3*7) 1 + 2 (3*5) 1
But would be true for any of the mk. Therefore, x satisfies all of the
equations.
One of the most useful results of number theory is in speeding up some operations in
the RSA public-key scheme, since it allows to perform calculations modulo factors,
and then combine the answers to get the actual result. Since the computational cost
is proportional to size, such manipulating is faster than working in the full size
modulus, ie when M is 150 digits or more. However it is necessary to know
beforehand the factorization of M. operations in n-tuples.
See worked examples in Stallings section 8.4.
Ancient Chinese Pirates:
A band of 17 pirates stole a sack of gold coins.
When they tried to divide the fortune into
equal portions, 3 coins remained. In the
ensuing brawl over who should get the extra
coins, one pirate was killed. The wealth was
redistributed, but this time an equal division
left 10 coins. Again an argument developed in
which another pirate was killed, but now the
fortune could be evenly distributed.
What was the least number of coins which could have been stolen?
What are all possible numbers of coins which could have been stolen?
If x is the number of coins, it has to satisfy the following modular equations:
x = 3 mod (17)
x = 10 mod (16)
x = 0 mod (15)
These numbers are relatively prime, so the Chinese Remainder Theorem says
there IS a solution mod 17*16*15 = 4080.
It might have been possible there that there is NO SOLUTION.
Write down the equations for yk:
x = 3 (mod 17)

(16x15) y1 =1 (mod 17)
x = 10 (mod 16) 
(17x15) y2 = 1 (mod 16)
x = 0 (mod 15)

(17x16) y3 = 1 (mod 15)
240 y1 = 1
255 y2 =1
272 y3 =1
Solve the equations for yk by whatever way is easiest (brute force or by finding
inverses):
(16*15) y1 = 1 (mod 17)  2 y1
=1
 y1 = 9 (mod 17)
(17*15) y2 = 1 (mod 16)  15 y2 = 1
 y2 = 15 (mod 16)
(17*16) y3 = 1 (mod 15)  2 y3
=1
 y3 = 8 (mod 15)
Construct the solution x (mod 17*16*15):
x
= a1M1y1 + a2M2y2 + … + anMnyn.
= 3 * (16*15) * 9 + 10 * (17*15) * 15 + 0 * (17*16) * 8
= 44730 = 3930 (mod 4080).
3930
= 231*17 + 3
= 3 (mod 17)
= 245*16 + 10
= 10 (mod 16)
= 262*15
= 0 (mod 15)
Therefore the solution works
The smallest number of coins which the pirates could have stolen: 3930.
Other possible numbers of coins could the pirates have stolen?
Any number satisfying 3930 + 4080n could have been stolen.
N = { 3930, 8010, 12090, 16170, …}
• Pairs representing nonnegative integers less than 12, when
represented with ordered pairs, first remainder of mod(3),
and the second remainder of mod(4),
0={0, 0}, 1={1, 1}, 2={ }.. 3={0, 3}
4={1, 0}, ………….
…
….
…
11={2, 3}
• By CRT, n-tuples on a processor will run quicker, ie, every
nonnegative integer less than 100; using moduli of pairwise
relatively prime integers: 99, 98, 97, 95.. Any integer <
89403930 can be represented uniquely. Suppose two
numbers 123684  (33, 8, 9, 89) and 413456 (32, 92, 42,
16).. Find remainder (congruencies) for each module and
(proceed with an operation) sum in each tuple and merge
the results to reach the final sum as 537140.. Unique
nonnegative solution.
• Particular good (applicable) choices of moduli with large
integers are of the form 2k -1, (binary), k a positive integer.
Binary operations, the second reason is that
gcd(2a -1, 2b -1) =2gcd(a,b) -1, (proof of which is in page 112 104 Number Theory Problems,
source in our webpage)
Systems of Linear Modular Equations:
a1x = b1 (mod m1)
a2x = b2 (mod m2)
….
anx = bn (mod mn)
Solve each equation aix = bi (mod mi) individually.
a1x = b1 (mod m1)

x = c1 (mod m1)
a2x = b1 (mod m2)

x = c2 (mod m2)
….
….
anx = b1n (mod mn)

x = cn (mod mn)
Solve the following set of simultaneous congruences:
i)
x = 5 (mod 6)
iii)
2x = 1 (mod 5)
x = 4 (mod 11)
3x = 9 (mod 6)
x = 3 (mod 17)
4x = 1 (mod 7)
5x = 9 (mod 11)
ii)
x = 5 (mod 11)
x = 14 (mod 29)
x = 15 (mod 21)
Question iii is a bit harder. How badly do you want an A+?
Homeworks:
Answer Brahmagupta’s question: (7th century AD)
An old woman goes to market and a horse steps on her basket
and crashes the eggs. The rider offers to pay for the damages
and asks her how many eggs she had brought. She does not
remember the exact number, but when she had taken them out
two at a time, there was one egg left. The same happened
when she picked them out three, four, five, and six at a time, but
when she took them seven at a time they came out even. What
is the smallest number of eggs she could have had?
What other possible number of eggs could she have?
[Hint: x = 1 (mod 2,3,4,5,6), x = 0 (mod 7).]
Pseudoprimes
• Suppose integers larger than 235, in our computers.. Moduli
of 235 -1, 234 -1, 233 -1, 231 -1, 229 -1, 223 -1 are pairwise
prime, giving in the possibility working with the integers (not
exceeding) 2184.
• An integer n is prime when not divisible by any prime smaller
than sqrt(n). ie 101, easy.. But inefficient.
• Chinese mathematicians: believed that p is prime whenever
congruence of 2p-1 = mod(p) holds but couldn’t show for
composite numbers.
• Fermat shows that the congruence holds when n is prime.
Fermat’s Little theorem, if p is prime and a is an integer not
divisible with p, then ap-1 = 1 mod(p), besides for every
integer a, ap = a mod(p).
• The corollary, if 2n-1 = 1 mod(n), n is always prime, is not
always correct.
• Composite positive integers for which 2n-1 = 1 mod(n) holds
are pseudoprime numbers, for example for n=341=11*31, n
is pseudoprime to base 2.
• Definition: for n, a>0 integers, and if n is composite such that
an-1 = 1 mod(n), then n is a pseudoprime to a.
• So if n satisfies, 2n-1 = 1 mod(n), n is either prime or
pseudoprime to the base 2, performing similar tests for the
other bases coprime to n.
• More than 40*106 prime numbers <1010-1, but only 14884
pseudoprime numbers to the base 2.
• Not easy to distinguish, due to the other composite numbers
that pass all the tests with bases; gcd(b, n)=1.
• Composite integer numbers n satisfying both bn-1 = 1 mod(n)
and gcd(b, n), are Carmichael numbers. (6k + 1)(12k +
1)(18k + 1), where k is prime, not proved for large numbers.
Example 561=3*11*17, if gcd(b, 561) = 1; then gcd(b, 3) =
gcd(b, 11) = gcd(b, 17) = 1; 561 passes fermat’s little thr, b3-1
= 1 mod(3), b11-1 = 1 mod(11), = b17-1 = 1 mod(17)
• b560 = b(2)280 = 1 mod(3), b560 = b(10)56 = 1 mod(11), b560 =
b(16)35 = 1 mod(17)
• b560 = 1 mod(561), for every b with gcd(b, 561) = 1. there
are many infinite number of carmichael numbers.
Magicicada Species: Wikipedia. New Lease of Life for Prime Numbers
Two discoveries in the last 30 years. One is amusing, the other important in the
business world.
Periodic cicadas - a type of insect found in North America - hibernate for many
years, lying dormant in the ground.
After a long period, they emerge to begin a new life.
It has been found that there are two distinct types of Cicada, those that remain
dormant for 13 years and those that choose to sleep for 17 years.
It is no coincidence that these two numbers are primes! This is explained in the
Guide entry on cicadas.
Euler Totient Function Φ(n)
• coprimes to n form a group
• reduced set of residues is those numbers (residues) which
are relatively prime to n
– eg for n=10, complete set of residues is
– {0,1,2,3,4,5,6,7,8,9}  reduced set of residues is {1,3,7,9}
• to compute Φ(n): exclude every fold of its primes.. and count
excluded ones..
Theorem 1. If p is prime then Φ(p) = p – 1.
– for p (p prime) ø(p) = p-1
– for p.q (p,q prime) ø(p.q) = ø(q) ø(q)=(p-1)(q-1)
Theorem 2. If gcd(n, m) = 1, then Φ(nm) = Φ(n) · Φ(m).
• Examples: Φ(3) and Φ(5) above.
• Relatively prime, so the theorem claims that Φ(15) = Φ(3) ·
Φ(5) = 2 · 4 = 8, which is correct.
Euler Totient Function Φ(n)
• Proof: Consider the # of integers that are not relatively prime
in {0…(pq-1)}
{p,2 p,..., (q-1)p}, and {q,2 q,..., (p-1)q},
therefore
Φ(p.q) = p.q-(p-1)-(q-1)-1
= p.q-p-q+1 = (p-1)(q-1)
• eg.
– Φ(37) = 36
– Φ(21) = (3–1)×(7–1) = 2×6 = 12
Theorem 3. If p is prime and n ≥ 1, then Φ(pn) = pn – pn-1.
• Examples: Consider p = 3 and n = 2. Integers relatively
prime to 9: {1, 2, 4, 5, 7, 8}<9.
• Φ(52) = 20, inspect: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
Theorem 4 (Euler's Theorem). For any integer n > 1, if gcd(a,
n) = 1, then aΦ(n) ≡ aΦ(n)mod N ≡ 1 (mod n)..
• Examples: For n = 10, Φ(10) = 4. gcd(a, n) = 1:
• a = 1:  14 = 1
• a = 3:  34 = 81
• a = 7:  74 = 2401

all of these are ≡ 1 (mod
10)
• a = 9:  94 = 6561
• a = 17:  174 = 83521
• a = 57:  574 = 10556001
• n=11, a=2, ø(11)=10  210=1024=1 mod(11)
Theorem 5 (Fermat's Theorem). If p is prime and gcd(a, p) =
1, then ap-1 ≡ 1 (mod p).
• From Euler's Theorem by applying Theorems 1 and 4.
• Example: p = 5 and a = 6, which is relatively prime to 5. 65-1
≡ 1 (mod 5) = 1296.
Theorem 6. For any n > 1, if gcd(a, n) = 1, then the equation
ax ≡ 1 (mod n) has a unique solution, modulo n. From the
gcd identity, there is a linear combination of a and n such
that 1 = ax + ny. Since n divides ny evenly, ax mod n must
be 1.
• Example:
• 3·0 ≡ 0 mod 5
• 3·1 ≡ 3 mod 5
• 3·2 ≡ 1 mod 5 (for a = 3, n = 5, the solution is x = 2)
• 3·3 ≡ 4 mod 5
• 3·4 ≡ 2 mod 5
Theorem 7. From CRT, if gcd(p, q) = 1, then for all integers x
and a, x ≡ a (mod p) and x ≡ a (mod q) if and only if x ≡ a
mod(pq). This is a corollary of the Chinese Remainder
Theorem.
• Example: gcd(3, 5) = 1
• 32 ≡ 2 mod 15
• 32 ≡ 2 mod 3
• 32 ≡ 2 mod 5
• Factoring a number n: n=a b c
• Relatively hard when compared to multiplying
the factors together to generate the number
• Prime factorisation of a number n
– eg. 91=7 13; 3600=24 32 52
• two numbers are relatively prime to each other if..
• Conversely; gcd  the common least powers of
prime factorizations.
– 300=21 31 52 18=21 32 hence GCD(18,300)=21 31
50=6
• Fermat’s Little Theorem: ap-1 = 1 mod p = 1
where p is prime and naturally gcd(a, p)=1
The RSA Algorithm
•
•
Rivest, Shamir, Adleman, MIT, 1978
Reference: From Cormen et al at end of handout]
1. Select two large primes p and q such that p ≠ q. Typical values might be
integers with 200 digits. RSA currently recommends a module of at
least 768 bits long.
2. Compute N = pq.
3. Select an exponent odd integer Ke relatively prime to Φ(N) = (p – 1)(q –
1), where Φ(N) is kept private. (Theorem 2 and 1). Finding p and q,
probabilistic primality test. However this can not be done for 400 digits
in a reasonable time.
4. Find Kd such that Ke*Kd ≡ 1 (mod Φ(N)). Theorem 6: a solution for Kd
exists and that given Ke and Φ(N), it is uniquely defined. (Euclid's
Algorithm), Kd and Ke are multiplicative inverses in modulo Φ(N), since
their product modulo Φ(N) is 1. Note that Kd is easy to compute only if
one knows the value of Φ(N) which is essentially the same as knowing
the values of P and Q.
5. Publish Ke and N as the public key.
6. Keep secret Kd, which along with N, is the private key.
The RSA Algorithm
Finally: Ke (made public) and Kd(kept private),
• If PT is a numeric encoding of a block of
plaintext, the cipher text is CT = PTKe mod N.
• PT = CTKd mod N. = (PT(Ke))Kd mod N.
• P = 5, Q=11. N = P*Q = 55
• Message length encoded < log(N)= 5 the length
of message. A larger message, break it is into
blocks.
• Φ(N) = (P-1)*(Q-1) = 40.
• Ke, relatively prime, gcd(Ke, Φ(N)): Ke=13
• Kd, such that Kd * Ke =1 mod(Φ(N)): Kd=37
The RSA Algorithm
• Multiplying P by Q is easy: the number of operations depends on the
number of bits in P and Q.
• For example, multiplying two 384-bit numbers takes approximately 3842 =
147,456 bit operations
• If one knows only N, factoring P and Q is hard: in essence, the # of
operations depends on the value of N.
The simplest method for factoring a 768-bit number takes about 2384  3.94
*10115 trial divisions, 285  3.87  1025 trial divisions, 241  219,000,000,000
trial divisions
• A quick factoring algorithm of large number M hasn’t been done yet. And
also the absence or impossibility of such an algorithm not proven yet.
• Peter Shor has devised a very fast factoring algorithm for a quantum
computer, if anyone manages to build one.
• ***Prove that MKeKd mod(N) = M mod(N). In other words Pub(Priv(M)) =
Priv(Pub(M)) = M.
• It turns out that there are feasible attacks on RSA. To guarantee security,
a very large modulus must be used and some preprocessing of the
message should be done.
Euler’s theorem
• For every a and n relatively prime,
aΦ(n) =1 mod(n).
• For example, n=10, a=3, or n=11, a=21
• For n prime, Φ(n) = n-1; from Fermat’s thr
a(n-1) = 1 mod(n)  a(n) = n mod(n).
• It holds for any integer n, relatively prime numbers
R={x1, …, xΦ(n)} to n, aR,
S=a{x1mod(n), …, xΦ(n) mod(n)} = {ax1, …, a xΦ(n)},
any of axi must be relatively prime to n, and less then n,
since R and a are both prime to n.
• There are no duplicates in S and
πi=1:Φ(n) axi mod(n) =
 aΦ(n) πi=1:Φ(n) xi
=
 aΦ(n) =1 mod(n) 
πi=1:Φ(n) xi, …
πi=1:Φ(n) xi mod(n)
aΦ(n)+1 =a mod(n).
Euler’s theorem
• Corollary, n=pq, both primes, and 0<m<n,
mΦ(n)+1 = m(p-1)(q-1)+1 =m mod(n) holds.
If gcd(m, n) =1, by virtue of Euler’s thr it holds.
Suppose gcd(m,n) not one, then n= pq, gcd(m, n)=1, means that, neither
p nor q divides m.
If m=c1p, (or m=c2q, where c is integer and c>0) then m and n cannot be
relative therefore gcd(m,n) =not 1.
• If gcd(m, q)=1, from Euler’s thr
mΦ(q)
= 1 mod(q),
[mΦ(q)]Φ(p)
= 1 mod(q)
mΦ(n)
= 1 mod(q)
mΦ(n)
= 1 +kq
mΦ(n)+1
= m + kcpq = m + kcn
mΦ(n)+1
= m mod(n)
• Or an alternative way,
mkΦ(n)+1 =[[mΦ(n) ]k m ]mod(n)= [1k m] mod(q) =m mod(n)
Primality Testing
• traditionally sieve using trial division
– ie. divide by all numbers (primes) in turn less than the square root of the
number
– only works for small numbers
• Miller Rabin Algorithm based on Fermat’s Theorem a(n-1) =1 mod(n) = 1
if n is prime.
• Consider an odd number n>2, then n-1 is even number, therefore
equal to 2kq with k>0, q must be odd. Keep dividing n-1 by 2, k
divisions, you’ve got the q, determine a number 1<a<n-1, compute
2kq power of the number a, and check the equality to n-1 (line 5), or to
1 (line 3):
• TEST (n) is:
1. Find integers k, q, k > 0, q odd, so that (n–1)=2kq;
2. Select a random integer a, 1<a<n–1;
3. if aq mod n = 1 then return (“maybe prime");
4. for j = 0 to k – 1 do
j
5. if (a2 q mod n = n-1)
then return(" maybe prime ")
6. return ("composite")
Primality Testing
If n is prime there is a smallest value of j, 0jk, such that
(a)^(2jq)mod(n) = 1.
For j=0,
For 1jk,
aq-1 = 0,
or
n|(aq-1).
a^(2jq)mod(n) = 1

(a^(2(j-1)q)mod(n)-1)*(a^(2(j-1)q)mod(n)+1) = 1
n divides either side, by assumption j is the smallest
such that n does not divide (a^(2(j-1)q)mod(n)-1),
therefore n|(a^(2(j-1)q)mod(n)-1).
Or equivalently a^(2(j-1)q)mod(n)=(-1)mod(n)=n-1;
line5.
Probabilistic Considerations
• Millar-Rabin test returns inconclusive for (n-1)/4 < ¼
• if Miller-Rabin returns “composite” the number is definitely
not prime otherwise is a prime or a pseudo-prime chance
it detects a pseudo-prime is < ¼
• Therefore if test returns inclusive t times in succession ..
then probability n is prime is 1-4-t…
• Prime distribution: Considering distribution of the primes,
“prime number theorem states that the primes near n are
spaced on the average one every ln(n) integers”, so the is
on the order of ln(n), even integers and fold of 5 are
rejected.. So, 0.4ln(n), for example if the order of prime
number is 2200, 0.4ln(n)=55 trials. Closely located
ones, 1.000.000.000.061, 1.000.000.000.063 are primes.
Probabilistic Considerations of Millar-Rabin
• n=29, 2kq=28= 22 7, k=2;
• a=10;
– j=0, 107mod(29)=17, neither 28, nor 1..
– j=1, (107)2 mod(29)=28, inconclusive, may be pr..
• a=2,
– j=0, 27mod(29)=12,
– j=1, 214mod(29)=28, inc..
• For all a’s this will give inconclusive, so n is a prime..
• n=13*17=221, 2kq=220= 22 55, k=2;
• a=5;
– j=0, 555mod(221)=112..
– j=1, (555)2 mod(221)=168, n returns composite..
• But if a would have been chosen as 21,
– j=0, 217mod(221)=200..
– j=1, 2114mod(221)=220, returns inclusive, which points 221 as prime..
• In fact of the 220 integers, 1, 21, 47, 174, 200, 220 return inc..
Primitive Roots
Primitive Roots
• From Euler’s theorem have aø(n)mod n=1
• consider ammod n=1, and (a, n) relative prime
GCD(a,n)=1
– at least one positive m<n satisfying ammod n=1, for example m =
ø(n) or may be smaller, this is called the order of a (mod n)..
– once powers reach m, cycle will repeat
• if smallest is m= ø(n) then corresponding a is called a
primitive root
• To check if a number x is primitive root, it suffices to check
xm=1 mod p, ***
– the order of any x coprime to p has to be a divisor of (p − 1) since xp1=1 mod p, *** following words are not clear yet to me too but the
statement written is valid** --- if n is not a primitive root, then there
exists a strict positive divisor m of p-1, such that p-1, xm=1 mod p,
so there the statement we made suffices.. ---
• if p is prime, then successive powers of a "generate" the
group mod p
Primitive Roots
• Example: from previous table,
p=19, p-1 = 2.32, divisors 1, 2, 3, 6, 9, 18,
check a=10, 102=5, 103=12, 103=5, 106=11,
109=18, 1018=1 mod(19),
so the smallest power of x, such that xm=1 mod p is 18,
hence ordp(a) = ord19(10)=18.
• Check if the rule applies to a=5,
53=11, 59=1,.. Then will recycle the
numbers periodically since 1018= 1 mod(19)
.
Primitive Roots
• Similarly, you can do the rest of the homework by
yourselves. The complete list of primitive roots is:
–
–
–
–
–
–
–
mod 3 : 2
mod 5 : 2, 3
mod 7 : 3, 5
mod 11 : 2, 6,
mod 13 : 2, 6,
mod 17 : 3, 5,
mod 19 : 2, 3,
7, 8
7, 11
6, 7, 10, 11, 12
10, 13, 14, 15
• Once you have found '(p − 1) many primitive roots mod p,
you are done, because mod p there are exactly '(p − 1)
distinct primitive roots.
• CHECK THIS.
Discrete Logarithms or Indices
• Input: p - prime number, a- primitive root of p, b - a residue mod p.
• Goal: Find k such that ak = b( mod p). (In other words, find the position of
y in the large list of {a, a2, . . . , aq-1}.
• 14 is a primitive root of 19.
• The powers of 14 (mod 19) are in order: 14 6 8 17 10 7 3 4 18 5 13 11 2
9 12 16 15 1
• For example L14(5) = 10 mod 19, because 1410 = 5( mod 19).
• the inverse problem to exponentiation is to find the discrete logarithm
of a number modulo p
• that is to find x where ax = b mod p
• written as x=loga b mod p or x=inda,p(b)
• if a is a primitive root then always exists, otherwise may not
– x = log3 4 mod 13 (x st 3x = 4 mod 13) has no answer
– x = log2 3 mod 13 = 4 by trying successive powers
• whilst exponentiation is relatively easy, finding discrete logarithms is
generally a hard problem
Diffie-Hellman, 1976, Section 10.2 of Stallings
• Based on the difficulty of computing discrete logarithms of large numbers.
• No known successful attack strategies.
• Two numbers public: a prime p, a primitive root q of P.
• User A chooses a random integer XA < q and computes YA = qXamod(p) for
secret A (known only to itself) and similarly user B chooses XB < q and
computes YB = qXbmod(p)..
• Each exchanges YA and YB, while XA, XB remains private
• Parties A and B compute K = YBXamod(p) and K= YAXbmod(p), respectively,
• K= (YB)Xa mod p = (qXb)Xa mod p = (qXa)Xb mod p = (YA)Xb mod p
• Attacking the secret key of user A for example will require opponent to
calculate
XA= indb,p(YA)= dlogb,p(YA)or the other way around.
• Example p= 353 and a primitive root of 353, q = 3. Suppose A and B choose
XA=97, XA= 233.
• YA = 397mod(353) = 40, YB = 3233mod(353) = 248
• K= 160.. Attacker must 3Xamod(353) = 40 or 3Xbmod(353)=248..
• RSA is more convenient because there is no need to distribute keys.
• DES is within two orders of magnitude faster.
• A viable combination is to distribute the secret keys using RSA, and
then, for the bulk data to use DES.
• Similar combination is implemented in the Pretty Good Privacy
(PGP) method.
• A number of public-key ciphers are based on the use of an abelian
group. For example, Diffie-Hellman key exchange involves
multiplying pairs of nonzero integers modulo a prime number p.
Keys are generated by exponentiation over the group, with
exponentiation defined as repeated multiplication.
Elliptic Curves Chapter 10.3 and 10.4..
• The same level of security but shorter key are possible.
• An equation in two variables. For cryptography, the variables
and coefficients are restricted to elements in a finite field,
which results in the definition of a finite abelian group.
• Elliptic curves are not ellipses. They are so named because
described by cubic equations, similar to the circumference of an
ellipse. In general, cubic equations for elliptic curves take the
form of y2 + axy + by = x3 + cx2 + dx + e..
• Limiting attention (Stallings) to y2 = y3 + ax + b is sufficient. y
= sqrt(y3 + ax + b)
El Gamal public-key cryptosystem
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Secure against CT only attacks.
Each party (say Bob) chooses the following parameters.
p, large prime number, q- primitive root of p, made public.
a random a {2, 3, . . . , p − 1}, private
¯= qa(mod p), made public.
• Encrypting: Choose a random k {1, 3, . . . , p − 1} (a).
Suppose message is a number x < p.
• Epublic−k(x) = {qk(mod p), x · ¯k( mod p)}.
• Two numbers, the first one hides k, and the second one the
message.
• Decrypting: Dprivate−k(y1, y2) = y2 · (y1a)-1(mod p)
• y2 · (y1a)-1 = x · ¯k(qak)-1 = x · = x · (qak) · (qak)-1(mod p) = x
• Check example next slight.
El Gamal public-key cryptosystem
• Example:
– p = 43, q=3 primitive root of p, Alice’s choice of secret
key is a=7,
– ¯ = qa( mod p) = 37( mod 43) = 37,
– Bob picks a random key k=26, and his message x=14,
y1= 326 = 15 mod(43), y2= 3726 14 = 31 mod(43),
– CT= {15, 43}
large prime number, q- primitive root of p, made public.
• Alice: 31 · (157)-1 = 14( mod 43).
• El Gamal encryption is randomized, depends on
random k. So the same x has many encryptions.
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Authentication Server
Based on the fact that there exists a trustworthy authentication server.
The authentication server provides a secure way for pair of processes to
obtain secret keys.
Needham & Schroeder suggested two mechanisms to construct such a
server:
– Authentication with secret keys.
– Authentication with public keys.
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Kerberos is based on the secret-key method.
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Secret Key Authentication
A,B : the processes.
S : the server
N : a nonce
K : a key.
Only the server knows K and K
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A S:A,B,Na
SA:{Na, B, Kab, {Kab, A, timestamp}Kb}Ka
AB:{Kab, A, timestamp}Kb
BA:{Nb}Kab
AB:{Nb-1}Kab
• Digital Signatures
• Enable to verify that a message was originally
produced by the signatory.
• Enable to verify that a message content was not
subsequently altered.
• Handwritten signatures cannot entirely provide that.
• Hard to detect forged signatures.
• Hard to prevent alteration of the document.
• Digital Signature with Public Key
• For each message we add a signature which is
constructed as follows:
– Compute a digest function of the message (like hash function) to
reduce length.
– Encrypt the result using our private key.
• The receiver
– Deciphers the signature using our public key.
– Computes the digest function on the document.
– Compares the two to validate the document.