PUBLIC KEY
CRYPTOGRAPHY
Prime - totient - coprime –
relative prime - RSA
prasaugus
Friday, 28
March 2014
1
Prime numbers
Divisible by 1 and itself
Friday, 28 March 20142
 Any number
other than prime
are broken to
prime.
 A group of
smaller prime
number
 Factorization
involves set of
prime numbers
to bring back the
number
 Prime factors are
unique
Prime Number
What is it made of ?
Factorization
Friday, 28 March 20143
Discrete Logarithm Problem
 46 mod 12 Ξ 10
 3n mod 17 Ξ Equally
likely
with in 17
 Modulus of any
given number with
any exponent is
equally likely
Friday, 28 March 20144
Friday, 28 March 20145
 32 mod 17 Ξ 9
 33 mod 17 Ξ 10
 34 mod 17 Ξ 13
 35 mod 17 Ξ 5
 36 mod 17 Ξ 15
 37 mod 17 Ξ 11
 38 mod 17 Ξ 16
 39 mod 17 Ξ 14
 310 mod 17 Ξ 8
 311 mod 17 Ξ 7
Equally likely……!
Discrete
Logarithm
problem
(one way
function)
Is to find the exponent given
the resultant value
Eg : given 7 to find the
exponent of 3
You may find it easy with
smaller prime numbers….
It would take years to find if the
prime number was 100 digits
long…..
Strength of this one way
function is the time needed to
compute
Friday, 28 March 20146
Sender
Intruder/eavesdropper
Receiver
Public distribution of generator ‘g’ and prime number ‘p’
gn mod p Ξ c
315 mod 17 Ξ 6 313 mod 17 Ξ 12
3, 17
3, 17
3,17
g = 3
p = 17
Spr= 15
Rpu= 12
Rpr= 13
Spu= 6
Sender selects his private ‘n’ to generate his public key and
distributes it to all
Receiver keeps sender’s public key and selects his private
key ‘n’ and sends back his public key to all
THE CRUX NOW!!Sender = Rpu
Spr mod 17 = Actual Secret Key 10
Receiver = Spu
Rpr mod 17 = Actual Secret Key 10
Friday, 28 March 20147
The Story behind the logic
1215 mod 17 Ξ 10
12 Ξ 313 mod 17
313^15 mod 17 Ξ 10
Spr= 15
Rpu= 12
Rpr= 13
Spu= 6
Sender Receiver
613 mod 17 Ξ 10
6 Ξ 315 mod 17
315^13 mod 17 Ξ 10
Without any of the private keys, intruder or eavesdropper
cannot access the secret key.
It requires huge computation power to find it
 Thanks to Diffie –
Hellman
 Who devised the
algorithm to
share keys in
public
Public Key
Cryptography
Thus the sharing of the keys
between any unknown person
is made
Friday, 28 March 20148
 Complexities
 What if there are
multiple
receivers
 Key
management
problem
 Computation
overhead
Friday, 28 March 20149
 Ronald Rivest
 Adi Shamir
 Leonard
Adleman
You know us all by name
Friday, 28 March 201410
Sender Receiver
Intruder
R wants to send ‘89’
p1 = 53
p2 = 59
n = 53 * 59
n = 3127
(n) = 3016 by totient
e = 3
d = 2011
Exponent e
- Odd
- Not a factor of (n)
Calculated by modular
Inverse using Euclidean
Algorithm
d = (k * (n) + 1)/e
Hide eth
except
n = 3127
e = 3
n = 3127
e = 3
n = 3127
e = 3
n = 3127
e = 3
c = 1394
1394d Ξ 89 mod 3127
Message = 89
893 mod 3127 Ξ 1394
Crypt c = 1394
c = 1394
Friday, 28 March 201411
 In 1970 James
Alice, a British
Engineer, devise
d a plan.
 ‘A’ Sends an
open lock to
all, let those who
wish to send
message may
lock it and send
back.
 ‘A’ opens all
locks with his one
and only key
RSA
Foundation of RSA
To resolve computational
complexityy
Key Management
Friday, 28 March 201412
 Multiplication is
easy to perform
 (computer takes
less than
seconds to do it)
 Instead finding
factors of given
numbers is hard.
 (for larger
numbers days
and years)
factorizor
RSA
Foundation of RSA
To resolve computational
complexityy
Key Management
Friday, 28 March 201413
  PHI function is
to measure the
breakability of
the number;
 Where it is less
than ‘n’ and not
a factor of ‘n’
 (Prime) = P-1
 (7) =6
RSA
Foundation of RSA
To resolve computational
complexityy
Key Management
Totient of a number
 When n is a product of two primes, in
arithmetic operations modulo n, the
exponents behave modulo the totient
φ(n) of n
 15 = 3 x 5
 φ(15) = 8
 Relation
(43)5 mod 15 Ξ 4(3x5)mod 8 mod 15 Ξ 47 mod 15
Friday, 28 March 201414
Connection bw PHI φ function
and Modular exponentiation
 Connection
mφ(n) Ξ 1 mod n
let us assume any two numbers such that
they do not share any common factors
m = 5 and n = 8
5φ(8) Ξ 1 mod 8
Friday, 28 March 201415
Connection bw PHI φ function
and Modular exponentiation
 Breakthrough
a. 1k = 1
b. m* 1k = m
based on the above propositions
mφ(n) Ξ 1 mod n can be written as
mk*φ(n) Ξ 1 mod n using (a)
m*mk*φ(n) Ξ m mod n using (b)
Finally we get
mk*φ(n)+1 Ξ m mod n
Friday, 28 March 201416
Friday, 28 March 201417
Sender Receiver
Intruder
R wants to send ‘89’
p1 = 53
p2 = 59
n = 53 * 59
n = 3127
(n) = 3016 by totient
e = 3
d = 2011
Exponent e
- Odd
- Not a factor of (n)
Calculated by modular
Inverse using Euclidean
Algorithm
d = (k * (n) + 1)/e
Hide eth
except
n = 3127
e = 3
n = 3127
e = 3
n = 3127
e = 3
n = 3127
e = 3
c = 1394
1394d Ξ 89 mod 3127
Message = 89
893 mod 3127 Ξ 1394
Crypt c = 1394
c = 1394
QUERIES ?
Just a sec …
Khan Videos on Cryptography
https://engineering.purdue.ed
u/kak/compsec/NewLectures/
Lecture12.pdf
Friday, 28 March 201418
Friday, 28 March 201419
 Confidential
communication
 An individual
can use (e,n)
and (d, n) as
public and
private keys
respectively.
 If the message is
long, it could be
used as block
cipher to reduce
the size
Usage of RSA
Choice of values, keys, primes
THANKS
EVERYONE
prassanna john paul
prasaugus@gmail.com
Friday, 28
March 2014
20

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Cryptography - key sharing - RSA

  • 1. PUBLIC KEY CRYPTOGRAPHY Prime - totient - coprime – relative prime - RSA prasaugus Friday, 28 March 2014 1
  • 2. Prime numbers Divisible by 1 and itself Friday, 28 March 20142
  • 3.  Any number other than prime are broken to prime.  A group of smaller prime number  Factorization involves set of prime numbers to bring back the number  Prime factors are unique Prime Number What is it made of ? Factorization Friday, 28 March 20143
  • 4. Discrete Logarithm Problem  46 mod 12 Ξ 10  3n mod 17 Ξ Equally likely with in 17  Modulus of any given number with any exponent is equally likely Friday, 28 March 20144
  • 5. Friday, 28 March 20145  32 mod 17 Ξ 9  33 mod 17 Ξ 10  34 mod 17 Ξ 13  35 mod 17 Ξ 5  36 mod 17 Ξ 15  37 mod 17 Ξ 11  38 mod 17 Ξ 16  39 mod 17 Ξ 14  310 mod 17 Ξ 8  311 mod 17 Ξ 7 Equally likely……! Discrete Logarithm problem (one way function) Is to find the exponent given the resultant value Eg : given 7 to find the exponent of 3 You may find it easy with smaller prime numbers…. It would take years to find if the prime number was 100 digits long….. Strength of this one way function is the time needed to compute
  • 6. Friday, 28 March 20146 Sender Intruder/eavesdropper Receiver Public distribution of generator ‘g’ and prime number ‘p’ gn mod p Ξ c 315 mod 17 Ξ 6 313 mod 17 Ξ 12 3, 17 3, 17 3,17 g = 3 p = 17 Spr= 15 Rpu= 12 Rpr= 13 Spu= 6 Sender selects his private ‘n’ to generate his public key and distributes it to all Receiver keeps sender’s public key and selects his private key ‘n’ and sends back his public key to all THE CRUX NOW!!Sender = Rpu Spr mod 17 = Actual Secret Key 10 Receiver = Spu Rpr mod 17 = Actual Secret Key 10
  • 7. Friday, 28 March 20147 The Story behind the logic 1215 mod 17 Ξ 10 12 Ξ 313 mod 17 313^15 mod 17 Ξ 10 Spr= 15 Rpu= 12 Rpr= 13 Spu= 6 Sender Receiver 613 mod 17 Ξ 10 6 Ξ 315 mod 17 315^13 mod 17 Ξ 10 Without any of the private keys, intruder or eavesdropper cannot access the secret key. It requires huge computation power to find it
  • 8.  Thanks to Diffie – Hellman  Who devised the algorithm to share keys in public Public Key Cryptography Thus the sharing of the keys between any unknown person is made Friday, 28 March 20148  Complexities  What if there are multiple receivers  Key management problem  Computation overhead
  • 9. Friday, 28 March 20149  Ronald Rivest  Adi Shamir  Leonard Adleman You know us all by name
  • 10. Friday, 28 March 201410 Sender Receiver Intruder R wants to send ‘89’ p1 = 53 p2 = 59 n = 53 * 59 n = 3127 (n) = 3016 by totient e = 3 d = 2011 Exponent e - Odd - Not a factor of (n) Calculated by modular Inverse using Euclidean Algorithm d = (k * (n) + 1)/e Hide eth except n = 3127 e = 3 n = 3127 e = 3 n = 3127 e = 3 n = 3127 e = 3 c = 1394 1394d Ξ 89 mod 3127 Message = 89 893 mod 3127 Ξ 1394 Crypt c = 1394 c = 1394
  • 11. Friday, 28 March 201411  In 1970 James Alice, a British Engineer, devise d a plan.  ‘A’ Sends an open lock to all, let those who wish to send message may lock it and send back.  ‘A’ opens all locks with his one and only key RSA Foundation of RSA To resolve computational complexityy Key Management
  • 12. Friday, 28 March 201412  Multiplication is easy to perform  (computer takes less than seconds to do it)  Instead finding factors of given numbers is hard.  (for larger numbers days and years) factorizor RSA Foundation of RSA To resolve computational complexityy Key Management
  • 13. Friday, 28 March 201413   PHI function is to measure the breakability of the number;  Where it is less than ‘n’ and not a factor of ‘n’  (Prime) = P-1  (7) =6 RSA Foundation of RSA To resolve computational complexityy Key Management
  • 14. Totient of a number  When n is a product of two primes, in arithmetic operations modulo n, the exponents behave modulo the totient φ(n) of n  15 = 3 x 5  φ(15) = 8  Relation (43)5 mod 15 Ξ 4(3x5)mod 8 mod 15 Ξ 47 mod 15 Friday, 28 March 201414
  • 15. Connection bw PHI φ function and Modular exponentiation  Connection mφ(n) Ξ 1 mod n let us assume any two numbers such that they do not share any common factors m = 5 and n = 8 5φ(8) Ξ 1 mod 8 Friday, 28 March 201415
  • 16. Connection bw PHI φ function and Modular exponentiation  Breakthrough a. 1k = 1 b. m* 1k = m based on the above propositions mφ(n) Ξ 1 mod n can be written as mk*φ(n) Ξ 1 mod n using (a) m*mk*φ(n) Ξ m mod n using (b) Finally we get mk*φ(n)+1 Ξ m mod n Friday, 28 March 201416
  • 17. Friday, 28 March 201417 Sender Receiver Intruder R wants to send ‘89’ p1 = 53 p2 = 59 n = 53 * 59 n = 3127 (n) = 3016 by totient e = 3 d = 2011 Exponent e - Odd - Not a factor of (n) Calculated by modular Inverse using Euclidean Algorithm d = (k * (n) + 1)/e Hide eth except n = 3127 e = 3 n = 3127 e = 3 n = 3127 e = 3 n = 3127 e = 3 c = 1394 1394d Ξ 89 mod 3127 Message = 89 893 mod 3127 Ξ 1394 Crypt c = 1394 c = 1394
  • 18. QUERIES ? Just a sec … Khan Videos on Cryptography https://engineering.purdue.ed u/kak/compsec/NewLectures/ Lecture12.pdf Friday, 28 March 201418
  • 19. Friday, 28 March 201419  Confidential communication  An individual can use (e,n) and (d, n) as public and private keys respectively.  If the message is long, it could be used as block cipher to reduce the size Usage of RSA Choice of values, keys, primes