Public Key Cryptography
All secret key algorithms & hash algorithms do the same thing
but public
key algorithms look very different
from each other.
What is common among all of them is that
each participant has two keys, public and private,
and most of them are based on modular arithmetic.
x mod n is the remainder
of x
when divided by n.
e.g.,
8 mod 10 = 8, 18 mod 10 =
8, 24 mod 10 = 4
8 mod 7 = 1, 18 mod 7 =
4, 24 mod 7 = 3
Ø Addition:
Example: addition mod 10
8 + 8 = 6,
1 + 9 = 0,
7 + 6 = 3
See Fig. 6-1 for addition mod 10 Table:
Encryption: Addition mod 10 can be used for encryption of
digits.
Add k, a secret
key between 1-9, to each digit.
Example: if k = 7, then 1987 is encrypted to 8654.
Decryption: Add -k, the additive inverse of k,
to each digit.
An additive
inverse of x is the number you'd
have to add to x to get 0.
Example: if k = 7, then -k
is 3 since 7+3 = 0
Thus
8654 will is decrypted to 1987.
In the
above table (Fig.
6-1), each "0" is the intersection of
k and -k, e.g., 0 is the intersection of 3 and 7.
Ø Multiplication:
Example: multiplication
mod 10 :
8 x 8 = 4, 1 x
9 = 9 , 7 x 6 = 2
See Fig. 6-2 for multiplication mod 10 Table:
Encryption: multiplication by 1, 3, 7, 9 works as a
cipher
since it performs 1-1 mapping.
Example: if k = 7, then 1987 is encrypted to 7369
Decryption: is done by multiplying each digit by k-1 ,
the multiplicative
inverse of k.
It is
the number to multiply by k to get 1.
Example: if k = 7, then k-1 is 3 since
7x3 = 1
In the
above table (Fig.
6-2),
each
"1" is the intersection of k
and k-1.
Note that only {1,3,7,9} have multiplicative inverse mod 10.
Ø
What
is so special about the set {1,3,7,9}?
These
numbers are relatively
prime to 10,
i.e.,
they do not share with 10 any common factors other than 1.
Note
that 9 is not a prime number but it is relatively prime
to 10.
Ø
How
many numbers less than n are relatively prime
to n?
This quantity is referred to as:
Ø(n) and is called the totient function.
· If
n is prime:
then
{1,2,
..., n-1} are all relatively prime and
thus Ø(n) = n-1.
· If n = p.q where p
and q
are two distinct primes,
then Ø(n) = (p-1)(q-1).
Example: for n = 10 = 2.5, Ø(10)
=(2-1).(5-1)=1.4=4,
which is the set {1,3,7,9}.
Ø Exponentation:
Example: exponentiation mod 10
4 2 = 6, 8 8 = 6, 19 =
1 , 76 = 9
See Fig.
6-3 for exponentiation mod 10 Table:
Amazing
fact about Ø(n):
x m mod n = x m mod Ø(n)
mod n
Since Ø(10)=4, in Fig. 6-3:
x m mod 10 = x m mod 4
mod 10
the ith
column is identical to the i+4th column,
e.g., 1st =5th =9th
and 3rd = 7th =11th.
Special case:
if m = 1 mod Ø(n),
then for any number x,
x m mod n =
x mod n.
Example: For n =10, Ø(10)=4
Since m = 9 is
= 1 mod 4:
39 mod
10 = 3 mod 10 = 3 &
69 mod
10 = 6 mod 10 = 6
in
general:
for any x : x9 mod 10 = x mod 10 = x
An
exponentiative inverse of e is the number d such that:
e.d = 1 mod Ø(n)
Example: For n= 10, Ø(10)=4:
e=3 and d=7 are exponentiative inverses since:
3.7 = 21 = 1 mod 4
Encrypt/Decrypt:
· To encrypt m:
compute c = me mod
n
· To decrypt c:
compute m = cd mod
n
Example:
encrypt m = 8: c = 83 = 2
decrypt c = 2: m = 27 = 8
Sign/Verify:
· To sign m:
compute s = md mod
n
·
To verify s:
compute m = se mod n
Example:
sign m = 8: s = 87 = 2
verify s = 2:
m = 23 = 8
In public cryptography:
<e, n>
is public key
&
<d, n> is private key
Key length:
Variable (long for security, short for efficiency).
Most common values is 512
& 1024 bits.
Block size:
plain text
length is variable but less than key length &
cipher
text length equals key length.
Thus RSA is used for encrypting small amount of data,
e.g., a secret key and then
use the secret key cryptography for
encrypting/decrypting large amount of data.
RSA Algorithm:
Generate public & private keys pair:
1. Choose two large primes p and q.
(Typically 256 bits each & keep them
secret).
2. Compute n = p.q &
Ø(n) = (p-1)(q-1).
(It is very hard to factor n into p &
q ).
3. Choose a number e that
is relatively prime to Ø(n).
4. Find a number d that is the
exponentiative inverse of e
i.e., e.d = 1 mod Ø(n).
5. The public key: <e,n> & the private key: <d,n>.
Encrypt/Decrypt:
To encrypt a message m ( less than n):
c = me mod n
To decrypt c:
m = cd mod n
This works since:
cd mod n = (me)d mod n
= me.d mod n
= m mod n // since e.d = 1 mod Ø(n)
= m
// since
m
< n
Sign/Verify:
To sign a message m ( less
than n):
s = md mod n
To verify s:
m = se mod n
This also works since:
se mod n
= me.d mod n
= m mod n = m
Why is RSA Secure:
Every one knows the public key: <e, n>.
To find the private key
<d,n>
you need to know Ø(n) since e.d = 1 mod Ø(n).
To know Ø(n) you need
to know p and q since
Ø(n) = (p-1).(q-1).
Thus to break RSA you
should know:
how
to factor n to find p and q.
Factoring a big number like n is hard. The best technique to factor 512 bit number takes 30,000
MIPS-years!
Efficiency of RSA Operations:
Exponentiation
How to compute 12354 mod 678?
1232 = 123.123 = 15129 = 213 mod 678
1233 = 123.213 =
26199 = 435 mod 678
1234 = 123.435 =
53505 = 621 mod 678
......
12354 = ......
= 87 mod 678
This requires 54
small number multiplications
and 54 small number divisions.
How to
compute 12332 mod 678?
1232 = 123.123 =
15129 = 213 mod 678
1234
= 213.213 = 45369 = 621 mod 678
1238
= 621.621 = 385641 = 537 mod 678
12316 =
537.537 = 288369 = 219 mod 678
12332 =
219.219 = 47961 = 501 mod 678
This requires 5
multiplications
and 5 divisions instead of
32.
To efficiently compute
12354 : 54
is represented in binary as:
1
1
0
1
1
0
|
|
|
| |
(((( (1232)123
) 2
)2123 )2123 )2
This requires 8
multiplications and 8
divisions instead of 32.
Each 1 requires two multipliactions and two
divisions
and each 0 requires one multipliaction
and one division.
Thus in the above we have three 1s and two 0s
and that yields: 3.2 + 2.1 = 8.
Note that we ignore the leading 1.
Another example: y14
, 14 is represented in binary as:
1
1
1
0
|
|
|
(( (
y2) y )2y
)2
This
requires 5 multiplications
and 5 divisions instead
of 32.
Generating RSA Keys
Finding e:
Two
popular values for e are: 3 and 65537 (216 +
1).
These make
public key operations on message m faster
(encryption and signature
verification is me):
-
m3 requires 2 multiplications & 2
divisions.
-
m65537 requires 17 multiplactions & 17
divisions
since the binary value of 65537 is 100..01 (15 zeros).
Finding
n:
If e
= 3:
Choose random numbers x and y then:
p = (2x+1)*3+2 &
q = (2y+1)*3+2
If e = 65537:
Randomly choose p and q and make sure that
they are not 1 mod 65537
(the probability of rejection is
1 in 216).
Once
we selected p and q, then:
n =
p.q &
Ø(n) = (p-1)(q-1).
Finding d:
How to
fine d such that e.d = 1 mod Ø(n)
?
Use
The RSA keys:
public key: <3 | 65537, n> private key: <d , n>.
Alice and Bob agree on: p
(large prime) & g <
p.
Pick SA (512-bit random
number)
Pick SB (512-bit
random number)
Compute TA =
( gSA) mod
p
Compute TB = (gSB) mod p
TA >>> <<<
TB
Compute X = TB SA mod p
Compute Y = TA SB mod p
X is the same as Y!
why?
X = TBSA = gSBSA
Y = TASB = gSASB
No one can compute g (SASB)
by
knowing g (SA ) & g (SB )
The bucket Brigade/Man-in-theMiddle Attach
Pick SA
Pick SX
Pick SB
Compute:
TA = gSA mod
p
TX = gSX mod
p
TB = gSB mod p
TA >>
TA .. TX >> TX
TX <<
TX
.. TB << TB
Compute: KAX = TX
SA mod p KAX
= TA SX mod p
KBX = TX SB
mod p
KBX
= TB SX mod p
Possible Defense
· Each person i picks Si and computes Ti = gSi mod p and