1. Advanced Algorithms Analysis and DesignBy
Dr. Nazir Ahmad Zafar
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

2. Lecture No. 41 RSA Cryptosystem String Matching
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

3. Fermat Theorem Statement
If p is prime, a is positive integer not divisible by p,
ap-1 = 1 mod p OR ap = a mod p
Proof
Consider the set, Zp = {0,1,…, p –1}
Multiplying each element of Zp by “a mod p”, the result is a set, A, of all the elements of Zp with a different sequence, where A = Zp
A = {0, a mod p, 2a mod p……(p-1)a mod p}
{0, a mod p, 2a mod p……(p-1)a mod p} = {0,1,…, p –1} Since A = Zp
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

4. Fermat Theorem
If all the elements are multiplied together, except 0, on both sides we should
{a mod p * 2a mod p… *(p-1) a mod p} mod p = (p-1) mod p OR
a p-1 (p-1)! mod p = (p-1)! mod p
Since (p-1)! is relatively prime to p. So It can be cancelled from both sides
ap-1 mod p ≡ 1 OR
ap-1 ≡ 1 mod p OR
ap ≡ a mod p
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

5. Euler’s Theorem: Generalization of Fermat’s
Statement
If a and n are relatively prime then
a(n) + 1 = a mod n OR a(n) = 1 mod n
Proof
If n = prime, then (n) = n – 1
By Fermat’s Theorem an-1 = a(n) = 1 mod n
If n is a positive integer, then (n) = number of positive integers less than n, relatively prime to n.
Consider such positive integers as follows:
S1 = {x1, x2, . . ., x(n) }
Now multiply each element with a mod n
S2 = {a x1 mod n, a x2 mod n, . . ., a x(n) mod n}
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

6. Euler’s Theorem The set S2 is a permutation of S1 because:
1. a is relatively prime to n.
2. xi is relatively prime to n.
3. Therefore axi is also relatively prime to n.
Hence each axi mod n has value less than n
Hence every element of S2 is relatively prime to n and less than n.
The number of elements of S2 equal to that of S1
Moreover S2 contains no duplicates. It is because if axi mod n = axj mod n, then xi = xj
But S1 has no duplicates
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

7. Euler’s Theorem On multiplying the terms of S1 and S2
 ( axi mod n) =  xi OR
 (axi) = (  xi ) mod n OR
a = 1 mod n OR a = a mod n, Proved
Corollary:
Given primes p and q. Let m and n are integers such that n = p*q and 0 < m < n then
m(n)+1 = m mod n OR m(n) = 1 mod n
(n)
(n)
i=1
i=1
(n)
(n)
i=1
i=1
(n)
(n) + 1
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

8. RSA Cryptosystem Encryption:
Any number m, (m < n), can be encrypted.
ciphertext c = me mod n
Decryption:
cd mod n gives us back m.
Proof
To prove that cd mod n is equal to m:
cd mod n = (me)d mod n
= mde mod n
Since de = 1 mod (n)  de = k(n) + 1
cd = mde = mk(n) +1
By the above corollary to Euler’s theorem,
cd = mde = mk(n) +1 = m mod n = m, since m < n
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

9. Example 7: RSA Cryptosystem
Encrypt message STOP using RSA cryptosystem with p = 43, q = 59 and e = 13, n = pq = 2537,
Solution
gcd(e, (p-1)(q-1)) = 1, encryption can be done
Translate STOP in numerical values, blocks of 4
Encrypt
C = Me mod 2537 = M13 mod 2537
After computing using fast modular multiplication
mod 2537 = 2081; mod 2537 = 2181
The encrypted message is:
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

10. Example 8: RSA Cryptosystem
Decrypt if encrypted using RSA
Public key = (e, n) = (13, = 2537)
Solution
p = 43, p-1 = 42, q = 59, q-1 = 58, e = 13
d = e-1 mod (p-1).(q-1) = 13-1 mod = 937
Decrypt
M = C937 mod 2537 = C937 mod 2537
After computing using fast modular multiplication
mod 2537 = 0704; mod 2537 = 1115
The decrypted message is:
Translating back to English: HELP
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

11. String Matching
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

12. String Matching Problem
We assume that the text is an array T [1 .. n] of length n and that the pattern is an array P[1 .. m] of length m ≤ n.
We further assume that the elements of P and T are characters drawn from a finite alphabet Σ.
For example, we may have Σ = {0, 1} or Σ = {a, b, , z}.
The character arrays P and T are often called strings of characters.
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

13. String Matching Problem
We say that pattern P occurs with shift s in text T (or, equivalently, that pattern P occurs beginning at position s + 1 in text T) if
0 ≤ s ≤ n - m and T [s s + m] = P[1 .. m] i.e T [s + j] = P[ j], for 1 ≤ j ≤ m).
If P occurs with shift s in T, we call s a valid shift;
otherwise, we call s an invalid shift.
String Matching Problem
The string-matching problem is “finding all valid shifts with which a given pattern P occurs in a given text T”.
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

14. Example: String Matching Problem13
Text T a b c a b a a b c a b a c
s = 3
Pattern P a b a a
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

15. Definitions and Notations
Terminology Σ*
The set of all finite-length strings formed using characters from the alphabet Σ. ε
The zero-length empty string, also belongs to Σ*.
|x|
The length of a string x. xy
The concatenation of two strings x and y has length |x| + |y| and consists of the characters from x followed by the characters from y.
w  x
A string w is a prefix of a string x, if x = wy for some string y  Σ*. If w  x, then |w| ≤ |x|.
w  x
A string w is a suffix of a string x, if x = yw for some y  Σ*. If w  x that |w| ≤ |x|.
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

16. 1. Naive Approach The idea is based on Brute Force Approach.
The naive algorithm finds all valid shifts using a loop that checks the condition P[1 .. m] = T[s s + m] for each of the n - m + 1 possible values of s.
It can be interpreted graphically as sliding a “template“ containing the pattern over the text, noting for which shifts all of the characters on the template equal the corresponding characters in the text.
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

17. 1. Naive String Matching Algorithm
NAIVE-STRING-MATCHER(T, P)
1 n ← length[T]
2 m ← length[P]
3 for s ← 0 to n - m
do if P[1 .. m] = T[s s + m]
then print "Pattern occurs with shift" s
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

18. Naive String Matching Algorithm
Worst case Running Time
Outer loop: n – m + 1
Inner loop: m
Total ((n - m + 1)m)
Best-case: n-m
Note
Not an optimal procedure for String Matching problem.
It has high running time for worst case.
The naive string-matcher is inefficient because information gained about the text for one value of s is entirely ignored in considering other values of s.
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

19. 2. The Rabin-Karp Algorithm
Let us assume that Σ = {0, 1, 2, , 9}, so that each character is a decimal digit.
A string of k consecutive characters is viewed as representing a length-k decimal number.
Given a pattern P[1 .. m], let p denote its corresponding decimal value and a text T [1 .. n], we let ts denotes the decimal value of the length-m substring T[s s + m], for s = 0, 1, ..., n - m.
Now, ts = p if and only if T [s s + m] = P[1 .. m]; thus, s is a valid shift if and only if ts = p.
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

20. 2. The Rabin-Karp Algorithm
We can compute p in time Θ(m) using Horner's rule
p = P[m] + 10 (P[m - 1] + 10(P[m - 2] + · · · + 10(P[2] + 10P[1]) )).
Example: Horner's rule
“345” = (4 + 10(3)) = (4 + 30) = = 345
The value t0 can be similarly computed from T [1 .. m] in time Θ(m).
To compute the remaining values t1, t2, , tn-m in time Θ(n - m), it suffices to observe that ts+1 can be computed from ts in constant time.
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

21. 2. The Rabin-Karp Algorithm
Subtracting 10m-1 T[s + 1] removes the high-order digit from ts, multiplying the result by 10 shifts the number left one position, and adding T [s + m + 1] brings in the appropriate low-order digit.
ts+1 = (10(ts – T[s + 1] 10m-1 ) + T[s + m + 1])
The only difficulty with this procedure is that p and ts may be too large to work with conveniently.
Fortunately, there is a simple cure for this problem compute p and the ts's modulo a suitable modulus q.
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design