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

2. Lecture No. 39 Number Theoretic Algorithms (Theorems and Algorithms)
Dr. Nazir A. Zafar Advanced Algorithms Analysis and Design

3. Today Covered Some More Proofs GCD as a Linear Combination
Finding GCD, a Recursive Theorem
Euclid’s Algorithm
Extended Euclid’s Algorithm
Time Complexity of Euclid’s Algorithm
Residues and Reduced set of Residues
Groups and Rings
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design

4. Method of Proof by Contraposition
Steps in proving by contraposition
Express the statement to be proved in the form:
 x  D, P(x)  Q(x)
Rewrite the statement in the contrapositive form
 x  D,  Q(x)   P(x)
Prove the contrapositive by direct proof
Suppose that x is an arbitrary but particular element of D such that Q(x) is false
Show that P(x) is false
Dr. Nazir A. Zafar

5. Examples: Proof by Contraposition
Prove that for all integers n, if n2 is even then n is also even
Proof
Express the above statement in the form:
 x  D, P(x)  Q(x)
Suppose that
D = Z,
Even(n, 2)  n2 is even
Even(n)  n is even
We have to prove that
 n  Z, Even(n, 2)  Even(n)
Dr. Nazir A. Zafar

6. Examples: Proof by Contraposition
Contraposition of the above statement
 n  Z,  Even(n)   Even(n, 2) is even
Now we prove above contrapositive by direct proof
Suppose that n is an arbitrary element of Z such that,  Even(n) (n is not even) i.e., n is odd
n2 = n.n = odd. odd = odd
n2 is odd
 Even(n, 2) is even
Hence,  n  Z,  Even(n)   Even(n, 2) is even
Therefore,  n  Z, Even(n, 2)  Even(n) is even
Hence  n  Z, if n2 is even then n is even
Dr. Nazir A. Zafar

7. Examples: Proof by Contraposition
Prove that for all integers n, if n2 is divisible by 7 then n is divisible by 7.
Proof
Express the above statement in the form:
 x  D, P(x)  Q(x)
Suppose that
D = Z,
Div(n, 2, 7)  n2 is divisible by 7
Div(n, 7)  n is divisible by 7
We have to prove that
 n  Z, Div(n, 2, 7)  Div(n, 7)
Dr. Nazir A. Zafar

8. Examples: Proof by Contraposition
Contraposition of the above statement
 n  Z,  Div(n, 7)   Div(n, 2, 7)
Now we prove above contrapositive by direct proof
Suppose that n is an arbitrary element of Z such that,  Div(n, 7) (n is not divisible by 7)
n does contain any factor of 7
n2 does contain any factor of 7
Hence,  n  Z,  Div(n, 7)   Div(n, 2, 7)
Therefore,  n  Z, Div(n, 2, 7)  Div(n, 7)
Hence,  n  Z, if n2 is divisible by 7 then n is divisible by 7.
Dr. Nazir A. Zafar

9. Lemma 1
Statement : The square of an odd integer is of the form 8m + 1 for some integer m.
Proof:
Suppose n is an arbitrary odd integer.
By quotient remainder theorem any integer has the form
4m, 4m + 1, 4m + 2 OR 4m+3
Now since n is an odd integer, hence n can be represented as
4m + 1 OR 4m+3
Now we have to prove that squares of 4m + 1 and 4m + 3 are of the form 8m + 1.
Dr. Nazir A. Zafar

10. Lemma 1
Case 1
Square of 4m + 1
(4m + 1)2 = 16m2 + 8m + 1 = 8(2m2 + m) + 1
= 8m’ + 1, where m ‘ = (2m2 + m)
Case 2
Square of 4m + 3
(4m + 3)2 = 16m2 + 24m + 9 = 8(2m2 + 3m + 1) + 1
= 8m’’ + 1, where m’’ = (2m2 + 3m + 1)
Hence any odd integer has the form 8m + 1 for some m
Dr. Nazir A. Zafar

11. Theorem 1
Statement:
If a and b are any integers, not both zero, then gcd(a, b) is the smallest positive element of the set {ax + by : x, y  Z} of linear combinations of a and b.
Proof
Let s be the smallest positive such linear combination of a and b, i.e.
s = ax + by, for some x, y  Z
By quotient remainder theorem
a = qs + r = qs + a mod s, where q = ⌊a/s⌋.
a mod s = a – qs = a - q(ax + by) = a (1 - qx) + b(-qy)
Dr. Nazir A. Zafar

12. Theorem 1 Hence a mod s is a linear combination of a and b.
But, since a mod s < s, therefore, a mod s = 0
Now a mod s = 0  s | a
Similarly we can prove that, s | b.
Thus, s is a common divisor of both a and b,
Therefore, s  gcd(a, b) (1)
We know that if d | a and d | b then d | ax + by for all x, y integers.
Since gcd(a, b)| a and gcd(a, b) | b, hence gcd(a, b) | s and s > 0 imply that
gcd(a, b) ≤ s. (2)
By (1) and (2), gcd(a, b) = s
Dr. Nazir A. Zafar

13. Corollary
Statement;
For all integers a and b and any nonnegative integer n, gcd(an, bn) = n gcd(a, b).
Proof
If n = 0, the corollary is trivial.
If n > 0, then gcd(an, bn) is the smallest positive element of the set {anx + bny}, i.e.
gcd(an, bn) = min {anx + bny} = min{n.{ax + by}}
= n. min{ax + by}
n times smallest positive element of set {ax + by}.
Hence gcd(an, bn) = n.gcd(x, y)
Dr. Nazir A. Zafar

14. Relatively Prime Integers
Two integers a, b are said to be relatively prime if their only common divisor is 1, i. e, if gcd(a, b) = 1.
Generalized Form of Relatively Prime Integers
We say that integers n1, n2, ..., nk are pairwise relatively prime if, whenever i ≠ j, we have gcd(ni, nj) = 1.
Dr. Nazir A. Zafar

15. Lemma 2
Statement
For any integers a, b, and p, if both gcd(a, p) = 1 and gcd(b, p) = 1, then gcd(ab, p) = 1.
Proof
As, gcd(a, p) = 1, there exist integers x, y such that
ax + py = 1 (1)
gcd(b, p) = 1, there exist integers x’, y’ such that bx’ + py’ = 1 (2)
Multiplying equations (1) and (2) and rearranging, ab(x x′) + p(ybx′ + y′ax + pyy′) = 1, abx’’ + py’’ = 1
Since 1 is a positive linear combination of ab and p,
Hence gcd(ab , p) = 1,which completes the proof
Dr. Nazir A. Zafar

16. Lemma 3
Statement
For all primes p and all integers a, b, if p | ab, then p | a or p | b (or p divides both and b).
Proof
Let P = set of all primes; Z = set of all integers
P(p, ab)  p | ab; Q(p, a, b)  p | a or p | b
Express above statement to be proved in the form:
 a, b, p, P(p, ab)  Q(p, a, b)
 p  P, a, b  Z, p | ab  (p | a or p | b)
Assume for the purpose of contradiction that p | ab but that p ∤ a and p ∤ b.
Dr. Nazir A. Zafar

17. Lemma 3 Now p ∤ a  gcd(a, p) = 1 And, p ∤ b  gcd(b, p) = 1
Since only divisors of p are 1 and p, and by assumption p divides neither a nor b.
Above Lemma 2, states that for any integers a, b, and p, if both gcd(a, p) = 1 and gcd(b, p) = 1, then gcd(ab, p) = 1.
Now, gcd(ab, p) = 1, contradicts our assumption that p | ab, since p | ab implies gcd(ab, p) = p.
This contradiction completes the proof.
Dr. Nazir A. Zafar

18. Theorem 2: GCD Recursion Theorem
Statement
For any nonnegative integer a and any positive integer b, gcd(a, b) = gcd(b, a mod b).
Proof
If we will be able to prove that gcd(a, b) and gcd(b, a mod b) divide each other, It will complete the proof of the theorem. This is because both are nonnegative.
Case 1
We first show that gcd(a, b) | gcd(b, a mod b).
If we let d = gcd(a, b).
By quotient remainder theorem:
(a mod b) = a - qb, where q = ⌊a/b⌋.
Dr. Nazir A. Zafar

19. Theorem 2: GCD Recursion Theorem
Now d = gcd(a, b) 
d | a and
d | b,
Hence, d | (a – qb),
(this is because, a – qb is a linear combination of a and b, where x = 1, y = -q)
And consequently d | (a mod b),
this is because (a mod b = a – qb)
Now, d | b and d | (a mod b), implies that:
d | gcd(b, a mod b)
Hence gcd(a, b) | gcd(a, a mod b). (A)
Dr. Nazir A. Zafar

20. Theorem 2: GCD Recursion Theorem
Case 2
We now show that: gcd(a, a mod b) | gcd(a, b).
If we let, d = gcd(b, a mod b), then
d | b and
d | (a mod b).
By quotient remainder theorem
a = qb + (a mod b), where q = ⌊a/b⌋,
a is a linear combination of b and a mod b,  d | a
Now, d | a and d | b  d | gcd(a, b)
Hence, gcd(a, a mod b) | gcd(a, b) (B)
By (A) and (B):
gcd(a, b) = gcd(b, a mod b).
Dr. Nazir A. Zafar

21. Example: Compute gcd (1970, 1066)
a = 1970, b = 1066
1970 = 1 x = gcd(1066, 904), R = 904
1066 = 1 x = gcd(904, 162), R = 162
904 = 5 x = gcd(162, 94), R = 94
162 = 1 x = gcd(94, 68), R = 68
94 = 1 x = gcd(68, 26), R = 26
68 = 2 x = gcd(26, 16), R = 16
26 = 1 x = gcd(16, 10), R = 10
16 = 1 x = gcd(10, 6), R = 6
10 = 1 x = gcd(6, 4), R = 4
6 = 1 x = gcd(4, 2), R = 2
4 = 2 x = gcd(2, 0), R = 0
Hence gcd(1970, 1066) = 2
Dr. Nazir A. Zafar

22. Euclid’s Algorithm EUCLID(a, b) 1 if b = 0 2 then return a
3 else return EUCLID(b, a mod b)
Example
Compute the gcd of 30 and 21
Solution
EUCLID(30, 21) = EUCLID(21, 9) = EUCLID(9, 3)
= EUCLID(3, 0) = 3
Here, there are three recursive invocations of EUCLID.
The correctness of EUCLID follows from Theorem 2
And the fact that if the algorithm returns a in line 2, then b = 0, and so gcd(a, b) = gcd(a, 0) = a
Dr. Nazir A. Zafar

23. Euclid’s Algorithm Note: The algorithm cannot recurse indefinitely
This is because the second argument strictly decreases in each recursive call
And this second argument is also always nonnegative.
Hence it must be 0 after some number of calls
Therefore, EUCLID always terminates with the correct answer.
Dr. Nazir A. Zafar

24. Running Time of Euclid’s Algorithm
We analyze the worst-case running time of EUCLID as a function of the size of a and b.
We assume without loss of generality that a > b ≥ 0.
This assumption justified because if b > a ≥ 0, then EUCLID(a, b) makes recursive call EUCLID(b, a).
That is, if first argument is less than second one, EUCLID spends one recursive call swapping a, b
Similarly, if b = a > 0, the procedure terminates after one recursive call, since a mod b = 0.
The overall running time of EUCLID is proportional to the number of recursive calls it makes.
Our analysis makes use of the Fibonacci numbers Fk, defined earlier in the first part of our course
Dr. Nazir A. Zafar

25. Running Time of Euclid’s Algorithm
Statement
If a > b ≥ 1 and the invocation EUCLID(a, b) takes k ≥ 1 recursive calls, then a ≥ Fk+2 and b ≥ Fk+1.
Proof
The proof is by induction on k.
Case 1
For base case, let k = 1. Then, b ≥ 1 = F2, and since a > b, we must have a ≥ 2 = F3.
Hence the statement is true for k = 1
Please note that, b > (a mod b), in each recursive call, i.e., first argument is strictly larger than the second and hence the assumption that a > b therefore holds for each recursive call.
Dr. Nazir A. Zafar

26. Running Time of Euclid’s Algorithm
Case 2
Now suppose that the lemma is true for k – 1 i.e., if a > b ≥ 1 and invocation EUCLID(a, b) takes k-1 ≥ 1 recursive calls, then a ≥ Fk+1 and b ≥ Fk.
Case 3
Now we have to prove that statement is true for k i.e. if a > b ≥ 1 and invocation EUCLID(a, b) takes k ≥ 1 recursive calls, then a ≥ Fk+2 and b ≥ Fk+1.
Since k > 0, and b > 0, and EUCLID(a, b) calls EUCLID(b, a mod b) recursively, which in turn makes k - 1 recursive calls.
Since we know that statement is true for k-1, hence b ≥ Fk+1, and (a mod b) ≥ Fk.
Dr. Nazir A. Zafar

27. Running Time of Euclid’s Algorithm
Now we have
b + (a mod b) = b + (a - ⌊a/b⌋ b) (1)
Since, a > b > 0, therefore, ⌊a/b⌋ ≥ 1
 ⌊a/b⌋ b ≥ b
 b - ⌊a/b⌋ b  0
 a + b - ⌊a/b⌋ b  0 + a
 b + (a - ⌊a/b⌋ b)  a
By (1), b + (a mod b) = b + (a - ⌊a/b⌋ b) ≤ a
b + (a mod b) ≤ a
Thus, a ≥ b + (a mod b) ≥ Fk+1 + Fk = Fk+2 .
Hence, a ≥ Fk+2, It completes proof of the theorem
Dr. Nazir A. Zafar

28. Extended Euclid’s Algorithm
EXTENDED-EUCLID(a, b)
1 if b = 0
then return (a, 1, 0)
(d’, x’, y’)  EXTENDED-EUCLID(b, a mod b)
(d, x, y)  (d’, y’, x’ -  a/b y’)
return (d, x, y)
Proof of Correctness
d’ = bx’+ (a mod b)y’
d = bx’+ (a -  a/bb)y’ gcd(a, b) = gcd(b, a mod b)
d = ay’ + b(x’ -  a/by’)
Dr. Nazir A. Zafar

29. Reduced set of residues mod n
Complete set of residues is: n-1
Reduced set of residues consists of all those numbers (residues) which are relatively prime to n
And it is denoted by
Zn* = {k : gcd(k, n) = 1, 0  k < n}
The number of elements in reduced set of residues is called the Euler Totient Function (n)
Example
For n = 10, find reduced list of residues of n
All residues: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Reduced residues (primes) = {1, 3, 7, 9}, (n) = 4
Dr. Nazir A. Zafar

30. Group Definition of a Group
Group is a set, G, together with a binary operation : G * G  G, usually denoted by a*b, such that the following properties are satisfied :
Associativity :
(a*b)*c = a*(b*c) for all a, b, c  G
Identity :
 e  G, such that e*g = g = g*e for all g  G.
Inverse :
For each g  G, there exists the g’, inverse of g, such that g’*g = g*g’ = e
Dr. Nazir A. Zafar

31. The Multiplicative Group Z*n
Zn* = {k : gcd(k, n) = 1, 1  k < n}
For any positive integer n, Zn* forms a group under multiplication modulo n.
Proof:
Binary Operation
Let a, b Zn*, gcd(a, n) = 1; gcd(b, n) = 1
gcd(ab, n) = gcd(a, n)*gcd(b,n) = 1*1 = 1
Associativity holds,
1 is the identity element.
inverse of each element exits
Hence (Zn* ,*) forms a group.
Dr. Nazir A. Zafar

32. Rings
Definition
A ring is a set R with two binary operations + : R × R → R and · : R × R → R (where × denotes the Cartesian product), called addition and multiplication, such that:
(R, +) is an abelian group with identity element 0
(a + b) + c = a + (b + c)
0 + a = a + 0 = a
For every a in R, there exists an element denoted −a, such that a + −a = −a + a = 0
a + b = b + a
Dr. Nazir A. Zafar

33. Rings Definition (Contd..) (R, ·) is a monoid with identity element 1:
(a·b)·c = a·(b·c)
1·a = a·1 = a
Multiplication distributes over addition:
a·(b + c) = (a·b) + (a·c)
(a + b)·c = (a·c) + (b·c)
Note
Ring addition is commutative so that a + b = b + a
But ring with multiplication is not required to be commutative i.e. a·b need not equal b·a.
Rings that satisfy commutative property for multiplication are called commutative rings.
Not all rings are commutative.
Dr. Nazir A. Zafar

34. Rings Rings need not have multiplicative inverses either.
An element a in a ring is called a unit if it is invertible with respect to multiplication
An element a is called invertible under multiplication if there is an element b in the ring such that a·b = b·a = 1,
This b is uniquely determined by a and we write a−1 = b.
Lemma
The set of all units in R forms a group under ring multiplication
Dr. Nazir A. Zafar

35. Example: Rings Example
Prove that Z (+, *) ( the set of integers) is a ring.
Solution
+ and * are binary operation on Z because sum and product of two integers are also an integer
Now,  a, b, c  Z
(a + b) + c = a + (b + c),
0 + a = a + 0 = a
a + (−a) = (−a) + a = 0
a + b = b + a
Hence (Z, +) is an abelian group with identity element 0
Dr. Nazir A. Zafar

36. Example: Rings Since,  a, b, c  Z (a·b)·c = a·(b·c) 1·a = a·1 = a
Hence (Z, ·) is a monoid with identity element 1
Finally  a, b, c  Z
a·(b + c) = (a·b) + (a·c)
(a + b)·c = (a·c) + (b·c)
i.e., multiplication is distributive over addition
Hence we can conclude that Z (+, *) is a ring
Dr. Nazir A. Zafar