1. CS104 : Discrete Structures
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CS104 : Discrete Structures
Chapter III
Proof Techniques
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

2. Rules of Inference – Valid Arguments in Propositional Logic
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Rules of Inference – Valid Arguments in Propositional Logic
Argument: An argument is a sequence of statements that end with a conclusion.
Valid: An argument is valid if and only if it is impossible for all premises (preceding statements) to be true and the conclusion to be false. By valid, we mean that the conclusion of the argument must follow from the truth of the premises of the argument.
Consider the arguments:
“If you have a current password, then you can log onto the network.”
“You have a current password.”
Therefore,
“You can log onto the network.”
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

3. Rules of Inference – Valid Arguments in Propositional Logic
4/16/2017
Rules of Inference – Valid Arguments in Propositional Logic
The conclusion “You can log onto the network” must be true when the premises “If you have a current password, then you can log onto the network” and “You have a current password” are true.
Let p= “You have a current password.”
and q=“You can log onto the network.” Then the argument has the form
where  is the symbol that denotes “therefore.”
The statement ((p → q) ⌃p) → q is a tautology.
p → q p
 q
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

4. Rules of Inference – Rule for Propositional Logic
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Rules of Inference – Rule for Propositional Logic
The argument form with premises p1, p2, …., pn and conclusion q is valid, when (p1⌃p2⌃….⌃pn) → q is a tautology.
To show that an argument is valid, instead of showing by truth table, we can establish the validity of some relatively simple argument forms, called rules of inference, which can be used as building blocks to construct more complicated valid argument forms.
The tautology ((p → q) ⌃p) → q is the basis of the rule of inference called modus ponens or law of detachment.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

5. Rules of Inference – Rules for Propositional Logic
4/16/2017
Rules of Inference – Rules for Propositional Logic
Example: Suppose that the conditional statement “If it snows today, then we will go skiing” and its hypothesis “It is snowing today”, are true. Then by modus ponens, it follows that the conclusion of the conditional statement, “We will go skiing” is true.
Q1: Determine whether the argument given here is valid and determine whether its conclusion must be true because of the validity of the argument.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

6. Rules of Inference – Rules for Propositional Logic
4/16/2017
Rules of Inference – Rules for Propositional Logic
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

7. Rules of Inference – Rule for Propositional Logic
4/16/2017
Rules of Inference – Rule for Propositional Logic
Example: State which rule of inference is the basis of the following argument: “It is below freezing now. Therefore, it is either below freezing or raining now”.
Sol: Let p= “It is below freezing now.” and q = “It is raining now.” Then this argument is of the form: p
p v q
This is an argument that uses the addition rule.
Q2: State which rule of inference is the basis of the following argument: “It is below freezing and raining now. Therefore, it is below freezing”.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

8. Introduction to Proofs
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Introduction to Proofs
A theorem is a statement that can be shown to be true (usually important statement)
Less important theorem sometimes are called propositions
A proof is a sequence of statements (valid argument) to show that a theorem is true
The statements to be used in proofs include:
Axioms (statement assumed to be true without proof)
Ex: If x is positive integer then x+1 is positive integer.
Hypothesis (premises) of the theorem
Previously proven theorems
Rules of inference used to draw conclusions and to move from one step to another
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

9. Introduction to Proofs
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Introduction to Proofs
Axioms
Rules of
inference
Hypothesis
New theorem
proven theorems
A less important theorem that is helpful in the proof of other results is called a lemma
A corollary is a theorem that can be established directly from a theorem that has been proved
A conjuncture is a statement that is being proposed to be a true statement, usually on the basis of some partial evidence
When a proof of a conjuncture is found, the conjuncture becomes a theorem
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

10. Introduction to Proofs
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Introduction to Proofs
Example 1: If I have a car (C) I will drive to Makkah (M). My boss gave me 60,000 (G) or Fired me (F). If I have SR60,000 (H) then I have a car (C). My boss did not fire me. Therefore I will drive to Makkah (M).
GF Hypothesis
F Hypothesis
G Disjunctive syllogism rule using 1 and 2
GH Axiom 
H C Hypothesis
G C Hypo. syllogism using 4,5
C Modus ponens using 3 and 6
C M Hypothesis
M Modus ponens using 7 and 8
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

11. Methods of proving theorems: Direct proofs
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Methods of proving theorems: Direct proofs
A direct proof of a conditional statement pq is constructed when the first step is the assumption that p is true; subsequent steps are constructed using rules of inference, with the final step showing that q must also be true
In a direct proof, we assume that p is true and use axioms, definitions, and previously proven theorems, together with rules of inference, to show that q must also be true
Def: The integer n is even if there exists an integer k such that n = 2k, and n is odd if there exists an integer k such that n = 2k + 1.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

12. Methods of proving theorems: Direct proofs
4/16/2017
Methods of proving theorems: Direct proofs
Example 2: Use a direct proof to show that “if n is even then n2 is even”
Proof: Assume that n is even (hypothesis)
=> n = 2k where k is integer (definition of even number)
=> n2 = (2k)2 = 4k2 = 2(2k2) (By squaring)
Since r = 2K2 is integer (Axiom)
=> n2 = 2r is even
Q 3: Use a direct proof to show that “if n is odd then n2 is odd”
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

13. Methods of proving theorems: Proof by contraposition
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Methods of proving theorems: Proof by contraposition
An indirect proof of a conditional statement p q is a direct proof of its contraposition q  p.
Example 3: Use an indirect proof to show that if a and b are integers, and (a + b) ≥ 15, then a ≥ 8 or b ≥ 8.
Proof: The contraposition of (a + b ≥ 15)  (a ≥ 8) v (b ≥ 8) is (a < 8)  (b < 8)  (a + b < 15)
Suppose (a < 8)  (b < 8) (hypothesis).
=> (a ≤ 7)  (b ≤ 7),
=> (a + b) ≤ 14,
=> (a + b) < 15.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

14. Methods of proving theorems: Proof by contraposition
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Methods of proving theorems: Proof by contraposition
Example 4: Use an indirect proof to show that “if n2 is even then n is even”
Proof: The contraposition is “if n is not even then n2 is not even”
Assume that n is not even i.e., n is odd (hypothesis)
=> n = 2k+1 where k is integer (definition of odd number)
=> n2 = (2k + 1)2
= 4k2 + 4k + 1
= 2(2k2 + 2k) + 1 = 2r + 1, where r = 2k2 + 2k
Since r is integer (Axiom) => n2 is not even
Q 4: Use an indirect proof to show that “if n is odd then n2 is odd”
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

15. Methods of proving theorems: Proof by contraposition
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Methods of proving theorems: Proof by contraposition
Example 5: Prove that if n = ab then a  n or b  n where a and b are positive integers
Proof: Let p=“an”, q=“bn” and r=“n=ab”
We want to prove that r pq
=> The contraposition is (pq )   r (By definition)
=> p  q   r (De Morgan’s law)
Now, assume that an and bn (p  q)
=> a.b  n.n = n (by multiplying above twos)
=> ab  n
=>  r
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

16. Methods of proving theorems: Vacuous Proofs
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Methods of proving theorems: Vacuous Proofs
Vacuous Proofs: A conditional
statement p  q is TRUE
if p is FALSE. If we can show
that p is False, then we have
a proof, called vacuous proof,
of the conditional statement p  q
Example 6: Prove that if x2  0 then 1=2 where x is a real number
Proof: Since x2  0 for every real number then the implication is vacuously true
Example 7: Prove that if he is alive and he is dead then the sun is ice cold.
Proof: Since the hypothesis is always false the implication is vacuously true. p q
p  q F T
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

17. Methods of proving theorems: Trivial Proofs
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Methods of proving theorems: Trivial Proofs
Trivial Proofs: A conditional
statement p  q is TRUE if q
is TRUE. If we can show that
q is TRUE, then we have a
proof, called trivial proof, of
the conditional statement p  q
Example 8: Prove that if x=2 then x2  0 for all real numbers
Proof: Since x2  0 is true then the implication is trivially true. (we didn’t use the fact x=2)
Q 5: Use a trivial proof to show that if n > 1 then n2 ≥ n for all integers p q
p  q F T
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

18. Methods of proving theorems: Proofs by Contradiction
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Methods of proving theorems: Proofs by Contradiction
Proof by Contradiction: To prove a proposition p, assume not p and show a contradiction.
Example 9: Use a proof by contradiction to show that 2 is irrational
Proof: Let 2 is rational
=> 2 = a/b for some integers a and b (b0) (relatively prime).
(Definition of rational numbers)
=> 2 = a2/b2 (Squaring both sides)
=> 2b2 = a2
=> a2 is even (Definition of even numbers)
=> a is even (a = 2k for some k)
=> 2b2 = a2 = (2k)2 = 4k2
=> b is even (Definition of even numbers)
But if a and b are both even, then they are not relatively prime!
Hence, 2 is irrational
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

19. Methods of proving theorems: Proofs by Contradiction
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Methods of proving theorems: Proofs by Contradiction
Why is this method valid ?
The contradiction forces us to reject our assumption because our other steps based on that assumption are logical and justified. The only “mistake” that we could have made was the assumption itself.
Be careful!
Sometimes the contradiction comes from a mistake in the steps of the proof and not from the assumption. This makes the proof invalid.
Example 10: Prove that 1=2
Proof: Suppose that 21 and a=b for some a.
=> 2b b [multiply by b]
=> a+b b [2b=b+b=a+b by hypothesis]
=> (a-b)(a+b)  b(a-b) [multiply by a-b]
=> a2-b2  ab-b2
=> a2 ab [subtract b2 from both sides]
=> a b which contradicts our assumption that a=b
Hence it follows that 1=2
Can you find the error 
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

20. Methods of proving theorems: Proofs by Contradiction
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Methods of proving theorems: Proofs by Contradiction
To prove a conditional statement p  q by contradiction we prove that p  q F is true which is equivalent to p  q .
Example 11: Use a proof by contradiction to show that If 3n+2 is odd then n is odd
Proof: Suppose that 3n+2 is odd and n is even [p  q]
=> n = 2r [hypothesis q, definition of even numbers]
=> 3n = 6r [multiply 1 by 3]
=> 3n+2=2+6r [add 2 to both sides]
=> 3n+2=2(1+3r)
=> 3n+2=2k [let k=1+3r]
=> Thus 3n+2 is even which is false (a contradiction !)
Therefore the implication is true.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

21. Methods of proving theorems: Proofs of Equivalence
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Methods of proving theorems: Proofs of Equivalence
Proofs of Equivalence: To prove p q we have to prove p  q and q  p
Example 12: prove “n is even if and only if n2 is even”
“if n is even then n2 is even” proved in example 2
“n2 is even then n is even” proved in example 4
Therefore, n is even if and only if n2 is even
Proving equivalence of several propositions: If we want to prove that p1 p2  p3 …  pn
Then it is sufficient to prove p1  p2,, p2  p3 … pn  p1
Disproof by Counterexample
Example 13: Prove that “For all real numbers x2 > x” is false
Proof: X=0.5 is a counterexample since 0.52 > 0.5 is not true
Q 6: Prove that “If n is not positive, then n2 is not positive” is false
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

22. Methods of proving theorems: Proof by Cases
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Methods of proving theorems: Proof by Cases
Proof by Cases: A proof by cases must cover all possible cases that arise in a theorem. Each case may cover an infinite number of instances
Example 15: Prove "if n is an integer then n2 ≥ n".
Proof: We can prove that n2 ≥ n for every integer by considering three cases, when n = 0, when n ≥ 1, and when n ≤ -1.
Case 1: When n = 0, since 02 = 0, 02 >= 0. So, n2 ≥ n is true.
Case 2: When n ≥ 1, From n ≥ 1, we get n2 ≥ n (by multiplying with n).
Case 3: When n ≤ -1, since n is negative, n2 is positive, so n2 ≥ n.
Hence, if n is an integer then n2 ≥ n.
Q 7: Prove that |xy| = |x||y|, where x and y real numbers.
Hint: Consider all 4 cases: x and y positive or negative
Common errors with exhaustive proof and proof by cases:
Draw conclusion from non-exhaustive examples
Not covering all possible cases
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

23. Methods of proving theorems: Existence Proofs
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Methods of proving theorems: Existence Proofs
Many theorems state that an object with certain properties exists, i.e., xP(x) where P is a predicate. A proof of such a theorem is called an existence proof. There are two kinds:
Constructive Existence Proof: The proof is established be giving example a such that P(a) is true
Example 16: Prove "there is a positive integer that can be written as the sum of cubes in two different ways.“
Proof: Consider 1729=103+93= Finding such examples may require computer assistance.
Non-constructive Existence Proof: The proof is established by showing that an object a with P(a) is true must exist without explicitly demonstrating one. Proofs by contradiction are usually used in such cases.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

24. Methods of proving theorems: Existence Proofs
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Methods of proving theorems: Existence Proofs
Example 17: Let x1,x2,..,xn be positive integers such that their average is m. prove that there exists xi such that xi ≥ m
Proof: Suppose that there is no such number, i,e.,
x1  m, x2  m, … , xn  m
By adding these inequalities we get: x1+ x2+…+ xn  nm
Dividing by n: (x1+ x2+…+ xn)/n  m
But since the average is defined as (x1+ x2+…+ xn)/n
Then we have m  m which is a contradiction.
Therefore there must be a number xi such that xi ≥ m. But we can not specify which number is that.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

25. Methods of proving theorems: Uniqueness Proofs
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Methods of proving theorems: Uniqueness Proofs
Some theorems state that there is exactly one element with a certain property. A proof of such a theorem is called a uniqueness proof.
Strategy here is (1) show that an element x with the desired property exists (2) show that any other y (y != x) does not have the property, i.e., if x and y both have the property, then x must equal y.
Example 18: Prove that the equation 3x+5 = 9 has a unique solution.
Proof: (1) There exists a solution namely x = 4/3
(2) Suppose that y and z are solution then
3y+5 = 9 = 3z+5
So 3y = 3z
Dividing by 3 we get y = z
This proves that the solution is unique
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

26. Methods of proving theorems: Mathematical Induction
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Methods of proving theorems: Mathematical Induction
Principle of Mathematical Induction:
Let P(n) be a statement for all the positive integers (n = 1, 2, 3, . .). If the following two properties hold:
P(1) is true.
P(k+1) is true if P(k) is true for each positive integer k.
Then P(n) is true for all n.
First part is a simple proposition we call the base step
Second part is an inductive step. Start by assuming P(k) is true, and show that P(k+1) is also true
The assumption that P(k) is true called the inductive hypothesis
So, we prove that
(P(1)  k (P(k)  P(k+1)))  (n P(n))
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

27. Methods of proving theorems: Mathematical Induction
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Methods of proving theorems: Mathematical Induction
Example 19: Suppose we have an infinite ladder, and we want to know whether we can reach every step on this ladder. We know two things:
We can reach the first rung of the ladder
If we can reach a particular rung of the ladder, then we can reach the next rung.
How does Induction Work?
Consider the above example. The rules for reaching steps can help you remember how induction works
Statements (1) and (2) are the basic step and inductive step respectively of the proof that P(n) is true for all positive integers n, where P(n) is the statement that we can reach the nth rung of the ladder.
Consequently, we can invoke the mathematical induction to conclude that we can reach every rung of the ladder
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

28. Methods of proving theorems: Mathematical Induction
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Methods of proving theorems: Mathematical Induction
Example 20 (A Summation Problem): Prove that for any integer n ≥ 1: · · · + n = n(n+1)/2 .
Proof: Let P(n) be the proposition that the sum of the first n positive integers is n(n+1)/2. Then to proof that P(n) is true for all n ≥ 1, we have to show that P(1) is true and P(k+1) is true if P(k) is true for k ≥ 1.
Basic step: P(1) is true, because 1 = 1.(1+1)/2
Inductive step: Let us assume that it is true for n = k, that is, …..+k = k(k+1)/2
Now, if we can prove that it is true for n = k+1 also, then it can be said that it is true for all n.
For, · · · + k + (k + 1)
= k(k + 1)/2 + (k + 1)
= (k + 1)(k/2 + 1)
= (k + 1)(k + 2)/2
=> P(k + 1) is also true, hence P(n) is true for all integer n.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

29. Methods of proving theorems: Mathematical Induction
4/16/2017
Methods of proving theorems: Mathematical Induction
Example 21: Use induction to prove that the sum of the first n odd integers is n2.
Proof: Let P(n) be the proposition that the sum of the first n odd integers is n2. Then to proof that P(n) is true for all n ≥ 1, we have to show that P(1) is true and P(k+1) is true if P(k) is true for k ≥ 1.
Basic step: P(1) is true, because the sum of the first 1 odd integer is 12.
Inductive step: Assume P(k): the sum of the first k odd integers is k2, that is, … + (2k - 1) = k2
Now, if we can prove that it is true for n = k+1 also, then it can be said that it is true for all n.
For, … + (2k-1) + (2k+1)
= k2 + (2k + 1)
= (k+1)2
=> P(k + 1) is also true, hence P(n) is true for all integer n.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

30. Methods of proving theorems: Mathematical Induction
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Methods of proving theorems: Mathematical Induction
Q 8: Use induction to prove that
11! + 22! + … + nn! = (n+1)! - 1, n
Q 9: Use induction to prove that for all n,
Q 10: Use induction to prove that for all n,
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

31. Methods of proving theorems: Mathematical Induction
4/16/2017
Methods of proving theorems: Mathematical Induction
Example 22: Use induction to prove the inequality n < 2n, n > 0.
Proof: Let P(n) be the proposition that n < 2n
Basic step: P(1) is true, because 1 < 21 = 2.
Inductive step: Assume P(k) is true, that is, k < 2K
Now, if we can prove that it is true for n = k+1 also, then it can be said that it is true for all n.
For, k < 2K
=> k + 1 < 2K + 1
≤ 2K + 2K
=2. 2K
= 2K+1
=> k + 1 < 2K+1
=> P(k + 1) is also true, hence P(n) is true for all integer n.
Q 11: Use induction to prove the inequality 2n < n! n > 3.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

32. Methods of proving theorems: Recursive Definitions
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Methods of proving theorems: Recursive Definitions
Recursion: Sometimes it is difficult to define an object explicitly. However, it may be easy to define this object in terms of itself. This process is called recursion.
Recursive defined functions: We use two steps to define a function with the nonnegative integers as its domain:
Basis Step: Specify the value of the function at zero.
Recursive Step: Give a rule for finding its value at an integer from its values at smaller integers.
Such definition is called a recursive or inductive definition.
Example 25: The definition of factorial function:
n! = 1 · 2 · 3 · … · (n-1) · n, n  1
But equivalently, we could define it like this:
Recursive Step
Base Step
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

33. Methods of proving theorems: Recursive Definitions
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Methods of proving theorems: Recursive Definitions
Example 26: The definition of Fibonacci Numbers:
Base Step
Recursive Step
Is there a non-recursive definition for the Fibonacci Numbers?
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

34. Methods of proving theorems: Recursive Definitions
4/16/2017
Methods of proving theorems: Recursive Definitions
Example 27: Suppose that f is defined by
f(0) = 3,
f(n+1) = 2f(n) + 3.
Find f(1), f(2), f(3), and f(4).
Solution: From the recursive definition, it follows that
f(1) = 2f(0) + 3 = = 9,
f(2) = 2f(1) + 3 = = 21,
f(3) = 2f(2) + 3 = = 45,
f(4) = 2f(3) + 3 = = 93.
Example 28: Give a recursive definition of
Solution: The first and second part of the recursive definition are
Base Step
Recursive Step
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

35. Methods of proving theorems: Recursive Definitions
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Methods of proving theorems: Recursive Definitions
Example 29: Give an inductive definition of S = {x: x is a multiple of 3}
Solution:
3  S
x,y  S  x + y  S
x,y  S  x - y  S
No other numbers are in S.
Q 12: Find the Fibonacci numbers f(2), f(3), f(4), f(5), and f(6).
Q 13: Give an inductive definition of an.
Base Step
Recursive Step
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

36. Methods of proving theorems: Recursive Definitions
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Methods of proving theorems: Recursive Definitions
Recursively Defined Sets and Structures: Sets can be defined recursively. Recursive definition of sets have two part – basis step and recursive step.
Basis Step: An initial collection of elements is specified.
Recursive Step: Rules for forming new elements in the set from those already known to be in the set are provided.
Example 30: Consider subset S of the set of integers defined by
Basis Step: 3 S.
Recursive Step: If x, y  S then x + y  S .
The new elements found to be in S are 3 by the basis step,
3+3=6 at the first application of the recursive step,
3+6=6+3=9 at the second application of the recursive step,
6+6=12 at the third application of the recursive step, and so on.
We will show later that S is set of all positive multiples of 3.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

37. Recursive Definitions: Recursively Defined Sets
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Recursive Definitions: Recursively Defined Sets
Example 31: The set of Natural Numbers N can be defined recursively as follows:
1N [basis step]
If x N then x+1N [Recursive step]
Lets try to constructs the set using this definition
1 in N [basis step]
1+1=2 in N [Recursive step]
2+1=3 in N [Recursive step]
Etc…
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Prepared by Dr. Zakir H. Ahmed

38. Prepared by Dr. Zakir H. Ahmed
Algorithms
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Prepared by Dr. Zakir H. Ahmed

39. Algorithms: Definitions
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Algorithms: Definitions
The word ‘algorithm’ comes from the name of a Persian author, Abu Ja’far Mohammad Ibn Musa Al Khowarizmi (825 AD).
Definition: An algorithm is a finite set of instructions that, if followed, carries out a particular task. In addition, all algorithms must satisfy the following criteria:
Input: Zero or more quantities are externally supplied.
Output: At least one quantity is produced.
Definiteness: Each instruction is clear and unambiguous.
Finiteness: If we trace out the instructions of an algorithm, then for all cases, the algorithm terminates after a finite number of steps.
Effectiveness: Every instruction must be very basic so that it can be carried out, in principle, by a person using only pen and paper. It must be feasible.
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Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

40. Prepared by Dr. Zakir H. Ahmed
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Study of Algorithms
How to devise algorithms?
Creating an algorithm is an art which may never be fully automated.
How to validate algorithms?
Once an algorithm is devised, it is necessary to show that it computes the correct answer for all possible legal inputs.
A program can be written and then be verified.
How to test a program?
Debugging is the process of executing programs on sample data sets to determine whether the faulty results occur and, if so, to correct them.
Profiling is the process of executing a correct program on data sets and measuring the time and space it takes to compute the results.
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

41. Pseudocode Conventions
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Pseudocode Conventions
Comments begin with // and continue until the end of line.
Blocks are indicated with matching braces: {and}.
An Identifier begins with a letter: max.
Assignment of values to variables is done using assignment statement : Variable:=expression.
Logical operators: and, or and not are provided.
Relational operators: <, ≤, =, ≠, ≥ and > provided.
Elements of arrays are accessed using: [ and ].
While loop:
while (condition) do {
statements; }
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

42. Pseudocode Conventions
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Pseudocode Conventions
For loop:
for variable:=value1 to value2 step step do {
Statements; }
Repeat-until loop:
repeat
} until (condition)
Conditional statement:
if (condition) then (statement);
if (condition) then (statement 1);
else (statement 2);
4/16/2017
Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

43. Pseudocode Conventions
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Pseudocode Conventions
Case statement:
case
{ : (condition 1): (statement 1);
………………….
: (condition n): (statement n);
: else: (statement n+1); }
Input and output are done using: read and write.
There is only one type of procedure: Algorithm.
An algorithm consists of a heading and a body.
The heading of an algorithm takes the form
Algorithm Name ((parameter list))
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Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

44. Prepared by Dr. Zakir H. Ahmed
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Finding Maximum Value
Input: A sequence of n numbers (a1, a2,…, an).
Output: Maximum of (a1, a2,…, an).
Algorithm Maximum(A, n)
{ Max:=A[1];
for i:=2 to n do
{ if ( Max < A[i]) then
Max:=A[i]; }
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Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

45. Prepared by Dr. Zakir H. Ahmed
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Sorting Problem
Input: A sequence of n numbers (a1, a2,…, an).
Output: A permutation of n numbers (reordering) (a'1, a'2,…, a'n) of the input sequence such that a'1≤ a'2 ≤… ≤ a'n.
Bubble Sort:-
It is a popular sorting algorithm
It swaps adjacent elements that are out of order
Insertion sort:-
Insert an element to a sorted array such that the order of the resultant array be not changed.
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Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

46. Prepared by Dr. Zakir H. Ahmed
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Bubble Sort Algorithm
Bubble sort:
Algorithm BubbleSort (A, n) {
for i:=1 to n-1 do
for j := n downto i+1 do
if ( A[j]<A[j-1]) then
exchange A[j] ↔ A[j-1]; }
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Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

47. Prepared by Dr. Zakir H. Ahmed
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Bubble Sort Algorithm
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Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

48. Insertion Sort Algorithm
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Insertion Sort Algorithm
Insertion sort:
Algorithm InsertionSort (A, n)
{ for i:=2 to n do
{ key:=A[i];
// Insert A[i] into the sorted sequence A[1…i-1].
j:=i-1;
while ( (j>0) and (A[j]>key) ) do
{ A[j+1]:=A[j];
j:=j-1; }
A[j+1]:=key;
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Prepared by Dr. Zakir H. Ahmed
Prepared by Dr. Zakir H. Ahmed

49. Prepared by Dr. Zakir H. Ahmed
End of Chapter III
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Prepared by Dr. Zakir H. Ahmed