Vinod HBachelor of Computer Application

Semester - 5BC0052 – Theory of Computer

ScienceFall 2012 Assignment Set – 1

Roll No: 521110833Center Code: 03011

2BC0052 – Theory of Computer Science Set -1

1. Show that the relation a ≡ b(mod m) is an equivalence relation

Let a Є Z.

Reflexive: Since n divides a - a = 0, we have a ≡ a mod m.

Symmetric: Let a ≡b mod m.

=> m divides a - b

=> m divides - (a - b)

=> m divides b - a

=> b ≡ a mod m.

Transitivity: Let a, b, c Є Z such that a ≡ b mod n, b ≡ c mod m.

=> m divides a - b, and m divides b - c

=> m divides ( a - b ) + ( b - c)

=> m divides a - c

=> a ≡ c mod m.

Hence the relation is an equivalence relation.

Example

Suppose m = 5. Then

[0] = {x / x ≡0 mod 5} = {x / 5 divides x – 0 = x} = {…, -10, -5, 0, 5, 10, …}

[1] = {x / x ≡1 mod 5} = {x / 5 divides x - 1}= {…, -9, -4, 1, 6, …}

[2] = {x / x ≡2 mod 5} = {x / 5 divides x -2} = {…, -8, -3, 2, 7, 12, …}

[3] = {x / x ≡3 mod 5} = { x / 5 divides x -3} = {…, -7, -2, 3, 8, 13, …}

[4] = {x / x ≡4 mod 5} = { x / 5 divides x - 4} = {…, -6, -1, 4, 9, 14, …}

Also it is clear that [0] = [5] = [10] = …

[1] = [6] = [11] = …

[2] = [7] = [12] = …

[3] = [8] = [13] = …

[4] = [9] = [14] = …

Therefore the set of equivalence classes is given by {[0], [1], [2], [3], [4]}.

2. Using the definition of order show that x2 + 2x + 1 is O(x2)

The functions f and g referred to in the definition of O-notation are defined as follows. For real numbers

x, f(x) = x2 + 2x + 1 and g(x) = x2

3BC0052 – Theory of Computer Science Set -1

For all real numbers x > 1, | x2 + 2x +1 | = x2 + 2x + 1

≤ x2 + 2x2 + x2

≤ 4x2

≤ 4 | x2 |

Therefore, |f(x) | ≤ 4 | g(x) | for all x > 1 or | f(x) | ≤ C | g(x) | for all x > k

Where C = 4 and k = 1

Hence x2 + 2x +1 is O(x2)

Example

Use the definition of order to show that 5x3 – 3x + 4 is O(x3).

Solution: The functions f and g referred to in the definition of O-notation are defined as follows.

3. Prove by the method of contradiction that √2 is not a rational number

A rational number is of the form p/q where, q ≠ 0 and p, q are not having any common factors.

Assume that √ 2 is a rational number. So it can be written as

If p is even, then it can be written as p = 2k. Therefore 4k2 = 2q2. Therefore q is even.

This is a contradiction to our assumption that p and q have no common factors. Therefore √ 2 is not a

rational number.

Example

Give a proof by contradiction of “if 3n + 2 is odd, then n is odd”.

4BC0052 – Theory of Computer Science Set -1

Solution: Let p: 3n+2 is odd

q: n is odd.

To construct a proof by contradiction, assume that both p and ~ q are true. That is, assume that 3n + 2 is

odd and that n is not odd.

Since n is not odd, it is even.

Now we can show that if n is even, then 3n + 2 is even.

(Verification: n is even n = 2k for some integer k. Substituting 2k for n,

we get 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1) 3n + 2 is even).

Now the statement “3n + 2 is even” is ~ p. Now since p and ~ p are true, we have a contradiction.

This completes the proof by contradiction, proving that if 3n + 2 is odd, then n is odd.

4. Prove by mathematical induction 12+ 22 + 32 + …. + n2 = n (n + 1)(2n + 1)

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(i) Base Step: Let n = 0. Then the sum on the left is zero, since there is nothing to add. The

expression on the right is also zero.

If n=1, left side. = 12=1.

Right Side =

Hence the result is true for n=1

(ii) Induction Hypothesis: Assume that the result to be true for n=m

Then 12 + 22 + … + m2 = 12 + 22 + 32 + … + m2 + (m + 1)2 =

Adding the (m + 1)th term i.e. (m + 1)2 to both side of the above equation, we get

Therefore the result is true for n = m+1. Hence by mathematical induction the given result is true for all

positive integers n.

5BC0052 – Theory of Computer Science Set -1

5. Prove that “The sum of the degrees of the vertices of a graph G is twice the number of edges”

The sum of the degrees of the vertices of a graph G is twice the number of edges. That is, = 2e. (Here e is the number of edges).

Proof: (The proof is by induction on the number of edges ‘e’).

Case-(i): Suppose e = 1. Suppose f is the edge in G with f = uv.

Then d(v) = 1, d(u) = 1. Therefore

= + d (u) + d (v) = 0 + 1 + 1 = 2 = 2 1

= 2 (number of edges).

Hence the given statement is true for n = 1.

Now we can assume that the result is true for e = k - 1.

Take a graph G with k edges.

Now consider an edge ‘f’ in G whose end points are u and v.

Remove f from G.

Then we get a new graph G* = G - {f}.

Suppose d*(v) denotes the degree of vertices v in G*.

Now for any x {u, v}, we have d(x) = d*(x), and

d*(v) = d(v) -1, d*(u) = d(u) - 1.

Now G* has k - 1 edges. So by induction hypothesis

= 2(k - 1).

Now 2(k - 1) = = + d*(u) + d*(v)

= + (d(u) - 1) + (d(v) - 1)

= + d(u) + d(v) - 2 = - 2

2(k - 1) + 2 = 2k =

Hence by induction we get that “the sum of the degrees of the vertices of the graph G is twice the

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numbers of edges”.

6. Prove that “T is a tree there is one and only one path between every pair of vertices

Part 1: Suppose T is a tree. Then T is a connected graph and contains no circuits.Since T is connected, there exists at least one path between every pair of vertices in T. Suppose that between two vertices a and b of T, there are two distinct paths. Now, the union of these two paths will contain a circuit in T, a contradiction (since T contains no circuits). This shows that there exists one and only one path between a given pair of vertices in T. Part 2: Let G be a graph. Assume that there is one and only one path between every pair of vertices in G. This shows that G is connected. If possible suppose that G contains a circuit. Then there is at least one pair of vertices a, b such that there are two distinct paths between a and b. But this is a contradiction to our assumption.

So G contains no circuits. Thus G is a tree