Pt1420 Unit 2 Math Paper

Homework-1

Problem-1:

Produce a simple formula for

∑_(i=a)^n▒〖i 〗

where a, n € Z and 1 ≤a≤n.

ANS:

Given Equation we have to find out the summation of natural numbers starting from ‘a’ to ‘n’.

Which can be written as below,

(1) ∑_(i=a)^n▒〖i 〗 = (∑_(i=1)^n▒〖i 〗) – (∑_(i=1)^(a-1)▒〖i 〗)

As we generally know the equation for sum of ‘N’ natural numbers is

(2) ∑_(i=1)^n▒〖i 〗 = n(n+1)/2

From above equations we can able to apply formula -2 in

formula-1 we can get,

∑_(i=a)^n▒〖i 〗 = (∑_(i=1)^n▒〖i 〗) – (∑_(i=1)^(a-1)▒〖i 〗)
= ( n(n+1)/2 ) – ( (a-1)(a-1+1)/2 )
= ( (n2 +n)/2 ) – ( (a2 –a)/2 )
= ( (n2 +n–a2 +a)/2 )
= ( (n2 –a2)+(n+a) )/2
= (n+a)(n–a+1)/2
So finally the deducted formula is as below,

∑_(i=a)^n▒〖i 〗= (n+a)(n–a+1)/2

Problem-2
Given that Algorithm 1 will perform f1(n) = n2+4n steps and Algorithm
2 will perform f2(n) = 39n + 4 steps for a problem of size n, for what
values of n is Algorithm 1 faster than Algorithm 2?
ANS:
Given ,
Algorithm-1 => f1(n)=n2+4n
Algorithm-2 => f2(n)=39n+4
In order for Algorithm-1 to be faster than Algorithm2, The number of steps in Algorithm-1 are less than the no of steps in Algoithm-2
So it deducts as,
Speed of Algorithm-1 > Speed of Algorithm-2
No. of steps of Algorithm-1 < No.of steps of Algorithm-2
i.e. f1(n) < f2(n)
=> n2+4n < 39n+4
=> n2+4n-39n-4 n2-35n-4n0.
So both inequalities (2) and (3) are not satisfied simul-taneously, So it is clear that f(n) ≠ Θ(g(n)).Which clearly implies that,
=>f(n) ∈ Θ(g(n)) is Incorrect.
=>f(n) Θ(g(n))

∑_( i=1)^( n)▒i2 Θ(n2).

Problem-4:
Prove or disprove ∑_( i=1)^( n)▒i0 ∈ Θ(n2).

ANS:
Let f(n) = ∑_( i=1)^( n)▒i0
= n
So f(n) = n
And assume
g(n) = n2

Inorder to show that ∑_( i=1)^( n)▒i0 ∈ Θ(n2).

We have to show,

C1.g(n)≤f(n) ≤C2.g(n) for all n≥n0

C1(n2) ≤ n ≤ C2(n2) for all n≥n0 ….(1)

To satisfy this inequality (1) simultaneously, we have to find the value of C1,C2 and ,n0 using the following inequality

C1(n2) ≤ n ………(2)

n ≤ C2(n2) ………(3)

Inequality (2) is not satisfied for any values of C1,n0 for n≥n0

Inequality (3) is satisfied for a value of C2=1 and n0=1.

So both inequalities (2) and (3) are not satisfied simul-taneously, So it is clear that f(n) ≠ Θ(g(n)).Which clearly implies that,

=>f(n) ∈ Θ(g(n)) is Incorrect.

=>f(n) Θ(g(n))

∑_( i=1)^( n)▒i0 Θ(n2).

Problem-5:

Write pseudo-code for each of the following two algorithms to raise an

integer to an integer power. In both cases, assume n = 2m. Analyze

the run-time complexity of each algorithm in therms of the number of

multiplications performed (determine _(n)).

(a) Na¨ıve:

xn = x × xn−1

x0 = 1

(b) Repeated Squaring:

xn = (xn/2)2

x0 = 1

ANS:

Here firstly generating pseudo code for both algorithms,

Algorithm-(1)

void Naive(int x, int n)
if n > 1 then
value = x×Naive(x,n-1);
endif

if n = 1 then
return x
endif
if n=0 then
return 1
endif

return value

In the above case the Algorithm-1
The problem size reduces from n to n – 1 at every stage, and at every stage two arithmetic operations are involved (one multiplication and one subtraction).

Thus total number of operations needed to execute the function for any given n, can be expressed as sum of 2 operations and the total number of operations needed to execute the function for n-1. Also when n=1, it just needs one operation to execute the function

In other words T(n) can be expressed as sum of T(n-1) and two operations using the following recurrence relation:

T(n) = T(n – 1 ) + 2

T(1) = 1

We need to solve this to express T(n) in terms of n. The solution to the recurrence relation proceeds as follows. Given the relation

T(n) = T(n – 1 ) + 2 ..(1)

we try to reduce the right hand side till we get to T(1) , whose solution is known to us. We do this in steps. First of all we note (1) will hold for any value of n. Let us rewrite (1) by replacing n by n-1 on both sides to yield

T(n – 1 ) = T(n – 2 ) + 2 …(2)

Substituting for T(n – 1 ) from relation (2) in relation (1)

yields

T(n ) = T(n – 2 ) + 2 (2) …(3)

Also we note that

T(n – 2 ) = T(n – 3 ) + 2 …(4)

Substituting for T(n – 2 ) from relation(4) in relation (3) yields

T(n ) = T(n – 3 ) + 2 (3) … (5)

Following the pattern in relations (1) , (3) and (5), we can write a generalized relation

T(n ) = T(n – k ) + 2(k) …(6)

To solve the generalized relation (6), we have to find the value of k. We note that T(1) , that is the number of operations needed to raise a number x to power 1 needs just one operation. In other words

T(1) = 1 ...(7)

We can reduce T(n-k) to T(1) by setting
n – k = 1
and solve for k to yield
k = n – 1
Substituting this value of k in relation (6), we get

T(n) = T(1) + 2(n – 1 )

Now substitute the value of T(1) from relation (7) to yield the solution

T(n) = 2n– 1 ..(8)

When the right hand side of the relation does not have any terms involving T(..), we say that the recurrence relation has been solved. We see that the algorithm is linear in n.

Algorithm-(2)
void RepeatedSquaring(int x,int n)
if n > 1 then
value = RepeatedSquaring (x,n/2)× RepeatedSquaring (x,n/2);
endif

if n = 1 then
return x
endif

if n=0 then
return 1
endif
return value

In the above Algorithm-2,
At every step the problem size reduces to half the size. When the power is an odd number, an additional multiplication is involved. To work out time complexity, let us consider the worst case, that is we assume that at every step an additional multiplication is needed. Thus total number of operations T(n) will reduce to number of operations for n/2, that is T(n/2) with three additional arithmetic operations (the odd power case). We are now in a position to write the recurrence relation for this algorithm as
T(n) = T(n/2) + 3 ..(1)
T(1) = 1 ..(1)
To solve this recurrence relation, we note that T(n/2) can be expressed as
T(n/2) = T(n/4) + 3 ..(2)
Substituting for T(n/2) from (2) in relation (1) , we get
T(n) = T(n/4) + 3(2)
By repeated substitution process explained above, we can solve for T(n) as follows
T(n) = T(n/2) + 3
= [ T(n/4) + 3 ] + 3
= T(n/4) + 3(2)
= [ T(n/8) + 3 ] + 3(2)
= T(n/8) + 3(3)
Now we know that there is a relationship between 3 and 8, and we can rewrite the recurrence relation as
T(n) = T( n/ 23 ) + 3(3)
We can continue the process one step further, and rewrite the relation as
T(n) = T( n/ 24 ) + 3(4)
Now we can see a pattern running through the various relations and we can write the generalized relation as
T(n) = T( n/ 2k) + 3(k)
Since we know that T(1) = 1, we find a substitution such that the first term on the right hand side reduces to 1. The following choice will make this possible
2k= n
We can get the value of k by taking log of both sides to base 2, which yields
k = log n
Now substituting this value in the above relation, we get
T(n) = 1 + 3 log n
Thus we can say that this algorithm runs in LOGARITHMIC TIME, and obviously is much more efficient than the previous algorithm.

Problem-6:

For the following pseudo-code:
1 for i = 1 to n
2 for j = 1 to i
3 for k = 1 to j
4 print ”Hel lo , world ”
(a) Exactly how many times is the print statement on line ?? executed?
Your answer should be a function of n.
(b) What is the run-time complexity of the code?

ANS:

for i = 1 to n
for j = 1 to i
for k = 1 to j
print ”Hello , world”
In the above pseudo code the there are 3 forloops nested so the possible no of iterations are occuring can be determined in the function of ‘n’ as below,
The innermost loop will execute k=1,..,j means the no of inner loop iterations are,
∑_( k=1)^( j)▒〖(1〗) = j
The inner 2nd loop will execute as follows,
∑_( j=1)^( i)▒〖( j〗 ) = i(i+1)/2=(i2+i)/2
The outer loop will execute as follows,
∑_( i=1)^( n)▒ ((i2+i)/2)
( ∑_( i=1)^( n)▒i2 + ∑_( i=1)^( n)▒i)/2
( ( (n(n+1)(2n+1))/6 ) + ( n(n+1)/2 ) ) / 2
(2n3+6n2+4n+3) / 12
So finally it is a function of ‘n’ gives the no of times the print statement will be executed.
F(n)= (2n3+6n2+4n+3) / 12

The run time complexity of above code is O(n3). Since it is a function of degree ‘3’.