Question

What is the time-complexity of the algorithm abc?

Procedure abc(n: integer) s := 0
i :=1
while i ≤ n

s := s+1

i := 2*i return s

consider the following algorithm:

Procedure foo(n: integer) m := 1
for i := 1 to n

for j :=1 to i2m:=m*1

return m

c.) Find a formula that describes the number of operations the algorithm foo takes for every input n? d.)Express the running time complexity of foo using big-O/big-

Answer 1

Solution:

(1)

Explanation:

Finding time complexity of abc algorithm:

=>We can see in the abc() method, there is a while loop starting from i = 1 and ending with condition i $\leq$ n doubling the value of i each time using i = 2i

=>Number of times loops will be executed = log(n) times

=>Time complexity of abc algorithm = O(logn)

(2)

(c)

Explanation:

=>In procedure foo() we can see that there are 2 for loops.

=>Outer for loop is independent for loop loop and will run n times from i = 1 to i = n incrementing the value of i by 1 each time.

=>Inner for loop is dependent for loop and will run from j = 1 to j $\leq$ i incrementing the value of j by 1 each time.

=>Total number of times loops will be executed = 1 + 2 + 3 +....+ n

=>Total number of times loops will be executed = n(n+1)/2

=>Let say cost of comparisons and other operation takes time = C where C is some constant.

=>Cost of all operations = n(n+1)/2 + C

(d)

Explanation:

=>Cost of all operations = n(n+1)/2 + C

=>Hence time complexity in big-O = O(n^2)

I have explained each and every part with the help of statements attached to it.