For each of the following six program fragments: a) Give an analysis of the running time (Big-Oh will do). b) Implement the code in the language of your choice, and give the running time for several values of N. c) Compare your analysis with the actual running times. (1) sum

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Computers and Technology

Ashleigh

14 February, 17:19

For each of the following six program fragments: a) Give an analysis of the running time (Big-Oh will do). b) Implement the code in the language of your choice, and give the running time for several values of N. c) Compare your analysis with the actual running times. (1) sum

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Answers (1)

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Hailey Neal · Answer 1

1) no of computations are:

sum = 0 - (1 time)

i=0 (1 times)

i
++sum (n times)

++i (n times)

Total = 3n+2

O (n) is the big O notation

2) no of computations are:

sum = 0 - (1 time)

i=0 (1 times)

i
++sum (n^2 times)

++i (n times)

j=0 (n times)

j
++j (n^2 times)

Total = 3n^2 + 3n+2

O (n^2) is the big O notation as we considered the maximum possible computation

3) no of computations are:

sum = 0 - (1 time)

i=0 (1 times)

i
++sum (n^3 times)

++i (n times)

j=0 (n times)

j
++j (n^3 times)

Total = 3n^3 + 3n+2

O (n^3) is the big O notation as we considered the maximum possible computation

4) no of computations are:

sum = 0 - (1 time)

i=0 (1 times)

i
++i (n times)

In the "j" th loop

j=0 (n times)

j
Similarly + +j (n^2 times when i=n-1)

Hence + +sum (n^2 times)

Total = 3n^2 + 3n+2

O (n^2) is the big O notation as we considered the maximum possible computation

6) no of computations are:

sum = 0 - (1 time)

i=0 (1 times)

i
++i (n times)

In the "j" th loop

j=0 (n times)

j
Similarly + +j (n^3 times when i=n-1)

In the "k" th loop

k=0 (max of n^3 times when j=n^3)

k
Similarly + +k (max of n^4 times when j=n^3)

Finally + +sum (n^4 times when j=n^3)

Total = 3n^4 + 3n^3+3n+2

O (n^4) is the big O notation as we considered the maximum possible computation

5) no of computations are:

sum = 0 - (1 time)

i=0 (1 times)

i
++i (n times)

In the "j" th loop

j=0 (n times)

j
Similarly + +j (n^3 times when i=n-1)

In the "k" th loop

k=0 (max of n^3 times when j=n^3)

k
Similarly + +k (max of n^4 times when j=n^3)

Finally + +sum (n^4 times when j=n^3)

Total = 3n^4 + 3n^3+3n+2

O (n^4) is the big O notation as we considered the maximum possible computation