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Nastasia [14]
3 years ago
11

Permission must be sought in order to use a play that is in the public domain. True or False?

Computers and Technology
2 answers:
Kruka [31]3 years ago
7 0

Answer:

False

Explanation:

Pubic Domain is PUBLIC which means you do not need permission to use a play

sertanlavr [38]3 years ago
5 0

Answer:

False

Explanation:

Permission can't be sought in order to use a play that is in the public domain.

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For each of the following six program fragments: a) Give an analysis of the running time (Big-Oh will do). b) Implement the code
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Answer:

1) no of computations are :

sum =0 - (1 time )

i=0 ( 1 times)

i<n ( n times )

++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<n ( n times )

++sum(n^2 times)

++i (n times)

j=0 ( n times)

j<n (n^2 times)

++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<n ( n times )

++sum(n^3 times)

++i (n times)

j=0 ( n times)

j<n (n^3 times)

++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<n ( n times )

++i (n times)

In the "j" th loop

j=0 ( n times)

j<i (this executes for max of n^2 times when i=n-1)

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<n ( n times )

++i (n times)

In the "j" th loop

j=0 ( n times)

j<i*i(this executes for max of n^3 times when i=n-1)

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<j(max of n^4 times when j=n^3)

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<n ( n times )

++i (n times)

In the "j" th loop

j=0 ( n times)

j<i*i(this executes for max of n^3 times when i=n-1)

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<j(max of n^4 times when j=n^3)

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

8 0
3 years ago
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