What is the effect in the time required to solve a prob - lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos - sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]

A. log n

B. log log n

C. 100 n

D. n log n

E. n2

F. n3

G. 2n

f (2n) - f (n) = log2

b. lg (lg2+lgn) - lglgn

c. f (2n) / f (n) = 2

d. 2nlg2+nlgn

e. f (2n) / (n) = 4

f. f (2n) / f (n) = 8

g. f (2n) / f (n) = 2

Step-by-step explanation:

What is the effect in the time required to solve a prob - lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos - sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]

from a

f (n) = logn

f (2n) = lg (2n)

f (2n) - f (n) = log2n-logn

lo (2*n) = lg2+lgn-lgn

f (2n) - f (n) = lg2+lgn-lgn

f (2n) - f (n) = log2

2. f (n) = lglgn

F (2n) = lglg2n

f (2n) - f (n) = lglg2n-lglgn

lg2n=lg2+lgn

lg (lg2+lgn) - lglgn

3. f (n) = 100n

f (2n) = 100 (2n)

f (2n) / f (n) = 200n/100n

f (2n) / f (n) = 2

the time will double

4. f (n) = nlgn

f (2n) = 2nlg2n

f (2n) - f (n) = 2nlg2n-nlgn

f (2n) - f (n) = 2n (lg2+lgn) - nlgn

2nLg2+2nlgn-nlgn

2nlg2+nlgn

5. we shall look for the ratio

f (n) = n^2

f (2n) = 2n^2

f (2n) / (n) = 2n^2/n^2

f (2n) / (n) = 4n^2/n^2

f (2n) / (n) = 4

the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare

6. n^3

f (n) = n^3

f (2n) = (2n) ^3

f (2n) / f (n) = (2n) ^3/n^3

f (2n) / f (n) = 8

the ratio will be times 8 the initial

7.2n

f (n) = 2n

f (2n) = 2 (2n)

f (2n) / f (n) = 2 (2n) / 2n

f (2n) / f (n) = 2