Question

Question: 0.1. In each of the following situations, indicate whether $\mathbf{\text{f=O(g),orf=Ω(g),}}$or both (in which case $\mathit{f}\mathbf{}\mathbf{=}\mathbf{\odot }\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{)}$

Short answer

Answer

$$\begin{gathered} \text{Theindicationofthefollowingsituationsareasfollows,} \\ \text{a)}\text{Thefunctionsare}f\left(n\right)=n-100\text{andg}\left(n\right)=n-200\text{,whichmatches}f\left(n\right)=\Theta g\left(n\right)\text{andcanbewrittenas}n-100\text{=Θ}n-200\text{.} \\ \text{b)}{\text{Thefunctionsaref(n)}}^{1/2}\text{andg}\left(n\right)=n^{2/3}\text{,whichmatches}f\left(n\right)=\Theta g\left(n\right)\text{andcanbewrittenas.} \\ \text{c)}\text{Thefunctionsare}f\left(n\right)=100n+logn\text{andg}\left(n\right)=100n+\log n^2\text{,whichmatches}f\left(n\right)=\Theta g\left(n\right){\text{andcanbewrittenasn}}^{2/3}{\text{=O(n}}^{2/3}\text{).} \\ \text{d)}\text{Thefunctionsare}f\left(n\right)=nlogn\text{and,whichmatches}f\left(n\right)=\Theta g\left(n\right)\text{andcanbewritten}logn=\Theta \left(n+{\left(logn\right)}^2\right)\text{.} \\ \text{e)}\text{Thefunctionsare}f\left(n\right)=log2n\text{andg}\left(n\right)=log3n\text{,whichmatches}ogn=\Theta \left(n+{\left(logn\right)}^2\right)\text{andcanbewrittenas}ogn=\Theta \left(n+{\left(logn\right)}^2\right)\text{.} \\ \text{f)}\text{Thefunctionsare}f\left(n\right)=log10n\text{andg(n)=}n{\log }^2\text{n,whichmatchesandcanbewrittenas}{n}^{1.01}{\text{=nlog}}^2\text{n} \\ \text{g)}\text{Thefunctionsare}f\left(n\right)=n^{1.01}\text{and}g(n)=n{\log }^{2}n,\text{,whichmatches}{n}^{1.01}=n{\log }^{2}n, \end{gathered}$$

$$\begin{gathered} {\text{h)Thefunctionsaref(n)=n}}^2/lognandg\left(n\right)=n{\left(logn\right)}^2\text{,whichmatchf(n)=Ω}\left(g\left(n\right)\right)\text{andcanbewrittenas}\raisebox{1ex}{$n^2$}\!\left/ \!\raisebox{-1ex}{$logn$}\right.=\Omega n{\left(logn\right)}^2. \\ \text{i)Thefunctionsaref}\left(n\right)={\left(logn\right)}^{logn}andg\left(n\right)=\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$logn$}\right.,\text{whichmatchf(n)=Ω}\left(g\left(n\right)\right){\text{andcanbewrittenaslogn}}^{logn}=\Omega n/\left(logn\right) \\ \text{j)Thefunctionsaref}\left(n\right)=n^{1.01}andg\left(n\right)=\raisebox{1ex}{${\displaystyle n}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle logn}$}\right.,whichmatchf\left(n\right)=\Omega \left(g\left(n\right)\right){\text{andcanbewrittenasn}}^{1.01}=nlo{g}^{2}n. \\ \text{K)Thefunctionsare}f\left(n\right)=\sqrt{n}andg\left(n\right)={\left(logn\right)}^3,\text{whichmatch}f\left(n\right)=\Omega \left(g\left(n\right)\right)\text{andcanbewrittenas}\sqrt{n}=\Omega {(logn)}^3. \\ {\text{I)Thefunctionsaref(n)=n}}^{\frac{1}{2}}andg\left(n\right)=5^{lo{g}_{2}n},\text{whichmatch}f\left(n\right)=\Omega \left(g\left(n\right)\right)\text{andcanbewrittenas}{n}^{\frac{1}{2}}=O5(lo{g}_{2}m). \\ \text{m)Thefunctionsaref}\left(n\right)=n{2}^{n}andg\left(n\right)=3^n,\text{whichmatch(n)=O}\left(g\left(n\right)\right)\text{andcanbewrittenas}n{2}^n=3^n. \\ \text{n)Thefunctionsaref}\left(n\right)=2^{n}andg\left(n\right)=2^{n+1}\text{,whichmatch(}n)=O\left(g\left(n\right)\right)\text{andcanbewrittenas}{2}^n=⊝2^{n+1}. \end{gathered}$$$$\begin{gathered} \text{o)}\text{Thefunctionsare}f\left(n\right)=n!\text{and}g\left(n\right)=2^n\text{,whichmatch}f\left(n\right)=\Omega g\left(n\right)\text{andcanbewrittenas}n!=\Omega \left({\left(2\right)}^n\right)\text{.} \\ \text{p)}\text{Thefunctionsare}f\left(n\right)={\left(\log \text{n}\right)}^{\log \text{n}}\text{and}g\left(n\right)=2{\left({\log }_{2}n\right)}^2\text{,whichmatch}f\left(n\right)=Og\left(n\right)\text{andcanbewrittenas}{\left(ogn\right)}^{logn}=O2{\left({\log }_{2}n\right)}^2\text{.} \\ \text{q)}\text{Thefunctionsare}f\left(n\right)=\sum _{i=1}^n{i}^k\text{and}g\left(n\right)=n^{k+1}\text{,whichmatch}f\left(n\right)=\Theta g\left(n\right)\text{andcanbewrittenas}\sum _{i=1}^n=\Theta \left(n^{k+1}\right)\text{.} \end{gathered}$$

Step-by-step solution

Step 1: Introduction of the concept

An event's indicator function is a random variable that takes the value 1 when the event occurs and 0 when the event does not.

Step 2: Solution (a)

The given functions are:

$f\left(n\right)=n-100$and$g\left(n\right)=n-200$

Because $f=O\left(g\right)$and $\mathit{g}\mathbf{=}\mathit{O}\left(f\right)\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{f}\mathbf{=}\mathit{\Omega }\left(g\right)$are , the function mentioned above corresponds to the case.

1. $\mathit{f}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$denotes when comparing the computing speed of two functions, f(n) and g(n); it is said that f(n) is no faster than g(n).
2. $\mathit{f}\mathbf{=}\mathit{\Omega }\mathbf{\left(}\mathit{g}\mathbf{\right)}$ denotes when comparing the computing speed of two functions, f(n) and g(n); it is said that g(n) is no faster than f(n).

Thus, is the case that corresponds to the above function, and the function can be expressed as $\mathit{n}\mathbf{-}\mathbf{100}\mathbf{=}\mathit{\Theta }\left(n-200\right)$.

Step 3: Solution (b)

The given functions are:

.$f{\left(n\right)}^{1/2}\text{andg}\left(n\right)=n^{2/3}$

Because while collating the powers ${\left(n\right)}^{1/2}<n^{2/3}$, the above-mentioned function corresponds to the case $f\left(n\right)=\Theta g\left(n\right)$.

1. When comparing two functions f(n) and g(n), the big-O notation $f\left(n\right)=Og\left(n\right)$states that g(n) has a faster computational speed than f(n).
2. According to the principles for simplifying functions,${\left(n\right)}^{1/2}$ is dominatedby $n^{2/3}$; hence, g(n) is better than f(n).

Thus, $f\left(n\right)=\Theta g\left(n\right)$is the case that matches the preceding function, and the function can be represented as ${\left(n\right)}^{1/2}=O\left(n^{2/3}\right)$.

Step 4: Solution (c)

The given functions are:

$f\left(n\right)=100n+logn\text{andg}\left(n\right)=100n+\log n^2$

Because $\mathit{f}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$and localid="1658215662597" $\mathit{g}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{f}\mathbf{=}\mathit{\Omega }\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$ are O(n), the above-mentioned function corresponds to the case$f\left(n\right)=\Theta g\left(n\right)$.

1. $\mathit{f}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$denotes when comparing thecomputing speed of two functions, f(n) and g(n), it is said that f(n) is no faster than g(n).
2. denotes when comparing the computing speed of two functions, f(n) and g(n), it is said that g(n) is no faster than f(n).
3. Any polynomial dominates any logarithmaccording to the criteria for simplifying functions. It means that 100n dominates log n in f(n)and n dominates in g(n).

Thus$f\left(n\right)=\Theta g\left(n\right)$, is the case that corresponds to the above function, and the function can be expressed as$\mathbf{100}\mathit{n}\mathbf{+}\mathit{l}\mathit{o}\mathit{g}\mathit{n}\mathbf{\text{⊙g}}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathit{n}\mathbf{+}\mathit{l}\mathit{o}\mathit{g}{\mathit{n}}^{\mathbf{2}}$.

Step 5: Solution(d)

The given functions are:

$f\left(n\right)=nlogn\mathrm{and}g(n)=10n\ln 10n$.

Because $\mathbf{}\mathit{f}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$and $\mathit{g}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{f}\mathbf{=}\mathit{\Omega }\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$ , the above-mentioned function corresponds to the case$f\left(n\right)=\Theta g\left(n\right)$.

1. denotes when comparing the computing speed of two functions, f(n) and g(n), it is said that f(n) is no faster than g(n).
2. denoteswhen comparing the computing speed of two functions, f(n) and g(n), it is said that g(n) is no faster than f(n).
3. Any polynomial dominates any logarithm according to the criteria for simplifying functions.

Thus, $f\left(n\right)=\Theta g\left(n\right)$is the case that corresponds to the above function, and the function can be expressed aslocalid="1658217762370" $\mathit{n}\mathit{l}\mathit{o}\mathit{g}\mathit{n}\mathbf{\text{=⊙}}\mathbf{10}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{10}\mathit{n}$.

Step 6: Solution (e) 

The given functions are:

$f\left(n\right)=log2n\text{andg}\left(n\right)=log3n$.

Because $\mathit{f}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$and $\mathit{g}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{f}\mathbf{=}\mathit{\Omega }\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$, the above-mentioned function corresponds to the case$f\left(n\right)=\Theta g\left(n\right)$.

1. denotes when comparing the computing speed of two functions, f(n) and g(n), it is said that f(n) is no faster than g(n).
2. denoteswhen comparing the computing speed of two functions, f(n) and g(n), it is said that g(n) is no faster than f(n).
3. Any polynomial dominates any logarithmaccording to the criteria for simplifying functions.

Thus,$f\left(n\right)=\Theta g\left(n\right)$is the case that corresponds to the above function, and the function can be expressed as$\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{2}\mathbf{}\mathit{n}\mathbf{}\mathbf{=}\mathbf{\odot }\mathbf{(}\mathbf{log}\mathbf{}\mathbf{3}\mathit{n}\mathbf{)}$ .

Step 7: Solution (f)

The given functions are:

$\mathit{f}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$and$\mathit{g}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{f}\mathbf{=}\mathit{\Omega }\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$.

Because and are, the above-mentioned function corresponds to the case$f\left(n\right)=\Theta g\left(n\right)$.

1. $\mathbf{}\mathit{f}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$denotes when comparing the computing speed of two functions, f(n) and g(n), it is said that f(n) is no faster than g(n).
2. $\mathbf{}\mathit{f}\mathbf{=}\mathit{\Omega }\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$ denotes when comparing the computing speed of two functions, f(n) and g(n), it is said that g(n) is no faster than f(n).

Thus,$f\left(n\right)=\Theta g\left(n\right)$ is the case that corresponds to the above function, and the function can be expressed as $\mathbf{10}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{n}\mathbf{}\mathbf{=}\mathbf{\odot }\mathbf{(}\mathbf{log}\mathbf{}\mathbf{(}{\mathit{n}}^{\mathbf{2}}\mathbf{\left)}\mathbf{\right)}$.

Step 8: Solution (g)

The given functions are:

$f\left(n\right)=n^{1.01}\text{and}g(n)=n{\log }^{2}n$.

The above-mentioned function corresponds to the case$f\left(n\right)=\Omega \left(g\left(n\right)\right)$.

1. The values of the functions f(n) and g(n)should be compared if they are divided by n, but the comparison will take longer.
2. However, according to the simplification rules, the function with the highest power value wins; hence,f(n) is better than g (n).
3. indicates that when comparing two functions,f(n) and g(n),it is said that f(n) is not dominated by g(n).

Thus, $f\left(n\right)=\Omega \left(g\left(n\right)\right)$is the case that corresponds to the above function, and the function can be expressed as${\mathit{n}}^{\mathbf{1}\mathbf{.}\mathbf{01}}\mathbf{=}\mathit{n}\mathbf{}\mathit{l}\mathit{o}{\mathit{g}}^{\mathbf{2}}\mathit{n}$ .

Step 9: Solution (h)

The given functions are:

$f(n)=n^2/lognandg\left(n\right)=n{\left(logn\right)}^2$ .

The above-mentioned function corresponds to the case $f\left(n\right)=\Omega \left(g\left(n\right)\right)$.

1. The values of the functions f(n) and g(n)should be compared if they are divided by , but the comparison by the functions n and there are divisible.
2. However, according to the simplification rules, the function with the highest power value wins; hence,f(n) is better than g (n).
3. indicates when comparing two functions,f(n) and g(n), it is said that f(n) is not dominated by g(n).

Thus, $f\left(n\right)=\Omega \left(g\left(n\right)\right)$is the case that corresponds to the above function, and the function can be expressed as$\mathbf{}\raisebox{1ex}{${\mathbf{n}}^{\mathbf{2}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{n}$}\right.\mathbf{}\mathbf{=}\mathit{\Omega }\mathit{n}{\left(logn\right)}^{\mathbf{2}}$.

Step 10: Solution (i)

The given functions are:

and $f\left(n\right)=n^{1.01}andg\left(n\right)={\left(\log \right)}^{10}$.

The above-mentioned function corresponds to the case$f\left(n\right)=\Omega \left(g\left(n\right)\right)$.

1. The values of the functions f(n) and g(n)should be compared if they are divided by n, but the comparison will take longer.
2. However, according to the simplification rules, the function with the highest power value wins; hence,f(n) is better than g (n).
3. $\mathit{f}\mathbf{=}\mathit{\Omega }\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$ indicates when comparing two functions f(n) and g(n), it is said that f(n) is not dominated by g(n).

Thus, $f\left(n\right)=\Omega \left(g\left(n\right)\right)$is the case that corresponds to the above function, and the function can be expressed as ${\mathit{n}}^{\mathbf{1}\mathbf{.}\mathbf{01}}\mathbf{=}\mathit{n}\mathit{l}\mathit{o}{\mathit{g}}^{\mathbf{2}}\mathbf{}\mathit{n}\mathbf{}\mathbf{.}$.

Step 11: Solution (j)

The given functions are:

$\left(n\right)={\left(logn\right)}^{logn}andg\left(n\right)=\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$logn$}\right.$ .

The above-mentioned function corresponds to the case.role="math" localid="1658221596761" $f\left(n\right)=n^{\log \log n^{}}$

1. The function f(n) can be expressed as $f\left(n\right)=\Omega \left(g\left(n\right)\right)$, and the function with the highest power value wins; therefore,f(n) beats g(n).
2. indicates that when comparing two functions,f(n) and g(n), it is said that f(n) is not dominated by g(n).

Thus,$f\left(n\right)=\Omega \left(g\left(n\right)\right)$ is the case that corresponds to the above function, and the function can be expressed as${\left(logn\right)}^{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{n}}\mathbf{}\mathbf{=}\mathit{\Omega }\raisebox{1ex}{$\mathbf{n}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{n}$}\right.$.

Step 12: Solution (k)

The given functions are:

and .$f\left(n\right)=\sqrt{n}andg\left(n\right)={\left(logn\right)}^3$

The above-mentioned function corresponds to the case$f\left(n\right)=\Omega \left(g\left(n\right)\right)$.

1. The values of the functions f(n) and g(n)should be compared if they are divided by n, but the comparison will take longer.
2. However, according to the simplification rules, the function with the highest power value wins; hence,f(n) is better than g (n).
3. indicates that when comparing two functions,f(n) and g(n), it is said that f(n) is not dominated by g(n).

Thus, $f\left(n\right)=\Omega \left(g\left(n\right)\right)$is the case that corresponds to the above function, and the function can be expressed as$\sqrt{n}=\Omega {(logn)}^3.$.

Step 13: Solution (l)

The given functions are:

.$f\left(n\right)={\left(logn\right)}^{logn}andg\left(n\right)=\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$logn$}\right.$

The above-mentioned function corresponds to the case$f(n)=\Omega \left(g\left(n\right)\right)$.

1. Because while collating the powers , the function g(n) is written as .
2. When comparing two functions, f(n) and g(n), the big-O notation states that g(n) has a faster computational speed than f(n).
3. According to the principles for simplifying functions, is dominatedby ; hence,g(n) is better than f(n).

Thus$f(n)=\Omega \left(g\left(n\right)\right)$, is the case that matches the preceding function, and the function can be represented as$\mathit{l}\mathit{o}\mathit{g}{\mathit{n}}^{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{n}}\mathbf{=}\mathit{\Omega }\mathit{n}\mathbf{/}\left(logn\right)$ .

Solution (m) 

The given functions are:

$f\left(n\right)=n{2}^{n}andg\left(n\right)=3^n$$f\left(n\right)=n{2}^{n}andg\left(n\right)=3^n$.

The above-mentioned function corresponds to the case$F(n)=O\left(g\left(n\right)\right)$.

1. Here, the comparing powers.
2. When comparing two functions, f(n) and g(n), the big-O notation states that g(n) has a faster computational speed than f(n).
3. According to the principles for simplifying functions, is dominated by ; hence,g(n) is better than f(n).

Thus, $F(n)=O\left(g\left(n\right)\right)$is the case that matches the preceding function, and the function can be represented as $n{2}^n=3^n$.

Step 15: Solution (n)

The given functions are:

$f\left(n\right)=2^{n}andg\left(n\right)=2^{n+1}$.

Because and are , the above-mentioned function corresponds to the case$F(n)=O\left(g\left(n\right)\right)$.

1. denotes when comparing the computing speed of two functions, f(n) and g(n), it is said thatf(n) is no faster than g(n).
2. denotes when comparing the computing speed of two functions, f(n) and g(n), it is said that g(n) is no faster than f(n).

Thus, localid="1658226767960" $F(n)=O\left(g\left(n\right)\right)$is the case that corresponds to the above function, and the function can be expressed as${\mathbf{2}}^{\mathbf{n}}\mathbf{=}\mathbf{⊝}{\mathbf{2}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{}\mathbf{.}\mathbf{}$.

Step 16: Solution (o)

The given functions are:

a$f\left(n\right)=n!\text{and}g\left(n\right)=2^n$

The above-mentioned function is in the case of $f\left(n\right)=\Omega g\left(n\right)$.The value of is and is known.

1. denotes when comparing the computing speed of two functions, f(n) and g(n), it is said thatf(n) is no faster than g(n).

Thus, $f\left(n\right)=\Omega g\left(n\right)$is the case that corresponds to the above function, and the function can be expressed as$\mathit{n}\mathbf{!}\mathbf{=}\mathit{\Omega }\mathbf{\left(}{\mathbf{\left(}\mathbf{2}\mathbf{\right)}}^{\mathbf{n}}\mathbf{\right)}\mathbf{\text{.}}$.

Step 17: Solution (p)

The given functions are:

$f\left(n\right)={\left(\log \text{n}\right)}^{\log \text{n}}\text{and}g\left(n\right)=2{\left({\log }_{2}n\right)}^2$.

The above-mentioned function corresponds to the case$f\left(n\right)=Og\left(n\right)$.

1. Thefunction f(n)can be expressed as , and the function g(n)can be expressed as .
2. Here, the function with the power value succeeds. Thus,f(n) is a higher level than g(n)
3. When comparing two functions, f(n) and g(n), the big-O notation states that f(n) has a faster computational speed than g(n).

Thus,$f\left(n\right)=Og\left(n\right)$ is the case that matches the preceding function, and the function can be represented as${\left(ogn\right)}^{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{n}}\mathbf{=}\mathit{O}\mathbf{2}{\left(lo{g}_{2}n\right)}^{\mathbf{2}}\mathbf{}$

Step 18: Solution Explanation (q)

The given functions are:

$f\left(n\right)=\sum _{i=1}^n{i}^k\text{and}g\left(n\right)=n^{k+1}$

Because $\mathbf{}\mathit{f}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$and $\mathit{g}\mathbf{=}\mathit{O}\mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{f}\mathbf{=}\mathit{\Omega }\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}$, the above-mentioned function corresponds to the case$f\left(n\right)=\Theta g\left(n\right)$.

1. denotes when comparing thecomputing speed of two functions, f(n) and g(n), it is said that g(n) is higher-level than g(n).
2. denotes when comparing the computing speed of two functions, f(n) and g(n), it is said that g(n) is higher-level than f(n).

Thus, $f\left(n\right)=\Theta g\left(n\right)$is the case that corresponds to the above function, and the function can be expressed as$\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\mathbf{=}\mathit{\Theta }\left(n^{k+1}\right)\mathbf{\text{.}}$.