Theta complexity of the expression
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Why is the complexity of expression is given as follows?
- $(1 + 2 + 3 + … + (n – 2) + (n – 1) + n)! = Theta((n^2)!)$
- $1 + 2 + 2^2 + … + 2^{n-2} + 2^{n-1} + 2^n = Theta(2^{n + 1})$
- $log(1) + log(2) + … + log(n – 1) + log(n) = Theta(n log(n))$
algorithms asymptotics
edited Dec 19 '18 at 16:19
SmileyCraft
3,591517
asked Dec 19 '18 at 16:09
JuneJune
122
$endgroup$
add a comment |
$begingroup$
Why is the complexity of expression is given as follows?
- $(1 + 2 + 3 + … + (n – 2) + (n – 1) + n)! = Theta((n^2)!)$
- $1 + 2 + 2^2 + … + 2^{n-2} + 2^{n-1} + 2^n = Theta(2^{n + 1})$
- $log(1) + log(2) + … + log(n – 1) + log(n) = Theta(n log(n))$
algorithms asymptotics
edited Dec 19 '18 at 16:19
SmileyCraft
3,591517
asked Dec 19 '18 at 16:09
JuneJune
122
$endgroup$
$begingroup$
$$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
$endgroup$
– Don Thousand
Dec 19 '18 at 16:27
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It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
$endgroup$
– Mason
Dec 19 '18 at 16:30
add a comment |
$begingroup$
Why is the complexity of expression is given as follows?
- $(1 + 2 + 3 + … + (n – 2) + (n – 1) + n)! = Theta((n^2)!)$
- $1 + 2 + 2^2 + … + 2^{n-2} + 2^{n-1} + 2^n = Theta(2^{n + 1})$
- $log(1) + log(2) + … + log(n – 1) + log(n) = Theta(n log(n))$
algorithms asymptotics
edited Dec 19 '18 at 16:19
SmileyCraft
3,591517
asked Dec 19 '18 at 16:09
JuneJune
122
$endgroup$
Why is the complexity of expression is given as follows?
- $(1 + 2 + 3 + … + (n – 2) + (n – 1) + n)! = Theta((n^2)!)$
- $1 + 2 + 2^2 + … + 2^{n-2} + 2^{n-1} + 2^n = Theta(2^{n + 1})$
- $log(1) + log(2) + … + log(n – 1) + log(n) = Theta(n log(n))$
algorithms asymptotics
algorithms asymptotics
edited Dec 19 '18 at 16:19
SmileyCraft
3,591517
asked Dec 19 '18 at 16:09
JuneJune
122
edited Dec 19 '18 at 16:19
SmileyCraft
3,591517
asked Dec 19 '18 at 16:09
JuneJune
122
edited Dec 19 '18 at 16:19
SmileyCraft
3,591517
edited Dec 19 '18 at 16:19
SmileyCraft
3,591517
edited Dec 19 '18 at 16:19
SmileyCraft
3,591517
3,591517
asked Dec 19 '18 at 16:09
JuneJune
122
asked Dec 19 '18 at 16:09
JuneJune
122
asked Dec 19 '18 at 16:09
JuneJune
122
122
$begingroup$
$$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
$endgroup$
– Don Thousand
Dec 19 '18 at 16:27
$begingroup$
It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
$endgroup$
– Mason
Dec 19 '18 at 16:30
add a comment |
$begingroup$
$$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
$endgroup$
– Don Thousand
Dec 19 '18 at 16:27
$begingroup$
It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
$endgroup$
– Mason
Dec 19 '18 at 16:30
$begingroup$
$$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
$endgroup$
– Don Thousand
Dec 19 '18 at 16:27
$begingroup$
$$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
$endgroup$
– Don Thousand
Dec 19 '18 at 16:27
$begingroup$
It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
$endgroup$
– Mason
Dec 19 '18 at 16:30
$begingroup$
It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
$endgroup$
– Mason
Dec 19 '18 at 16:30
add a comment |
1 Answer
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The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.
For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.
The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.
answered Dec 19 '18 at 16:29
SmileyCraftSmileyCraft
3,591517
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
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oldest
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$begingroup$
The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.
For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.
The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.
answered Dec 19 '18 at 16:29
SmileyCraftSmileyCraft
3,591517
$endgroup$
add a comment |
$begingroup$
The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.
For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.
The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.
answered Dec 19 '18 at 16:29
SmileyCraftSmileyCraft
3,591517
$endgroup$
add a comment |
$begingroup$
The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.
For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.
The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.
answered Dec 19 '18 at 16:29
SmileyCraftSmileyCraft
3,591517
$endgroup$
The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.
For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.
The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.
answered Dec 19 '18 at 16:29
SmileyCraftSmileyCraft
3,591517
answered Dec 19 '18 at 16:29
SmileyCraftSmileyCraft
3,591517
answered Dec 19 '18 at 16:29
SmileyCraftSmileyCraft
3,591517
answered Dec 19 '18 at 16:29
SmileyCraftSmileyCraft
3,591517
3,591517
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$begingroup$
$$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
$endgroup$
– Don Thousand
Dec 19 '18 at 16:27
$begingroup$
It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
$endgroup$
– Mason
Dec 19 '18 at 16:30