Non-polynomial difference: $T(n) = 2T(n/2) + n log n$?
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It is said that we can not apply Master Theorem to $T(n) = aT(n/b) + f(n)$ if there is a non-polynomial difference between $f(n)$ and $n^{log_{b}a}$. Polynomial difference means: f(n) / n^log_b(a) = n^c for any real number c. However, in $T(n) = 2T(n/2) + n log n$, $f(n) / n^{log_{b}a} = log n$ which is not $n^c$, right? That means there's non-polynomial difference, right? Then why can we apply MT to this expression?
PLEASE PLEASE NOTE: I am not asking about $T(n) = 2T(n/2) + n/log n$. I am asking about $T(n) = 2T(n/2) + n log n$ and I want to know whether $f(n) / n^{log_{b}a} = log n = n^c$.
algorithms
edited Dec 26 '18 at 10:11
dantopa
6,67442245
asked Sep 22 '16 at 22:59
Lipeng WanLipeng Wan
42
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add a comment |
$begingroup$
It is said that we can not apply Master Theorem to $T(n) = aT(n/b) + f(n)$ if there is a non-polynomial difference between $f(n)$ and $n^{log_{b}a}$. Polynomial difference means: f(n) / n^log_b(a) = n^c for any real number c. However, in $T(n) = 2T(n/2) + n log n$, $f(n) / n^{log_{b}a} = log n$ which is not $n^c$, right? That means there's non-polynomial difference, right? Then why can we apply MT to this expression?
PLEASE PLEASE NOTE: I am not asking about $T(n) = 2T(n/2) + n/log n$. I am asking about $T(n) = 2T(n/2) + n log n$ and I want to know whether $f(n) / n^{log_{b}a} = log n = n^c$.
algorithms
edited Dec 26 '18 at 10:11
dantopa
6,67442245
asked Sep 22 '16 at 22:59
Lipeng WanLipeng Wan
42
$endgroup$
$begingroup$
Hint: The condition on $f(n)$ should be polynomially bounded growth function. Please review the statement of the master theorem and the definition of polynomially bounded.
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– Dosetsu
Sep 22 '16 at 23:56
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Research is done before I ask the question. MT: en.wikipedia.org/wiki/Master_theorem . Polynomial Bounded: math.stackexchange.com/questions/336374/… Please tell me where I can find the answer, thx.
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– Lipeng Wan
Sep 24 '16 at 15:28
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"Please tell me where I can find the answer" Maybe the shortest route is to forget the so-called M****r theorem and to consider $$t_k=2^{-k}T(2^k)$$ then $$t_k=t_{k-1}+klog2$$ hence $$t_k=t_0+log2sum_{i=1}^ki=Theta(k^2)$$ that is, $$T(n)=Theta(n(log n)^2)$$ You see, no mysteries, no cases, no nothing except mathematics... :-)
$endgroup$
– Did
Sep 24 '16 at 16:15
add a comment |
$begingroup$
It is said that we can not apply Master Theorem to $T(n) = aT(n/b) + f(n)$ if there is a non-polynomial difference between $f(n)$ and $n^{log_{b}a}$. Polynomial difference means: f(n) / n^log_b(a) = n^c for any real number c. However, in $T(n) = 2T(n/2) + n log n$, $f(n) / n^{log_{b}a} = log n$ which is not $n^c$, right? That means there's non-polynomial difference, right? Then why can we apply MT to this expression?
PLEASE PLEASE NOTE: I am not asking about $T(n) = 2T(n/2) + n/log n$. I am asking about $T(n) = 2T(n/2) + n log n$ and I want to know whether $f(n) / n^{log_{b}a} = log n = n^c$.
algorithms
edited Dec 26 '18 at 10:11
dantopa
6,67442245
asked Sep 22 '16 at 22:59
Lipeng WanLipeng Wan
42
$endgroup$
It is said that we can not apply Master Theorem to $T(n) = aT(n/b) + f(n)$ if there is a non-polynomial difference between $f(n)$ and $n^{log_{b}a}$. Polynomial difference means: f(n) / n^log_b(a) = n^c for any real number c. However, in $T(n) = 2T(n/2) + n log n$, $f(n) / n^{log_{b}a} = log n$ which is not $n^c$, right? That means there's non-polynomial difference, right? Then why can we apply MT to this expression?
PLEASE PLEASE NOTE: I am not asking about $T(n) = 2T(n/2) + n/log n$. I am asking about $T(n) = 2T(n/2) + n log n$ and I want to know whether $f(n) / n^{log_{b}a} = log n = n^c$.
algorithms
algorithms
edited Dec 26 '18 at 10:11
dantopa
6,67442245
asked Sep 22 '16 at 22:59
Lipeng WanLipeng Wan
42
edited Dec 26 '18 at 10:11
dantopa
6,67442245
asked Sep 22 '16 at 22:59
Lipeng WanLipeng Wan
42
edited Dec 26 '18 at 10:11
dantopa
6,67442245
edited Dec 26 '18 at 10:11
dantopa
6,67442245
edited Dec 26 '18 at 10:11
dantopa
6,67442245
6,67442245
asked Sep 22 '16 at 22:59
Lipeng WanLipeng Wan
42
asked Sep 22 '16 at 22:59
Lipeng WanLipeng Wan
42
asked Sep 22 '16 at 22:59
Lipeng WanLipeng Wan
42
42
$begingroup$
Hint: The condition on $f(n)$ should be polynomially bounded growth function. Please review the statement of the master theorem and the definition of polynomially bounded.
$endgroup$
– Dosetsu
Sep 22 '16 at 23:56
$begingroup$
Research is done before I ask the question. MT: en.wikipedia.org/wiki/Master_theorem . Polynomial Bounded: math.stackexchange.com/questions/336374/… Please tell me where I can find the answer, thx.
$endgroup$
– Lipeng Wan
Sep 24 '16 at 15:28
$begingroup$
"Please tell me where I can find the answer" Maybe the shortest route is to forget the so-called M****r theorem and to consider $$t_k=2^{-k}T(2^k)$$ then $$t_k=t_{k-1}+klog2$$ hence $$t_k=t_0+log2sum_{i=1}^ki=Theta(k^2)$$ that is, $$T(n)=Theta(n(log n)^2)$$ You see, no mysteries, no cases, no nothing except mathematics... :-)
$endgroup$
– Did
Sep 24 '16 at 16:15
add a comment |
$begingroup$
Hint: The condition on $f(n)$ should be polynomially bounded growth function. Please review the statement of the master theorem and the definition of polynomially bounded.
$endgroup$
– Dosetsu
Sep 22 '16 at 23:56
$begingroup$
Research is done before I ask the question. MT: en.wikipedia.org/wiki/Master_theorem . Polynomial Bounded: math.stackexchange.com/questions/336374/… Please tell me where I can find the answer, thx.
$endgroup$
– Lipeng Wan
Sep 24 '16 at 15:28
$begingroup$
"Please tell me where I can find the answer" Maybe the shortest route is to forget the so-called M****r theorem and to consider $$t_k=2^{-k}T(2^k)$$ then $$t_k=t_{k-1}+klog2$$ hence $$t_k=t_0+log2sum_{i=1}^ki=Theta(k^2)$$ that is, $$T(n)=Theta(n(log n)^2)$$ You see, no mysteries, no cases, no nothing except mathematics... :-)
$endgroup$
– Did
Sep 24 '16 at 16:15
$begingroup$
Hint: The condition on $f(n)$ should be polynomially bounded growth function. Please review the statement of the master theorem and the definition of polynomially bounded.
$endgroup$
– Dosetsu
Sep 22 '16 at 23:56
$begingroup$
Hint: The condition on $f(n)$ should be polynomially bounded growth function. Please review the statement of the master theorem and the definition of polynomially bounded.
$endgroup$
– Dosetsu
Sep 22 '16 at 23:56
$begingroup$
Research is done before I ask the question. MT: en.wikipedia.org/wiki/Master_theorem . Polynomial Bounded: math.stackexchange.com/questions/336374/… Please tell me where I can find the answer, thx.
$endgroup$
– Lipeng Wan
Sep 24 '16 at 15:28
$begingroup$
Research is done before I ask the question. MT: en.wikipedia.org/wiki/Master_theorem . Polynomial Bounded: math.stackexchange.com/questions/336374/… Please tell me where I can find the answer, thx.
$endgroup$
– Lipeng Wan
Sep 24 '16 at 15:28
$begingroup$
"Please tell me where I can find the answer" Maybe the shortest route is to forget the so-called M****r theorem and to consider $$t_k=2^{-k}T(2^k)$$ then $$t_k=t_{k-1}+klog2$$ hence $$t_k=t_0+log2sum_{i=1}^ki=Theta(k^2)$$ that is, $$T(n)=Theta(n(log n)^2)$$ You see, no mysteries, no cases, no nothing except mathematics... :-)
$endgroup$
– Did
Sep 24 '16 at 16:15
$begingroup$
"Please tell me where I can find the answer" Maybe the shortest route is to forget the so-called M****r theorem and to consider $$t_k=2^{-k}T(2^k)$$ then $$t_k=t_{k-1}+klog2$$ hence $$t_k=t_0+log2sum_{i=1}^ki=Theta(k^2)$$ that is, $$T(n)=Theta(n(log n)^2)$$ You see, no mysteries, no cases, no nothing except mathematics... :-)
$endgroup$
– Did
Sep 24 '16 at 16:15
add a comment |
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$begingroup$
Hint: The condition on $f(n)$ should be polynomially bounded growth function. Please review the statement of the master theorem and the definition of polynomially bounded.
$endgroup$
– Dosetsu
Sep 22 '16 at 23:56
$begingroup$
Research is done before I ask the question. MT: en.wikipedia.org/wiki/Master_theorem . Polynomial Bounded: math.stackexchange.com/questions/336374/… Please tell me where I can find the answer, thx.
$endgroup$
– Lipeng Wan
Sep 24 '16 at 15:28
$begingroup$
"Please tell me where I can find the answer" Maybe the shortest route is to forget the so-called M****r theorem and to consider $$t_k=2^{-k}T(2^k)$$ then $$t_k=t_{k-1}+klog2$$ hence $$t_k=t_0+log2sum_{i=1}^ki=Theta(k^2)$$ that is, $$T(n)=Theta(n(log n)^2)$$ You see, no mysteries, no cases, no nothing except mathematics... :-)
$endgroup$
– Did
Sep 24 '16 at 16:15