Analytic Method for Solving $8n^2 > 64nlog_2{n}$
I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64nlog_2{n}$ and asked what is the greatest input size $n$ where $n in mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)
The problem can be mathematically formulated as solving the following inequality:
$$tag{1}quad 8n^2 < 64nlog_2{n}.$$
I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.
$$8n^2 < 64nlog_2{n}implies n < 8log_2{n}implies n < log_2{n^8}implies 2^{n} < n^8.$$
Anyone know how to analytically derive the result of $n<44?$
inequality alternative-proof
edited Nov 25 at 5:14
Tianlalu
3,03021038
asked Nov 25 at 5:13
Dunka
1,42232043
|
show 1 more comment
I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64nlog_2{n}$ and asked what is the greatest input size $n$ where $n in mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)
The problem can be mathematically formulated as solving the following inequality:
$$tag{1}quad 8n^2 < 64nlog_2{n}.$$
I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.
$$8n^2 < 64nlog_2{n}implies n < 8log_2{n}implies n < log_2{n^8}implies 2^{n} < n^8.$$
Anyone know how to analytically derive the result of $n<44?$
inequality alternative-proof
edited Nov 25 at 5:14
Tianlalu
3,03021038
asked Nov 25 at 5:13
Dunka
1,42232043
I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
– mathworker21
Nov 25 at 5:17
@mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
– Dunka
Nov 25 at 5:28
well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
– mathworker21
Nov 25 at 5:30
You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
– eSurfsnake
Nov 25 at 6:42
Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command-8/Log[2]*LambertW[-1,-Log[2]/8]to WA, one get something reasonable $approx 43.55926$.
– achille hui
Nov 25 at 8:36
|
show 1 more comment
I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64nlog_2{n}$ and asked what is the greatest input size $n$ where $n in mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)
The problem can be mathematically formulated as solving the following inequality:
$$tag{1}quad 8n^2 < 64nlog_2{n}.$$
I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.
$$8n^2 < 64nlog_2{n}implies n < 8log_2{n}implies n < log_2{n^8}implies 2^{n} < n^8.$$
Anyone know how to analytically derive the result of $n<44?$
inequality alternative-proof
edited Nov 25 at 5:14
Tianlalu
3,03021038
asked Nov 25 at 5:13
Dunka
1,42232043
I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64nlog_2{n}$ and asked what is the greatest input size $n$ where $n in mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)
The problem can be mathematically formulated as solving the following inequality:
$$tag{1}quad 8n^2 < 64nlog_2{n}.$$
I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.
$$8n^2 < 64nlog_2{n}implies n < 8log_2{n}implies n < log_2{n^8}implies 2^{n} < n^8.$$
Anyone know how to analytically derive the result of $n<44?$
inequality alternative-proof
inequality alternative-proof
edited Nov 25 at 5:14
Tianlalu
3,03021038
asked Nov 25 at 5:13
Dunka
1,42232043
edited Nov 25 at 5:14
Tianlalu
3,03021038
asked Nov 25 at 5:13
Dunka
1,42232043
edited Nov 25 at 5:14
Tianlalu
3,03021038
edited Nov 25 at 5:14
Tianlalu
3,03021038
edited Nov 25 at 5:14
Tianlalu
3,03021038
3,03021038
asked Nov 25 at 5:13
Dunka
1,42232043
asked Nov 25 at 5:13
Dunka
1,42232043
asked Nov 25 at 5:13
Dunka
1,42232043
1,42232043
I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
– mathworker21
Nov 25 at 5:17
@mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
– Dunka
Nov 25 at 5:28
well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
– mathworker21
Nov 25 at 5:30
You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
– eSurfsnake
Nov 25 at 6:42
Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command-8/Log[2]*LambertW[-1,-Log[2]/8]to WA, one get something reasonable $approx 43.55926$.
– achille hui
Nov 25 at 8:36
|
show 1 more comment
I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
– mathworker21
Nov 25 at 5:17
@mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
– Dunka
Nov 25 at 5:28
well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
– mathworker21
Nov 25 at 5:30
You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
– eSurfsnake
Nov 25 at 6:42
Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command-8/Log[2]*LambertW[-1,-Log[2]/8]to WA, one get something reasonable $approx 43.55926$.
– achille hui
Nov 25 at 8:36
I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
– mathworker21
Nov 25 at 5:17
I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
– mathworker21
Nov 25 at 5:17
@mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
– Dunka
Nov 25 at 5:28
@mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
– Dunka
Nov 25 at 5:28
well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
– mathworker21
Nov 25 at 5:30
well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
– mathworker21
Nov 25 at 5:30
You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
– eSurfsnake
Nov 25 at 6:42
You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
– eSurfsnake
Nov 25 at 6:42
Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command
-8/Log[2]*LambertW[-1,-Log[2]/8] to WA, one get something reasonable $approx 43.55926$.– achille hui
Nov 25 at 8:36
Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command
-8/Log[2]*LambertW[-1,-Log[2]/8] to WA, one get something reasonable $approx 43.55926$.– achille hui
Nov 25 at 8:36
|
show 1 more comment
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I think these finite things are inherently dirty. I don't view anything that can be checked by a computer (in finite time) as clean
– mathworker21
Nov 25 at 5:17
@mathworker21 maybe clean was the wrong word to use, but I wonder if someone could work it out with pure reason rather than just testing cases with a computer like I did.
– Dunka
Nov 25 at 5:28
well, i don't know how $44$ is supposed to come out of pure reason. you're going to have to do some computation at some point. so why not just have a computer to all the computation. it's arbitrary...
– mathworker21
Nov 25 at 5:30
You're really looking for where they are equal, right? So $n=log_2n^8$. Try raisng e to both sides, so $e^n=n^8$?
– eSurfsnake
Nov 25 at 6:42
Notice $$8n^2 = 64nlog_2(n)iff e^{frac{log 2}{8} n} = n iff -frac{log 2}{8} = -frac{log 2}{8} n e^{-frac{log 2}{8} n}$$ The root for above equation can be expressed in terms of suitable choosen branch of Lambert W function $$n = -frac{8}{log 2} Wleft(-frac{log 2}{8}right)$$ The relevant branch seems to be the branch '-1', If one throw the command
-8/Log[2]*LambertW[-1,-Log[2]/8]to WA, one get something reasonable $approx 43.55926$.– achille hui
Nov 25 at 8:36