Multiplying exponents, solving for n
When solving for n in this equation I get stuck.
Question: What is the smallest value of n such that an algorithm with running time of $\ 100n^2 $ runs faster than an algorithm whose running time is $\ 2^n $ on the same machine?
Straight out of CLRS chapter 1. Class starts in 2 months wanted to get a head start.
My approach:
$\ 100n^2 = 2^n $
$\ \sqrt(100n^2) = \sqrt(2^n) $
$\ 10n = (2^n)^{1/2} $
$\ 10n = (2^{n/2}) $
Is this last step correct? I know I add exponents when multiplying but this is raising an exponent to an exponent so I should multiply. I'm still unsure how to bring the n down out of the exponent on the two so I can solve for it.
$\endgroup$ 55 Answers
$\begingroup$Sometimes in cases like this, it's useful to simply try a few and see what happens:
Trying $n=20$ gives $$10n=200$$ and $$2^{n/2}=2^{10}=1024.$$
Since the rate of increase of $n\mapsto 10n$ is constant, while $n\mapsto 2^{n/2}$ grows more and more quickly for larger $n,$ it follows that $10n<2^{n/2}$ whenever $n\ge 20,$ so we're looking for some $n<20.$
Trying $n=14$ gives us $$10n=140$$ and $$2^{n/2}=128,$$ so by similar reasoning, we need some $n\ge 14$.
From there, the solution's fairly quickly found to be $n=15$.
Note: $n=15$ is not the solution to the equation $100n^2=2^n$, but it is the least $n$ for which $100n^2<2^n$.
$\endgroup$ $\begingroup$One of the answers says that problem can be stated as an inequality:
$$100n^2<2^n$$
Take $\log_{2}$ of both sides:
$$\log_2{(100n^2)} < n$$
By logarithm rules:
$$\log_2{(100)} + \log_2{(n^2)} < n$$
You may remove the squared term (I'm not sure if this matters):
$$\log_2{(100)} + 2\log_2{(n)} < n$$
Rearrange by subtracting $ 2\log_2{(n)}$ from both sides:
$$\log_2{(100)} < n- 2\log_2{(n)}$$
Square both sides:
$$(\log_2{(100)})^2 < (n- 2\log_2{(n)})^2$$
Expanding out the brackets, this may be expressed as a quadratic:
$$ n^2 - 4\log_2{(n)}n + 4(\log_2{(n)})^2 - (\log_2{(100)})^2 > 0$$
Now we must solve for $n$.
one of the solutions is $ n > 14.3247$ which is what one of the other answers shows in the linear plot.
This might be a very convoluted way to go about it, _'m not sure. But the other answer involved plotting and looking to narrow down. That might be found with newton iteration algorithm as well. But this quadratic way I think is a direct way to get the answer. The other root is less than 1 so I guess it can be ignored.
I'm very happy to be shown if there's a simpler way!
$\endgroup$ $\begingroup$This is correct. You can then start guessing to find it by hand. Clearly for $n=20$ left hand side is much smaller than right hand side. For $n=16$ similarly, $n=15$ seems okay, since $2^7=128$ and $\sqrt 2>1.4$, and $n=14$ is too small for similar reasons.
$\endgroup$ $\begingroup$The equation $10 n = 2^{n/2}$ can't be solved in terms of elementary functions: you have to use the Lambert W function. The two real solutions are $-2 LambertW(-\ln(2)/20)/\ln(2)$ and $-2 LambertW(-1,-\ln(2)/20)/\ln(2)$. But for what you want to do, this is not very relevant. "Guess and check" is the way to go.
$\endgroup$ $\begingroup$I used Wolfram to narrow down on the approximate solution. I started with:
plot 2^x - 100 *x^2 , x=1 to 100and looked at the graph to narrow the value of n down to:
plot 2^x - 100 *x^2 , x=14.3245 to 14.325after a couple steps.
This was after trying to reduce it to the simplest terms though my knowledge doesn't make it to Lambert functions.
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