I am practicing problems on asymptotic analysis and I am stuck with this problem.
Is log(n!) = O((log(n))^2) ?
I am able to show that
log(n!) = O(n*log(n))
(log 1 + log 2 + .. + log n <= log n + log n + ... + log n)
and
(log(n))^2 = O(n*log(n))
(log n <= n => (log n)^2 <= n*logn )
I am not able to proceed further. Any hint or intuition on how to proceed further? Thanks
Accoriding to Stirling's Approximation:
log(n!) = n*log(n) - n + O(log(n))
So clearly upper bound for log(n!) will be O(nlogn)
Lower bound can be calculated by removing first half of the equation as:
log(1) + ... + log(n/2) + ... + log(n) = log(n/2) + ... + log(n)
= log(n/2) + ... + log(n/2)
= n/2 * log(n/2)
So Lower bound is also nlogn. Clearly answer would be NO
I think I got the answer to my own question. We will prove the following facts:
1) n*log(n) is a tight bound for log(n!)
2) n*log(n) is a upper bound for (log(n))^2
3)n*log(n) is not a lower bound for (log(n))^2
For proof of (1) see this.
Proof(2) & (3) is provided in the question itself.
growth rate of log n < growth rate of n.
So growth rate of log(n)^2 < growth rate of n*log(n).
So log(n)^2 = o(n*log(n)) (Here I have used little-o to denote that growth rate of n*log(n) is strictly greater than growth rate of log(n)^2
So the conclusion is that log(n!) = big-omega(log(n^2))
Correct me if I have made any mistake
Related
What is O(log(n!)) and O(n!)? I believe it is O(n log(n)) and O(n^n)? Why?
I think it has to do with Stirling's approximation, but I don't get the explanation very well.
Am I wrong about O(log(n!) = O(n log(n))? How can the math be explained in simpler terms? In reality I just want an idea of how this works.
O(n!) isn't equivalent to O(n^n). It is asymptotically less than O(n^n).
O(log(n!)) is equal to O(n log(n)). Here is one way to prove that:
Note that by using the log rule log(mn) = log(m) + log(n) we can see that:
log(n!) = log(n*(n-1)*...2*1) = log(n) + log(n-1) + ... log(2) + log(1)
Proof that O(log(n!)) ⊆ O(n log(n)):
log(n!) = log(n) + log(n-1) + ... log(2) + log(1)
Which is less than:
log(n) + log(n) + log(n) + log(n) + ... + log(n) = n*log(n)
So O(log(n!)) is a subset of O(n log(n))
Proof that O(n log(n)) ⊆ O(log(n!)):
log(n!) = log(n) + log(n-1) + ... log(2) + log(1)
Which is greater than (the left half of that expression with all (n-x) replaced by n/2:
log(n/2) + log(n/2) + ... + log(n/2) = floor(n/2)*log(floor(n/2)) ∈ O(n log(n))
So O(n log(n)) is a subset of O(log(n!)).
Since O(n log(n)) ⊆ O(log(n!)) ⊆ O(n log(n)), they are equivalent big-Oh classes.
By Stirling's approximation,
log(n!) = n log(n) - n + O(log(n))
For large n, the right side is dominated by the term n log(n). That implies that O(log(n!)) = O(n log(n)).
More formally, one definition of "Big O" is that f(x) = O(g(x)) if and only if
lim sup|f(x)/g(x)| < ∞ as x → ∞
Using Stirling's approximation, it's easy to show that log(n!) ∈ O(n log(n)) using this definition.
A similar argument applies to n!. By taking the exponential of both sides of Stirling's approximation, we find that, for large n, n! behaves asymptotically like n^(n+1) / exp(n). Since n / exp(n) → 0 as n → ∞, we can conclude that n! ∈ O(n^n) but O(n!) is not equivalent to O(n^n). There are functions in O(n^n) that are not in O(n!) (such as n^n itself).
I tried to solve this problem many hours and I think the solution is O(log n/[log (log n)^2]). but I'm not sure.Is this solution correct?
Expand the equation:
T(n) = (T(n/(log^2(n)*log(n/log^2(n))^2) + Theta(1)) Theta(1) =
T(n/(log^4(n) + 4 (loglog(n))^2 - 4log(n)loglog(n)) + 2 * Theta(1)
We know n/(log^4(n) + 4 (log(log(n)))^2 - 4log(n)log(log(n)) is greater than n/log^4(n) asymptotically. As you can see, each time n is divided by log^2(n). Hence, we can say if we compute the height of dividing n by log^2(n) up to reaching to 1, it will be a lower bound for T(n).
Hence, the height of the expansion tree will be k such that
n = (log^2(n))^k = lof^2k(n) => (take a log)
log(n) = 2k log(log(n)) => k = log(n)/(2 * log(log(n)))
Therefore, T(n) = Omega(log(n)/log(log(n))).
For the upper bound, as we know that n/(i-th statement) < n/log^i(n) (instead of applying log^2(n), we've applied log(n)), we can say the height of division of n by log(n) will be an upper bound for T(n). Hence, as:
n = log^k(n) => log(n) = k log(log(n)) => k = log(n) / log(log(n))
we can say T(n) = O(log(n) / log(log(n))).
Why is the average case time complexity of tree sort O(n log n)?
From Wikipedia:
Adding one item to a binary search tree is on average an O(log n)
process (in big O notation), so adding n items is an O(n log n)
process
But we don't each time add an item to a tree of n items. We start with an empty tree, and gradually increase the size of the tree.
So it looks more like
log1 + log2 + ... + logn = log (1*2*...*n) = log n!
Am I missing something?
The reason why O(log(n!)) = O(nlog(n)) is a two-part answer. First, expand O(log(n!)),
log(1) + log(2) + ... + log(n)
We can both agree here that log(1), log(2), and all the numbers up to log(n-1) are each less than log(n). Therefore, the following inequality can be made,
log(1) + log(2) + ... + log(n) <= log(n) + log(n) + ... + log(n)
Now the other half of the answer depends on the fact that half of the numbers from 1 to n are greater than n/2. This means that log(n!) would be greater than n/2*log(n/2) aka the first half of the sum log(n!),
log(1) + log(2) + ... + log(n) => log(n/2) + log(n/2) + ... + log(n/2)
The reason being that the first half of log(1) + log(2) + ... + log(n) is log(1) + log(2) + ... + log(n/2), which is less than n/2*log(n/2) as proven by the first inequality so by adding the second half of the sum log(n!), it can be shown that it is greater than n/2*log(n/2).
So with these two inequalities, it can be proven that O(log(n!)) = O(nlog(n))
O(log(n!)) = O(nlog(n)).
https://en.wikipedia.org/wiki/Stirling%27s_approximation
(Answers must be 30 characters.)
I have solved all of them however i have been told there are some mistakes, can somebody please help me
n^4 - 10^3 n^3 + n^2 + 4n + 10^6 = O(n^4)
10^5 n^3 + 10^n = O(10^n)
10 n^2 + n log n + 30 √n = O(n^2)
25^n = O(1)
n^2+ n log n + 7 n = O(n^2)
(n^3 + 10) (n log n+ 1) / 3 = O(n^4 log n)
20 n^10 + 4^n = O(4^n)
n^2 log n^3 + 10 n^2 = O(n^2 log n)
10^20 = O(1)
n^2 log (6^2)n = O(n^2 log n)
n log(2n) = O(n log n)
30 n + 100 n log n + 10 = O(n log n)
(n+√n) log n^3 = O(n+√n log n)
n (n + 1) + log log n = O(n^2)
4n log 5^(n+1) = O(n log 5^n)
3^(n+4) = O(3^n)
n^2 log n^2 + 100 n^3 = O(n^3)
(n log n) / (n + 10) = O(n^2 log n)
5n + 8 n log(n) + 10n^2 = O(n^2)
2n^3 + 2n^4 + 2^n + n^10 = O(2^n)
Hints:
if you have n on the left, you should have it on the right
there should not be any + operations on the right
log(x^y) can be simplified
Most of your answers look correct, but you have 25^n = O(1) which looks wrong (unless it's 0.25^n), and also you have (n log n) / (n + 10) = O(n^2 log n) which does not look like the tightest possible bound (I'm assuming you want the tightest possible upper bound function). Also you should never have to add functions in your big-O, unless your original function is taking the sum or max of two functions or something and the two functions have cris-crossing different growth rates at different values of n as n goes to infinity. And that very rarely happens.
I am to show that log(n!) = Θ(n·log(n)).
A hint was given that I should show the upper bound with nn and show the lower bound with (n/2)(n/2). This does not seem all that intuitive to me. Why would that be the case? I can definitely see how to convert nn to n·log(n) (i.e. log both sides of an equation), but that's kind of working backwards.
What would be the correct approach to tackle this problem? Should I draw the recursion tree? There is nothing recursive about this, so that doesn't seem like a likely approach..
Remember that
log(n!) = log(1) + log(2) + ... + log(n-1) + log(n)
You can get the upper bound by
log(1) + log(2) + ... + log(n) <= log(n) + log(n) + ... + log(n)
= n*log(n)
And you can get the lower bound by doing a similar thing after throwing away the first half of the sum:
log(1) + ... + log(n/2) + ... + log(n) >= log(n/2) + ... + log(n)
= log(n/2) + log(n/2+1) + ... + log(n-1) + log(n)
>= log(n/2) + ... + log(n/2)
= n/2 * log(n/2)
I realize this is a very old question with an accepted answer, but none of these answers actually use the approach suggested by the hint.
It is a pretty simple argument:
n! (= 1*2*3*...*n) is a product of n numbers each less than or equal to n. Therefore it is less than the product of n numbers all equal to n; i.e., n^n.
Half of the numbers -- i.e. n/2 of them -- in the n! product are greater than or equal to n/2. Therefore their product is greater than the product of n/2 numbers all equal to n/2; i.e. (n/2)^(n/2).
Take logs throughout to establish the result.
Sorry, I don't know how to use LaTeX syntax on stackoverflow..
See Stirling's Approximation:
ln(n!) = n*ln(n) - n + O(ln(n))
where the last 2 terms are less significant than the first one.
For lower bound,
lg(n!) = lg(n)+lg(n-1)+...+lg(n/2)+...+lg2+lg1
>= lg(n/2)+lg(n/2)+...+lg(n/2)+ ((n-1)/2) lg 2 (leave last term lg1(=0); replace first n/2 terms as lg(n/2); replace last (n-1)/2 terms as lg2 which will make cancellation easier later)
= n/2 lg(n/2) + (n/2) lg 2 - 1/2 lg 2
= n/2 lg n - (n/2)(lg 2) + n/2 - 1/2
= n/2 lg n - 1/2
lg(n!) >= (1/2) (n lg n - 1)
Combining both bounds :
1/2 (n lg n - 1) <= lg(n!) <= n lg n
By choosing lower bound constant greater than (1/2) we can compensate for -1 inside the bracket.
Thus lg(n!) = Theta(n lg n)
Helping you further, where Mick Sharpe left you:
It's deriveration is quite simple:
see http://en.wikipedia.org/wiki/Logarithm -> Group Theory
log(n!) = log(n * (n-1) * (n-2) * ... * 2 * 1) = log(n) + log(n-1) + ... + log(2) + log(1)
Think of n as infinitly big. What is infinite minus one? or minus two? etc.
log(inf) + log(inf) + log(inf) + ... = inf * log(inf)
And then think of inf as n.
Thanks, I found your answers convincing but in my case, I must use the Θ properties:
log(n!) = Θ(n·log n) => log(n!) = O(n log n) and log(n!) = Ω(n log n)
to verify the problem I found this web, where you have all the process explained: http://www.mcs.sdsmt.edu/ecorwin/cs372/handouts/theta_n_factorial.htm
http://en.wikipedia.org/wiki/Stirling%27s_approximation
Stirling approximation might help you. It is really helpful in dealing with problems on factorials related to huge numbers of the order of 10^10 and above.
This might help:
eln(x) = x
and
(lm)n = lm*n
If you reframe the problem, you can solve this with calculus! This method was originally shown to me via Arthur Breitman https://twitter.com/ArthurB/status/1436023017725964290.
First, you take the integral of log(x) from 1 to n it is n*log(n) -n +1. This proves a tight upper bound since log is monotonic and for every point n, the integral from n to n+1 of log(n) > log(n) * 1. You can similarly craft the lower bound using log(x-1), as for every point n, 1*log(n) > the integral from x=n-1 to n of log(x). The integral of log(x) from 0 to n-1 is (n-1)*(log(n-1) -1), or n log(n-1) -n -log(n-1)+1.
These are very tight bounds!