How do I solve the following as f(n)=n! does not as to my knowledge apply to any of the cases of master theorem.
T (n) = 16T (n/4) + n!
David Eisenstat is partially correct. Case 3 does apply, but T(n) = theta(n!), not O(n!).
T(n) = 16T(n/4) + n!
Case 3 of the Master Theorem (AKA Master Method) applies. a = 16, b = 4, f(n) = n!. n^(log [base(b)] a) = n^2. f(n) is n!. Since n! is omega(f(n)) i.e. n! omega n^2 AND af(n/b) <= cf(n) for a large n, T(n) is theta(n!).
For reference, consult #10 here : http://www.csd.uwo.ca/~moreno/CS433-CS9624/Resources/master.pdf
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I struggle to fill this table in even though I took calculus recently and good at math. It is only specified in the chapter how to deal with lim(n^k/c^n), but I have no idea how to compare other functions. I checked the solution manual and no info on that, only a table with answers which provides little insight.
When I solve these I don't really think about limits -- I lean on a couple facts and some well-known properties of big-O notation.
Fact 1: for all functions f and g and all exponents p > 0, we have f(n) = O(g(n)) if and only if f(n)p = O(g(n)p), and likewise with o, Ω, ω, and Θ respectively. This has a straightforward proof from the definition; you just have to raise the constant c to the power p as well.
Fact 2: for all exponents ε > 0, the function lg(n) is o(nε). This follows from l'Hôpital's rule for limits: lim lg(n)/nε = lim (lg(e)/n)/(ε nε−1) = (lg(e)/ε) lim n−ε = 0.
Fact 3:
If f(n) ≤ g(n) + O(1), then 2f(n) = O(2g(n)).
If f(n) ≤ g(n) − ω(1), then 2f(n) = o(2g(n)).
If f(n) ≥ g(n) − O(1), then 2f(n) = Ω(2g(n)).
If f(n) ≥ g(n) + ω(1), then 2f(n) = ω(2g(n)).
Fact 4: lg(n!) = Θ(n lg(n)). The proof uses Stirling's approximation.
To solve (a), use Fact 1 to raise both sides to the power of 1/k and apply Fact 2.
To solve (b), rewrite nk = 2lg(n)k and cn = 2lg(c)n, prove that lg(c) n − lg(n) k = ω(1), and apply Fact 3.
(c) is special. nsin(n) ends up anywhere between 0 and n. Since 0 is o(√n) and n is ω(√n), that's a solid row of NO.
To solve (d), observe that n ≥ n/2 + ω(1) and apply Fact 3.
To solve (e), rewrite nlg(c) = 2lg(n)lg(c) = 2lg(c)lg(n) = clg(n).
To solve (f), use Fact 4 and find that lg(n!) = Θ(n lg(n)) = lg(nn).
How would I determine the time complexity using the Master Theorem for the given problem?
T(n) = aT(n/b) + O(n^d)
T(n) = 4 T(n/2) + n*log(n)
a = 4, b = 2, d = 1
1. O(n^d) if d > logb(a)
2. O(n^d logn) if d = logb(a)
3. O(n^logb(a)) if d < logb(a)
In my case:
log2(4) = 2 --> d < log2(4)
T(n) = O(n^logb(a))
= O(n^2)
Is this correct?
edit:
I have made an different approach now, resulting in the same as before though, and following my professors instruction.
T(n) = a*T(n/b) + f(n)
T(n) = 4 T(n/2) + n*log(n)
1. Ө(n^(logba)) , if --> f(n) є O(n^(logba-ε)),
T(n) = { 2. Ө(n^(logba) log2n) , if --> f(n) є Ө(n^(logba))
3. Ө(f(n)) , if --> f(n) є Ω(n^(logba+ε))
First I look at the 2nd case
f(n) = n*log(n)
logb^a = log2^4 = 2
n^log24 = n^2
The second case does not apply because:
n*log(n) ∉ Θ(n^logba) —> the runtime of f(n) is smaller, because n*log(n) < n^2
Look at the first case if the runtime of f(n) is smaller than n^2
1. Ө(n^(logba)) , if --> f(n) є O(n^(logba-ε))
O(n^(logba-ε)) = O(n^(log24-ε))= O(n^2-ε)
n*log(n) ∈ O(n^(log24-ε)) --> works
f(n) has O(n^2-ε) as upper limit for worst case scenario,
case of Master—Theorems applies:
T(n) = Θ(n^log24) = Θ(n^2)
T(n) = Θ(n^2)
The version of the master theorem that you've stated here specifically assumes that the additive term has the form O(nd). In this case, the additive term is of the form n log n, which is not O(n1). This means that you cannot apply the master theorem as you have above.
A useful technique when working with recurrences like these is to sandwich the recurrence between two others. Notice, for example, that T(n) is bounded by these recurrences:
S(n) = 4S(n / 2) + n
R(n) = 4T(n / 2) + n1+ε, where ε is some small positive number.
Try solving each of these recurrences using the master theorem and see what you get, keeping in mind that your recurrence solves to a value sandwiched between them.
Hope this helps!
So I was wondering if the following recurrence would be considered to fall under case 3 of the Master Theorem: T(n)=4T(n/2) + 10000 - 5000sin(n).
So I've labeled my answer as the following...
A = 4, B = 2, F(N) = 10000 - 5000sin(n)
n^k = n^2
So when comparing F(n) to n^k we can see that f(n) grows faster than n^k, implying that this is case 3 of the master theorem. Is this correct?
Since -1 ≤ sin(n) ≤ 1, 5000 ≤ f(n) ≤ 15000, which is O(1) since it never grows with input size.
This is case three because f(n) is constant time (O(1) == n0) which is asymptotically less than nlog24. So by the master theorem your recurrence relation T(n) = Θ(n2).
T(n) = 16T(n/4) + n!
I know it can be solved using Master theorem, but I don't know how to handle
f(n) = n!
This is case three of Master Theorem.
Since T(n) = 16T(n/4) + n!
Here f(n) = n!.
a = 16 and b = 4, so logb a = log4 16 = 2.
Master Theorem states that the complexity T(n) = Θ(f(n)) if
c > logb a where f(n) ∈ Ω(nc) .
Since f(n) = n! > nc for some value of n > n0 the statement f(n) ∈ Ω (nc) is true. Thus the statement
c > logb a =2 is also true. Hence by the third case of Master Thoerem the complexity T(n) = Θ(f(n)) = Θ(n!).
Can we solve this
T(n) = 2T( n/2 ) + n lg n recurrence equation master theorem I am coming from a link where he is stating that we can't apply here master theorem because it doesn't satisfied any of the 3ree case condition. On the other hand he has taken a another example
T(n) = 27T(n/3) + Θ(n^3 lg n) and find the closed solution theta(n^3logn) For solving this he used 2nd case of master theorem If f(n) = Θ(nlogba (lg n)k ) then T(n) ∈ Θ(nlogba (lg n)k+1) for some k >= 0 Here my confusion arises why not we can apply 2nd case here while it is completely fit in 2nd case.
My thought: a = 2 , b =2; let k =1 then
f(n) = theta(n^log_2 2 logn) for k= 1 so T(n) = theta(nlogn) But he as mentioned on this we can't apply master theorem I m confused why not.
Note: It is due to f(n) bcz in T(n) = 2T( n/2 ) + n lg n f(n) = nlog n and in T(n) = 27T(n/3) + Θ(n^3 lg n) *f(n) = theta(n^3log n)* Please Correct me if I am wrong here.
Using case 2 of master theorem I find that
T(n) = Theta( n log^2 (n))
Your link states that the case 2 of theroem is :
f(n) = Theta( n log_b(a))
While from several other links, like the one from mit, the case is :
f(n) = Theta( n log_b(a) log_k(n)) for k >= 0