How do i solve this recurance relation? T(n) = 4T(n/3) + lg n
I know that Master-Theorem-Case 1 applies but i dont understand why. The way i approach this until now is this one.
a=4, b=3, f(n)=lg n.
Is lg n = (lg10 n) or (lg2 n) and i know that because of (lge n) it doesn´t really matter but i still dont understand why it doesn´t matter if it is lg10 or lg2. I could calculate (lg10 n) / (lg2 n) or sth. and for some reason it doesn´t matter but why?... but lets go on.
n^log3^4 ~ 1.26 but what is lg n in terms of n^someting.
Another example so maybe you understand me.
If i had f(n) = square root of n and not lg n it would be f(n) = n^0.5.
So n^1.26 <= n^0.5 for an e > 0. For e = 1, first case, f(n) becomes element of n^logb^(a-e) = n^log3^(4-1) = n^log3^3. Is n^0.5 element of n^1? Yes? because it is smaller?, so this leads to n^logb^a, or T(N) = O(N^logba) or O(n^log3^4).
If this is correct how do i follow this way for f(n) = lg n?
I hope you understood my question, i cannot format properly all the n^logba stuff.
No. The growth rate of logarithmic function is less than any polynomial function with the exponential greater than 0. That is to say even something like x^0.0000001 will eventually grow faster than log x.
So in this case its O(n^log_3 4).
Related
I don't understand how the runtime of the algorithm can be log(log(n)). Can someone help me?
s=1
while s <= (log n)^2 do
s=3s
Notation note: log(n) indicates log2(n) throughout the solution.
Well, I suppose (log n)^2 indicates the square of log(n), which means log(n)*log(n). Let us try to analyze the algorithm.
It starts from s=1 and goes like 1,3,9,27...
Since it goes by the exponents of 3, after each iteration s can be shown as 3^m, m being the number of iterations starting from 1.
We will do these iterations until s becomes bigger than log(n)*log(n). So at some point 3^m will be equal to log(n)*log(n).
Solve the equation:
3^m = log(n) * log(n)
m = log3(log(n) * log(n))
Time complexity of the algorithm can be shown as O(m). We have to express m in terms of n.
log3(log(n) * log(n)) = log3(log(n)) + log3(log(n))
= 2 * log3(log(n)) For Big-Oh notation, constants do not matter. So let us get rid of 2.
Time complexity = O(log3(log(n)))
Well ok, here is the deal: By the definition of Big-Oh notation, it represents an upper bound runtime for our function. Therefore O(n) ⊆ O(n^2).
Notice that log3(a) < log2(a) after a point.
By the same logic we can conclude that O(log3(log(n)) ⊆ O(log(log(n)).
So the time complexity of the algorithm : O(log(logn))
Not the most scientific explanation, but I hope you got the point.
This follows as a special case of a more general principle. Consider the following loop:
s = 1
while s < k:
s = 3s
How many times will this loop run? Well, the values of s taken on will be equal to 1, 3, 9, 27, 81, ... = 30, 31, 32, 33, ... . And more generally, on the ith iteration of the loop, the value of s will be 3i.
This loop stops running at soon as 3i overshoots k. To figure out where that is, we can equate and solve:
3i = k
i = log3 k
So this loop will run a total of log3 k times.
Now, what do you think would happen if we used this loop instead?
s = 1
while s < k:
s = 4s
Using similar logic, the number of loop iterations would be log4 k. And more generally, if we have the following loop:
s = 1
while s < k:
s = c * s
Then assuming c > 1, the number of iterations will be logc k.
Given this, let's look at your loop:
s = 1
while s <= (log n)^2 do
s = 3s
Using the reasoning from above, the number of iterations of this loop works out to log3 (log n)2. Using properties of logarithms, we can simplify this to
log3 (log n)2
= 2 log3 log n
= O(log log n).
Why is the growth of n^1.001 greater than n log n in Big O notation?
The n^0.001 doesn't seem significant...
For any exponent (x) greater than 1, nx is eventually greater than n * log(n). In the case of x = 1.001, the n in question is unbelievably large. Even if you lower x to 1.01, nx doesn't get bigger than n * log(n) until beyond n = 1E+128 (but before you reach 1E+256).
So, for problems where n is less than astronomical, n1.001 will be less than n * log(n), but you will eventually reach a point where it will be greater.
In case someone is interested, here is a formal proof:
For the sake of simplicity, let's assume we are using logarithms in base e.
Let a > 1 be any exponent (e.g., a = 1.001). Then a-1 > 0. Now consider the function
f(x) = x^(a-1)/log(x)
Using L'Hôpital's rule it is not hard to see that this function is unbounded. Moreover, computing the derivative of f(x), one can also see that the function is increasing for x > exp(1/(a-1)).
Therefore, there must exist an integer N such that, for all n > N, is f(n) > 1. In other words
n^(a-1)/log(n) > 1
or
n^(a-1) > log(n)
so
n^a > n log(n).
This shows that O(n^a) >= O(n log(n)).
But wait a minute. We wanted >, not >=, right? Fortunately this is easy to see. For instance, in the case a = 1.001, we have
O(n^1.001) > O(n^1.0001) >= O(n log(n))
and we are done.
I'm unsure of the general time complexity of the following code.
Sum = 0
for i = 1 to N
if i > 10
for j = 1 to i do
Sum = Sum + 1
Assuming i and j are incremented by 1.
I know that the first loop is O(n) but the second loop is only going to run when N > 10. Would the general time complexity then be O(n^2)? Any help is greatly appreciated.
Consider the definition of Big O Notation.
________________________________________________________________
Let f: ℜ → ℜ and g: ℜ → ℜ.
Then, f(x) = O(g(x))
⇔
∃ k ∈ ℜ ∋ ∃ M > 0 ∈ ℜ ∋ ∀ x ≥ k, |f(x)| ≤ M ⋅ |g(x)|
________________________________________________________________
Which can be read less formally as:
________________________________________________________________
Let f and g be functions defined on a subset of the real numbers.
Then, f is O of g if, for big enough x's (this is what the k is for in the formal definition) there is a constant M (from the real numbers, of course) such that M times g(x) will always be greater than or equal to (really, you can just increase M and it will always be greater, but I regress) f(x).
________________________________________________________________
(You may note that if a function is O(n), then it is also O(n²) and O(e^n), but of course we are usually interested in the "smallest" function g such that it is O(g). In fact, when someone says f is O of g then they almost always mean that g is the smallest such function.)
Let's translate this to your problem. Let f(N) be the amount of time your process takes to complete as a function of N. Now, pretend that addition takes one unit of time to complete (and checking the if statement and incrementing the for-loop take no time), then
f(1) = 0
f(2) = 0
...
f(10) = 0
f(11) = 11
f(12) = 23
f(13) = 36
f(14) = 50
We want to find a function g(N) such that for big enough values of N, f(N) ≤ M ⋅g(N). We can satisfy this by g(N) = N² and M can just be 1 (maybe it could be smaller, but we don't really care). In this case, big enough means greater than 10 (of course, f is still less than M⋅g for N <11).
tl;dr: Yes, the general time complexity is O(n²) because Big O assumes that your N is going to infinity.
Let's assume your code is
Sum = 0
for i = 1 to N
for j = 1 to i do
Sum = Sum + 1
There are N^2 sum operations in total. Your code with if i > 10 does 10^2 sum operations less. As a result, for enough big N we have
N^2 - 10^2
operations. That is
O(N^2) - O(1) = O(N^2)
Assume that function f is in the complexity class O(N (log N)2), and that for N = 1,000 the program runs in 8 seconds.
How to write a formula T(N) that can compute the approximate time that it takes to run f for any input of size N???
Here is the answer:
8 = c (1000 x 10)
c = 8x10^-4
T(N) = 8x10-4* (N log2 N)
I don't understand the first line where does the 10 come from?
Can anybody explain the answer to me please? Thanks!
I don't understand the first line where does the 10 come from? Can
anybody explain the answer to me please? Thanks!
T(N) is the maximum time complexity. c is the constant or O(1) time, which is the portion of the algorithm's speed which is not affected by the size of the input. The 10 comes from rounding to simplify the math. It's actually 9.965784, which is log2 of 1000, e.g.
N x log2 N is
1000 x 10 or
1000 x 9.965784
O(N (log N)^2) describes how the runtime scales with N, but it's not a formula for calculating runtime in seconds. In fact, Big-O notation doesn't generally give the exact scaling function itself, but an upper bound on it as N becomes large. See here (there's a nice picture showing this last point).
If you're interested in a function's runtime in practice (particularly in the non-asymptotic regime, i.e. small N), one option is to actually run the function and measure it. Do this for multiple values of N, chosen on some grid (possibly with nonlinear spacing). Then, you can interpolate between these points.
Define S(N)=N(log N)^2
If you can assume that S(N) bounds your program for all N >= 1000
Then you can bound your execution time by good'ol rule of three:
S(1000) - T(1000)
S(N) - T(N)
T(N) <= S(N)* T(1000)/S(1000) for all N >=1000
S(1000) approx 10E4
T(1000) = 8
T(N) <= N(log N)^2 * 8 / 10E4
How can I order the following functions by rate of growth? n^(logn), 3^n, (logn)^n, n choose n-4, and n^3 ?
What I have is: n^3, n choose n-4, n^logn, 3^n, (logn)^n but I'm not sure if this is right.
Your ordering looks correct to me.
n^3 is obviously the smallest polynomial in the list.
n choose (n-4) is n! / ((n-4)! 4!) = n (n-1) (n-2) (n-3) / 4!. It's O(n^4), and is the second smallest function.
n^log n = exp((log n)^2) is not even exponential, it's quasi-polynomial.
3^n is classical exponential.
(log n)^n obviously grows faster than 3^n since it has both base and power increasing as n grows. By the way, it's still exponential because (log n)^n = exp(n log log n) = O(exp(n^2)), for example.