The Master theorem's generic form mentions that:
it is assumed that all subproblems are essentially the same size
The Akra–Bazzi method is applied when:
the sub-problems have substantially different sizes
But what are the criteria for substantially different? For example I have a recurrence relation like:
T(n) = T(n/4) + T(3n/4) + cn
(c is some constant)
Can I still use the master theorem to solve this relation (for instance approximating it as T(n) = 2T(3n/4) + cn)? Or, in other words, are these subproblem sizes "essentially the same" or are they already "substantially different"?
Assuming c is some constant, you have: T(n) = T(n/4) + T(3n/4) + O(n)
Solving this with the Akra-Bazzi method gives O(n^2)
Solving it by assuming T(n) = 2T(3n/4) + O(n) gives O(n^2.4094) (exp. rounded to 4 dp)
So just by trying it out, you can confirm that they are already substantially different.
Related
Suppose we have been given a recurrence relation T(n) = 2T(n/3) + n. And need to find the time complexity.
My problem is why I am getting different answers by master theorem and using recursion tree method.
By Master Theorem for equation T(n) = aT(n/b) + O(nklogpn)
Here a=2,b=3,p=0,k=1 and as a < bk here i.e 2 < 31 by masters theorem it gives us T(n) = O(n) time complexity.
Here is my recursion tree method
And at the end I seriously don't know what did I found out. It looks something like O(n.2logn).
Surely its not the answer and I have messed it up. But I don't get where I am wrong? What is right approach?
Also I wanted to ask if recurrence relations T(n) = 2T(n/3) + n and T(n) = 2T(n/3) + C are both same? where C is a constant in 2nd equation.
How to calculate time complexity of recurrence relation f(n) = f(n/2) + f(n/3). We have base case at n=1 and n=0.
How to calculate time complexity for general case i.e f(n) = f(n/x) + f(n/y), where x<n and y<n.
Edit-1 :(after first answer posted) every number considered is integer.
Edit-2 :(after first answer posted) I like the answer given by Mbo but is it possible to answer this without using any fancy theorem like master theorem etc.Like by making tree etc.
However users are free to answer the way they like and i will try to understand.
In "layman terms" you can get dependence with larger coefficient:
T(n) = T(n/2) + T(n/2) + O(1)
build call tree for n=2^k and see that the last tree level contains 2^k items, higher level 2^k-1 items, next one 2^k-2 and so on. Sum of sequence (geometric progression)
2^k + 2^k-1 + 2^k-2 + ... + 1 = 2^(k+1) = 2*n
so complexity for this dependence is linear too.
Now get dependence with smaller (zero) second coefficient:
T(n) = T(n/2) + O(1)
and ensure in linear complexity too.
Seems clear that complexity of recurrence in question lies between complexities for these simpler examples, and is linear.
In general case recurrences with complex branching might be solved with Aktra-Bazzi method (more general approach than Master theorem)
I assume that dependence is
T(n) = T(n/2) + T(n/3) + O(1)
In this case g=1, to find p we should numerically solve
(1/2)^p + (1/3)^p = 1
and get p~0.79, then integrate
T(x) = Theta(x^0.79 * (1 + Int[1..x]((1/u^0.79)*du))) =
Theta(x^0.79 * (1 + 4.8*x^0.21 - 4.8) =
Theta(x^0.79 + 4.8*x) =
Theta(x)
So complexity is linear
In this recurrence relation T(n)=T(n/4)+T(3n/4)+c , I am having just confusion that what is the relation of this recurrence relation with the best and worst case analysis since we have to solve both the sub-problems which are of size n/4 and 3n/4 so what is the terminology of worst case or best case analysis here ?
Moreover we should use here theta(log n ) our O(log n ) ,although seeing the below link I found O(log n ) more applicable but still couldn't get why are we not using theta(log n ) here .
How to solve the recursive complexity T(n) = T(n/4)+T(3n/4)+cn
T(n) = T(n/4) + T(3n/4) + CONST <= 2T(3n/4) + CONST
We will use case 1 of master theorem with:
a = 2, b = 4/3.
c = log_{4/3}(2) ~= 0.4
CONST is in O(n^0.4)
Thus, from master theorem, one cad derive that 2T(3n/4) + CONST is in Theta(logn), and since T(n) <= 2T(3n/4) + CONST, we can say that T(n) is in O(logn).
By following the same idea, but with lower bound:
T(n) >= T(3n/4) + CONST ...
And using master theorem again, we can tell that T(n) is also in Omega(logn).
Since T(n) is both O(logn) and Omega(logn), it is also Theta(logn).
As for your question, you can use either big-O or Theta notation, whatever you prefer. As you can see, proving Theta requires a bit more work, but it is also more informative, as it tells you the bound you found is tight.
These types of recurrences can be easily solved with Akra-Bazzi theorem (and if you have looked at the question you linked, someone showed you a solution to a similar problem).
So 1/4^p + (3/4)^p = 1, where p = 1. In your case g(u) = c, so the integral
So int c/u^2 du from 1 to x which is equal to -1/u evaluated from 1 to x. This is equal to -1/x + 1. Now when you multiply it by x and you will get that the complexity is O(n) and not O(log n) as other people suggested.
The Master Method is a direct way to get the solution. The Master Method works only for following type of recurrences or for recurrences that can be transformed to following type.
T(n) = a T(n / b) + f(n) where a ≥ 1, b > 1, and f(n) = Θ(nc).
There are following three cases:
If c < logb(a) then T(n) = Θ(nlogb(a)).
If c = logb(a) then T(n) = Θ(nc log(n)).
If c > logb(a) then T(n) = Θ(f(n)).
In the Master Method, if f(n) contains some term of log(n), is it possible to solve this by the Master Method?
for example in
T(n)=4T(n/2)+n^2logn
Here master's theorem appplicable or not
It is not really possible to tell directly whether or not the Master Method works for some logarithmic function. This would depend on the specific recurrence you're trying to solve. It all depends on how f grows in comparison to nlogb a.
In the example given by JPC (where T(n) = 4T(n/2) + log(n)), it is indeed possible. However, also consider the example T(n) = 2T(n/5) + log(n). In this recurrence it is harder to determine whether nlog5 2 grows faster than log(n). If the logarithmic function f(n) gets more complex (e.g. log3(n/2)), it becomes even harder.
In short, it may be hard to determine for logarithmic functions how they grow when compared to an exponential function when the exponent is less than 1 (for exponents >= 1, log(n) is always faster). If it doesn't seem to work for you, you'll have to use other techniques to solve the recurrence.
I understand that it is similar to the Fibonacci sequence that has an exponential running time. However this recurrence relation has more branches. What are the asymptotic bounds of T(n) = 2T(n-1) + 3T(n-2)+ 1?
Usually, you will need to make some assumptions on T(0) and T(1), because there will be exponentially many of them and their values may determine the functional form of T(n). However in this case it doesn't seem to matter.
Then, recurrence relations of this form can be solved by finding their characteristic polynomials. I found this article: http://www.wikihow.com/Solve-Recurrence-Relations
I obtained the characteristic polynomial with roots 3 and 1, so that guesses the form T(n) = c_1*3^n + c_2. In particular, T(n) = 1/2*3^n - 1/4 satisfies the recurrence relation, and we can verify this.
1/2*3^n - 1/4 = 2*T(n-1) + 3*T(n-2) + 1
= 2*(1/2*3^(n-1) - 1/4) + 3*(1/2*3^(n-2) - 1/4) + 1
= 3^(n-1) - 1/2 + 1/2*3^(n-1) - 3/4 + 1
= 3/2*3^(n-1) - 1/4
= 1/2*3^n - 1/4
Hence it would give that T(n) = Theta(3^n). However, this may not be the only function that satisfies the recurrence and other possibilities will also depend on what you defined the values T(0) and T(1), but they should all be O(3^n).
This type of recurrences are called: non-homogeneous recurrence relations and you have to solve in the beginning homogeneous recurrence (the one without a constant at the end). If you are interested, read the math behind it.
I will show you an easy way. Just type your equation in wolfram-alpha and you will get:
,
which is clearly an exponential complexity: O(3^n)