I am having trouble understanding the concept of recurrences. Given you have T(n) = 2T(n/2) +1 how do you calculate the complexity of this relationship? I know in mergesort, the relationship is T(n) = 2T(n/2) + cn and you can see that you have a tree with depth log2^n and cn work at each level. But I am unsure how to proceed given a generic function. Any tutorials available that can clearly explain this?
The solution to your recurrence is T(n) ∈ Θ(n).
Let's expand the formula:
T(n) = 2*T(n/2) + 1. (Given)
T(n/2) = 2*T(n/4) + 1. (Replace n with n/2)
T(n/4) = 2*T(n/8) + 1. (Replace n with n/4)
T(n) = 2*(2*T(n/4) + 1) + 1 = 4*T(n/4) + 2 + 1. (Substitute)
T(n) = 2*(2*(2*T(n/8) + 1) + 1) + 1 = 8*T(n/8) + 4 + 2 + 1. (Substitute)
And do some observations and analysis:
We can see a pattern emerge: T(n) = 2k * T(n/2k) + (2k − 1).
Now, let k = log2 n. Then n = 2k.
Substituting, we get: T(n) = n * T(n/n) + (n − 1) = n * T(1) + n − 1.
For at least one n, we need to give T(n) a concrete value. So we suppose T(1) = 1.
Therefore, T(n) = n * 1 + n − 1 = 2*n − 1, which is in Θ(n).
Resources:
https://www.cs.duke.edu/courses/spring05/cps100/notes/slides07-4up.pdf
http://www.cs.duke.edu/~ola/ap/recurrence.html
However, for routine work, the normal way to solve these recurrences is to use the Master theorem.
Related
Is this recurrence relation O(infinity)?
T(n) = 49*T(n/7) + n
There are no base conditions given.
I tried solving using master's theorem and the answer is Theta(n^2). But when solving with recurrence tree, the solution comes to be an infinite series, of n*(7 + 7^2 + 7^3 +...)
Can someone please help?
Let n = 7^m. The recurrence becomes
T(7^m) = 49 T(7^(m-1)) + 7^m,
or
S(m) = 49 S(m-1) + 7^m.
The homogeneous part gives
S(m) = C 49^m
and the general solution is
S(m) = C 49^m - 7^m / 6
i.e.
T(n) = C n² - n / 6 = (T(1) + 1 / 6) n² - n / 6.
If you try the recursion method:
T(n) = 7^2 T(n/7) + n = 7^2 [7^2 T(n/v^2) + n/7] + n = 7^4 T(n/7^2) + 7n + n
= ... = 7^(2i) * T(n/7^i) + n * [7^0 + 7^1 + 7^2 + ... + 7^(i-1)]
When the i grows n/7^i gets closer to 1 and as mentioned in the other answer, T(1) is a constant. So if we assume T(1) = 1, then:
T(n/7^i) = 1
n/7^i = 1 => i = log_7 (n)
So
T(n) = 7^(2*log_7 (n)) * T(1) + n * [7^0 + 7^1 + 7^2 + ... + 7^(log_7(n)-1)]
=> T(n) = n^2 + n * [1+7+7^2+...+(n-1)] = n^2 + c*n = theta(n^2)
Usually, when no base case is provided for a recurrence relation, the assumption is that the base case is something T(1) = 1 or something along those lines. That way, the recursion eventually terminates.
Something to think about - you can only get an infinite series from your recursion tree if the recursion tree is infinitely deep. Although no base case was specified in the problem, you can operate under the assumption that there is one and that the recursion stops when the input gets sufficiently small for some definition of "sufficiently small." Based on that, at what point does the recursion stop? From there, you should be able to convert your infinite series into a series of finite length, which then will give you your answer.
Hope this helps!
I have a Computer Science Midterm tomorrow and I need help determining the complexity of a particular recursive function as below, which is much complicated than the stuffs I've already worked on: it has two variables
T(n) = 3 + mT(n-m)
In simpler cases where m is a constant, the formula can be easily obtained by writing unpacking the relation; however, in this case, unpacking doesn't make the life easier as follows (let's say T(0) = c):
T(n) = 3 + mT(n-m)
T(n-1) = 3 + mT(n-m-1)
T(n-2) = 3 + mT(n-m-2)
...
Obviously, there's no straightforward elimination according to these inequalities. So, I'm wondering whether or not I should use another technique for such cases.
Don't worry about m - this is just a constant parameter. However you're unrolling your recursion incorrectly. Each step of unrolling involves three operations:
Taking value of T with argument value, which is m less
Multiplying it by m
Adding constant 3
So, it will look like this:
T(n) = m * T(n - m) + 3 = (Step 1)
= m * (m * T(n - 2*m) + 3) + 3 = (Step 2)
= m * (m * (m * T(n - 3*m) + 3) + 3) + 3 = ... (Step 3)
and so on. Unrolling T(n) up to step k will be given by following formula:
T(n) = m^k * T(n - k*m) + 3 * (1 + m + m^2 + m^3 + ... + m^(k-1))
Now you set n - k*m = 0 to use the initial condition T(0) and get:
k = n / m
Now you need to use a formula for the sum of geometric progression - and finally you'll get a closed formula for the T(n) (I'm leaving that final step to you).
How is the following recurrence relation solved:
T(n) = 2T(n-2)+O(1)
What I have tried so far is:
O(1) is lower or equal than a constant c.
So
T(n) <= 2T(n-2) + c
T(n) <= 4T(n-4) + 2c
T(n) <= 8T(n-6) + 3c
.
.
.
So a pattern is emerging... the general term is:
T(n) <= 2^k*T(n-2k) + kc
But I dont know how to continue from there.Any advice is appreciated.
Assuming your generalization for k is true[1]
T(n) <= 2^k*T(n-2k) + kc
For k=n/2 you get:
T(n) <= 2^n/2 * T(0) + n/2 * c = 2^(n/2) + n/2*c
Which is in O(sqrt(2^n))
Formal proof can be done with induction, with the induction hypothesis of:
T(n) <= 2^(n/2) + c*n
And step of:
T(n) = 2T(n-2) + c = (induction hypothesis)
T(n) = 2* 2^((n-2)/2) + (n-2)*c + c
T(n) = 2^ (n/2 - 2/2 + 1) + (n-1)*c
And indeed:
T(n) = 2^(n/2) + (n-1)*c <= 2^(n/2) + c*n
(1) It is not, it ignores the fact that the constant is multiplied in the loop.
I have been asked to analyze the asymptotic time complexity of the following recursion function:
for-all k ≥ 1:
T(n) = n + T(n/2) + T(n/4) + T(n/8) + .... + T(n/2^k)
I was able to prove that:
T(n) = O(n⋅log n) and T(n) = Ω(n),
but I am looking for a tighter bound (Big Theta).
First of all:
I understand "for-all k >= 1" this way: for k = 1 to k = m where 2m-1 ≤ n ≤ 2m.
So basicly m = log₂(n) holds.
Have a look at my calculation:
T(n) = n + Σk=1,...,m T(n/2k)
= n + T(n/2) + Σk=2,...,m T(n/2k)
= n + n/2 + 2⋅Σk=2,...,m T(n/2k)
= ...
= n + Σk=1,...,m k⋅n/2k
= n + n⋅Σk=1,...,m k/2k
= n + n⋅(2 - 2-mm - 21-m)
≤ n + 2⋅n
= 3n
So T(n) is in Θ(n).
Notice:
You can also approximate Σk=1,...,m k/2k by the integral s(m) = ∫1m k/2k dk.
And here limm → ∞s(m) = 2 also holds.
How would I go about solving this kind of recurrence using the Master Theorem?
T(n) = 4T(n/2) + n2 + logn
I have no idea how to go about doing this, but I'm pretty sure it is possible to solve it using Master Theorem. Do I have to ignore one of the terms? Any help is appreciated, thanks.
The Master Theorem works for functions that can be written as
T(n) = aT(n / b) + f(n)
Here, you have that a = 4, b = 2, and f(n) = n2 + log n. Notice that we're grouping "n2 + log n" together as the f(n) term, rather than treating it as two separate terms.
Now that we've done that, we can apply the Master Theorem directly. Notice that logb a = log2 4 = 2 and that f(n) = Θ(n2), so by the Master Theorem this solves to Θ(n2 log n). The reason this works is that n2 + log n = Θ(n2), and the Master Theorem only cares about the asymptotic complexity of f(n). In fact, any of these recurrences can be solved the same way:
T(n) = 4T(n / 2) + n2 + 137n + 42
T(n) = 4T(n / 2) + 5n2 + 42n log n + 42n + 5 log n + 106
T(n) = 4T(n / 2) + 0.5n2 + n log137 n + n log n + n2 / log n + 5
Hope this helps!