I have been trying to solve a recurrence relation.
The recurrence is T(n) = T(n/3)+T(2n/3)+n^2
I solved the the recurrence n i got it as T(n)=nT(1)+ [ (9/5)(n^2)( (5/9)^(log n) ) ]
Can anyone tell me the runtime of this expression?
I think this recurrence works out to Θ(n2). To see this, we'll show that T(n) = Ω(n2) and that T(n) = O(n2).
Showing that T(n) = Ω(n2) is pretty straightforward - since T(n) has an n2 term in it, it's certainly Ω(n2).
Let's now show that T(n) = O(n2). We have that
T(n) = T(n / 3) + T(2n / 3) + n2
Consider this other recurrence:
S(n) = S(2n / 3) + S(2n / 3) + n2 = 2S(2n / 3) + n2
Since T(n) is increasing and T(n) ≤ S(n), any upper bound for S(n) should also be an upper-bound for T(n).
Using the Master Theorem on S(n), we have that a = 2, b = 3/2, and c = 2. Since logb a = log3/2 2 = 1.709511291... < c, the Master Theorem says that this will solve to O(n2). Since S(n) = O(n2), we also know that T(n) = O(n2).
We've shown that T(n) = Ω(n2) and that T(n) = O(n2), so T(n) = Θ(n2), as required.
Hope this helps!
(By the way - (5 / 9)log n = (2log 5/9)log n = 2log n log 5/9 = (2log n)log 5/9 = nlog 5/9. That makes it a bit easier to reason about.)
One can't tell about runtime from the T(n) OR the time complexity!It is simply an estimation of running time in terms of order of input(n).
One thing which I'd like to add is :-
I haven't solved your recurrence relation,but keeping in mind that your derived relation is correct and hence further putting n=1,in your given recurrence relation,we get
T(1)=T(1/3)+T(2/3)+1
So,either you'll be provided with the values for T(1/3) and T(2/3) in your question OR you have to understand from the given problem statement like what should be T(1) for Tower of Hanoi problem!
For a recurrence, the base-case is T(1), now by definition its value is as following:
T(1) = T(1/3) + T(2/3) + 1
Now since T(n) denotes the runtime-function, then the run-time of any input that will not be processed is always 0, this includes all terms under the base-case, so we have:
T(X < 1) = 0
T(1/3) = 0
T(2/3) = 0
T(1) = T(1/3) + T(2/3) + 1^2
T(1) = 0 + 0 + 1
T(1) = 1
Then we can substitute the value:
T(n) = n T(1) + [ (9/5)(n^2)( (5/9)^(log n) ) ]
T(n) = n + ( 9/5 n^2 (5/9)^(log n) )
T(n) = n^2 (9/5)^(1-log(n)) + n
We can approximate (9/5)^(1-log(n)) to 9/5 for asymptotic upper-bound, since (9/5)^(1-log(n)) <= 9/5:
T(n) ~ 9/5 n^2 + n
O(T(n)) = O(n^2)
Related
I watched a video where they prove T(n)= T(n-1) + n is O(n^2)
I have the following expressions which are:
T(1) = 4
T(N) = T(N – 1) + N + 3, N > 1
My question is, is the expression above solved the same way, even though there is a +3 after N.
The question is a bit messed up, but i hope you get the point. If there are questions i will try to explain better.
In a word is T(N) = T(N – 1) + N + 3 = O(n^2)
T(n) = T(n-1) + n-1 + 4 => given equation by adding 1 and subtracting 1
T(n) = T(n-1) + n-1 + T(1) ...(1)
Now, T(1) = constant.
Therefore, from eq(1),
T(n) = T(n-1) + (n-1) ...(2)
Eq(2) reduces to T(n) = T(n-k) + n*k - k*(k+1)/2 ...(3)
Upon substituting (n-k)=1 or k=(n-1) in eq(3),
we get,
T(n) = T(1) + n*(n-1) - (n-1)(n)/2
T(n) = n*(n-1)/2 => O(n^2)
PS: If we won't neglect T(1) in eq(1), final equation we get is T(n) = n*(n-1)/2 + T(1) + 4*k => T(n) = n*(n-1)/2 + 4 + 4*(n-1) which still gives O(n^2) as final answer.
i want to know what the Time Complexity of my recursion method :
T(n) = 2T(n/2) + O(1)
i saw a result that says it is O(n) but i don't know why , i solved it like this :
T(n) = 2T(n/2) + 1
T(n-1) = 4T(n-1/4) + 3
T(n-2) = 8T(n-2/8) + 7
...... ………….. ..
T(n) = 2^n+1 T (n/2^n+1) + (2^n+1 - 1)
I think you have got the wrong idea about recursive relations. You can think as follows:
If T(n) represents the value of function T() at input = n then the relation says that output is one more double the value at half of the current input. So for input = n-1 output i.e. T(n-1) will be one more than double the value at half of this input, that is T(n-1) = 2*T((n-1)/2) + 1
The above kind of recursive relation should be solved as answered by Yves Daoust. For more examples on recursive relations, you can refer this
Consider that n=2^m, which allows you to write
T(2^m)=2T(2^(m-1))+O(1)
or by denoting S(m):= T(2^m),
S(m)=2 S(m-1) + O(1),
2^m S(m)=2 2^(m-1)S(m-1) + 2^(m-1) O(1)
and finally,
R(m) = R(m-1) + 2^(m-1) O(1).
Now by induction,
R(m) = R(0) + (2^m-1) O(1),
T(n) = S(m) = 2^(1-m) T(2^m) + (2 - 2^(m-1)) O(1) = 2/n T(n) + (2 - n/2) O(1).
There are a couple of rules that you might need to remember. If you can remember these easy rules then Master Theorem is very easy to solve recurrence equations. The following are the basic rules which needs to be remembered
case 1) If n^(log b base a) << f(n) then T(n) = f(n)
case 2) If n^(log b base a) = f(n) then T(n) = f(n) * log n
case 3) 1) If n^(log b base a) >> f(n) then T(n) = n^(log b base a)
Now, lets solve the recurrence using the above equations.
a = 2, b = 2, f(n) = O(1)
n^(log b base a) = n = O(n)
This is case 3) in the above equations. Hence T(n) = n^(log b base a) = O(n).
How to get big-O for this?
T(N) = 2T(N − 1) + N, T(1) = 2
I got two variants of answer O(2^N) or O(N^2), but I am not sure how to solve it correctly
Divide T(N) by 2^N and name the result:
S(N) = T(N)/2^N
From the definition of T(N) we get
S(N) = S(N-1) + N/2^N (eq.1)
meaning that S(N) increases, but quickly converges to a constant (since N/2^N -> 0). So,
T(N)/2^N -> constant
or
T(N) = O(2^N)
Detailed proof
In the comment below Paul Hankin suggests how to complete the proof. Take eq.1 and sum from N=2 to N=M
sum_{N=2}^M S(N) = sum_{N=2}^M S(N-1) + sum_{N=2}^M N/2^N
= sum_{N=1}{M-1} S(N) + sum_{N=1}^{M-1} (N-1)/2^{N-1}
thus, after canceling terms with indexes N = 2, 3, ..., M-1, we get
S(M) = S(1) + sum_{N=1}^M N/2^N - M/2^M
and since the series on the right converges (because its terms are bounded by 1/N^2 for N>>1 which is known to converge), S(M) converges to a finite constant.
It's a math problem and Leandro Caniglia is right.
let b(n) = T(n) / 2^n
thus b(n) = b(n-1) + n / 2^n = b(n-2) + n / 2^n + (n-1) / 2^(n-1) ....
i / 2^i is less than 1 for every integer i
So the sum of them has limit and must smaller than some constant.
thus b(n) < C.
thus T(n) < 2^n * C.
It is obvious that T(n) >= 2^n.
So T(n) is O(2^n)
Check by plugging the answer in the equation.
2^N = 2.2^(N-1) + N = 2^N + N
or
N^2 = 2 (N-1)^2 + N
Keeping only the dominant terms, you have
2^N ~ 2^N
or
N^2 ~ 2 N^2.
Conclude.
I'm familiar with solving recurrences with iteration:
t(1) = c1
t(2) = t(1) + c2 = c1 + c2
t(3) = t(2) + c2 = c1 + 2c2
...
t(n) = c1 + (n-1)c2 = O(n)
But what if I had a recurrence with no base case? How would I solve it using the three methods mentioned in the title?
t(n) = 2t(n/2) + 1
For Master Theorem I know the first step, find a, b, and f(n):
a = 2
b = 2
f(n) = 1
But not where to go from here. I'm at a standstill because I'm not sure how to approach the question.
I know of 2 ways to solve this:
(1) T(n) = 2T(n/2) + 1
(2) T(n/2) = 2T(n/4) + 1
now replace T(n/2) from (2) into (1)
T(n) = 2[2T(n/4) + 1] + 1
= 2^2T(n/4) + 2 + 1
T(n/4) = 2T(n/8) + 1
T(n) = 2^2[2T(n/8) + 1] + 2 + 1
= 2^3T(n/8) + 4 + 2 + 1
You would just keep doing this until you can generalize. Eventually you will spot that:
T(n) = 2^kT(n/2^k) + sum(2^(k-1))
You want T(1) so set n/2^k = 1 and solve for k. When you do this you will find that, k = lgn
Substitute lgn for k you will end up with
T(n) = 2^lgnT(n/2^lgn) + (1 - 2^lgn) / (1 - 2)
2^lgn = n so,
T(n) = nT(1) + n - 1
T(n) = n + n - 1 where n is the dominant term.
For Master Theorem its really fast
Consider, T(n) = aT(n/b) + n^c for n>1
There are three cases (note that b is the log base)
(1) if logb a < c, T(n)=Θ(n^c),
(2) if logb a = c, T (n) = Θ(n^c log n),
(3) if logb a > c, T(n) = Θ(n^(logb a)).
In this case a = 2, b = 2, and c = 0 (n^0 = 1)
A quick check shows case 3.
n^(log2 2)
note log2 2 is 1
So by master theorem this is Θ(n)
Apart from the Master Theorem, the Recursion Tree Method and the Iterative Method there is also the so
called "Substitution Method".
Often you will find people talking about the
substitution method, when in fact they mean the iterative method (especially on Youtube).
I guess this stems from the fact that in the iterative method you are also substituting
something, namely the n+1-th recursive call into the n-th one...
The standard reference work about algorithms
(CLRS)
defines it as follows:
Substitution Method
Guess the form of the solution.
Use mathematical induction to find the constants and show that the solution works.
As example let's take your recurrence equation: T(n) = 2T(ⁿ/₂)+1
We guess that the solution is T(n) ∈ O(n²), so we have to prove that
T(n) ≤ cn² for some constant c.
Also, let's assume that for n=1 you are doing some constant work c.
Given:
T(1) ≤ c
T(n) = 2T(ⁿ/₂)+1
To prove:
∃c > 0, ∃n₀ ∈ ℕ, ∀n ≥ n₀, such that T(n) ≤ cn² is true.
Base Case:
n=1: T(1) ≤ c
n=2: T(2) ≤ T(1) + T(1) + 1 ≤ 4c
(≤c) (≤c) (cn²)
Induction Step:
As inductive hypothesis we assume T(n) ≤ cn² for all positive numbers smaller than n
especially for (ⁿ/₂).
Therefore T(ⁿ/₂) ≤ c(ⁿ/₂)², and hence
T(n) ≤ 2c(ⁿ/₂)² + 1 ⟵ Here we're substituting c(ⁿ/₂)² for T(ⁿ/₂)
= (¹/₂)cn² + 1
≤ cn² (for c ≥ 2, and all n ∈ ℕ)
So we have shown, that there is a constant c, such that T(n) ≤ cn² is true for all n ∈ ℕ.
This means exactly T(n) ∈ O(n²). ∎
(for Ω, and hence Θ, the proof is similar).
I need to solve: T(n) = T(n-1) + O(1)
when I find the general T(n) = T(n-k) + k O(1)
what sum is it? I mean when I reach the base case: n-k=1; k=n-1
Is it "sum k, k=1 to n"? but the result of this sum is n(n-1)/2 and I know that the result is O(n).
So I know that I don't need a sum with this relation but what sum is correct for this recurrence relation?
Thanks
If we make the (reasonable) assumption that T(0) = 0 (or T(1) = O(1)), then we can apply your
T(n) = T(n - k) + k⋅O(1) to k = n and obtain
T(n) = T(n - n) + n⋅O(1) = 0 + n⋅O(1) = O(n).
Edit: if you insist on representing the recurrence as a sum, here it is:
T(n) = T(n - 1) + O(1) = T(n - 2) + O(1) + O(1) = ... = Σk = 1,...n O(1) = n⋅O(1) = O(n)