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See title. I'm trying to apply the method from this question: Easy: Solve T(n)=T(n-1)+n by Iteration Method. What I have so far is this, but I don't know how to proceed from here on:
T(n) = T(n-1) + n2
T(n-1) = T(n-2) + (n-1)2 = T(n-2) + n2 - 2n + 1
T(n-2) = T(n-3) + (n-2)2 = T(n-3) + n2 - 4n + 4
T(n-3) = T(n-4) + (n-3)2 = T(n-4) + n2 - 6n + 9
Substituting the values of T(n-1), T(n-2) and T(n-3) into T(n) gives:
T(n) = T(n-2) + 2n2 - 2n + 1
T(n) = T(n-3) + 3n2 - 6n + 5
T(n) = T(n-4) + 4n2 - 12n + 14
Now I have to find a pattern but I don't really know how to do that. What I got is:
T(n) = T(n-k) + kn2 - ...???
I'd start with a guess that since each T (n) is equal to the previous one plus a square, T (n) is something cubic.
Write T (n) = an^3 + bn^2 + cn + d.
Substitute this into T (n) = T (n - 1) + n^2 and solve for a, b, c.
And obviously T (0) = d.
If you combine the equations you yourself write:
T(n)=
=n^2+T(n-1)=
=n^2+(n-1)^2+T(n-2)=
=n^2+(n-1)^2+(n-2)^2+T(n-3)=
=n^2+(n-1)^2+(n-2)^2+(n-3)^2+...+2^2+1^2+T(0)=
=n(n+1)(2n+1)/6+T(0) //based on well known formula for S(x^2, x=1..n)
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How do I determine the running time (in terms of Big-Theta) for the algorithm of input size n that satisfies recurrence relation T(n) = T(n-1)+n where n >= 1 and with initial condition T(1) = 1?
Edit: I was practicing a past exam paper. Got stuck on this question. Need guidance
Look at it this way: T(n) = T(n-1) + n = T(n-2) + (n-1) + n = T(n-3) + (n-2) + (n-1) + n. Which means if n >= 1 then you will get something like T(n) = 1 + 2 + 3 + ... + n. If you work out the pattern of this series you will see that (n+1)n/2. Therefore, Ө(n^2)
I thought it would be something like this...
T(n) = 2T(n-1) + O(n)
= 2(2T(n-2)+(n-1)) + (n)
= 2(2(2T(n-3)+(n-2))+(n-1))+(n)
= 8T(n-3) + 4(n-2) + 2(n-1) + n
Which ends up being something like the summation of 2i * (n-i), and my book says this ends up being O(2n). Could anybody explain this to me? I don't understand why it's 2n and not just O(n) as the (n-i) will continue n times.
This recurrence has already been solved on Math Stack Exchange. As I solve this recurrence, I get:
T(n) = n + 2(T(n-1))
= n + 2(n - 1 + 2T(n-2)) = 3n - 2 + 2^2(T(n-2))
= 3n - 2 + 4(n - 2 + 2(T(n-3))) = 7n - 10 + 2^3(T(n-3))
= 7n - 10 + 8(n - 3 + 2(T(n-4))) = 15n - 34 + 2^4(T(n-4))
= (2^4 - 1)n - 34 + 2^4(T(n-4))
...and so on.
Effectively the recurrence boils down to:
T(n) = (2n+1) * T(1) − n − 2
See the Math Stack Exchange link for how we arrive at this solution. Taking T(1) to be constant, the dominating factor in the above recurrence is (2(n + 1)).
Therefore, the rate of growth of given recurrence is O(2n).
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Find the tight asymptotic bound:
T(n) = 1 if n = 1
2T(n/4) + T(n/2) + n2 if n > 1
I tried drawing a recurrence tree. First row I had n2, second row I had (3/8)(n4), third row I had (27/1024)(n8).. Don't know how to continue from there.
Thanks in advance.
I think the Master Theorem can be applied in this problem. It is much easier to understand than recurrence tree.
Write the recurrence formula in another form. Add T(n/2) to both the left and right side.
T(n) + T(n/2) = + 2T(n/4) + 2T(n/2) + n2
T(n) + T(n/2) = + 2[ T(n/4) + T(n/2) ] + n2
Define G(n) = T(n) + T(n/2), then
G(1) = Θ(1)
G(n) = 2G(n/2) + n2, if n > 1
Apply the Master Theorem to the above new recurrence formula
G(n) = Θ(n2)
that is
T(n) + T(n/2) = Θ(n2)
for n > 1
T(n) = 2 * T(n/4) + T(n/2) + n2
= T(n/4) + [ T(n/2) + T(n/4) ] + n2
= T(n/4) + Θ(n2/4) + n2
= T(n/4) + Θ(n2)
Apply the Master Theorem to the above new recurrence formula
T(n) = Θ(n2)
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I am having some issues on how to solve recurrence relations.
T(n) = T(n/2) + log2(n), T(1) = 1, where n is a power of 2
This is a homework problem, so don't just give me the answer. I was just wondering how to start the problem.
In class we went over the Master theorem. But I don't think that would be the best way to solve this particular relation.
I don't really know how to start the problem... should I just be going
T(n) = T(n/2) + log_base2(n)
T(n/2) = [T(n/4)+log_base2(n/2)]
T(n) = [T(n/4)+log_base2(n/2)] + log_base2(n)
And just keep working my way down to get something I can see makes a basic equation?
This recurrence solves to Θ((log n)2). Here are two ways to see this.
Some Substitutions
If you know that n is a perfect power of two (that is, n = 2k), you can rewrite the recurrence as
T(2k) = T(2k-1) + k
Let's define a new recurrence S(k) = T(2k). Then we get that
S(k) = S(k - 1) + k
If we expand out this recurrence, we get that
S(k) = S(k - 1) + k
= S(k - 2) + (k - 1) + k
= S(k - 3) + (k - 2) + (k - 1) + k
= S(k - 4) + (k - 3) + (k - 2) + (k - 1) + k
...
= S(0) + 1 + 2 + 3 + ... + k
= S(0) + Θ(k2)
Assuming S(0) = 1, then this recurrence solves to Θ(k2).
Since S(k) = T(2k) = T(n), we get that T(n) = Θ(k2) = Θ(log2 n).
Iterating the Recurrence
Another option here is to expand out a few terms of the recurrence and to see if any nice patterns emerge. Here’s what we get:
T(n) = T(n / 2) + lg n
= T(n / 4) + lg (n / 2) + lg n
= T(n / 8) + lg (n / 4) + lg (n / 2) + lg n
...
Eventually, after lg n layers, this recurrence bottoms out and we’re left with this expression:
lg n + lg (n / 2) + lg (n / 4) + ... + lg (n / 2lg n)
Using properties of logarithms, we can rewrite this as
lg n + (lg n - 1) + (lg n - 2) + (lg n - 3) + ... + (lg n - lg n)
Or, written in reverse, this is the sum
0 + 1 + 2 + 3 + ... + lg n
That sum is Gauss’s sum up to lg n, which evaluates to (lg n)(lg n + 1) / 2 = Θ((log n)2).
Hope this helps!
If n is a power of 2 then you can just expand out the recurrence and solve exactly, using that lg(a/b) = lg(a) - lg(b).
T(n) = lg(n) + lg(n/2) + lg(n/4) + ... + lg(1) + 1
= (lg(n) - 0) + (lg(n) - 1) .... + (lg(n) - lg(n)) + 1
= lg(n)*lg(n) - lg(n)*(lg(n)+1)/2 + 1
= lg(n)*lg(n)/2 - lg(n)/2 + 1
This can be done with the Akra-Bazzi theorem. See the third example in http://people.mpi-inf.mpg.de/~mehlhorn/DatAlg2008/NewMasterTheorem.pdf.
This can be solved with Master theorem. Your a=1 and b=2 and f(n) = log(n). Then c = log2(1) = 0. Because of your c and f(n) you fall into the second case (where k=1).
So the solution is Θ(log2 n)
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I am looking to prove that T(n)=T(n/2)+sqrt(n) is O(sqrt(n)) given T(1)=1
using only induction.
It is easy to solve using the Master theorem but this is not the case.
I tried to assume
T(n/2) < c*sqrt(n/2)
but didnt get very far with the rest of the proof.
Thank you all in advance for your answers.
Edit:
my line of solution (after the assumption above) is:
T(n) <= c*sqrt(n/2)+sqrt(n) = sqrt(n)(c/sqrt(2)+1) <= sqrt(n)(c+1)
I dont know how to move from this to the required
T(n)<=c*sqrt(n)
ok, you're close. So basically, as I mentioned in the comment, base case is simple. For induction case, you want to show that T(n) is O(sqrt(n)) given that T(n/2) is O(sqrt(n/2)).
So, it goes like this:
T(n) = T(n/2) + sqrt(n) ; this is just your recurrence
< c sqrt(n/2) + sqrt(n) ; since T(n/2) is O(sqrt(n))
; wlog here, assume c > 4
= c sqrt(n) / sqrt(2) + sqrt(n)
= (c/sqrt(2) + 1) sqrt(n)
observe that for c > 4, c / sqrt(2) + 1 < c, so
(c/sqrt(2) + 1) sqrt(n) < c sqrt(n)
so
T(n) < c sqrt(n)
Therefore, T(n) is O(sqrt(n))
So there's a couple key points here that you missed.
The first is that you can always increase the c to whatever value you want. This is because big O only requires <. if it's < c f(n) then it is < d f(n) where d > c.
The second is to note that the line f(c) = c/sqrt(2) + 1 intersects with the line f(c) = c at about c = sqrt(2) / (sqrt(2)-1) = 3.4143 (or so), so all you have to do is force c to be > this value in order to get (c/sqrt(2) + 1) < c. 4 certainly works, so that's where the 4 comes from.
In retrospect, I should have given the key points as hints. My fault. Sorry!
One line of thinking which may help is to expand the recurrence recursively. You get
T(n) = sqrt(n) + sqrt(n/2) + sqrt(n/4) + ... + sqrt(n/(2^k)) + ... + sqrt(1)
= sqrt(n) + sqrt(n)/sqrt(2) + sqrt(n)/sqrt(4) + ... + sqrt(n)/sqrt(2^k) + ... + sqrt(1)
= sqrt(n) * (1 + sqrt(1/2) + sqrt(1/2)^2 + ... + sqrt(1/2)^k + ...)
<= sqrt(n) * ∑(k=0 to ∞) sqrt(1/2)^k
= sqrt(n) * 1/(1 - sqrt(1/2))
Since 1/(1-sqrt(1/2)) is a finite constant (it's about 3.4), T(n) must be O(sqrt(n)). You can use this information to prove it using standard induction.