I have this recurrence relation
T(n) = T(n-1) + n, for n ≥ 2
T(1) = 1
Practice exercise: Solve recurrence relation using the iteration method and give an asymptotic running time.
So I solved it like this:
T(n) = T(n-1) + n
= T(n-2) + (n - 1) + n
= T(n-3) + (n - 2) + (n - 1) + n
= …
= T(1) + 2 + … (n - 2) + (n - 1) + n **
= 1 + 2 + … + (n - 2) + (n - 1) + n
= O(n^2)
I have some questions:
1)How I can find asymptotic running time?
**2)At this state of problem T(1) means that there was n that when it was subtracted with a number it gave the result 1, right?
3)What if T(0) = 1 and what if T(2) = 1?
Edit: 4) Why n ≥ 2 is useful?
I need really to understand it for my Mid-Term test
T(n) = T(n-1) + n, for n ≥ 2
T(1) = 1
If T(x) represents the running time:
You have already found the asymptotic running time, O(n^2) (quadratic).
If the relation is changed to T(0) = 1 or T(2) = 1, then the running time is still quadratic. The asymptotic behavior does not change if you add a constant or multiply by a constant, and changing the initial condition only adds a constant to the following terms.
n ≥ 2 is present in the relation so that T(n) is defined at exactly once for every positive n. Otherwise, both lines would apply to T(1). You cannot compute T(1) from T(0) using T(n) = T(n-1) + n. Even if you could, T(1) would be defined in two different (and potentially inconsistent) ways.
Related
.0 < c < 1 ,T(n) = T(cn) + T((1 - c)n) + 1
Base level:
if(n<=1) return;
data type - positive integers
I have to find the Big-Theta function of this recursive function.
I've tried to develop the recursive equation but it gets complicated from level to level and no formation is seen.
I also tried this -
assume that c<(1-c).
so -
2T(cn) + 1 <= T(cn) + T((1-c)n)+1 <= 2T((1-c)n)+1
It gave me some lower bound and upper bound but not a theta bound :(
As c approaches either 0 or 1, the recursion approaches T(n) = T(n-1) + 2 (assuming that T(0) = 1 as well). This has as a solution the linear function T(n) = 2n - 1 for n > 0.
For c = 1/2, the recursion becomes T(n) = 2T(n/2) + 1. It looks like T(n) = 2n - 1 is a solution to this for n > 0.
This seems like strong evidence that the function T(n) = 2n - 1 is a solution for all c: it works on both ends and in the middle. If we sub in...
2n - 1 = 2cn - 1 + 2(1-c)n - 1 + 1
= 2cn - 1 + 2n - 2cn - 1 + 1
= 2n - 1
We find that T(n) = 2n - 1 is a solution for the general case.
Quicksort's recurrence equation is
T(n) = T(n/2) + T(n/2) + theta(n)
if pivot always divides the original array into two same-sized sub arrays.
so the overall time complexity would be O(nlogn)
But what if the ratio of the two sub-lists is always 1:99?
The equation definitely would be T(n) = T(n/100) + T(99n/100) + theta(n)
But how can I derive time complexity from the above equation?
I've read other answer which describes that we can ignore T(n/100) since T(99n/100) will dominate the overall time complexity.
But I quite cannot fully understand.
Any advice would be appreciated!
Plug T(n) = n log(n) + Θ(n) and you get
n log(n) + Θ(n) = n/100 log(n/100) + 99n/100 log(99n/100) + Θ(n/100) + Θ(99n/100) + Θ(n)
= n log(n)/100 + 99n log(n)/100 - n/100 log(100) - 99n/100 log(99/100) + Θ(n)
= n log(n) + Θ(n)
In fact any fraction will work:
T(n) = T(pn) + T((1-p)n) + Θ(n)
is solved by O(n log(n)).
I have a question in my data structure course homework and I thought of 2 algorithms to solve this question, one of them is O(n^2) time and the other one is:
T(n) = 3 * n + 1*1 + 2*2 + 4*4 + 8*8 + 16*16 + ... + logn*logn
And I'm not sure which one is better.
I know that the sum of geometric progression from 1 to logn is O(logn) because I can use the geometric series formula for that. But here I have the squares of the geometric progression and I have no idea how to calculate this.
You can rewrite it as:
log n * log n + ((log n) / 2) * ((log n) / 2) + ((log n) / 4) * ((log n) / 4) ... + 1
if you substitute (for easier understanding) log^2 n with x, you get:
x + x/4 + x/16 + x/64 + ... + 1
You can use formula to sum the series, but if you dont have to be formal, then basic logic is enough. Just imagine you have 1/4 of pie and then add 1/16 pie and 1/64 etc., you can clearly see, it will never reach whole piece therefore:
x + x/4 + x/16 + x/64 + ... + 1 < 2x
Which means its O(x)
Changing back the x for log^2 n:
T(n) = O(3*n + log^2 n) = O(n)
I'm trying to prove that this formula (n2+1)/(n+1) is O(n)
As you know, we need to come up with n0 and C.
So I'm confused a little bit about how to choose an appropriate C since the equation here is division.
So with C=1, (n2+1) / (n+1) / n
(n2+n) / (n+n) / n >= (n2+1) /(n+1)
but I'm stuck here in how to simplify the division here.
As n tends to infinity your original equation becomes n^2/n which is equivalent to O(n)
Choosing c = 1:
(n^2 + 1)/(n + 1) <= 1*n definition of Big-Oh with c = 1
n^2 + 1 <= n^2 + n multiplying both sides by n + 1
1 <= n subtracting n^2 from both sides
n >= 1 rearranging
Therefore, the choice n0 = 1 works for c = 1.
Prove that
1 + 1/2 + 1/3 + ... + 1/n is O(log n).
Assume n = 2^k
I put the series into the summation, but I have no idea how to tackle this problem. Any help is appreciated
This follows easily from a simple fact in Calculus:
and we have the following inequality:
Here we can conclude that S = 1 + 1/2 + ... + 1/n is both Ω(log(n)) and O(log(n)), thus it is Ɵ(log(n)), the bound is actually tight.
Here's a formulation using Discrete Mathematics:
So, H(n) = O(log n)
If the problem was changed to :
1 + 1/2 + 1/4 + ... + 1/n
series can now be written as:
1/2^0 + 1/2^1 + 1/2^2 + ... + 1/2^(k)
How many times loop will run? 0 to k = k + 1 times.From both series we can see 2^k = n. Hence k = log (n). So, number of times it ran = log(n) + 1 = O(log n).