Describe the recurrence running time T(n) on an input of size n?
A divide and conquer algorithm takes an array of n elements and divides into three sub arrays of size n/4 each taking Θ(n) time to do the subdivision. The time taken to combine outputs of each sub-problem is Θ(1).
I came with this recurrence relation, but it's not correct
T(n) = 3T(n/4) + Θ(1)
Can someone knows what I am doing wrong on this?
You missed taking Θ(n) time to do the subdivision part.
So relation should include subdivision + working on smaller parts + combining
T(n)= Θ(n) + 3T(n/4) + Θ(1) = 3T(n/4) + Θ(n)
Related
I'm trying to solve the recurrence equation of a divide and conquer algorithm knowing that the time required to decompose the copy into sub copies is in O(n^2/log n) and the time required for recombination is in O (n^2) .the number of sub-examples is a of size n/4 which makes T(n)=a*T(n/4) + n^2/log n + n^2
so how to find the recurrence equation expressing the execution time T(n) in a function of the size n of the copy
thanks a lot
after analyzing the algorithm complexity I have a few questions:
For the best case complexity - the recurrence relation is T(n) = T(n/2) + dn which implies that the complexity is Θ(n).
So by the master theory I can clearly see why this is true , but when I draw the algorithm recursive calls as a tree I don't fully understand the final result. (It seems like I have one branch in height of log(n) which in each level I operate a partition O(n) - so it suppose to be nlog(n) .
(just for memorizing - this is very similar to the best case of mergeSort algorithem , but here we ignore the unwanted sub-array after partition).
Thanks!
It is as Yves Daoust wrote. Image it with real numbers, i.e. n=1024
T(n) = T(n/2) + dn
T(1024) = T(512) + 1024
T(512) = T(256) + 512
....
T(2) = T(1) + 2 -> this would be the last operation
Therefore you get 1024+512+256+...+1 <= 2048, which is 2n
You must think about that dn is as big as n, but in recurrence relation the n is not global variable, it is local variable based on method you call.
So there is log(n) calls but they do not take n-time everyone, they take less and less time.
here we have an algorithm
T(n) = n-1 + T(i-1) + T(n-i)
T(1) = 1
How to calculate it's time complexity?
i is between 1 and n
I can recognise this as quick sort algorithm (randomized quick sort).
I am sure the question somehow missed the summation part.
Okay! you can use substitution method over here..check with O(n^2). you will get the idea that O(n^2) is the worst case time complexity.
The average case is a bit tricky. Then the pivot can be any element from 1 to n. Then analyse it. Here also you can apply substituin with T(n)=O(nlogn).
I think we should solve it like this
if i = 2 then we have
T(n) = n + T(n-2) = Theta(n^2)
if i = n/2 then we have
T(n) = n-1 + T(n/2 -1) + T(n/2) = Theta(n.logn)
then we have upper bound O(n^2) and algorithm is in order of O(n^2)
Following are two recurrence relations
T(n)= T(n/2)+T(n/2) + C
T(n)= T(n/2)*T(n/2) + C
Will both the have the same time complexity? Can I write both recurrence relations like this?
T(n) = 2T(n/2) + C
(1) is obviously the same as (3): T(n/2) + T(n/2) = 2 T(n/2). That's elementary math.
(1) is not the same as (2), and it shouldn't be difficult to see that the solution to these relations is completely different. (1) = (3) means that for data that's twice as large, the complexity measure is about twice as large — linear complexity. (2) means that for data that's twice as large, the complexity is squared — exponential complexity.
Assume we constructed a quicksort and the pivot value takes linear time. Find the recurrence for worst-case running time.
My answer:
T(n)= T(n-1) + T(1) + theta(n)
Worst case occurs when the subarrays are completely unbalanced.
There is 1 element in one subarray and (n-1) elements in the other subarray.
theta(n) because it takes running time n to find the pivot.
Am I doing this correctly?
Your recurrence is mostly correct, but you don't actually have two recursive calls made. In the worst-case for quicksort, the pivot will be the largest or smallest element in the array, so you'll recur on one giant array of size n - 1. The other subarray has length 0, so no recursive calls are made. To top everything off, the total work done is Θ(n) per level, so the recurrence relation would more appropriately be
T(n) = T(n - 1) + Θ(n)
This in turn then solves to Θ(n2).
Hope this helps!
you cannot observe, because according to my research T(N)= T(N-K)+T(K-1)+n
we cannot observe exact value until we have
value of k,
T(n) = T(an/(a+b)) + T(bn/(a+b)) + n
Where a/(a+b) and b/(a+b) are fractions of array under consideration