When analyzing a recurrence of the form T(n) = a T(nb) + θ(nc), under which of the following conditions can we conclude that “most of the work occurs at the leaves of the recursion tree”?
Case 1: f(n) is O(nlogba - ε). Since the leaves grow faster than f, asymptotically all of the work is done at the leaves, and so T(n) is Θ(nlogb a).
Case 2: f(n) is Θ(nlogba). The leaves grow at the same rate as h, so the same order of work is done at every level of the tree. The tree has O(lg n) levels, times the work done on one level, yielding T(n) is Θ(nlogb a lg n).
Case 3: f(n) is Ω(nlogba + ε). In this case we also need to show that af(n/b)≤kf(n) for some constant k and large n, which means that f grows faster than the number of leaves. Asymptotically all of the work is done at the root node, so T(n) is Θ(f(n)).
Yeah Sure !!!
@Akriti, ma'am this might clear your concept.
T(n)=2T(n/2) + n2.
The recursion tree for this recurrence is of the following form:
and the children / leaves work O(n).
It is Case 2.
@jason,i am understanding all the things ,you are sayng but .just tell me this-"
"Means , the root works θ(nc) and the children especially the Leaves work Θ(nlogba)."
the recurrence is T(n) = a T(nb) + θ(nc) which means n is divided into n/b at each level.right?and at each level nc work is done..right?so how is that leaves do more or less work?overall work at all the levls will be same.right??
and pls dun call me ma'am..i am also a student like you..;):-P
MAY be i am getting u now.suppose T(n) =2T(N/2) + n2
n/2 n/2 (n/2)2 + (n/2)2
at,next level (n/4)2 + (n/4)2 + (n/4)2 + (n/4)2
hence,at leaves,work done is less.
am i right??
X->YZ , Y->XZ , ...