八卦树的摊销复杂度是多少？

Question

I tried to understand the amortized complexity and did several searches on the net, but I couldn't find a good resource yet.

So can any one explain what amortized complexity is and how it becomes O(lg n) in splay tree per operation?

Answer 1

Any operation performed on splay trees whether insert delete..etc the cost is dominated by the splay operation. hence accounting for the cost of only the splay operation which is the rotations performed on the node to be splayed.

The amortized function is given by a=c+3Rfinal(v)-3Rinitial(v)

where a is the amortized cost, c is the actual cost and Rfinal is the rank after the splay operation and Rinitial is the rank of the node before rotation.(rank of any node is given by the height of its subtree ie log|S| where S is the number of nodes rooted under it)

Now consider the worst case where node to be splayed, is the leaf hence its initial rank is given by 0. After splaying it to the top ie as the root node its rank becomes log n where n is the total number of nodes in the tree.

 a<= 2+3logn-0
 O(logn).