Why is amortized analysis of splay tree only focusing on the splay operation and not accounting for the downwards search

Question

Each dictionary operation in a splay tree uses a splay operation to bring a node to the root of the tree. The amortized efficiency of this splay operation is typically analyzed with the potential method and described in many sources online (including the wikipedia) page. The amortized time of this splay operation is then reported as O(m lg n).

However, I nowhere find an actual analysis of complete dictionary operations, such as insert, delete, ... Each of these operations uses, besides a splay operation, also a downward search through the tree to find the correct position of the node to insert or delete. Only after you have found that node, you can start the splay operation.

People tend to make statements like:

• "the complexity of splay tree operation is the same as that of the associated splay"
• ["For our analysis, we note that the time for performing a search, insertion, or deletion is proportional to the time for the associated splaying"], p. 456 of the book of Goodrich titled "Data structures and algorithms in C++"

I have two questions:

• How is one able to make this conclusion that the time to perform a search is proportional to the time for the splay? This kind of implies that the time for the downwards traversal to the node is also proportional to the tie of the splay?
• What is the amortized time efficiency of a downwards traversal? Is it a constant, simply because you don't change the structure of the tree by simply doing a downwards traversal (so your potential stays the same)? And isn't this constant than = N, since this is the worst case?

How is one able to make this conclusion?

Answer 1

How is one able to make this conclusion that the time to perform a search is proportional to the time for the splay? This kind of implies that the time for the downwards traversal to the node is also proportional to the tie of the splay?

The splay phase operates on each of the nodes traversed during the search phase. Since the work done at each node during the search phase is constant, we infer that over any sequence of operations, search = O(splay), hence O(search + splay) = O(splay).

What is the amortized time efficiency of a downwards traversal? Is it a constant, simply because you don't change the structure of the tree by simply doing a downwards traversal (so your potential stays the same)? And isn't this constant than = N, since this is the worst case?

Yes, if it were possible to search without splaying afterward. For the reason previously discussed, we treat them as inseparable, so effectively we multiply by a constant the amortization credits used by a lengthy splay to cover the search too.