


















Note:


Example:
r(T) = lg(15) + lg(11) + 2lg(5) + 3lg(3)
Notes:
Before splay(3): r(T) = lg(15) + lg(11) + 2lg(5) + 3lg(3) After splay(3): r'(T) = lg(15) + lg(11) + lg(9) + lg(5) + 3lg(3) Change in balance: Δ(r(T)) = r'(T)  r(T) = lg(9)  lg(5) 



(This is trivial because we added in the difference so the total will never become negative...)

To answer this question, we must look at the amortized cost of a zigzig, a zigzag, a zig operation
Because: splay(x) = a sequence of these operations


The proof for the zigzag and zig step are similar to the zigzig step (see Goodrich for details)

δ = r( new tree )  r( old tree ) = r'(a) + r'(b) + .... + r'(x) + r'(y) + r'(z)  r(a) + r(b) + .... + r(x) + r(y) + r(z) = (r'(x) + r'(y) + r'(z))  (r(x) + r(y) + r(z)) 
Because r(v) = lg(n(v)), we have that:
Therefore:
δ = (r'(x) + r'(y) + r'(z))  (r(x) + r(y) + r(z)) ^^^^ ^^^^ = (r'(y) + r'(z))  (r(x) + r(y)) ^^^^ < (r'(x) + r'(z))  (r(x) + r(y)) = (r'(x) + r'(z))  r(x)  r(y) ^^^^ < (r'(x) + r'(z))  r(x)  r(x) < (r'(x) + r'(z))  2r(x) ........(2) 


Proof:

The amortized cost to splay a node x at depth d:


From Porposition 10.4: r'(T)  r(T) ≤ 3r(root)  3r(x)  d + 2 <==> r'(T)  r(T) + d ≤ 3r(root)  3r(x) + 2 <==> d + Δ(T) ≤ 3r(root)  3r(x) + 2 ==> d + Δ(T) ≤ 3r(root) + 2 = 3*log(n(root)) + 2 = 3*log(2n+1) + 2 ==> d + Δ(T) = O(log(n)) 

But the change is very small (1 node compared to n nodes).

I'll omit it; the most difficult part of the proof have been discussed...