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\begin{center} Lecture 2: Splay Trees \end{center}


Binary Search Trees

Splay Trees

Sleator and Tarjan, “Self Adjusting Binary Search Trees” JACM 32(3) 1985

Key design principle: Be lazy! That is, don't work until you must.

Path shortening:

Re-balancing (via Rotations):

Key operation: Splaying a node $x$:

Figures of rotations

Implementing $Search(x)$ operation:

(Will talk about $Insert(x)$ and $Delete(x)$ later.)


Our potential function: $\Phi=\sum r(x)$.

Some intuition:

The Access Lemma

Key technical fact: \begin{center} Access Lemma: Amortized time to splay a node $x$ given root $t$ is at most $$3(r(t)-r(x))+1 = O(\log(s(t)/s(x))).$$ (Recall: Amortized cost of an operation $=$ real cost $+ \ \Delta \Phi$. ) \end{center}

Note: $x$ is the root of the tree after the splay operation and the rank of the root is always the same $\log W$, where $W=\sum_x w_x$ is the total weight of the nodes.

$\Rightarrow$ So, rank $r'(x)$ after the splay is equal to $r(t)$.

$\Rightarrow$ $r(t)-r(x)=r'(x)-r(x)$ -- the amortized cost of a splay is within a constant of its change in rank.


One immediate consequence of Access Lemma:

Important detail: To make the above analysis work also for unsuccessful Search(x) operations, we need to splay the last node we visited in that search.

Proof of the Access Lemma

       r                r' 

z x / \ / \ y D A y / \ ====> / \ x C B z / \ / \ A B C D



Actual Math (requires more care):

Bounding Overall Change of Potential

Important Caveat: The Access Lemma bounds only the amortized cost of each operation.

$\Rightarrow$ If there is a sequence of $m$ operations, its total real cost is at most \[ O(m \log n) - \left(\Phi_m - \Phi_0\right)=O(m \log n) + \Phi_0 - \Phi_m, \] where $\Phi_m$ is the potential of our data structure at the end of processing the whole sequence and $\Phi_0$ is its initial potential.

Key question: How large can $\Phi_0-\Phi_m$ be?


Also: Note that potential of empty tree is $0$.

$\Rightarrow$ If we start with empty tree, no problem (but for that we need inserts...)

Applications of Access Lemma

By applying the above analysis with different weights we can derive a lot of interesting facts about performance of splay trees. (Key fact to keep in mind: the behavior of the splay tree is completely independent of our choice of the weights!)

Update Operations

We have two remaining operations we need to implement: $Insert(x)$ and $Delete(x)$. There are two ways to implement them:

  1. Insert/delete the item as in a standard BST and then just splay the inserted element/parent of deleted element.
    • Splaying ensures that our work is “paid for” by the resulting potential change.
    • But showing that this is the case requires a little bit of calculations.
  2. Alternatively, consider the following operations:
    • $Split(x)$ splits the BST into two BST, one containing all the elements smaller or equal to $x$ and one containing all the elements larger than $x$.

      $\Rightarrow$ Can be achieved by simply splaying $x$ and thus takes amortized $O(\log n)$ time (corresponds to taking all weights be equal to $1$ in the analysis).

    • $Join$ takes two BSTs $S$ and $T$ such that all elements in $S$ are not larger than all elements in $T$ and creates a single BST contains elements of both $S$ and $T$.

      $\Rightarrow$ Can be achieved by splaying first the largest element $x$ of $S$ (note that then this element becomes a root with no right child) and adding the root of $T$ as the right child of $x$.

      $\Rightarrow$ Joining increases only the potential of the new root, and only by at most $O(\log n)$.

      $\Rightarrow$ Thus, again, the whole operation takes $O(\log n)$ amortized time (if we take all $w_x$ equal to $1$ in the analysis).

    Note: For most balanced trees such $Split(x)/Join$ operations do not take only $O(\log n)$ time!
  3. To implement $Insert(x)$:
    • Perform $Split(x')$, where $x'$ is the largest element not larger than $x$. (Finding $x'$ can be easily amortized into the subsequent splay.)
    • Create a new BSTs with $x$ as a root and the two BSTs obtained from splitting be its subtrees.

      $\Rightarrow$ Total potential increased by $O(\log n)$, i.e., the increase of potential of the root. (Remember: we are performing analysis with all weights being $1$ here.)

      $\Rightarrow$ Total amortized cost is $O(\log n)$.

  4. To implement $Delete(x)$:
    • Perform $Split(x)$

      $\Rightarrow$ $x$ will be the root of one of the resulting BSTs and will have no right child.

    • Remove $x$ and perform $Join$ on the resulting two BSTs.
    • $O(\log n)$ amortized cost, as desired.

Possible Heuristics