1. 15-211 Fundamental Data Structures and Algorithms Jeffrey Stylos February 8 th, 2005 Splay Trees Adapted from slides by Peter Lee October 4, 2001

2. Today  Review HW2  Review of binary trees  Splay trees

3. Announcements  HW2 is out!  Due next Thursday at midnight!  Get started now!  Quiz 2 next week, during recitation

4. Objects in calendar are closer than they appear. — Jim Duncan

5. And now, a few words about HW2  [Homework 2 write-up]

6. Let’s talk about BlackBoard  What you always wanted to know about the discussion board (but were afraid to ask)

7. Last Time…

8. Binary trees A binary tree is either empty (we'll write nil for clarity), or looks like (x,L,R) where x is an object, and L, R are binary subtrees

9. Binary search trees (BSTs) A binary tree T is a binary search tree iff flat(T) is an ordered sequence. Equivalently, in (x,L,R) all the nodes in L are less than x, and all the nodes in R are larger than x.

10. flat(T) = 2,3,4,5,6,7,9 Example 5 3 6 7 249

11. search(x,nil) = false search(x,(x,L,R)) = true search(x,(a,L,R)) = search(x,L) x<a search(x,(a,L,R)) = search(x,R) x>a Binary search How does one search in a BST? Inductively:

12. Log-time vs linear time  log k (N) = O(N) for any constant k.  I.e, logarithms grow very slowly.

13. flat(T) = 2,3,4,5,6,7,9 A bad tree 234567 9

14. Balanced trees  Intuitively, one way to ensure good behavior is to make “balanced” search trees.  If always balanced, then operations such as search and insert will require only O(log(N)) comparisons.

15. flat(T) = 2,3,4,5,6,7,9 Example 5 3 6 7 249

16. Forcing good behavior It is clear (?) that for any n inputs, there always is a BST containing these elements of logarithmic depth. But if we just insert the standard way, we may build a very unbalanced, deep tree. Can we somehow force the tree to remain shallow? At low cost?

17. Today  Review HW2  Review of binary trees  Splay trees

18. Splay Trees

19. Binary search trees  Simple binary search trees can have bad behavior for some insertion sequences.  Average case O(log N), worst case O(N).  AVL trees maintain a balance invariant to prevent this bad behavior.  Accomplished via rotations during insert.  Splay trees achieve amortized running time of O(log N).  Accomplished via rotations during find.

20. Never do today…  If you are a reasonably lazy person…  Consider the problem of serving a six-course dinner.  You could spend 10 minutes washing the dishes right after each course.  Or you could let the dishes sit and spend an hour tomorrow (or the next day) washing them.  Either way, you will spend one hour doing dishes.

21. A better example  If you are a reasonably smart person…  Consider the problem of getting groceries.  You could spend 30 minutes going to the Giant Eagle every time you run out of something.  Or you could wait until you have run out of 10 items, and then make a single 2- hour trip to the Giant Eagle.  In this case, you would save 3 hours!

22. Amortized running time  The analysis that allows us to conclude that we spend the same (or less) amount of time over a sequence of operations is called amortized analysis.  If we say that the amortized running time of a sequence of operations is O(f(N)):  Some operations might be more than O(f(N)), other less.  But the average over the entire sequence is O(f(N)).

23. Splay trees  Splay trees provide a guarantee that any sequence of M operations (starting from an empty tree) will require O(Mlog N) time.  Hence, each operation has amortized cost of O(log N).  It is possible that a single operation requires O(N) time.  But there are no bad sequences of operations on a splay tree.

24. A basic observation  Let’s suppose that a node requires O(N) time to find.  If and when this happens, we must move it somewhere closer to the root so that if we access it again, it will definitely require less than O(N) time.  If we don’t do this, then O(log N) amortized behavior is not possible.

25. Danny Sleator  The inventor of splay trees.  Winner of Kannelakis award.  See his splay tree demo at http://www.cs.cmu.edu /~sleator http://www.cs.cmu.edu /~sleator  (And definitely a procrastinator.)

26. The basic idea  Every time a node is accessed, move it to the root (“splay” it)  The move can be accomplished by performing “AVL” rotations.  Practical benefits:  In practice, nodes are often accessed multiple times.  AVL rotations make trees more balanced.

27. Using rotations  Suppose we perform a find operation, which accesses node n.  Splay n, moving it up to the root.  Doing this requires O(d) time, where d is the depth of n. Hence O(N) (worst case)  But a subsequent access of n will require only O(1) time, and the tree will be better balanced, too.

28. Splaying, case 1  There are four cases to consider.  Case 1: The node is already the root.  Nothing to do.

29. Splaying, case 2  Case 2: Accessed node’s parent is the root.  Perform a single rotation. Z Y X ZY X

30. Splaying, cases 3 and 4  The next two cases cover the situation in which the accessed node has a grandparent.

31. Splaying, case 3  Case 3: Zig-zag (left).  Perform an AVL double rotation. a Z b X Y1Y1 Y2Y2 a Z b XY1Y1 Y2Y2

32. Splaying, case 4  Case 4: Zig-zig (left).  Special rotation. a Z b Y W X a Z b Y W X

33. Symmetry  And there are symmetric cases for zig-zag and zig-zig to the right.

34. Splay tree example 2 5 3 4 6 1 0 Insert {0, 1, 2, 3, 4, 5, 6}, then find 6 2 5 3 6 4 1 0 zig-zig right

35. Splay tree example, cont’d 2 5 3 6 4 1 0 6 5 3 2 4 1 0 zig-zag right

36. Splaying example, cont’d zig-zag right 6 5 3 2 4 1 0 0 5 3 2 4 1 6

37. Result of splaying  Access of 6 required N nodes visited and modified.  But now accessing 5 requires only N/2 nodes visited and modified.  Will also bring all nodes up to N/4 of root. (Try it!)  And all nodes are shallower.  Every access will tend to improve the tree for future operations. 0 5 3 2 4 1 6

38. Using splay operations  …to implement “find(x)”:  Splay x  Look at root (compare to x)  …to implement “insert(x)”:  Splay x  Look at root If it exists, do nothing Else, make root the child of x, call x the new root

39. Using splay operations (cont.)  …to implement “delete(x)”:  Splay x  Look at root (compare to x) If it doesn’t exist, do nothing  Delete root  If either child of root is null make the other child new root  Splay largest key of left-branch (we could also splay smallest of right) Can do this by splaying the non-existing element, because of the way splay works Everything in right-branch tree will be smaller than everything in left-branch tree  New left-branch tree will have empty right-branch (since we splayed the largest value) Set left-branch tree's right-branch to be the right-branch tree

40. Using splay operations (cont.)  …to implement “delete(x)”:  Splay x  Look at root (compare to x) If it doesn’t exist, do nothing  Delete root  Splay largest b f a c x de

41. Nice properties of splay trees  If you have a small working set, the operations’ costs are proportional to the size of the working set  Why?  If you splay the elements in order, the total cost is O(N)  Why?

42. Analysis of splay trees  The analysis of the running time of splay trees is quite difficult.  Any single find or insert might take O(N) time.  But any sequence of M operations, starting from an empty tree, will take only O(Mlog N) time.  In practice, splay trees work extremely well.

43. Splaying summary  Splaying has the effect of moving the accessed node to the root.  It also reduces the depth of almost all of the nodes along the access path.

44. All done for today!