15-750 Graduate Algorithms Feb 1, 2002
Daniel Sleator
* splay trees
----------------------------------------------------------------------------
-- definition of splay trees
-- analysis of splay trees
Splay Trees (self-adjusting search trees)
-----------------------------------------
These notes just describe the bottom-up splaying algorithm, the proof of
the access lemma, and a few applications.
Every time a node is accessed in a splay tree, it is moved to the root
of the tree. The amortized cost of the operation is O(log n). Just
moving the element to the root by rotating it up the tree does not have
this property. Splay trees do movement is done in a very special way
that guarantees this amortized bound.
I'll describe the algorithm by giving three rewrite rules in the form of
pictures. In these pictures, x is the node that was accessed (that will
eventually be at the root of the tree). By looking at the local
structure of the tree defined by x, x's parent, and x's grandparent we
decide which of the following three rules to follow. We continue to
apply the rules until x is at the root of the tree:
y x
Zig (terminal case): / ====> \
x y
z z
/ / x
Zig-zag: y ====> x ====> / \
\ / y z
x y
z x
/ y \
Zig-zig: y ====> / \ ====> y
/ x z \
x z
Notes (1) Each rule has a mirror image variant, which covers all the cases.
(2) The zig-zig rule is the one that distinguishes splaying from
just rotating x to the root of the tree.
(3) Top-down splaying is much more efficient in practice. Code for
doing this is on my web site (www.cs.cmu.edu/~sleator).
Here are some examples:
6 0 3
/ \ / \
5 5 / \
/ / \ / \
4 splay(0) 3 6 splay(3) 0 5
/ =======> / \ =======> \ / \
3 1 4 1 4 6
/ \ \
2 2 2
/
1
/
0
To analyze the performance of splaying, we start by assuming that each
node x has a weight w(x) > 0. These weights can be chosen arbitrarily.
For each assignment of weights we will be able to derive a bound on the
cost of a sequence of accesses. We can choose the assignment that gives
the best bound. By giving the frequently accessed elements a high
weight, we will be able to get tighter bounds on the running time.
Note that the weights are only used in the analysis, and do not change
the algorithm at all.
(A commonly used case is to assign all the weights to be 1.)
Before we can state our performance lemma, we need to define two more
quantities. The size of a node x (denoted s(x)) is the total weight of
all the nodes in the subtree rooted at x. The rank of a node x (denoted
r(x)) is the floor(log_2) of the size of x. Restating these:
s(x) = Sum (over y in the subtree rooted at x) of w(y)
r(x) = floor(log(s(x)))
For each node x, we'll keep r(x) tokens on that node. (Alternatively,
the potential function will just be the sums of the ranks of all the
nodes in the tree.)
Here's an example to illustrate this: Here's a tree, labeled with sizes
on the left and ranks on the right.
9 o 3 o
/ /
8 o 3 o
/ \ / \
6 o o 1 2 o o 0
/ \ / \
2 o o 3 1 o o 1
/ / / /
1 o o 2 0 o o 1
\ \
o 1 o 0
Notes about this potential:
(1) Doing a rotation between a pair of nodes x and y only effects
the ranks of the nodes x and y, and no other nodes in the tree.
Furthermore, if y was the root before the rotation, then the
rank of y before the rotation equals the rank of x after the
rotation.
(2) Assuming all the weights are 1, the potential of a balanced tree
is O(n), and the potential of a long chain (most unbalanced
tree) is O(n log n).
(3) In the banker's view of amortized analysis, we can
think of having r(x) tokens on node x.
Access lemma: The number of splaying steps done when splaying
node x in a tree with root t is at most 3(r(t)-r(x))+1.
Proof:
As we do the work, we must pay one token for each splay step we do.
Furthermore we must make sure that there are always r(x) tokens on node
x as we do our restructuring. We are going to allocate 3(r(t) - r(x)) +1
tokens to do the splay. Our job in the proof is to show that this is
enough.
First we need the following observation about ranks, called the Rank
Rule. Suppose that two siblings have the same rank, r. Then the parent
has rank at least r+1. This is because if the rank is r, then the size
is at least 2^r. If both siblings have size at least 2^r, then the
total size is at least 2^(r+1) and we conclude that the rank is at least
r+1. We can represent this with the following diagram:
x
/ \ Then x >= r+1
r r
Conversly, suppose we find a situation where a node and its parent have
the same rank, r. Then the other sibling of the node must have rank < r.
So if we have three nodes configured as follows, with these ranks:
r
/ \ Then x < r
x r
Now we can go back to proving the lemma. The approach we take is to
show that the 3(r'(x) - r(x)) tokens are sufficient to pay for the a
zig-zag or a zig-zig steps. And that 3(r'(x) - r(x)) +1 is sufficient
to pay for the zig step. (Here r'() represents the rank function after
the step, and r() represents the rank function before the step.)
When we sum these costs to compute the amortized cost for the entire
splay operation, they telescope to:
3(r(t) - r(x)) +1.
Note that the +1 comes from the zig step, which can happen only once.
Here's an example, the labels are ranks:
2 o 2 x
/ / \
2 o 0 o o 2
/ /
2 o splay(x) 2 o
/ ==========> / \
2 x 1 o o 0
/ \ /
0 o o 1 0 o
/
o 0
----------------- ----------------
Total: 9 7
We allocated: 3(2-2)+1 = 1
extra tokens from restructuring: 9-7 = 2
------
3
There are 2 splay steps. So 3 > 2, and we have enough.
It remains to show these bounds for the individual steps. There are
three cases, one for each of the types of splay steps.
Zig:
r o o r
/ ==> \
a o o b <= r
The actual cost is 1, so we spend one token on this.
We take the tokens on a and augment them with another
r-a and put them on b. Thus the total number of tokens
needed is 1+r-a. This is at most 1+3(r-a).
Zig-zag: We'll split it into 2 cases:
Case 1: The rank does not increase between the starting
node and ending node of the step.
r o o r
/ / \
r o ===> a o o b
\
r o
By the Rank Rule, one of a or b must be < r, so
one token is released from the data structure.
we use this to pay for the work.
Thus, our allocation of 3(r-r) = 0 is sufficient
to pay for this.
Case2: (The rank does increase)
r o o r
/ / \
b o ===> c o o d
\
a o
The tokens on c can be supplied from those on b.
(There are enough there cause b >= c.)
Note that r-a > 0. So:
use r-a (which is at least 1) to pay for the work
use r-a to augment the tokens on a to make enough for d
Summing these two gives: 2(r-a). But 2(r-a) <= 3(r-a), which
completes the zig-zag case.
Zig-zig: Again, we split into two cases.
Case 1: The rank does not increase between the starting node and the
ending node of the step.
r o r o o r
/ / \ \
r o =====> r o o d ====> o c <= r
/ \
r o (d o c <= r
/ \
a o o d <= r
use r-a (which is at least 1) to pay for the work
use r-a to boost the tokens on a to cover those needed for d
use r-a to boost the tokens on b to cover those needed for c
Summing these gives 3(r-a), which completes the analyzis
of the zig-zig case.
This completes the proof of the access lemma. QED
Balance Theorem: A sequence of m splays in a tree of n nodes takes time
O(m log(n) + n log(n)).
Proof: We apply the access lemma with all the weights equal to 1. For
a given splay, r(t) <= log(n), and r(x) >= 0. So the amortized cost of
the splay is at most:
3 log(n) + 1
We now switching to the world of potentials (potential = total tokens in
the tree). To bound the cost of the sequence we add this amount for
each splay, then add the initial minus the final potential. The initial
potential is at most n log(n), and the final potential is at least 0.
This gives a bound of:
m (3 log(n) + 1) + n log(n) = O(m log(n) + n log(n))
Q.E.D.
In the nect lecture, we'll prove more interesting results by using
non-uniform weights.
15-750 Graduate Algorithms
Daniel Sleator
* splay trees continued
--------------------------------------------------------------------------
In this lecture we'll see how the access lemma can be applied to prove
more advanced properties of splay trees. We take advantage of the fact
that we can assign arbitrary positive values to the weights of the
nodes.
In the last lecture, we proved the following:
Access lemma: The number of splaying steps done when splaying
node x in a tree with root t is at most 3(r(t)-r(x))+1.
Actually, something a bit stronger is true. By defining rank to be the
log of the weight of a node (instead of the floor of the log), we can
prove a stronger lemma:
Strong Access Lemma: The number of rotations done when splaying
node x in a tree with root t is at most 3(r(t)-r(x))+1.
Here r(t) and r(x) denote the continuous definition of rank. The bound
in this lemma is approximately a factor of two tighter than the previous
one. Throughout these notes, we'll use the strong access lemma and the
continuous definition of rank. The main advantage here is simply that
it means we can stop writing floors, because in these notes we're only
trying to prove big-oh bounds.
Let's rewrite the access lemma a bit. Let w(x) be the weight of a node
x. r(x) is log(w(x) + total weight of other descendants of x). Since
the weight of the descendants of x is a positive number we have:
r(x) >= log(w(x))
So the amortized cost of a splay is bounded by 3(r(t) - log(w(x))) + 1.
The theorems below concern bounds on the running times of sequencess of
accesses in splay trees. (The set of nodes in these trees is static.)
We'll use the rotation bound of the strong access lemma as our measure of
the running time of a sequence of operations.
Recall the following consequence of the definition of amortized cost:
total real cost = total amortized cost + initial potential - final potential
To bound the cost of a sequence, we'll bound the total amortized cost
(using the access lemma) and bound the total decrease in potential that
could result from the sequence.
NOTATION: There will be n items in the tree, numbered 1...n.
The letter i will indicate an index into this list of items.
We'll be considering a sequence of m accesses, and the index j
will refer to one of these acceses. Let i[j] refer to the index
of the jth access.
BALANCE THEOREM: The total access time is O(m + (n+m) log(n))
PROOF: Assign a weight of 1/n to each item. Then the total weight is 1
and the rank of the root is 0. So the access lemma bounds the amortized
cost of an access by 3log(n)+1. Summing this over all accesses gives
O(m + m log(n)). For any node i:
0 >= r(i) >= log(1/n)
Write this as:
0 <= -r(i) <= log(n)
Thus, the net decrease in the potential is at most log(n) per node, and
totals to n log(n). QED.
STATIC OPTIMALITY THEOREM: Let q(i) be the number of times item i is
accessed. Let m = Sum q(i) be the number of accesses. Assume that
q(i) > 0 for each i. Then the total cost of the sequence is:
m
O(m + Sum log ------- )
j q(i[j])
PROOF: Choose w(i) (the weight of item i) to be q(i)/m. The total
weight of all items is 1. So the amortized cost of an access to element
i is at most:
3(r(root) - log(w(i))) + 1 = 3 (- log(q(i)/m)) +1 = 3 log(m/q(i))+1
Summing this, we see that the amortized cost satisfies the bound stated
in the theorem. It remains to bound the decrease in potential. Each
item contributes its rank to the potential. We have the following
bounds on ranks:
0 >= r(i) = log(size(i)) >= log(w(i))
Rewriting this:
0 <= -r(i) <= log(1/w(i))
The the maximum amount that a node's potential can decrease is just
log(1/w(i)), which is log(m/q(i)). Summing this we see:
total potential decrease <= Sum log(m/q(i))
i
Since each element is accessed at least once, the terms in this
summation are a subset of the terms in the bound we're tryig to prove in
the theorem. So this additional cost can be absorbed into the big-oh.
QED.
NOTE 1: The reason we need to assume that each element is accessed at
least once is to guard against the possibility that we start out with a
very bad tree which contains millions of elements near the top of the
tree that are never accessed, and we are forced to incur a cost to get
by them.
NOTE 2: This is called the static optimatilty theorem because the static
tree that is constructed to be optimal for this specific access sequence
will incur a cost of O(log(m/q(i))) to access the item i.
STATIC FINGER THEOREM: Let f be a specific element called the finger.
Then the following is a bound on the cost of splaying a sequence:
O(m + n log(n) + Sum log (|f - i[j]| + 1))
j
NOTE: |f-i| is the distance in the symmetric ordering of the items
between the finger and item i.
PROOF: Assign the weights w(i) as follows:
1
w(i) = ----------------
2
(|f-i| + 1)
Since Sum 1/n^2 (n from 1 to infinity) is a constant (pi^2/6) we see
that the sum of all the weights is bouned by a constant (at most twice
this). Let's look at the amortized cost of an access. Recall:
3(r(t) - log(w(x))) + 1.
The term r(t), meaning the rank of the root, is just bounded by a
constant, and sums to O(m). The 1 term also sums to O(m). The
-log(w(x)) term can be written:
1
-log ( ---------------- )
2
(|f-i| + 1)
= 2 log (|f-i|+1)
Combining these estimates shows that the amortized cost is bounded by
the summation stated in the theorem. It remains to bound the decrease
in potential. What bound can we put on the rank of an item i?
2 pi^2 1
log( ------ ) >= r(i) >= 2 log (-------)
6 |f-i|+1
Negating this we get:
2 pi^2
- log( ------ ) <= -r(i) <= 2 log (|f-i|+1)
6
So the potential change of a node is bounded by O(log n).
This contributes O(n log n) to the total real cost. QED.
Some other theorems that we're not going to prove:
WORKING SET THEOREM: Let t(j) be the number of accesess of different
items that occured between access j and the previous access of the same
item. The the cost of the sequence is bounded by:
O(m + n log(n) + Sum log (t(j) + 1))
j
NOTE: this is proven by using a weight assignment very similar to the
one used in the proof of the static finger theorem. Imagine that we're
maintaining a move-to-front list (the accessed element moves to the
front). When accessing an item the number of elements it has to pass
when it moves to the front is t(j). If an item is in position p,
(positions numbered 1, 2, 3...) in the list, then it has a weight of
1/p^2.
The following theorems do not follow from the access lemma.
SEQUENTIAL ACCESS THEOREM: The cost of the access sequence that accesses
each of the n items in the treey in symmetric (left-to-right) order is
O(n).
This is generalized by the following theorem, the proof of which is
very difficult:
DYNAMIC FINGER THEOREM: Using the notation of the static finger theorem,
a bound on the splaying cost is:
O(m + n + Sum log (|i[j-1] - i[j]| + 1))
j
The two theorems above are special cases of the following conjecture:
DYNAMIC OPTIMAILTY CONJECTURE: Consider the following class of
binary-tree based algorithms for processing a sequence of requests:
on each access we pay the distance from the root to the accessed node.
Arbitrary rotations are also allowed, at a cost of 1 per rotation.
For a sequence Sigma, let C(T,Sigma) be the cost of the optimal algorithm
in this class for processing a sequence Sigma starting from tree T.
Then the cost of splaying the sequence (starting from any tree) is
O(C(T,Sigma) + n).
What about dynamic operations that change the set of elements in the
tree? Here's how we can do Insertion, Join, and Deletion.
Insertion: Say we want to insert a new item e. Proceed like ordinary
binary tree insertion, but stop short of linking the new item e into the
tree. Let the node you were about to link e to be called p. We splay p
to the root. Now construct a tree according to one of the following
pictures:
e e
/ \ / \
p p
/ \ / \
depending on if the item in e is before or after p.
Let's analyze this when all the weights are 1. The amortized cost of
the splay is O(log n) according to the access lemma. Then we have to
pay the additional amortized cost which equals the potential of the new
root. This is log(n+1). So the amortized cost of the whole thing is
O(log(n)).
An alternative insertion algorithm is to do ordinary binary tree
insertion, and then simply splay the item to the root. This also has
efficient amortized cost, but the analysis is a tiny bit more
complicated.
Join: Here we have two trees, say, A and B. We want to put them
together with all the items in A to the left of all the items in B.
This is caled the "Join" operation. This is done as follows. Splay the
rightmost node of A. Now make B the right child of this node. The
amortized cost of the operation is easily seen to be O(log n) where n is
the number of nodes in the tree after joining. (Just assign uniform
weights to all the nodes, as we did in the Insertion analysis.)
Deletion: To delete a node, x, simply splay it to the root and then Join
its left and right subtrees together as described above. The amortized
cost is easily seen to be O(log n), where n is the size before the
deletion.