15-750 Graduate Algorithms January 25, 2001 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.