CMSC 341Skip Lists8/3/2007UMBC CMSC 341 SkipList2Looking Back at Sorted Lists Sorted Linked ListWhat is the worst case performance of find( ), insert( )? Sorted ArrayWhat is the worst case performance of find( ), insert( )?8/3/2007UMBC CMSC 341 SkipList3An Alternative Sorted Linked List What if you skip every other node? Every other node has a pointer to the next and the one after that Find : follow “skip” pointer until target < this.skip.element Resources Additional storage Performance of find( )?8/3/2007UMBC CMSC 341 SkipList4Skipping Every 2ndNodeThe value stored in each node is shown below the node and corresponds to the the position of the node in the list.It’s clear that find( ) does not need to examine every node. It can skip over every other node, then do a final examination at the end. The number of nodes examined is no more than n/2 + 1.For example the nodes examined finding the value 15 would be2, 4, 6, 8, 10, 12, 14, 16, 15 -- a total of 16/2 + 1 = 9.8/3/2007UMBC CMSC 341 SkipList5Skipping Every 2ndand 4thNodeThe find operation can now make bigger skips than the previous example. Every 4thnode is skipped until the search is confined between two nodes of size 3. At this point as many as three nodes may need to be scanned. It’s also possible that some nodes may be examined more than once. The number of nodes examined is no more than n / 4 + 3.Again, look at the nodes examined when searching for 15.8/3/2007UMBC CMSC 341 SkipList6New and Improved Alternative Add hierarchy of skip pointers every 2i-th node points 2inodes ahead For example, every 2ndnode has a reference 2 nodes ahead; every 8thnode has a reference 8 nodes ahead8/3/2007UMBC CMSC 341 SkipList7Skipping Every 2i-th nodeSuppose this list contained 32 nodes and we want to search for some value in it. Working down from the top, we first look at node 16 and have cut the search in half. When we look again one level down in either the right or left half, we have cut the search in half again. We continue in this manner until we find the node being sought (or not).This is just like binary search in an array. Intuitively we can understand why the max number of nodes examined is O(lg N).8/3/2007UMBC CMSC 341 SkipList8Some Serious Problems This structure looks pretty good, but what happens when we insert or remove a value from the list? Reorganizing the the list is O(N).  For example, suppose the first element of the list was removed. Since it’s necessary to maintain the strict pattern of node sizes, it’s easiest to move all the values toward the head and remove the end node. A similar situation occurs when a new node is added.8/3/2007UMBC CMSC 341 SkipList9Skip Lists Concept:A skip list maintains the same distribution of nodes, but without the requirement for the rigid pattern of node sizes 1/2 have 1 pointer 1/4 have 2 pointers 1/8 have 3 pointers … 1/2ihave i pointers It’s no longer necessary to maintain the rigid pattern by moving values around for insert and remove. This gives us a high probability of still having O(lg N) performance. The probability that a skip list will behave badly is very small.8/3/2007UMBC CMSC 341 SkipList10A Probabilistic Skip ListThe number of forward reference pointers a node has is its “size”.The distribution of node sizes is exactly the same as the previous figure, the nodes just occur in a different pattern.8/3/2007UMBC CMSC 341 SkipList11Inserting a Node When inserting a new node, we choose the size of the node probabilistically. Every skip list has an associated (and fixed) probability, p, that determines the distribution of node sizes. A fraction, p, of the nodes that have at least r forward references also have r + 1 forward references.8/3/2007UMBC CMSC 341 SkipList12Skip List Insert To insert node: Create new node with random size. For each pointer, i , connect to next node with at least i pointers.int generateNodeSize(double p, int maxSize){int size = 1;while (drand48() < p) size++;return (size >maxSize) ? maxSize : size;}8/3/2007UMBC CMSC 341 SkipList13An Aside on Node Distribution Given an infinitely long skip list with associated probability p, it can be shown that 1 – p nodes will have just one forward reference. This means that p(1 – p) nodes will have exactly two forward references and in general pk(1 – p) nodes will have k + 1 forward reference pointers. For example, with p = 0.50.5 (1/2 of the nodes will have exactly one forward reference)0.5 (1 – 0.5) = 0.25 (1/4 of the nodes will have 2 references)0.52(1 – 0.5) = 0.125 (1/8 of the nodes will have 3 references)0.53(1 – 0.5) = 0.0625 (1/16 of the nodes will have 4 references) Work out the distribution for p = 0.25 (1/4) for yourself.8/3/2007UMBC CMSC 341 SkipList14Determining the Size of the Header Node The size of the header node (the number of forward references it has) is the maximum size of any node in the skip list and is chosen when the empty skip list is constructed (i.e. it must be predetermined) Dr. Pugh has shown that the maximum size should be chosen as log 1/pN. For p = ½, the maximum size for a skip list with 65,536 elements should be no smaller than log 265536 = 16.8/3/2007UMBC CMSC 341 SkipList15Performance Considerations The expected time to find an element (and therefore to insert or remove) is O( lg N ). It is possible for the time to be substantially longer if the configuration of nodes is unfavorable for a particular operation. Since the node sizes are chosen randomly, it is possible to get a “bad” run of sizes. For example, it is possible that each node will be generated with the same size, producing the equivalent of an ordinary linked list. A “bad” run of sizes will be less important in a long skip list than in a short one. The probability of poor performance decreases rapidly as the number of nodes increases.8/3/2007UMBC CMSC 341 SkipList16More performance The probability that an operation takes longer than expected is function of the associated probability p. Dr. Pugh calculated that with p = 0.5 and 4096 elements, the probability that the actual time will exceed the expected time by more than a factor of 3 is less than one in 200 million. The relative time and space performance depends on p. Dr. Pugh suggests p = 0.25 for most cases. If the predictability ofperformance is important, then he suggests using p =