CS 206 Introduction to Computer Science II 04 / 22 / 2009 Instructor: Michael Eckmann

Michael Eckmann - Skidmore College - CS Spring 2009 Today’s Topics Questions? Comments? B-Trees

We would rather do many many calculations/read/writes to RAM in order to save us from having to access the disk. The worst case height for an AVL tree is 1.44 log n which is close to optimal (least) height for a set of nodes, but we'll see that for large amounts of data that reside on disk, it will not perform as well as we'd like. The worst case height of an optimal BST is log n. Log n is the best we can do with binary trees and that's not good enough, because we want a small constant number of disk accesses. So, instead of using a Binary tree, we'll use an M-ary tree (where M>2.)‏ The least height of an M-ary tree is log M n. Example on the board for M=5 B-Trees are discussed in section 19.8 in your text. Leading up to B-Trees

B-Trees were created to take that discrepancy in processing time into account (when we have large volumes of data to process.)‏ A B-tree can guarantee only a few disk accesses. An M-ary B-Tree has the following properties –data items stored in leaves only –interior (non-leaf) nodes store a maximum of M-1 keys –root is a leaf or has from 2 to M children –all non-leaf nodes have between the ceil(M/2) up to M children –all leaves are at same depth –all leaves have between the ceil(L/2) up to L data items L means... (see next few slides)‏ all non leaf nodes have at least half of M children, so for large M this will guarantee that the M-ary tree will not approach anything close to a binary tree (why is that a good thing?)‏ B-Trees

Each node represents a disk block (the amount of data read in one disk access). –example: if a disk block is 8k each node holds M-1 keys and M branches (links)‏ –let's say our keys are of size 32 bytes each –and a link is 4 bytes How would we determine M? –the largest value such that a node doesn't hold more than 8k And we determine L by the number of data items (records) we can store in one block. –if our records are 256 bytes, how many could we store in one block of 8k? B-Trees

Each node represents a disk block (the amount of data read in one disk access). –example: if a disk block is 8k (=8192 bytes)‏ each node holds M-1 keys and M branches (links)‏ –let's say our keys are of size 32 bytes each –and a link is 4 bytes How would we determine M? –the largest value such that a node doesn't hold more than 8k 32*(M-1) + 4*M = 8192 M = 228 (is the largest M such that we don't go over 8192)‏ And we determine L by the number of data items (records) we can store in one block. –if our records are 256 bytes, how many could we store in one block of 8k? 8192/256 = 32 From the rules of our B-Tree, each leaf then has to have between 16 and 32 records and each non-leaf node (except the root) has to have at least 114 children (up to 228 children). B-Trees

Example: –From the rules of our B-Tree, each leaf then has to have between 16 and 32 records and each non-leaf node (except the root) has to have at least 114 children (up to 228 children). –Picking the worst case B-tree for our example, that is the one with the least number of children per node to give us our highest possible B-Tree for 10,000,000 records, we'll have at most 625,000 leaves which means the height of our B-tree is 4. The worst height of an M-ary B-tree is approx. log M/2 n. Why? B-Trees

Each disk read will get a block which is a whole node. When we read an interior node's data from disk we get keys and links. When we read a leaf node's data from disk we get up to L data records. Insert examples –normal insert –may need to split a leaf into two option of putting a child up for adoption to a neighbor –may need to split parents may need to split the root (root will then have 2 children)‏ Splits are infrequent (For every split there will be L/2 nonsplits, on average)‏ Heightening of the tree is even more infrequent ( the only way the tree gets higher is when insertion leads to splitting the root.)‏ –notice: for a tree with four levels, the root was only split 3 times during all those inserts. And for M & L as we set in the example it occurred 3 times in 10,000,000 inserts. The crux is, splits and heightening lead to more processing when they happen but they don't happen often. B-Trees

Let's see some insertions into an existing B-Tree –a 5-ary B-tree, M=5, (with L=5 too) will be shown on the board. First verify visually that it is indeed a 5-ary B-tree with L=5. Insert 57 and rearrange the leaf Insert 55. The leaf is full (it has L=5 items already). With the 55 it would have L+1. (L+1) / 2 is > L/2 so we can split it into 2 leaves. B-Trees

What would be the first step in deletion of a node? Do we have to check anything after the deletion of a node? B-Trees

Delete examples: –normal delete –may need to adopt if leaf goes below minimum ok if the neighbor is not already at minimum or if neighbor has minimum too, then join the two leaves together to get one full leaf –parent loses a child –if parent is now below min, then continue up if root ever loses a child and would cause only 1 remaining child, reduce the height of the tree by one and make that child the root B-Trees

Anyone remember MergeSort? What kind of algorithm was that? Quicksort

Anyone remember MergeSort? What kind of algorithm was that? –Divide and Conquer It divided the list in half and did MergeSort on each half then combined the 2 halves --- how did it do this? The divide part of MergeSort is simple. The conquer part of MergeSort is more time consuming. Quicksort

Quicksort is a Divide and Conquer sort algorithm as well. But instead of dividing the list into same size halves we divide it in some other way into two sublists. The DIVIDE part of Quicksort is more complex than MergeSort but the CONQUER part of Quicksort is much simpler than MergeSort. Quicksort

Quicksort algorithm 1) if size of list, L is 0 or 1, return 2) pick some element in list as a pivot element 3) divide the remaining elements (minus the pivot) of L into two groups, L 1, those with elements less than the pivot, and L 2, those with elements greater than or equal to the pivot 4) return (Quicksort(L 1 ) followed by pivot, followed by Quicksort(L 2 ))‏ Depending on which is the pivot element, the sizes of the two sides could differ greatly. Compared to mergeSort, Quicksort does not guarantee equal size portions to sort (which is bad.) But, the divide stage can be done in-place (without any additional space like another array.)‏ Quicksort

To pick some element in list as a pivot element we can either –pick the first (bad if list is almost sorted, why?)‏ –pick a random one (random # generation is time consuming)‏ –a good way is to pick the pivot is the median of 3 elements (say the median of the first, middle and last element) not much extra work the almost sorted case isn't a problem for this Divide strategy – how to divide our list into two sublists of less than pivot and greater than pivot (assume all elements distinct for now)‏ The strategy about to be described gives good results. Quicksort

Divide strategy 1) swap the pivot with the last in the list 2) start index i pointing to first in list and index j to next to last element 3) while (element at i < pivot)‏ increment i 4) while (element at j >= pivot)‏ decrement j 5) if (i pivot and element at j is < pivot so, we swap them and repeat from step 3. 6) when i > j, we swap the pivot that is in the last place with the element at i. Quicksort

Notice that in the best case, if Quicksort could divide the list in equal portions at each level then we would have the fewest recursion levels O(log 2 n)‏ The work to be done on each level is on the order of n. So in the best case quicksort is O(n log 2 n)‏ Any ideas on what it'd be in the worst case? Quicksort

Let's write Quicksort –We can make quicksort be a recursive method that takes in an array the starting index of the data to be sorted the number of elements of the data to be sorted –quicksort will call a method to partition the elements find a pivot and divide a (portion of a) list into elements less than pivot, followed by pivot, followed by elements greater than pivot This method will take in –an array –the starting index of the data to be sorted –the number of elements of the data to be sorted and it will return the pivot index and alter the order of the elements of a subset of the array passed in Quicksort

A typical speedup for Quicksort is to do the following: –when we get down to some small number of elements (say 10) in our list, instead of using quicksort on them, we do insertion sort. –Let's use that xSort applet to visualize insertion sort. How would we alter the code we just wrote to do insertion sort when the number of elements to sort is small? Quicksort