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Harvard Advanced Algorithms 2 | Advanced Algorithms (COMPSCI 224), Lecture 2 | Fusion Trees
Citation Via Youtube - Harvard University
Harvard Advanced Algorithms 2
This lesson covers fusion trees.
Harvard Advanced Algorithms
Last class Van Deboez trees and wifast trees were used and the same bounds were found.
Fredmand Willard Jess 93. Made dynamic by Anderson Thorpe 93' O log wn plus log log updates. Also Raman ESA 96 O log wn. Updates expected time. Query forf all O log wn. Instead
using arrays use hash tables. You get O of n space instead of O of u space.
The idea of fusion trees is to have k bigger than 2 instead of binary. 2, 3, 4 trees or b trees are also used. Every node is going to have approixately K trees. The fusion tree
is sorted. It also has branching off trees that are in a range. A binary tree just has a single node and everything to the left is smaller, to the right is bigger. In K trees
the numbers are sorted and its 1-5 or 5 to the next number in the sub trees.
K is 1 5th of feta of the node. Log base w 5th of n. Equals one fith over log n over w. Equals log wn. The amount of space to represent a node is
w to the 6th fifths.
Harvard Advanced Algorithms 2
Word Level Parallelism Basic Indredients Multiplication Sketch Compression Most Signifigent Set Bit (MSB) in feta 1 time, constant time. The basic issue if how do we search a single fusion tree node in constant time? All k keys can't fit into a single node. With any algorithm it takes w find the index of an algorithm. Loops to w bitwise.
Lets focus on building a single fusion tree node containg x o x less than x X less than x minus 1. We can mask but we don't have too. To find y xi upper carrot q. This tells you the bits that are different. Compute msb xi plus 1. Upper carrot q. Take the more signifigant of the 2. That's how we use sketching to find where we should recurse. to. There isn't a C operation that makes sketches for us.
Problem 1 how do we form sketches? One way is to mask out the b sub i's important bits. Perfect compression would allow us to get down to w5th. We're allowed to get and be somewhat looser. O we'll compress down to O r4th bit. Bits. The important bits will be represeted with some known amount of 0 spacing in between them. We have to decide which index the information is going to land in constant time.
Suppose x times m is multiplied. x r-1 x equals zero. The sum of xb times b 2i.
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