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CSE - 5311 Advanced Algorithms

Explore the key concepts of Red-Black Trees, including structural properties, rotations, insertion, hashing, and the Union-Find algorithm. Dives into balancing, insertion rules, and operation complexities.

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CSE - 5311 Advanced Algorithms

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  1. CSE - 5311AdvancedAlgorithms Instructor : Gautam Das Submitted by Raja Rajeshwari Anugula & Srujana Tiruveedhi

  2. Contents • RED – BLACK Trees • History • Properties • Rotations • Insertion • Hashing • Union Find Algorithm

  3. RED BLACK TREES • RB trees is a Binary Tree with one extra bit of storage per node; its color, which can be either RED or BLACK. • Each node of the tree contains fields color, key, left, right, parent. • Red-Black Trees ensure that longest path is no more than twice as long as any other path so that the tree is approximately BALANCED.

  4. RED BLACK TREES STRUCTURAL PROPERTIES: • Every node is colored red or black. • The Root is Black. • Every “leaf” (Nil) is colored black. • Both children of a red node are black. • Every simple path from a child of node X to a leaf has the same number of black nodes.

  5. RED BLACK TREES Points to remember: • This number is known as the black-height of X(bh(X)). • A RB tree with n internal nodes has a height of almost 2log(n+1). • Maximum path length is O(log n). • Finding an element is real quick in RB trees, i.e,., it takes O(log n) time. • Insertion and Deletion take O(log n) time.

  6. RED BLACK TREES ROTATIONS • Insertion and Deletion modify the tree, the result may violate the properties of red black trees. To restore these properties rotations are done. • We can have either LEFT rotation or RIGHT rotation by which we must change colors of some of the nodes in the tree and also change the pointer structure.

  7. y x x y RED BLACK TREES LEFT ROTATE RIGHT ROTATE a c a b b c • When we do a LEFT rotation on a node x we assume that its right child y is not nil ,i.e x may be any node in the tree whose right child is not Nil. • It makes y the new root of the sub tree, with x as y’s left child and y’s left child as x’s right child.

  8. RED BLACK TREES • The Idea to insert is that we traverse the tree to see where it fits , assuming that it fits at the end, and initially color the inserted node RED, then traverse up again. • Coloring rule while insertion • If the father node and uncle node of the inserted node are red then make father and uncle as BLACK and grand father as RED.

  9. RED BLACK TREES C Case 1a: Father and Uncle are Red , problem node is right child C Recolor it to BLACK After Recoloring D A D A a d e e B B d a b c b c

  10. RED BLACK TREES Case 1b: Father and Uncle are Red , problem node is left child C Recolor it to BLACK C After Recoloring B D B D c A c d e d e A a b a b

  11. RED BLACK TREES Case 2a: Father red and Uncle Black, problem node is left child B C After Rotation D C B A c d A a b c D a b d

  12. RED BLACK TREES Case 2b: Father red and Uncle Black, problem node is right child C C After Rotation D A B D d B c d A a c b a b apply 2a for the above tree

  13. RED BLACK TREES Example: 11 Apply Case 1b 11 2 14 2 14 7 15 7 15 1 1 8 5 8 5 Insert Node 4 4 4

  14. RED BLACK TREES 11 11 Apply Case 2b 2 14 7 14 7 15 8 15 1 2 8 5 1 5 4 4

  15. RED BLACK TREES 11 Apply Case 2a 7 7 14 2 11 8 15 5 14 2 8 1 15 4 5 1 4

  16. HASHING • Has table is an effective data structure for implementing dictionaries. • Although searching for an element in hash table in the worst case is Θ(n) time, under reasonable assumptions the expected time to search for an element is O(1). • With hashing this element is stored in slot h(k) i.e we use a hash function h to compute the slot from the key k. (h maps the universe U of keys into the slots of a hash table T[0…m-1]) h :U {0,1,2…..m-1} • Two keys may hash to the same slot. This is called collision.

  17. UNION FIND Basics: • Applications involve grouping n elements into a collection of disjoint sets. • The important operations are • MAKE-SET(x): Creates a new set • UNION(x,y): Unites the dynamic sets that contain x and y into a new set that is the union of these two sets. • FIND-SET(x): Returns a pointer to the representative of the set containing x • The number of union operations is atmost n-1.

  18. MAKE-SET OPERATION • Makes a singleton set • Every set should have a name which should be any element of the set Make-Set(1) Make-Set(2) * * * * * Make-Set(n)

  19. UNION OPERATION Initially each number is a set by itself. From n singleton sets gradually merge to form a set. After n-1 union operations we get a single set of n numbers. UNION :Merge two sets and create a new set 3 1 4 2

  20. FIND OPERATION • Every set has a name. • Thus Find(number) returns name of the set. • It can be performed any number of times on the sets. • The time taken for a find operation is O(n) whereas for Union operation it is O(1).

  21. LINKED LIST REPRESENTATION • Every Set is represented as linked list where the first element is the name of the set. • We have the array of elements which have pointers to the elements in the linked lists. 5 2 1 3 4 7 ‘3’ is head of this set 3 4 7 5 1 2 ‘5’ is head if this set 10 10 ’10’ is head of this singleton set

  22. LINKED LIST REPRESENTATION • For ‘n’ Unions and ‘m’ Finds, then time taken is n+mn. • If we have a pointer from each element in the set to the head, then the time to find operation is O(1). NOTE: If m is large, Find : O(n+mn) Union: O(1) ‘3’ is the head ‘5’ is the head

  23. LINKED LIST REPRESENTATION Each element pointing to the head i,e ‘3’ in this example The 2 sets are being merged by connecting 5 and 7 In this case the union takes O(n2+m) time.

  24. If we assume that the smaller set is attached to the end of the larger set in Union operation, then UnionO(n) and Find O(1) • But in the AMORTIZED ANALYSIS, Average time taken for union is O(log n). • So ‘n’ Unions and ‘m’ Finds take (nlog n +m) time. • To guarantee O(log n) time for Union, instead of pointing each element to the main head, point the Heads’ of the individual sets to a main head.

  25. If we consider path lengths to combine two trees(L1 is the path length of tree1 and L2 is the path length of tree2), then • If L1>L2 or L2>L1, path length doesn’t change i.e. It is still the longer path. • If L1=L2, then the path length is • L2+1 if the head of tree1 is pointed to head of tree2 • L1+1 if the head of tree2 is pointed to head of tree1

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