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CS 332: Algorithms

CS 332: Algorithms. Skip Lists. Administration. Hand back homework 3 Hand back exam 1 Go over exam. Review: Red-Black Trees. Red-black trees : Binary search trees augmented with node color Operations designed to guarantee that the height h = O(lg n )

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CS 332: Algorithms

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  1. CS 332: Algorithms Skip Lists David Luebke 16/3/2014

  2. Administration • Hand back homework 3 • Hand back exam 1 • Go over exam David Luebke 26/3/2014

  3. Review: Red-Black Trees • Red-black trees: • Binary search trees augmented with node color • Operations designed to guarantee that the heighth = O(lg n) • We described the properties of red-black trees • We proved that these guarantee h = O(lg n) • Next: describe operations on red-black trees David Luebke 36/3/2014

  4. Review: Red-Black Properties • The red-black properties: 1. Every node is either red or black 2. Every leaf (NULL pointer) is black • Note: this means every “real” node has 2 children 3. If a node is red, both children are black • Note: can’t have 2 consecutive reds on a path 4. Every path from node to descendent leaf contains the same number of black nodes 5. The root is always black David Luebke 46/3/2014

  5. Review: RB Trees: Rotation • Our basic operation for changing tree structure is called rotation: • Preserves BST key ordering • O(1) time…just changes some pointers y x rightRotate(y) x C A y leftRotate(x) A B B C David Luebke 56/3/2014

  6. Review: Red-Black Trees: Insertion • Insertion: the basic idea • Insert x into tree, color x red • Only r-b property 3 might be violated (if p[x] red) • If so, move violation up tree until a place is found where it can be fixed • Total time will be O(lg n) David Luebke 66/3/2014

  7. rbInsert(x) treeInsert(x); x->color = RED; // Move violation of #3 up tree, maintaining #4 as invariant: while (x!=root && x->p->color == RED) if (x->p == x->p->p->left) y = x->p->p->right; if (y->color == RED) x->p->color = BLACK; y->color = BLACK; x->p->p->color = RED; x = x->p->p; else // y->color == BLACK if (x == x->p->right) x = x->p; leftRotate(x); x->p->color = BLACK; x->p->p->color = RED; rightRotate(x->p->p); else // x->p == x->p->p->right (same as above, but with “right” & “left” exchanged) Case 1 Case 2 Case 3

  8. rbInsert(x) treeInsert(x); x->color = RED; // Move violation of #3 up tree, maintaining #4 as invariant: while (x!=root && x->p->color == RED) if (x->p == x->p->p->left) y = x->p->p->right; if (y->color == RED) x->p->color = BLACK; y->color = BLACK; x->p->p->color = RED; x = x->p->p; else // y->color == BLACK if (x == x->p->right) x = x->p; leftRotate(x); x->p->color = BLACK; x->p->p->color = RED; rightRotate(x->p->p); else // x->p == x->p->p->right (same as above, but with “right” & “left” exchanged) Case 1: uncle is RED Case 2 Case 3

  9. if (y->color == RED) x->p->color = BLACK; y->color = BLACK; x->p->p->color = RED; x = x->p->p; Case 1: “uncle” is red In figures below, all ’s are equal-black-height subtrees Review: RB Insert: Case 1 new x C C case 1 A D A D y B B x           Change colors of some nodes, preserving #4: all downward paths have equal b.h. The while loop now continues with x’s grandparent as the new x David Luebke 96/3/2014

  10. if (x == x->p->right) x = x->p; leftRotate(x); // continue with case 3 code Case 2: “Uncle” is black Node x is a right child Transform to case 3 via a left-rotation B A x x     Review: RB Insert: Case 2 C C case 2 y A B y     Transform case 2 into case 3 (x is left child) with a left rotation This preserves property 4: all downward paths contain same number of black nodes David Luebke 106/3/2014

  11. x->p->color = BLACK; x->p->p->color = RED; rightRotate(x->p->p); Case 3: “Uncle” is black Node x is a left child Change colors; rotate right A x   Review: RB Insert: Case 3 C B case 3 B A y x  C      Perform some color changes and do a right rotation Again, preserves property 4: all downward paths contain same number of black nodes David Luebke 116/3/2014

  12. Red-Black Trees • Red-black trees do what they do very well • What do you think is the worst thing about red-black trees? • A: coding them up David Luebke 126/3/2014

  13. Skip Lists • A relatively recent data structure • “A probabilistic alternative to balanced trees” • A randomized algorithm with benefits of r-b trees • O(lg n) expected time for Search, Insert • O(1) time for Min, Max, Succ, Pred • Much easier to code than r-b trees • Fast! David Luebke 136/3/2014

  14. So these all take O(1)time in a linked list. Can you think of a wayto do these in O(1) timein a red-black tree? Goal: make these O(lg n) time in a linked-list setting Linked Lists • Think about a linked list as a structure for dynamic sets. What is the running time of: • Min() and Max()? • Successor()? • Delete()? • How can we make this O(1)? • Predecessor()? • Search()? • Insert()? David Luebke 146/3/2014

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