CS 473 Algorithms I Lecture X Augmenting Data

CS 473 -Algorithms I Lecture X Augmenting Data Structures CS 473 Lecture X 1

How to Augment a Data Structure FOUR STEP PROCEDURE: 1. Choose an underlying data structure (UDS) 2. Determine additional info to be maintained in the UDS 3. Verify that additional info can be maintained for the modifying operations on the UDS 4. Develop new operations Note: Order of steps may vary and be intermixed in real design. CS 473 Lecture X 2

Example Ø Design of our order statistic trees: 1. Choose Red Black (R-B) TREES 2. Additional Info: Subtree sizes 3. INSERT, DELETE => ROTATIONS 4. OS-RANK, OS-SELECT Ø Bad design choice for OS-TREES: 2. Additional Info: Store in each node its rank in the subtree • OS-RANK, OS-SELECT would run quickly but; • Inserting a new minimum element would cause a change to this info in every node of the tree. CS 473 Lecture X 3

Augmenting R-B Trees: Theorem: • Let f be a field that augments a R-B Tree T of n nodes • Suppose that f[x] for a node x can be computed using only • The info in nodes, x, left[x], right[x] • f [ left[x] ] and f [ right[x] ] Proof Main idea: • Changing f[x] => Update only f[p[x]] but nothing else • Updating f[p[x]] => Update only f[p[p[x]]] but nothing else • And so on up to the tree until f[root[x]] is updated • When f[root] is updated, no other node depends on new value • So the process terminates • Changing an f field in a node costs O(lgn) time since the height of a R-B tree is O(lgn) CS 473 Lecture X 4

INTERVALS: DEFINITIONS DEFINITION: A Closed interval • An ordered pair of real numbers [t 1, t 2] with t 1 ≤ t 2 • [t 1, t 2] = { t ∈ R : t 1 ≤ t≤ t 2} INTERVALS: • Used to represent events that each occupy a continuous period of time • We wish to query a database of time intervals to find out what events occurred during a given interval • Represent an interval [t 1, t 2] as an object i, with the fields low[i] = t 1 & high[i] = t 2 • Intervals i & i' overlap if i ∩ i' ≠ ∅ that is low[i] ≤ high [i'] AND low[i'] ≤ high [i] CS 473 Lecture X 5

INTERVALS: DEFINITIONS Any two intervals satisfy the interval trichotomy That is exactly one of the following 3 properties hold a) i and i' overlap i i' b) high[i] < low[i'] i c) high[i'] < low[i] i' CS 473 i i' i' i Lecture X 6

INTERVAL TREES Ø Maintain a dynamic set of elements with each element x containing an interval int[x] Ø Support the following operations: • INSERT(T, x) : Adds an element x whose int field contains an interval to the tree • DELETE(T, x): Removes the element x from the tree T • SEARCH(T, i): Returns a pointer to an element x in T such that ü int[x] overlaps with i' ü NIL if no such element in the set. CS 473 Lecture X 7

INTERVAL TREES(Cont. ) Ø S 1: Underlying Data Structure • Choose R-B Tree • Each node x contains an interval int[x] • Key of x=low[ int[x] ] Ø Inorder tree walk of the tree lists the intervals in sorted order by low endpoints Ø S 2: Additional information Store in each node x the maximum endpoint “max[x]” in the subtree rooted at the node CS 473 Lecture X 8

[7, 10] EXAMPLE [5, 11] [17, 19] [4, 8] [15, 18] [21, 23] int [17, 19] max 23 [5, 11] [21, 23] 18 23 [4, 8] [15, 18] 8 18 [7, 10] 10 CS 473 Lecture X 9

INTERVAL TREES(cont. ) Ø S 3: Maintaining Additional Info(max[x]) • max[x] = minimum {high[int[x]], max[left[x]], max[right[x]]} • Thus, by theorem INSERT & DELETE run in O(lgn) time CS 473

INTERVAL TREES(cont. ) Ø INSERT OPERATION • Fix subtree Max’s on the way down • As traverse path for INSERTION while comparing “new low” to that of node intervals • Use “new high” to update “max” of nodes as appropriate • Restore balance with rotations; updating of “max” fields for rotation No change [11, 35] 35 X [6, 20] Y 20 14 Z Right Rotate 30 [6, 20] X Y 35 Z [11, 35] 14 19 • Thus, fixing “max” fields for rotation takes O(1) time. CS 473 35 19 14

INTERVAL TREES(cont. ) Ø S 4: Developing new operations INTERVAL-SEARCH(T, i) x root[T] while x ≠NIL and i ∩int[x] = ∅ do if left[x] ≠NIL and max[left[x]] < low[i] then x left[x] else x right[x] return x CS 473

INTERVAL TREES(cont. ) Time: O(lgn) Ø Starts with x at the root and proceeds downward • On a single path, until • EITHER an overlapping interval is found • OR x becomes NIL Ø Each iteration takes O(1) time Ø Height of the tree = O(lgn) CS 473

Correctness of the Search Procedure Ø Key Idea: Need to check only 1 of the node’s 2 children Ø Theorem • Case 1: If search goes right then Either overlap in the right subtree or no overlap • Case 2: If search goes left then Either overlap in the left subtree or no overlap CS 473

Correctness of the Search Procedure Case 1: Go Right • If overlap in right, then done • Otherwise (if no overlap in RIGHT) – Either left[x] = NIL No overlap in LEFT – OR left[x] ≠ NIL and max[left[x]] < low[i] For each interval i’’ in LEFT high[i’’] <= max[left[x]] < low[i] Therefore, No overlap in LEFT CS 473

Correctness of the Search Procedure Case 2: GO LEFT • If overlap in left, then done • Otherwise (if no overlap in LEFT) – low[i] <= max[left[x]] =high[i’] for some i’ in LEFT – Since i & i’ don’t overlap and low[i] <= high[i’] We have high[i] < low [i’] (Interval Trichotomy) – Since tree is sorted by lows we have high[i] < low[i’]<Any lows in RIGHT – Therefore, no overlap in RIGHT CS 473

Pictorial View of Case 1 & Case 2 i’ i i’’ max[left[x]] Case 1 t max[left[x]] Case 2 i’’: any interval in left i’’: any interval in right i’ in left such that high[i’]=max[left[x]] CS 473

Interval Trees • How to enumarate all intervals overlapping a given interval – Can do in O(klgn) time, where k = # of overlapping intervals – Find and Delete overlapping intervals one by one; • When done reinsert them • Theoritical Best is O(k+lgn) CS 473

How to maintain a dynamic set of numbers that support min-gap operations MIN-GAP(Q): retuns the magnitude of the difference of the two closest numbers in Q Example: Q={1, 5, 9, 15, 18, 22} MIN-GAP(Q) = 18 -15 = 3 1. Underlying Data Structure: • A R-B Tree containing the numbers keyed on the numbers 2. Additional Info at each Node: min-gap[x]: minimum gap value in the subtree TX rooted at x min[x] : minimum value (key) in TX max[x] : maximum value (key) in TX These values are ∞ if x is a leaf node CS 473

3. Maintaining the Additional Info min[x] = min[left[x]] if left[x] NIL key[x] otherwise min-gap[x] = Min CS 473 min-gap[left[x]] min-gap[right[x]] key[x] – max[left[x]] min[right[x]] – key[x] Each field can be computed from info in the node & its children Hence, by theorem, they would be maintained during insert & delete operation without affecting the O(lgn) running time

How to maintain a dynamic set of numbers that support min-gap operations(cont. ) • The reason for defining the min & max fields is to make it possible to compute min-gap from the info at the node & its children Develop the new operation: MIN-GAP(Q) • MIN-GAP(Q) simply returns the min-gap value of the root – It is an O(1) time operation It is also possible to find the two closest numbers in O(lgn) time CS 473

How to maintain a dynamic set of numbers that support min-gap operations(cont. ) CLOSEST-NUMBERS(Q) x root[Q] gapmin min-gap[x] while x ≠ NIL do if gapmin = min-gap[left[x]] then x left[x] elseif gapmin = min-gap[right[x]] x right[x] elseif gapmin = key[x] - max[left[x]] return { key[x], max[left[x]] } else return { min[right[x]], key[x] } CS 473

How to find the overlap of rectilinearly rectangles • Given a set R of n rectilinearly oriented rectangles – i. e sides of all rectangles are paralled to the x & y axis • Each rectangle r R is represented with 4 values – xmin[r], xmax[r], ymin[r], ymax[r] • Give an O(nlgn)-time algorithm To decide whether R contains two rectangle that overlap CS 473

OVERLAP(R) TY ∅ SORT xmin & xmax values of rectangles in R for each extremum x value in the sorted order do r rectangle [x] yint [ ymin[r], ymax[r] ] if x = xmin[r] then v INTERVAL-SEARCH(TY, yint) if v ≠NIL then return TRUE else z MAKE-NEW-NODE() left[z] right[z] p[z] NIL int[z] yint INSERT(TY, z) else /* x=xmax[r] */ DELETE(TY, yint) return FALSE CS 473