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An Improved Algorithm for the Rectangle Enclosure Problem Anatoli Uchitel From an article By D.T. Lee and F.P. Preparata (March 8, 1981)

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Introduction w This essay gives an algorithm for solving the Rectangle Enclosure problem in time and O(n) space w n is the number of rectangles and q is the number of the required rectangle pairs

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Topics of Presentation w Introduction w Rectangle Enclosure and Dominance w A Two-Dimensional Subproblem w Time and Space Analysis

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Introduction Given a set of n rectangles in the plane, with sides parallel to the coordinate axes, find all q pairs of rectangles such that one rectangle of the pair encloses the other w What is the Rectangle Enclosure problem ? Given a set of n rectangles in the plane, with sides parallel to the coordinate axes, find all q pairs of rectangles such that one rectangle of the pair encloses the other w The suggested solution achieves time using O(n) space

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Rectangle Enclosure and Dominance w We begin by transforming the rectangle enclosure problem into an equivalent one, which is easier to describe and understand w Let be a set of iso- oriented rectangles in the plane w For each i=1,…,n

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Transforming to Point Dominance problem w iff all the following hold:

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w This is equivalent to: which expresses the relation “<“ of dominance between two four-dimensional points w Thus, after mapping each from R to its corresponding point in 4D we need to solve

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The Point Dominance Problem in 4D w Definition: p
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Solving the dominance problem w Input : a set Required Output : all pairs of points (p,q) where p < q w Notation : the i-th coordinate of a point p will be donated by w Preliminary step : sort the elements of S by values in increasing order If two points have equal coordinates sort by etc. w Strategy : Divide & Conquer

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Algorithm Dominance w D1. ( Devide ) Partition S into S1 S2 where, w D2. ( Recur ) Solve the point-dominance problem on S 1 and S 2 separately w D3. ( Merge ) Find all pairs where,

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Explaining the correctness of the algorithm w All dominance pairs within S 1 and S 2 will be found in D2. ( Recur ) w If there exists a dominance pair in which one point is from S 1 ( say p ) and the other is from S 2 ( say q ), than it must be that p < q, since by construction

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Implementing D3 step - Merge w If and than by construction w Thus iff for l=2,3,4 w We have reduced the 4D problem to a 3D problem w let be the median of w Strategy : Divide & Conquer again

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Why does that work? w By division : S = S 11 S 12 S 21 S 22 w Within each S ij the dominance pairs are found at step D2 since they are contained in S 1 and S 2 w About Dominant pairs where the points are from different subsets Sij we should notice that we should only examine cases where the dominant point is from a higher indexed set since the points are sorted by increasing order of u 1 ( if u 1 are the same a secondary sort by u 2 is done etc. )

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The way all cases are processed w The solution includes the following cases: (S 21,S 22 ) and (S 11,S 12 ) are processed in D2 since and (S 11,S 21 ) and (S 12,S 22 ) are processed in M2 ( Recur in Merge ) (S 11,S 22 ) is processed in M3 ( Combine in Merge ) (S 12,S 21 ) need not be considered because for each and we have and

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Implementation of the Combine w The key operation of the entire task is therefore the implementation of step M3 ( the combine ) w Again we reduce the dimension of the problem, since the Combine step is a 2D merge problem ( in u 3 and u 4 )

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A Two-Dimensional Subproblem w Initial preprocessing : Construct a quadruply threaded list ( QTL ) with bidirectional links, for S. w for each p in S there is a node containing all the coordinate values and 8 pointers NEXTi,PREVi for i=1,2,3,4 that describe the ordering by the appropriate coordinates. w The construction of the QTL takes O(nlogn) time and O(n) space

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Explaining the algorithm for Combine w We traverse the points in S 22 by the sort sorting order that u 3 constrains ( NEXT 32 ) w For each such point q in S 22 we find all the points in S 11 which is dominated by it in u 3. This is done by traversing S11 by the sort sorting order that u 3 constrains ( NEXT31 )

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w For each such point p we insert u 4 (p) into the sorted list L. w When we reach p such that u 3 (p) > u 3 (q) we print all the points is S 11 which are dominated by q, by traversing L from the beginning until reaching a point p’ for which u 4 (p’) > u 4 (q)

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Algorithm Combine J1 = BEG31; j2 = BEG32; // initialization while ( j2 != null){ if ((j1!=null) && (u3[j1] <= u3[j2])){ // if j2 dominates j1 by u3 L->insert_maintaining_order(u4[j1]); j1=NEXT3[j1]; } else { // j2 doesn’t dominate j1 l = BEGL; // print all the points in S11 which are dominated in both u3 and u4 by j2 while ((l!=null) and (u4[j2]>=u4[l])){ print(j2,l); l = NEXTL[l]; } j2 = NEXT3[j2]; }

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Example

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The Crucial task of the procedure w How can we insert u 4 [j 1 ] into L maintaining sorted order efficiently? w The naive approach would yield time w Using AVL Trees we can achieve O(|S 1 |log|S 11 |) w We can take advantage of the fact that the points are known in advance to achieve O(|S 1 |) total time

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Creating Schedule Of Insertion Into L in O(|S 11 |) w The Basic Idea : Let’s take a look at the point p in S 11 which has the highest u 3 value. We reach p after we processed all the other points from S 11. It should be placed in L after PRED4[L] w Thus we can use PRED4 from the QTL in order to determine the schedule of insertion into L

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Algorithm insertion schedule l = last(u3 list) while ( PRED3[l] != BEG3 ) { NEXT4[PRED4[l]] = NEXT4[l]; PRED4[NEXT4[l]] = PRED4[l]; l = PRED3[l]; } w Final L is PREV4

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Example

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Time and Space Analysis w Space All the processing is done in a place proportional by a constant to the QTL arrays, yielding O(n) space complexity w Time Notations: D(n) = The time complexity of the dominance algorithm M 3 (r,s) = The time complexity of the merge algorithm M 2 (r’,s’) = The time complexity of the combine algorithm

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Time Complexity - Dominance w Let |S|=n be an even number for simplicity w D(n) = 2D(n/2) + M 3 (n/2,n/2) + O(n) D2. Recur step for S1 and S2 D3. The Merge step on (S1,S2) D1. The split of S to S1 and S2

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Time Complexity - Merge w Let r be an even number for simplicity w M3(r,s) = M3(r/2,m) + M3(r/2,s-m) + M2(r/2, max(m,s-m)) + O(r+s) M2. Recur step for and M3. Combine on M1. Splitting S 1 and S 2 using the QTL

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Time Complexity - Combine w M 2 (r’,s’) = T( insertion schedule ) + T( traversing S 11 and S 22 elements ) + T( insertions to L ) + T( printing of dominance pairs ) = O(r’) + O(r’+s’) + O(r’) + O(q) = O(r’+s’+q) w Since the q dominance pairs are printed only once in the entire algorithm we can add O(q) to the final time complexity and say that M 2 (r’,s’) = O(r’+s’) for the calculations

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Solving the equations M 3 (r,s) = M 3 (r/2,m) + M 3 (r/2,s-m) + O(r+s) Since the first parameter of M3 is always divided by 2, we can make a bound M3(r,s) = O((r+s)log(r)) = O((r+s)log(r+s)) Thus, D(n) = 2D(n/2) + O(nlogn),which yields D(n) = Keeping in mind the debt for printing the pairs of dominance points we get that the final time complexity is

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Next Steps There are variations of this algorithm which improve the time complexity on account of the space complexity

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