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Computer Science
Biomedical Informatics

k-Enclosing Axis-Parallel Square

Profile image of Partha GoswamiPartha Goswami

2011, Lecture Notes in Computer Science

https://doi.org/10.1007/978-3-642-21931-3_7
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6 pages

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Abstract

Let P be a set of n points in the plane. Here an optimization technique is used to solve some optimization problems. A simple deterministic algorithm is proposed to compute smallest square containing at least k points of P. The time and space complexities of the algorithm are O(n log 2 n) and O(n) respectively. For large values of k, the worst case time complexity of the algorithm is O(n + (n − k) log 2 (n − k)) using O(n) space which is the best known bound for worst case time complexity. Then an algorithm is designed to locate smallest rectangle containing at least k points of P for large values of k and all values of k.

Figures (3)
x or y coordinates. Let x(p) and y(p) denote the x-coordinate and the y-coordinate of any point p respectively. The size of a square is represented by the length of it’s side. We have the following observation. Observation 1: At least one pair of opposite sides of S;, must contain points from P. For each optimization problem, there is a corresponding de- cision problem that asks whether there is a feasible solution for some particular measure [32]. The decision version of the problem of locating S;, can be stated as “ given a length a, does there exist a square of size a that encloses at least & points of P?” . The following section describes an O((n—k) log(n—k)) algorithm to solve the decision problem for given k and a.
x or y coordinates. Let x(p) and y(p) denote the x-coordinate and the y-coordinate of any point p respectively. The size of a square is represented by the length of it’s side. We have the following observation. Observation 1: At least one pair of opposite sides of S;, must contain points from P. For each optimization problem, there is a corresponding de- cision problem that asks whether there is a feasible solution for some particular measure [32]. The decision version of the problem of locating S;, can be stated as “ given a length a, does there exist a square of size a that encloses at least & points of P?” . The following section describes an O((n—k) log(n—k)) algorithm to solve the decision problem for given k and a.
We represent each W; as a vertical strip parallel to the y- axis (see Figure 3). All the vertical strips (W's) are arranged along the x-axis such that med(W,)’s fall on the z-axis and the median values are in nonincreasing order along the x-axis. Again the elements of each W; are arranged in nondecreasing order parallel to the y-axis. At initial step of iteration, all me- dians med(W 1), med(W2),..., med(W,) are in nonincreasing order. This ordering may change in subsequent iterations due to pruning of W,’s. Therefore at each iteration, we need to rearrange W,’s such that med(W,)’s are in nonincreasing order. Let Y,, W2,..., Y, be an arrangement of the sequences in A, such that med(V 1) > med(W2) >... > med(W,p). At jth iteration we find an index c such that S~‘_, |W;| is half of the size of A;. Observe that the size of A; is at most + of the size of A;_;. Consider med(W.) as a and compute Maz- square(a). If Maz-square(a) encloses at least & points of P, then size of 5; is less than or equal to a and we can ignore the elements in A, greater than med(W.). Note that all the med(-) values corresponding to W,, W2,...,W._; are greater than med(W.). Therefore for each 7, 1 <i < cwecan delete upper half of Y;. In case, Maz-square(a) encloses less than & points, we similarly delete lower half of each ©; for c <i <b. Now continue with the subsequent iterations until we end up at an iteration z such that size of A. is constant. Lemma 1: At every iterative step the size of the current solution space is reduced by a factor of +. Proof: At j*” iteration, either we discard upper half of W1, Wo,...,Y-_1 or lower half of ©., U.41,..., Uy. As the total number of elements in the sequences 1, Wo, ..., Vey is 5 of size of A;, we can discard at least + elements of A,. Similar amount of elements is discarded for pruning of lower half.
We represent each W; as a vertical strip parallel to the y- axis (see Figure 3). All the vertical strips (W's) are arranged along the x-axis such that med(W,)’s fall on the z-axis and the median values are in nonincreasing order along the x-axis. Again the elements of each W; are arranged in nondecreasing order parallel to the y-axis. At initial step of iteration, all me- dians med(W 1), med(W2),..., med(W,) are in nonincreasing order. This ordering may change in subsequent iterations due to pruning of W,’s. Therefore at each iteration, we need to rearrange W,’s such that med(W,)’s are in nonincreasing order. Let Y,, W2,..., Y, be an arrangement of the sequences in A, such that med(V 1) > med(W2) >... > med(W,p). At jth iteration we find an index c such that S~‘_, |W;| is half of the size of A;. Observe that the size of A; is at most + of the size of A;_;. Consider med(W.) as a and compute Maz- square(a). If Maz-square(a) encloses at least & points of P, then size of 5; is less than or equal to a and we can ignore the elements in A, greater than med(W.). Note that all the med(-) values corresponding to W,, W2,...,W._; are greater than med(W.). Therefore for each 7, 1 <i < cwecan delete upper half of Y;. In case, Maz-square(a) encloses less than & points, we similarly delete lower half of each ©; for c <i <b. Now continue with the subsequent iterations until we end up at an iteration z such that size of A. is constant. Lemma 1: At every iterative step the size of the current solution space is reduced by a factor of +. Proof: At j*” iteration, either we discard upper half of W1, Wo,...,Y-_1 or lower half of ©., U.41,..., Uy. As the total number of elements in the sequences 1, Wo, ..., Vey is 5 of size of A;, we can discard at least + elements of A,. Similar amount of elements is discarded for pruning of lower half.

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