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The closest pair of points problem or closest pair problem is a problem of computational geometry: given ''n'' points in metric space, find a pair of points with the smallest distance between them. The closest pair problem for points in the Euclidean plane〔M. I. Shamos and D. Hoey. ("Closest-point problems." ) In ''Proc. 16th Annual IEEE Symposium on Foundations of Computer Science'' (FOCS), pp. 151—162, 1975 (DOI (10.1109/SFCS.1975.8 ))〕 was among the first geometric problems which were treated at the origins of the systematic study of the computational complexity of geometric algorithms. A naive algorithm of finding distances between all pairs of points and selecting the minimum requires time. It turns out that the problem may be solved in time in a Euclidean space or Lp space of fixed dimension d. In the algebraic decision tree model of computation, the algorithm is optimal. The optimality follows from the observation that the element uniqueness problem (with the lower bound of for time complexity) is reducible to the closest pair problem: checking whether the minimal distance is 0 after the solving of the closest pair problem answers the question whether there are two coinciding points. In the computational model which assumes that the floor function is computable in constant time the problem can be solved in time.〔S. Fortune and J.E. Hopcroft. "A note on Rabin's nearest-neighbor algorithm." Information Processing Letters, 8(1), pp. 20—23, 1979〕 If we allow randomization to be used together with the floor function, the problem can be solved in time.〔S. Khuller and Y. Matias. A simple randomized sieve algorithm for the closest-pair problem. Inf. Comput., 118(1):34—37,1995〕〔(【引用サイトリンク】title=Rabin Flips a Coin )〕 ==Brute-force algorithm== The closest pair of points can be computed in O(''n''2) time by performing a brute-force search. To do that, one could compute the distances between all the pairs of points, then pick the pair with the smallest distance, as illustrated below. ''minDist'' = infinity for ''i'' = 1 to length(''P'') - 1 for ''j'' = ''i'' + 1 to length(''P'') let ''p'' = ''P''(), ''q'' = ''P''() if ''dist''(''p'', ''q'') < ''minDist'': ''minDist'' = ''dist''(''p'', ''q'') ''closestPair'' = (''p'', ''q'') return ''closestPair'' 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「closest pair of points problem」の詳細全文を読む スポンサード リンク
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