Abstract

Let k and n be positive integers, n > k . Define r ( n , k ) to be the minimum positive value of | a 1 + ⋯ + a k − b 1 − ⋯ − b k | where a 1 , a 2 , … , a k , b 1 , b 2 , … , b k are positive integers no larger than n . Define R ( n , k ) to be − log r ( n , k ) . It is important to find tight bounds for r ( n , k ) and R ( n , k ) , in connection to the sum-of-square-roots problem, a famous open problem in computational geometry. The current best lower bound and upper bound are far apart. In this paper, we prove an upper bound of 2 O ( n / log n ) for R ( n , k ) , which is better than the best known result O ( 2 2 k log n ) whenever n ≤ c k log k for some constant c . In particular, our result implies an algorithm subexponential in k (i.e. with time complexity 2 o ( k ) ( log n ) O ( 1 ) ) to compare two sums of k square roots of integers of value o ( k log k ) . We then present an algorithm to find r ( n , k ) exactly in n k + o ( k ) time and in n ⌈ k / 2 ⌉ + o ( k ) space. As an example, we are able to compute r ( 100 , 7 ) exactly in a few hours on a single PC. The numerical data indicate that the root separation bound is very far away from the true value of r ( n , k ) .

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