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A Chip Search Problem on Binary Numbers

机译:二进制数的筹码搜索问题

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Suppose that we have an unknown point d in the interval and an unbounded reservoir of chips on both points 0 and 1. In every step, we can either move two pebbles from points x and y to (x+y)/2, or we can ask whether dx, but only if there is currently a pebble at x. Our aim is to determine an interval of length 2 sup -n including d, and we are interested in the exact number of necessary moves in the worst case, especially for the firt values of n. First we analyze a natural GREEDY strategy solvign this problem in the other hand, n sup 2/12 is a lower bound. Our analysis allows to compute the exact worst-case number of moves g(n) that GREEDY takes for the first values of n, although a nice general expressio for this is missing. GREEDY sends pebbles only to points being binary approximatios of the target d. Moreover, our analysi will show that GREEDY is optimal among all strategie sharing this property. Hence any such strategy needs sita(n sup 2) moves in the worst case. It might seem that sending pebbles to other points brings no advnatages. So it is surprising that, without the mentioned restriction, we can achieve an asymptotic result of moves, by an acceleration technique. Hence GREEDY is not optimal in genral, nevertheless it seems to be the best choice for small n, since it is simple, and we have no strategy beating g(n) if n<20. An open problem is to determine the maximum n for which g(n) moves are optimal. In a final section we discuss a way to save tests if d is very small, and we briefly consider a variant of the problem where a test at x destroys a pebble lying there.
机译:假设我们在间隔中有一个未知的点d,并且在点0和1上都有一个无穷大的筹码池。在每一步中,我们可以将两个小卵石从点x和y移动到(x + y)/ 2,或者可以询问d x,但前提是当前x处只有一个卵石。我们的目标是确定包括d在内的长度2 sup -n的间隔,并且我们对最坏情况下的必要移动的确切数量感兴趣,尤其是对于n的点火值。首先,我们分析了解决该问题的自然贪婪策略,另一方面,n sup 2/12是一个下界。我们的分析允许计算GREEDY对于n的第一个值所采取的精确的最坏情况下的移动g(n),尽管缺少很好的通用表达。 GREEDY仅将小卵石发送到目标d的二进制近似点。此外,我们的分析将表明,在共享该属性的所有策略中,GREEDY是最佳的。因此,任何这种策略在最坏的情况下都需要sita(n sup 2)移动。似乎将鹅卵石发送到其他点不会带来任何好处。因此令人惊讶的是,在没有上述限制的情况下,我们可以通过加速技术获得运动的渐近结果。因此,GREEDY一般而言并不是最优的,尽管如此,它似乎是小n的最佳选择,因为它很简单,并且如果n <20,我们没有策略击败g(n)。一个开放的问题是确定g(n)移动最佳的最大n。在最后一节中,我们讨论了如果d非常小则保存测试的方法,并且我们简要地考虑了问题的一种变体,其中x处的测试会破坏那里的卵石。

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