On-Line q-adic Covering by the Method of the n-th Segment and its Application to On-Line Covering by Cubes

We prove that in Euclidean d-space every sequence of cubes with total volume 2 + 3 is able to cover on-line the unit cube. The proof is based on an on-line q-adic method of covering the unit segment by segments of lengths of the form q , where q 2 and r 1 are integers. The fact that this method is q-adic means that every segment has to be placed in such a way that both end-points are at points that are multiples of the length of the segment. MSC 1991: 52C17, 68Q20 We say that a sequence Q1; Q2; : : : of subsets of Euclidean space E d permits a covering of ae set C E if there exist rigid motions 1; 2; : : : such that C is contained in the union of sets 1Q1; 2Q2; : : :. A survey of results about covering of convex bodies by sequences of convex bodies is provided by Groemer [2]. An on-line covering problem consists in nding the motions 1; 2; : : :, under the condition that we are given every set Qi, with i > 1, only after the motion i 1 has been provided. We learn the set Q1 at the beginning. Problems of the on-line covering by convex bodies are considered in a number of papers. The rst result about on-line covering the unit cube by cubes was obtained by Kuperberg [5]. This result was improved in [3,4] and the present paper provides a further improvement. The reader may compare the above on-line assumption with that in the area of bin-packing problems (see the survey paper [1]). 1 Supported in part by Komitet Bada n Naukowych (Committee of Scienti c Research), Grant Number 2 2005 92 03. 0138-4821/93 $ 2.50 c 1996 Heldermann Verlag, Berlin 52 J. Januszewski, M. Lassak, G. Rote, G. Woeginger: On-Line q-adic Covering In this paper we rst deal with a specially restricted one-dimensional covering problem and then apply it to higher-dimensional problems. By [x; y] we denote the closed segment with end-points x and y, where x < y. The symbol (x; y) denotes the corresponding open segment. Let q be an integer greater than or equal to 2. By the on-line q-adic covering problem of the unit segment [0; 1] we mean the problem of on-line covering the segment [0; 1] by a sequence of closed segments Si of lengths i, where i 2 fq ; q ; : : :g, and where the motions i are restricted so that every segment iSi is of the form [ci i; (ci + 1) i] with ci 2 f0; : : : ; 1 i 1g for i = 1; 2; : : :. The 2-adic covering problem was posed by Kuperberg [6]. 1. The method of the n-th segment Let n 2 be an integer. At every moment of the covering process we determine the greatest number b 2 [0; 1] such that the whole segment [0; b] is covered. Assume that we are given a segment S from our sequence and let q r be the length of this segment. Denote by a the greatest number among 0; : : : ; q 1 such that aq r b. If the interval [(a+ k 1)q ; (a+ k)q ], where k 2 f1; 2; : : :g, is a subset of [0; 1], then we call it the k-th segment. Let c be the greatest number among a; : : : ; a+ n 1 such that the segment T = [cq ; (c + 1)q ] is contained in [0; 1] and such that T is not yet totally covered. The segment S is put on T , i. e., we provide a motion such that S = T . (In other words, we successively try the n-th segment, the (n 1)-st segment, and so on, until we nd an interval not totally covered.) We nish the process when the whole segment [0; 1] is covered. The above described on-line algorithm for q-adic covering is called the n-th segment method. We will mainly be interested in the cases n = q+1, n = q+2, and n = q+3, because these three values give the best performance ratio for all values of q. Let us rst introduce some notation that will be used in the sequel. We imagine the segment [0; 1] laid out horizontally with 0 on the left end. For every integer i > 1 we denote by bi the greatest number in the segment [0; 1] such that the segment [0; bi] is covered by the union of segments hSh for h < i. Moreover, let b1 = 0. We call bi the current bottom; we choose this name because our notion serves here a similar purpose as the notion of the current bottom in some other papers where a cube or a box is lled from bottom to top. We start with a somewhat technical lemma. Lemma 1. Consider the n-th segment method for some n > q. Let p be a multiple of q w for some w such that 0 < p < 1. Assume that at some point of the algorithm, the interval [0; p] has not been completely covered yet. For j 0 let j be the number of segments of length q w j which the algorithm has placed to the right of p. If 0; : : : ; ` n 2 for some ` 0, then at most one of these ` + 1 numbers is equal to n 1. If, in addition, one of the numbers 0; : : : ; ` equals n 1, then the interval [p; p + q ] is completely covered. Proof. Assume that j = n 1 for some j and 0; : : : ; j 1 n 2. We want to show that 0 = = j 1 = n 2 and that the whole interval [p; p+ q ] is covered. Since p is a multiple of q w j , the equation j = n 1 can only hold if the n 1 segments of length J. Januszewski, M. Lassak, G. Rote, G. Woeginger: On-Line q-adic Covering 53 q w j to the right of p are the adjacent intervals [p; p+ q w j ]; : : : ; [p+(n 2)q w j ; p+ (n 1)q w j ]. Since n 1 q, this means that the interval [p; p+ q w ] is completely covered. If we now look at j 1, we see that the possible positions for the segments of length q w j+1 are the n 1 intervals [p; p+q w ]; : : : ; [p+(n 2)q w ; p+(n 1)q w ]. Among them, the rst interval is not possible because it has been covered completely by smaller segments. Thus we conclude that j 1 = n 2, and we know that [p; p+ q w ] is completely covered. In this way we proceed inductively to j 2; : : : ; 0, and the lemma is proved. During the course of the algorithm, the algorithm makes progress by gradually advancing the current bottom bi. The following lemma relates this progress to the sum of the lengths of segments which have been used to cover the ground to the left of bi. Lemma 2. Consider the n-th segment method for some n with q < n 2q. Assume that i 1 and bi < bi+1 < 1 and put b = bi+1 bi. Denote by l the total length of those among the segments 1S1; : : : ; iSi which have non-empty intersection with (bi; bi+1). We have l < 1 + 1 q 1 + 1 n 1 1 (q 1)(n 1) b: (1) Proof. Denote by w the smallest positive integer such that a segment of length q w has been used for the covering of the segment (bi; bi+1). From the description of the method we see that q w < b nq : Thus we can represent b in the form