首页> 外文期刊>Journal of Combinatorial Theory, Series A >The maximum number of times the same distance can occur among the vertices of a convex n-gon Is O(n log n)
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The maximum number of times the same distance can occur among the vertices of a convex n-gon Is O(n log n)

机译:凸n边形的顶点之间出现相同距离的最大次数为O(n log n)

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Proof: Denote by f( n) the maximum number of times the unit distance can occur among n points ill convex position ill the plane. Let p(1), p(2),...., p(n), in this cyclic order, be the vertices of a convex polygon, for which the maximum is attained. Let G denote the geometric graph obtained by connecting two points of P by a straight-line segment (edge) if and only if their distance is one. Pick a point p(i) antipodal to p(1), i,e., assume that there are two parallel lines, t and t', passing through p(1), and p(i), resp., such that all elements of P are in the strip between them. We claim that all but at most 2n edges of G cross p(1) p(i). To verify this. suppose without loss of generality that t and t' are parallel to the x-axis, and that no edge of G is parallel to the y-axis. Color any edge of G red if its slope is positive and blue otherwise. Assign every red edge lying in the closed half-plane to the left (right) of p(1) p(i) to its left ( right) endpoint. It is easy to see that every element of P is assigned to at most one red edge. Therefore, the number of red edges not crossing p(1) p(i) is at most n. The same is true for the blue edges, which proves the claim. We can assume without loss of generality that i > n/2, otherwise the numbering of the vertices can be reversed. Take a point p(j) antipodal to p([i/2]). As above, there are at most 2n edges of G, which do not cross p([i/2]) p(j). Every edge of G, crossing both p1 pi and p([i/2]) p(j), connects a pair of points in P-1:= {p(2), p(3),.. P[i/2]-1}boolean OR { Pi+1, Pi+2,..., Pj-1} or in P-2 :={p([i/2]+1), p([i/2]+2),..., Pi-1} boolean OR {Pj+1, Pj+2, ..., P-n}. Thus, we have f(n) = E(G) less than or equal to f( P-1) + f( P-2 ) + 4n. Using the facts that P-1 + P-2 = n - 4 and min (P-1, P-2 ) greater than or equal to (n-7/)(4), the theorem follows by induction. It is an exciting open problem to decide whether f(n)= O(n) holds. The best known general lower bound, f(n) greater than or equal to 2n - 7, is due to Edelsbrunner and Hajnal [EH]. [References: 2]
机译:证明:用f(n)表示在n个点,在平面上的凸位置之间可出现单位距离的最大次数。令p(1),p(2),...,p(n)以此循环顺序为凸多边形的顶点,为此可获得最大值。令G表示当且仅当P的两个点的距离为1时,才通过直线段(边缘)连接两个点而获得的几何图。选择一个与p(1)对映的点p(i),即假设有两条平行线t和t'经过p(1)和p(i),分别是P的所有元素都在它们之间的条带中。我们要求G的最多2n个边缘都与p(1)p(i)交叉。为了验证这一点。在不失一般性的前提下,假设t和t'平行于x轴,并且G的任何边都不平行于y轴。如果G的任何斜率是正的,则将其任何颜色都为红色,否则为蓝色。将位于闭合半平面中的每个红色边缘分配给p(1)的左(右)p(i)的左(右)端点。很容易看出,P的每个元素最多被分配一个红色边缘。因此,未穿过p(1)p(i)的红色边缘的数量最多为n。蓝色边缘也是如此,这证明了要求。我们可以不失一般性地假定i> n / 2,否则可以颠倒顶点的编号。将p(j)与p([i / 2])对等。如上所述,G最多有2n个边缘,它们不与p([i / 2])p(j)交叉。 G的每个边缘都穿过p1 pi和p([i / 2])p(j),连接P-1中的一对点:= {p(2),p(3),.. P [i / 2] -1}布尔值OR {Pi + 1,Pi + 2,...,Pj-1}或在P-2中:= {p([i / 2] +1),p([i / 2 ] +2),...,Pi-1}布尔值或{Pj + 1,Pj + 2,...,Pn}。因此,我们有f(n)= E(G)小于或等于f( P-1 )+ f( P-2 )+ 4n。利用 P-1 + P-2 = n-4并且min( P-1 , P-2 )大于或等于(n-7 /)(4)的事实,定理通过归纳法得出。决定是否成立f(n)= O(n)是一个令人兴奋的开放问题。大于或等于2n-7的最著名的一般下界f(n)是由Edelsbrunner和Hajnal [EH]引起的。 [参考:2]

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