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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Source. The Theorem, p. 112, unnumbered, of Paul Erdős and Peter Fishburn, A postscript on distances in convex n-gons, Discrete Comput. Geom. 11 (1994), 111--117, doi:10.1007/BF02573998, as named on the source card; labels and pages are the print's own.

Statement

Setting (p. 112). d(x,y)d(x,y) is the Euclidean distance in the plane. A run of a convex nn-gon from a vertex x0x_0 is a sequence x0,x1,…,xkx_0,x_1,\ldots,x_k of successively adjacent vertices, going clockwise or counterclockwise from x0x_0, with d(x0,x1)<d(x0,x2)<⋯<d(x0,xk)d(x_0,x_1)<d(x_0,x_2)<\cdots<d(x_0,x_k); its length is kk. g(n)g(n) is the minimum over all convex nn-gons of the maximum run length of the nn-gon.

Theorem (p. 112). "For all n≥4n \geq 4, g(n)=⌊(n+3)/3⌋g(n) = \lfloor (n+3)/3 \rfloor."

In the corpus's words: for every n≥4n\ge4, every convex nn-gon has a run of length at least ⌊(n+3)/3⌋=⌊n/3⌋+1\lfloor(n+3)/3\rfloor=\lfloor n/3\rfloor+1, and for every n≥4n\ge4 some convex nn-gon has no longer run. The abstract (p. 111) states the same as g(n)=⌊n/3⌋+1g(n)=\lfloor n/3\rfloor+1 for n≥4n\ge4.

Context the paper gives around the statement (p. 112): Moser's 1952 proof yields g(n)≥⌊(n+2)/3⌋g(n)\ge\lfloor(n+2)/3\rfloor for all n≥2n\ge2, with equality for n≤5n\le5, while g(6)=3g(6)=3; so Moser's bound is exact except at n=3tn=3t with t≥2t\ge2. The proof finds the starting vertex x0x_0 of a run of length at least ⌊(n+3)/3⌋\lfloor(n+3)/3\rfloor among the vertices on the smallest circle enclosing the polygon or the vertices adjacent to them, and the paper notes that since that circle can be found in O(n)O(n) time, a similar result holds for finding such a run.

Read depth. Claims checked: the statement, the definitions it uses, the example of Section 2 and the argument of Section 3 were read clause by clause on the printed pages 111--116. Nothing here is independently reviewed.

Proof pointer

Upper bound, Section 2 (pp. 112--113), for n≥5n\ge5. Start from an isosceles triangle abcabc with apex angle β<π/4\beta<\pi/4 at bb, add a vertex xx just above aa near line abab and a vertex yy just above cc near line cbcb, and choose nonnegative integers A,BA,B with 2A+B=n−52A+B=n-5: BB vertices go near the middle of acac, symmetric about the axis of the triangle, and 2A2A near bb, half beside each of baba and bcbc, all placed so that the polygon stays convex. The paper reads off that the longest run has length max⁡{A+2,B+2}\max\{A+2,B+2\}, and minimizing over the admissible (A,B)(A,B) gives ⌊n/3⌋+1\lfloor n/3\rfloor+1 (p. 113). For n=4n=4 the upper bound is the equality with Moser's bound recorded on p. 112.

Lower bound, Section 3 (pp. 113--115). Let CC be the smallest circle enclosing the polygon PP. For two vertices x,yx,y on CC and a closed circular sector cut off by the chord xyxy that is at most a half-disk, the vertices of PP in it, in order from xx to yy, form a run from xx and, reversed, a run from yy (an extension of Moser's Lemma 3, attributed to Moser's 1952 paper). If only two vertices lie on CC they span a diameter, and this gives g(P)≥⌊(n+1)/2⌋g(P)\ge\lfloor(n+1)/2\rfloor. Otherwise three vertices a,b,ca,b,c on CC span a triangle with no angle above π/2\pi/2, the polygon lies in the three caps cut off by its sides, one cap holds at least ⌊(n+5)/3⌋\lfloor(n+5)/3\rfloor vertices, and g(P)≥⌊(n+2)/3⌋g(P)\ge\lfloor(n+2)/3\rfloor. This equals ⌊(n+3)/3⌋\lfloor(n+3)/3\rfloor unless 3∣n3\mid n. For n=3tn=3t with t≥2t\ge2 the proof assumes no run of length t+1t+1, so each cap holds exactly t−1t-1 vertices besides a,b,ca,b,c; it takes α\alpha the largest angle of abcabc, so α≥π/3\alpha\ge\pi/3, lets xx and yy be the neighbours of aa, shows first that d(x,y)d(x,y) exceeds both d(x,a)d(x,a) and d(y,a)d(y,a), and then uses the first vertices where the runs from yy clockwise and from xx counterclockwise must stop to derive a cyclic left-to-right order xx before xx, a contradiction (p. 115).

Dependencies

Within the paper: the example of Section 2 and the argument of Section 3. Outside it: L. Moser, On the different distances determined by n points, Amer. Math. Monthly 59 (1952), 85--91, for his Lemma 3 and the smallest enclosing circle argument that the paper extends.

Bears on

  • Problem 982: a run of length kk from x0x_0 gives kk distinct distances from x0x_0, and the paper states the resulting bound f(n)≥⌊(n+3)/3⌋f(n)\ge\lfloor(n+3)/3\rfloor for n≥4n\ge4 on p. 116, paged on inequality_p116; the run theorem itself concerns runs, not the problem's count of distinct distances.