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General Balanced Subdivision of Two Sets of Points in the Plane

机译:平面中两组点的一般平衡细分

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摘要

Let R and B be two disjoint sets of red points and blue points, respectively, in the plane such that no three points of R U B are co-linear. Suppose ag ≤ |R| ≤ (a+1)g, bg ≤ |B| ≤ (b+1)g. Then without loss of generality, we can express |R| = a(g_1 + g_2) + (a + 1)g_3, |B| = bg_1 + (b + 1)(g_2 + g_3), where g = g_1 + g_2 + g_3, g_1 ≥ 0, g_2 ≥ 0, g_3 ≥ 0 and g_1 + g_2 +g_3 ≥ 1. We show that the plane can be subdivided into g disjoint convex polygons X_1∪…∪X_(g1) ∪Y_1∪…∪Y_(92)∪Z_1∪…∪ Z_(g3) such that every X_i contains a red points and b blue points, every Y_i contains a red points and b+1 blue points and every Z_i contains a + 1 red points and 6+1 blue points.
机译:令R和B分别是平面上的两个不相交的红点和蓝点集合,以使R U B的三个点都不共线。假设ag≤| R | ≤(a + 1)g,bg≤| B | ≤(b + 1)g。然后,不失一般性,我们可以表示| R |。 = a(g_1 + g_2)+(a +1)g_3,| B | = bg_1 +(b +1)(g_2 + g_3),其中g = g_1 + g_2 + g_3,g_1≥0,g_2≥0,g_3≥0且g_1 + g_2 + g_3≥1.我们证明了平面可以是细分为g个不相交的凸多边形X_1∪…∪X_(g1)∪Y_1∪…∪Y_(92)∪Z_1∪…∪Z_(g3),这样每个X_i包含一个红色点和b个蓝色点,每个Y_i包含一个红色点点和b + 1个蓝色点,每个Z_i包含+ 1个红色点和6 + 1个蓝色点。

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