Balanced intervals of two stes of points on a line or circle

Balanced In e als o Two Se s o Poin s on a line o ci cle A sushi Kaneko aand M. Kano b aDepa men o Compu e Science and Communica ion Enginee ing Kogakuin Uni e si y, Nishi-Shinjuku, Shinjuku-ku, Tokyo, 163-8677, Japan bDepa men o Compu e and In o ma ion Sciences Iba aki Uni e si y, Hi achi, Iba aki, 316-8511, Japan [email p o ec ed]a aki.ac.jp h p://go ogo o.cis.iba aki.ac.jp Abs ac Le n, m, k, h be posi i e in ege s such ha 1 ≤n≤m, 1 ≤k≤nand 1 ≤h≤m. Then we gi e a necessa y and su icien condi ion o e e y con igu a ion wi h n ed poin s and mblue poin s on a line o ci cle o ha e an in e al con aining p ecisely k ed poin s and hblue poin s. Key wo ds: balanced in e al, in e al, wo se s o poin s, line, ci cle 1. A balanced in e al on a line In his sec ion we shall p o e he ollowing he- o em. Theo em 1 Le n, m, k, h be in ege s such ha 1≤n≤m,1≤k≤nand 1≤h≤m. Then o any n ed poin s and mblue poin s on a line in gene al posi ion (i.e., no wo poin s lie on he same posi ion.), he e exis s an in e al ha con- ains p ecisely k ed poin s and hblue poin s i and only i j n k+ 1k+ 1(h−1) < m < j n k−1k(h+ 1), (1) whe e he igh mos e m is an in ini e numbe when k= 1. We begin wi h an example o ou heo em. Con- side a con igu a ion consis ing o 10 ed poin s and 20 blue poin s on a line in gene al posi ion. Then by he abo e heo em, we can easily show ha i k∈ {1,2,3,5,10}, hen such a con igu a ion has an in e al con aining exac ly k ed poin s and 2k blue poin s; o he wise (i.e., k∈ {4,6,7,8,9}) he e exis a con igu a ion ha has no such an in e al (Fig. 1). We call an in e al ha con ains gi en numbe o ed poin s and blue poin s a balanced in e al. (a): (b): Red poin s = Blue poin s = Fig. 1. (a): An in e al con aining 3 ed poin s and 6 blue poin s; (b): A con igu a ion ha has no in e al con aining exac ly 4 ed poin s and 8 blue poin s. Theo em 1 is an easy consequence o he ollow- ing i e lemmas. Fo a con igu a ion wi h ed and blue poin s on a line, we deno e by Rand B he se s o ed poin s and blue poin s, espec i ely. A con igu a ion X wi h n ed poin s and mblue poin s on he line is exp essed as {x1} ∪ {x2} ∪ · · · ∪ {xn+m}, 20 h EWCG Se ille, Spain (2004) 20 h Eu opean Wo kshop on Compu a ional Geome y whe e each xideno es a ed poin o a blue poin o de ed om le o igh . The con igu a ion Xis also exp essed as R(1) ∪B(1) ∪ · · · ∪ R(s)∪B(s), whe e R(i) and B(i) deno e disjoin subse s o R and B, espec i ely, and some o hem may be emp y se s. Fo a se Y, we deno e by |Y| he ca - dinali y o Y. Lemma 2 I m≤j n k+ 1k+ 1(h−1),(2) hen he e exis s a con igu a ion wi h n ed poin s and mblue poin s ha has no in e al con aining exac ly k ed poin s and hblue poin s. PROOF. Le =⌊n k−1⌋. Then n≤( +1)(k−1), and m≥ (h+ 1) by (4). Hence we can ob ain he ollowing con igu a ion wi h n ed poin s and m blue poin s: R(1) ∪B(1) ∪ · · · ∪ R( + 1) ∪B( + 1), whe e |R(i)| ≤ k−1 o e e y 1 ≤i≤ +1, |R(1)∪ · · ·∪R( +1)|=n,|B(i)|=h+1 o e e y 1 ≤i≤ ,|B( + 1)|=m−(h+ 1) ≥0 and |B(1) ∪ · · · ∪ B( + 1)|=m. Then his con igu a ion ob iously has no in e al con aining exac ly k ed poin s and hblue poin s since e e y in e al con aining k ed poin s mus include B(j) o some 1 ≤j≤ . Lemma 3 I m > j n k+ 1k+ 1(h−1),(3) hen e e y con igu a ion wi h n ed poin s and m blue poin s on a line has an in e al con aining ex- ac ly k ed poin s and a leas hblue poin s. PROOF. Le =⌊n k+1 ⌋. Le Xbe a con igu a ion wi h n ed poin s and mblue poin s. Suppose ha Xhas no desi ed in e al. Namely, we assume ha e e y in e al con aining exac ly k ed poin s has a mos h−1 blue poin s. Le 1, 2,··· , nbe he ed poin s o Xo - de ed om le o igh . Fo in ege s 1 ≤i < j≤n, le I(i, j) deno e an open in e al ( i, j), and le B(i, j) deno e he se o blue poin s con- ained in I(i, j). Fu he mo e, B(−∞, i) deno es he se o blue poin s con ained in he open in e al (−∞, i), and B(i, ∞) is de ined analogously. Then o any in ege 1 ≤s≤ −1, I(s(k+1),(s+1)(k+ 1)) con ains exac ly k ed poin s { j|s(k+1)+1 ≤ j≤(s+1)(k+1)−1)}, and hus |B(s(k+1),(s+ 1)(k+ 1))| ≤ h−1 by ou assump ion. Simila ly, an open in e al (−∞, k+1) con ains exac ly k ed poin s, and hus |B(−∞, k + 1)| ≤ h−1. Mo e- o e , since n < ( + 1)(k+ 1), I( (k+ 1),∞) has a mos k ed poin s, and hus B( (k+1),∞)≤h−1. The e o e |B| ≤ |B(−∞, k + 1) ∪B(k+ 1,2(k+ 1)) ∪ · · · ∪B( (k+ 1),∞)| ≤( + 1)(h−1). This con adic s (3). Consequen ly he lemma is p o ed. Lemma 4 I 2≤kand m≥jn k−1k(h+ 1),(4) hen he e exis s a con igu a ion wi h n ed poin s and mblue poin s on a line ha has no in e al con aining exac ly k ed poin s and hblue poin s. PROOF. Le =⌊n k−1⌋. Then n≤( +1)(k−1), and m≥ (h+ 1) by (4). Hence we can ob ain he ollowing con igu a ion wi h n ed poin s and m blue poin s: R(1) ∪B(1) ∪ · · · ∪ R( + 1) ∪B( + 1), whe e |R(i)| ≤ k−1 o e e y 1 ≤i≤ +1, |R(1)∪ · · ·∪R( +1)|=n,|B(i)|=h+1 o e e y 1 ≤i≤ ,|B( + 1)|=m−(h+ 1) ≥0 and |B(1) ∪ · · · ∪ B( + 1)|=m. Then his con igu a ion ob iously has no in e al con aining exac ly k ed poin s and hblue poin s since e e y in e al con aining k ed poin s mus include B(j) o some 1 ≤j≤ . Lemma 5 I 2≤kand m < jn k−1k(h+ 1),(5) hen e e y con igu a ion wi h n ed poin s and m blue poin s on a line has an in e al con aining ex- ac ly k ed poin s and a mos hblue poin s. PROOF. Le =⌊n k−1⌋. Le Xbe a con igu a- ion wi h n ed poin s and mblue poin s. Suppose ha Xhas no desi ed in e al. Namely, we assume Ma ch 25-26, 2004 Se ille (Spain) ha e e y in e al con aining exac ly k ed poin s has a leas h+ 1 blue poin s. Le 1, 2,··· , nbe he ed poin s o Xo de ed om le o igh . Fo in ege s 1 ≤i < j ≤n, le I[i, j], deno e a closed in e al [ i, j], and le B′(i, j) deno e he se o blue poin s con ained in I[i, j]. Then o any in ege 0 ≤s≤ −2, I[k+s(k− 1), k + (s+ 1)(k−1)] con ains exa ly k ed poin s { j|k+s(k−1) ≤j≤k+ (s+ 1)(k−1))}, and hus |B′(k+s(k−1), k + (s+ 1)(k−1))| ≥ h+ 1 by ou assump ion. Simila ly, we ha e |B′(1, k)| ≥ h+ 1. The e o e |B| ≥ |B′(1, k)∪B′(k, k + (k−1)) ∪ · · · ∪B′(k+ ( −2)(k−1), k + ( −1)(k−1))| ≥ (h+ 1). This con adic s (5). Consequen ly he lemma is p o ed. Lemma 6 Conside a con igu a ion wi h n ed poin s and mblue poin s on a line. Suppose ha he e exis s wo in e als Iand Jsuch ha bo h I and Jcon ain exac ly k ed poin s espec i ely, I con ains a mos hblue poin s, and ha Jcon ains a leas hblue poin s. Then he e exis s an in e al ha con ains exac ly k ed poin s and hblue poin s. PROOF. I he se s o ed poin s con ained in I and J, espec i ely, a e he same, hen he lemma immedia ely ollows. Thus we may assume ha I∩ R6=J∩R, whe e Rdeno e he se o n ed poin s. Wi hou loss o gene ali y, we may assume ha he le mos ed poin o Ilies o he le o J. We shall show ha we can mo e I o Js ep by s ep in such a way ha he numbe o ed poin s is a cons an kand he numbe o blue poin s changes ±1 a each s ep. We i s emo e he blue poin s le o he le mos ed poin o Ione by one, and hen add he consecu i e blue poin s lying o he igh o Ione by one, and deno e he esul ing in- e al by I1(Fig. 3). We nex simul aneously e- mo e he le mos ed poin o I1and add he ed poin lying o he igh o I1, and ge an in e al I2, which also con ains exac ly k ed poin s and whose blue poin s a e he same as hose in I1(Fig. 3). By epea ing his p ocedu e, we can ge an in e al whose ed poin se is equal o ha o J. The e- o e, we can mo e I o Jin he desi ed way. Con- sequen ly, we can ind he equi ed in e al, which con ains exac ly k ed poin s and hblue poin s. 2. A balanced in e al on a ci cle (a) (b) Red poin s = Blue poin s = Fig. 2. (a): An in e al con aining 4 ed poin s and 8 blue poin s; (b): A con igu a ion ha has no in e al con aining exac ly 4 ed poin s and 5 blue poin s. In his sec ion, we conside he ollowing heo- em, and gi e i s example in Figu e e ig:2 Theo em 7 Le n, m, k, h be in ege s such ha 1≤n≤m,1≤k≤nand 1≤h≤m. Then o any n ed poin s and mblue poin s on a ci cle in gene al posi ion (i.e., no wo poin s lie on he same posi ion.), he e exis s an in e al ha con- 20 h Eu opean Wo kshop on Compu a ional Geome y ains p ecisely k ed poin s and hblue poin s i and only i n k+ 1(h−1) < m < n k−1(h+ 1),(6) whe e he igh mos e m is an in ini e numbe when k= 1. Theo em 8 can be p o ed by showing simila lemmas as in he case o line. We conclude he pape by he nex conjec u e. Conjec u e 8 Le n, m, k, h be in ege s such ha 1≤n≤m,1≤k≤nand 1≤h≤m. Then o any n ed poin s and mblue poin s in he plane in gene al posi ion (i.e., no h ee poin s lie on he same.), he e exis s a wedge ha con ains p ecisely k ed poin s and hblue poin s i and only i n k+ 2(h−1) < m < n k−2(h+ 1),(7) whe e he igh mos e m is an in ini e numbe when k= 1. P 1 2 P (a) (b) 1 2 Fig. 3. Wedges con aining 4 ed poin s and 8 blue poin s. Re e ences [1] B´a ´any, I., and Ma ouˇsek, J.; Simul aneous pa i ions o measu es by k- ans, Disc e e Compu . Geom. 25 (2001)317–334. [2] Goodman, J. and O’Rou ke, J.; Handbook o Disc e e and Compu a ional Geome y, CRC P ess, (1997) . [3] Kaneko, A. and Kano, M.; Disc e e geome y on ed and blue poin s in he plane — A su ey, Disc e e and Compu a ional Geome y・The Goodman-Pollack Fes sch i ・Wi h con ibu ions by nume ous expe s (Sp inge ) (2003) 551-570.