Vol. 14 No. CH INESE QUARTERLY JOURNAL OF M ATHEM ATICS June 1999 On the M in imum Spann ing Tree Determ ined by n Poin ts in the Un it Square Ye J ichang ( ) Xu Yinfeng ( ) Xu Chengxian ( ) (X iπan J iaotong U niversity, X iπan, 710049) Abstract L et P n be a set of n points in the unit square S, l (P n) denoe the length of the m inim um spanning tree of P n, and C n = m axl (P n), n = P n Α S, 3, In th is paper, the exact value of C n for n =, 3, 4 and the corresponding configurations are given. A dditionally, the conjectures of the configuration for n=, 6, 7, 8, 9 are p roposed. Keywords m inim um spanning tree; m axim in p roblem; configuration 1. In troduction A m in im um spann ing tree (M ST ) is w idely app lied to the field s of com pu ter, comm unication, nerwork and so on. M any results have been obtaines, but few of them deal w ith the worstcase analysis for the given finite region. In fact, it is a m axim in p roblem ( see [1 ] [3 ]). T h is paper is devoted to the worstcase of M ST in the unit square. L et S denote the unit square, P n the set of n point in S, l (P n) be the length of a M ST of P ṅ T he distance betw een two points is of Euclidean sense. O ur p roblem is to determ ine C n defined as fo llow s C n = m ax l (P ) P n Α S n, n =, 3,. and the point set location of P 3 n for w h ich C n = l (P 3 n ), P 3 n Α S is called the configuration of C ṅ L et T n demote the set of all possible spanning trees w ith vertices P n and the length of t for t T n, l (t),w e have C n = m ax m in l (t) P n Α S t T n T 3 n is called an op tim al tree if the length of T 3 n equals C ṅ discussed. T h is paper is organized as follow s: in Section, the configration of C n in the general case is T he special cases for n=, 3, 4 are investigated in Section 3. F inally, the conjectures for n=, 6, 7, 8, 9 are p roposed in Section 4. Received Sep ṫ 8, 1998.
N o. ON TH E M IN IM UM SPANN IN G TR EE D ET ERM IN ED BY N PO IN T S 77. Genera l Ca se L et P 3 n Α S, C n= l (P 3 n ), CH (P 3 n ) be the convex hull of P 3 n and V (P 3 n ) = P 3 n CH (P 3 n ). Theorem 1 O n every edge of S there m ust be at least one point of P 3 n. Proof A s show n in F ig. 1 (a), suppose that no points in P 3 n is on A D and p 3 in P 3 n is the nearest point from AD. L et straigh t line lp 3 be parallel to DC and P θ 3 be the crossing point of AD and lp 3. If w e only move p 3 to P θ 3, it is obvious that l (P 3 n ) < l (P 3 n {p θ3 }g{p 3 }). T h is is a contradiction w ith the definition of P 3 n. F ig. 1 Theorem Points in V (P 3 n ) m ust be on the boundary of S. Proof Suppose that p 3 V (P 3 n ) and p 3 and p j be the two points in V (P 3 n ) that connect w ith p 3 is an interior point of S. By T heorem 1, let p i in CH (P 3 n ) and on the boundary of S. L et straigh t line lp 3 be the bisector of angle p ip 3 p j. If w e only move p 3 in direction as show n in F ig. (b). W rite p θ3 the crossing point that lp 3 and S, w e have l (P 3 n ) < l (P 3 n p θ3 gp 3 ), w h ich is a contradiction w ith the definition of P 3 n. Theorem 3 C n+ 1 C ṅ It is an obvioud conclusion, so the p roof is om itted. Theorem 4 lim C n n=. Proof D ivide the unit square into ( n - 1 ) sm all equal squares w ith edge lengh 1g(n- 1). T he total num ber of vertices of all sm all square is n. Consider them ST w ith the n vetices P n, it is obviius that l (P 3 n ) is n+ 1. It follow s from T heorem 3 that n lim C n=. T herefore l (P 3 n ) Ε l (P n ) = n + 1. 3. Spec ia l Ca ses T he objective of th is section is to determ ine the lication of P 3 n for n=, 3, 4.
CH 78 IN ESE QUA R T ERL Y JOU RNAL O F M A TH EM A T ICS V ol. 14 F ig. Theorem For n=, 3, 4 the location of P 3 n is show n in F ig.. Proof For n= the location is trivial and C =. It is sufficiency only to p rove n= 3 and n= 4. F irst consider the case n= 3,w e shall p rove that the location P 3 3 as show n in F ig. (b) is op tim al and C 3= 1+. F ig. 3 Form T heorem 1 and T heorem w e know that CH (P 3 3 ) is a triangle and there m ust be one point located at a corner of the square. L et pointsa, v, v 3 be in P 3 3, as show n in F ig. 3 ( a). W e claim that triangle A v v 3 is an iso sceles triangle. Conversely suppo se that triangle A v v 3 is no t an iso sceles triangle. If vv 3 is the longest edge, i. e. g A v3 g < g vv 3 g and gva g< gvv 3g (ga B g denotes the Euclidean distance betw een A and B ), then l ( P 3 3 ) = ga v 3g+ ga v g. M oving the point v 3 on the edge D C to v 3 sligh tly enough w e get P θ 3 3 = {A, v, v 3 3 } such that ga v 3 3 g gv 3 3 v g and gva g gv 3 3 v g. So w e have l (P θ 3 3 ) > l (P 3 3 ). T h is is a contradiction w ith the definition of P 3 3. Sim ilar w e can give a contradiction that either A v or A v 3 is the longest one. So the triangle A v v 3 is an isosceles triangle. Suppose ga v 3 g = ga v g, w e can p rove that triangle A v v 3 is an equilateral triangle and l (P 3 3 ) = ( 6 - ). If point v 3 is in D E, as show n in F ig. 3 (b), it is obvious that gvv 3g > ga v 3g = ga v g. H ence,w e have
N o. ON TH E M IN IM UM SPANN IN G TR EE D ET ERM IN ED BY N PO IN T S 79 L et gd E g = gb E g = - EC, as show n in F ig. 3 (c), then ga v 3g + ga v g < ga E g + ga E g = ( 6 - ). 3, then ga E g = ga E g + g E E g = 6 -. If point v 3 is in gvv 3g < ga v 3g = ga v g, gd v 3g = gb v g > gd E g = gb E g = - 3, ga v 3g = (1 + (gd E g + ge v 3g) ) 1g, ga E g = (1 + gd E g ) 1g, ga E g + ge E g - ga v 3g - gvv 3g = (ge E g - gvv 3g) - (ga v 3g - ga E g) = ge v 3g - > ge v 3g - > ge v 3g ( - = ge v 3g ( - gd E g g ge v 3g + ge v 3g 1 + gd E g + 1 + (gd E g + ge v 3g) ge v 3g (gd E g + ge v 3g) 1 + gd E g 1 (gd E g + ge v 3g) ) 1 ( - 3 + gd v 3g) ) ge v 3g ( - 3-3 ) > ge v 3g ( - 1) > 0. So,w e have ga v 3g+ gv v 3g< ga E g+ ge E g= ( 6 - ). Suppose that ga v 3 g = g v v 3 g. F ig. 4 A s show n in F ig. 4, the definition of point E is as above and point M is the m idpoint of D C. If point v is on D E, w e have ga v 3g = gvv 3g < ga v g, ga v 3g < ga E g, ga v g + gvv 3g < ga E g + ga E g. If point v 3 is on EM, as show n in F ig. 4 (b), then ga v 3g= gv 3v g > ga v g, w e shall show that ga v 3g+ ga v g < gam g+ ga B g= 1+ L et gv3m g= y,w e have 0 < y < 3 -. 3, gd v 3g = 1 - y, gv 3C g = 1 + y, ga v 3g = 1 + gd v 3g = 1 + ( 1 - y ) = 1-4y + 4y, gcv g = ga v 3g - gv3cg = 1 + ( 1 - y ) - ( 1 + y ) = 1 - y,
CH 80 IN ESE QUA R T ERL Y JOU RNAL O F M A TH EM A T ICS V ol. 14 N o ticing the inequality w e get gam g - ga v 3g = gvb g = 1 - gcv g = 1-1 - y, ga v g = 1 + gb v g = 1 + (1-1 - y ) = 3 - y - 1 - y. 1 - y > 1 - y - y, 0 < y < 3 - - - 4y + 4y = 3, 4y (1 - y ) ( + - 4y + 4y ) > y (1 - y ), ga v g - ga B g = 3 - y - 1 - y - 1 = - y - 1 - y 1 + 3 - y - 1 - y < - y - 1 - y = 1 - y - 1 - y < 1 - y - (1 - y - y ) = y < y (1 - y ) So, gam g- ga v 3g> ga v g- ga B g, i. e. ga v g+ ga v 3g< gam g+ ga B g= 1+ O bserving that ( 6 - ) < 1+.. ) is follow s from above discussion that the location of P 3 3 is as show n in F ig. (b) and C 3= 1+. For n= 4 w e shall p rove that the location of P 3 4 is as show n in F ig. (c) and C 4= 3. square. Itπs know n from T heo rem 1 that there are at least tw o po in ts on the boundary of the F ig. F irst w e suppose that there are two points of P 3 4 on the boundary. By T heorem 1 and w e know that the two pointsm ust coincide w ith A and C as show n F ig. (a) and the other two points of P 3 4 are on the line segm ent A C. H ence l (P 3 4 ) < 3. Secondly, suppose that there are th ree points on the boundary of S. F rom T heorem 1 and,w e know that the hull of P 3 4 is a triangle and there is one point of P 3 4 in the interior of the
N o. ON TH E M IN IM UM SPANN IN G TR EE D ET ERM IN ED BY N PO IN T S 81 triangle as F ig. (b). L emma Given any triangle A B C, such that ga B g gb C g ga C g and D is in the interior of the triangle A B C. T hen the follow ing inequality holdṡ ga D g + gb D g + gcd g ga B g. Proof A s show n F ig. (c) w e construct an ellip se th rough point D w ith the foci at points A and B and a circle w ith the center C and the radium gcd g. W rite D and D the in tersection s of CA and the ellip se and CB and the circle respectively. It is obviou s that gcd g< gcd g< gca g. So ga D g + gb D g + gcd g ga D g + gb D g + gd C g = ga Cg + gb D g < ga Cg + ga B g ga B g. By the L emm a and the fact that the longest edge of the triangle A v 3v in F ig. (b) is less than,w e get l (P 3 4 ) < < 3. F inally suppose that the four points of P 3 4 are all on boundary of the unit square and there is at least one point on each edge. T he convex hull of P 3 4 is a quadrilateral or a degenerate quadrilateral w h ich is a righ t triangle. L et l1 l l3 l4 denote the edge length s of the quadrilateral. O bviously l1+ l+ l3+ l4 4. i. e. l1+ l+ l3 3. If l4 1 then 4 l1 + l + l3 + l4 l1 + l + l3 + 1, If l4< 1 then l1+ l+ l3< 3. So l1+ l+ l3= 3 if and only if l1= l= l3= l4= 1. T h is imp lies that the location of P 3 4 is F ig. (c). 4. Con jectures for n=, 6, 7, 8, 9 F ig. 1 W e guess that the op tim al locations of P 3 n (n=, 6, 7, 8, 9) as show n in F ig. 6.
CH 8 IN ESE QUA R T ERL Y JOU RNAL O F M A TH EM A T ICS V ol. 14 n=,a E = B E = A B, C = + 3sin (Πg1) = 3. 0376 ; n= 6,A E = B E = A B, EC = FC, C 6= 3. 1313 ; n= 7,D C = D E = GC = GE = GB = GF = FE = B C, C 7= 3. 3383 ; n = 8, A F = FG = GA = GB = GH = H B = H C = H E = EH = ED = D F = E F, C 8= 3. 6346631. n= 9, E, F, G, H are the m idpoints of CD, DA, A B, B C respectively and I is the crossing point w ith EG and H F, C 9= 4. T he values of C n (n=, 3,, 9) are listed in T able 1. Table 1 n 3 4 6 7 8 9 C n 1. 414. 1180 3 3. 03. 131 3. 338 3. 634 4. Conclusion and Remark W e have p resented four general results for the w hole construction on M axim um M ST determ ined by n points in the unit square ( see T heorem 1 to 4). But it is not enough to understand the location of P 3 n. Som e related p roblem s rem ain open. 1) W hat is the asymp totic value of C n? ) A re the longest edges in op tim al tree unique. 3) Does there exist an op tim al tree in w h ich the degree of the nodes is no more than 4? 4) How is a m axim um M ST determ ined by n points in other k inds of regions, such as unit circle, equ ilateral triangle an s so on? References [ 1 ] J. Schaer, The densest packing of 9 circles in a square, Canad M ath. Bull. 8 (196) 73-77. [ ] G. V alette, A better packing of the equal circles in a square, NorthHolland, D iscrete M ath. 76 (1989) 7-9. [ 3 ] D. Z. Du and F. K. Hw ang, A n new bound for the steiner ratio, T ranṡ Am eṙ M ath. Soc., 78 (1983) 137-148. [ 4 ] M ichel Gondran and M ichek M inoux, Graphs and algorithm s, John W iley and Sons, 1984.