- Email: [email protected]
On a “Complementary Problem” of Courant and Robbins
Location Science 6 (1998) 337±354
On a ``Complementary Problem'' of Courant and Robbins Jakob Krarup
*
Professor of Computer Science and Operation Research, Department of Computer Science, University of Copenhagen, Universitetsparken 1, DK-2100 Copenhagen, Denmark
Abstract For a given triangle ABC with \A > 120°, the Simpson variant of TorricelliÕs geometrical construction for solving a problem, allegedly ®rst formulated by Fermat in the early 1600s, will identify a point which incorrectly was claimed by Courant and Robbins (Courant, R., Robbins, H., 1941. What is Mathematics? Oxford University Press, Oxford.) to solve the socalled Complementary problem: minfBX CX ÿ AX : X 2 R2 g. The correct solution for any triangle is provided here. Ó 1999 Elsevier Science Ltd. All rights reserved.
1. Introduction For a given triangle ABC, the Fermat problem asks for a fourth point T minimizing the sum of the distances to the three given points. For any point X in the plane, let f X AX BX CX . A formal statement of the problem is then FERMAT(ABC): Find a point T such that f T AT BT CT minff X : X 2 R2 g: FERMAT(ABC) was solved geometrically by Torricelli around 1640. The Simpson variant (Simpson, 1750) of TorricelliÕs construction of T is illustrated in Fig. 1. ABC is the given triangle. ABC1 , AB1 C, A1 BC are the outward equilateral triangles. The three so-called Simpson lines AA1 , BB1 , CC1 do all intersect at T and are all at the same length f(T). The angle between any pair of neighboring Simpson lines is 60°. T is accordingly the point from which each of the three sides AB, AC, BC subtends an angle of 120°. A surprisingly simple proof of these and related properties can be found in Krarup and Vajda (1997).
*
E-mail. [email protected].
0966-8349/99/$ ± see front matter Ó 1999 Elsevier Science Ltd. All rights reserved. PII: S 0 9 6 6 - 8 3 4 9 ( 9 8 ) 0 0 0 4 3 - 6
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Fig. 1. SimpsonÕs construction of the fourth point T.
SimpsonÕs construction, however, does only provide the desired result if the given triangle has no angle exceeding 120°. Assume therefore that one angle, say \A, is strictly greater than 120°. As was noted by Torricelli but ®rst proved in Heinen (1834), A itself will then solve FERMAT(ABC) whereas the point T 0 at which the Simpson lines BB1 , CC1 , and the line containing AA1 now intersect will appear outside the given triangle as illustrated in Fig. 2. T 0 does not certainly solve FERMAT(ABC) but is seen to represent an optimal solution to FERMAT(AB1 C1 ). What else can in this case be deducted from the Torricelli±Simpson construction?
Fig. 2. The Simpson construction for triangles ABC, \A > 120 .
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In the book What is Mathematics?'' (Courant and Robbins, 1941) it is asserted that T 0 will solve what is termed ``A Complementary Problem'' which is to minimize the expression fBX CX ÿ AX : X 2 R2 g. The proof is ``. . . left as an exercise for the reader''. Actually, the interpretation by Courant and Robbins of T 0 is wrong. Correct geometrical and analytical solutions to the Complementary Problem for any data instance ABC will be provided in Section 2. The Complementary Problem can be viewed as a special case of the Weber problem provided that both positive and negative weights are allowed for. Aspects of this observation are brie¯y discussed in the concluding section concurrently with reference to the relevant literature.
2. Geometrical and analytical solutions Let us for any point X in the plane de®ne the objective function g(X), g X BX CX ÿ AX : For a given triangle ABC, let the Complementary Problem, henceforth referred to as CP(ABC), be the problem of ®nding a point V minimizing g(X), that is, CP(ABC): Find V such that g V BV CV ÿ AV minfg X : X 2 R2 g: To see that T 0 in Fig. 2 does not solve CP(ABC) is fairly straightforward. Disregarding the degenerate cases where A, B, C are collinear, T 0 and A must always be on the same side of the line containing B and C. If T 00 is the mirror image of T 0 with respect to that line, BT 0 CT 0 BT 00 CT 00 whereas AT 00 > AT 0 or g T 00 < g T 0 . T 00 is thus a better solution to CP(ABC) than T 0 . To solve CP(ABC) for any triangle, we shall throughout the remainder of the paper and without loss of generality ®rst assume that the given triangle is labeled such that AB 6 AC. Furthermore, no less than six cases in total need be distinguished. The determinants separating these six cases are: A, B, C are collinear or not, AB < AC or AB AC, \C 6 60° or \C > 60°. 2.1. A, B, C are non-collinear: the Restricted Complementary Problem Where should the point V minimizing g(X) be found in the general case? Part of the answer can be derived from what may be termed the Restricted Complementary Problem or RCP(ABC) for short. RCP(ABC) is identical to CP(ABC) with the additional constraint AX 6 AC. Let W be the point sought for: RCP(ABC): Find W such that g W BW CW ÿ AW minfg X : X 2 R2 ; AX 6 ACg: Theorem 1. For any triangle ABC with AB 6 AC, RCP(ABC) will either have the uniquely determined optimal solution W C if AB < AC or two optimal solutions, W B and W C, if AB AC.
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Fig. 3. An instance ABC, AB 6 AC, of RCP(ABC).
Proof. Fig. 3 shows a triangle ABC with AB 6 AC. Furthermore, a circle R having A as its center and with radius AC is drawn. B must be either within or on that circle. For any point X in the plane it follows from the triangle inequality applied to DBXC that BX CX P BC. BC is thus a lower bound on the ®rst two terms of the objective function g(X). Similarly, if X is either within or on the circle, AX cannot exceed AC which then is an upper bound on AX (or AC is a lower bound on AX, the third term of g(X)). For any X on the line segment BC, BX CX will equal the lower bound BC. For X C, both the lower bound BC on BX CX and the upper bound AC on AX will be reached. If AB < AC, B is an interior point in the circle and X C is the uniquely determined optimal solution to RCP(ABC) as asserted. For AB AC, both B and C are optimal solutions. h Fig. 3 is supplemented by the point B1 which, together with A and C, forms an equilateral triangle and hence must be on the circle R. Let C1 (not shown in Fig. 3) be the third point on the outward equilateral triangle ABC1 . Since AB 6 AC, C1 is an interior point if AB < AC or otherwise a point on R. No point on the Simpson lines BB1 and CC1 can thus be outside R and T 0 (see Fig. 2) must therefore be an interior point. This is yet another way to prove that Courant and Robbins' claim is incorrect. Theorem 1 will in some cases apply for CP(ABC) as well, for example, in degenerate cases where A, B, C are collinear. However, as will be elaborated on in the sequel, there also exist non-degenerate triangles ABC having one or in®nitely many optimal solutions to CP(ABC) outside the circle R. The idea of the following approach to solving CP(ABC) in general is to identify a set of points A0 which, together with B and C, de®nes a set of ``companion instances'' (not equivalent instances) of FERMAT(A0 BC) such that an optimal solution to CP(ABC) also solves any FERMAT(A0 BC). The solution properties of the latter are thereafter be exploited to obtain the ®nal result. The relationship between CP(ABC) and FERMAT(A0 BC) is established in Theorem 2 as illustrated in Fig. 4. LBC is the line containing B and C (recall that we
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Fig. 4. An illustration of Theorem 2 with V 6 C.
consider here non-degenerate triangles, still labeled such that AB 6 AC). D is the third vertex of an equilateral triangle having BC as one of its sides and placed on the same side of LBC as A. Theorem 2. (i) V solves CP(ABC))V solves FERMAT(A0 BC) where A0 is either V itself or any point on the line L containing A and V such that V is between A and A0 . (ii) If V and C are distinct points, the four points D, A, V, A0 are all collinear. Proof. (i): It follows from Theorem 1 that A itself can only solve instances of both RCP(ABC) and CP(ABC) if A and C coincide. Since A and C are distinct points, A and the optimizing point V must be distinct too and the line L containing A and V is uniquely de®ned. A0 is then either V itself or any point on L such that V is between A and A0 in Fig. 4. Let X be any point in the plane and f X BX CX A0 X . Since V is assumed to solve CP(ABC), g(X) P g(V), or BX CX ÿ AX P BV CV ÿ AV or BX CX P BV CV AX ÿ AV or BX CX A0 X P BV CV AX A0 X ÿ AV
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or f X P BV CV AA0 ÿ AV
since AX A0 X P AA0
or f X P BV CV AV A0 V ÿ AV or f X P BV CV A0 V f V : (ii): Since V and C are assumed to be distinct points, V cannot be within or on the circle R having A as its center and with radius AC (see Fig. 3). It follows then from Theorem 1 that neither A nor B can solve CP(ABC). V cannot be above the line LBC in Fig. 4 since the mirror image of V with respect to that line represents a better solution. Furthermore, neither A nor B can solve FERMAT(A0 BC) which implies that \A0 BC < 120° and \A0 CB < 120°. If FERMAT(A0 BC) has no vertex solution, V must be an interior point in DA0 BC. Hence, both A0 and V must be somewhere on L below or on LBC and within the region marked by heavy lines in Fig. 4. Suppose we choose a point A0 such \BA0 C 6 120°. We know from Fig. 2 that the point, now called V, solving FERMAT(A0 BC) must be on the Simpson line between A0 and D, the third vertex of the outward equilateral triangle having BC as one of its sides. The four points D, A, V, A0 must therefore all be collinear as asserted. h The second part of Theorem 2 tells us how to identify the line L containing A, D, V, A0 . We note ®rst that Theorem 2 applies regardless of whether A and D coincide or not. Thus, unless A and D do coincide (a special case to be dealt with separately later on), L is uniquely determined by A and D. The ground is now prepared for the geometrical construction of V in the general case. For ease of reference, we de®ne a coordinate p system (see pFig. 5) such that the coordinates of A, B, C, are: A : p; q, B : ÿ 12 3; 0, C : 12 3; 0. Without loss of generality, we assume q P 0. BCD is an equilateral triangle. Its circumscribing circle is a unit circle having its center at S with coordinates 0; 12. M is the midpoint between B and D. Since AB 6 AC, p must be non-positive and A can accordingly be anywhere in one of the four regions Q1; . . . Q4 including the lines de®ning them. (i) AB < AC, \C 6 60°. We will show that C is the uniquely determined optimal solution to CP(ABC) if AB < AC and \C 6 60°. If V is not a vertex of triangle ABC, the four points D, A, V, A0 must be collinear in an optimal solution to both CP(ABC) and the corresponding (set of) FERMAT(A0 BC), cf. Theorem 2. V and hence A0 as well must then be either on or below the x-axis. Unless triangle A0 BC has one angle greater than or equal to 120°, in which case V is the corresponding vertex, V is the point from which each of the three sides A0 B,
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Fig. 5. C solves CP(ABC) if AB < AC and \C 6 60 .
A0 C, BC subtends an angle of 120°. Thus, V must lie somewhere on the arc BC not containing D of the circle circumscribing DBCD. Let X be the corresponding set of points, that is, 2 1 2 1; y 6 0g: X f x; y : x y ÿ 2 A 2 Q1: Assume that A is a point in region Q1 (notation: A 2 Q1) then no point on X is outside the circle R (cf. Fig. 3) having A as its centre and with radius AC. Hence, Theorem 1 applies. A 2 Q2: No line containing A and D can intersect X except at B or C if \B or \C equals 60°. V must therefore be one of the vertices of nABC. V A is impossible since A and C are distinct points. B can also be excluded since g B BC ÿ AB < BC ÿ AC g C: Thus, C is optimal and L is the line containing A and C. A 2 Q3: A cannot be on the y-axis since AB < AC. A and D are thus distinct points de®ning the line L which intersects X at a point Y; see Fig. 6. Clearly, Y itself will solve FERMAT(YBC) since \BYC 120°. f(Y) BY CY must equal DY since DY is one of the Simpson lines. We obtain then: g Y ÿ g C BY CY ÿ AY ÿ BC ÿ AC DY ÿ AY ÿ CD ÿ AC AD AC ÿ CD > 0 (ii)
AB < AC,
\C > 60°
since MBCD is equilateral since A and D are distinct points
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Fig. 6. C solves CP(ABC) if AB < AC and A 2 Q3.
A with coordinates p; q must be somewhere in the interior part of Q4, the unbounded shaded area in Fig. 7 excluding the two lines de®ning it; hence, p < 0. Fig. 7 shows how V is determined as the intersection between X and the line L de®ned by A and D. Let a 12 2q ÿ 3=p be the slope of L and let (xV , yV ) be the coordinates of V. (xV , yV ) must then satisfy: yV ÿ 1:5 axV ; yV ÿ 122 x2V 1. These equations have two solutions, one of which, corresponds to the coordinates of D. The other solution is xV ÿ2a= 1 a2 ;
yV 1=2 3 ÿ a2 1 a2
1
(iii) AB AC, \C > 60° The construction in Fig. 7 does also apply when AB AC. The slope a of L, however, is unde®ned since L here coincides with the y-axis so the general expression (1) cannot be used. L intersects X at a point V with coordinates 0; ÿ 12 which, of course, also obtains as the limiting values of (xV , yV ) for a ! 1 in Eq. (1). We have so far covered all cases where the solution to CP(ABC) is uniquely determined. What remain are the cases where AB AC and \C 6 60°. (iv) AB AC, \C < 60° (iv) can be viewed as an extension of (i) for which the same line of reasoning applies. Thus, C is an optimal solution. Moreover, Theorem 1 asserts that not only C but also B is an optimal solution. (v) AB AC, \C 60°: nABC is equilateral
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Fig. 7. V solves CP(ABC) if AB < AC and \C > 60 .
Because the solution to a companion instance of FERMAT(A0 BC) is still uniquely determined, it appears that CP(ABC) in this case has in®nitely many optimal solutions. We prefer therefore to treat AB AC, \C 60° as a limiting case of AB 6 AC, \C P 60°. We have previously used A and D for de®ning the line L containing both V and A0 unless L and X do not intersect. Since MABC now is equilateral, A and D will coincide and L can be any line through A intersecting X. Fig. 8 shows the equilateral triangle ABC and the point Y which is any point on X. (iv) extends easily to (v) and tells us that optimal solutions to CP(ABC) can be found at both B and C with g B g C BC ÿ AC 0. To verify that CP(ABC) actually has in®nitely many optimal solutions, it suces to show that g(Y) 0 for any Y as de®ned or, in other words, that BY CY AY . Since Y also solves FERMAT(YBC), F Y BY CY must equal AY since AY is one of the Simpson lines. The ®ndings can now be summarized in Theorem 3. Theorem 3. For a given triangle ABC of non-collinear points within p a coordinate system p de®ned such that A, B, C have coordinates A : p; q; B : ÿ 12 3; 0 and C : ÿ 12 3; 0 respectively, an optimal solution V to CP(ABC) will be either a single point or a set of points:
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Fig. 8. nABC is equilateral: CP(ABC) has in®nitely many optimal solutions.
Case
The point set V solving CP(ABC)
(i) AB < AC; \C 6 60 (ii) AB < AC; \C > 60 (iii) AB AC;
\C > 60
(iv) AB AC; \C < 60 (v) AB AC; \C 60
V {C} V 2 f xV ; yV g where xV ÿ2a= 1 a2 ; yV 12 3 ÿ a2 = 1 a2 and a 12 2q ÿ 3=p the slope of L V 2 f 0; ÿ 12g V 2 fB; Cg 1 V 2 X f x; y : x2 y ÿ2 2 1; y 6 0g
Disregarding alternate optima, Theorem 3 asserts that C solves CP(ABC) whenever \C 6 60°. The line L is thus the line containing A and C which may (\C 60°) or may not (\C < 60°) contain D as well. For any point A0 on L such that AA0 P AC, C is also seen to solve FERMAT(A0 BC) since \A0 CB P 120°, cf. (Heinen, 1834). Theorem 2 does thus apply in all non-degenerate cases, the only modi®cation being that A, V ( C), and A0 but not D are collinear if \C < 60°. 2.2. A, B, C are collinear To claim that the correct solution to CP(ABC) is provided for any triangle ABC, AB 6 AC, we need also to consider, for the sake of completion, the degenerate cases where A, B, C are collinear. Let X be a point in the plane at a distance d, d P 0, from C.
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1. 2. 3. 4. 5.
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A B C : g X 2d ÿ d d ) V fAg fBg fCg; A B; B 6 C : g X BX d ÿ AX d ) V fCg; A 6 B, A C: cannot occur since AB 6 AC; A 6 B, B C: since AX d P AB, g X 2d ÿ Ax AB P d ) V fCg; A, B, C are distinct points C is between A and B : cannot occur since AB 6 AC; B is between A and C : g X ÿ g C BX d ÿ AX ÿ BC ÿ AC or g X ÿ g C AB BX ÿ AX AB ÿ AB d since BC ÿ AC ÿAB or g X ÿ g C P d since AB BX P AX ) V fCg
A is between B and C: It follows from cases (i) and (iv) in Theorem 3 that V {C} if AB < AC or V fB; Cg if AB AC. For the degenerate cases where A, B, C are collinear, AB 6 AC, we have thus shown that C is the optimal solution to CP(ABC) with B as an alternate optimum if AB AC. 2.3. The ``worst-case'' instance of CP(ABC) The construction in Fig. 7 shows that the Restricted Complementary Problem RCP(ABC) considered in Theorem 1 and the corresponding CP(ABC) can have dierent solutions. How far outside the circle R can we ®nd the point V solving CP(ABC)? To answer this question, we need only to consider the case where \C > 60 . Let L, V and X be as de®ned in Fig. 7 and let V0 be the intersection of L and that part of R which is on the same side of BC as X. It is seen that VV0 attains its maximum value for AB AC in which case V has coordinates 0; ÿ 12. Now, if AC approaches in®nity, the ``bottom part'' of R will approach the straight line BC. In the limit, all points on that line segment will therefore represent optimal solutions to RCP(ABC). p For V and C we ®nd: g V 2 ÿ 12 ÿ AV 0 , g C 3 ÿ AC. Hence, if B and C are distinct, the relative deviation g V ÿ g C=BC is for all triangles ABC bounded p from below by 1 ÿ 12 3 and this bound is approached for AC ! 1. ``\B \C, AC ! 1'' can thus be viewed as a worst-case instance of CP(ABC) in the sense that g V ÿ g C=BC then approaches its maximum value. 3. Notes The Weber Problem, known under several of other names including the more obvious The Euclidean Minisum Distance Facility Location Problem, asks for a point T minimizing the sum of the weighted Euclidean distances from itself to n given points A1 ; . . . ; An , each associated with a given positive weight.
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For all j, let xj ; yj be the coordinates of Aj and let wj , wj > 0, be the corresponding weight. Furthermore, let x; y be the coordinates of T representing the optimal location. The problem can then be stated as ( q) n X 2 2 2 wj x ÿ xj y ÿ yj : WEBER : min f x; y x;y2R2
j1
With weights w1 ; . . . ; wn , WEBER( A1 ; w1 ; . . . ; An ; wn ) appears to be optimally solvable via a mechanical analogue device called the Varignon frame after its originator, Pierre Varignon (1654±1722), and described in his treatise ``Projet d'une mouvelle m ecanique'', published in 1687. Fig. 9 shows the Varignon frame in the general case with n demand points. A board has n holes corresponding to the n ®xed points. The n strings are tied together in a knot at one end, and the loose ends are passed through the holes. The loose ends are attached to physical weights w1 ; . . . ; wn below the board. If the experiment is implemented in a ``perfect environment'', i.e. in®nitely thin strings, very small holes, no friction etc., then the ®nal position of the knot will be at the optimal location sought. To see this, let dj be the Euclidean distance between x; y and xj ; yj and let Hj be the angle between the x-axis and the line containing x; y and xj ; yj .
Fig. 9. The Varignon frame with n demand points.
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An optimal location of x; y can be characterized in two dierent ways. For the symbolic model (2) the partial derivatives of f x; y must both equal zero. For the analogue model, the resultant force exerted on the knot must equal zero. For the partial derivatives we obtain of x; y=ox Rj wj xj ÿ x=dj 0 or Rj wj cosHj 0; of x; y=oy Rj wj yj ÿ y=dj 0 or Rj wj sin Hj 0: The rightmost expressions are recognized as the force components in the x and y directions, respectively. The two requirements to be satis®ed by an optimal solution are thus identical. To see that the Varignon frame also provides the desired result in the case where the ®nal position of the knot coincides with one of the holes is less obvious. As discussed in detail in Wesolowsky (1993) who also formally states the necessary and sucient conditions for the optimum to occur at a hole, the diculty is that the partial derivatives do not exist at the holes. However, a simple argument reveals that a hole is an optimum location if the weight attached to the string through that hole is greater than or equal to the sum of all other weights in the system. With reference to the Varignon frame, the essence of Theorem 2 is that a negative force of ÿ1 can be replaced by an oppositely directed force of 1 in the search for an equilibrium position of the knot. Another observation is illustrated in Fig. 10 which is a sort of extension of Fig. 4. Fig. 10 shows the solution to an instance of FERMAT(A0 BC) where no angle exceeds 120° and where the optimizing point called V is found as the intersection of the arc X of the circle circumscribing nBCD and the line L containing A0 and D. Let LA0 be the half-line of L starting at V and containing A0 . Let LB and LC be analogously de®ned. It is seen that the geometrical construction of V remains unaltered if another position for A0 is chosen on LA0 . Likewise, B and C can be replaced by B0 and
Fig. 10. V solves in®nitely many instances of FERMAT(A0 B0 C0 ) and CP(A"B0 C0 ).
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C0 on LB and LC , respectively, without aecting the position of V. Thus, V solves ``in®nitely many instances'' of FERMAT(A0 B0 C0 ) regardless of the exact positions of the three points on the corresponding half-lines. For triangles A0 BC having no angles exceeding 120°, the same extension to ``in®nitely many instances'' does also apply for CP and allows us, so to speak, to ``reverse'' Theorem 2. Corollary. V solves FERMAT(A0 BC) ) V solves CP(A"BC) where A" is any point on the line L containing A0 and D such that D is between A0 and A". Proof. Suppose that a point W 6 V solves CP(A"BC). By Theorem 2, (ii), A", D, W must be collinear which implies that W must be on L. Since DA0 BC has no angle exceeding 120°, \BVC must equal 120°. W must then be on X as well and cannot be distinct from V. h It is seen that WEBER((A1 , 1), (A2 , 1), (A3 , 1)) is identical to FERMAT(ABC) with A A1 , B A2 , C A3 , that is, FERMAT(ABC) can be considered as the unweighted case of WEBER( ) for n 3. Likewise, for CP(ABC) we can consider CP(ABC) as the problem of locating a single facility V such that it is as close as possible to clients B and C and at the same time as far as possible away from client A. For CP(ABC) we can thus de®ne an equivalent Weber problem denoted by WEBER( A; ÿ1; B; 1; C; 1) where A, B, C represent the n 3 clients to be served and ÿ1, 1, 1 are the corresponding weights. CP(ABC) reduces accordingly to a special case of WEBER( A1 ; w1 ; . . . ; An ; wn ) where n can be any positive number and where each weight wj can be positive or negative. The earliest paper considering WEBER( ) with positive and negative weights appears to be Tellier (1985), who investigates cases with two attraction points and one repulsion point represented by positive and negative weights, respectively. A detailed account of TellierÕs ®ndings is given in Tellier and Polanski (1989). TellierÕs trigonometrical approach is illustrated in Fig. 11. ABC is the given triangle where A is the repulsion point associated with a repulsive force q corresponding to the negative weight wA ÿq. B and C are attraction points associated with positive weights or attractive forces wB , wC , respectively. We assume wB wC P q since the optimal location otherwise is at in®nity, and furthermore, that no attractive force exceeds the sum of the two other forces since the corresponding attraction point otherwise will be the optimal solution. Let V be the optimizing point. Under the tacit assumption that V is the point where the net force is zero, angles (1 2), 1 are determined by the cosine relations, cf. Fig. 11b; the sign should be changed in the ®rst of these expressions in Tellier and Polanski (1989). Note that angles 1 and 2 are in no way related to the positions of A, B, C. To determine where the line containing A and V in Fig. 11a actually is, we need to ®x angle 3 in Fig. 11c such that D and E coincide with angles 1 and 2 preserved as shown. Angles 4, 5, 6 in Fig. 11a are then immediately at hand.
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Fig. 11. The trigonometrical solution to WEBER( A; wA ; B; wB ; C; wC ) (Tellier, 1985).
To work as desired, however, the various angles should be compatible with each other. This requirement is illustrated in Tellier and Polanski (1989) as in Fig. 11d where the repulsion point must belong to one of the regions R1 and R2 or one the lines de®ning them. If the repulsion point is placed in the ``incompatibility zone'', one of the two attraction points will solve the problem and rules are given to decide which one of them. Although CP(ABC) is a special case of the problem studied in Tellier and Polanski (1989) and thus amenable for the trigonometrical solution proposed, our approach does complement their analysis rather than duplicate their eorts, since we provide both the coordinates of V and a geometrical construction.
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FERMAT(ABC) and CP(ABC) may appear to be merely variations of a theme, yet there is one important feature that separates them. FERMAT( ) is a convex problem which implies that there is no local minimum dierent from the global one. CP( ), on the other hand, is in general non-convex as pointed out in Chen et al. (1992), Drezner and Wesolowsky (1991), and Tellier and Polanski (1989). Besides analyzing the properties of the Weber problem in the plane when some of the weights are negative, Drezner and Wesolowsky (1991) devise exact algorithms when the distances are rectilinear or squared Euclidean. If the sum of all weights is positive, it is further shown for the more dicult Euclidean distance problem that the optimizing point must be within a circle slightly larger than the smallest circle containing all demand points. Finally, based on this result, three heuristic procedures are proposed. FERMAT(ABC) and CP(ABC) are both examples of continuous location problems in the sense that the optimizing point sought can be placed anywhere in the plane. Focussing on continuous models, how would one proceed to more complex cases? One possibility is to seek the geometrical constructions extended to cases where ``more than three clients are to be served'' for various combinations of positive and negative weights. The road ahead could then be to move from the triangles previously considered to a square, a rectangle, a general quadrilateral, a pentagon and so forth. As Tellier and Polanski put it (1989): ``The triangle case greatly diers from the more complex polygon case because it involves between-the-forces angles that are uniquely determined. When a polygon has more than three vertices, the angles ensuring an equilibrium of the forces cannot be determined without considering the form of the polygon. In the triangle case, this is possible.'' As regards WEBER( ) with positive and negative weights, (Chen et al., 1992) should not be ignored. WEBER( ) is ®rst shown to be a so-called d.-c. program (that is, a global optimization problem with both the objective function and the constraints expressed as dierences of convex functions) reducible to a problem of concave minimization over a convex set. Based on the theory of d.-c. programming, an exact algorithm is then devised. The analysis is complemented by extensive computational experiments which show that instances of WEBER( ) with n up to 1000 are solvable in at most two minutes. The concluding remark: ``So d.-c. and concave programming appear to be powerful new tools for continuous location theory'' seems to be well justi®ed. The approach for solving CP(ABC), the contents of the present paper, is extended in Jalal and Krarup (1997) to cover any combination of positive and negative weights associated with the vertices of a given triangle. Among the byproducts are surprisingly simple proofs of various geometrical constructions spanning the period 1640±1997. Furthermore, PtolemyÕs Theorem (around A.D. 150) is ``rediscovered'' and an alternative proof of HeinenÕs observation (1834) is given. Alternatively, one can look at discrete formulations where the set of potential sites for the facilities to be placed is ®nite and often represented by the vertices of a
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network. A fruitful way at times is to identify what is termed well-solved special cases. For example, if a problem de®ned on a network is known to be NP-hard in general, it may be polynomially solvable (and hence well-solved) if the underlying network exhibits a special structure. Thus, the network could be, say, a simple path, a tree, or a cycle, or it could be identi®ed in terms of more complex forbidden subgraphs. This is the line of approach taken in Burkard and Krarup (1998) for the 1-median problem in a network with both positive and negative weights. It is shown that the 1-median can be found in linear time if the underlying graph is a cactus, that is, a connected graph in which no two cycles have more than one point in common. The motiviation for the present study was the interest taken in FERMAT(ABC) and the subsequent confrontation with the incorrect solution to the Complementary Problem. As regards realistic applications, however, we believe that models capable of handling arbitrary weights may well provide useful decision support, for example, to companies operating in a competitive environment or to monopolistic companies wishing to increase or decrease the number of existing facilities. It should ®nally be noted that among the papers referred to in the text, we have been unable to retrieve copies of Simpson (1750) and Tellier (1985) or to provide other bibliographical data than those listed below.
Acknowledgements The original proof of the ®rst part of Theorem 2, based on a mechanical analogue device, was not fully convincing although the correct conclusion was drawn. I am indebted to A. Hertz and D. Kobler, Ecole Polytechnique Federale de Lausanne, who provided the present version. Likewise, the critical remarks made by G. Rote, Technische Universit at Graz, are much appreciated. Finally, sincere thanks are due to an anonymous referee whose meticulously prepared report was most valuable. References Burkard, R.E., Krarup, J., 1998. A linear algorithm for the ``pos/neg-weighted'' 1-median problem on a cactus. Computing 60, 193±215. Chen, P.-C. et al., 1992. WeberÂs problem with attraction and repulsion. Journal of Regional Science 32, 467±486. Courant, R., Robbins, H., 1941. What is Mathematics? Oxford University Press, Oxford. Drezner, Z., Wesolowsky, G.O., 1991. The Weber problem on the plane with some negative weights. INFOR 29, 87±99. Heinen, F., 1834. Uber Systeme von Kr aften. Gymnasium zu Cleve, pp. 18±19. Jalal, G., Krarup, J., 1997. Geometrical solution to the Fermat problem with arbitrary weights. Working Paper, DIKU (Department of Computer Science, University of Copenhagen). Krarup, J., Vajda, S., 1997. On Torricelli's geometrical solution to a problem of Fermat. IMA Journal of Mathematics Applied in Business and Industry 8, 215±224. Simpson, T., 1950. The Doctrine and Application of Fluxions. London. Tellier, L.-N., 1985. Economie spatiale: rationalite economique de l'espace habite. Chicoutimi, Quebec: Gaetan Morin.
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Tellier, L.-N., Polanski, B., 1989. The Weber problem: frequency of dierent solution types and extension to repulsive forces and dynamic processes. Journal of Regional Science 29, 387±405. Wesolowsky, G.O., 1993. The Weber problem: history and perspectives. Location Science 1, 5±23.
On a ``Complementary Problem'' of Courant and Robbins Jakob Krarup
*
Professor of Computer Science and Operation Research, Department of Computer Science, University of Copenhagen, Universitetsparken 1, DK-2100 Copenhagen, Denmark
Abstract For a given triangle ABC with \A > 120°, the Simpson variant of TorricelliÕs geometrical construction for solving a problem, allegedly ®rst formulated by Fermat in the early 1600s, will identify a point which incorrectly was claimed by Courant and Robbins (Courant, R., Robbins, H., 1941. What is Mathematics? Oxford University Press, Oxford.) to solve the socalled Complementary problem: minfBX CX ÿ AX : X 2 R2 g. The correct solution for any triangle is provided here. Ó 1999 Elsevier Science Ltd. All rights reserved.
1. Introduction For a given triangle ABC, the Fermat problem asks for a fourth point T minimizing the sum of the distances to the three given points. For any point X in the plane, let f X AX BX CX . A formal statement of the problem is then FERMAT(ABC): Find a point T such that f T AT BT CT minff X : X 2 R2 g: FERMAT(ABC) was solved geometrically by Torricelli around 1640. The Simpson variant (Simpson, 1750) of TorricelliÕs construction of T is illustrated in Fig. 1. ABC is the given triangle. ABC1 , AB1 C, A1 BC are the outward equilateral triangles. The three so-called Simpson lines AA1 , BB1 , CC1 do all intersect at T and are all at the same length f(T). The angle between any pair of neighboring Simpson lines is 60°. T is accordingly the point from which each of the three sides AB, AC, BC subtends an angle of 120°. A surprisingly simple proof of these and related properties can be found in Krarup and Vajda (1997).
*
E-mail. [email protected].
0966-8349/99/$ ± see front matter Ó 1999 Elsevier Science Ltd. All rights reserved. PII: S 0 9 6 6 - 8 3 4 9 ( 9 8 ) 0 0 0 4 3 - 6
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Fig. 1. SimpsonÕs construction of the fourth point T.
SimpsonÕs construction, however, does only provide the desired result if the given triangle has no angle exceeding 120°. Assume therefore that one angle, say \A, is strictly greater than 120°. As was noted by Torricelli but ®rst proved in Heinen (1834), A itself will then solve FERMAT(ABC) whereas the point T 0 at which the Simpson lines BB1 , CC1 , and the line containing AA1 now intersect will appear outside the given triangle as illustrated in Fig. 2. T 0 does not certainly solve FERMAT(ABC) but is seen to represent an optimal solution to FERMAT(AB1 C1 ). What else can in this case be deducted from the Torricelli±Simpson construction?
Fig. 2. The Simpson construction for triangles ABC, \A > 120 .
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In the book What is Mathematics?'' (Courant and Robbins, 1941) it is asserted that T 0 will solve what is termed ``A Complementary Problem'' which is to minimize the expression fBX CX ÿ AX : X 2 R2 g. The proof is ``. . . left as an exercise for the reader''. Actually, the interpretation by Courant and Robbins of T 0 is wrong. Correct geometrical and analytical solutions to the Complementary Problem for any data instance ABC will be provided in Section 2. The Complementary Problem can be viewed as a special case of the Weber problem provided that both positive and negative weights are allowed for. Aspects of this observation are brie¯y discussed in the concluding section concurrently with reference to the relevant literature.
2. Geometrical and analytical solutions Let us for any point X in the plane de®ne the objective function g(X), g X BX CX ÿ AX : For a given triangle ABC, let the Complementary Problem, henceforth referred to as CP(ABC), be the problem of ®nding a point V minimizing g(X), that is, CP(ABC): Find V such that g V BV CV ÿ AV minfg X : X 2 R2 g: To see that T 0 in Fig. 2 does not solve CP(ABC) is fairly straightforward. Disregarding the degenerate cases where A, B, C are collinear, T 0 and A must always be on the same side of the line containing B and C. If T 00 is the mirror image of T 0 with respect to that line, BT 0 CT 0 BT 00 CT 00 whereas AT 00 > AT 0 or g T 00 < g T 0 . T 00 is thus a better solution to CP(ABC) than T 0 . To solve CP(ABC) for any triangle, we shall throughout the remainder of the paper and without loss of generality ®rst assume that the given triangle is labeled such that AB 6 AC. Furthermore, no less than six cases in total need be distinguished. The determinants separating these six cases are: A, B, C are collinear or not, AB < AC or AB AC, \C 6 60° or \C > 60°. 2.1. A, B, C are non-collinear: the Restricted Complementary Problem Where should the point V minimizing g(X) be found in the general case? Part of the answer can be derived from what may be termed the Restricted Complementary Problem or RCP(ABC) for short. RCP(ABC) is identical to CP(ABC) with the additional constraint AX 6 AC. Let W be the point sought for: RCP(ABC): Find W such that g W BW CW ÿ AW minfg X : X 2 R2 ; AX 6 ACg: Theorem 1. For any triangle ABC with AB 6 AC, RCP(ABC) will either have the uniquely determined optimal solution W C if AB < AC or two optimal solutions, W B and W C, if AB AC.
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Fig. 3. An instance ABC, AB 6 AC, of RCP(ABC).
Proof. Fig. 3 shows a triangle ABC with AB 6 AC. Furthermore, a circle R having A as its center and with radius AC is drawn. B must be either within or on that circle. For any point X in the plane it follows from the triangle inequality applied to DBXC that BX CX P BC. BC is thus a lower bound on the ®rst two terms of the objective function g(X). Similarly, if X is either within or on the circle, AX cannot exceed AC which then is an upper bound on AX (or AC is a lower bound on AX, the third term of g(X)). For any X on the line segment BC, BX CX will equal the lower bound BC. For X C, both the lower bound BC on BX CX and the upper bound AC on AX will be reached. If AB < AC, B is an interior point in the circle and X C is the uniquely determined optimal solution to RCP(ABC) as asserted. For AB AC, both B and C are optimal solutions. h Fig. 3 is supplemented by the point B1 which, together with A and C, forms an equilateral triangle and hence must be on the circle R. Let C1 (not shown in Fig. 3) be the third point on the outward equilateral triangle ABC1 . Since AB 6 AC, C1 is an interior point if AB < AC or otherwise a point on R. No point on the Simpson lines BB1 and CC1 can thus be outside R and T 0 (see Fig. 2) must therefore be an interior point. This is yet another way to prove that Courant and Robbins' claim is incorrect. Theorem 1 will in some cases apply for CP(ABC) as well, for example, in degenerate cases where A, B, C are collinear. However, as will be elaborated on in the sequel, there also exist non-degenerate triangles ABC having one or in®nitely many optimal solutions to CP(ABC) outside the circle R. The idea of the following approach to solving CP(ABC) in general is to identify a set of points A0 which, together with B and C, de®nes a set of ``companion instances'' (not equivalent instances) of FERMAT(A0 BC) such that an optimal solution to CP(ABC) also solves any FERMAT(A0 BC). The solution properties of the latter are thereafter be exploited to obtain the ®nal result. The relationship between CP(ABC) and FERMAT(A0 BC) is established in Theorem 2 as illustrated in Fig. 4. LBC is the line containing B and C (recall that we
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Fig. 4. An illustration of Theorem 2 with V 6 C.
consider here non-degenerate triangles, still labeled such that AB 6 AC). D is the third vertex of an equilateral triangle having BC as one of its sides and placed on the same side of LBC as A. Theorem 2. (i) V solves CP(ABC))V solves FERMAT(A0 BC) where A0 is either V itself or any point on the line L containing A and V such that V is between A and A0 . (ii) If V and C are distinct points, the four points D, A, V, A0 are all collinear. Proof. (i): It follows from Theorem 1 that A itself can only solve instances of both RCP(ABC) and CP(ABC) if A and C coincide. Since A and C are distinct points, A and the optimizing point V must be distinct too and the line L containing A and V is uniquely de®ned. A0 is then either V itself or any point on L such that V is between A and A0 in Fig. 4. Let X be any point in the plane and f X BX CX A0 X . Since V is assumed to solve CP(ABC), g(X) P g(V), or BX CX ÿ AX P BV CV ÿ AV or BX CX P BV CV AX ÿ AV or BX CX A0 X P BV CV AX A0 X ÿ AV
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or f X P BV CV AA0 ÿ AV
since AX A0 X P AA0
or f X P BV CV AV A0 V ÿ AV or f X P BV CV A0 V f V : (ii): Since V and C are assumed to be distinct points, V cannot be within or on the circle R having A as its center and with radius AC (see Fig. 3). It follows then from Theorem 1 that neither A nor B can solve CP(ABC). V cannot be above the line LBC in Fig. 4 since the mirror image of V with respect to that line represents a better solution. Furthermore, neither A nor B can solve FERMAT(A0 BC) which implies that \A0 BC < 120° and \A0 CB < 120°. If FERMAT(A0 BC) has no vertex solution, V must be an interior point in DA0 BC. Hence, both A0 and V must be somewhere on L below or on LBC and within the region marked by heavy lines in Fig. 4. Suppose we choose a point A0 such \BA0 C 6 120°. We know from Fig. 2 that the point, now called V, solving FERMAT(A0 BC) must be on the Simpson line between A0 and D, the third vertex of the outward equilateral triangle having BC as one of its sides. The four points D, A, V, A0 must therefore all be collinear as asserted. h The second part of Theorem 2 tells us how to identify the line L containing A, D, V, A0 . We note ®rst that Theorem 2 applies regardless of whether A and D coincide or not. Thus, unless A and D do coincide (a special case to be dealt with separately later on), L is uniquely determined by A and D. The ground is now prepared for the geometrical construction of V in the general case. For ease of reference, we de®ne a coordinate p system (see pFig. 5) such that the coordinates of A, B, C, are: A : p; q, B : ÿ 12 3; 0, C : 12 3; 0. Without loss of generality, we assume q P 0. BCD is an equilateral triangle. Its circumscribing circle is a unit circle having its center at S with coordinates 0; 12. M is the midpoint between B and D. Since AB 6 AC, p must be non-positive and A can accordingly be anywhere in one of the four regions Q1; . . . Q4 including the lines de®ning them. (i) AB < AC, \C 6 60°. We will show that C is the uniquely determined optimal solution to CP(ABC) if AB < AC and \C 6 60°. If V is not a vertex of triangle ABC, the four points D, A, V, A0 must be collinear in an optimal solution to both CP(ABC) and the corresponding (set of) FERMAT(A0 BC), cf. Theorem 2. V and hence A0 as well must then be either on or below the x-axis. Unless triangle A0 BC has one angle greater than or equal to 120°, in which case V is the corresponding vertex, V is the point from which each of the three sides A0 B,
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Fig. 5. C solves CP(ABC) if AB < AC and \C 6 60 .
A0 C, BC subtends an angle of 120°. Thus, V must lie somewhere on the arc BC not containing D of the circle circumscribing DBCD. Let X be the corresponding set of points, that is, 2 1 2 1; y 6 0g: X f x; y : x y ÿ 2 A 2 Q1: Assume that A is a point in region Q1 (notation: A 2 Q1) then no point on X is outside the circle R (cf. Fig. 3) having A as its centre and with radius AC. Hence, Theorem 1 applies. A 2 Q2: No line containing A and D can intersect X except at B or C if \B or \C equals 60°. V must therefore be one of the vertices of nABC. V A is impossible since A and C are distinct points. B can also be excluded since g B BC ÿ AB < BC ÿ AC g C: Thus, C is optimal and L is the line containing A and C. A 2 Q3: A cannot be on the y-axis since AB < AC. A and D are thus distinct points de®ning the line L which intersects X at a point Y; see Fig. 6. Clearly, Y itself will solve FERMAT(YBC) since \BYC 120°. f(Y) BY CY must equal DY since DY is one of the Simpson lines. We obtain then: g Y ÿ g C BY CY ÿ AY ÿ BC ÿ AC DY ÿ AY ÿ CD ÿ AC AD AC ÿ CD > 0 (ii)
AB < AC,
\C > 60°
since MBCD is equilateral since A and D are distinct points
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Fig. 6. C solves CP(ABC) if AB < AC and A 2 Q3.
A with coordinates p; q must be somewhere in the interior part of Q4, the unbounded shaded area in Fig. 7 excluding the two lines de®ning it; hence, p < 0. Fig. 7 shows how V is determined as the intersection between X and the line L de®ned by A and D. Let a 12 2q ÿ 3=p be the slope of L and let (xV , yV ) be the coordinates of V. (xV , yV ) must then satisfy: yV ÿ 1:5 axV ; yV ÿ 122 x2V 1. These equations have two solutions, one of which, corresponds to the coordinates of D. The other solution is xV ÿ2a= 1 a2 ;
yV 1=2 3 ÿ a2 1 a2
1
(iii) AB AC, \C > 60° The construction in Fig. 7 does also apply when AB AC. The slope a of L, however, is unde®ned since L here coincides with the y-axis so the general expression (1) cannot be used. L intersects X at a point V with coordinates 0; ÿ 12 which, of course, also obtains as the limiting values of (xV , yV ) for a ! 1 in Eq. (1). We have so far covered all cases where the solution to CP(ABC) is uniquely determined. What remain are the cases where AB AC and \C 6 60°. (iv) AB AC, \C < 60° (iv) can be viewed as an extension of (i) for which the same line of reasoning applies. Thus, C is an optimal solution. Moreover, Theorem 1 asserts that not only C but also B is an optimal solution. (v) AB AC, \C 60°: nABC is equilateral
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Fig. 7. V solves CP(ABC) if AB < AC and \C > 60 .
Because the solution to a companion instance of FERMAT(A0 BC) is still uniquely determined, it appears that CP(ABC) in this case has in®nitely many optimal solutions. We prefer therefore to treat AB AC, \C 60° as a limiting case of AB 6 AC, \C P 60°. We have previously used A and D for de®ning the line L containing both V and A0 unless L and X do not intersect. Since MABC now is equilateral, A and D will coincide and L can be any line through A intersecting X. Fig. 8 shows the equilateral triangle ABC and the point Y which is any point on X. (iv) extends easily to (v) and tells us that optimal solutions to CP(ABC) can be found at both B and C with g B g C BC ÿ AC 0. To verify that CP(ABC) actually has in®nitely many optimal solutions, it suces to show that g(Y) 0 for any Y as de®ned or, in other words, that BY CY AY . Since Y also solves FERMAT(YBC), F Y BY CY must equal AY since AY is one of the Simpson lines. The ®ndings can now be summarized in Theorem 3. Theorem 3. For a given triangle ABC of non-collinear points within p a coordinate system p de®ned such that A, B, C have coordinates A : p; q; B : ÿ 12 3; 0 and C : ÿ 12 3; 0 respectively, an optimal solution V to CP(ABC) will be either a single point or a set of points:
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Fig. 8. nABC is equilateral: CP(ABC) has in®nitely many optimal solutions.
Case
The point set V solving CP(ABC)
(i) AB < AC; \C 6 60 (ii) AB < AC; \C > 60 (iii) AB AC;
\C > 60
(iv) AB AC; \C < 60 (v) AB AC; \C 60
V {C} V 2 f xV ; yV g where xV ÿ2a= 1 a2 ; yV 12 3 ÿ a2 = 1 a2 and a 12 2q ÿ 3=p the slope of L V 2 f 0; ÿ 12g V 2 fB; Cg 1 V 2 X f x; y : x2 y ÿ2 2 1; y 6 0g
Disregarding alternate optima, Theorem 3 asserts that C solves CP(ABC) whenever \C 6 60°. The line L is thus the line containing A and C which may (\C 60°) or may not (\C < 60°) contain D as well. For any point A0 on L such that AA0 P AC, C is also seen to solve FERMAT(A0 BC) since \A0 CB P 120°, cf. (Heinen, 1834). Theorem 2 does thus apply in all non-degenerate cases, the only modi®cation being that A, V ( C), and A0 but not D are collinear if \C < 60°. 2.2. A, B, C are collinear To claim that the correct solution to CP(ABC) is provided for any triangle ABC, AB 6 AC, we need also to consider, for the sake of completion, the degenerate cases where A, B, C are collinear. Let X be a point in the plane at a distance d, d P 0, from C.
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1. 2. 3. 4. 5.
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A B C : g X 2d ÿ d d ) V fAg fBg fCg; A B; B 6 C : g X BX d ÿ AX d ) V fCg; A 6 B, A C: cannot occur since AB 6 AC; A 6 B, B C: since AX d P AB, g X 2d ÿ Ax AB P d ) V fCg; A, B, C are distinct points C is between A and B : cannot occur since AB 6 AC; B is between A and C : g X ÿ g C BX d ÿ AX ÿ BC ÿ AC or g X ÿ g C AB BX ÿ AX AB ÿ AB d since BC ÿ AC ÿAB or g X ÿ g C P d since AB BX P AX ) V fCg
A is between B and C: It follows from cases (i) and (iv) in Theorem 3 that V {C} if AB < AC or V fB; Cg if AB AC. For the degenerate cases where A, B, C are collinear, AB 6 AC, we have thus shown that C is the optimal solution to CP(ABC) with B as an alternate optimum if AB AC. 2.3. The ``worst-case'' instance of CP(ABC) The construction in Fig. 7 shows that the Restricted Complementary Problem RCP(ABC) considered in Theorem 1 and the corresponding CP(ABC) can have dierent solutions. How far outside the circle R can we ®nd the point V solving CP(ABC)? To answer this question, we need only to consider the case where \C > 60 . Let L, V and X be as de®ned in Fig. 7 and let V0 be the intersection of L and that part of R which is on the same side of BC as X. It is seen that VV0 attains its maximum value for AB AC in which case V has coordinates 0; ÿ 12. Now, if AC approaches in®nity, the ``bottom part'' of R will approach the straight line BC. In the limit, all points on that line segment will therefore represent optimal solutions to RCP(ABC). p For V and C we ®nd: g V 2 ÿ 12 ÿ AV 0 , g C 3 ÿ AC. Hence, if B and C are distinct, the relative deviation g V ÿ g C=BC is for all triangles ABC bounded p from below by 1 ÿ 12 3 and this bound is approached for AC ! 1. ``\B \C, AC ! 1'' can thus be viewed as a worst-case instance of CP(ABC) in the sense that g V ÿ g C=BC then approaches its maximum value. 3. Notes The Weber Problem, known under several of other names including the more obvious The Euclidean Minisum Distance Facility Location Problem, asks for a point T minimizing the sum of the weighted Euclidean distances from itself to n given points A1 ; . . . ; An , each associated with a given positive weight.
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For all j, let xj ; yj be the coordinates of Aj and let wj , wj > 0, be the corresponding weight. Furthermore, let x; y be the coordinates of T representing the optimal location. The problem can then be stated as ( q) n X 2 2 2 wj x ÿ xj y ÿ yj : WEBER : min f x; y x;y2R2
j1
With weights w1 ; . . . ; wn , WEBER( A1 ; w1 ; . . . ; An ; wn ) appears to be optimally solvable via a mechanical analogue device called the Varignon frame after its originator, Pierre Varignon (1654±1722), and described in his treatise ``Projet d'une mouvelle m ecanique'', published in 1687. Fig. 9 shows the Varignon frame in the general case with n demand points. A board has n holes corresponding to the n ®xed points. The n strings are tied together in a knot at one end, and the loose ends are passed through the holes. The loose ends are attached to physical weights w1 ; . . . ; wn below the board. If the experiment is implemented in a ``perfect environment'', i.e. in®nitely thin strings, very small holes, no friction etc., then the ®nal position of the knot will be at the optimal location sought. To see this, let dj be the Euclidean distance between x; y and xj ; yj and let Hj be the angle between the x-axis and the line containing x; y and xj ; yj .
Fig. 9. The Varignon frame with n demand points.
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An optimal location of x; y can be characterized in two dierent ways. For the symbolic model (2) the partial derivatives of f x; y must both equal zero. For the analogue model, the resultant force exerted on the knot must equal zero. For the partial derivatives we obtain of x; y=ox Rj wj xj ÿ x=dj 0 or Rj wj cosHj 0; of x; y=oy Rj wj yj ÿ y=dj 0 or Rj wj sin Hj 0: The rightmost expressions are recognized as the force components in the x and y directions, respectively. The two requirements to be satis®ed by an optimal solution are thus identical. To see that the Varignon frame also provides the desired result in the case where the ®nal position of the knot coincides with one of the holes is less obvious. As discussed in detail in Wesolowsky (1993) who also formally states the necessary and sucient conditions for the optimum to occur at a hole, the diculty is that the partial derivatives do not exist at the holes. However, a simple argument reveals that a hole is an optimum location if the weight attached to the string through that hole is greater than or equal to the sum of all other weights in the system. With reference to the Varignon frame, the essence of Theorem 2 is that a negative force of ÿ1 can be replaced by an oppositely directed force of 1 in the search for an equilibrium position of the knot. Another observation is illustrated in Fig. 10 which is a sort of extension of Fig. 4. Fig. 10 shows the solution to an instance of FERMAT(A0 BC) where no angle exceeds 120° and where the optimizing point called V is found as the intersection of the arc X of the circle circumscribing nBCD and the line L containing A0 and D. Let LA0 be the half-line of L starting at V and containing A0 . Let LB and LC be analogously de®ned. It is seen that the geometrical construction of V remains unaltered if another position for A0 is chosen on LA0 . Likewise, B and C can be replaced by B0 and
Fig. 10. V solves in®nitely many instances of FERMAT(A0 B0 C0 ) and CP(A"B0 C0 ).
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C0 on LB and LC , respectively, without aecting the position of V. Thus, V solves ``in®nitely many instances'' of FERMAT(A0 B0 C0 ) regardless of the exact positions of the three points on the corresponding half-lines. For triangles A0 BC having no angles exceeding 120°, the same extension to ``in®nitely many instances'' does also apply for CP and allows us, so to speak, to ``reverse'' Theorem 2. Corollary. V solves FERMAT(A0 BC) ) V solves CP(A"BC) where A" is any point on the line L containing A0 and D such that D is between A0 and A". Proof. Suppose that a point W 6 V solves CP(A"BC). By Theorem 2, (ii), A", D, W must be collinear which implies that W must be on L. Since DA0 BC has no angle exceeding 120°, \BVC must equal 120°. W must then be on X as well and cannot be distinct from V. h It is seen that WEBER((A1 , 1), (A2 , 1), (A3 , 1)) is identical to FERMAT(ABC) with A A1 , B A2 , C A3 , that is, FERMAT(ABC) can be considered as the unweighted case of WEBER( ) for n 3. Likewise, for CP(ABC) we can consider CP(ABC) as the problem of locating a single facility V such that it is as close as possible to clients B and C and at the same time as far as possible away from client A. For CP(ABC) we can thus de®ne an equivalent Weber problem denoted by WEBER( A; ÿ1; B; 1; C; 1) where A, B, C represent the n 3 clients to be served and ÿ1, 1, 1 are the corresponding weights. CP(ABC) reduces accordingly to a special case of WEBER( A1 ; w1 ; . . . ; An ; wn ) where n can be any positive number and where each weight wj can be positive or negative. The earliest paper considering WEBER( ) with positive and negative weights appears to be Tellier (1985), who investigates cases with two attraction points and one repulsion point represented by positive and negative weights, respectively. A detailed account of TellierÕs ®ndings is given in Tellier and Polanski (1989). TellierÕs trigonometrical approach is illustrated in Fig. 11. ABC is the given triangle where A is the repulsion point associated with a repulsive force q corresponding to the negative weight wA ÿq. B and C are attraction points associated with positive weights or attractive forces wB , wC , respectively. We assume wB wC P q since the optimal location otherwise is at in®nity, and furthermore, that no attractive force exceeds the sum of the two other forces since the corresponding attraction point otherwise will be the optimal solution. Let V be the optimizing point. Under the tacit assumption that V is the point where the net force is zero, angles (1 2), 1 are determined by the cosine relations, cf. Fig. 11b; the sign should be changed in the ®rst of these expressions in Tellier and Polanski (1989). Note that angles 1 and 2 are in no way related to the positions of A, B, C. To determine where the line containing A and V in Fig. 11a actually is, we need to ®x angle 3 in Fig. 11c such that D and E coincide with angles 1 and 2 preserved as shown. Angles 4, 5, 6 in Fig. 11a are then immediately at hand.
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Fig. 11. The trigonometrical solution to WEBER( A; wA ; B; wB ; C; wC ) (Tellier, 1985).
To work as desired, however, the various angles should be compatible with each other. This requirement is illustrated in Tellier and Polanski (1989) as in Fig. 11d where the repulsion point must belong to one of the regions R1 and R2 or one the lines de®ning them. If the repulsion point is placed in the ``incompatibility zone'', one of the two attraction points will solve the problem and rules are given to decide which one of them. Although CP(ABC) is a special case of the problem studied in Tellier and Polanski (1989) and thus amenable for the trigonometrical solution proposed, our approach does complement their analysis rather than duplicate their eorts, since we provide both the coordinates of V and a geometrical construction.
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FERMAT(ABC) and CP(ABC) may appear to be merely variations of a theme, yet there is one important feature that separates them. FERMAT( ) is a convex problem which implies that there is no local minimum dierent from the global one. CP( ), on the other hand, is in general non-convex as pointed out in Chen et al. (1992), Drezner and Wesolowsky (1991), and Tellier and Polanski (1989). Besides analyzing the properties of the Weber problem in the plane when some of the weights are negative, Drezner and Wesolowsky (1991) devise exact algorithms when the distances are rectilinear or squared Euclidean. If the sum of all weights is positive, it is further shown for the more dicult Euclidean distance problem that the optimizing point must be within a circle slightly larger than the smallest circle containing all demand points. Finally, based on this result, three heuristic procedures are proposed. FERMAT(ABC) and CP(ABC) are both examples of continuous location problems in the sense that the optimizing point sought can be placed anywhere in the plane. Focussing on continuous models, how would one proceed to more complex cases? One possibility is to seek the geometrical constructions extended to cases where ``more than three clients are to be served'' for various combinations of positive and negative weights. The road ahead could then be to move from the triangles previously considered to a square, a rectangle, a general quadrilateral, a pentagon and so forth. As Tellier and Polanski put it (1989): ``The triangle case greatly diers from the more complex polygon case because it involves between-the-forces angles that are uniquely determined. When a polygon has more than three vertices, the angles ensuring an equilibrium of the forces cannot be determined without considering the form of the polygon. In the triangle case, this is possible.'' As regards WEBER( ) with positive and negative weights, (Chen et al., 1992) should not be ignored. WEBER( ) is ®rst shown to be a so-called d.-c. program (that is, a global optimization problem with both the objective function and the constraints expressed as dierences of convex functions) reducible to a problem of concave minimization over a convex set. Based on the theory of d.-c. programming, an exact algorithm is then devised. The analysis is complemented by extensive computational experiments which show that instances of WEBER( ) with n up to 1000 are solvable in at most two minutes. The concluding remark: ``So d.-c. and concave programming appear to be powerful new tools for continuous location theory'' seems to be well justi®ed. The approach for solving CP(ABC), the contents of the present paper, is extended in Jalal and Krarup (1997) to cover any combination of positive and negative weights associated with the vertices of a given triangle. Among the byproducts are surprisingly simple proofs of various geometrical constructions spanning the period 1640±1997. Furthermore, PtolemyÕs Theorem (around A.D. 150) is ``rediscovered'' and an alternative proof of HeinenÕs observation (1834) is given. Alternatively, one can look at discrete formulations where the set of potential sites for the facilities to be placed is ®nite and often represented by the vertices of a
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network. A fruitful way at times is to identify what is termed well-solved special cases. For example, if a problem de®ned on a network is known to be NP-hard in general, it may be polynomially solvable (and hence well-solved) if the underlying network exhibits a special structure. Thus, the network could be, say, a simple path, a tree, or a cycle, or it could be identi®ed in terms of more complex forbidden subgraphs. This is the line of approach taken in Burkard and Krarup (1998) for the 1-median problem in a network with both positive and negative weights. It is shown that the 1-median can be found in linear time if the underlying graph is a cactus, that is, a connected graph in which no two cycles have more than one point in common. The motiviation for the present study was the interest taken in FERMAT(ABC) and the subsequent confrontation with the incorrect solution to the Complementary Problem. As regards realistic applications, however, we believe that models capable of handling arbitrary weights may well provide useful decision support, for example, to companies operating in a competitive environment or to monopolistic companies wishing to increase or decrease the number of existing facilities. It should ®nally be noted that among the papers referred to in the text, we have been unable to retrieve copies of Simpson (1750) and Tellier (1985) or to provide other bibliographical data than those listed below.
Acknowledgements The original proof of the ®rst part of Theorem 2, based on a mechanical analogue device, was not fully convincing although the correct conclusion was drawn. I am indebted to A. Hertz and D. Kobler, Ecole Polytechnique Federale de Lausanne, who provided the present version. Likewise, the critical remarks made by G. Rote, Technische Universit at Graz, are much appreciated. Finally, sincere thanks are due to an anonymous referee whose meticulously prepared report was most valuable. References Burkard, R.E., Krarup, J., 1998. A linear algorithm for the ``pos/neg-weighted'' 1-median problem on a cactus. Computing 60, 193±215. Chen, P.-C. et al., 1992. WeberÂs problem with attraction and repulsion. Journal of Regional Science 32, 467±486. Courant, R., Robbins, H., 1941. What is Mathematics? Oxford University Press, Oxford. Drezner, Z., Wesolowsky, G.O., 1991. The Weber problem on the plane with some negative weights. INFOR 29, 87±99. Heinen, F., 1834. Uber Systeme von Kr aften. Gymnasium zu Cleve, pp. 18±19. Jalal, G., Krarup, J., 1997. Geometrical solution to the Fermat problem with arbitrary weights. Working Paper, DIKU (Department of Computer Science, University of Copenhagen). Krarup, J., Vajda, S., 1997. On Torricelli's geometrical solution to a problem of Fermat. IMA Journal of Mathematics Applied in Business and Industry 8, 215±224. Simpson, T., 1950. The Doctrine and Application of Fluxions. London. Tellier, L.-N., 1985. Economie spatiale: rationalite economique de l'espace habite. Chicoutimi, Quebec: Gaetan Morin.
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Tellier, L.-N., Polanski, B., 1989. The Weber problem: frequency of dierent solution types and extension to repulsive forces and dynamic processes. Journal of Regional Science 29, 387±405. Wesolowsky, G.O., 1993. The Weber problem: history and perspectives. Location Science 1, 5±23.