On Reconfiguring Radial Trees

07 PAPER Special Section on Discrete Mathematics and Its Applications On Reconfiguring Radial Trees Yoshiyuki KUSAKARI a), Member SUMMARY A linkage is a collection of line segments, called bars, possibly joined at their ends, called joints. We consider flattening a tree-like linkage, that is, a continuous motion of their bars from an initial configuration to a final configuration looking like a straight line segment, preserving the length of each bar and not crossing any two bars. In this paper, we introduce a new class of linkages, called radial trees, and show that there exists a radial tree which cannot be flattened. key words: linkage, reconfiguration, straightening, flattening, monotone tree, radial tree. Introduction A linkage is a collection of line segments, called bars, possibly joined at their ends, called joints. A reconfiguration of a linkage is a continuous motion of their bars, or equivalently a continuous motion of their joints, that preserves the length of each bar. Applications of this problem include robotics, hydraulic tube bending, and the study of macromolecule folding [3], [7]. A linkage is called planar if all bars are in the plane R with no intersection. A reconfiguration of a linkage is called planar if all bars are in the plane during the motion, and is called non-crossing if every two bars do not cross each other during the motion. In this paper, we consider only a planar reconfiguration of a planar linkage, and we may omit the word planar. Furthermore, we consider only a non-crossing reconfiguration, and we may omit the word non-crossing. For such planar reconfiguration problems, there is a fundamental question: whether any polygonal chain can be straightened. This problem has been known as The Carpenter s Rule Problem [3], [7], [8], and had been open from the 970 s to the 990 s. However, Connelly et al. have answered this question affirmatively: they showed that any polygonal chain can be straightened, and they gave a method for straightening polygonal chains [3]. Streinu gave another method for straightening polygonal chains [8]. Both methods above work well for convexifying closed polygonal chains (polygons). For a non-crossing planar reconfiguration of tree-like linkages, several negative results are known, that is, there exist trees which cannot be flattened [], [4], [6]. Figure Manuscript received August 3, 005. Manuscript revised November 7, 005. Final manuscript received December, 005. The author is with the Department of Electronics and Information Systems, Faculty of Systems Science and Technology, Akita Prefectural University, Honjo-shi, 05-0055 Japan. a) E-mail: kusakari@akita-pu.ac.jp DOI: 0.093/ietfec/e89 a.5.07 Fig. A locked tree []. Fig. A monotone tree [5]. illustrates a tree which cannot be flattened []. In Ref. [4], Connelly et al. gave a method for proving that some trees are locked, that is, cannot be flattened. On the other hand, an affirmative result was reported for reconfiguring tree linkages: Kusakari et al. showed that any monotone tree can be flattened, and gave a method for flattening monotone trees [5]. Figure illustrates a monotone tree [5]. Recently, the complexity of flattening tree linkages has been studied: Alt et al. showed that deciding lockability for trees is PSPACEcomplete []. However, there exist a few characterizations of tree linkages which can be flattened. Thus, it is desired to characterize a class of trees which can be flattened. In this paper, we define a new class of trees, called radially monotone trees or radial trees, which is a natural modification of the class of monotone trees, and show that there exists a radial tree which cannot be flattened. Figure 3 illustrates a locked radial tree. An early version of the paper was presented at a conference [6]. The remainder of this paper is organized as follows. In Sect., we give some preliminary definitions. In Sect. 3, we give a method to construct a locked radial tree. In Sect. 4, we show that the tree constructed in Sect. 3 is simple and radial. In Sect. 5, we present a theorem that the tree constructed in Sect. 3 cannot Copyright c 006 The Institute of Electronics, Information and Communication Engineers

08 Fig. 3 An overview of a locked radial tree. definition of monotone trees. A polygonal chain P is radially monotone for a point p (or, for short, radial for a point p ) if the intersection of P and any circle with the same center p is either a single point or empty. A tree T is radially monotone or radial if T is a rooted tree and every root-leaf polygonal chain in T is radially monotone for the root r. A radial tree is illustrated in Fig. 3. On the other hand, the tree illustrated in Fig. is not radial. Furthermore, every locked tree in [], [], [4] is not radial. Note that an x-monotone tree may not be radial, and a radial tree may not be x-monotone. For three points p, p, p 3 R, the angle p p p 3 is measured counterclockwise at point p from the direction of p p to the direction of p p 3, and ranges in [0, ). For two bars j = ( j 0, j ), j = ( j, j ) B joined with joint j, the angle j 0 j j is denoted by θ( j, j ). The slope s( j )of bar j = ( j 0, j ) is the angle measured counterclockwise at the parent joint j 0 from +x direction to the direction j0 j, andrangesin[0, ). The length of bar b B is denoted by b. For two points p, p R, the ray starting from p and passing through p is denoted by R(p, p ). For a point p R and a direction d [0, ), the ray starting from p and going in the direction d is denoted by R p (d). The circle centered at p R with radius a R is denoted by O p (a), and the circle with diameter p p is denoted by O(p, p ). be flattened. In Sect. 6, we prove lemmas whose proofs are omitted in Sect. 5. We conclude in Sect. 7.. Preliminaries Let L = (J, B) denote a linkage consisting of a joint set J and a bar set B. A structural graph of a linkage L is denoted by G(L). An embedding of a structural graph G(L) is called a configuration of linkage L. A linkage L is calleda(rooted) tree linkage or a (rooted) tree if the structural graph G(L) is a (rooted) tree. Let T = (J, B) be such a rooted tree linkage, and let r J be the root of T. A bar b B is denoted by ( j s, j t )if j s J is the parent of j t J. For any joint j J, a bar emanating from j, b = ( j, j ), is called a child bar of joint j. Aleaf is a joint having no child bar. For any joint j J {r}, a bar entering to j, b = ( j, j), is called a parent bar of joint j. For any joint j J {r}, a parent bar of j is unique, and is denoted by j. A joint j J is internal if j is neither the root nor a leaf. A flattened configuration of a rooted tree linkage is one in which the parent bar j of j makes angle with each child bar of j for every internal joint j, and the angle between each pair of child bars of j is zero for every non-leaf joint j J. Flattening a tree linkage T is a reconfiguration of T from an initial configuration to a flattened configuration. As an initial configuration of a tree linkage, we first define a monotone tree [5]. A polygonal chain P is x-monotone if the intersection of P and any vertical line is either a single point or a line segment if the intersection is not empty. A tree T is x-monotone if T is a rooted tree and every root-leaf polygonal chain in T is x-monotone. (See Fig..) Next, we define radial trees by slightly modifying the 3. Constructing a Locked Radial Tree In this section, we construct a radial tree which cannot be flattened, i.e., we construct a locked radial tree. Figure 3 illustrates such a locked radial tree. 3. Outline of Construction In this subsection, we show how to construct a locked radial tree. We first define some terms on a locked radial tree. Figure 3 does not illustrate a strict locked radial tree. Some bars should be overlapped and some joints should be touched each other in a strict locked radial tree. The locked tree T in Fig. 3 contains six congruent components C 0, C,, C 5, all of which are joined at the root r of T. More generally, one can construct a locked radial tree by such n(> 4) congruent components, each of which is called a C i component and is often denoted by C i,fori, 0 i n. Each C i component consists of three subcomponents: a V i component, anl i component and a Γ i component. TheseV i, L i and Γ i components are often denoted by V i, L i and Γ i, respectively. For each i,0 i n, these V i, L i and Γ i are incident to the root r counterclockwise in this order. Furthermore, L i is wrapped by V i and Γ i, as illustrated in Fig. 4. The V i component has two bars v = (v 0,v ) and v = (v,v ), joined with the internal joint v, whose angle θ(v, v ) is equal to + n (= + 6 ), and V i looks like the letter V, as illustrated in Fig. 5. The L i component has two bars l = (l 0, l )andl = (l, l ), joined with the internal joint l, whose angle θ(l, l ) is equal to 3,andL i looks like the letter L, as illustrated

KUSAKARI: ON RECONFIGURING RADIAL TREES 09 Fig. 4 Component C i. Fig. 8 Position of v +. 3. Detail of Construction Fig. 5 Subcomponent V i. Fig. 6 Subcomponent L i. Fig. 7 Subcomponent Γ i. in Fig. 6. The Γ i component has four bars γ = (γ 0,γ ), γ = (γ,γ ), γ 3 = (γ,γ 3 )andγ 4 = (γ 3,γ 4 ), and two internal joints γ,γ 3. Γ i looks like the letter Γ, as illustrated in Fig. 7. The angles γ 0 γ γ, γ 0 γ γ 3 and γ 0 γ 3 γ 4 are, and 3, respectively. In this subsection, we focus on a single C i component, and may often omit the index i for simplicity. Furthermore, subcomponents, bars, and joints in C i or C i+ are designated by the corresponding objects with symbol or +, respectively. For example, Γ i, Γ i and Γ i+ are denoted by Γ, Γ and Γ +, respectively. Moreover, the bar in Γ i+ corresponding to the bar γ in Γ i is denoted by γ +. We use similar notation for the others. For two points p, p R, we denote by p p the line segment connecting p and p, and by p p the length of the segment p p. We will define an initial configuration Ci of the C i component. A joint j J in C i is denoted by j in Ci,anda bar b B in C i is denoted by b in Ci.Fori, 0 i n, we draw all figures Ci simultaneously so that each pair of corresponding bars in consecutive components makes angle n, and the index i increases counterclockwise. Without loss of generality, we may assume that the length γ is equal to, and the slope s(γ ) is equal to. Recall that we draw all bars corresponding to γ for all Ci simultaneously. Then, for each i, 0 i n, we draw line segments γ = γ γ on each ray R γ (0) from the point γ, so that the angle θ(γ, γ ) is equal to. We choose the length γ long enough, so that the ray R(r, (γ ) ) intersects γ. Thus, we choose the length γ so as to satisfy γ > tan( n ). Next, we choose a point γ4 on the bar (γ ) +,sothat the following inequality holds: ( ) ( tan < γ + n γ 4 < min tan n ), sin ( n ). () The condition tan( n ) < γ+ γ 4 < tan( n ) ensures that γ 4 lies between points X and Y, which are defined later. (See Fig. 8.) Moreover, the condition tan( n ) < γ+ γ 4 < sin( n ) ensures that γ4 lies between points W and Y, which are also defined later. (See Fig. 0.) Thus, inequality () ensures that γ4 is contained in both XY and WY. One can find the point γ3, on the bar γ, satisfying

0 γ4 γ 3 r =. Actually, the point γ 3 is the intersection of the bar γ and the circle O(r,γ4 ). Then, we draw two line segments γ4 r and γ 4 γ 3. (See Fig. 0.) We finally drop a perpendicular from γ to rv (= r(γ 4 ) ), and let l be the foot of the perpendicular. We construct C i on the figure Ci. For a joint j and a point p, the notation j = p means that joint j is configured at point p. For a bar b and a line segment s, the notation b = s also means that bar b is configured at line segment s. The initial configuration of the C i component is obtained as follows. (See Figs. 4 7, and 0.) For the V i component, let v = v (= (γ 4) ), v = γ, v = v (= rv ), and v = v (= v γ ). For the L i component, let l = l, l = l (= γ ), l = l (= rl ), and l = l (= l l ). For the Γ i component, γ = γ, γ = γ, γ 3 = γ3, γ 4 = γ4 (= v+ ), γ = γ (= rγ ), γ = γ (= γ γ ), γ 3 = γ 3 (= γ γ 3 ), and γ 4 = γ 4 (= γ3 γ 4 ). From this construction, one observes that v = v, v = v, l = l, l = l, γ = γ, γ = γ, γ 3 = γ 3,and γ 4 = γ 4. Note that all lengths γ 3, v, l and l are determined if γ + γ 4 is determined. 4. The Initial Configuration In this section, we show that the tree constructed in the previous section is simple and radial. From now on, we often do not distinguish the linkage and its configuration, and may often omit the symbol *. 4. Simplicity The slope s(γ + ) is equal to n and is smaller than if n > 4. The length γ is greater than tan n from the construction. Therefore, the ray R γ ( ) should cross γ +. Let X be the intersection point of the ray R γ ( ) with γ +,andlety be the intersection point of the ray R γ () with γ +, as illustrated in Fig. 8. Since the length γ + is equal to and Xrγ + is equal to γ rγ + (= n ), the length γ+ X is equal to tan( n ). Furthermore, one can observe γ + Y = tan( n ) as follows: since the hypotenuses are common and rγ = rγ + (= ), the two right triangles ryγ and ryγ + are congruent, and hence γ ry = Yrγ + = n. Thus, the point γ 4 is contained in the open line segment XY since the condition tan( n ) < γ+ γ 4 < tan( n ) holds by construction. Therefore, one can observe that the two line segments γ4 r(= (v ) + )and γ4 γ 3 (= γ 4 ) can be drawn without (properly) crossing any other line segments. Furthermore, one can easily observe that any pair of line segments in Ci can be drawn without (properly) crossing even if the pair contains neither (v ) + nor γ 4. Therefore, the tree constructed in Sect. 3 is simple. 4. Radial Monotonicity For any joint j J, a subtree of T rooted at j is denoted Fig. 9 Fan F j [θ,θ ]. by T( j). A fan p jp is the set of points swept out by a ray starting j moving counterclockwise from the direction jp to the direction jp, and contains points on both R( j, p )and R( j, p ). A fan p jp may be denoted by F j [θ,θ ], where θ = rjp and θ = rjp.(seefig.9.) The following lemmas hold. Lemma : (i) A tree T = (J, B) is radial if and only if, for any joint j J, all points p (except for j) contained in the subtree T( j) are properly outside of the circle O r ( rj ). (ii) A tree T = (J, B) isradialifandonlyif,foranyjoint j J {r}, every child bar b = ( j, j ) B is contained in the fan F j [, 3 ]. Proof. Both (i) and (ii) are obvious from the definition of radial trees. Note that, for every bar b = (r, j) emanating from the root r, the slope s(b) can take any value in [0, ) evenift is radial. Lemma : (i) The V i component is radial for the root r. (ii) The L i component is radial for the root r. (iii) The Γ i component is radial for the root r. Proof. (i) One can easily observe that the angle θ(v, v ) is greater than or equal to.(seefig.5.) Thus, the bar v is contained in the fan F v [, 3 ], and hence the path (r =)v 0 v v is radial by Lemma (ii). (ii) From the construction of the L i component, the angle θ(l, l ) is equal to 3. Thus, the bar l is contained in F l [, 3 ], and hence the path (r =)l 0l l is radial by Lemma (ii). (iii) Since the Γ i component has two leaves, it is sufficient to show that both path P and path P are radial, where P = γ 0 γ γ and P = γ 0 γ γ 3 γ 4. From the construction, θ(γ, γ ) =, and hence the path P is radial by Lemma (ii). Furthermore, one observe that the bar γ 3 is contained in F γ [, 3 ]andthebarγ 4 is contained in F γ3 [, 3 ]since θ(γ, γ 3 ) = and γ 0γ 3 γ 4 = 3. Thus, by Lemma (ii), the path P is radial. Every C i component is radial since all subcomponents are radial by Lemma above, and hence the tree constructed in Sect. 3 is radial.

KUSAKARI: ON RECONFIGURING RADIAL TREES 5. Lockability In this section, we present a theorem that the tree constructed in Sect. 3 cannot be flattened. Note that the method of the proof described in [4] cannot directly apply to our tree. For the initial configuration, the following lemma holds. Lemma 3: (i) rγ l <, (ii) l γ γ <, (iii) rγ 4 γ 3 <,and (iv) γ 3 γ 4 v + <. The following four inequalities hold: Proof. Since the sum rγ l + l γ γ is equal to rγ γ (= ), both (i) and (ii) immediately follow. Moreover, since the sum rγ 4 γ 3 + γ 4 γ 3 r + γ 3 rγ 4 is equal to and the angle γ 4 γ 3 r is equal to, (iii) follows. Thus, we only prove (iv) below. A quadrangle is a convex polygon with four vertices A, B, C, D counterclockwise, and is denoted by ABCD. For the quadrangle rγ 3 γ 4 γ +, it follows that γ 4γ 3 r = rγ + γ 4(= ) from our construction of the tree, and hence the sum γ + γ 4γ 3 + γ 3 rγ + is equal to. Let Z and W be vertices of a rectangle rzwγ + such that the vertex Z is on γ and the vertex W is on γ +, as illustrated in Fig. 0. Then, the angle γ Zr is equals to γ rγ + (= n ), since the equation Zrγ + γ Zr = Zrγ + γ rγ + = holds. Therefore, the equation γ + W = rz = follows. The sin( n ) condition γ + v+ < holds from Eq. (), and hence sin( n ) v+ is on the open line segment γ + W. Since both O(r,v+ )and O(r, W) have the same chord rγ + and the radius of O(r,v+ ) is smaller than the radius of O(r, W), γ 3 lies between points γ and Z from the construction. On the other hand, one can easily observe that γ 3 γ 4 v + = γ 3rγ + since the equation γ + γ 4γ 3 + γ 3 γ 4 v + = γ+ γ 4γ 3 + γ 3 rγ + = holds. Thus, the angle γ 3 γ 4 v + is equal to Zrγ+ and is smaller than. For each C i, the angle v rv + is called the angle of C i and may be denoted by C i. A reconfiguration widens C i if it makes the angle C i increase, and squeezes C i if it makes the angle C i decrease. The following lemmas hold. Lemma 4: There exists a widened C i component if and only if there exists a squeezed C j component, where 0 i, j n andi j. Proof. n Since the sum C i is equal to, the claim i=0 immediately follows. From Lemma 3 and 4, one can observe the following Lemma holds, a proof of which will be given in Sect. 6. Lemma 5: (i) No reconfiguration can squeeze any C i component. (ii) No reconfiguration can widen any C i component. Thus, the following theorem holds. Theorem : There exists a radial tree which cannot be flattened. 6. Proof of Lemma 5 In this section, we prove Lemma 5 whose proof is omitted in Sect. 5. We first give some additional notation. For any point p R, the x-,y-coordinate of p is denoted by x(p),y(p),respectively. Recall that the circle centered at p R with radius a R is denoted by O p (a). An inner open region bounded by O p (a) is denoted by Ô p (a), and an outer closed region bounded by O p (a) is denoted by Ŏ p (a). Note that Ô p (a) does not include any points on the boundary O p (a), but Ŏ p (a) does. An arc on O p (a) from a point q to a point q counterclockwise is denoted by A p (a; q, q ). Below, we regard the reconfiguration as a continuous function on time t [0, ), where the initial configuration is the image of this reconfiguration at time t = 0. The configuration of any object o at time t is denoted by o(t). For example, a configuration of joint v at time t is denoted by v (t). Moreover,C i (0) = Ci for every i, 0 i n. We use similar notation for the others. We may assume, without loss of generality, that the root r is located on the origin of the xy-plane, and the bar v of C 0 is fixed during any time t [0, ) for any reconfiguration. Thus, for any time t [0, ) of any reconfiguration, the following equation holds: v (t) = v (0)(= v ). () Therefore, the following equations also hold: { v0 (t) = v 0 (0)(= v 0 = r), v (t) = v (0)(= v ). Fig. 0 Position of γ 3. We shall prove some lemmas before proving Lemma 5.

In order to prove these lemmas, we consider infinitely small motion of linkage. We say that a motion of linkage L = (J, B)isinfinitely small for ε>0andt > 0 if every joint j J is contained in Ô j(0) (ε) during period [0, t). Since every reconfiguration is continuous, for any small real number ε>0, there exists a time δ>0such that the reconfiguration is infinitely small for ε and δ, i.e., every joint j J is properly contained in Ô j(0) (ε) during period [0,δ). Below, we only consider such infinitely small motion for a sufficiently small ε and a corresponding short time δ, sothatnocritical event occur, like s(γ 4 (t)) s(v + (t)) =. A crescent region CR j (ε) is the region defined by Ô j (ε) Ŏ j ( j ), where j is the parent joint of j. Thefollowing lemmas hold. Lemma 6: For any reconfiguration, the following membership holds during any time t [0,δ): l (t) CR l (ε). to l on A r( l ; l, l ). Then, at any time t [0,δ), joint l (t) is contained in R. One can easily observe that the region R contained in Ŏ l ( l ) Ô l ( l + ε). Thus, the claim holds. Lemma 7: For any reconfiguration, the following inequality holds at any time t [0,δ): v rγ (t) v rγ (0). Proof. By Eq. (), it is sufficient to show that the slope s(γ (t)) is greater than or equal to. Suppose for a contradiction that s(γ (t)) is smaller than for some time t [0,δ). Let γ (t)γ be the upper tangent of point γ (t) andcircle O γ (ε), and let F γ (t) beafanrγ (t)γ, as illustrated in Fig.. Proof. From the assumption of infinitely small motion, l (t) is contained in Ô l (ε). Thus, we shall only show that l (t) Ŏ l ( l ). Let l be the intersection of O l (ε) ando r( l ) satisfying s(rl ) < s(rl ). (See Fig..) Since l 0(= r) isfixedfor any reconfiguration and any time, l (t) should lie on O r ( l ). Moreover, the angle v (t)rl (t) should be greater than or equal to zero at any time t [0,δ) from Eq. () and the condition of a non-crossing reconfiguration. Thus, l (t) should lie on A r ( l ; l, l ) during time t [0,δ). Furthermore, joint l (t) should lie on O l (t)( l ). Let R be the region swept by a part of Ŏ l (t)( l ) with center from l Fig. Feasible position of l in time [0,δ). Fig. Impossible motion of γ.

KUSAKARI: ON RECONFIGURING RADIAL TREES 3 Then, one can observe that l (t) should be contained in F γ (t) from the condition of a non-crossing reconfiguration since no critical events occur during [0,δ). On the other hand, by Lemma 6, l (t) should be contained in CR l (ε). We shall show below that the equation F γ (t) CR l (ε) = holds. Let γ and γ be two intersections of O γ (ε) and O r ( γ ) satisfying s(rγ ) < s(rγ ), and let T l be the tangent of O l ( l )atl, as illustrated in Fig.. Then, T l is perpendicular to bar l. Moreover, by Lemma 3, the angle rγ l is smaller than and the angle l γ γ is also smaller than, and hence the equation F γ (0) CR l (ε) = {γ } holds at time t = 0. Furthermore, γ (t) should lie on A r ( γ ; γ,γ ), and hence x(γ (t)) is greater than zero since the slope s(γ (t)) is smaller than. This means that F γ (t) CR l (ε) = at time t > 0. Lemma 8: For any reconfiguration, the following inequality holds during the time t [0,δ): s(γ (t)) 0. Proof. By the proof of Lemma 7, one can observe that γ (t) should lie on A r ( γ ; γ,γ ). The y-coordinate of any point on A r ( γ ; γ,γ ) is smaller than or equal to y(γ )(= ). Moreover, γ (t) should be above v whose y-coordinate is equal to y(v )(= ) from Eq. (), and γ (t) should be contained in Ô γ (ε). Thus, the claim should hold from the condition of a non-crossing reconfiguration. Lemma 9: For any reconfiguration, if γ is fixed during the time t [0,δ) then the following inequality holds: γ rv+ (t) γ rv+ (0). Proof. Assume that γ is fixed during the time t [0,δ) in this proof. Then, it is sufficient to show that s(v + (t)) s(v + (0)). Suppose for a contradiction s(v + (t)) < s(v + (0)) for some time t [0,δ). Let γ3 be the intersection of O γ3 (ε)and O γ ( γ 3 ) satisfying s(γ γ 3 ) > 0, as illustrated in Fig. 3. Since γ and γ are fixed, γ 3 (t) should lie on A γ ( γ 3 ; γ3,γ 3 ). Similarly to the proof of Lemma 6, one can easily observe that the following membership holds: γ 4 (t) CR γ4 (ε). Let F v + (t) bethefanrv + (t)v+ (t). Then, similarly to the proof of Lemma 7, one can observe that F v + (t) CR γ4 (ε) = as follows. By Lemma 3, the angle rv + γ 3 is smaller than and the angle γ 3 γ+ v+ is also smaller than, and hence the equation F v + (0) CR γ4 (ε) = {γ4 } holds at time t = 0. However, since the slope s(v + (t)) is smaller than s(v + ), the equation F v + (t) CR γ4 (ε) = follows at time t > 0. We are now ready to prove Lemma 5. Proof of Lemma 5 By Lemma 4, it is sufficient to show Fig. 3 Feasible position of γ 4 in time [0,δ). only (i), i.e., we shall only show that no reconfiguration can squeeze any C i component. By Lemma 9, the angle v rv+ (t) is greater than or equal to v rv+ (0) if γ is fixed during the time t [0,δ). Moreover, since the inequality x(γ (t)) x(γ ) holds by Lemma 7 and the inequality s(γ (t)) 0 holds by Lemma 8, the x-coordinate x(γ 3 (t)) is smaller than or equal to x(γ3 ). This means that v + (t) should be contained in the region obtained by shifting CR γ4 (ε) tothe x direction. Therefore, the angle v rv+ (t) is greater than or equal to v rv+ (0) even if γ (t) rotates about the center r counterclockwise. Thus, the claim holds. 7. Conclusion In this paper, we show that there exists a tree linkage which is radially monotone and cannot be flattened. The following future works remain: () find new characterization of a class of tree linkages which can be flattened, () find a method for flattening a class of tree linkages other than monotone trees, and (3) find othernecessary or sufficient conditions for linkages to be reconfigured to some regular form, say, straightened, flattened, or convexified,. Acknowledgments We would like to thank Dr. Takao Nishizeki and Dr. Akiyoshi Shioura in Tohoku University for their helpful comments, which improve the presentation of this paper. We also wish to thank anonymous referees for their valuable comments, which improve the clarity of this paper. References [] H. Alt, C. Knauer, G. Rote, and S. Whitesides, On the complexity of the linkage reconfiguration problem in Towards a Theory of

4 Geometric Graphs, pp. 4, American Mathematical Society, 004. A preliminary version appeared as, The complexity of (un)folding, Proc. 9th ACM Symp. Computational Geometry, pp.64 70, 003. [] T. Biedl, E. Demaine, M. Demaine, S. Lazard, A. Lubiw, J. O Rourke, S. Robbins, I. Streinu, G. Toussaint, and S. Whitesides, A note on reconfiguring tree linkages: Trees can lock, Discrete Applied Mathematics, vol.54, pp.9 3, 00. A preliminary version appeared in Proc. 0th Canadian Conference on Computational Geometry, 998. [3] R. Connelly, E. Demaine, and G. Rote, Straightening polygonal arcs and convexifying polygonal cycles, Discrete and Computational Geometry, vol.30, pp.05 39, 003. A preliminary version appeared in Proc. 4st IEEE Symp. Foundations of Computer Science, pp.43 44, 000. [4] R. Connelly, E. Demaine, and G. Rote, Infinitesimally locked selftouching linkages with applications to locked trees, in Physical Knots: Knotting, Linking, and Folding Geometric Objects in R 3, pp.87 3, American Mathematical Society, 00. A preliminary version appeared in Proc. th Annual Fall Workshop on Computational Geometry, 00. [5] Y. Kusakari, M. Sato, and T. Nishizeki, Planar reconfiguration of monotone trees, IEICE Trans. Fundamentals, vol.e85-a, no.5, pp.938 943, May 00. [6] Y. Kusakari, On reconfiguring radial trees, Proc. JCDCG00, Lecture Notes Comput. Sci. 866, pp.8 9, Springer-Verlag, 003. [7] J. O Rourke, Folding and unfolding in computational geometry, Proc. JCDCG 98, Lecture Notes Comput. Sci. 763, pp.58 66, Springer-Verlag, 000. [8] I. Streinu, A combinatorial approach to planar non-colliding robot arm motion planning, Proc. 4st IEEE Symp. Foundations of Computer Science, pp.443 453, 000. Yoshiyuki Kusakari received the B.E. degree in information engineering, and the M.S. and Ph.D. degrees in information science from Tohoku University, Japan, in 993, 995 and 998, respectively. He joined Akita Prefectural University in 000, and is currently Associate Professor of Faculty of Systems Science and Technology. His research interests include graph theory and computational geometry.