Flexibility of Bricards planar linkages and other structures via resultants and computer algebra
aa r X i v : . [ m a t h . M G ] A ug Flexibility of Bricard’s linkages and otherstructures via resultants and computeralgebra
Robert H. Lewis
Fordham University, New York, NY 10458, USAhttp://fordham.academia.edu/RobertLewis
Evangelos A. Coutsias ∼ coutsias Abstract
Flexibility of structures is extremely important for chemistry and robotics. Followingour earlier work, we study flexibility using polynomial equations, resultants, and asymbolic algorithm of our creation that analyzes the resultant. We show that thesoftware solves a classic arrangement of quadrilaterals in the plane due to Bricard.We fill in several gaps in Bricard’s work and discover new flexible arrangements thathe was apparently unaware of. This provides strong evidence for the maturity ofthe software, and is a wonderful example of mathematical discovery via computerassisted experiment.
Keywords: resultant, polynomial, flexible, octahedron, quadrilateral, computeralgebra.
This project results from the convergence of four topics: systems of polyno-mial equations, flexibility of two and three dimensional objects, computationalchemistry, and computer algebra. It also has application to robotics [19], [22].We have developed software to detect flexibility in certain structures thatare generically rigid. It is based on symbolic computation of polynomials andrational functions, not numerical computing. We previously reported on earlier
Preprint submitted to Elsevier 19 May 2018 tages of this research in [16] and [11]. Since then, the software has beenenormously improved in both power and efficiency, to the point where it notonly discovers the previously known modes of flexibility of a classic structuredue to R. Bricard, but discovers new modes apparently unknown to him.We are mostly concerned with the framework in Figure 1. It is a system ofseven bars, joined at the nine junctions shown by rotational joints, allowingfree rotation in the plane. It is generically rigid.
This follows from a gen-eral theorem in kinematics [12] by which the mobility (number of degrees offreedom of relative motion) of a linkage system is given by the relation M = 3( n − g −
1) + g X i =1 f i where n is the number of members, g is the number of joints and f i is themobility at joint i . For the system in Figure 1, comprised entirely of rigid rodswith rotatable joints (with n = 7, g = 9 and f , ..., f = 1) this gives M = 0.When M > M = 0 it is genericallyrigid or determined . We wish to discover cases, by means of particular relationsexisting between its edges, that determinacy (rigidity) ceases to hold. Thenthe framework will be deformable (flexible).Flexibility is an intuitive concept. Imagine a triangle made of three stiff rodsjoined with movable hinges. The formula above confirms the clear intuitionthat the structure is obviously rigid. In the same way, a quadrilateral in theplane is obviously flexible, and for it M >
0. ( M = 1).In computational chemistry, protein folding has been a major research topicfor a number of years [21], [6], [7]. Molecules can fold because they are flexible.Simple examples are easily built from a few plastic balls and rods. In 1812,Cauchy considered flexibility of three dimensional polyhedra with triangularfacets (similar to a geodesic dome) where each joint can pivot or hinge. Heproved that if the polyhedron is convex it cannot be flexible; it must be rigid[4]. In 1896 Bricard [2] tried to find non-convex flexible polyhedra by lookingat one of the smallest possible cases, octahedra. He partially succeeded: hisflexible octahedra are not imbedible in actual three-space because some oftheir facets intercross. He also described the system of three quadrilaterals inthe plane (Figure 1) whose motion is algebraically equivalent to the octahedra.People came to believe that there were no flexible polyhedra at all. But in1978 Robert Connelly, building on Bricard, astonished them by finding a non-convex one [5], and soon models appeared of a simpler flexible structure [9],[18].Our approach is to describe the geometry of the object or molecule with a set2 ig. 1. Bricard’s quadrilaterals, showing labeled sides and base angles. of multivariate polynomial equations. Solving a system of multivariate poly-nomial equations is a classic, difficult problem. The approach via resultantswas pioneered by Bezout [1], Dixon [3], [10], [13], Sylvester [8], and others.The resultant appears as a factor of the determinant of a matrix containingmultivariate polynomials. Computing it can be quite a challenge [17], but wedeveloped methods to do so [15]. Once we have the resultant, we described[16] an algorithm we call Solve that examines the resultant and determinesways that the structure can be flexible.We discovered in this way some of the conditions of flexibility for Bricard’sarrangement of quadrilaterals in [2] that is algebraically equivalent to theoctahedra.
Solve was greatly improved by Fox [11] and more recently againby us. It is at least 500 times faster on Bricard’s quadrilaterals than in 2008,and now finds all three of Bricard’s ways the quadrilaterals can flex (this wasnot true in [16]). Surprisingly,
Solve has discovered new flexible arrangementsof the quadrilaterals that were apparently not anticipated by Bricard.The main point of this paper is that our new algebraic and algorithmic solutionof Bricard’s quadrilaterals demonstrates that the software has matured to thepoint where one can confidently use it on more complex structures, such asmolecules.
All computations here were done with Lewis’s computer algebra system
Fer-mat [14], which excels as polynomial and matrix computations [20].As explained in the introduction, we are primarily concerned here with the3nalysis of the flexibility of a certain structure of Bricard consisting of sevenrigid rods forming three quadrilaterals in the plane (Figure 1). We need toestablish that our software can find all the flexible cases. It order to show whywe are confident in this, we will present proofs paralleling some of those ofBricard [2], but ours are quite different and more algebraic.Other than historical motivation, why should we concentrate on this arrange-ment of rods? As remarked in the introduction, a quadrilateral in the plane
AD, DC, CB, BA is obviously flexible ( M = 1; see Figure 1). Imagine that AB is fixed. As AD pivots about A , the angles α and γ take on a continuumof values. If we add GF and F E we have two nested quadrilaterals, and thestructure remains obviously flexible ( M = 1), and β also takes on a continuumof values, unless we set F = A , so GF = GA and F E = AE, s = s , s = e ;then β would be constantly π during the flex. That is a degenerate case. Theaddition of one more rod or “brace” HI produces a third quadrilateral EHIB .The structure is now generically rigid ( M = 0). However, it can be made flex-ible in several ways. A degenerate way to do so is to simply place HI on topof EB , so HI = EB, s = s = 0 , s = b − e . We are not concerned with suchdegeneracies here. Far more interesting is to choose the lengths of the rods(sides) so that each quadrilateral is a parallelogram. Obviously, the system isthen flexible. This is one of the cases we analyze below (section four).This is our goal: non-generic flexibility.
The system of Figure 1 is one of thesimplest to examine for non-degenerate flexibility, and was thought by Bricardto be “equivalent” to two octahedra in three dimensions.Our strategy is to describe the arrangement by a system of three polynomialequations, where the parameters are the lengths of the sides, and the threevariables represent the tangents of certain angles in the structure. Usingresultants, we show that flexibility implies that each of the three tangents isa rational function of the other two, and analyze when one tangent can be arational function of only one other. We thereby derive the three flexible casesthat were defined by Bricard, but with new subcases.As in Bricard’s paper on flexible octahedra, elementary geometry and trigonom-etry lead to a system of three polynomial equations in three variables t , t , t and fifteen parameters (details in next paragraph), namely a t t + d t + 2 c t t + b t + e = 0 (1) a t t + d t + 2 c t t + b t + e = 0 (2) a t t + d t + 2 c t t + b t + e = 0 (3)The fifteen parameters are themselves simple polynomial functions of the sides Informally, degeneracy means a side is 0, or an angle is constant during the flex. Technically, the tangent of one-half the angle.
4f the flexing quadrilaterals, in such a way that c c c = 0. The other param-eters might be 0. The t i are half-angle tangents of angles α, β, γ in the quadri-laterals; see Figure 1. (Cotangents could also be used, which has the effect ofreplacing t i with t − i . This will be used in Theorem 3.) The seven rigid rodsare AD, DC, CB, AB, GF, F E, HI . The joints allow each rod to pivot freelyin the plane. AB remains fixed on the x − axis during pivoting, with A at theorigin. We allow negative values for s , s , s , or s , so points G, F, H, I mightbe below the x -axis. Angles θ and θ will be discussed later in Theorem 4.The equations arise from Figure 1 using basic geometry and trigonometry.For example, the coordinates of the point D are ( s cos( α ) , s sin( α )). C is( b + s cos( γ ) , s sin( γ )). Therefore s = ( b + s cos( γ ) − s cos( α )) + ( s sin( γ ) − s sin( α )) One obtains three equations of this kind (using also s and s ) and threeobvious equations of type sin ( x ) + cos ( x ) = 1. Then use the well-known halfangle tangent substitutions sin( α ) = 2 t / (1 + t )cos( α ) = (1 − t ) / (1 + t )and so on (with β, γ ; t , t ) to form the three equations (1) − (3). The fifteenparameters become a = ( − s + e − s + s )( − s + e + s + s ) b = ( s + e − s + s )( s + e + s + s ) c = − s s d = ( − s + e + s − s )( − s + e − s − s ) e = ( s + e − s − s )( s − s + s + e ) a = ( − b + e + s − s − s )( − b + e + s − s + s ) b = ( − b + e − s + s − s )( − b + e − s − s − s ) c = − s s d = ( − b + e + s + s − s )( − b + e + s + s + s ) e = ( − b + e − s + s − s )( − b + e − s + s + s ) a = ( b + s − s − s )( b + s − s + s ) b = ( b + s + s + s )( b + s − s + s ) c = − s s d = ( − b + s + s + s )( − b + s + s − s ) e = ( − b + s − s − s )( − b + s − s + s ) (4)None of the sides s i , e, b is 0. b = e, s = s , s = s , s = s . For convenience,we also define s ≡ e, s ≡ b , and we also refer to s − s , s − s , and s − s as “sides”. 5s we discussed above, the arrangement of quadrilaterals in Figure 1 is gener-ically rigid. That means, in spite of the flexible joints, if numerical values wereassigned arbitrarily for the eleven sides, the angles α, β, γ would be uniquelydetermined. The main task of this paper is
Problem 1: Find conditions on the sides under which the quadrilat-eral arrangement becomes flexible.
Flexibility is marked mathematically by the three angles, and their half-angletangents t i , each taking on uncountably many values. As remarked above, ifall three quadrilaterals are parallelograms, the arrangement is flexible. Thismeans that under the substitutions s = s , b = s , s = s , s = e, s = s , s = b − e , not only are there common roots to the system of equations(1) − (3), but there is a continuum of common roots; each t i is a never-constantcontinuous function t i : I → R . Never-constant means there is no open inter-val over which t i is constant. Allowing that would create degenerate cases,which we do not discuss here. (Some are discussed in [16].) We therefore havesecondarily: Problem 1 ′ : Find all conditions on the sides under which the quadri-lateral arrangement becomes non-degenerate flexible. To understand flexibility, we follow Bricard and ask
Problem 2: When is one of these variables, t , say,(1) a rational function of another t j , or(2) a rational function of both of the other ones t , t ? Using resultants, we will show in our Main Theorem (section 5) that flexibilityalways implies the second case. The first case is referred to as splitting. To make sense of “rational function” we must discuss the ground field, GF .Let F be a field. In many of our algebraic results, F could be any field ofcharacteristic not 2. However, eventually we will evaluate expressions like thosein (4) by substituting each parameter with an element of F . Therefore F = asubfield of R is appropriate. We do not allow the sides to be arbitrary complexnumbers.Then given F , we may first think of the ground ring as F [ a , . . . , e ] and theground field GF as F ( a , . . . , e ), the field of rational functions over F of the fif-teen parameters. However, the ground field is really GF = F ( s , s , . . . , s , e, b ), At least up to sign or supplement. Some assignments would be impossible. One may fairly ask for the motivation for Problem 2. Recall that a similar ques-tion about the roots of a polynomial is the basic idea in Abel’s analysis of theunsolvability of the quintic. a , . . . , e must sometimes bethought of as even larger polynomials in the eleven s , s , . . . , s , e, b .The evaluation homomorphism, obtained by substituting parameters with val-ues in F , can be thought of as a map from GF to itself. Also, when we speakof finding a solution to the system (1) − (3), we understand as usual that thecommon root may lie in an extension field of the ground field, for example, aradical extension.We can now specify what we mean in Problem 1 by “find conditions on thesides under which the quadrilateral arrangement becomes flexible.” We meanfind substitutions of the form s i = p ( s , s , . . . , ˆ s i , . . . ), where p ∈ GF , sothat the t i are continuous functions from some interval to R . We will showthat this notion of “condition” does indeed lead to both old and new flexibilearrangements. That in turn suggests: Problem 3: Can all flexible cases be represented by a table of sub-stitutions in this sense?
We will see in the conclusion that, very surprisingly, this is false, and weconjecture a modification of it.The rest of the paper is organized as follows: In section three we developthree lemmas to identify when an equation splits. The various split cases aresummarized in section four. In section five we present the main theorem, whichsolves Problem 2 and says that if no equation splits, then every t i is a rationalfunction of the other two. In section six we complete the theory of the non-splitcase. In section seven we describe the software results and two surprising newflexible cases for the quadrilaterals that were apparently unknown to Bricard.In all of the following definitions, lemmas, and theorems we assume flexibility.Some of them are true without this assumption, but we are not concernedwith that. We say that one of the equations (1) − (3) splits or decomposesif one of the t i in it can be expressed as a rational function of the other one. For simplicity, let’s concentrate on solving for t in equation (1). Suppose in(1) we have a = d = 0. Then (1) reduces to2 c t t + b t + e = 0 (4)7ince c cannot be 0, we can solve this for t and obtain a rational function,so (1) would split. This example is an important case in the following lemmas.Assuming that a = 0 or d = 0, it is natural to solve for t using the quadraticformula. We have t = − c t ± q c t − ( a t + d )( b t + e ) a t + d (4) Definition 2:
The polynomial under the square root sign in (3) is called F ( t ) . Lemma 1: t is a rational function of t if and only if F ( t ) is a perfect squarein F [ s , s , . . . , s , e, b ] [ t ] .Proof: If a = 0 and d = 0 the result is immediate. So assume a = 0 or d = 0.The “if” part of the statement is obvious. To prove the converse, assumethat there is rational function t = f /g with f, g ∈ F [ s , s , . . . , s , e, b ] [ t ].Inserting this into (3) and clearing denominators yields f · ( a t + d ) = g · ( − c t ± q F ( t ) )Multiply it out, collect terms, and solve for F ( t ): F ( t ) = p /q where p and q are polynomials in F [ s , s , . . . , s , e, b ] [ t ]. But that ring is aUFD. By a standard argument with irreducible polynomials, q divides p , sowe are done. ✷ Lemma 2: If F ( t ) is a perfect square in F [ s , s , . . . , s , e, b ] [ t ] then a b = 0 and d e = 0 .Proof: Note that F ( t ) = − a b t + ( c − a e − b d ) t − d e is a polynomial in t . If this is truly a quadratic in t and a perfect squarethen its discriminant must be 0. But when the parameters a , b , . . . , e arereplaced with their expressions (4) in terms of the eleven sides s , s , . . . , e , thediscriminant simplifies enormously to 256 e s s s . Therefore, the discrimi- We recommend a computer algebra system for this computation. However,Bricard did not have one! F ( t ) is linear in t . Thus, a = 0or b = 0.We now have that F ( t ) = ( c − a e − b d ) t − d e If this is a perfect square, then it equals some ( At + B ) . As there is no linearterm in F ( t ), it must be that A = 0 or B = 0.We will prove that A = 0 leads to a contradiction. We showed above that thereare two cases to consider, a = 0 or b = 0. Assume that a = 0. Then A = 0implies that c = b d . Since a = ( − s + e − s + s )( − s + e + s + s ) = 0this in turn leads to two cases. If − s + e − s + s = 0, then s + s = e + s .When this is plugged into the definitions of c , b and d , we see after somecomputation that c = b d reduces to s + s = s . But then e = 0, impossible.The second alternative, − s + e + s + s = 0, leads in the same way to thesame contradiction that e = 0.If b = 0, the argument is analogous. Now A = 0 implies that c = a e . b = ( s + e − s + s )( s + e + s + s ) = 0 so we again have two subcases.Each leads to the contradiction e = 0.Therefore, A = 0 is impossible so it must be that B = 0. Therefore d e = 0. ✷ Lemma 3:
In equation ( i ) , i = 1 , , , consider the six ways to choose a pairof { a i , b i , d i , e i } , the four parameters that might be 0. In all six cases, if thatpair of parameters is 0, the equation splits.Proof: We illustrate with i = 1. The case of { a , d } was shown in the aboveexample (3) with t a rational function of t . { a , b } is analogous, solving for t .For { b , e } we have a t t + d t + 2 c t t = 0Since t is a function taking on a range of values, it may be divided out andwe obtain a t t + d t + 2 c t = 0whence we may solve for t as t = − c t a t + d This is valid unless both d and a are 0. But if that were true, we would have2 c t t = 0, which is impossible. The case { d , e } is analogous, solving for t .9ow consider the case { a , e } . (1) reduces to d t + 2 c t t + b t = 0If d = 0 or b = 0 we are done. Otherwise, by the quadratic formula, t = − c t ± q ( c − d b ) t d so splitting depends on analysis of the polynomial under the radical. This issimply F ( t ) from Lemma 1 and Lemma 2. We proceed as follows. From therelations (4) we see that a and e are each the product of two linear polyno-mials in the eleven parameters s , . . . , e . Thus a = 0 = e leads to four cases,each of which is a system of two linear equations. This system may be solved,allowing some of the s i to be replaced with others. This greatly simplifies theexpressions in (4) for c , d , and b . Two cases lead to the contradiction e = 0.In the other two, we have F ( t ) = 16 e s t . Therefore, F ( t ) is a perfectsquare and we are done.The final case { b , d } is similar to { a , e } . (1) reduces to a t t + 2 c t t + e = 0If a = 0 we are done. Otherwise, by the quadratic formula, t t = − c ± q c − a e a c − a e = F ( t ) /t . Once again we use the relations (4) to produce fourcases. As before, the solution of two linear systems leads to the contradiction e = 0; in the other two we have F ( t ) = 16 e s t . Therefore, c − a e is aperfect square and we are done. ✷ t a rational function of t The cases and subcases in the previous section may seem bewildering. We havewritten a program in a computer algebra system to summarize the details ofthe four split cases for t a rational function of t . Recall from Lemma 2 thatwhen this occurs, we have a b = 0 and d e = 0. This leads to four cases a = 0 and d = 0; a = 0 and e = 0; b = 0 and d = 0; b = 0 and e = 0.As in the proof of Lemma 3, cases { a , e } and { b , d } , we use relations (4)to produce systems of two linear equations. This yields substitutions for one s i in terms of others, and produces four main cases, each with two subcases.The table shows the resulting F ( t ) and t in terms of t .10 = 0 , d = 0 : s = s , s = e : F ( t ) = 16 e s t , t = ( s t + et − s + e ) / (2 s t ) s = − s , s = e : F ( t ) = 16 e s t , t = ( s t − et − s − e ) / (2 s t ) a = 0 , e = 0 : s = s , s = e : F ( t ) = 16 e s t , t = t s + es − e or t = t s = s , s = − e : F ( t ) = 16 e s t , t = t s + es − e or t = t b = 0 , d = 0 : s = − s , s = e : F ( t ) = 16 e s t , t = − /t or t = e + s t ( e − s ) s = − s , s = − e : F ( t ) = 16 e s t , t = − /t or t = e + s t ( e − s ) b = 0 , e = 0 : s = s , s = − e : F ( t ) = 16 e s t , t = 0 (degenerate) or t = − s t / ( s t + et − s + e ) s = − s , s = − e : F ( t ) = 16 e s t , t = − s t / ( s t − et − s − e ) or t = 0 (degenerate)Some of the cases above lead to degenerate solutions, such as s = s , s = e .This is a “kite”, which was discussed in [16]. On the other hand, kites canbe part of a non-degenerate configuration if other conditions hold. Bricard [2]distinguished two types of (non-degenerate) split solutions. He was focusedon the octahedra. His Case two corresponds to two quadrilaterals similar, thethird a parallelogram.
Case three corresponds to all three quadrilaterals beingparallelograms.
Case one is non-split, which we now address.11
The main theoremTheorem 1:
Assuming flexibility, if none of the equations (1) − (3) split, theneach of the variables t i is a rational function of the other two. The main step in the proof of Theorem 1 is the following lemma. Althoughconsidered to be well known, we can find neither proof nor even precise state-ment of it, so we include it for completeness.
Lemma 4:
Let f and g be univariate polynomials over some field, say f = a n x n + . . . + a and g = b m x m + . . . + b , where a n b m a b = 0 . Let S be theirSylvester Resultant matrix, N × N , where N = n + m . If the rank of S = N − ,then there exists a polynomial h ( x ) of degree , whose coefficients are rationalfunctions of { a i , b j } , satisfied by all the common roots of f and g.Proof: In other words, x is rational function of the coefficients { a i , b j } . Notethat 0 is not a common root. We may assume N ≥ N −
1, we may perform row and column operations on S until we have S ′ = . . . c . . . c . . . c . . . . . . . . . . . . . . . . . . c r . . . where r = N −
1. All of the c i are rational combinations of the originalcoefficients { a i , b j } . For a common root x , the column vector (cid:20) x N − x N − . . . x x (cid:21) T is in the kernel of the original S . Since column swaps may have been made,the transformed vector p = (cid:20) x e x e . . . x e N − (cid:21) T is in the kernel of S ′ . The exponents are a permutation of the set { , , , . . . , N − } . If no column swaps were made, the permutation is the identity map and e N − = 0.If we multiply the matrix S ′ by the vector p we must get 0. That produces N − c i can be 0 as0 is not a common root). We distinguish three cases, according to e N − = 0 , k >
1. In the first case, one of the equations is x + c N − = 0, done. In thesecond case, one of the equations is 1 + c N − x = 0, done. In the third case,two of the equations are x + c j x k = 0 and 1 + c i x k = 0. Solve for x k in oneequation, plug into the other, done. ✷ Remark:
This theorem can be generalized to the situation where the rank ofthe Sylvester matrix is < N −
1, but we don’t need that here.To use Lemma 4, we apply the Sylvester resultant method to equations (2)and (3) to eliminate t . This Sylvester matrix is 4 × a t + d c t b t + e a t + d c t b t + e a t + d c t b t + e a t + d c t b t + e (5) Lemma 5:
The rank of the Sylvester matrix (7) is 3 almost everywhere.Proof:
Recall that the t i are functions of time that are not 0 on any nontrivialinterval.As equations in t , (2) and (3) are quadratic. The leading coefficients are a i t + d i and the “constant” terms are b i t + e i (where t is t or t ). Noneof these can vanish, as then that equation would split (see Lemma 3). Thehypotheses of Lemma 4 are satisfied.Since for all values of t and t in some interval equations (2) and (3) havecommon root(s), the rank is either 0, 1, 2, or 3. The rank is obviously not 0,as for example c = 0. If the rank were 1, then every 2 × × a t + d ) .This cannot be 0, as none of the equations split (see Lemma 3).If the rank were 2, then every 3 × a t + d c t a t + d c t b t + e a t + d c t Its determinant is ( a t + d )( − c a t t + c a t t + c d t − c d t ). If this is0, the second factor must be 0 (by Lemma 3). Examining the second factor,we distinguish 3 cases: • a = 0 and a = 0: Then again by Lemma 3, d = 0 and d = 0. Weimmediately solve for t as a rational function of t , contradiction. • a = 0: In the second factor, solve for t as a function of t (to the firstpower only) and t . Plug this into equation (1) to obtain( c a a t + c a d t + c d a t + 2 c c a t + c d d ) t - ( a d t + b a t − d d + e a ) c t = 0Unless the coefficient of t = 0, we can solve for t as a rational functionof t , contradiction. Therefore both expressions in parentheses are 0. Theseare both polynomial functions of t so their coefficients relative to t mustbe 0. We immediately see then that a a = 0 (coefficient of t ) and d d = 0(coefficient of t ). Again by Lemma 3, that yields only two possibilities: a = 0 , d = 0 or a = 0 , d = 0. We are soon led to contradictions, such as a must be 0, in both cases. The details are left to the reader. • a = 0: Exactly like the previous case, only solve for t instead of t .This competes the proof that the rank of Sylvester matrix is 3, except forisolated times when t or t could be 0. ✷ .The proof of Theorem 1 is now easy: since the rank of the Sylvester matrixis 3, use Lemma 4 to produce t as a rational function of t , t . By symmetry,any t i is a rational function of the other two. ✷ The proof of Lemma 5 allows us to deduce another result that will soon be ofinterest:
Lemma 6:
With the notation of Lemma 5, the rank of the Sylvester matrix(7) is at most 2 (almost everywhere) iff equations (2) and (3) are multiples ofeach other, by a nonzero rational function of t and t .Proof: The “if” part is obvious.Suppose the rank is at most 2. In the proof of Lemma 5 we used the 3 × a t + d )(2 c t ) − ( a t + d )(2 c t ) = 0If we also consider the minor formed by rows 2, 3, 4 and columns 1, 2, 4, weobtain (2 c t )( b t + e ) − (2 c t )( b t + e ) = 0Thus, c t c t times equation (2) equals equation (3). ✷ Corollary 1:
If none of the equations (1) − (3) split, then for almost all valuesof t , t , equations (2) and (3) , thought of as equations in t , do not have tworoots in common.Proof: If they had two roots in common, they would be multiples of eachother and the Sylvester rank would be no more than 2, contradicting Lemma3.Of course, the analogous statements can be made about the other pairs of(1) − (3) and the other t i . ✷ Since none of the equations (1) − (3) split, we may use the quadratic formulato solve for, say, t and t in terms of t : t = − c t ± q F ( t ) a t + d (5) t = − c t ± q F ( t ) a t + d (5)From Theorem 1 we know that t is a rational function of t and t . Therefore − c t ± q F ( t ) a t + d = φ t , − c t ± q F ( t ) a t + d where φ denotes a rational function.Expand φ and collect terms. This yields an expression P q F ( t ) + Q q F ( t ) = L + M q F ( t ) F ( t )15nd after squaring, eventually F ( t ) F ( t ) = R S for some polynomials P, Q, L, M, R, S . But again, as in Lemma 1, we are overa UFD, so we have proven
Theorem 2:
With the notation of (6) − (6) , in the non-split case the product F ( t ) F ( t ) is a perfect square, but neither F ( t ) nor F ( t ) is a perfect square. ✷ Obviously, the same statement is true for the analogous polynomials F ( t ) , F ( t ), F ( t ) , F ( t ) . Recall from (3) that the F polynomials are in general quartic with no cubicor linear terms: F ( t ) = − a b t + ( c − a e − b d ) t + d e (5) F ( t ) = − a b t + ( c − a e − b d ) t + d e (5)However, it is possible that, say, a = 0, reducing F ( t ) to a quadratic. Let usabbreviate F ( t ) ≡ F, F ( t ) ≡ F . We distinguish three cases: • Both F and F are quartic. • Both F and F are quadratic. • One of F and F is quartic and one is quadratic.Bricard seems to have missed the possibility of the third case, which we call“quart-quad.” As he is mostly concerned with octahedra, perhaps he elim-inated that case by some three dimensional argument. We used Solve andfound no non-split solutions of quart-quad. Motivated by these experiments,we found a purely algebraic proof of the next Theorem:
Theorem 3:
If one of F and F is quartic and one is quadratic, then we havea split case.Proof: Suppose without loss of generality that F is quadratic, so a b = 0. F and F can be factored in some extension field, yielding F = p ( t − α )( t − α ) (5) F = q ( t − β )( t + β )( t − β )( t + β ) (5)Since neither F nor F is a perfect square but their product is, relabeling ifnecessary we must have that α = β , α = − β , β = 0. Thus in F , d e = 0. The notation follows Bricard’s. Strictly speaking, this is not one function F or F being applied to different t i as the parameters vary. F · F is a perfect square (in the polynomial ring) we have F = s t F ,where s is some polynomial. Therefore, √ F = st √ F .Since a b = 0 and d e = 0, there are four cases. Let’s consider first a = 0and d = 0. Then we have from (6), (6) t = ( − c t + √ F ) /d (6) t = ( − c t + √ F ) / ( a t ) (7)In the second equation, replace √ F with s t √ F . Simplifying, we have a t t = − c + s √ F . Now solve for √ F in the other equation, plug into the above. Weget s d t = s c t + a t t + c (7)So t is a linear function of t or t . Now, we know from Theorem 1 that any t i is a rational function of the others, so (6) may not seem surprising. However,the important fact is that all the exponents are 1.Solve (6) for t and plug that into equation (2). Collecting terms yields: m t + m t + m = 0 (7)The m i are polynomials in t , up to degree 4.Suppose first that no m i = 0. The key point is that this equation is quadraticin t , which is because (6) is linear in each t i . Since it is quadratic in t we canapply the same logic used in the proof of Lemma 5, Lemma 6, and Corollary1 to the pair of equations (6) and (3). There must be common root(s). Ifthe rank of the Sylvester matrix is three, then t is a rational function of t (earlier theorem), so we have a split case. If the rank is less than three, thenthe two equations are multiples of each other (by a polynomial in the elevenparameters). However, this is impossible because the constant (degree zero)terms in the two equations are m = s d d t + c a t + 2 sc c d t + s d e b c , and d t + e Note that the coefficient of t in the first is 2 sc c d but there is no t termin the second. Thus, sc c d = 0, but sc c d can’t be 0 unless d = 0, whichimplies splitting, since a = 0.Now suppose that some m i = 0. These are equations in t alone, so m i = 0implies every coefficient in it is 0. As shown above, this is impossible for m (without splitting). The vanishing of m is irrelevant, as (6) remains quadratic.If only m = 0, then t is a rational function of t , hence a splitting. This provesthe first case, that a = 0 = d implies splitting.17or the other three cases, one of them is just as above, but the other twoseem harder, because we no longer have the simple monomial denominators of(6). However, recall that in forming equations (1) − (3) we may use cotangentas well as tangent, which means we can replace t i with t − i . That has theeffect of switching a i ↔ e i and b i ↔ d i , which reinstates the needed monomialdenominators. ✷ Suppose now that both F and F are quartic. In a splitting field we have F = p ( t − α )( t + α )( t − α )( t + α ) F = q ( t − β )( t + β )( t − β )( t + β )But since F F is a perfect square, each α i must equal some ± β j . Therefore, F and F are multiples of each other. The same is true if both F and F arequadratic. This is the key fact in the proof of: Theorem 4:
Referring to Figure 1 for the angles θ , θ opposite to α , in a non-split flexible case we have that cos( θ ) = ± cos( θ ) . The same cosine relationis true for the angles CDG and
IHE opposite to γ (technically π − γ ) and for HIB and
AGF opposite to β (and π − β ).Proof: We emphasize that this is true throughout the flex. Unlike Bricard’srather specialized geometric argument, we give an algebraic proof.Since it is a non-split case we know from above that F and F are multiplesof each other; let r be the ratio. Comparing coefficients, d e − r d e = 0 b a − r b a = 0( c − a e − b d ) − r ( c − a e − b d ) = 0Draw lines GE and DB . GE is on two triangles, one containing α , one con-taining θ . DB is on two triangles, one containing α , one containing θ . Fromthe law of cosines we deduce: s + e − e s cos( α ) − ( s + s − s s cos( θ )) = 0 s + b − b s cos( α ) − ( s + s − s s cos( θ )) = 0If we consider the cosines as abstract variables, (18) − (22) is a system of fivepolynomial equations. Plug in relations (4) for a , . . . , d . Using resultants, wecan eliminate any four variables. If we eliminate s , s , r , cos( α ), the resultantis quite simple and has these factors:cos( θ ) , s , b, s , s , s , e, cos ( θ ) − cos ( θ )As the resultant must vanish, at least one of these factors must be 0. cos( θ )18an’t be 0, as then cos( α ) would be a constant. The only choice is thatcos ( θ ) = cos ( θ ).The other cases are similar. ✷ Using Theorem 4 we can form a system of six equations to effectively describethe non-split case. Assume first that cos( θ ) = cos( θ ). If we eliminate cos( θ )from equations (21) and (22), we obtain an expression involving cos( α ) thatmust be 0, of the form A cos( α ) + B . As we assume non-degeneracy, this canonly be true if A and B are both 0. We repeat the argument with the twoother quadrilaterals, yielding the following six equations (set each to 0): es s s − bs s s , (23) − s s s − s s s + s s s + s s s − e s s − s s s + s s s + b s s ,bs s s + es s s − bs s s ,s s s + s s s − s s s − s s s + b s s + s s s − s s s − e s s + 2 bes s − b s s ,es s s − bs s s − es s s ,s s s + s s s − s s s − s s s + e s s − bes s + b s s + s s s − s s s − e s s Minor variations result by using cos( θ ) = − cos( θ ), etc. The program
Solve was described in [16] and [11]. Here is a brief description.Let res be the resultant of a system of equations defining a structure, such as(1) − (3). res is a polynomial in one of the angles, say t , and the fifteen param-eters a , b , . . . , e , or alternatively, in the eleven side parameters s , s , . . . , e .If the structure is flexible, then infinitely many values of t satisfy the poly-nomial. The only way this can be is if every coefficient of t k vanishes. Solve examines these coefficients finding ways to kill them one-by-one, usually start-ing at the top coefficient. Whenever a way is found to kill the coefficient of t k ,that substitution is put on a stack and applied to res , creating a polynomial res ′ of fewer terms and one fewer parameter. Then Solve calls itself on the newpolynomial res ′ . This is essentially an enormous tree search. Many heuristicsand techniques are used to keep the search manageable yet effective.The output of the algorithm is a list of tables consisting of substitutions of theform s i = p ( s , s , . . . , ˆ s i , . . . ), where p ∈ GF . Here is a simple example. If res were ( s s − s s ) t + ( s − s ) t + s − s , one solution would be the tableof the three relations s = s , s = s , s = s .The relations may be described as follows: Partition the set of N parameters19 ig. 2. Flexing of case one quadrilaterals, forming “isohexagon.” into nonempty subsets X = { x i } ni =1 , Y = { y j } mj =1 , n + m = N . Each relationis an equation y j = g ( x i , x i , . . . ) where g is a rational function. A collectionof m of these for j = 1 , . . . , m is a solution table if res evaluated at them allis 0. In the example above X = { s , s , s } and Y = { s , s , s } . Problem 3: Can all flexible cases be represented by a table of rela-tions in the above sense?
To apply this to the quadrilaterals of Figure 1, we eliminate two of the threeangles in equations (1) − (3). In terms of the eleven side parameters, res has190981 terms . From 2006 to 2011, Solve(res) found many flexible cases ofBricard’s types two and three, and many degenerate cases [16], [11]. Improve-ments by 2012 yielded the first non-split cases, Bricard’s case one. These wereall what we call isohexagons . Here is an actual table as computed by
Solve : s = ( b s ) / ( b − e ) s = s ( e − b ) /es = ( e s ) /bs = s ( b − e ) /bs = ( e s ) / ( b − e ) s = ( b s ) /e Note that s is negative; that just means that point H is below the x -axis. Theyare called isohexagons because a hexagon with equal opposite sides appearsin the midst of the flex. Here are two images (Figure 2), taken from a modelmade by plugging in numbers for the sides. The hexagon is outlined in blue.Note that it does not consist entirely of sides s i .These isohexagons exhibit a curious kind of symmetry or “quasi-similarity.”Let r = ( b − e ) /e . By late 2013, this computation takes
Fermat lower lef t lower right largee b − e = r e b = (1 + r ) es s = − r s s = (1 + r ) s s s = r s s = (1 + r ) s s s = r s s = (1 + r ) s Each column lists the four sides of one of the quadrilaterals. Note that the sidesof the lower right and large quadrilaterals are multiples of those of the lowerleft, but in an odd shifting pattern; none of these quadrilaterals is similar toanother. From this and other examples we see that the following isohexagonproperty holds: for every one of the twelve sides, say x , there is a side ona different quadrilateral, say y , such that x/y or y/x equals r, r, − r , or − (1 + r ). Bricard, who studied only octahedra, remarks that in case one thereis an odd symmetry also.Thus, the isohexagon may be thought of as the three-quadrilaterals analogueof the octahedra case one. We therefore conjectured that all case one exampleswere isohexagons. Surprisingly, this is false. To see why, recall the six equations(23). All six of these must be 0. Form a single polynomial f = c t + c t + c t + c t + c t + c (8)where { c i } are the six equations in (23) and t is an abstract variable. Execute Solve(f ) , adding code to suppress split cases. This finishes very quickly with136 tables, several of which are striking, such as s = s ( − b e s + b s + b e − b e ) / ( e s − e s ) s = s ( e − b ) /es = ( e s ) / ( b − e ) s = ( e s ) / ( b − e ) s = ( − b s ) /es = s ( b e s − b s − b e + b e ) / ( e s − e s )This is not an isohexagon, as it does not satisfy the isohexagon property.Substituting numerical values, we created a model of this case. Figure 3 showstwo snapshots during the flex.The last case above seems rather complicated due to the s and s equations.Notice that s /s = s /s . To experiment, we removed the s and s equations,added s = s s /s , and plugged the resulting table into the resultant res of21 ig. 3. Flexing of case one quadrilaterals, not an isohexagon. res ′ of 8803 terms.Further analysis of this polynomial revealed a surprising “irrational case”: s = ( − b s ) / ( e − b ) ,s = s ( e − b ) /e,s = e s / ( e − b ) ,s = e s / ( e − b ) ,s = b s /es = ( − b s ) / ( e − b ) s = ( e s + e s − b e s + b s − e + 2 b e − b e ) /e (Use the last equation to replace s in the earlier relations.) This arises because s occurs with only even degree in res ′ . In the definition of table, the sets X and Y are X = { b, e, s , s } Y = { s , s , s , s , s , s , s } Figure 4 is an image of an instantiation of this, plugging in real numbers forthe X (and then Y ) parameters.The polynomial in the definition of s is easily seen to be not a perfect square.Therefore, the answer to Problem 3 is “no.”The three structures in Figures 2 − Solve ( f ) for f defined in (7). However, all three now show up with the latest version of Solve(res) . This is because when
Solve encounters a polynomial (like res ′ ) inwhich a variable (here s ) occurs with every exponent a multiple of n (here2), the exponents are divided by n and the algorithm continues.22 ig. 4. Flexing of case one quadrilaterals, irrational relationship. This problem of the flexible planar linkages was posed by Bricard in his mem-oir on the flexible octahedron. He seemed to imply that the two problemswould have completely analogous resolutions given that they are described bysystems of equations of identical form. As Bricard pointed out in his memoir,“it ought to be possible to analyze these equations by purely algebraic means,however the amount of computation required would be daunting”. He pro-ceeded therefore to analyze his equations geometrically, arriving at his wellknown three classes of flexible octahedra. Here, with the help of computeralgebra, which a hundred years after Bricard is now a mature field, we wereable to carry out this “daunting” task for the planar mechanism case, and wererewarded by a surprising divergence from Bricard’s conclusions. Although theseparation to three classes according to the type of splitting is identical forthe two problems, the underlying geometric differences led to unexpectedlyrich properties for the structures of case one, the case of no splitting, with noanalogs in the octahedron. The other two split cases seem to be completelyanalogous for the two problems.Earlier we defined three problems:
Problem 1: Find conditions on the sides under which the quadrilateral arrange-ment becomes flexible.
This has been solved.
Problem 1 ′ : Find all conditions on the sides under which the quadrilateralarrangement becomes non-degenerate flexible. We do not have a mathematicalproof that our list is complete. However we have found cases analogous to allof Bricard’s cases for the articulated octahedron, and discovered unexpectedlyrich properties for case one, where our algebraic analysis led to two quitedifferent types of flexible structures with no apparent analogy to Bricard’sthree-dimensional results. 23 roblem 2: When is one of these variables, t , say, a rational function ofanother t j , or a rational function of both of the other ones t , t ? This hasbeen solved. Our proofs are new and algebraic.
Problem 3: Can all flexible cases be represented by a table of relations in[our] sense?
No. However, we believe that our algorithm needs to be mod-ified only by changing the definition of table to allow relations of the form s ni = p ( s , s , . . . , ˆ s i , . . . ).The great success we have had on this project bodes well for future work withmore complex structures, as equations describing those structures are alsoquadratic, based on distances and angles.Work is ongoing applying these methods directly to the octahedra and to thecyclo-octane molecule. References [1] P. Bikker, On Bezout’s method for computing the resultant, RISC-Linz ReportSeries, Johannes Kepler University, Linz, Austria (1995).[2] R. Bricard, M´emoire sur la th´eorie de l’octa`edre articul´e, J. Math. Pures Appl.3 (1897) 113 - 150 (English translation: http://arxiv.org/abs/ 1203.1286).[3] L. Buse, M. Elkadi, B. Mourrain, Generalized resultants over unirationalalgebraic varieties. J. Symbolic Comp. 29 (2000) 515 - 526.[4] A. Cauchy, Deuxi`eme M´emoire sur les polygones et les poly`edres, J. ´EcoleImp´eriale Polytechnique, XVI Cahier (1813) 87 - 99.[5] R. Connelly, A counterexample to the rigidity conjecture for polyhedra. Publ.Math. I. H. E. S. 47 (1978) 333 - 338.[6] E. A. Coutsias, C. Seok, M. J. Wester, K. A. Dill, Resultants and loop closure,Int. J. of Quantum Chem. 106 (2005) no. (1) 176 - 189.[7] E. A. Coutsias, C. Seok, M. J. Jacobson, K. A. Dill, A Kinematic view of loopclosure, Journal of Comput. Chem. 25 (2004) no. (4) 510 - 528.[8] D. Cox, J. Little, D. O’Shea, Using Algebraic Geometry, Graduate Texts inMathematics 185, Springer-Verlag, New York, 1998.[9] P. R. Cromwell, Polyhedra, Cambridge University Press, New York, 1997, 222- 224.[10] A. L. Dixon, The eliminant of three quantics in two independent variables, Proc.London Math. Society, 6 (1908) 468 - 478.
11] S. Fox, R. H. Lewis, Algebraic detection of flexibility of polyhedral structureswith applications to robotics and chemistry, Fordham Undergraduate ResearchJournal 2 (2012) 46 - 49.[12] K. H. Hunt, Kinematic Geometry of Mechanisms, Oxford, Cambridge, 1978.[13] D. Kapur, T. Saxena, L. Yang, Algebraic and geometric reasoning using Dixonresultants, in: Proc. of the International Symposium on Symbolic and AlgebraicComputation, A.C.M. Press, 1994, pp. 99 - 107.[14] R. H. Lewis, Computer algebra system Fermat. http://home.bway.net/ lewis.[15] R. H. Lewis, Heuristics to accelerate the Dixon resultant, Math. and Comput.in Simul. 77 (2008) 400 - 407.[16] R. H. Lewis, E. A. Coutsias, Algorithmic search for flexibility using resultantsof polynomial systems, in: F. Botana, T. Recio (Eds), Automated Deduction inGeometry, Lecture Notes in Comp. Sci., Vol. 4869, Springer, Berlin, 2007, pp.68 - 79.[17] R. H. Lewis, P. Stiller, Solving the recognition problem for six lines using theDixon resultant, Math. and Comput. in Simul. 49 (1999) 205 - 219.[18] I. Maksimov, Polyhedra with bendings and Riemann surfaces, UspekhiMatemat. Nauk 50 (1995) 821 - 823.[19] D. Manocha, J. Canny, Efficient techniques for multipolynomial resultantalgorithms, in: ISSAC ’91, Proc. of the 1991 Int. Symp. on Symb. and Algebr.Comput. ACM, New York, 1991, pp. 86 - 95.[20] D. Robertz, V. Gerdt, Comparison of software systems athttps://home.bway.net/lewis/fermat/gcdcomp (2004).[21] M. Thorpe, M. Lei, A. J. Rader, D. J. Jacobs, L. Kuhn, Protein flexibility anddynamics using constraint theory, J. of Mol. Graph. and Model. 19 (2001) 60 -69.[22] D. Walter, M. L. Husty. A spatial nine-bar linkage, possible configurations andconditions for paradoxical mobility, 13th National Conference on Mechanismsand Machines (NaCoMM07), IISc, Bangalore, India, 2007, pp. 195 - 201.11] S. Fox, R. H. Lewis, Algebraic detection of flexibility of polyhedral structureswith applications to robotics and chemistry, Fordham Undergraduate ResearchJournal 2 (2012) 46 - 49.[12] K. H. Hunt, Kinematic Geometry of Mechanisms, Oxford, Cambridge, 1978.[13] D. Kapur, T. Saxena, L. Yang, Algebraic and geometric reasoning using Dixonresultants, in: Proc. of the International Symposium on Symbolic and AlgebraicComputation, A.C.M. Press, 1994, pp. 99 - 107.[14] R. H. Lewis, Computer algebra system Fermat. http://home.bway.net/ lewis.[15] R. H. Lewis, Heuristics to accelerate the Dixon resultant, Math. and Comput.in Simul. 77 (2008) 400 - 407.[16] R. H. Lewis, E. A. Coutsias, Algorithmic search for flexibility using resultantsof polynomial systems, in: F. Botana, T. Recio (Eds), Automated Deduction inGeometry, Lecture Notes in Comp. Sci., Vol. 4869, Springer, Berlin, 2007, pp.68 - 79.[17] R. H. Lewis, P. Stiller, Solving the recognition problem for six lines using theDixon resultant, Math. and Comput. in Simul. 49 (1999) 205 - 219.[18] I. Maksimov, Polyhedra with bendings and Riemann surfaces, UspekhiMatemat. Nauk 50 (1995) 821 - 823.[19] D. Manocha, J. Canny, Efficient techniques for multipolynomial resultantalgorithms, in: ISSAC ’91, Proc. of the 1991 Int. Symp. on Symb. and Algebr.Comput. ACM, New York, 1991, pp. 86 - 95.[20] D. Robertz, V. Gerdt, Comparison of software systems athttps://home.bway.net/lewis/fermat/gcdcomp (2004).[21] M. Thorpe, M. Lei, A. J. Rader, D. J. Jacobs, L. Kuhn, Protein flexibility anddynamics using constraint theory, J. of Mol. Graph. and Model. 19 (2001) 60 -69.[22] D. Walter, M. L. Husty. A spatial nine-bar linkage, possible configurations andconditions for paradoxical mobility, 13th National Conference on Mechanismsand Machines (NaCoMM07), IISc, Bangalore, India, 2007, pp. 195 - 201.