3264 Conics in a Second

TL;DR

This paper demonstrates how enumerative and numerical algebraic geometry complement each other by computing the 3264 conics tangent to five given conics, using HomotopyContinuation.jl for efficient solutions.

Contribution

It introduces a web interface leveraging HomotopyContinuation.jl to compute all solutions to a classical enumerative geometry problem efficiently.

Findings

Successfully computed all 3264 solutions for a specific instance.

Identified an example with all solutions being real.

Showcased the integration of theoretical and numerical methods.

Abstract

Enumerative algebraic geometry counts the solutions to certain geometric constraints. Numerical algebraic geometry determines these solutions for any given instance. This article illustrates how these two fields complement each other. Our focus lies on the 3264 conics that are tangent to five given conics in the plane. We present a web interface for computing them. It uses the software HomotopyContinuation.jl, which makes this process fast and reliable. We discuss an instance where all 3264 solutions are real.

Equations

$$A(x,y)=a_1{x}^2+a_{2}xy+a_3{y}^2+a_{4}x+a_{5}y+a_6.$$

$$U(x,y)=u_1{x}^2+u_{2}xy+u_3{y}^2+u_{4}x+u_{5}y+u_6,$$

$$\begin{array}[]{rl}&\frac{\partial A}{\partial x}\cdot\frac{\partial U}{\partial y}-\frac{\partial A}{\partial y}\cdot\frac{\partial U}{\partial x}\,=\,2(a_{1}u_{2}-a_{2}u_{1})x^{2}\\ &\,+4(a_{1}u_{3}-a_{3}u_{1})xy+\cdots+(a_{4}u_{5}-a_{5}u_{4}).\end{array}$$

$$\begin{array}[]{l}A(x,y)\,=\,a_{1}x^{2}+a_{2}xy+a_{3}y^{2}+a_{4}x+a_{5}y+a_{6},\\[2.70003pt] B(x,y)\,=\,b_{1}x^{2}+b_{2}xy+b_{3}y^{2}+b_{4}x+b_{5}y+b_{6},\\[2.70003pt] C(x,y)\,=\,c_{1}x^{2}+c_{2}xy+c_{3}y^{2}+c_{4}x+c_{5}y+c_{6},\\[2.70003pt] D(x,y)\,=\,d_{1}x^{2}+d_{2}xy+d_{3}y^{2}+d_{4}x+d_{5}y+d_{6},\\[2.70003pt] E(x,y)\,=\,e_{1}x^{2}+e_{2}xy+e_{3}y^{2}+e_{4}x+e_{5}y+e_{6}.\end{array}$$

$$\begin{array}[]{rl}\mathcal{T}(A,U)\,=&\mkern-12.0mu256a_{1}^{4}a_{3}^{2}u_{3}^{2}u_{6}^{4}-128a_{1}^{4}a_{3}^{2}u_{3}u_{5}^{2}u_{6}^{3}\\ &\mkern-12.0mu+\,16a_{1}^{4}a_{3}^{2}u_{5}^{4}u_{6}^{2}+\cdots+a_{5}^{4}a_{6}^{2}u_{1}^{2}u_{2}^{4}.\end{array}$$

$$\mathcal{T}(A,U)=\mathcal{T}(B,U)=\cdots =\mathcal{T}(E,U)=0.$$

$$U(x,y)=(x,y,1)\cdot {\ell }^T\ell \cdot (x,y,1{)}^T,$$

$$\mathrm{rank}\left(\begin{array}{ccc}2u_1& u_2& u_4\\ u_2& 2u_3& u_5\\ u_4& u_5& 2u_6\end{array}\right)\le 1.$$

$$\begin{array}[]{l}\text{\em Find all solutions $U$ of the equations (\ref{eq:formulation1})}\\ \text{\em such that the matrix in (\ref{eq:3by3}) has rank $\geq 2$.}\end{array}$$

$$\begin{array}[]{l}\text{\em Find an explicit instance $A,B,C,D,E$ such}\\ \text{\em that the $3264$ solutions $U$ to (\ref{ourproblem}) are all real.}\end{array}$$

$$\begin{array}[]{l}\begin{bmatrix}a_{1}&a_{2}&a_{3}&a_{4}&a_{5}&a_{6}\\ b_{1}&b_{2}&b_{3}&b_{4}&b_{5}&b_{6}\\ c_{1}&c_{2}&c_{3}&c_{4}&c_{5}&c_{6}\\ d_{1}&d_{2}&d_{3}&d_{4}&d_{5}&d_{6}\\ e_{1}&e_{2}&e_{3}&e_{4}&e_{5}&e_{6}\end{bmatrix}\,\,=\,\,\begin{bmatrix}\frac{10124547}{662488724}&\frac{8554609}{755781377}&\frac{5860508}{2798943247}&\frac{-251402893}{1016797750}&\frac{-25443962}{277938473}&1\\[5.0pt] \frac{520811}{1788018449}&\frac{2183697}{542440933}&\frac{9030222}{652429049}&\frac{-12680955}{370629407}&\frac{-24872323}{105706890}&1\\[5.0pt] \frac{6537193}{241535591}&\frac{-7424602}{363844915}&\frac{6264373}{1630169777}&\frac{13097677}{39806827}&\frac{-29825861}{240478169}&1\\[5.0pt] \frac{13173269}{2284890206}&\frac{4510030}{483147459}&\frac{2224435}{588965799}&\frac{33318719}{219393000}&\frac{92891037}{755709662}&1\\[5.0pt] \frac{8275097}{452566634}&\frac{-19174153}{408565940}&\frac{5184916}{172253855}&\frac{-23713234}{87670601}&\frac{28246737}{81404569}&1\end{bmatrix}.\end{array}$$

$$\left(\begin{array}{cc}2u_1& u_2\\ u_2& 2u_3\end{array}\right).$$

$$P^5=L^5=1,P^{4}L=P{L}^4=2,P^3{L}^2=P^2{L}^3=4.$$

$$C=2P+2L.$$

$$\mathcal{T}=16\cdot u_6^2(4u_3{u}_6-u_5^2{)}^2\cdot a_1^4{a}_2^3\mathrm{mod}⟨a_2,a_3^3,a_4,a_5,a_6⟩.$$

$$|a_1|\gg |a_3|\gg \max \{|a_2|,|a_4|,|a_5|,|a_6|\}.$$

$$\begin{array}[]{rl}C^{5}\mkern-18.0mu&=32(L+P)^{5}\\ &=32(L^{5}+5L^{4}P+10L^{3}P^{2}+10L^{2}P^{3}+5LP^{4}+P^{5})\\ &=32(1+5\cdot 2+10\cdot 4+10\cdot 4+5\cdot 2+1)\\ &=32\cdot 102=3264\,.\end{array}$$

$$\left[\begin{array}{ccc}A(x_1,y_1)& U(x_1,y_1)& (\frac{\partial A}{\partial x}\frac{\partial U}{\partial y}-\frac{\partial A}{\partial y}\frac{\partial U}{\partial x})(x_1,y_1)\\ B(x_2,y_2)& U(x_2,y_2)& (\frac{\partial B}{\partial x}\frac{\partial U}{\partial y}-\frac{\partial B}{\partial y}\frac{\partial U}{\partial x})(x_2,y_2)\\ C(x_3,y_3)& U(x_3,y_3)& (\frac{\partial C}{\partial x}\frac{\partial U}{\partial y}-\frac{\partial C}{\partial y}\frac{\partial U}{\partial x})(x_3,y_3)\\ D(x_4,y_4)& U(x_4,y_4)& (\frac{\partial D}{\partial x}\frac{\partial U}{\partial y}-\frac{\partial D}{\partial y}\frac{\partial U}{\partial x})(x_4,y_4)\\ E(x_5,y_5)& U(x_5,y_5)& (\frac{\partial E}{\partial x}\frac{\partial U}{\partial y}-\frac{\partial E}{\partial y}\frac{\partial U}{\partial x})(x_5,y_5)\end{array}\right].$$

$$z_{k+1}=N_F(z_k),\text{ where }{N}_F(x)=x-J_F(x{)}^{-1}F(x)$$

$$\Vert z_{k+1}-\zeta \Vert \le \frac{1}{2}\Vert z_k-\zeta {\Vert }^2.$$

$$\gamma (F,z)$$

$$\alpha (F,z)$$

$$H(x,t)=F_{t\cdot (A,B,C,D,E)+(1-t)\cdot (A^{\prime },B^{\prime },C^{\prime },D^{\prime },E^{\prime })}(x).$$

$$\tilde{H}(x,t)=t{F}_{(A,B,C,D,E)}+(1-t)F_{(A^{\prime },B^{\prime },C^{\prime },D^{\prime },E^{\prime })}.$$

$$\{F_{(A,B,C,D,E)}(x)\mid (A,B,C,D,E)\text{ are conics}\}.$$

Full text

We provide an animation showing all the 3264 real conics of this instance at this URL:

[TABLE]

The construction of our example originates from an arrangement of double lines, which we call the pentagon construction. One can see the pentagon in the middle of Figure 3. There are points where the red conics seems to intersect a blue line but they are actually points where the red conic touches one branch of a blue hyperbola. See [Sot] for further details.

Later we shall discuss the algebro-geometric meaning of the pentagon construction, and we present a rigorous computer-assisted proof that indeed all of the $3264$ conics tangent to our five conics are real. But, first, let us introduce our web browser interface.

Do It Yourself

In this section we invite you, the reader, to chose your own instance of five conics. We offer a convenient way for you to compute the $3264$ complex conics that are tangent to your chosen conics. Our web interface for solving instances of Steiner’s problem is found at

[TABLE]

Here you can type in your own $30=5\times 6$ coefficients for the conics in (4). After specifying five conics, you press a button and this calls the numerical algebraic geometry software HomotopyContinuation.jl. This is the open source Julia package described in [BT18]. Those playing with the web interface need not worry about the inner workings. But, if you are curious, please read our section titled How Does This Work?

Shortly after the user submits their instance, by entering real coefficients, the web interface reports whether the instance was generic enough to yield $3264$ distinct complex solutions. These solutions are computed numerically. The browser displays the number of real solutions, along with a picture of the instance and a rotating sample of real solutions. As promised in our title, the computation of all solutions takes only around one second.

Remark 2**.**

We always assume that the five given conics are real and generic. This ensures that there are $3264$ complex solutions, and these conics are tangent to the given conics at $5\times 3264$ distinct points. The number of real solutions is even, and our web interface displays them sequentially. For every real solution, the points of tangency on the given conics are also real. This fact uses the genericity assumption, since two particular real conics can be tangent at two complex conjugate points. For instance, the conics defined by $x^{2}-y^{2}+1$ and $x^{2}-4y^{2}+1$ are tangent at the points $(i:0)$ and $(-i:0)$ where $i=\sqrt{-1}$.

Figure 4 shows what the input and the visual output of our web interface looks like. The user inputs five conics and the system shows these in blue. After clicking the “compute” button, it responds with the number of complex and real conics that were found. The $3264$ conics, along with all points of tangency, are available to the user upon request. The real conics are shown in red, as seen on the right in Figure 4.

When seeing this output, the user might ask a number of questions. For instance, among the real conics, how many are ellipses and how many are hyperbolas? Our web interface answers this question. The distinction between ellipses and hyperbolas is characterized by the eigenvalues of the real symmetric matrix

[TABLE]

If the two eigenvalues of this matrix have opposite signs, then the conic is a hyperbola. If they have the same sign, then the conic is an ellipse. Among the ellipses, we might ask for the solution which looks most like a circle. Our program does this by minimizing the expression $\,(u_{1}-u_{3})^{2}+u^{2}_{2}$. Users with a numerical analysis background might be interested in maximizing the distance to the degenerate conics. Equivalently, we ask: among all $3264$ solutions, which $3\times 3$ matrix in (7) has the smallest condition number?

You can adapt all this for your favorite geometry problems. As pointed out above, the Julia package HomotopyContinuation.jl is available to everyone – follow the link at [BT18]. This may enable you to solve your own polynomial systems in record time.

Chow Rings and Pentagons

We next present the approach to deriving the number $3264$ that would be taught in an algebraic geometry class, along the lines of the article [BKT08]. Thereafter we explain the geometric degeneration we used to construct the fully real instance in Proposition 1.

Steiner phrased his problem as that of solving five equations of degree six on the five-dimensional space $\mathbb{P}^{5}_{\mathbb{C}}$. The incorrect count occurred because of the locus of double conics in $\mathbb{P}^{5}_{\mathbb{C}}$. This is a surface of extraneous solutions. One fixes the problem by replacing $\mathbb{P}^{5}_{\mathbb{C}}$ with another five-dimensional manifold, namely the space of complete conics. This space is the blow-up of $\mathbb{P}^{5}_{\mathbb{C}}$ at the locus of double lines. It is a compactification of the space of nonsingular conics that has desirable geometric properties. A detailed description of this construction, suitable for a first course in algebraic geometry, can be found in [BKT08, §5.1].

In order to answer enumerative geometry questions about the space of complete conics, one considers its Chow ring, as explained in [BKT08, §5.2]. Elements in the Chow ring of the space of complete conics correspond to subvarieties of this space. More precisely, to classes of subvarieties. Two subvarieties belong to the same class if and only if they are rationally equivalent. Rational equivalence is a technical concept. We refer interested readers to the textbook by Eisenbud and Harris [EH16]. The Chow ring for the space of complete conics is worked out in [EH16, §8.2.4]. Nevertheless, the idea behind studying Chow rings is crystal clear: taking intersections of varieties is translated to multiplication in the Chow ring. In the remainder of this section we will see this in action.

The Chow ring of the space of complete conics contains two special classes $P$ and $L$. The class $P$ encodes the conics passing through a fixed point, while the class $L$ encodes the conics tangent to a fixed line. The following relations hold in the Chow ring:

[TABLE]

These relations are derived in [BKT08, §4.4–5.3]. For instance, the first equation means that, if we take five general conics passing through a fixed point, then the intersection contains one point (namely the point we fixed in the first place). See [BKT08, Table 3] for the geometric meaning of the other equations.

We write $C$ for the class of conics that are tangent to a given conic. In the Chow ring, we have

[TABLE]

This identity is derived in [BKT08, equation (8)]. Our preferred proof is to inspect the first three terms in the expression (5) for the tact invariant $\mathcal{T}(A,U)$:

[TABLE]

This has the following intuitive interpretation. We assume that the given fixed conic $A$ satisfies

[TABLE]

Thus the conic $A$ is close to $x^{2}-\epsilon y^{2}$, where $\epsilon$ is a small quantity. The process of letting $\epsilon$ tend to zero is understood as a degeneration in the sense of algebraic geometry. With this, the condition for $U$ to be tangent to $A$ degenerates to $\,u_{6}^{2}\cdot(4u_{3}u_{6}-u_{5}^{2})^{2}=0$.

The first factor $u_{6}$ represents all conics that pass through the point $(0,0)$. The second factor $4u_{3}u_{6}-u_{5}^{2}$ represents all conics tangent to the line $\{x=0\}$. The Chow ring classes of these factors are $P$ and $L$. Each of these arises with multiplicity $2$, as seen from the exponents. The desired intersection number is now obtained from the Binomial Theorem:

[TABLE]

The final step in turning this into a rigorous proof of Chasles’ result is carried out in [BKT08, §7].

The degeneration idea in (11) can be used to construct real instances of Steiner’s problem whose $3264$ solutions are all real. Fulton first observed this and communicated it to Sottile, who then wrote down Fulton’s proof in detail [Sot95, Sot]. Ronga, Tognoli and Vust [RTV97] independently gave a proof. Apparently, they did not know about Fulton’s ideas.

Fix a convex pentagon in $\mathbb{R}^{2}$ and one special point somewhere in the relative interior of each edge. Consider all conics $C$ such that, for each edge of the pentagon, $C$ either passes through the special point or is tangent to the line spanned by the edge. By the count above, there are $(L+P)^{5}=102$ such conics $C$. If the pentagon is chosen sufficiently asymetric, then the $102$ conics are all real. We now replace each pointed edge by a nearby hyperbola, satisfying (11). For instance, if the edge has equation $x=0$ and $(0,0)$ is its special point then we take the hyperbola $x^{2}-\epsilon y^{2}+\delta$, where $\epsilon>\delta>0$ are very small. After making appropriate choices of these parameters along all edges of the pentagon, each of the $102$ conics splits into $32$ conics, each tangent to the five hyperbolas. Here ’splits’ means, if the process is reversed, then the $32$ different conics collapse into one solution of multiplicity 32. By construction, all $3264$ conics are real.

The argument shows that there exists an instance in the neighborhood of the pentagon whose $3264$ conics are all real, but it does not say how close they should be. Serious hands-on experimentation was necessary for finding the instance in Proposition 1.

We next present an alternative formulation of Steiner’s conic problem. The idea is to remember the five points of tangency on each solution conic. The five sextics in (6) did not involve these points. They were obtained directly from the tact invariant. The next system of equations avoids the use of the tact invariant. It uses five copies of the equations (1)–(3), each with a different point of tangency $(x_{i},y_{i})$, for $i=1,2,3,4,5$. The ten equations from (1) and (2) are quadrics. The five equations from (3) are cubics. Altogether, we get the following system of $15$ equations that we display as a $5\times 3$ matrix $F_{(A,B,C,D,E)}$:

[TABLE]

Each matrix entry is a polynomial in the $15$ variables $u_{1},\ldots,u_{5},x_{1},y_{1}\ldots,x_{5},y_{5}$. The parameters of this system are the coefficients of the conics $A,B,C,D,E$. The system of five equations seen in (6) is obtained by eliminating the $10$ variables $x_{1},y_{1},x_{2},y_{2},x_{3},y_{3},x_{4},y_{4},x_{5},y_{5}$ from the new system $F_{(A,B,C,D,E)}(x)$ introduced in (12).

At first glance, it looks like the new formulation (12) is worse than the one in (6). Indeed, the number of variables has increased from $6$ to $15$, and the Bézout number has increased from $\,6^{5}=7776\,$ to $\,2^{10}3^{5}=248832$. However, the new formulation is better suited for the numerical solver that powers our website. We explain this in the last section.

Approximation and Certification

Steiner’s conic problem amounts to solving a system of polynomial equations. Two formulations were given in (6) and (12). But what does ’solving’ actually mean? One answer is suggested in the textbook by Cox, Little and O’Shea [CLO15]: Solving means computing a Gröbner basis $\mathcal{G}$. Indeed, crucial invariants, like the dimension and degree of the solution variety, are encoded in $\mathcal{G}$. The number of real solutions is found by applying techniques like deriving Sturm sequences from the polynomials in $\mathcal{G}$. Yet, Gröbner bases can take a very long time to compute. We found them impractical for Steiner’s problem.

Computing $3264$ conics in a second requires numerical methods. Our encoding of the solutions are not Gröbner bases but numerical approximations. How does one make this rigorous? This question can be phrased as follows. Suppose $u_{1},\ldots,u_{6}$ are the true coordinates of a solution and $u_{1}+\Delta u_{1},\ldots,u_{6}+\Delta u_{6}$ are approximations of those complex numbers. How small must the entries of $\Delta u_{1},\ldots,\Delta u_{6}$ be before it is justified to call them approximations? This question is elegantly circumvented by using Smale’s definition of approximate zero [BCSS98, Definition 1 in §8].

In short, an approximate zero of a system $F(x)$ of $n$ polynomials in $n$ variables is any point $z\in\mathbb{C}^{n}$ such that Newton’s method when applied to $z$ converges quadratically fast towards a zero of $F$. Here is the precise definition.

Definition 3** (Approximate zero).**

Let $J_{F}(x)$ be the $n\times n$ Jacobian matrix of $F(x)$. A point $z\in\mathbb{C}^{n}$ is an approximate zero of $F\,$ if there exists a zero $\zeta\in\mathbb{C}^{n}$ of $F$ such that the sequence of Newton iterates

[TABLE]

starting at $z_{0}=z$ satisfies or all $k=1,2,3,\ldots$ that

[TABLE]

If this holds, then we call $\zeta$ the associated zero of $z$. Here $\|x\|:=(\sum_{i=1}^{n}x_{i}\overline{x_{i}})^{\frac{1}{2}}$ is the standard norm in $\mathbb{C}^{n}$, and the zero $\zeta$ is assumed to be nonsingular, i.e. ${\rm det}(J_{F}(\zeta))\not=0$.

The reader should think of approximate zeros as a data structure for representing solutions to polynomial systems, just like a Gröbner basis is a data structure. Different types of representations of data provide different levels of accessibility to the desired information. For instance, approximate zeros are not well suited for computing algebraic features of an ideal. But they are a powerful tool for answering geometric questions in a fast and reliable manner.

Suppose that $z$ is a point in $\mathbb{C}^{n}$ whose real and imaginary part are rational numbers. How can we tell whether $z$ is an approximate zero of $F$? This is not clear from the definition.

It is possible to certify that $z$ is an approximate zero without dealing with the infinitely many Newton iterates. We next explain how this works. This involves Smale’s $\gamma$-number and Smale’s $\alpha$-number:

[TABLE]

Here $D^{k}F(z)$ denotes the tensor of order-$k$ derivatives at the point $z$, the tensor $J_{F}(z)^{-1}D^{k}F(z)$ is understood as a multilinear map $A:(\mathbb{C}^{n})^{k}\to\mathbb{C}^{n}$, and the norm of this map is $\|A\|:=\max_{\|v\|=1}\|A(v,\ldots,v)\|$.

Shub and Smale [SS93] derived an upper bound for $\gamma(F,z)$ which can be computed exactly. Based on the next theorem [BCSS98, Theorem 4 in Chapter 8], one can thus decide algorithmically if $z$ is an approximate zero, using only data of the point $z$ itself.

Theorem 4** (Smale’s $\alpha$-theorem).**

If $\alpha(F,z)<0.03$, then $z$ is an approximate zero of $F(x)$. Furthermore, if $y\in\mathbb{C}^{n}$ is any point with $\|y-z\|<(20\,\gamma(F,z))^{-1}$, then $y$ is also an approximate zero of $F$ with the same associated zero $\zeta$ as $z$.

Actually, Smale’s $\alpha$-theorem is more general in the sense that $\alpha_{0}=0.03$ and $t_{0}=20$ can be replaced by any two positive numbers $\alpha_{0}$ and $t_{0}$ that satisfy a certain list of inequalities.

Hauenstein and Sottile [HS12] use Theorem 4 in an algorithm, called alphaCertified, that decides if a point $z\in\mathbb{C}^{n}$ is an approximate zero and if two approximate zeros have distinct associated solutions. An implementation is publicly available. Furthermore, if the polymomial system $F$ has only real coefficients, then alphaCertified can decide if an associated zero is real. The idea behind this is as follows: Let $z\in\mathbb{C}^{n}$ be an approximate zero of $F$ with associated zero $\zeta$. If the coefficients of $F$ are all real, then the Newton operator $N_{F}(x)$ from Definition 3 satisfies $N_{F}(\overline{x})=\overline{N_{F}(x)}$. Hence $\overline{z}$ is an approximate zero of $F$ with associated zero $\overline{\zeta}$. If $\|z-\overline{z}\|<(20\,\gamma(F,z))^{-1}$, then, by Theorem 4, the associated zeros of $z$ and $\overline{z}$ are equal. This means $\zeta=\overline{\zeta}$.

A fundamental insight is that Theorem 4 allows us to certify candidates for approximate zeros regardless of how they were obtained. Typically, candidates are found by inexact computations using floating point arithmetic. We do not need to know what happens in that computation, because we can certify the result a posteriori. Certification constitutes a rigorous proof of a mathematical result. Let us see this in action.

Proof of Proposition 1.

Fix five non-degenerate conics with rational coefficients listed after equation (9). We apply HomotopyContinuation.jl [BT18] to compute $3264$ solutions in a second in 64-bit floating point arithmetic. The output is inexact. Each coefficient $u_{i}$ of each true solution $U$ is a complex number that is algebraic of degree $3264$ over $\mathbb{Q}$. The floating point numbers that represent these coefficients are rational numbers and we treat them as elements of $\mathbb{Q}$.

Our proof starts with the resulting list of $3264$ vectors $x\in\mathbb{Q}^{15}$ corresponding to the $15$ variables of (12). The computation was mentioned to make the exposition more friendly. It is not part of the proof.

We are now given $3264$ candidates for approximate zeros of the polynomial system in (12). These candidates have rational coordinates. We use them as input to the software alphaCertified from [HS12]. That software performs exact computations in rational arithmetic. Its output shows that the $3264$ vectors $x$ are approximate zeros, that their associated zeros $\zeta$ are distinct, and that they all have real coordinates. This is shown in Figure 5. The output data of alphaCertified is available for download through the arXiv version of this article. ∎

This was a rigorous proof of Proposition 1, just as trustworthy as a computer-assisted proof by symbolic computation (e.g. Gröbner bases and Sturm sequences) might have been. Readers who are experts in algebra should not get distracted by the appearance of floating point arithmetic: it is not part of the proof! Floating point numbers are only a tool for obtaining the $3264$ candidates. The actual proof is carried out by exact symbolic computations.

How Does This Work?

In this section we discuss the methodology and software that powers the web interface (10).

We use the software HomotopyContinuation.jl that was developed by two of us [BT18]. This is a Julia [BEKV17] implementation of a computational paradigm called homotopy continuation. The reasons we chose Julia as the programming language are threefold: the first is that Julia is open source and free for anyone to use. The second is that Julia can be as fast as well written C. For instance, we use Julia’s JIT compiler for fast evaluation of polynomials. Finally, the third reason is that, despite its high performance, Julia still provides an easy high-level syntax. This makes our software accessible for users from many backgrounds.

Homotopy continuation works as follows: We wish to find a zero in $\mathbb{C}^{n}$ of a system $F(x)$ of $n$ polynomials in $n$ variables. Let $G(x)$ be another such system with a known zero $G(\zeta)=0$. We connect $F$ and $G$ in the space of polynomial systems by a path $t\mapsto H(x,t)$ with $H(x,0)=G(x)$ and $H(x,1)=F(x)$.

The aim is to approximately follow the solution path $x(t)$ defined by $H(x(t),t)=0$. For this, the path is discretized into steps $\,t_{0}=0<t_{1}<\cdots<t_{k}=1$. If the discretization is fine enough then $\zeta$ is also an approximate zero of $H(x,{t_{1}})$. Hence, by Definition 3, applying the Newton operator $N_{H(x,t_{1})}(x)$ to $\zeta$, we get a sequence $\zeta_{0},\zeta_{1},\zeta_{2},...$ of points that converges towards a zero $\xi$ of $H(x,t_{1})$. If $t_{2}-t_{1}$ and $\|\zeta_{i}-\xi\|$ are small enough, for some $i\geq 0$, then the iterate $\zeta_{i}$ is an approximate zero of $H(x,t_{2})$.

We may repeat the procedure for $H(x,t_{2})$ and starting with $\zeta_{i}$. Inductively, we find an approximate zero of $H(x,t_{j})$ for all $j$. In the end, we obtain an approximate zero for the system $F(x)=$ $H(x,1)$. Most implementations of homotopy continuation, including Bertini [BHSW06] and HomotopyContinuation.jl [BT18], use heuristics for setting both the step sizes $t_{j+1}-t_{j}$ and the number of Newton iterations.

Our homotopy for Steiner’s conic problem computes zeros of the system $F_{(A,B,C,D,E)}(x)$ from (12). We prefer formulation (12) over (6) because the equations in the former formulation have lower degrees and fewer terms. It is known that high degrees and many terms introduce numerical instability in the evaluation of polynomials. We use the homotopy

[TABLE]

The conic $tA+(1-t)A^{\prime}$ is defined by the coefficients $ta_{i}+(1-t)a_{i}^{\prime}$, where $a_{i}$ and $a_{i}^{\prime}$ are the coefficients of $A$ and $A^{\prime}$. This is called a parameter homotopy in the literature. Geometrically, (13) is a straight line in the space of quintuples of conics. An alternative would have been

[TABLE]

The advantage of (13) over (14) is that the path stays within the space of structured systems

[TABLE]

The structure of the equations is preserved. The system in (13) has $3264$ solutions for almost all $t$, whereas (14) has $7776$ solutions for random $t$. In the language of algebraic geometry, we prefer the flat family (13) over (14), which is not flat.

The last missing piece in our Steiner homotopy is a start system. That is, we need five explicit conics together with all 3264 solutions. For this, we construct a generic instance $(A,B,C,D,E)$ by randomly sampling complex coefficients for the conics. Then, we compute the 3264 solutions using standard homotopy continuation techniques [SW05]. Those 3264 solutions are saved and used for further computations. This initial computation is significantly more expensive than tracking 3264 solutions along the homotopy (13) but it only has to be done once.

In closing, we emphasize the important role played by enumerative geometry for solving polynomial systems. It gives a criterion for deciding if the initial numerical computation found the correct number of solutions. This is why numbers like $3264$ are so important, and why a numerical analyst might care about Chow rings and the pentagon construction. We conclude that enumerative algebraic geometry and numerical algebraic geometry complement each other.