Eccentric pie charts and an unusual pie cutting

TL;DR

This paper introduces eccentric pie charts, a generalization of traditional pie charts, and presents a method for calculating sector areas using polynomial approximations and homotopy continuation, demonstrated on a pie cutting problem.

Contribution

It develops a novel approach for analyzing eccentric pie charts by transforming the problem into polynomial systems and solving with homotopy continuation and Newton iteration.

Findings

Method effectively computes sector areas in eccentric pie charts.

Approach applicable to various nonlinear, non-polynomial systems.

Illustrated on a specific pie cutting problem.

Abstract

The eccentric pie chart, a generalization of the traditional pie chart is introduced. An arbitrary point is fixed within the circle and rays are drawn from it. A sector is bounded by a pair of neighboring rays and the arc between them, The sector's area, aimed to be equal to a given proportion, is calculated from some well known equations in coordinate geometry. The resulting system of polynomial and trigonometric equations can be approximated by a fully polynomial system, once the non-polynomial functions are approximated by their Taylor series written up to the first few terms. The roots of the polynomial system have been found by the homotopy continuation method, then used as starting points of a Newton iteration for the original (non-polynomial) system. The method is illustrated on a special pie cutting problem, and is applicable to a wide class of nonlinear, non-polynomial systems.

Equations

$$\frac{1}{2}\left|\det \left(\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_0& y_0& 1\end{array}\right)\right|=\frac{1}{2}\left|x_1{y}_2+x_2{y}_0+x_0{y}_1-x_2{y}_1-x_0{y}_2-x_1{y}_0\right|.$$

$$x_1{x}_2+y_1{y}_2=\cos \beta ,$$

$$2\left(\lambda \pi -\frac{\beta }{2}+\frac{\sin \beta }{2}\right)=|x_0(y_1-y_2)+x_1(y_2-y_0)+x_2(y_0-y_1)|,$$

$$x_1^2+y_1^2=1,$$

$$x_2^2+y_2^2=1,$$

$$\arccos (x)=\sum _{n=0}^{\infty }\frac{{\arccos }^{(n)}(a)}{n!}(x-a{)}^n,$$

$$\arccos (x)=\frac{\pi }{2}-\sum _{n=0}^{\infty }\frac{(2n)!}{4^n(n!{)}^2(2n+1)}{x}^{2n+1}=\frac{\pi }{2}-x-\frac{1}{6}{x}^3-\frac{3}{40}{x}^5+O(x^7),(|x|<1)$$

$$(x_1{x}_2+y_1{y}_2{)}^2+{\sin }^2\beta =1,$$

$$(x_2{x}_3+y_2{y}_3{)}^2+{\sin }^2\phi =1,$$

$$2{\lambda }_1\pi -\beta +\sin \beta =|x_0(y_1-y_2)+x_1(y_2-y_0)+x_2(y_0-y_1)|,$$

$$2{\lambda }_2\pi -\phi +\sin \phi =|x_0(y_2-y_3)+x_2(y_3-y_0)+x_3(y_0-y_2)|,$$

$$x_1^2+y_1^2=1,$$

$$x_2^2+y_2^2=1,$$

$$x_3^2+y_3^2=1,$$

$$(x_1-x_0)(x_2-x_0)+(y_1-y_0)(y_2-y_0)=\cos \left(\frac{2\pi }{3}\right)\sqrt{[(x_1-x_0{)}^2+(y_1-y_0{)}^2][(x_2-x_0{)}^2+(y_2-y0{)}^2]},$$

$$(x_2-x_0)(x_3-x_0)+(y_2-y_0)(y_3-y_0)=\cos \left(\frac{2\pi }{3}\right)\sqrt{[(x_2-x_0{)}^2+(y_2-y_0{)}^2][(x_3-x_0{)}^2+(y_4-y0{)}^2]}$$

$$(x_1{x}_2+y_1{y}_2{)}^2+s_{\beta }^2=1,$$

$$(x_2{x}_3+y_2{y}_3{)}^2+s_{\phi }^2=1,$$

$${\left(2{\lambda }_1\pi -\beta +s_{\beta }\right)}^2=[x_0(y_1-y_2)+x_1(y_2-y_0)+x_2(y_0-y_1){]}^2,$$

$${\left(2{\lambda }_2\pi -\phi +s_{\phi }\right)}^2=[x_0(y_2-y_3)+x_2(y_3-y_0)+x_3(y_0-y_2){]}^2,$$

$$[(x_1-x_0)(x_2-x_0)+(y_1-y_0)(y_2-y_0){]}^2=\frac{1}{4}[(x_1-x_0{)}^2+(y_1-y_0{)}^2][(x_2-x_0{)}^2+(y_2-y0{)}^2],$$

$$[(x_2-x_0)(x_3-x_0)+(y_2-y_0)(y_3-y_0){]}^2=\frac{1}{4}[(x_2-x_0{)}^2+(y_2-y_0{)}^2][(x_3-x_0{)}^2+(y_4-y0{)}^2]$$

$$\cos \beta =x_1{x}_2+y_1{y}_2$$

$$\cos \phi =x_2{x}_3+y_2{y}_3$$

$$\beta =\frac{\pi }{2}-(x_1{x}_2+y_1{y}_2{)}^{-}\frac{1}{6}(x_1{x}_2+y_1{y}_2{)}^3-\frac{3}{40}(x_1{x}_2+y_1{y}_2{)}^5,$$

$$\phi =\frac{\pi }{2}-(x_2{x}_3+y_2{y}_3{)}^{-}\frac{1}{6}(x_2{x}_3+y_2{y}_3{)}^3-\frac{3}{40}(x_2{x}_3+y_2{y}_3{)}^5,$$

$$x_0$$

$$x_1$$

$$y_1$$

$$x_2$$

$$y_2$$

$$x_3$$

$$y_3$$

$$\beta$$

$$\phi$$

$$s_{\beta }$$

$$s_{\phi }$$

Full text

Keywords: eccentric pie chart, area-proportional diagram, pie cutting, multivariate polynomial system, homotopy continuation method

1 Introduction

Assume that a circular pie is to be distributed among three children such that their shares are proportional to the children’s time spent with assisting, 40, 35 and 25 minutes. There is a three-blade pie (or pizza) cutter in the kitchen, but it is designed for equal slices: the angle of every pair of blades is $120^{\circ}$. Where to locate the cutter in order to get slices of area 40%, 35% and 25% of the whole pie? The answer is shown in Figure 1, the detailed solution is given in Section 3.

Figure 1. 40-35-25% pie-cutting with a regular 3-blade cutter

Pie chart is more than two hundred years old. According to Spence [13] and Tufte [14, page 44], Playfair [12, Chart 2 on page 49] invented it. The popularity of the traditional pie chart is rooted in its simplicity and efficiency in visualization. For an arbitrary set of $n\geq 2$ positive numbers $\lambda_{i},i=1,\ldots,n$ such that $\lambda_{1}+\lambda_{2}+\ldots+\lambda_{n}=1$, the corresponding circle sectors in the pie chart visualize the relative magnitude of numbers $\lambda_{i}$ in three equivalent ways: (i) areas, (ii) central angles, (iii) arc lengths are proportional to the numbers $\lambda_{i}$.

The pizza theorem [9, 11, 15] states that the $n$-blade ($n\geq 4$ is even) equiangular cutter, wherever it is located, halves the circle’s area by summing the areas of the alternate eccentric sectors, see Figure 2.

Figure 2. The total area of grey sectors is equal to the total area of white sectors (pizza theorem for $n=8$)

Area-proportionality can be an important feature in case of Venn diagrams, e.g., for the visualization of biological lists [5], where the sizes of the sets (represented by circles) as well as the overlaps correspond to the cardinalities in the data sets.

The rest of the paper is structured as follows. The eccentric pie chart is introduced in Section 2. It is a modification of the traditional pie chart such that the center, the common point of the sectors is moved from the circle’s center. The areas represent the relative magnitude of numbers $\lambda_{i}$. The area of a sector can be calculated with the help of a circular sector and two triangles, resulting in a system of nonlinear polynomial and trigonometric equations. Systems of nonlinear equations are usually hard to solve [10, Chapter 11]. If all equations are polynomial, homotopy continuation methods [7, 3] can be applied. HOM4PS-3 [3] is used in this paper. Several geometric problems, such as Littlewood’s problem on seven mutually touching infinite cylinders [1], or Steiner’s conic problem [2] lead to polynomial equations. The equations of the eccentric pie chart include non-polynomials, however, an approximation by the Taylor series, written up to the first few terms, is polynomial as in [6, Example 4.4] and [8, 16]. The roots of the approximating polynomial system can be found by the homotopy method, then they are used as starting points of a Newton iteration for the approximated system of non-polynomial equations.

The solution of the 40-35-25% pie-cutting problem is presented in Section 3. The problem leads to a system of 11 polynomial equations of 11 variables. Section 4 concludes.

2 Eccentric pie charts

Consider the point $(x_{0},y_{0})$ inside the unit circle, and draw rays from it. A sector is bounded by a pair of neighboring rays and the arc between them (Figure 3). We focus on the area of the sectors. Unlike in case of the traditional pie chart, the angle of the neighboring rays and the arc length are not proportional to the sector’s area any more.

Figure 3. An eccentric pie chart

There are infinitely many eccentric pie charts representing the same set of proportions (percentages). In fact the degree of freedom of eccentric pie charts is 2, if rotations of the circle are not distinguished. Once the starting point, e.g. (0,1), on the boundary of the circle is fixed, the point $(x_{0},y_{0})$ can be located anywhere inside the unit circle, there exists exactly one eccentric pie chart representing the given set of proportions counterclockwise, and another one clockwise. Figure 4 shows nine eccentric pie charts ($x_{0},y_{0}\in\{-1/2,0,1/2\}$), all of them visualize the 20-30-15-25-10% counterclockwise. The first ray is between points $(x_{0},y_{0})$ and (0,1). The pie chart in the middle is the traditional pie chart.

Figure 4. Eccentric pie charts with areas 20-30-15-25-10%, where $x_{0},y_{0}\in\{-1/2,0,1/2\}$

The areas of eccentric sectors can be calculated from the area of a circular sector and two triangles (Figures 5 and 6). Let $(x_{1},y_{1})$ and $(x_{2},y_{2})$ be two points on the boundary of the unit circle centered at the origin. The area of a circular sector is equal to $\frac{\beta}{2}$, where $\beta$ is the central angle. It is also well known that $x_{1}x_{2}+y_{1}y_{2}=\cos\beta$. The area of the triangle $(x_{1},y_{1}),(x_{2},y_{2}),(0,0)$ is $\frac{\sin\beta}{2}$, if $\beta<\pi$ (Figure 5), and $\frac{\sin(2\pi-\beta)}{2}$, if $\beta>\pi$ (Figure 6).

Figure 5. The calculation of the eccentric sector’s area ($\beta<\pi$)

Figure 6. The calculation of the eccentric sector’s area ($\beta>\pi$)

The area of a triangle can be directly calculated from the coordinates of its vertices.

Lemma 2.1**.**

(See, e.g., [4, Problem 52 on page 34] or [19, formula (4.7.2) on page 212] ) The area of triangle $(x_{1},y_{1}),(x_{2},y_{2}),(x_{0},y_{0})$ is equal to

[TABLE]

Now let us prescribe that the area of the eccentric sector $(x_{1},y_{1}),(x_{2},y_{2}),(x_{0},y_{0})$ with central angle $\beta<\pi$ must be $\lambda\pi$, where $\lambda<1$ is given. The following equations can be written:

[TABLE]

The system of equations above includes both polynomials and trigonometric expressions, which is hard to solve. Since there exist powerful algorithms [3, 7] for solving polynomial systems, our aim is to build a polynomial system, which is close to the non-polynomial system (2)-(5). Replace $\sin\beta$ by the new variable $s_{\beta}$, and consider the non-polynomial equation $\beta=\arccos(x_{1}x_{2}+y_{1}y_{2})$ from (2).

Lemma 2.2**.**

(See, e.g., [19, Section 1.9.4.2 on page 50 and Section 1.9.6.5 on page 61]) The Taylor series of function $\arccos$ around $a$ and [math].

[TABLE]

especially with $a=0$

[TABLE]

Figure 7 shows that the Taylor series around $a=0$ up to 6 terms approximates the $\arccos$ function well if $|x|<0.8$. If $|x|$ is close to 1, the Taylor series around $a=0.9$ up to 6 terms provides a good approximation (Figure 8). Then one of the equations (6),(7) with $x=x_{1}x_{2}+y_{1}y_{2}$ results in a multivariate polynomial. The absolute values in (3) can be eliminated by taking the squares of both sides of the equation (false roots are possible, and they have to be filtered out).

\frac{\pi}{2}-x-\frac{1}{6}x^{3}-\frac{3}{40}x^{5}$$\arccos(x)

Figure 7. $\arccos(x)$ and its Taylor series around $a=0$ up to 6 terms

($x^{2}$ and $x^{4}$ have zero coefficients)

0.45-2.29(x-0.9)-5.43(x-0.9)^{2}-27.75(x-0.9)^{3}$$-173.84(x-0.9)^{4}-1218.58(x-0.9)^{5}$$\arccos(x)

Figure 8. $\arccos(x)$ and its Taylor series around $a=0.9$ up to 6 terms

3 Pie cutting with a multi-blade cutter

Let us solve the 40-35-25% pie-cutting problem from the beginning of the paper. The geometrical problem is transformed into an algebraic one. A system of equations is developed, then the solutions are filtered.

Let us have a unit circle with its center in the origin. Denote by $(x_{0},y_{0})$ the coordinates of the 3-blade cutter’s center. We can assume that $y_{0}=0$ due to rotational symmetry. $-1<x_{0}<1$ is assumed in an implicit way: no equation is generated, but after the system of equations is solved, roots not satisfying this double inequality are filtered out.

Denote by $(x_{1},y_{1}),(x_{2},y_{2})$ and $(x_{3},y_{3})$ the coordinates of the three points, where the boundary of the circle and the blades meet, as in Figure 9. Let $\beta$ denote the central angle of the first eccentric sector (of area $\lambda_{1}\pi=0.4\pi)$ bounded by line sections $[(x_{0},0),(x_{1},y_{1})]$ and $[(x_{0},0),(x_{2},y_{2})]$ and the arc between them. Similarly, let $\varphi$ denote the central angle of the second eccentric sector (of area $\lambda_{2}\pi=0.35\pi)$ bounded by line sections $[(x_{0},0),(x_{2},y_{2})]$ and $[(x_{0},0),(x_{3},y_{3})]$ and the arc between them. The third eccentric sector’s area is $\lambda_{3}\pi=0.25\pi$, and its central angle is $2\pi-\beta-\varphi.$

Figure 9. Pie-cutting with a regular 3-blade cutter

Following Section 2 and including the regularity of 3-blade cutter (pairwise angles are $\frac{2\pi}{3}$) we have the following equations:

[TABLE]

where $\lambda_{1}=0.4,\lambda_{2}=0.35,y_{0}=0$ and $\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}.$

Introduce variables $s_{\beta},s_{\varphi}$ to replace $\sin\beta$ and $\sin\varphi$, respectively. In order to avoid absolute values in (10)-(11), take the squares of both sides. This step may bring false solutions, and we will see that it does so indeed. The square root in (15)-(16) can be eliminated likewise, with another possibility to have false solutions. Equations (8)-(11) and (15)-(16) are replaced by polynomial equations

[TABLE]

Finally approximate the equations

[TABLE]

or, equivalently, $\beta=\arccos(x_{1}x_{2}+y_{1}y_{2})$ and $\varphi=\arccos(x_{2}x_{3}+y_{2}y_{3})$ by the Taylor series of function $\arccos$ around zero up to the fifth power as in (7). Here we expect that angles $\beta$ and $\varphi$ are between $\arccos(0.8)\approx 0.6435\approx 37^{\circ}$ and $\pi-\arccos(0.8)\approx 2.498\approx 143^{\circ}$. Would this assumption fail, we can try the other Taylor series around 0.9 as in (6) and Figure 8.

[TABLE]

The polynomial system (12)-(14),(17)-(22),(25)-(26) has 11 equations and 11 variables:$x_{0},x_{1},y_{1},x_{2},y_{2},x_{3},y_{3},\beta,\varphi,s_{\beta},s_{\varphi}$. Homotopy algorithm HOM4PS-3 [3] found 28224 roots, 720 of them are real. However, most of them are false solutions to the geometric problem, due to several reasons. Many roots do not fulfil $|x_{0}|,|s_{\beta}|,|s_{\varphi}|\leq 1$. Some roots satisfy (19)/(20) but not (10)/(10), or, similarly, satisfy (21)/(22) but not (15)/(16). Furthermore, $s_{\beta}\not\approx\sin\beta$, but instead of that $s_{\beta}\approx-\sin\beta.$

After all 4 solutions remain to use as starting points of a Newton-iteration for the system of equations (12)-(14), (17)-(22), (23)-(24). Maple’s fsolve refines the solution with an arbitrary accuracy. However we observed that the roots calculated from the polynomial system were already within an error of 0.002 for all variables, which is due to that the Taylor series provided a good polynomial approximation of the function $\arccos$.

The four solutions are essentially the same: one solution, given below up to 10 correct digits,

[TABLE]

has already been shown in Figures 1 and 9, the others are its reflections on the vertical and/or horizontal axes.

Note that $\lambda_{i}$ ($i=1,2,3$) cannot be arbitrary in this problem. Even if the the center $(x_{0},y_{0})$ is located at (1,0), i.e., on the circle’s border, the largest sector’s area is at most $\pi-2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)\approx 0.9423\pi.$

4 Conclusions

The eccentric pie chart, a generalization of the traditional pie chart, visualizes the proportions is infinitely many ways, with degree of freedom 2. This variety suggests that additional information can also be taken into consideration, such as the constraints on the angles of the rays in the 40-35-25% pie-cutting problem.

The method of solving nonlinear, non-polynomial systems through a polynomial approximation, presented in Section 2 and illustrated on an example in Section 3, seems applicable in larger systems, too. The replacement of a non-polynomial function by its Taylor series has practical limitations, because a polynomial system can also be hopelessly hard to solve.