How to Run a Centipede: a Topological Perspective

Abstract

In this paper we study the topology of the configuration space of a device with d legs (“centipede”) under some constraints, such as the impossibility to have more than k legs off the ground. We construct feedback controls stabilizing the system on a periodic gait and defined on a ‘maximal’ subset of the configuration space.

A centipede was happy quite!

Until a toad in fun

Said, “Pray, which leg moves after

which?”

This raised her doubts to such a pitch,

She fell exhausted in the ditch

Not knowing how to run.

Katherine Craster

1 Appendix

A Discrete autonomous control

Below we describe an interesting discrete dynamical system associated with our construction above. It addresses a somewhat different problem - not the stabilization on a single attractor, but rather generating a simple flow with piece-wise linear trajectories, but its nice mathematical features compelled us to present it here.

We construct a flow through the union of the pyramids Pyr _i such that on each of them this flow enters only through its entrance face, $F: = Fc_i^ +$ and leaves through the exit face, $G: = Fc_i^ -$.

Both faces are cones over certain (d − 2)-dimensional cubes (corresponding to the legs i, (i − 1) and i, (i + 1) being simultaneously leaders, in the proper order). The flow we are looking for should move from the entrance face F through the (d — 1)-cube B := Cb _-i of the whole pyramid and then further to the exit face G.

Let us define two natural maps from the (open) entrance face F to the (open) base cube B and then from B to the (open) exit face G. Each such map can be transformed into a (continuous) flow by connecting the preimage and its image by a straight line within the pyramid. (Thus each trajectory of such a flow within Pyr _i will be the union of two straight segments.)

The most natural way to do it is by using the so-called blow-up/blow down rational transformations [5]. We present these transformations explicitly below for the cases d = 3 and d ≥ 4. (The essential distinction of these two cases is explained by the fact that for d = 3 the entrance/exit faces are the usual triangles and, therefore, they allow additional symmetry transformations unavailable for d ≥ 4.)

Case d = 3

The entrance/exit faces F and G are usual triangles and the base cube B is a usual square. Let us identify the entrance triangle F with the triangle with the vertices (0, 0), (1, 0), (1, 1) in ℝ²; the base square B with the square whose vertices are (0, 0), (1, 0), (0, 1), (1, 1) and, finally, the exit triangle G with the triangle with the vertices (0, 0), (0, 1), (1, 1).

The blow-up map Φ : $(x,y) \to (x,\frac{y}{x})$ sends F to B. (It sends the pencil of lines through the origin to the pencil of horizontal lines.) Its inverse blow-down map Ψ : (s, t) → (st, t) maps B to G. It sends the pencil of vertical lines to the pencil of lines through the origin. Their composition χ = Ψ ◯ Φ : (x, y) → $(y,\frac{y}{x})$ sends F to G, see Fig. 1

To get the whole discrete dynamical system assume that the three (since d = 3) pyramids Pyr₁, Pyr₂, Pyr₃ are cyclically ordered as 1 < 2 < 3 < 1 by the choice of Γ_γ. Denote their entrance faces as F ₁, F₂, F₃ and their exit faces as G ₁, G₂, G₃. Notice that F ₁ = G₂, F₂ = G₃, F₃ = G₁. Assume now that we apply our transformation χ three times consecutively, i.e first from F ₁ to G ₁ = F ₂, then from F ₂to G ₂ = F ₃, and, finally back to G ₃ = F₁. The resulting self-map Θ: F ₁ → F ₁ is classically referred to as the Poincare return map of the dynamical system. To calculate it explicitly we need to find a suitable affine transformation A sending G back to F in the above example. Then we get the self-map Θ by composing χ with A and taking the 3-rd power of the resulting composition. As such a map A one can choose A : (u, v) → (1 - u, 1 - v) which implies that the required Poincare return map is the third power of Θ = A ◯ χ where:

$$\Theta :(x,y) \to \left( {1 - y,1 - \tfrac{y}{x}} \right).$$

Lemma 1. The above map Θ has a unique fixed point within the triangle F ₁ and its fifth power is identity.

Proof. The system of equations defining fixed points reads as

$$\left\{ \begin{gathered} x = 1 - y, \hfill \\ y = 1 - \tfrac{y}{x} \hfill \\ \end{gathered} \right.$$

and its two solutions are ${\psi _1} = \frac{{2\sqrt {2 - 1} }}{2},{y_1}\frac{{3 - 2\sqrt 2 }}{2}{\kern 1pt} \;and\;{\psi _2} = - \frac{{1 + 2\sqrt 2 }}{2},\;{y_2} = \frac{{3 + 2\sqrt 2 }}{2}$. One can easily check that only the first solution belongs to F ₁ . Direct calculations show that

$$\begin{gathered} {\Theta ^2}:(x,y) \to \left( {\tfrac{y}{x},\tfrac{{y(1 - x)}}{{x(1 - y)}}} \right),\quad {\Theta ^3}:(x,y) \to \left( {\tfrac{{x - y}}{{x(1 - y)}},\tfrac{{x - y}}{{(1 - x)}}} \right) \hfill \\ {\Theta ^4}:(x,y) \to \left( {\tfrac{{1 - x}}{{1 - y}},1 - x} \right),\quad {\Theta ^5}:(x,y) \to (x,y). \hfill \\ \end{gathered} $$

The Poincare return map is thus equals to Θ³ : (x, y) → $\left( {\frac{{x - y}}{{x(1 - y)}},\frac{{x - y}}{{(1 - x)}}} \right)$.

Case d ≥ 4

Analogously, we have d pyramids each being a cone over a .(d - 1)-cube. Their entrance and exit faces are cones over a square respectively. The map Φ sends the open entrance face F to the open base .(d - 1)-cube B and the map Ψ sends the open base cube B to the open exit face G. They can be given explicitly as follows. Let us identify F with the domain {0 < Ψ ₂ < Ψ₁ < 1;0 < Ψ ₃ < Ψ₁ < 1;… 0 < Ψ _{d- 1} < Ψ₁ < 1}, i.e. with the cone over the square {0 < Ψ ₂ < 1, 0 < Ψ _d-1 < 1} with the vertex at the origin. The base B will be identified with the cube {0 < Ψ ₁ < 1, 0 < Ψ ₂ < 1, 0 < Ψ _d−1 < 1}, and, finally, the exit face G with {0 < Ψ ₁ < Ψ ₂ < 1; 0 < Ψ ₃ < Ψ ₂ < 1,….,0 < Ψ _d-1 < Ψ ₂ < 1}. Then the blow-up map Φ and the blow-down map Ψ can be chosen as follows:

$$\begin{gathered} \Phi :({\psi _1},{\psi _2}, \ldots ,{\psi _{d - 1}}) \to \left( {{\psi _1},\frac{{{\psi _2}}}{{{\psi _1}}},\frac{{{\psi _3}}}{{{\psi _1}}}, \cdots ,\frac{{{\psi _{d - 1}}}}{{{\psi _1}}}} \right) \hfill \\ \Psi :({y_1},{y_2}, \ldots ,{y_{d - 1}}) \to ({y_1}{y_2},{y_2}, \ldots ,{y_{d - 1}}{y_2}). \hfill \\ \end{gathered} $$

Their composition χ : F → G coincides with

$$\chi :({\psi _1},{\psi _2}, \ldots ,{\psi _{d - 1}}) \to \left( {{\psi _2},\frac{{{\psi _2}}}{{{\psi _1}}},\frac{{{\psi _2}{\psi _3}}}{{{\psi _1}}}, \ldots ,\frac{{{\psi _2}{\psi _{d - 1}}}}{{\psi _1^2}}} \right).$$

An appropriate linear map A sending G back to F is just a cyclic permutation of coordinates:

$$A:({z_1},{z_2}, \ldots ,{z_{d - 1}}) \to ({z_2},{z_3}, \ldots ,{z_1}).$$

Thus we get the composition Θ = A ◯ χ : F → F (whose d-th power is the Poincare return map) given by:

$$\Theta :({\psi _1},{\psi _2}, \ldots ,{\psi _{d - 1}}) \to \left( {\frac{{{\psi _2}}}{{{\psi _1}}},\frac{{{\psi _2}{\psi _3}}}{{{\psi _1}}}, \ldots ,\frac{{{\psi _2}{\psi _{d - 1}}}}{{\psi _1^2}},{\psi _2}} \right).$$

Proposition 2. The above map Θ has a curve of fixed points parameterized by ((t, t ², t ²,…,t ²), t ∊ ℝ. Moreover, for any d ≥ 3 one has that Θ^d-1 = id.

Proof. Indeed, the system of equations defining fixed points reads as

$${\psi _1} = \frac{{{\psi _2}}}{{{\psi _1}}},\quad {\psi _2} = \frac{{{\psi _2}{\psi _3}}}{{\psi _1^2}},\quad {\psi _3} = \frac{{{\psi _2}{\psi _4}}}{{\psi _1^2}},\quad \ldots {\psi _{d - 2}} = \frac{{{\psi _2}{\psi _{d - 1}}}}{{\psi _1^2}},\quad {\psi _{d - 1}} = {\psi _2}.$$

which immediately implies $\psi _1^2 = {\psi _2} = {\psi _3} = ... = {\psi _{d - 1}}$. To show that Θ^d-1 = id notice that since Θ is a monomial map it suffices to show that M ^d-1 _d = id _d-1where M _d is the matrix of exponents of the map Θ and id _d-1 is the identity matrix of size d - 1. (Indeed, the matrix of exponents for Θⁱ coincides with $M_d^i$.) This is done in the following lemma.

Lemma 2. The characteristic polynomial of the (d - 1) × (d - 1)-matrix M _d equals (−1)^d (1 - t ^d-1). Therefore, by the Hamilton-Cayley theorem $M_d^{d - 1} = i{d_{d - 1}}$.

Proof. Looking at the exponents of Θ we see that the matrix M _d has the form

$${M_d} = \left( {\begin{array}{*{20}{c}} { - 1}&1&0&0& \ldots &0 \\ { - 2}&1&1&0& \ldots &0 \\ { - 2}&1&0&1& \ldots &0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ { - 2}&1&0&0& \ldots &1 \\ 0&1&0& \ldots & \ldots &0 \end{array}} \right)$$

To make our calculations easy we introduce two families of (k × k)-matrices D _k and E _k given by:

$${D_k} = \left( {\begin{array}{*{20}{c}} 1&1&0&0& \ldots &0 \\ 1&{ - t}&1&0& \ldots &0 \\ 1&0&{ - t}&1& \ldots &0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 1&0&0& \ldots &{ - t}&1 \\ 1&0&0& \ldots & \ldots &{ - t} \end{array}} \right),\quad {E_k} = \left( {\begin{array}{*{20}{c}} 2&1&0&0& \ldots &0 \\ 2&{ - t}&1&0& \ldots &0 \\ 2&0&{ - t}&1& \ldots &0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 2&0&0& \ldots &{ - t}&1 \\ 0&0&0& \ldots &0&{ - t} \end{array}} \right),$$

Expanding by the first row one obtains the following recurrences

$$Det({D_k}) = {( - t)^{k - 1}} - Det({D_{k - 1}})\quad Det({E_k}) = 2{( - t)^{k - 1}} - Det({E_{k - 1}})$$

resulting in the formulas

$$\begin{gathered} Det({D_k}) = {( - 1)^{k - 1}}({t^{k - 1}} + {t^{k - 2}} + \ldots + 1), \hfill \\ Det({E_k}) = {( - 1)^{k - 1}}2({t^{k - 1}} + {t^{k - 2}} + \ldots + t). \hfill \\ \end{gathered} $$

Expanding now the characteristic polynomial Ch _d (t) of M _d by the first row (after the sign change in the first row) we get the relation

$$ - C{h_d}(t) = (t + 1)\left[ {(1 - t){{( - t)}^{d - 3}} - Det({D_{d - 3}})} \right] - Det({E_{d - 2}}).$$

Substituting of the expressions for Det(D _d-3) and Det(E _d-2) in the latter formula one gets Ch _d (t) = (-1)^d (1 - t ^d-1).

This completes the proof.

Corollary 1. The Poincare return map equals Θ^d‚= Θ.

Acknowledgements The authors want to thank Profs. D. Koditschek and F. Cohen for important discussions of the topic. Support from AFOSR through MURI FA9550-10-1-0567 (CHASE) is gratefully acknowledged.