The Robot Rendezvous Problemfrancis/Bode.pdf · 2014-12-18 · Steve Morse gave a seminar at U of T...

The Robot Rendezvous Problem

Bruce Francis

1

Thursday, 18 December, 14

Thank you very much, Yutaka.

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I have Parkinson’s disease. To help me deliver the lecture as smoothly as possible, I’ve written text on the slides and will read it. You should read it along with me (not aloud). Let’s practice by an example ...

3


A friend advised me not to begin my lecture with a joke, and also not to pontificate during my lecture.

As the Pope and I have the same name (Francis) I may unavoidably or at least accidentally pontificate. But I assure you that I will not begin my lecture with a joke.

4


This lecture is targeted at non-experts.My goal is simply to be interesting.

There will be some math, because that’sthe nature of the subject and also because that’s what I like to do.

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One possible motivationfor the subject of this lecture

Each year, 1.3 million people worldwide die in car accidents; 33,000 Americans die each year in car accidents, on average 93 deaths per day.

90% of all these accidents are caused by human errors.

[from The Washington Post, May 30, 2014]

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It would seem that there’s only one practical solution: Replace the driver by a computer, i.e., roboticize the car.

7


The general topic of this lecture is distributedrobotics. Here are two modest examples so that you have some pictures in mind ...

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Four wheeled toy robots used for a graduate thesison formation control -- form a square. We’ll seethat each of these robots can be approximated by a model of the form and are therefore amenable to the theory of this lecture. (Laura Krick co-supervised by Mireille Broucke and me)

q = u

9


Two jeep-type robots. The first is driven by aperson, the second is unmanned. The second isrequired to follow the path of the first with a constant time delay. (Hien Goi co-supervised by Tim Barfoot and me)

10


Why I chose the general topic of distributed robotics for my Bode lecture

We all like to work on hot topics. Certainly distributed robotics has been hot for well over a decade. It’s a followup to decentralized control of the 1970s, and the results of distributed robotics are more profound, in my opinion.

11


The robot rendezvous problem

Get N identical mobile robots with only onboard sensors to move to a common location using distributed control.

This is also called an agreement problem. It is atheoretical problem; in practice it could be usedto get the robots to gather near to each other. It isrelated to the problem of electing a leader.

12


Having chosen the topic of distributed robotics, why I chose the rendezvous problem

A new field is started by the formulation of new problems. Distributed robotics began with flocking and rendezvous. These tasks are important because they are the most basic coordination tasks for a network of mobile robots. Also, they are physical illustrations of emergent behaviour: global consensus from local interactions. Finally, I chose rendezvous over flocking because, as I see it, flocking is still, after 15 years, an unsolved problem.

13


How I was introduced to distributed robotics ...

14


1999

In 1999 I gave my usual graduate course on linear control theory. Tim Barfoot, now a colleague, was a student in the course and showed me his research project: mobile robots that could form a circle in order to function as an antenna.

15


2001

Steve Morse gave a seminar at U of T on the flocking problem. It was very exciting. It was later expanded into the award-winning paper

“Coordination of groups of mobile autonomous agents using nearest neighbor rules”

Jadbabaie, Lin, Morse, 2003

The most highly cited TAC paper on distributed robotics.

16


2003

Block Island Workshop on Cooperative Control

organized by Steve Morse, Naomi Leonard, Vijay Kumar

presenters: Brian Anderson, Ronald Arkin, Murat Arcak, Ella Atkins, Tucker Balch, John Baillieul, Randy Beard, Andrea Bertozzi, Mireille Broucke, Francesco Bullo, Sheryl Coombs, Ian Couzin, Thomas Curtin, Raff D’Andrea, Ted Davison, Magnus Egerstedt, Daniel Grunbaum, Ali Jadbabaie, Eric Klavins, P.S. Krishnaprasad, Julia Parrish, Kevin Passino, Daniela Rus, Shankar Sastry, Rodolphe Sepulchre, Jean-Jacques Slotine, Claire Tomlin, Richard Yang, organizers

I left the workshop keen on working in this field.

17


Nov 2003

Naomi Leonard gave a colloquium at U of T on adaptive ocean sampling. This had the first motivation I had seen for the goal of rendezvous: get the submarines to surface together for re-charging of batteries.

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2005 - 2007

I developed and gave a graduate course on distributed robotics.

“The best way to learn something is to teach it.”

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During my course development, I came upon some of the even earlier papers, e.g.,

“Four bugs on a square,”

Martin Gardner, 1957.

“Why the ant trails look so straight and nice,”

Freddy Bruckstein, 1993.

20


There is now a large literature. E.g., at least three books:

Bullo, Cortes, Martinez 2007Ren, Beard 2007Mesbahi, Egerstedt 2010

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That was a general introduction. We comenow to the first of three main parts.

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Part 1: Mobile robot models

This part is included to assuage my feelings of guilt for a modeling assumption, namely,that an acceptable mobile robot model is

Also, for this part I was inspired by Justh and Krishnaprasad, CDC, 2003 (2007 Bode Lecture)

23

q = u


(x, y, ✓)

Here’s an example of a mobile robot. You can see the omni-directional camera.The variables that describethe position are .

24


Even though I used to be guilty of this, I now have strong objection to saying, without justification, let’s model this robot by

q = u

q = (x, y)

In this part of the lecturewe’ll deal with this objection.

25


Key assumption

A robot has only four things onboard: an omni-directional cameraa computera motor drivewheel encoders.

No leaders. No communication.

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27

But, you may ask, aren’t these conditions too limiting? For example, can’t Raff D’Andrea do a lot more than rendezvous without these constraints?Yes.But you can’t develop a theory without constraints. And let’s not kid ourselves, Raff does have a theory -- he just hasn’t written it down. You can’t win the world robot soccer competition four times without a theory.


The robots move on a large flat floor, which we model by the (x,y)-plane or, equivalently, the complex plane.

28


Our immediate tasks

1. Define the unicycle.

2. Show that a large class of mobile robots can be modeled as unicycles.

3. Show that the unicycle can be linearized as

4. Discuss if a controller for can be implemented on the unicycle.

q = u

q = u

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The (mathematical) unicycle

path

✓

x

y speed = v

The unicycle is a pointthat moves in the plane.At every instant in timethe unicycle has a position, a velocity, and a headingangle.

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x = v cos(✓)

y = v sin(✓)

✓ = !

x, y, ✓

v,!

The equations of the unicycle are

The states are and the inputs are

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vector of velocities

vector of positions

x = v cos(✓)

y = v sin(✓)

✓ = !

(v,!) (x, y, ✓)uni

The block diagram is therefore

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2. Show that a large class of mobile robots can be modeled as unicycles.

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designed and built by Jacob Apkarian, Quanser, Toronto

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K is designed so that over a wide range of inputs

(v,!) ⇡ (vd,!d)

uni(vd,!d) (x, y, ✓)

K mot

dyn enc uniPWM

(v,!)(vd,!d) (x, y, ✓)

In this way we get a unicycle:

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3. Show that the unicycle can be linearized as q = u

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The originators of the idea are apparentlyYun Xiaoping, Yoshio Yamamoto“On feedback linearization of mobile robots”Technical Report, Dept. Comp. Inf. Sci., U. Penn., 1992

“The method is motivated from vehicle maneuvering. When operating a vehicle, a driver looks at a point or an area in front of the vehicle.”

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The simplest way to do this is in terms ofcomplex variables.

Unicycle position z = x+ jy

Unicycle heading vector r = e

j✓

A point just ahead q = z + "r

Thus q = z + "r

Sub in and simplify: q = vr + "!jr

Define u = right-hand side.

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z

r

"r p = z + "r

u = vr + "!jr

q

The model is a linearization of the unicycle at a point just ahead in the direction of r.

q = u

q = u

Summary

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given u, ✓, can solve for v, !


pu 1

s

u(v,!)

uniq

q

(x, y, ✓)

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✓


There are limits to this linearization technique that we don’t need to go into.

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Certain mobile robots can be modeled as unicycles, and hence linearized as

Recap

q = u

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K mot

dyn enc uniPWMqu

(vd,!d) (v,!) (x, y, ✓)

q = u

✓


4. Say that a controller for can only sometimes be implemented on the unicycle.

q = u

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If we linearize the unicycle and then want to use a certain control law, u = ..., we have to show that it is implementable on the real robot with the onboard camera as the only sensor. This is usually not the case.

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Summary of Part 1

The common model for a mobile robot in 2Dis a kinematic integrator. I reviewed onepossible way to get this model. In using thismodel for the rendezvous problem (to be donenext), note that it is the points ahead thatactually rendezvous, not the robots themselves.

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Part 2: Rendezvous theorems

Controllers for rendezvous are based on the strategy of pursuit.

We shall do three theorems. The first usesonly the theory of eigenvalues.

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Consider N mobile robots numbered 1 to Nthat can move around in the complex plane.

Let qi denote the position of robot i. Supposerobot N pursues robot N � 1 according to

the equation

and so on down to

q2 = q1 � q2

q1 = qN � q1.

qN = qN�1 � qN

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13

4

The corresponding visibility graph (who can see whom) is (for N = 5)

2

This is called cyclic pursuit, for an obvious reason. Each robot has a neighbour set (the robots it can see) consisting of one robot.

5

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Write all the equations like this:

q1 = q5 � q1

q2 = q1 � q2

q3 = q2 � q3

q4 = q3 � q4

q5 = q4 � q5

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2

66664

q1q2q3q4q5

3

77775=

2

66664

q5q1q2q3q4

3

77775�

2

66664

q1q2q3q4q5

3

77775

Write these equations as a single vector equation:

51


q =

2

66664

q1q2q3q4q5

3

77775,

Define the vector and the matrix

Then the model for the combined robots is

U =

2

66664

0 0 0 0 11 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 0

3

77775

q = (U � I)q

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The centroid at time t is , where denotes the vector of 1’s.

11

51T q(t)

d

dt1T q(t) = 1T q(t)

= 1T (U � I)q(t)

= (1T � 1T )q(t)

= 0

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The eigenvalues of U � I lie on the circle centre �1, radius 1.One eigenvalue is at the origin and the others are stable.An eigenvector for the zero eigenvalue is 1.

N = 5

54

Ker (U � I) = span{1}

Ker = nullspace


Theorem 1

In cyclic pursuit, the centroid of the robots’

positions is stationary and the robots

asymptotically rendezvous at this centroid.

(“Formations of vehicles in cyclic pursuit”Marshall, Broucke, Francis, 2004)

55


More generally, suppose each robot has a fixedneighbour set and suppose each robot pursues the centroid of its neighbour set. The overall systemwill have the form q = Mq

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Theorem 2Under the pursuit law where each vehicle pursues the centroid of its neighbours, and assuming the neighbour sets do not change with time, the robots rendezvous iff the rank deficiency of M equals 1.

(“Local control strategies for groups of mobile autonomous agents”Lin, Broucke, Francis, 2004)

57


“Suppose that a large group of soldiers are scattered in a foggy battlefield, where visibility is limited to, say, 20 metres. Is it possible for the soldiers to gather silently at a single location?”

The problem of Yamashita et al.Ando, Oasa, Suzuki, Yamashita,“Distributed memoryless point convergencealgorithm for mobile robots with limited memory,” 1999

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The problem involveslimited range cameras.

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Nn = {m : kqm � qnk 1}

The neighbour set of robot n isn

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Find a control law so that the robots rendezvous.

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The centroid pursuit law doesn’t work.

t = 0

t = 0+

62


Yamashita et al. proposed the circumcentre pursuit law.

63


The circumcentre pursuit law does work.

64


Theorem 3Under the circumcentre control law, ifthe visibility graph is initially connected,the robots will rendezvous.

In this context, the visibility graph is symmetric rather than directed.

65


The proof in continuous time requires non-smooth analysis because the circumcentre law is not Lipschitz.

(“State agreement for continuous-time coupled nonlinear systems”Lin, Maggiore, Francis, 2007)

66


Lots of distributed robotics has been developed by computer science researchers. One notable such person is Masafumi Yamashita.

67


Masafumi YamashitaTheoretical Computer Science Group,Dept. of Computer Science and Communication Engineering,Kyushu University

68


I corresponded with him in preparation for this lecture. Here is some of what he had to say.

69


I asked, “Who first formulated the agreement problem?” (agreement is another term for rendezvous)

He replied,“Attaining agreement among processes (or sites) in a distributed environment is a fundamental problem in distributed computing. The problem has numerous applications, e.g., in load balancing, mutual exclusion (for accessing a common resource), etc. Research on agreement dates back to (at least) 1970s.”

70


Then I asked, “I believe you were the first to give a solution, namely, the circumcentre algorithm. Is that correct? If so, can you tell me how you did it?”

He replied, “The basic idea of the algorithm is to (i) place a restriction on the distance the robots canmove in one step, so that every pair of robots mutually visible at any moment will remain mutually visible at the next moment, and (ii) ensure that the convex hull of the robots’ positions converges to a point.”

71


This idea of using the convex hull was later exploredin depth by Luc Moreau (2005).

72


Can the circumcentre law be implemented on a unicycle with only an onboard camera? (Don’t forget, the unicycle model has been linearized about a point ahead.)

Homework for Part 2

73


Part 3: The case N = infinity

74


I have a friend, Avraham Feintuch, whom I’ve known since 1978. (I got my PhD in 1975.) Abie is in the math department at Ben Gurion University. His subject is operator theory, which includes matrix theory for infinite matrices.

How I got onto this topic

75


Abie has family in Toronto and he visits once or twice per year. When he does so, we always discuss these three topics: religion, Middle East politics, and control theory. Not necessarily in that order.

76


Abie and I decided to work together (as we had in the 1980s) and he asked me to suggest a problem. Finitely many robots doesn’t need operator theory. I remembered a very nice paper

“String stability of interconnected systems,” Swaroop and Hedrick, 1996

that had infinitely many vehicles and that arose from the PATH project in California. The paper asks this: Consider a semi-infinite chain of cars traveling at the same speed in a single lane. If the lead vehicle brakes abruptly, will there be a collision?

77


In fact, there have been a number of papers involving infinitely many vehicles.

78


Abie and I took as one goal to answer the question whether Theorem 1 (cyclic pursuit) holds for N = infinity.

Consider serial pursuit:

qn = qn�1 � qn

�1 < n < 1

n

n� 1

79


...

q�1 = q�2 � q�1

q0 = q�1 � q0

q1 = q0 � q1...

Exactly as for finite N , write all the equations like this:

80


Assemble as vectors:

The horizontal lines separate components n < 0

from n � 0.

81

2

6666664

...q�1

q0q1...

3

7777775=

2

6666664

...q�2

q�1

q0...

3

7777775�

2

6666664

...q�1

q0q1...

3

7777775


q =

2

666666664

...q�2

q�1

q0q1...

3

777777775

, U =

2

66666666664

......

......

.... . . 0 0 0 0 0 . . .. . . 1 0 0 0 0 . . .. . . 0 1 0 0 0 . . .. . . 0 0 1 0 0 . . .. . . 0 0 0 1 0 . . .

......

......

...

3

77777777775

Now define the vector and matrix

The vertical and horizontal lines in U separate columns

and rows n < 0 from n � 0.

82


q = (U � I)q

The combined robot model is

exactly as for finite N.

It will turn out that the robots willrendezvous provided the initial positions satisfy a suitable condition.

83


Some notation

Here, n runs over all integers, negative, zero,and positive.

84

A sequence {xn} in C belongs to the space `

2if it is

square-summable, i.e.,

Pn |xn|2 < 1


If {xn} is in `

2, if you plot xn for all n,

and if you draw arrows from xn to xn�1 for all n,

since xn goes to zero as n goes to ±1,

you will see a sort-of loop.

02

3

1

4

-1-2

to the originto the origin

So this is a little similar

to cyclic pursuit.

85


Note that if q(0) 2 `2, then the centroid of {qn(0)}

is the origin.

limN!1

1

2N + 1

NX

n=�N

qn(0)

!= 0

86


Theorem 4

If the initial positions are square-summable,

then the robots rendezvous at the origin,

which is the initial centroid.

87


88

The proof for finite N doesn’t carry over.Let’s see why.


For a finite N, we only had to look at the spectrum of the matrix.

q = (U � I)q, spectrum of U � I

N = 5

Ker (U � I) = span {1}

89

The spectrum decomposesinto {0} plus a part bounded to the left of theimaginary axis.


For infinite N, there’s no such decomposition.

q = (U � I)q, spectrum of U � I

N = 1Ker (U � I) = 0

90


91

So the proof has to be more subtle.


1. The initial states can’t just be bounded.

2. Think of stability of a bi-infinite chain ofdriven cars.

3. An interesting current problem: rendezvous for quadrotors. Rosa, Maggiore, Scardovi, 2014 CDC

4. I want to check out Peter Caines’s work.

Final remarks

92


The thank-you section

93


Collaborator on infinite robot theory Avraham Feintuch

94


Advisors on technical points

Tim BarfootAli JadbabaieNancy KopellManfredi MaggioreAngela SchölligMark Spong

95


Of course, any errors or omissions are their fault.

96


Comments after reading a draftor hearing a dry run

Mireille BrouckeAbie FeintuchJames ForbesSteve MorseMalcolm Smith

97


Suggestions on lecture style

Raff D’AndreaFlorian DörflerJessy GrizzleSteve MorseMathukumalli Vidyasagar

98


“I know what I would want to hear from your Bode Lecture. I would want to hear about you, and some of your research and personal anecdotes, in addition to a scholarly treatment.”

“I think you should give a proper technical talk and not go for laughs. At the CDC you would be addressing hundreds of persons who would be looking for an inspiration, and not jokes.”

“Definitely do try [to be funny]. Definitely do!! Be human! Just don't cross yourself.”

99


Thanks for attending and listening.

100


Appendix: Abie’s and my proof

101


time invariant, spatially invariant

z-transform

qn(t) = qn�1(t)� qn(t)

Q(z, t) =X

n

qn(t)z�n

@

@tQ(z, t) = (z�1 � 1)Q(z, t)

@

@tQ(ej!, t) = (e�j! � 1)Q(ej!, t)

102


@

@tQ(ej!, t) = (e�j! � 1)Q(ej!, t)

= (cos! � 1� j sin!)Q(e

j!, t)

Q(ej!, t) = e(cos!�1�j sin!)tQ(ej!, 0)

qn(t) =1

2⇡

Z ⇡

�⇡Q(ej!, t)ej!nd!

|qn(t)| 1

2⇡

Z ⇡

�⇡

��Q(ej!, t)�� d!

kq(t)k1 1

2⇡

Z ⇡

�⇡

��Q(ej!, t)�� d!

103


@

@tQ(ej!, t) = (e�j! � 1)Q(ej!, t)

= (cos! � 1� j sin!)Q(e

j!, t)

Q(ej!, t) = e(cos!�1�j sin!)tQ(ej!, 0)

qn(t) =1

2⇡

Z ⇡

�⇡Q(ej!, t)ej!nd!

|qn(t)| 1

2⇡

Z ⇡

�⇡

��Q(ej!, t)�� d!

kq(t)k1 1

2⇡

Z ⇡

�⇡

��Q(ej!, t)�� d!

104


✓

1p2⇡

Z ⇡

�⇡e2(cos!�1)td!

◆1/2

! 0

thx to WolframAlpha

kq(t)k1 1

2⇡

Z ⇡

�⇡e(cos!�1)t|Q(ej!, 0)|d!

✓1p2⇡

Z ⇡

�⇡|Q(ej!, 0)|2d!

◆1/2

= (a Bessel fn) ⇥ kq(0)k2

105


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The Robot Rendezvous Problemfrancis/Bode.pdf · 2014-12-18 · Steve Morse gave a seminar at U of T...

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