Algorithms for Walking, Running, Swimming, Flying, and Manipulation
© Russ Tedrake, 2024
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Note: These are working notes used for a course being taught at MIT. They will be updated throughout the Spring 2024 semester. Lecture videos are available on YouTube.
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In chapter 2, we spent some time thinking about the phase portrait of the simple pendulum, and concluded with a challenge: can we design a nonlinear controller to reshape the phase portrait, with a very modest amount of actuation, so that the upright fixed point becomes globally stable? With enough actuators and unbounded torque, feedback-cancellation solutions (e.g., invert gravity) can work well, but can also require an unnecessarily large amount of control effort. The energy-based swing-up control solutions presented for the Acrobot and cart-pole systems are considerably more appealing, but required some cleverness and might not scale to more complicated systems. Here we investigate another approach to the problem, using computational optimal control to synthesize a feedback controller directly.
In this chapter, we will introduce optimal control. Reinforcement
learning is the machine learning name for optimal control; we'll
discuss that machine learning perspective later in the notes. Optimal
control is powerful for a number of reasons. First and foremost, it is very
general - allowing us to specify the goal of control equally well for
fully-actuated or underactuated, linear or nonlinear, deterministic or
stochastic, and continuous or discrete systems. Second, it permits concise
descriptions of potentially very complex desired behaviors, specifying the
goal of control as a scalar objective (plus a list of constraints).
Finally, and most importantly, optimal control is very amenable to
numerical solutions.
The fundamental idea in optimal control is to formulate the goal of control as the long-term optimization of a scalar cost function. Let's introduce the basic concepts by considering a system that is even simpler than the simple pendulum.
Consider the "double integrator" system $$\ddot{q} = u, \quad |u| \le 1.$$ If you would like a mechanical analog of the system (I always do), then you can think about this as a unit mass brick moving along the $x$-axis on a frictionless surface, with a control input which provides a horizontal force, $u$. The task is to design a control system, $u = \pi(\bx,t)$, $\bx=[q,\dot{q}]^T$ to regulate this brick to $\bx = [0,0]^T$.
In order to formulate this control design problem using optimal control, we must define a scalar objective which scores the long-term performance of running each candidate control policy, $\pi(\bx,t)$, from each initial condition, $(\bx_0,t_0)$, and a list of constraints that must be satisfied. For the task of driving the double integrator to the origin, one could imagine a number of optimal control formulations which would accomplish the task, e.g.:
Note that the input limits, $|u|\le1$ are also required to make this problem well-posed; otherwise both optimizations would result in the optimal policy using infinite control input to approach the goal infinitely fast. Besides input limits, another common approach to limiting the control effort is to add an additional quadratic cost on the input (or "effort"), e.g. $\int \left[ \bu^T(t) {\bf R} \bu(t) \right] dt,$ ${\bf R}\succ0$. This could be added to either formulation above. We will examine many of these formulations in some detail in the examples worked out at the end of this chapter.
Optimal control has a long history in robotics. It started out motivated by achieving peak performance in niche applications like programming minimum-time paths for a pick-and-place robot in a factory. But it is the generality of the approach (one can program almost any task using similar tools) that has made optimization-based and now machine-learning-based control dominate the way that state-of-the-art robot control systems are developed.
For more intuition, let's do an informal derivation of the solution to the minimum time problem for the double integrator with input constraints: \begin{align*} \minimize_{\pi} \quad & t_f\\ \subjto \quad & \bx(t_0) = \bx_0, \\ & \bx(t_f) = {\bf 0}, \\ & \ddot{q}(t) = u(t), \\ & |u(t)| \le 1. \end{align*} What behavior would you expect an optimal controller to exhibit?
Your intuition might tell you that the best thing that the brick can do, to reach the goal in minimum time with limited control input, is to accelerate maximally towards the goal until reaching a critical point, then hitting the brakes in order to come to a stop exactly at the goal. This is called a bang-bang control policy; these are often optimal for systems with bounded input, and it is in fact optimal for the double integrator although we will not prove it until we have developed more tools.
Let's work out the details of this bang-bang policy. First, we can figure out the states from which, when the brakes are fully applied, the system comes to rest precisely at the origin. Let's start with the case where $q(0) < 0$, and $\dot{q}(0)>0$, and "hitting the brakes" implies that $u=-1.$ Integrating the equations, we have \begin{gather*} \ddot{q}(t) = u = -1, \\\dot{q}(t) = \dot{q}(0) - t, \\ q(t) = q(0) + \dot{q}(0) t - \frac{1}{2} t^2. \end{gather*} Substituting $t = \dot{q}(0) - \dot{q}$ into the solution reveals that the system trajectories are parabolic arcs: \[ q = -\frac{1}{2} \dot{q}^2 + c_{-}, \] with $c_{-} = q(0) + \frac{1}{2}\dot{q}^2(0)$.
Similarly, the solutions for $u=1$ are $q = \frac{1}{2} \dot{q}^2 + c_{+}$, with $c_{+}=q(0)-\frac{1}{2}\dot{q}^2(0)$.
Perhaps the most important of these trajectories are the ones that pass directly through the origin (e.g., $c_{-}=0$). Following our initial logic, if the system is going slower than this $\dot{q}$ for any $q$, then the optimal thing to do is to slam on the accelerator ($u=-\text{sgn}(q)$). If it's going faster than the $\dot{q}$ that we've solved for, then still the best thing to do is to brake; but inevitably the system will overshoot the origin and have to come back. We can summarize this policy with: \[ u = \begin{cases} +1, & \text{if } (\dot{q} < 0 \text{ and } q \le \frac{1}{2} \dot{q}^2) \text{ or } (\dot{q}\ge 0 \text{ and } q < -\frac{1}{2} \dot{q}^2), \\ 0, & \text{if } q=0 \text{ and } \dot{q}=0, \text{ or} \\ -1, & \text{otherwise.} \end{cases} \]
We can illustrate some of the optimal solution trajectories:
And for completeness, we can compute the optimal time to the goal by solving for the amount of time required to reach the switching surface plus the amount of time spent moving along the switching surface to the goal. With a little algebra, you will find that the time to the goal, $J(\bx)$, is given by \[ J(\bx) = \begin{cases} 2\sqrt{\frac{1}{2}\dot{q}^2-q} - \dot{q}, & \text{for } u=+1 \text{ regime}, \\ 0, & \text{for } u=0, \\ \dot{q} + 2\sqrt{\frac{1}{2}\dot{q}^2+q}, & \text{for } u=-1, \end{cases} \] which is plotted here:
Notice that the "time-to-the-origin function" is continuous over state (though not smooth), even though the policy is discontinuous.
As we begin to develop theoretical and algorithmic tools for optimal control, we will see that some formulations are much easier to deal with than others. One important example is the dramatic simplification that can come from formulating objective functions using additive cost, because they often yield recursive solutions. In the additive cost formulation, the long-term objective for a trajectory can be written as $$\int_0^T \ell(x(t),u(t)) dt,$$ where $\ell()$ is the instantaneous cost (also referred to as the "running cost"), and $T$ can be either a finite real number or $\infty$. We will call a problem specification with a finite $T$ a "finite-horizon" problem, and $T=\infty$ an "infinite-horizon" problem. Problems and solutions for infinite-horizon problems tend to be more elegant, but care is required to make sure that the integral converges for the optimal controller (typically by having an achievable goal that allows the robot to accrue zero cost, or by having costs that discount/decay to zero as time increases).
The "quadratic cost" formulation we considered in the double integrator example above is clearly written as an additive cost. At first glance, our "minimum-time" formulation doesn't appear to be of this form, but we actually can write it as an additive cost problem using an infinite horizon and the instantaneous cost $$\ell(x,u) = \begin{cases} 0 & \text{if } x=0, \\ 1 & \text{otherwise.} \end{cases}$$
We will examine a number of approaches to solving optimal control problems throughout the next few chapters. For the remainder of this chapter, we will focus on additive-cost problems and their solution via dynamic programming.
For systems with continuous states and continuous actions, dynamic programming is a set of theoretical ideas surrounding additive-cost optimal control problems. For systems with a finite, discrete set of states and a finite, discrete set of actions, dynamic programming also represents a set of very efficient numerical algorithms which can compute optimal feedback controllers. Many of you will have seen it introduced before as a tool for graph search.
Imagine you have a directed graph with states (or nodes) $\{s_1,s_2,...\} \in S$ and "actions" associated with edges labeled as $\{a_1,a_2,...\} \in A$, as in the following trivial example:
Let us also assume that each edge has an associate weight or cost, using $\ell(s,a)$ to denote the cost of being in state $s$ and taking action $a$. Furthermore we will denote the transition "dynamics" using \[ s[n+1] = f(s[n],a[n]). \] For instance, in the graph above, $f(s_1,a_1) = s_2$.
There are many algorithms for finding (or approximating) the optimal
path from a start to a goal on directed graphs. In dynamic programming,
the key insight is that we can find the shortest path from every node by
solving recursively for the optimal cost-to-go (the cost that will
be accumulated when running the optimal controller), which we'll denote
$J^*(s)$, from every node to the goal. One such algorithm starts by
initializing an estimate $\hat{J}^*(s)=0$ for all $s_i$, then proceeds with
an iterative algorithm which sets \begin{equation} \forall i \quad
\hat{J}^*(s_i) \Leftarrow \min_{a \in A} \left[ \ell(s_i,a) +
\hat{J}^*\left({f(s_i,a)}\right) \right]. \label{eq:value_update}
\end{equation} This notation assumes that the update $f(s,a)$ and cost
$\ell(s,a)$ are defined for every state-action pair; we can add self
transitions at each state for any missing actions which have zero cost for
the goal state and sufficiently high cost anywhere else. In software,
$\hat{J}^*(s)$ can be represented as a vector with dimension equal to the
number of discrete states, and equation \eqref{eq:value_update} gives us an
update for the entire vector. This algorithm, appropriately known as
value iteration, is guaranteed to converge to the optimal
cost-to-go up to a constant factor, $\hat{J}^* \rightarrow J^* + c$
Value iteration is an amazingly simple algorithm, but it accomplishes something quite amazing: it efficiently computes the long-term cost of an optimal policy from every state by iteratively evaluating the one-step cost. If we know the optimal cost-to-go, then it's easy to extract the optimal policy, $a = \pi^*(s)$: \begin{equation} \pi^*(s_i) = \argmin_a \left[ \ell(s_i,a) + J^*\left( f(s_i,a) \right) \right]. \label{eq:policy_update} \end{equation} It's a simple algorithm, but playing with an example can help our intuition.
Imagine a robot living in a grid (finite state) world
As always, I encourage you to run the notebook and play with the code yourself:
You can run value iteration for the double integrator (using
barycentric interpolation to discretize the continuous state space, as
we'll describe below) in
Please do take some time to try different cost functions by editing the code yourself.
Let's take a minute to appreciate how amazing this is. Our solution to finding the optimal controller for the double integrator wasn't all that hard, but it required some mechanical intuition and solutions to differential equations. The resulting policy was non-trivial -- bang-bang control with a parabolic switching surface. The value iteration algorithm doesn't use any of this directly -- it's a simple algorithm for graph search. But remarkably, it can generate effectively the same policy with just a few moments of computation.
It's important to note that there are some differences between the computed policy and the optimal policy that we derived, due to discretization errors. We will ask you to explore these in the exercises.
The real value of this numerical solution, however, is unlike our analytical solution for the double integrator, we can apply this same algorithm to any number of dynamical systems virtually without modification. Let's apply it now to the simple pendulum, which was intractable analytically.
You can run value iteration for the simple pendulum (again using
barycentric interpolation) in
Again, you can easily try different cost functions by editing the code yourself.
I find the graph search algorithm extremely satisfying as a first step, but also become quickly frustrated by the limitations of the discretization required to use it. In many cases, we can do better; coming up with algorithms which work more natively on continuous dynamical systems. We'll explore those extensions in this section.
It's important to understand that the value iteration equations,
equations (\ref{eq:value_update}) and (\ref{eq:policy_update}), are more
than just an algorithm. They are also sufficient conditions for
optimality: if we can produce a $J^*$ and $\pi^*$ which satisfy these
equations, then $\pi^*$ must be an optimal controller. There are an
analogous set of conditions for the continuous systems. For a system
$\dot{\bx} = f(\bx,\bu)$ and an infinite-horizon additive cost
$\int_0^\infty \ell(\bx,\bu)dt,$ we have: \begin{gather} 0 = \min_\bu \left[
\ell(\bx,\bu) + \pd{J^*}{\bx}f(\bx,\bu) \right], \label{eq:HJB} \\
\pi^*(\bx) = \argmin_\bu \left[ \ell(\bx,\bu) + \pd{J^*}{\bx}f(\bx,\bu)
\right]. \end{gather} Equation \ref{eq:HJB} is known as the
Hamilton-Jacobi-Bellman (HJB) equation.
Consider a system $\dot{\bx}=f(\bx,\bu)$ and an infinite-horizon additive cost $\int_0^\infty \ell(\bx,\bu)dt$, with $f$ and $\ell$ continuous functions, and $\ell$ a strictly-positive-definite function that obtains zero only at a unique state $\bx^*$. Suppose $J(\bx)$ is a solution to the HJB equation: $J$ is continuously differentiable in $\bx$ and satisfies \[ 0 = \min_{\bu \in U} \left[\ell(\bx,\bu) + \pd{J}{\bx}f(\bx,\bu) \right],\quad \text{for all } \bx. \] Further assume that $J(\bx)$ is positive definite and that $\pi(\bx)$ is the minimizer for all $\bx$. Then, under some technical conditions on the existence and boundedness of solutions, we have that $J(\bx) - J(\bx^*)$ is the optimal cost-to-go and $\pi$ is an optimal policy.
Given an open subset $\Omega\subset\mathbb R^n$, with a selected element $x^*$, a subset $U\subset\mathbb R^m$, continuous functions $f:~\Omega\times U\to\mathbb R^n$, $g:~\Omega\times U\to[0,\infty)$, continuously differentiable function $V:~\Omega\to[0,\infty)$, and a function $\mu:~\Omega\to U$, such that
Proof. First, observe that, for all piecewise continuous $(x,u)$ satisfying (\ref{e1}),
To finish the proof, for an arbitrary piecewise continuous solution of (\ref{e1}), equation (\ref{e3}), together with non-negativity of $g_V$, implies \[ J(x,u)=\int_0^Tg(x(t),u(t))dt\ge V(x_0)-V(x(T)).\] When $J(x,u)<\infty$, applying this to $T=T_k$, where $T_k$ are described in observation (II), and taking the limit as $k\to\infty$ (and $x(T_k)\to x^*$) yields $J(x,u)\ge V(x_0)-V(x^*)$. $\Box$
It is possible to prove sufficiency under different assumptions, too. The particular assumptions here were chosen to ensure that $J(\bx(0)) < \infty$ implies that $\bx(t) \rightarrow \bx^*$. As Sasha says, "without something like this, all sorts of counter-examples emerge."
As a tool for verifying optimality, the HJB equations are actually surprisingly easy to work with: we can verify optimality for an infinite-horizon objective without doing any integration; we simply have to check a derivative condition on the optimal cost-to-go function $J^*$. Let's see this play out on the double integrator example.
Consider the problem of regulating the double integrator (this time without input limits) to the origin using a quadratic cost: $$ \ell(\bx,\bu) = q^2 + \dot{q}^2 + u^2. $$ I claim (without derivation) that the optimal controller for this objective is $$\pi(\bx) = -q - \sqrt{3}\dot{q}.$$ To convince you that this is indeed optimal, I have produced the following cost-to-go function: $$J(\bx) = \sqrt{3} q^2 + 2 q \dot{q} + \sqrt{3} \dot{q}^2.$$
Taking \begin{gather*} \pd{J}{q} = 2\sqrt{3} q + 2\dot{q}, \qquad \pd{J}{\dot{q}} = 2q + 2\sqrt{3}\dot{q}, \end{gather*} we can write \begin{align*} \ell(\bx,\bu) + \pd{J}{\bx}f(\bx,\bu) &= q^2 + \dot{q}^2 + u^2 + (2\sqrt{3} q + 2\dot{q}) \dot{q} + (2q + 2\sqrt{3}\dot{q}) u \end{align*} This is a convex quadratic function in $u$, so we can find the minimum with respect to $u$ by finding where the gradient with respect to $u$ evaluates to zero. \[ \pd{}{u} \left[ \ell(\bx,\bu) + \pd{J}{\bx} f(\bx,\bu) \right] = 2u + 2q + 2\sqrt{3}\dot{q}. \] Setting this equal to $0$ and solving for $u$ yields: $$u^* = -q - \sqrt{3} \dot{q},$$ thereby confirming that our policy $\pi$ is in fact the minimizer. Substituting $u^*$ back into the HJB reveals that the right side does in fact simplify to zero. I hope you are convinced!
Note that evaluating the HJB for the time-to-go of the minimum-time problem for the double integrator will also reveal that the HJB is satisfied wherever that gradient is well-defined. This is certainly mounting evidence in support of our bang-bang policy being optimal, but since $\pd{J}{\bx}$ is not defined everywhere, it does not actually satisfy the requirements of the sufficiency theorem as stated above. Since cusps are commonplace in optimal value functions, one could say that the assumption in the sufficiency theorem that $\pd{J}{\bx}$ is defined everywhere makes our current statement very weak.
We still face a few barriers to actually using the HJB in an algorithm. The first barrier is the minimization over $\bu$. When the action set was discrete, as in the graph search version, we could evaluate the one-step cost plus cost-to-go for every possible action, and then simply take the best. For continuous action spaces, in general we cannot rely on the strategy of evaluating a finite number of possible $\bu$'s to find the minimizer.
All is not lost. In the quadratic cost double integrator example above, we were able to solve explicitly for the minimizing $\bu$ in terms of the cost-to-go. It turns out that this strategy will actually work for a number of the problems we're interested in, even when the system (which we are given) or cost function (which we are free to pick, but which should be expressive) gets more complicated.
Recall that I've already tried to convince you that a majority of the systems of interest are control affine, e.g. I can write \[ f(\bx,\bu) = f_1(\bx) + f_2(\bx)\bu. \] We can make another simplification by restricting ourselves to running cost functions of the form \[ \ell(\bx,\bu) = \ell_1(\bx) + \bu^T {\bf R} \bu, \qquad {\bf R}={\bf R}^T \succ 0. \] In my view, this is not very restrictive - many of the cost functions that I find myself choosing to write down can be expressed in this form. Given these assumptions, we can write the HJB as \[ 0 = \min_{\bu} \left[ \ell_1(\bx) + \bu^T {\bf R} \bu + \pd{J}{\bx} \left[ f_1(\bx) + f_2(\bx)\bu \right]\right]. \] Since this is a positive quadratic function in $\bu$, if the system does not have any constraints on $\bu$, then we can solve in closed-form for the minimizing $\bu$ by taking the gradient of the right-hand side: \[ \pd{}{\bu} = 2\bu^T {\bf R} + \pd{J}{\bx} f_2(\bx) = 0, \] and setting it equal to zero to obtain \[ \bu^* = -\frac{1}{2}{\bf R}^{-1}f_2^T(\bx) \pd{J}{\bx}^T.\] If there are linear constraints on the input, such as torque limits, then more generally this could be solved (at any particular $\bx$) as a quadratic program.
Exploiting the ability to solve for the optimal action in closed form
is also nice for generating benchmark problems for numerical optimal
control, as it can be used for "converse optimal control"
Consider a one-dimensional system of the form $\dot{x} = f_1(x) + xu,$ and the running cost function $\ell(x,u) = x^2 + u^2.$ The optimal policy is $u^* = -\frac{1}{2}x\pd{J^*}{x},$ leading to the HJB $$0 = x^2 - \frac{1}{4}x^2\left[\pd{J^*}{x}\right]^2 + \pd{J^*}{x} f_1(x).$$ Choosing $J^*(x) = x^2$, we find that this is the cost-to-go when $f_1(x) = -\frac{1}{2}x + \frac{1}{2}x^3$ and $u^* = -x^2.$
What happens in the case where our system is not control affine or if
we really do need to specify an instantaneous cost function on $\bu$
that is not simply quadratic? If the goal is to produce an iterative
algorithm, like value iteration, then one common approach is to make a
(positive-definite) quadratic approximation in $\bu$ of the HJB, and
updating that approximation on every iteration of the algorithm. This
broad approach is often referred to as differential dynamic
programming (c.f.
The other major barrier to using the HJB in a value iteration algorithm is that the estimated optimal cost-to-go function, $\hat{J}^*$, must somehow be represented with a finite set of numbers, but we don't yet know anything about the potential form it must take. In fact, knowing the time-to-goal solution for minimum-time problem with the double integrator, we see that this function might need to be non-smooth for even very simple dynamics and objectives.
One natural way to parameterize $\hat{J}^*$ -- a scalar valued-function defined over the state space -- is to define the values on a mesh. This approach then admits algorithms with close ties to the relatively very advanced numerical methods used to solve other partial differential equations (PDEs), such as the ones that appear in finite element modeling or fluid dynamics. One important difference, however, is that our PDE lives in the dimension of the state space, while many of the mesh representations from the other disciplines are optimized for two or three dimensional space. Also, our PDE may have discontinuities (or at least discontinuous gradients) at locations in the state space which are not known apriori.
These days, the other natural choice is to represent $J^*$ using a neural network, or some similarly richly parameterized function approximator. This is common practice in reinforcement learning, which we will discuss more later in the book.
If we approximate $J^*$ with a finitely-parameterized function $\hat{J}_\balpha^*$, with parameter vector $\balpha$, then this immediately raises many important questions. If the true cost-to-go function does not live in the prescribed function class -- e.g., there does not exist an $\balpha$ which satisfies the sufficiency conditions for all $\bx$ -- then we might instead write a cost function to match the optimality conditions as closely as possible. But our choice of cost function is important. Are errors in all states equally important? If we are balancing an Acrobot, presumably having an accurate cost-to-go in the vicinity of the upright condition is more important that some rarely visited state? Even if we can define a cost function that we like, then can we say anything about the convergence of our algorithm in this more general case?
Let us start by considering a least-squares approximation of the value iteration update.
Using the least squares solution in a value iteration update is sometimes referred to as fitted value iteration, where $\bx_k$ are some number of samples taken from the continuous space and for discrete-time systems the iterative approximate solution to \begin{gather*} J^*(\bx_0) = \min_{u[\cdot]} \sum_{n=0}^\infty \ell(\bx[n],\bu[n]), \\ \text{ s.t. } \bx[n+1] = f(\bx[n], \bu[n]), \bx[0] = \bx_0\end{gather*} becomes \begin{gather} J^d_k = \min_\bu \left[ \ell(\bx_k,\bu) + \hat{J}^*_\alpha\left({f(\bx_k,\bu)}\right) \right], \\ \balpha \Leftarrow \argmin_\balpha \sum_k \left(\hat{J}^*_\balpha(\bx_k) - J^d_k \right)^2. \label{eq:fitted_value_iteration} \end{gather} Since the desired values $J^d_k$ are depend on the current guess of the cost-to-go, we will apply this value iteration algorithm iteratively until (hopefully) we achieve some numerical convergence.
Note that the update in \eqref{eq:fitted_value_iteration} is not quite the same as doing least-squares optimization of $$\sum_k \left(\hat{J}^*_\balpha(\bx_k) - \min_\bu \left[ \ell(\bx_k,\bu) + \hat{J}^*_\alpha\left({f(\bx_k,\bu)}\right) \right] \right)^2,$$ because in this equation $\alpha$ has an effect on both occurrences of $\hat{J}^*$. In \eqref{eq:fitted_value_iteration}, we cut that dependence by taking $J_k^d$ as fixed desired values; this version performs better in practice. In the deep reinforcement learning literature, we cut the dependence in a similar way by using a "target network" with the weights, $\beta,$ that are updated more slowly than $\alpha:$ $$\sum_k \left(\hat{J}^*_\balpha(\bx_k) - \min_\bu \left[ \ell(\bx_k,\bu) + \hat{J}^*_\beta\left({f(\bx_k,\bu)}\right) \right] \right)^2.$$ For nonlinear function approximators, the update to $\alpha$ in \eqref{eq:fitted_value_iteration} is typically approximated by taking multiple steps of (stochastic) gradient descent.
In general, the convergence and accuracy guarantees for value iteration with generic function approximators are quite weak. But we do have some results for the special case of linear function approximators. A linear function approximator takes the form: \[ \hat{J}^*_\balpha(\bx) = \sum_i \alpha_i \psi_i(\bx) = \bpsi^T(\bx) \balpha, \] where $\bpsi(\bx)$ is a vector of potentially nonlinear features. Common examples of features include polynomials, radial basis functions, or most interpolation schemes used on a mesh. The distinguishing feature of a linear function approximator is the ability to exactly solve for $\balpha$ in order to represent a desired function optimally, in a least-squares sense. For linear function approximators, this is simply: \begin{gather*} \balpha \Leftarrow \begin{bmatrix} \bpsi^T(\bx_1) \\ \vdots \\ \bpsi^T(\bx_K)\end{bmatrix}^+ \begin{bmatrix} J^d_1 \\ \vdots \\ J^d_K \end{bmatrix}, \end{gather*} where the $^+$ notation refers to a Moore-Penrose pseudoinverse. One "theory of deep learning" argues that we can think of deep networks as linear function approximators (with randomly-initialized weights causing the last hidden layer to be a rich random basis vector, which is combined linearly into the outputs), which might justify how they can obtain zero training error quickly and then slowly adapt the features based on implicit regularization.
In finite state and action settings, fitted value iteration with
linear function approximators is known to converge to the globally
optimal $\balpha^*$
Imagine that we use a mesh to approximate the cost-to-go function
over that state space with $K$ mesh points $\bx_k$. We would like to
perform the value iteration update: \begin{equation} \forall k,
\hat{J}^*(\bx_k) \Leftarrow \min_\bu \left[ \ell(\bx_k,\bu) +
\hat{J}^*\left({f(\bx_k,\bu)}\right) \right],
\label{eq:mesh_value_iteration} \end{equation} but must deal with the
fact that $f(\bx_k,\bu)$ might not result in a next state that is
directly at a mesh point. Most interpolation schemes for a mesh can be
written as some weighted combination of the values at nearby mesh
points, e.g. \[ \hat{J}^*(\bx) = \sum_i \beta_i(\bx) \hat{J}^*(\bx_i),
\quad \sum_i \beta_i = 1 \] with $\beta_i$ the relative weight of the
$i$th mesh point. In
Of course the most popular function approximators today are based on
neural networks. In these much more general approximators, we cannot
solve the least-squares update \eqref{eq:fitted_value_iteration}
exactly, but instead attempt to minimize the loss function: with a few
steps of (stochastic) gradient descent on each iteration. In practice,
we can use any of the standard optimizers (the examples below uses
Adam
What sort of neural architectures work best for value iteration in
robotics? AlphaGo famously used convolutional
networks
When it comes to neural network software, I'll mostly use PyTorch throughout these notes. But for
small problems (with small neural nets and modest batch sizes during
training) like the ones we'll consider here, the cost of going back and
forth to PyTorch and especially the GPU outweighs most of the benefits.
I've also implemented a simple CPU-based
MultilayerPerceptron
in Systems
framework, and will use it when it
makes sense. My goal is to provide examples of both approaches so that
you can pick and choose.
Let us try reproducing our double-integrator value iteration examples using neural networks:
This is nice work on continuous neural fitted value iteration in
For solutions to systems with continuous-time dynamics, I have to uncover one of the details that I've so far been hiding to keep the notation simpler. Let us consider a problem with a finite-horizon: \begin{gather*} \min_{\bu[\cdot]} \sum_{n=0}^N \ell(\bx[n],\bu[n]), \\ \text{ s.t. } \bx[n+1] = f(\bx[n], \bu[n]), \bx[0] = \bx_0\end{gather*} When the horizon is finite, the optimal cost-to-go and the optimal policy, in general, can depend on time (if the post office closes at 5pm, then your actions at 4:55pm might be very different than your actions at 4pm).
The way that we compute this is by solving the time-varying cost-to-go function backwards in time: \begin{gather*}J^*(\bx,N) = \min_\bu \ell(\bx, \bu) \\ J^*(\bx,n-1) = \min_\bu \left[ \ell(\bx, \bu) + J^*(f(\bx,\bu), n) \right]. \end{gather*} Even in the infinite-horizon setting, one can actually understand the convergence of the value iteration update as solving this time-varying cost-to-go backwards in time until it reaches a steady-state solution (the infinite-horizon solution). This is one way to see why value iteration only converges if the optimal cost-to-go is bounded.
Now let's consider the continuous-time version. Again, we have a
time-varying cost-to-go, $J^*(\bx,t).$ Now $$\frac{dJ^*}{dt} =
\pd{J^*}{\bx}f(\bx,\bu) + \pd{J^*}{t},$$ and our sufficiency condition
is $$0 = \min_\bu \left[\ell(\bx, \bu) + \pd{J^*}{\bx}f(\bx,\bu) +
\pd{J^*}{t} \right].$$ But since $\pd{J^*}{t}$ doesn't depend on $\bu$,
we can pull it out of the $\min$ and write the (true) HJB:
$$-\pd{J^*}{t} = \min_\bu \left[\ell(\bx, \bu) + \pd{J^*}{\bx}f(\bx,\bu)
\right].$$ The standard numerical recipe
Probably most visible and consistent campaign using numerical HJB
solutions in applied control (at least in robotics) has come from Claire Tomlin's group
at Berkeley. Their work leverages Ian Mitchell's Level
Set Toolbox, which solves the Hamilton-Jacobi PDEs on a Cartesian
mesh using the technique cartooned above, and even includes the
minimum-time problem for the double integrator as a tutorial
example
There are many many nice extensions to the basic formulations that we've
presented so far. I'll try to list a few of the most important ones here.
I've also had a number of students in this course explore very interesting
extensions; for example
Throughout this chapter, we have focused primarily on minimizing an infinite-horizon sum of running costs. Whenever you write an infinite sum (or integral), one should immediately question whether that sum converges to a finite value. For many control problems, it does -- for instance, if we set the cost to be zero at the goal and the controller can drive the system to the goal in finite time, or exponentially at a rate that is faster than cost is accrued. However, once we enter the realm of approximate optimal control, we can run into problems. For instance, if we form a discrete state and action approximation of the dynamics, it may not be possible for the discretized system to arrive exactly at the zero-cost goal.
A common way to protect oneself from this is to add an exponential
discount factor to the integral cost. This is a natural choice since it
preserves the recursive structure of the Bellman equation. In discrete
time, it takes the form $$\min_{\bu[\cdot]} \sum_{n=0}^\infty \gamma^n
\ell(\bx[n], \bu[n]),$$ which leads to the Bellman equation $$J^*(\bx) =
\min_\bu \left[ \ell(\bx, \bx) + \gamma J^*(f(\bx,\bu) \right].$$ In
continuous time we have
A word of warning: this discount factor does change the optimal policy; and it is not without its pitfalls. We'll work out a specific case of the discounted cost for the LQR problem, and see that the discounted controller is exactly a controller which assumes that the plant is more stable than it actually is. It is possible to find an optimal controller for the discounted cost that results in an unstable controller.
A more robust alternative is to consider an average-cost formulations... (details coming soon!)
One of the most amazing features of the dynamic programming, additive cost approach to optimal control is the relative ease with which it extends to optimizing stochastic systems.
For discrete systems, we can generalize our dynamics on a graph by adding action-dependent transition probabilities to the edges. This new dynamical system is known as a Markov Decision Process (MDP), and we write the dynamics completely in terms of the transition probabilities \[\Pr(s[n+1] = s' | s[n] = s, a[n] = a). \] For discrete systems, this is simply a big lookup table. The cost that we incur for any execution of system is now a random variable, and so we formulate the goal of control as optimizing the expected cost, e.g. \begin{equation} J^*(s[0]) = \min_{a[\cdot]} E \left[ \sum_{n=0}^\infty \ell(s[n],a[n]) \right]. \label{eq:stochastic_dp_optimality_cond} \end{equation} Note that there are many other potential objectives, such as minimizing the worst-case error, but the expected cost is special because it preserves the dynamic programming recursion: \[ J^*(s) = \min_a E \left[\ell(s,a) + J^*(s')\right] = \min_a \left[ \ell(s,a) + \sum_{s'} \Pr(s'|s,a) J^*(s') \right].\] Remarkably, if we use these optimality conditions to construct our value iteration algorithm \[ \hat{J}(s) \Leftarrow \min_a \left[ \ell(s,a) + \sum_{s'} \Pr(s'|s,a) \hat{J}(s') \right],\] then this algorithm has the same strong convergence guarantees of its counterpart for deterministic systems. And it is essentially no more expensive to compute!
We will continue to develop a much more general toolkit for for stochastic and robust control later in the notes.
There is a particularly nice observation to be made here. Let's assume that we have discrete control inputs and discrete-time dynamics, but a continuous state space. Recall the fitted value iteration on a mesh algorithm described above. In fact, the resulting update is exactly the same as if we treated the system as a discrete state MDP with $\beta_i$ representing the probability of transitioning to state $\bx_i$! This sheds some light on the impact of discretization on the solutions -- discretization error here will cause a sort of diffusion corresponding to the probability of spreading across neighboring nodes.
For discrete MDPs, value iteration is a magical algorithm because it is simple, but known to converge to the global optimal, $J^*$. However, other important algorithms are known; one of the most important is a solution approach using linear programming. This formulation provides an alternative view, but may also be more generalizable and even more efficient for some instances of the problem.
Recall the optimality conditions from Eq. \eqref{eq:value_update}. If we describe the cost-to-go function as a vector, $J_i = J(s_i)$, then these optimality conditions can be rewritten in vector form as \begin{equation} \bJ = \min_a \left[ {\bf \ell}(a) + \bT(a) \bJ \right], \label{eq:vector_stochastic_dp} \end{equation} where $\ell_i(a) = \ell(s_i,a)$ is the cost vector, and $T_{i,j}(a) = \Pr(s_j|s_i,a)$ is the transition probability matrix. Let us take $\bJ$ as the vector of decision variables, and replace Eq. (\ref{eq:vector_stochastic_dp}) with the constraints: \begin{equation} \forall a, \bJ \le {\bf \ell}(a) + \bT(a) \bJ.\end{equation} Clearly, for finite $a$, this is finite list of linear constraints, and for any vector $\bJ$ satisfying these constraints we have $\bJ \le \bJ^*$ (elementwise). Now write the linear program: \begin{gather*} \maximize_\bJ \quad \bc^T \bJ, \\ \subjto \quad \forall a, \bJ \le {\bf \ell}(a) + \bT(a) \bJ, \end{gather*} where $c$ is any positive vector. The solution to this problem is $\bJ = \bJ^*$.
Perhaps even more interesting is that this approach can be
generalized to linear function approximators. Taking a vector form of
my notation above, now we have a matrix of features with $\bPsi_{i,j} =
\psi^T_j(\bx_i)$ and we can write the LP \begin{gather}
\maximize_\balpha \quad \bc^T \bPsi \balpha, \\ \subjto \quad \forall
a, \bPsi \balpha \le {\bf \ell}(a) + \bT(a) \bPsi \balpha. \end{gather}
This time the solution is not necessarily optimal, because $\bPsi
\balpha^*$ only approximates $\bJ^*$, and the relative values of the
elements of $\bc$ (called the "state-relevance weights") can determine
the relative tightness of the approximation for different features
There are some cases where the linear programming approach to dynamic programming can be extended to continuous states and actions, with strong guarantees. This approach follows naturally from our study of Lyapunov functions in an upcoming chapter, and builds on the computational tools we develop there. Feel free to skip ahead.
Let's consider the discrete time, state, and action version of value iteration: \begin{equation} \hat{J}^*(s_i) \Leftarrow \min_{a \in A} \left[ \ell(s_i,a) + \hat{J}^*\left({f(s_i,a)}\right) \right],\end{equation} which finds the optimal policy for the infinite horizon cost function $\sum_{n=0}^\infty \ell(s[n],a[n]).$
Surprisingly, adding a discount factor can help with this. Consider the infinite-horizon discounted cost: $\sum_{n=0}^\infty \gamma^n \ell(s[n],a[n])$, where $0 < \gamma \le 1$ is the discount factor. The corresponding value iteration update is \begin{equation} \hat{J}^*(s_i) \Leftarrow \min_{a \in A} \left[ \ell(s_i,a) + \gamma\hat{J}^*\left({f(s_i,a)}\right) \right].\end{equation} For simplicity, assume that there exists a state $s_i$ that is a zero-cost fixed point under the optimal policy; e.g. $\ell(s_i, \pi^*(s_i)) = 0$ and $f(s_i, \pi^*(s_i)) = s_i$.
The figure above shows an autonomous car moving at constant velocity $v>0$ on a straight road. Let $x$ be the (longitudinal) position of the car along the road, $y$ its (transversal) distance from the centerline, and $\theta$ the angle between the centerline and the direction of motion. The only control action is the steering velocity $u$, which is constrained in the interval $[u_{\text{min}}, u_{\text{max}}]$ (where $u_{\text{min}}<0$ and $u_{\text{max}}>0$). We describe the car dynamics with the simple kinematic model \begin{align*}\dot x &= v \cos \theta, \\ \dot y &= v \sin \theta, \\ \dot \theta &= u.\end{align*} Let $\bx = [x, y, \theta]^T$ be the state vector. To optimize the car trajectory we consider a quadratic objective function $$J = \int_{0}^{\infty} [\bx^T(t) \bQ \bx(t) + R u^2(t)] dt,$$ where $\bQ$ is a constant positive-semidefinite (hence symmetric) matrix and $R$ is a constant nonnegative scalar (note that $R=0$ is allowed here).
In this exercise we will see how seemingly simple cost functions can give surprising results. Consider the single-integrator system $\dot x = u$ with initial state $x(0)=0$. We would like to find the control signal $u(t)$ that minimizes the seemingly innocuous cost function $$J = \int_0^T x^2(t) + (u^2(t) - 1)^2 dt,$$ with $T$ finite. To this end, we consider a square-wave control parameterized by $\tau>0$: $$u_\tau(t) = \begin{cases} 1 &\text{if} & t \in [0, \tau) \cup [3 \tau, 5 \tau) \cup [7 \tau, 9 \tau) \cup \cdots \\ -1 &\text{if} & t \in [\tau, 3 \tau) \cup [5 \tau, 7 \tau) \cup [9 \tau, 11 \tau) \cup \cdots \end{cases}.$$
Consider the scalar control differential equation $$\dot{x} = x + u,$$ and the infinite horizon cost function $$J = \int_0^{\infty} [3x^2(t) + u^2(t)] dt.$$ As we will see in the chapter on linear-quadratic regulation, the optimal cost-to-go for a problem of this kind assumes the form $J^* = S x^2$. It is not hard to see that this, in turn, implies that the optimal controller has the form $u^* = - K x$.
In this exercise we analyze the performance of the value-iteration algorithm, considering its application to the minimum time problem for the double integrator. , you will find everything you need for this analysis. Take the time to go through the notebook and understand the code in it, then answer the following questions.
In this exercise we solve the dynamic programming using a linear program, considering its application to the grid world. , you will write a linear program for solving the optimal cost-to-go in the grid world.
In this excercise we will implement value iteration for a continuous time system with continuous actions and states. We will achieve by training a neural network to approximate our value function. You will work exclusively in to implement the algorithms described in the Continuous Dynamic Programming section.
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