A few weeks ago I got an email asking me about a game I worked on ages ago. I had to inform the sender that, unfortunately, my inspired dreams of creative development had not come to full fruition.

The message did get me thinking. What had I been doing back then, about 8 years ago? What problems did I run into? Why did I stop?

Now, I took a look at my old source code, and I’m pretty sure I stopped because of the mounting complexity of classes and cruft that happens in a growing codebase, especially when one has comparatively little idea of what one is doing. That, and I’m pretty sure this was around the time of my qualification exams, and well, I think that took over my life at the time.

Anyhow, I was thinking about this top-down 2D game, and the thoughts that kept circling around my brain centered around 2D character movement and collision. This was only magnified when I played some iPhone games to supplement my entertainment when I flew long distance over the holidays, and grew dismayed at some of the movement systems. I thought player movement was hard back then, and even today a fair number of games don’t really cut the mustard.

The Problem

So this post is borne out of mulling over this old problem of how to move a player around in a 2D game without them running into things. We want a few things:

• Nice continuous movement
• To not intersect static geometry
• To nicely slide along static geometry
• To not get hung up on edges, and nicely roll around them

And, as bonus objectives, we’d love to:

• Do all of these things efficiently
• Be able to handle non-tile maps with arbitrary geometry

The central issue with many games is that you have things you don’t want to run into. For example, below we have a screenshot from my old game-in-development Mystic Dave. We have a player character (Mystic Dave) and some objects (red) that we don’t want Dave to run into:

The player typically has a collision volume centered on their location. This collision volume is not allowed to intersect with any of the level collision geometry:

So what we’re really dealing with is moving this player collision volume around with the player, and making sure that it doesn’t intersect with anything.

Basic Player Movement

The player movement part is pretty simple. We can just look at whatever direction the player input is getting us to move in, and move ourselves in that direction:

void UpdatePlayer(Vec2 input_dir) {
   Vec2 pos_delta = input_dir * kPlayerStepSize;
   pos_player_ += pos_delta;
}

We’ll run into tuning problems later on if we don’t make this a function of the time step, so let’s go ahead and do that now:

void UpdatePlayer(Vec2 input_dir, float dt) {
   Vec2 pos_delta = input_dir * kPlayerStepSizePerTime * dt;
   pos_player_ += pos_delta;
}

This will give us some pretty basic movement:

That movement is a bit boring. It stutters around.

We can make it smoother by storing the player’s velocity and adding to that instead. We enforce a maximum velocity and apply some air friction to slow the player down as we stop our inputs:

void UpdatePlayer(Vec2 input_dir, float dt) {
   // Update the player's velocity
   vel_player_ += input_dir * kPlayerInputAccel * dt;
   
   // Clamp the velocity to a maximum magnitude
   float speed = Norm(vel_player_);
   if (speed > kPlayerMaxSpeed) {
       vel_player_ *= kPlayerMaxSpeed / norm;
   }

   // Update the player's position
   Vec2 pos_delta = vel_player_ * dt;
   pos_player_ += pos_delta;

   // Apply air friction
   vel_player_ *= kFrictionAir;
}

Alright, now we’re zooming around:

So how do we do the colliding part?

Naive Collision

The way I first approached player collision, way back in the day, was to say “hey, we have a top-down 2D tile-based game, so our collision objects are all squares, and we can just check for collision against squares”. A reasonable place to start.

Okay, so we have a rectangle for our player and consistently-sized squares for all of our static geometry. That can’t be too hard, right?

Checking for collisions between axis-aligned rectangles isn’t all that hard. You can use something like the separating axis theorem. A quick Google search gives you code to work with.

Our strategy now is to calculate the new player position, check it against our collision geometry for intersections, and to not move our player if that fails. If we do that, we find that we need (1) to zero out the player’s speed on collision, and (2) we still actually want to move the player, except we want to move them up against the colliding body:

Doing this part is harder. We can do it, but it takes some math. In the case of an axis-aligned rectangle, if we know which side our collision point is closest to, we can just shift our new position \(P’\) out of the closest face by half the player-rect’s width in that axis. The code looks something like:

void UpdatePlayer(Vec2 input_dir, float dt) {
   // Update the player's velocity, clamp, etc.
   ... same as before

   // Update the player's position
   Vec2 pos_delta = vel_player_ * dt;
   pos_player_ += pos_player_ + pos_delta;
   Box player_box_next = Box(pos_next, player_height, player_width);
   for (const Box& box : collision_objects_) {
      auto collision_info = CheckCollision(box, player_box_next)
      if (collision_info.collided) {
         // Collision!
         vel_player_ = Vec2(0,0);
         pos_player_ = box.center;
         if (collision_info.nearest_side == NORTH) {
            pos_player_ += Vec2(0, box.height/2 + player_height/2);
         } else if (collision_info.nearest_side == EAST) {
            pos_player_ += Vec2(box.width/2 + player_width/2, 0);
         } else if (collision_info.nearest_side == SOUTH) {
            pos_player_ -= Vec2(0, box.height/2 + player_height/2);
         } else if (collision_info.nearest_side == WEST) {
            pos_player_ -= Vec2(box.width/2 + player_width/2, 0);
         }
      }
   }

   // Apply air friction
   vel_player_ *= kFrictionAir;
}

So this works, but has some issues. First off, we’re iterating over all boxes. That’s a big inefficiency right off the bat, especially on a large level with a lot of filled tiles. That particular problem is pretty easy to avoid, typically by using a more efficient data structure for keeping track of our collision geometry, like a quad tree for general collision objects and searching around a simple tile-based index for something like Mystic Dave.

Second, we project out of the nearest face. This can lead to all sorts of glitches. If we start on one side of the box, but move such that our closest face is a different one, we get shunted out the wrong side:

This effect is particularly problematic when we have multiple collision bodies. The code above processes them all in series, which could be totally wrong. We want to collide with the first thing we’d hit first.

Alternatively we could use a while loop, and keep on checking for collisions after we get shunted around by our collision logic, but this could end up causing weird infinite loops:

After we process any collisions, we set the player’s speed to zero. This causes the player to just stop. There is no nice sliding along the surface, let alone nicely rolling around corners:

Thinking in Points

Okay, maybe we can do this differently. Let’s take a step back and refine our problem.

We have a point \(P\), which we are moving a small step to some \(P’\). We assume \(P\) is valid (the player is not colliding with anything).

We don’t want to move to \(P’\) if it collides with anything. Furthermore, we don’t want to move to \(P’\) if we collide with anything on the way there:

So far we’ve been talking about points (\(P\) and \(P’\)), but what complicates things are the bounding boxes around them. Can we do something to remove the boxes from the equation?

Let’s look at the set of points where \(P’\) collides with our collision volume:

We can change our strategy to simply test \(P’\), a single point, against this larger collision volume. By using it, we remove the complexity of the player’s bounding rectangle, and instead get to work directly with point-in-shape tests.

When checking for a collision, we just have to check whether the ray segment \(P\rightarrow P’\) intersects with the extended volumes, and if it does, only move the player up to the earliest intersection.

If we have multiple collision bodies, this approach will ensure that we stop at the right location:

So we are now working with extended volumes. These are obtained by using the Minkowski sum of a collision body with the player’s collision volume. The idea is quite general, and works for arbitrary polygons:

Computing the Minkowski sum of two polygons is actually pretty straightforward:

// Compute the Minkowski sum of two convex polygons P and Q, whose
// vertices are assumed to be in CCW order.
vector<Vec2i> MinkowskiSum(const vector<Vec2i>& P, const vector<Vec2i>& Q) {
    // Vertices (a,b) are a < b if a.y < b.y || (a.y == b.y && a.x < b.x)
    size_t i_lowest = GetLowestVertexIndex(P);
    size_t j_lowest = GetLowestVertexIndex(Q);

    std::vector<Vec2i> retval;

    size_t i = 0, j = 0;
    while (i < P.size() || j < Q.size()) {
        // Get p and q via cyclic coordinates
        // This uses mod to allow to indices to wrap around
        const Vec2i& p = GetCyclicVertex(P, i_lowest + i);
        const Vec2i& q = GetCyclicVertex(Q, j_lowest + j);
        const Vec2i& p_next = GetCyclicVertex(P, i_lowest + i + 1);
        const Vec2i& q_next = GetCyclicVertex(Q, j_lowest + j + 1);

        // Append p + q
        retval.push_back(p + q);

        // Use the cross product to gage polar angle.
        // This implementation works on integers to avoid floating point
        // precision issues. I just convert floats say, in meters, to
        // integers by multiplying by 1000 first so we maintain 3 digits
        // of precision.
        long long cross = Cross(p_next - p, q_next - q);

        if (cross >= 0 && i < P.size()) {
            i++;
        }
        if (cross <= 0 && j < Q.size()) {
            j++;
        }
    }

    return retval;
}

This post isn’t really about how the Minkowski sum works, so I won’t go too deep into it, but you basically store your polygon vertices in a counter clockwise order, start with a vertex on each polygon that is furthest in a particular direction, and then rotate that furthest direction around in a circle, always moving an index (or both if the directions match) to keep your indices matching that furthest direction.

Sliding and Rolling around Corners

Before we get too deep, let’s quickly cover how to slide along the collision face. Our code above was zeroing out the player’s velocity every time we hit a face. What we want to do instead is zero out the velocity into the face. That is, remove the component of the velocity that points into the face.

If we know the face unit normal vector \(u\), then we just project our velocity \(v\) into the direction perpendicular to \(u\).

Because running into things slows us down, we typically then multiply our resulting speed vector by a friction coefficient to mimic sliding friction.

// Lose all velocity into the face.
// This results in the vector along the face.
player_vel_ = VectorProjection(player_vel_, v_face);
player_vel_ *= player_friction_slide_;  // sliding friction

To roll around edges we can just trade out our rectangular player collision volume for something rounder. Ideally we’d use a circle, but a circle is not a polygon, so the Minkowski sum would result in a more awkward shape with circular segments. We can do that, but its hard.

We can instead approximate a circle with a polygon. I’m just going to use a diamond for now, which will give us some rolling-like behavior around box corners.

Thinking in Triangles

Now that we took that brief aside, let’s think again about what this Minkowski sum stuff means for the code we prototyped. We could maintain a list of these inflated collision objects and check our points against them every time step. That’s a bit better before, because we’re primarily doing point-in-polygon checks rather than polygon-polygon intersection, but it is still rather complicated, and we want to avoid having to repeatedly iterate over all objects in our level.

I had a big aha! moment that motivated this entire project, which elegantly avoids this list-of-objects problem and ties in nicely with a previous blog post.

What if we build a triangulation of our level that has edges everywhere the inflated collision polygons have edges, and we then mark all triangles inside said polygons as solid. Then all we have to do is keep track of the player’s position and which triangle they are currently in. Any time the player moves, we just check to see if they pass across any of the faces of their current triangle, and if they cross into a solid triangle, we instead move them up to the face.

TLDR: 2D player movement on a constrained Delaunay mesh with solid cells for Minkowksi-summed collision volumes.

To illustrate, here is the Mystic Dave screenshot with some inflated collision volumes:

And here is a constrained triangulation on that, with solid triangles filled in:

And here it all is in action:

We see the static collision geometry in red, the player’s collision volume in blue, and the inflated collision volumes in yellow. We render the constrained Delaunay triangulation, which has edges at the sides of all of the inflated collision volumes. For player movement we simply keep track of the player’s position, the player’s speed, and which triangle they are in (highlighted).

We don’t have to limit ourselves to tile-centered boxes for this. We don’t even have to limit ourselves to tile-aligned boxes:

Player movement now has to track both the player’s position and the triangle that the player is in. We can find the enclosing triangle during initialization, and can use the same techniques covered the Delaunay Triangulation post. I’m going to repeat that section here, since it is so relevant.

Suppose we start with some random face and extract its vertices A, B, and C. We can tell if a point P is inside ABC by checking whether P is to the left of the ray AB, to the left of the ray BC, and to the left of the ray CA:

Checking whether P is left of a ray AB is the same as asking whether the triangle ABP has a counter-clockwise ordering. Conveniently, the winding of ABP is given by the determinant of a matrix:

\[\det\left(\begin{bmatrix}A_x & A_y & 1 \\ B_x & B_y & 1 \\ P_x & P_y & 1\end{bmatrix}\right)\]

If this value is positive, then we have right-handed (counter-clockwise) winding. If it is negative, then we have left-handed (clockwise) winding. If it is zero, then our points are colinear.

If our current triangle has positive windings then it encloses our point. If any winding is negative, the we know we can cross over that edge to get closer to our point.

We simply iterate in this manner until we find our enclosing triangle. 

Every tick, when we update the player’s position, we run a similar procedure. We similarly are interested in whether the player’s new position has crossed any triangle edges.The basic concept is that, if the new position enters a new triangle, and the new triangle is solid, we need to prevent movement across the edge. If the new position enters a new triangle, and the new triangle is not solid, we can simply move the player and update the enclosing triangle.

Below we see the simplest case, where the move from P to P’ leads to a new triangle:

If we check the winding of CAP’ and find that it is negative, then we know that we have crossed AC.

There are a few corner cases that make simply accepting the new triangle the wrong thing to do. First off, a single move could traverse multiple triangles:

We need to make sure we iterate, and continue to check for edge transitions.

Second, we may cross multiple edges in the first triangle:

In the above example, the new point P’ is across both the line through BC and the line through CA. The order in which we choose to process these traversals matters, because some intermediate triangles may be solid.

The correct traversal order is to process them in the same order that the ray P -> P’ crosses them. In this case, we first cross CA:

Every time we cross a segment, we find the intersection with the edge and move our player up to that edge. If the new triangle is not solid, we can update our current triangle and iterate again, now with a segment from P” to P’.

If the new triangle is solid, we don’t update our triangle and we process a collision on the colliding face. This means removing all velocity into the face (perpendicular to it) and adding sliding friction along the face (parallel to it). We continue to iterate again, now with a reduced step in our new direction along the edge:

The code structure is thus a while loop that, in each iteration, checks for traversal across all three edges of the enclosing triangle, and retains the earliest collision (if any). If there is a collision, we move the player up to the crossed edge (P”) and potentially handle collision physics or update the enclosing triangle, and iterate again. If there is no collision, then we stay inside our current triangle and are done.

Finding the intersection point for a crossed segment (suppose it is AB) can be done as follows:

\[\begin{aligned}V & = B – A \\ W &= P – A \\ \alpha &= \frac{V \times W }{P’ \times V} \\ P^{\prime\prime} & = P + \alpha (P’ – P) \end{aligned}\]

The interpolant \(\alpha \in [0,1]\) tells us how far we’ve moved toward P’, so is a great way to track which edge we cross first. We first cross the edge with the lowest \(\alpha\).

The code for this is thus:

void UpdatePlayer(Vec2 input_dir, float dt) {
   // Update the player's velocity, clamp, etc.
   ... same as before

   // Update the player's position
   float dt_remaining = dt;
   while (dt_remaining > 0.0f) {
        Vec2f player_pos_delta = player_vel_ * dt_remaining;
        Vec2f player_pos_next = player_pos_ + player_pos_delta;

        // Exit if the step is too small, to avoid math problems.
        if (Norm(player_pos_delta) < 1e-4) {
            break;
        }

        // Check to see if we cross over any mesh edges
        QuarterEdge* qe_ab = qe_enclosing_triangle_->rot;
        QuarterEdge* qe_bc = qe_enclosing_triangle_->next->rot;
        QuarterEdge* qe_ca = qe_enclosing_triangle_->next->next->rot;

        // Get the vertices
        const auto& a = *(qe_ab->vertex);
        const auto& b = *(qe_bc->vertex);
        const auto& c = *(qe_ca->vertex);

        // The fraction of the distance we will move
        float interp = 1.0f; // (alpha)
        QuarterEdge* qe_new_triangle = nullptr;
        Vec2f v_face;

        if (GetRightHandedness(a, b, player_pos_next) < -1e-4) {
            // We would cross AB
            Vec2f v = b - a;
            Vec2f w = player_pos_ - a;
            float interp_ab = Cross(v, w) / Cross(player_pos_delta, v);
            if (interp_ab < interp) {
                interp = interp_ab;
                qe_new_triangle = qe_ab->rot;
                v_face = v;
            }
        }
        if (GetRightHandedness(b, c, player_pos_next) < -1e-4) {
            // We would cross BC
            Vec2f v = c - b;
            Vec2f w = player_pos_ - b;
            float interp_bc = Cross(v, w) / Cross(player_pos_delta, v);
            if (interp_bc < interp) {
                interp = interp_bc;
                qe_new_triangle = qe_bc->rot;
                v_face = v;
            }
        }
        if (GetRightHandedness(c, a, player_pos_next) < -1e-4) {
            // We would cross CA
            Vec2f v = a - c;
            Vec2f w = player_pos_ - c;
            float interp_ca = Cross(v, w) / Cross(player_pos_delta, v);
            if (interp_ca < interp) {
                interp = interp_ca;
                qe_new_triangle = qe_ca->rot;
                v_face = v;
            }
        }

        // Move the player
        // If we would have crossed any triangle, this merely moves
        // the player up to the edge instead
        player_pos_ += player_pos_delta * interp;
        dt_remaining *= (1.0f - interp);
        dt_remaining -= 1e-5;  // eps factor for numeric safety

        if (qe_new_triangle != nullptr) {
            // We would have crossed into a new triangle
            if (is_solid(qe_new_triangle)) {
                // The new triangle is solid, so do not change triangles
                // Process the collision
                player_vel_ = VectorProjection(player_vel_, v_face);
                player_vel_ *= player_friction_slide_; // sliding friction
            } else {
                // Accept the new triangle
                qe_enclosing_triangle_ = qe_new_triangle;
            }
        }
   }

   // Apply air friction
   vel_player_ *= kFrictionAir;
}

This code block isn’t drastically more complicated than what we had before. We get a lot for it – efficient collision physics against general static objects. There is a bit of repeated code, so you could definitely simplify this even further.

Conclusion

This post revisited the problem of 2D player collision against static geometry. We covered some basics of player movement, including integrating velocity over time, basic collision physics, and basic collision approaches based on geometric primitives. We used the Minkowski sum to inflate collision polygons, thereby changing polygon vs. polygon intersections into cheaper point vs. polygon intersections. We concluded by leveraging constrained Delaunay triangulations for our environment instead, which simplified the collision detection logic down to checking for when we cross enclosing triangle line segments.

Please note that I am not the first to come up with this. It turns out I independently discovered Nav Meshes. I since learned that they are used in motion planning for robotics and in path planning for games (notably, Starcraft II). I assume they are also used for collision physics in games, but I haven’t been able to verify that yet.

I implemented this demo in ROS 2 using C++ 20. This was a deviation from my previous posts, which are mostly created in Julia notebooks. This project had more game-like interactivity to it than I thought a Julia notebook could easily handle (I could be wrong) and I simply felt like I could get up and running faster with ROS 2.

The most complicated part to implement in this whole process turns out to be polygon unions. In order to get the constrained Delaunay triangulation to work, you want to union any overlapping collision polygons together. Polygon unions (and other Boolean operations between polygons, like clipping) is quite complicated. I tried to write my own, but ended up using Clipper2.

I hope you found this process interesting! I was a lot of fun to revisit.

Convex duality is the sort of thing where you see the symbolic derivation and it sort of makes sense, but then you forget the steps and a day or two later you just sort of have to trust yourself that the theorems and facts hold but forget where they really come from. I recently relearned convex duality from the great lectures of Stephen Boyd, and in so doing, also learned about a neat geometry interpretation that helps solidify things.

Working with Infinite Steps

Everything starts with an optimization problem in standard form:

\[\begin{matrix}\underset{\boldsymbol{x}}{\text{minimize}} & f(\boldsymbol{x}) \\ \text{subject to} & \boldsymbol{g}(\boldsymbol{x}) \leq \boldsymbol{0} \\ & \boldsymbol{h}(\boldsymbol{x}) = \boldsymbol{0} \end{matrix}\]

This problem has an objective \(f(\boldsymbol{x})\), some number of inequality constraints \(\boldsymbol{g}(\boldsymbol{x}) \leq \boldsymbol{0}\), and some number of equality constraints \(\boldsymbol{h}(\boldsymbol{x}) = \boldsymbol{0}\).

Solving an unconstrained optimization problem often involves sampling the objective and using its gradient, or several local values, to know which direction to head in to find a better design \(\boldsymbol{x}\). Solving a constrained optimization problem isn’t as straightforward – we cannot accept a design that violates any constraints.

We can, however, view our constrained optimization problem as an unconstrained optimization problem. We simply incorporate our constraints into our objective, setting them to infinity when the constraints are violated and setting them to zero when the constraints are met:

\[\underset{\boldsymbol{x}}{\text{minimize}} \enspace f(\boldsymbol{x}) + \sum_i \infty \cdot \left( g_i(\boldsymbol{x}) > 0\right) + \sum_j \infty \cdot \left( h_j(\boldsymbol{x}) \neq 0\right)\]

For example, if our problem is

\[\begin{matrix}\underset{x}{\text{minimize}} & x^2 \\ \text{subject to} & x \geq 1 \end{matrix}\]

which looks like this

then we can rewrite it as

\[\underset{x}{\text{minimize}} \enspace x^2+\infty \cdot \left( x < 1\right)\]

which looks like

We’ve simply raised the value of infeasible points to infinity. Problem solved?

By doing this, we made our optimization problem much harder to solve with local methods. The gradient of designs in the infeasible region is zero, so we cannot use those to know which way to progress toward feasibility. It is also discontinuous. Furthermore, infinity is rather difficult to work with – it just produces more infinities when added or multiplied to most numbers. This new optimization problem is perhaps simpler, but we’ve lost a lot of information.

What we can do instead is reinterpret our objective slightly. What if, instead of using this infinite step, we simply multiplied our inequality function \(g(x) = 1 – x\) by a scalar \(\lambda\)?

\[\underset{x}{\text{minimize}} \enspace x^2+ \lambda \left( 1 – x \right)\]

If we have a feasible \(x\), then \(\lambda = 0\) gives us the original objective function. If we have an infeasible \(x\), then \(\lambda = \infty\) gives us the penalized objective function.

We’re taking the objective and constraint, which are separate functions:

and combining them into a single objective based on \(\lambda\):

Let’s call our modified objective function the Lagrangian:

\[\mathcal{L}(x, \lambda) = x^2+ \lambda \left( 1 – x \right)\]

If we fix \(\lambda\), we can minimize our objective over \(x\). I am going to call these dual values:

We find that all of these dual values are below the optimal value of the original problem:

This in itself is a big deal. We can produce a lower bound on the optimal value simply by minimizing the Lagrangian. Why is this?

Well, it turns out that we are guaranteed to get a lower bound any time that \(\lambda \geq 0\). This is because a positive \(\lambda\) ensures that our penalized objective is at or below the original objective for all feasible points. Minimizing is thus guaranteed to give us a design at or below the original value.

This is such a big deal, that we’ll call this minimization problem the Lagrange dual function, and give it a shorthand:

\[\mathcal{D}(\lambda) = \underset{x}{\text{minimize}} \enspace \mathcal{L}(x, \lambda)\]

which for our problem is

\[\mathcal{D}(\lambda) = \underset{x}{\text{minimize}} \enspace x^2 + \lambda \left( 1 – x \right)\]

Okay, so we have a way to get a lower bound. How about we get the *best* lower bound? That gives rise to the Lagrange dual problem:

\[\underset{\lambda \geq 0}{\text{maximize}} \enspace \mathcal{D}(\lambda)\]

For our example problem, the best lower bound is given by \(\lambda = 2\), which produces \(x = 1\), \(f(x) = 1\). This is the solution to the original problem.

Having access to a lower bound can be incredibly useful. It isn’t guaranteed to be useful (the lower bound can be \(-\infty\)), but it can also be relatively close to optimal. For convex problems, the bound is tight – the solution to the Lagrange dual problem has the same value as the solution to the original (primal) problem.

A Geometric Interpretation

We’ve been working with a lot of symbols. We have some plots, but I’d love to build further intuition with a geometric interpretation. This interpretation also comes from Boyd’s lectures, and textbook.

Let’s work with a slightly beefier optimization problem with a single inequality constraint:

\[\begin{matrix}\underset{\boldsymbol{x}}{\text{minimize}} & \|\boldsymbol{A} \boldsymbol{x} – \boldsymbol{b} \|_2^2 \\ \text{subject to} & x_1 \geq c \end{matrix}\]

In this case, I’ll use

\[\boldsymbol{A} = \begin{bmatrix} 1 & 0.3 \\ 0.3 & 2\end{bmatrix} \qquad \boldsymbol{b} = \begin{bmatrix} 0.5 \\ 1.5\end{bmatrix} \qquad c = 1.5\]

Let’s plot the objective, show the feasible set in blue, and show the optimal solution:

Our Lagrangian is again comprised of two terms – the objective function \( \|\boldsymbol{A} \boldsymbol{x} – \boldsymbol{b} \|_2^2\) and the constraint \(\lambda (c – x_1)\). We can build a neat geometric interpretation of convex duality by plotting (constraint, objective) pairs for various values of \(\boldsymbol{x}\):

We get a parabolic region. This isn’t a huge surprise, given that we have a quadratic objective function.

Note that the horizontal axis is the constraint value. We only have feasibility when \(g(\boldsymbol{x}) \leq 0\), so our feasible designs are those left of the vertical line.

I’ve already plotted the optimal value – it is at the bottom of the feasible region. (Lower objective function values are better, and the bottom has the lowest value).

Let’s see if we can include our Lagrangian. The Lagrangian is:

\[\mathcal{L}(\boldsymbol{x}, \lambda) = f(\boldsymbol{x}) + \lambda g(\boldsymbol{x})\]

If we fix \(\boldsymbol{x}\), it is linear!

The Lagrange dual function is

\[\mathcal{D}(\lambda) = \underset{\boldsymbol{x}}{\text{minimize}} \enspace f(\boldsymbol{x}) + \lambda g(\boldsymbol{x})\]

If we evaluate \(\mathcal{D}\) for a particular \(\lambda\), we’ll minimize and get a design \(\hat{\boldsymbol{x}}\). The value of \(\mathcal{D}(\lambda)\) will be \(f(\hat{\boldsymbol{x}}) + \lambda g(\hat{\boldsymbol{x}})\). Remember, for \(\lambda \geq 0\) this is a lower bound. Said lower bound is the y-intercept of the line with slope \(-\lambda\) that passed through our (constraint, objective) at \(\hat{\boldsymbol{x}}\).

For example, with \(\lambda = 0.75\), we get:

We got a lower bound, and it is indeed lower than the true optimum. Furthermore, that line is a supporting hyperplane of our feasible region.

Let’s try it again with \(\lambda = 3\):

Our \(\lambda\) is still non-negative, so we get a valid lower bound. It is below our optimal value.

Now, in solving the Lagrange dual problem we get the best lower bound. This best lower bound just so happens to correspond to the \(\lambda\) that produces a supporting hyperplane at our optimum (\(\lambda \approx 2.15\)):

Our line is tangent at our optimum. This ties in nicely with the KKT conditions (out of scope for this blog post), which require that the gradient of the Lagrangian be zero at the optimum (for a problem with a differentiable objective and constraints), and thus the gradients of the objective and the constraints (weighted by \(\lambda\)) balance out.

Notice that in this case, the inequality constraint was active, with our optimum at the constraint boundary with \(g(\boldsymbol{x}) = 0\). We needed a positive \(\lambda\) to balance against the objective function gradient.

If we change our problem such that the unconstrained optimum is feasible, then we do not need a positive dual value to balance against the objective function gradient. Let’s adjust our constraint to max our solution feasible:

We now get an objective / constraint region with a minimum inside the feasible area:

Our best lower bound is obtained with \(\lambda = 0\):

This demonstrates complementary slackness, the idea that \(\lambda_i g_i(\boldsymbol{x})\) is always zero. If constraint \(i\) is active, \(g_i(\boldsymbol{x}) = 0\);\ and if constraint \(i\) is inactive, \(\lambda_i = 0\).

So with this geometric interpretation, we can think of choosing \(\lambda\) as picking the slope of a line that supports this multidimensional volume from below and to the left. We want the line with the highest y-intercept.

Multiple Constraints (and Equality Constraints)

I started this post off with a problem with multiple inequality and equality constraints. In order to be thorough, I just wanted to mention that the Lagrangian and related concepts extend the way you’d expect:

\[\mathcal{L}(\boldsymbol{x}, \boldsymbol{\lambda}, \boldsymbol{\nu}) = f(\boldsymbol{x}) + \sum_i \lambda_i g_i(\boldsymbol{x}) + \sum_j \nu_j h_j(\boldsymbol{x})\]

and

\[\mathcal{D}(\boldsymbol{\lambda}, \boldsymbol{\nu}) = \underset{\boldsymbol{x}}{\text{maximize}} \enspace \mathcal{L}(\boldsymbol{x}, \boldsymbol{\lambda}, \boldsymbol{\nu})\]

We require that all \(\lambda\) values be nonnegative, but do not need to restrict the \(\nu\) values for the equality constraints. As such, our Lagrangian dual problem is:

\[\underset{\boldsymbol{\lambda} \geq \boldsymbol{0}, \>\boldsymbol{\nu}}{\text{maximize}} \enspace \mathcal{D}(\boldsymbol{\lambda}, \boldsymbol{\nu})\]

The geometric interpretation still holds – it just has more dimensions. We’re just finding supporting hyperplanes.

I recently started working on new chapters for Algorithms for Optimization, starting with a new chapter on quadratic programs. We already have a chapter on linear programs, but quadratic programs are sufficiently different and are ubiquitous enough that giving them their own chapter felt warranted.

The chapter starts with a general formulation for quadratic programs and transforms the problem a few times into simpler and simpler formulations, until we get a nonnegative least squares problem. This can be solved, and then we can use that solution to back our way up the transform stack back to a solution of our original solution.

In texts, these problem transforms are typically solely the realm of mathematical manipulations. While math is of course necessary, I also tried hard to depict what these various problem representations looked like to help build an intuitive understanding of what some of the transformations were. In particular, there was one figure where I wanted to show how the equality constraints of a general-form quadratic program could be removed when formulating a lower-dimensional least squares program:

This post is about this figure, and more specifically, about the blue contour lines on the left side.

What is this Figure?

The figure on left shows the following quadratic program:

As we can see, the quadratic program is constrained to lie within a square planar region \(0 \leq x \leq 1\), \(0 \leq x_3 \leq 1\), and \(x_2 = x_3\). The feasible set is shown in blue.

The right figure on the right shows the least-squares program that we get when we factor out the equality constraint. No need to get into the details here, it isn’t really important for this post, but we can use our one equality equation to remove one variable in our original problem, producing a 2d problem without any equality constraints:

The original feasible set is rotated and stretched, and the objective function contours are rotated and stretched, but other than that we have a similar topology and a pretty standard problem that we can solve.

How do we Make the Figures?

I make the plots with PGFPlots.jl, which uses the pgfplots LaTeX package under the hood. It is a nice synergy of Julia coding and LaTeX within the LaTeX textbook.

Right Plot

The right plot is pretty straight forward. It has a contour plot for the objective function:

Plots.Contour(x -> norm(A_lsi*x - b_lsi, 2),
              xdomain, ydomain, xbins=600, ybins=600)

some over + under line plots and a fill-between for the feasible region:

x1_domain_feasible = (0.0, (0.5+√2)/2)
push!(plots,
      Plots.Linear(x1->max(-x1, x1 - sqrt(2)), x1_domain_feasible,
                   style="name path=A, draw=none, mark=none"))
push!(plots,
      Plots.Linear(x1->min(x1, 1/2-x1), x1_domain_feasible,
                   style="name path=B, draw=none, mark=none"))
push!(plots,
      Plots.Command("\\addplot[pastelBlue!50] fill between[of=A and B]"))

and a single-point scatter plot plus a label for the solution:

push!(plots,
      Plots.Scatter([x_val[1]], [x_val[2]],
      style="mark=*, mark size=1.25, mark options={solid, draw=black, fill=black}"))
push!(plots,
      Plots.Command("\\node at (axis cs:$(x_val[1]),$(x_val[2])) [anchor=west] {\\small \\textcolor{black}{\$y_2^\\star\$}}"))

These plots get added to an Axis and we’ve got our plot.

Left Plot

The left plot is in 3D. If we do away with the blue contour lines, its actually even simpler than the right plot. We simply have a square patch, a single-point scatter plot, and our label:

Axis([
   Plots.Command("\\addplot3[patch,patch type=rectangle, faceted color=none, color=pastelBlue!40] coordinates {(0,0,0) (1,0,0) (1,1,1) (0,1,1)}"),
   Plots.Linear3([x_val[1]], [x_val[2]], [x_val[3]], mark="*", style="mark size=1.1, only marks, mark options={draw=black, fill=black}"),
   Plots.Command("\\node at (axis cs:$(x_val[1]),$(x_val[2]),$(x_val[3])) [anchor=west] {\\small \\textcolor{black}{\$x^\\star\$}}"),
  ], 
  xlabel=L"x_1", ylabel=L"x_2", zlabel=L"x_3", width="1.2*$width", 
  style="view={45}{20}, axis equal",
  legendPos="outer north east"
)

The question is – how do we create that contour plot, and how do we get it to lie on that blue patch?

How do you Draw Contours?

Contour plots don’t seem difficult to plot, until you have to sit down and plot them. They are just lines of constant elevation – in this cases places where \(\|Ax – b\| = c\) for various constants \(c\). But, unlike line plots, where you can just increase \(x\) left to right and plot successive \((x, f(x))\) pairs, for contour plots you don’t know where your line is going to be ahead of time. And that’s assuming you only have one line per contour value – you might have multiple:

So, uh, how do we do this?

The first thing I tried was trying to write my function out the traditional way – as \((x_1, f(x_1))\) pairs. Unfortunately, this doesn’t work out, because the contours aren’t 1:1.

The second thing I briefly considered was a search method. If I fix my starting point, say, \(x_1 = 0, x_2 = 0.4\), I can get the contour level \(c = \|Ax – b\| \). Then, I could search toward the right in an arc to find where the next contour line was, maybe bisect over an arc one step distance away and proceed that way:

I think this approach would have worked for this figure, since there were never multiple contour lines for the same contour value, but it would have been more finicky, especially around the boundaries, and it would not have generalized to other figures.

Marching Squares

I happened upon a talk by Casey Muratori on YouTube, “Papers I have Loved“, a few weeks ago. One of the papers Casey talked about was Marching Cubes, A High Resolution 3D Surface Construction Algorithm. This 1987 paper by W. Lorenson and H. Cline presented a very simply way to generate surfaces (in 3D or 2D) in places of constant value: \(f(x) = c\).

Casey talked about it in the context of rendering metaballs, but when I found myself presented with the contour problem, it was clear that marching cubes (or squares, rather, since its 2d) was relevant here too.

The algorithm is a simple idea elegantly applied.

The contour plot we want to make is on a 2d surface in a 3d plot, but let’s just consider the 2d surface for now. Below we show that feasible square along with our objective function, \(f(x) = \|Ax – b\|\):

Our objective is to produce contour lines, the same sort as are produced when we make a PGFPlots contour plot:

We already know that there might be multiple connected contour lines for a given level, so a local search method won’t be sufficient. Instead, marching squares uses a global approach combined with local refinement.

We begin by evaluating a grid of points over our plot:

We’ll start with a small grid, but for higher-quality images we’d make it finer.

Let’s focus on the contour line \(f(x) = 3\). We now determine, for each vertex in our mesh, whether its value is above or below our contour target of 3. (I’ll plot the contour line too for good measure):

Marching squares will now tile the image based on each 4-vertex square, and the binary values we just evaluated. Any squares with all vertices equal can simply be ignored – the contour line doesn’t pass through them. This leaves the following tiles:

Now, each remaining tile comes in multiple possible “flavors” – the various ways that its vertices can be above or below the objective value. There are \(2^4\) possible tiles, two of which we’ve already discarded, leaving 14 options. The Wikipedia article has a great set of images showing all of the tile combinations, but here we will concern ourselves with two more kinds: tiles with one vertex below or one vertex above, and tiles with two neighboring vertices above and two neighboring vertices below:

Let’s start with the first type. These squares have two edges where the function value changes. We know, since we’re assuming our objective function is continuous, that our contour line must pass through those edges. As such, if we run a bisection search along each of those edges, we can quickly identify those crossing points:

We then get our contour plot, for this particular contour value, by drawing all of the line segments between these crossing points:

We can get higher resolution contour lines by using a finer grid:

To get our full contour plot, we simply run marching squares on multiple contour levels. To get the original plot, I run this algorithm to get all of the little tile line segments, then I render them as separate Linear3 line segments on top of the 3d patch. Pretty cool!

What does the Code Look Like?

The code for marching squares is pretty elegant.

We start off with defining our grid:

nbins = 41
x1s = collect(range(0.0, stop=1.0, length=nbins))
x2s = collect(range(0.0, stop=1.0, length=nbins))
is_above = falses(length(x1s), length(x2s))

Next we define a bisection method for quickly interpolating between two vertices:

function bisect(
    u_target::Float64, x1_lo::Float64, x1_hi::Float64, 
    x2_lo::Float64, x2_hi::Float64, k_max::Int=8)
    f_lo = norm(A*[x1_lo, x2_lo, x2_lo]-b)
    f_hi = norm(A*[x1_hi, x2_hi, x2_hi]-b)
 
    if f_lo > f_hi # need to swap low and high
        x1_lo, x1_hi, x2_lo, x2_hi = x1_hi, x1_lo, x2_hi, x2_lo
    end
 
    x1_mid = (x1_hi + x1_lo)/2
    x2_mid = (x2_hi + x2_lo)/2
    for k in 1 : k_max
        f_mid = norm(A*[x1_mid, x2_mid, x2_mid] - b)
        if f_mid > u_target
            x1_hi, x2_hi, f_hi = x1_mid, x2_mid, f_mid
        else
            x1_lo, x2_lo, f_lo = x1_mid, x2_mid, f_mid
        end
        x1_mid = (x1_hi + x1_lo)/2
        x2_mid = (x2_hi + x2_lo)/2
    end
 
    return (x1_mid, x2_mid)
end

In this case I’ve baked in the objective function, but you could of course generalize this and pass one in.

We then iterate over our contour levels. For each contour level (u_target), we evaluate the grid and then evaluate all of our square patches and add little segment plots as necessary:

for (i,x1) in enumerate(x1s)
    for (j,x2) in enumerate(x2s)
        is_above[i,j] = norm(A*[x1, x2, x2] - b) > u_target
    end
end
for i in 2:length(x1s)
    for j in 2:length(x2s)
        # bottom-right, bottom-left, top-right, top-left
        bitmask = 0x00
        bitmask |= (is_above[i,j-1] << 3)
        bitmask |= (is_above[i-1,j-1] << 2)
        bitmask |= (is_above[i,j] << 1)
        bitmask |= (is_above[i-1,j])
        plot = true
        x1_lo1 = x1_hi1 = x2_lo1 = x2_hi1 = 0.0
        x1_lo2 = x1_hi2 = x2_lo2 = x2_hi2 = 0.0
        if bitmask == 0b1100 || bitmask == 0b0011
            # horizontal cut
            x1_lo1, x1_hi1, x2_lo1, x2_hi1 = x1s[i-1], x1s[i-1], x2s[j-1], x2s[j]
            x1_lo2, x1_hi2, x2_lo2, x2_hi2 = x1s[i], x1s[i], x2s[j-1], x2s[j]
        elseif bitmask == 0b0101 || bitmask == 0b1010
            # vertical cut
            x1_lo1, x1_hi1, x2_lo1, x2_hi1 = x1s[i-1], x1s[i], x2s[j-1], x2s[j-1]
            x1_lo2, x1_hi2, x2_lo2, x2_hi2 = x1s[i-1], x1s[i], x2s[j], x2s[j]
        elseif bitmask == 0b1000 || bitmask == 0b0111
            # cut across bottom-right
            x1_lo1, x1_hi1, x2_lo1, x2_hi1 = x1s[i-1], x1s[i], x2s[j-1], x2s[j-1]
            x1_lo2, x1_hi2, x2_lo2, x2_hi2 = x1s[i], x1s[i], x2s[j-1], x2s[j]
        elseif bitmask == 0b0100 || bitmask == 0b1011
            # cut across bottom-left
            x1_lo1, x1_hi1, x2_lo1, x2_hi1 = x1s[i-1], x1s[i], x2s[j-1], x2s[j-1]
            x1_lo2, x1_hi2, x2_lo2, x2_hi2 = x1s[i-1], x1s[i-1], x2s[j-1], x2s[j]
        elseif bitmask == 0b0010 || bitmask == 0b1101
            # cut across top-right
            x1_lo1, x1_hi1, x2_lo1, x2_hi1 = x1s[i-1], x1s[i], x2s[j], x2s[j]
            x1_lo2, x1_hi2, x2_lo2, x2_hi2 = x1s[i], x1s[i], x2s[j-1], x2s[j]
        elseif bitmask == 0b0001 || bitmask == 0b1110
            # cut across top-left
            x1_lo1, x1_hi1, x2_lo1, x2_hi1 = x1s[i-1], x1s[i], x2s[j], x2s[j]
            x1_lo2, x1_hi2, x2_lo2, x2_hi2 = x1s[i-1], x1s[i-1], x2s[j-1], x2s[j]
        else
            plot = false
        end
        if plot
            x1a, x2a = bisect(u_target, x1_lo1, x1_hi1, x2_lo1, x2_hi1)
            x1b, x2b = bisect(u_target, x1_lo2, x1_hi2, x2_lo2, x2_hi2)
            push!(contour_segments, 
                Plots.Linear3([x1a, x1b], [x2a, x2b], [x2a, x2b],
                               style="solid, pastelBlue, mark=none")
            )
        end
    end
end

In this code, all we ever do is figure out which two line segments to bisect over, and if we get then, run the bisection and plot the ensuing line. Its actually quite neat and compact.

Conclusion

So there you have it – a home-rolled implementation for contour plotting turns out to be useful for embedding 2d contours in a 3d plot.

Note that the implementation here wasn’t truly generalWhich is fine - it was sufficient to generate the figure. As long as we are aware of where we cut corners, we are okay.. In this case, we didn’t handle some tile types, namely those where opposing corners have the same aboveness, but neighboring corners have differing aboveness. In those cases, we need to bisect across all four segments, and we typically need to remove ambiguity by sampling in the tile’s center to determine which way the segments cross.

I find it really satisfying to get to apply little bits of knowledge like this from time to time. Its part of computer programming that is endlessly rewarding. I have often been amazed by how a technique that I learned about in one context surprised me by being relevant in something else later on.

I recently read a fantastic blog post about Delaunay triangulations and quad-edge data structures, and was inspired to implement the algorithm myself. The post did a very good job of breaking the problem and the algorithm down, and had some really nice interactive elements that made everything make sense. I had tried to grok the quad-edge data structure before, but this was the first time where it really made intuitive sense.

This post covers more or less the same content, in my own words and with my own visualizations. I highly recommend reading the original post.

What is a Delaunay Triangulation

A triangulation is any breakdown of a set of vertices into a set of non-overlapping triangles. Some triangulations are better than others, and the Delaunay triangulation is a triangulation that makes each triangle as equilateral as possible. As a result, it is the triangulation that keeps all of the points in a given triangle as close to each other as possible, which ends up making it quite useful for a lot of spacial operations, such as finite element meshes.

As we can see above, there are a lot of bad triangulations with a lot of skinny triangles. Delaunay triangulations avoid that as much as possible.

One of the neatest applications of these triangulations is in motion planning. In constrained Delaunay triangulation, you for all walls in your map to be edges in your triangulation, and otherwise construct a normal Delaunay triangulation. Then, your robot can pass through any edge of sufficient length (i.e., if its a wide enough gap), and you can quickly find shortest paths through complicated geometry while avoiding regions you shouldn’t enter:

How to Construct a Delaunay Triangulation

So we want a triangulation that is as equilateral as possible. How, technically, is this achieved?

More precisely, a triangulation is Delaunay if no vertex is inside the circumcircle of any of the triangles. (A circumcircle of a triangle is the circle that passes through its three vertices).

One way to construct a Delaunay triangulation for a set of points is construct one iteratively. The Guibas & Stolfi incremental triangulation algorithm starts with a valid triangulation and repeatedly adds a new point, joining it into the triangulation and then changing edges to ensure that the mesh continues to be a Delaunay triangulation:

Here is the same sequence, with some details skipped and more points added:

This algorithm starts with a bounding triangle that is large enough to contain all future points. The triangle is standalone, and so by definition is Delaunay:

Every iteration of the Guibas & Stolfi algorithm inserts a new point into our mesh. We identify the triangle that new point is inside, and connect edges to the new point:

Note that in the special case where our new vertex is on an existing edge (or sufficiently close to an existing edge) we remove that edge and introduce four edges instead:

By introducing these new edges we have created three (or four) new triangles. These new triangles may no longer fulfill the Delaunay requirement of having vertices that are within some other triangle’s circumcircle.

Guibas & Stolfi noticed that the only possible violations occur between the new point and those opposite its surrounding face. That is, we need only walk around the edges of the triangle (or quadrilateral) that we inserted it into, and check whether we should flip those border edges in order to preserve the Delaunay condition.

For example, for the new vertex above, we only need to check the four surrounding edges:

Here is an example where we need to flip an edge:

Interestingly, flipping an edge simply adds a new spoke to our new vertex. As such, we continue to only need to check the perimeter edges around the hub. This guarantees that we won’t propagate out and have to spend a lot of time checking edges, but instead just iterate around our one new point.

That’s one iteration of the algorithm. After that you just start the next iteration. Once you’re done adding points you remove the bounding triangle and you have your Delaunay triangulation.

The Data Structure

So the concept is pretty simple, but we all know that the devil is in the details. What really excited me about the Delaunay triangulation was the data structure used to construct it.

If we look at the algorithm, we can see what we might need our data structure to support:

• Find the triangle enclosing a new point
• Add new edges
• Find vertices across from edges
• Walk around the perimeter of a face

Okay, so we’ve got two concepts here – the ability to think about edges in our mesh, but also the ability to reason about faces in our mesh. A face, in this case, is usually a triangle, but it could also sometimes be a quadrilateral, or something larger.

The naive approach would be to represent our triangulation (i.e. mesh) as a a graph. That is, a list of vertices and a way to represent edge adjacency. Unfortunately, this causes problems. The edges would need to be ordered somehow to allow us to navigate the mesh.

The data structure of choice is called a quad-edge tree. It gives us the ability to quickly navigate both clockwise or counter-clockwise around vertices, or clockwise or counter-clockwise around faces. At the same time, quad-edge trees make it easy for new geometry to be stitched in or existing geometry to be removed.

Many of the drawings and explanations surrounding quad-edge trees are a bit complicated and hard to follow. I’ll do my best to make this explanation as tractable as possible.

A quad-edge tree for 2D data represents a vertex mesh on a sphere:

Said sphere is mostly irrelevant for what we are doing. In fact, we can think of it as very large. However, there is one important consequence of being on a sphere. Now, every edge between two vertices is between two faces. For example, the simple 2D mesh above has two faces:

The quad-edge data structure can leverage this by associating each edge both with its two vertices and with its two faces, hence the name. Each real edge is represented as four directional edges in the quad-edge data structure: two directional edges between vertices and two directional edges between faces.

We can thus think of each face as being its own kind of special vertex:

As such, the quad-edge data structure actually contains two graphs – the directed graph of mesh edges and the dual graph that connects mesh faces. Plus, both of these graphs have inter-connectivity information in that each directed edge knows its left and right neighborsAs we'll see later, it will be enough to just store the counter-clockwise neighbor..

Okay, so pretty abstract. But, this gives us what we want. We can easily traverse the mesh by either flipping between the directed edges for a given mesh edge, or rotating around the base vertices for a given directed edge:

So now we are thinking about a special directed graph where each directed edge knows its left and right-hand neighbors and the other three directed edges that make up the edge in the original mesh:

struct QuarterEdge
    data::VertexIndex # null if a dual quarter-edge
    next_in_same_edge_on_right::QuarterEdgeIndex
    next_in_same_edge_on_left::QuarterEdgeIndex
    next_with_same_base_on_right::QuarterEdgeIndex
    next_with_same_base_on_left::QuarterEdgeIndex
end
struct Mesh
    vertices::Vector{Point}
    edges::Vector{QuarterEdge}
end

struct QuarterEdge
    data::VertexIndex # null if a dual quarter-edge
    rot::QuarterEdgeIndex # next in same edge on right
    next::QuarterEdgeIndex # next with same base on right
end

We can get the quarter-edge one rotation to our left within the same edge simply by rotating right three times. Similarly, we can get the quater-edge one rotation to our left with the same base vertex by following rot – next – rot:

Now we have our mesh data structure. When we add or remove edges, we have to be careful (1) to always construct new edges using four quarter edges whose rot entries point to one another, and (2) to update the necessary next entries both in the new edge and in the surrounding mesh. There are some mental hurdles here, but it isn’t too bad.

Finding the Enclosing Triangle

The first step of every Delaunay triangulation iteration involves finding the triangle enclosing the new vertex. We can identify a triangle using a quarter edge q whose base is associated with that face. If we do that, its three vertices, in counter-clockwise order, are:

• A = q.rot.data
• B = q.next.rot.data
• C = q.next.next.rot.data

(Note that in code this looks a little different, because in Julia we would be indexing into the mesh arrays rather than following pointers like you would in C, but you get the idea).

Suppose we start with some random face and extract its vertices A, B, and C. We can tell if our new point P is inside ABC by checking whether P is to the left of the ray AB, to the left of the ray BC, and to the left of the ray CA:

Checking whether P is left of a ray AB is the same as asking whether the triangle ABP has a counter-clockwise ordering. Conveniently, the winding of ABP is given by the determinant of a matrix:

\[\text{det}\left(\begin{bmatrix} A_x & A_y & 1 \\ B_x & B_y & 1 \\ P_x & P_y & 1 \end{bmatrix} \right)\]

If this value is positive, then we have right-handed (counter-clockwise) winding. If it is negative, then we have left-handed (clockwise) winding. If it is zero, then our points are colinear.

If our current face has positive windings then it encloses our point. If any winding is negative, the we know we can cross over that edge to get closer to our point.

We simply iterate in this manner until we find our enclosing face. (I find this delightfully clean.)

Is a Point inside a Circumcircle

Checking whether vertices are inside circumcircles of triangles is central to the Delaunay triangulation process. The inspiring blog post has great coverage of why the following technique works, but if we want to check whether a point P is inside the circumcircle that passes through the (counter-clockwise-oriented) points of triangle ABC, then we need only compute:

\[\text{det}\left(\begin{bmatrix} A_x & A_y & A_x^2 + A_y^2 & 1 \\ B_x & B_y & B_x^2 + B_y^2 & 1 \\ C_x & C_y & C_x^2 + C_y^2 & 1 \\ P_x & P_y & P_x^2 + P_y^2 & 1 \end{bmatrix}\right)\]

If this value is positive, then P is inside the circumcircle. If it is negative, then P is external. If it is zero, then P is on its perimeter.

Some Last Implementation Details

I think that more or less captures the basic concepts of the Delaunay triangulation. There are a few more minor details to pay attention to.

First off, it is possible that our new point is exactly on an existing edge, rather than enclosed inside a triangle. When this happens, or when the new point is sufficiently close to an edge, it is often best to remove that edge and add four new edges. I illustrate this earlier in the post. That being said, make sure to never remove / split an original bounding edge. It is, after all, a bounding triangle.

Second, after we introduce the new edges we have to consider flipping edges around the perimeter of the enclosing face. To do this, we simply rotate around until we arrive back where we started. The special cases here are bounding edges, which we never flip, and edges that connect to our bounding vertices, which we don’t flip if it would produce new triangles with clockwise windings:

Once you’re done adding vertices, you remove the bounding triangle and any edges connected to it.

And all that is left as far as implementation details are concerned is the implementation itself! I put this together based on that other blog post as a quick side project, so it isn’t particularly polished. (Your mileage may vary). That being said, its well-visualized and I generally find reference implementations to be nice to have.

You can find the code here.

Conclusion

I thought the quad-edge data structure was really interesting, and I bet with some clever coding they could be really efficient as well. I really liked how solving this problem required geometry and visualization on top of typical math and logic skills.

Delaunay triangulations have a lot of neat applications. A future step might be to figure out constrained Delaunay triangulations for path-finding, or to leverage the dual graph to generate Voronoi diagrams.

Hard copies of Algorithms for Decision Making (available in PDF form here for free) are shipping on August 16th! This book, the second one I have been fortunate to help write, was a large undertaking that spanned 4.5 years.

Back when Algorithms for Optimization came out, I wrote a blog post that detailed aspects of the special nature of our development setup. This blog post, 4 years later, is a look at how those aspects have changed for the second book.

But first, what is Algorithms for Decision Making, and how is it different than Algorithms for Optimization?

What Alg4DM is About

Algorithms for Optimization covers the sort of decision making theory that one thinks of when considering modern artificial intelligence. Think Go AIs, aircraft collision avoidance, and using drones for wildfire suppression.

The book’s central focus is on the Markov Decision Process, or MDP. The MDP framework and its variants are simply general ways to frame decision making problems. For a problem to be represented as an MDP, you have to specify the problem’s states, its actions, how actions take you to successor states, and what rewards (or costs) you incur when doing so. Simple. The game 2048 can be formulated as an MDP, but so too can the problem of balancing investment portfolios.

Without going into too much detail (there’s an entire book, after all), the core insight of the MDP framework is that problems can be solved with dynamic programming according to the Bellman equation. All sorts of techniques stem from this, from optimal solvers for small discrete problems to fancy reinforcement learning techniques for large deep neural network policies.

The book has the following progression:

• Starts with the basics of probability and utility, and introduces Bayesian networks and how to train them. (This is because BNs were used by Mykel for ACAS-X, and are a great way to represent transition models).
• MDPs, including all sorts of fancy things including policy gradient optimization and actor-critic methods.
• Model uncertainty, where we do not know the transition and reward functions ahead of time. There is some great content on imitation learning here too.
• POMDPs, which extend MDPs to domains where we don’t observe the full state.
• Multiagent systems, which involve problems with multiple decision-making agents. Think a bunch of robots jointly solving a problem together.

How Alg4DM is Different

Alg4DM took about 4.5 years between first commits to hard-copy. There were 3,417 commits to our git repository along with 826 filed issues. The printed book is 700 pages, with 225 color figures. In comparison, Alg4Opt is 500 pages, which makes the new book 40% longer.

While sheer size is a significant factor, there were other contributing complexities that made this book more challenging to write. I am proud to say that we still define the entire book in a LaTeX document and do all of figure generation and typesetting ourselves. We continued to use the Tufte LaTeX template. We typeset the book ourselves with LaTeX (as opposed to the publisher doing the typesetting), control our own lexing and styles, and have a suite of tests. We just had to overcome some additional hurdles to do so.

Interconnected Concepts (and Code)

Alg4DM crucially differs in how interconnected the material is. The chapters build on one another, with code from later chapters using or expanding on code from earlier chapters. These dependencies vastly increased the complexity of the work. We had to find unified representations that worked across the entire book, and any changes in one place could propagate to many others.

For example, the rollout method has an MDP version that accepts states and a POMDP method that accepts beliefs. Many POMDP methods call the state version, and thus depend on previous chapters.

Textbook-wide Codebase

When writing Alg4Opt, every chapter can be compiled alone, with standalone code. This made development pretty easy. We didn’t have any central shared code and could simply rely on pythontex blocks to execute Julia code for us:

For Alg4DM, we needed later chapters to use the code from earlier chapters. As such, we needed a new tactic.

The approach we settled on was to construct a Julia package that would contain all of our code. This package would include both utility methods like plotting functions and example problem formulations, but also all of the code presented in the textbook. Algorithm blocks like the one below would be automatically detected and their contents exported into our package:

Blocks like the one above existed in Alg4Opt and resulted in typeset code, but that code never actually ran. We usually had extra copies of every algorithm that were used to _actually_ run and produce our figures. By scraping the book for our code, we could ensure that the code that was typeset was the code that we were actually using to generate our figures and run our tests. Furthermore, we could still compile chapters individually, as each chapter could just load the Julia package.

The end result looked basically like this:

Furthermore, we improved our test setup by writing tests directly inside the .tex files alongside everything else. These test blocks would also be parsed out during extraction, and could then be used to test our code:

Pretty simple, and it solved a lot of problems.

Now, it wasn’t without its trade-offs. First, whenever I was working on the book, I would have to remember to first run the extraction script to make sure that I had the latest and greatest. This added overhead, and forgetting to do it caused me some hours of pain.

Second, our base support code was stored in a different repository than the LaTeX for the book. This meant that another author could update the base code and I would have to remember to check out the latest Julia package to get those updates. Again, a pain point. Now that I know how to have Julia packages that aren’t stored in the standard Julia location, we can actually move our Julia package into the main book repo and avoid this issue.

Third, loading one large central package significantly increased book, and individual chapter, compile time. For Alg4Opt, each chapter only included the packages and code it needed. For Alg4DM, every chapter includes the entire support package, which has some hefty dependencies (such as Flux.jl). While we could do more work to split out the biggest dependencies, this fundamental increase in size would definitely still be noticeable. (Yes, package precompilation helps, but only so much).

Overall, I think our solution ended up being sufficient, but there are some improvements we could make moving forward.

Bigger, Fancier Problems

Many of the example problems and implementations behind the figures in Alg4Opt were relatively simple. They were mostly all 2D optimization problems, which are very quick to run. We could compile the book quickly and regenerate the figures every time without much hassle.

In Alg4DM that was no longer the case. Training an agent policy to solve a POMDP problem takes time, even for something as simple as the mountain-car problem. Compiling the book from scratch, including recompilation of all figures from the original Julia code, would take a very, very long time. As such, there are multiple cases where we had to save the .tex code produced for a particular figure, comment out the code that generated it, and include that instead:

This change was pretty seamless and easy to execute. It did not disrupt our workflow much. Furthermore, because we were still producing the PDF from .tex input, the figure would still be recompiled with the latest colors and formatting. The primary incurred overhead was remembering when to regenerate a figure because something underneath changed, such as the algorithm, something it calls, or the problem definition it was run on.

Breaking it Down in ways it hasn’t been Broken Down Before

The content covered in the optimization book was more intuitive for me, in many ways. While there certainly was some synthesis / repackaging of ideas, the amount of ground that had to be covered with respect to the new book was far greater.

The decision making book covers a large span of academic literature. We’ve got Bayesian networks, deep learning, traditional MDPs and POMDPs, imitation learning, reinforcement learning, multiagent planning, Markov games, and more. We needed consistent nomenclature that didn’t have overlapping concepts, while staying as true as possible to the literature. (For example, we spent a long time deciding what the symbol for a conditional plan would be. That issue took over a year to resolve.)

Furthermore, as with the first book, the 2nd book tries to cover the principle components of the algorithmic ideas. That is, we break down the little nuggets of progress. An algorithm like SARSOP has a lot of ideas inside of it, and we break those apart rather than presenting the SARSOP algorithm as a whole. Similarly, 1D optimization is often solved with Brent’s method, but you won’t find Brent’s method in our first book. You will, however, find the bisection, secant, and inverse quadratic interpolation methods – all methods that are used inside of Brent’s method.

Perhaps the greatest contribution this text makes is how we cover policy gradient optimization. We tease many of the concepts apart and present them individually in a way I haven’t seen before. The restricted gradient update naturally evolves into the natural gradient update, which evolves into a trust region update, etc. That chapter, and the actor-critic chapter thereafter, where the hardest ones for me to write, but I am quite happy with the result.

Conclusion

And that’s it! The new book is something I worked very hard on and it came out great. Mykel and Kyle worked very hard on it to, we got a ton of feedback from all sorts of helpful individuals, and we had great support from MIT Press and their editors.

I’ve been really fortunate to have been able to write this book with Mykel and Kyle. Decision-making is Mykel’s specialty, and Kyle has a lot of valuable multiagent knowledge (and is extremely detail-oriented and perceptive). I learned a lot doing the course of Alg4DM’s development, and I hope that its readers can learn a lot as well.

Let’s work through how one might (in a broad sense) write a controller for guiding rockets under propulsive control for a pinpoint landing. Such a “planetary soft landing” is central to landing exploratory rovers and some modern rockets. It is also quite an interesting optimization problem, one that starts off seemingly hairy but turns out to be completely doable. There is something incredible about landing rockets, and there is something awesome and inspiring about the way this math problem lets us do just that.

Everything in this post is based off of Lars Blackmore’s paper, Lossless Convexification of Nonconvex Control Bound and Pointing Constraints of the Soft Landing Optimal Control Problem.

Problem Setup

We have a rocket. Our rocket is falling fast, and we wish to have it stop nice and gently on the surface of our planet. We will get it to stop gently by having the rocket use its thruster (or thrusters) to both slow itself down and navigate to a desired landing location.

As such, our problem is to find a thrust control profile \(\boldsymbol{T}_c\) to produce a trajectory from an initial position \(\boldsymbol{r}_0\) and initial velocity \(\dot{\boldsymbol{r}}_0\) to a final position at the landing pad with zero final velocity. As multiple trajectories may be feasible, we’ll choose the one that minimizes our fuel use.

We model the rocket as a point-mass. This is done both because it is a lot easier, but also because rockets are pretty stable and we want to maintain a vertical orientation, and any attitude control that we would need has much faster-acting dynamics that we could control separately.

Dynamics

Our state at a given point in time is given by both the physical state \(\boldsymbol{x} = [\boldsymbol{r}, \dot{\boldsymbol{r}}]\) consisting of the position and the velocity, and the rocket’s mass \(m\). I am going to do the derivation in 2D, but it is trivial to extend all of this to the 3D case.

Our physical system dynamics are:

\[\dot{\boldsymbol{x}}(t) = \boldsymbol{A} \boldsymbol{x}(t) + \boldsymbol{B} \left(\boldsymbol{g} + \frac{\boldsymbol{T}_c(t)}{m(t)}\right)\]

where \(\boldsymbol{g}\) is the gravity vector and our matrices are:

\[\boldsymbol{A} = \begin{bmatrix} \boldsymbol{0} & \boldsymbol{I} \\ \boldsymbol{0} & \boldsymbol{0} \end{bmatrix} \qquad \boldsymbol{B} = \begin{bmatrix} \boldsymbol{0} & \boldsymbol{0} \\ \boldsymbol{I} & \boldsymbol{0} \end{bmatrix}\]

for input vectors \([\boldsymbol{r}, \dot{\boldsymbol{r}}]\).

Our mass dynamics are \(\dot{m} = -\alpha \|\boldsymbol{T}_c(t)\|\), where the mass simply decreases in proportion to the thrust. Here \(\alpha\) is the constant thrust-specific fuel consumption that determines our rate of mass loss per unit of thrust.

Adding Constraints

We can take these dynamics and construct an optimization problem that finds a thrust control profile that brings us from our initial state to our final state with the minimum amount of fuel use. There are, however, a few additional constraints that we need to add.

First, we do not want our rocket to go below ground. The way this was handled in the paper was by adding a simple glide-slope constraint that prevents the rocket’s position from ever leaving a cone above the landing location. This ensure that we remain a safe distance above the ground. In 2D this is:

\[r_y(t) \geq \tan(\gamma_\text{gs}) r_x(t) \quad \text{for all } t\]

where \(r_x(t)\) and \(r_y(t)\) are our horizontal and vertical positions at time \(t\), and \(\gamma_\text{gs} \in [0,\pi/2]\) is a glide slope angle.

Next, we need to ensure that we do not use more fuel than what we have available. Let \(m_\text{wet}\) be our wet mass (initial rocket mass, consisting of the rocket plus all fuel), and \(m_\text{dry}\) be our dry mass (just the mass of the rocket – no fuel). Then we need our mass to remain above \(m_\text{dry}\).

Finally, we have some constraints with respect to our feasible thrust. First off, we cannot produce more thrust than our engine provides: \(\|\boldsymbol{T}_c\| \leq \rho_2\). Secondarily, we also cannot produce less thrust than a certain minimum: \(0 < \rho_1 \leq \|\boldsymbol{T}_c\|\). This lowerbound is caused by many rocket engines being physically unable to reduce thrust to zero without going out. Unfortunately, it is a non-convex constraint, and we will have to work around it later.

The last thrust constraint has to do with how far we can actuate our rocket engines. We are keeping our rocket vertically aligned during landing, and often have a limited ability to gimbal the engine. We introduce a thrust-pointing constraint that limits our deviation from vertical to within an angle \(\theta\):

\[ \boldsymbol{T}_{c,y}(t) \geq \|\boldsymbol{T}_c(t)\| \cos(\theta) \]

Our basic nonconvex minimum fuel landing problem is thus:

\[\begin{matrix}\text{minimize}_{\boldsymbol{T}_c} & \int_0^{t_f} \alpha \|\boldsymbol{T}_c(t)\| \> dt & & \\ \text{subject to} & \dot{\boldsymbol{x}}(t) = \boldsymbol{A} \boldsymbol{x}(t) + \boldsymbol{B} \left(\boldsymbol{g} + \frac{\boldsymbol{T}_c(t)}{m(t)}\right) & \text{for all } t \in [0, t_f] & \text{physical dynamics} \\ & \dot{m} = -\alpha \|\boldsymbol{T}_c(t)\| & \text{for all } t \in [0, t_f] & \text{mass dynamics} \\ & 0 < \rho_1 \leq \|\boldsymbol{T}_c\| \leq \rho_2 & \text{for all } t \in [0, t_f] & \text{thrust magnitude bounds} \\ & \boldsymbol{T}_{c,y}(t) \geq \|\boldsymbol{T}_c(t)\| \cos(\theta) & \text{for all } t \in [0, t_f] & \text{thrust pointing constraint} \\ & r_y(t) \geq \tan(\gamma_\text{gs}) r_x(t)& \text{for all } t \in [0, t_f] & \text{glide slope constraint} \\ & \boldsymbol{r}(0) = \boldsymbol{r}_0, \dot{\boldsymbol{r}}(0) = \dot{\boldsymbol{r}}_0 & & \text{physical initial conditions} \\ & m(0) = m_\text{wet} & & \text{mass initial condition} \\ & \boldsymbol{r}(t_f) = [0,0], \dot{\boldsymbol{r}}(t_f) = [0,0] & & \text{physical final conditions} \\ & m(t_f) \geq m_\text{dry} & & \text{mass final condition} \end{matrix}\]

Lossless Convexification

The basic problem formulation above is difficult to solve because it is nonconvex. Finding a way to make it convex would make it far easier to solve.

Our issues are the nonconvex thrust bounds and the division by mass in our dynamics.

To get around the thrust bounds, we introduce an additional slack variable, \(\Gamma(t)\). We use this slack variable in our thrust bounds constraint and mass dynamics, and constrain our actual thrust magnitude to lie below \(\Gamma\):

\[\dot{m}(t) = -\alpha \Gamma\]

\[\|\boldsymbol{T}\|_c(t) \leq \Gamma(t)\]

\[0 < \rho_1 \leq \Gamma(t) \leq \rho_2\]

\[ \boldsymbol{T}_{c,y}(t) \geq \cos(\theta) \Gamma(t) \]

On the one hand, we have introduced a new time-varying quantity, and have thus increased the complexity of our problem. However, by introducing this quantity, we have made our constraints convex. This trade-off will be well worth it.

We can get around the nonconvex division by mass with a change of variables:

\[\sigma \equiv \frac{\Gamma}{m} \qquad \boldsymbol{u} \equiv \frac{\boldsymbol{T}_c}{m} \qquad z \equiv \ln m\]

This simplifies our dynamics to:

\[\dot{\boldsymbol{x}}(t) = \boldsymbol{A} \boldsymbol{x}(t) + \boldsymbol{B} \left(\boldsymbol{g} + \boldsymbol{u}(t)\right)\]

\[ \dot{z}(t) = -\alpha \sigma(t) \]

Unfortunately, our thrust bounds constraint becomes nonconvex:

\[0 \leq \rho_1 e^{-z(t)} \leq \sigma(t) \leq \rho_2 e^{-z(t)}\]

We can convexify it by approximating the constraint with a 2nd-order Taylor expansion of \(e^{-z(t)}\):

\[\rho_1 e^{-z_o(t)} \left(1 – (z(t) – z_o(t)) + \frac{1}{2}(z(t) – z_o(t))^2\right) \leq \sigma(t) \leq \rho_2 e^{-z_o(t)} \left(1 – (z(t) – z_o(t)) \right)\]

where \(z_o(t) = \ln (m_\text{wet} – \alpha \rho_2 t)\).

The left-hand side forms a second-order cone constraint, which have the form \(\|x\|_2 \leq t\). Quadratic constraints can be converted to such second-order cone constraints.

We thus end up with the following convex problem:

\[\begin{matrix}\text{minimize}_{\boldsymbol{u}, \boldsymbol{\sigma}} & \int_0^{t_f} \sigma(t) \> dt & & \\ \text{subject to} & \dot{\boldsymbol{x}}(t) = \boldsymbol{A} \boldsymbol{x}(t) + \boldsymbol{B} \left(\boldsymbol{g} + \boldsymbol{u}(t)\right) & \text{for all } t \in [0, t_f] & \text{physical dynamics} \\ & \dot{z} = -\alpha \sigma(t) & \text{for all } t \in [0, t_f] & \text{mass dynamics} \\ & \|\boldsymbol{u}(t)\| \leq \sigma(t) & \text{for all } t \in [0, t_f] & \text{slack variable bound} \\ & \rho_1 e^{-z_o(t)} \left(1 – (z(t) – z_o(t)) + \frac{1}{2}(z(t) – z_o(t))^2\right) \leq \sigma(t) & \text{for all } t \in [0, t_f] & \text{thrust magnitude, lower} \\ & \sigma(t) \leq \rho_2 e^{-z_o(t)} \left(1 – (z(t) – z_o(t)) \right) & \text{for all } t \in [0, t_f] & \text{thrust magnitude, upper} \\ & u_y(t) \geq \sigma(t) \cos(\theta) & \text{for all } t \in [0, t_f] & \text{thrust pointing constraint} \\ & r_y(t) \geq \tan(\gamma_\text{gs}) r_x(t)& \text{for all } t \in [0, t_f] & \text{glide slope constraint} \\ & \boldsymbol{r}(0) = \boldsymbol{r}_0, \dot{\boldsymbol{r}}(0) = \dot{\boldsymbol{r}}_0 & & \text{physical initial conditions} \\ & z(0) = \ln m_\text{wet} & & \text{mass initial condition} \\ & \boldsymbol{r}(t_f) = [0,0], \dot{\boldsymbol{r}}(t_f) = [0,0] & & \text{physical final conditions} \\ & z(t_f) \geq \ln m_\text{dry} & & \text{mass final condition} \end{matrix}\]

It can be shown that if there is a feasible solution to the first problem, then that solution also defines a feasible solution to this new problem. Furthermore, this solution is an optimal solution of the new problem. (This is related to the fact that \(\boldsymbol{u}\) is not minimized – we only care about minimizing \(\boldsymbol{\sigma}\), but for physical feasibility we end up needing \(\boldsymbol{u} = \boldsymbol{\sigma}\)).

Discretization in Time

The final problem transformation involves moving from a continuous formulation to a discrete formulation, such that we can actually implement a numerical algorithm. We discretize time into \(n\) equal-width periods such that \(n \Delta t = t_f\):

\[t^{(k)} = k \Delta t, \qquad k = 0, \ldots, n\]

Our thrust profile is a continuous function of time. We construct it as a combination of \(n+1\) basis functions, \(\phi_{0:n}\):

\[\begin{bmatrix}\boldsymbol{u}(t) \\ \sigma(t)\end{bmatrix} = \sum_{k=0}^n \boldsymbol{p}^{(k)} \phi^{(k)}(t), \qquad t \in [0,t_f]\]

where the \(\boldsymbol{p}^{(k)} \in \mathbb{R}^3\) are mixing weights that we will optimize. Note that we did not have to have as many basis functions as discrete points – it just makes the math easier. Various basis functions can be used, but the paper uses the sawtooth:The sawtooth form makes for a continuous thrust profile. When I first tackled this problem as part of a class project, Bryant and I used a piecewise-constant basis function. That was simpler, but the control trajectory is discontinuous.

\[\phi_j(t) = \begin{cases} \frac{t_j-t}{\Delta t} & \text{for } t \in [t_{j-1}, t_j] \\ \frac{t – t_j}{\Delta t} & \text{for } t \in [t_{j}, t_{j+1}] \\ 0 & \text{otherwise} \end{cases}\]

Our system state is \([\boldsymbol{r}(t)^\top, \dot{\boldsymbol{r}}(t)^\top, z(t)]^\top\) with the continuous-time system dynamics:We have a linear system. The 'c' subscripts are for 'continuous'.

\[\begin{split} \begin{bmatrix} \dot{\boldsymbol{r}}(t) \\ \ddot{\boldsymbol{r}}(t) \\ \dot{z}(t) \end{bmatrix} & = \begin{bmatrix} \boldsymbol{0} & \boldsymbol{I} & 0 \\ \boldsymbol{0} & \boldsymbol{0} & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \boldsymbol{r}(t) \\ \dot{\boldsymbol{r}}(t) \\ z(t) \end{bmatrix} + \begin{bmatrix} \boldsymbol{0} & \boldsymbol{0} \\ \boldsymbol{I} & \boldsymbol{0} \\ 0 & -\alpha \end{bmatrix} \begin{bmatrix} \boldsymbol{u}(t) \\ \sigma(t) \end{bmatrix} + \begin{bmatrix} \boldsymbol{0} \\ \boldsymbol{g} \\ 0 \end{bmatrix} \\ \dot{\mathtt{x}(t)} & = \boldsymbol{A}_c \mathtt{x}(t) + \boldsymbol{B}_c \mathtt{u}(t) + \mathtt{g} \end{split} \]

In order to discretize, we would like to have a linear equation to get our state at every discrete time point. We know that the solution to the ODE \(\dot{x} = a x + b u\) is:

\[ x(t) = e^{at} x_0 + \int_0^t e^{A(t-\tau)} b u(\tau) \> d\tau \]

Similarly, the solution to our ODE is:

\[ \mathtt{x}(t) = e^{\boldsymbol{A_c}t} \mathtt{x}_0 + \int_0^t e^{\boldsymbol{A_c}(t-\tau)} \left(\boldsymbol{B}_c \mathtt{u}(\tau) + \mathtt{g}\right) \> d\tau \]

We can get our discrete update by plugging in \(\Delta t\):

\[ \mathtt{x}(\Delta t) = e^{\boldsymbol{A_c}\Delta t} \mathtt{x}_0 + \int_0^{\Delta t} e^{\boldsymbol{A_c}(\Delta t-\tau)} \left(\boldsymbol{B}_c \mathtt{u}(\tau) + \mathtt{g}\right) \> d\tau \]

If we’re going from \(k\) to \(k+1\), we get:

\[ \mathtt{x}^{(k+1)} = e^{\boldsymbol{A_c}\Delta t} \mathtt{x}^{(k)} + \int_0^{\Delta t} e^{\boldsymbol{A_c}(\Delta t-\tau)} \left(\boldsymbol{B}_c \mathtt{u}(\tau) + \mathtt{g}\right) \> d\tau \]

We then substitute in our basis function formulation:Note that only two components contribute to a given discrete time interval.

\[ \begin{split} \mathtt{x}^{(k+1)} &= \left[e^{\boldsymbol{A_c}\Delta t}\right] \mathtt{x}^{(k)} + \int_0^{\Delta t} e^{\boldsymbol{A_c}(\Delta t-\tau)} \left(\boldsymbol{B}_c \left( \boldsymbol{p}^{(k)} \phi^{(k)}(t) + \boldsymbol{p}^{(k+1)} \phi^{(k+1)}(t) \right) + \mathtt{g}\right) \> d\tau \\ & = \left[e^{\boldsymbol{A_c}\Delta t}\right] \mathtt{x}^{(k)} + \int_0^{\Delta t} e^{\boldsymbol{A_c}(\Delta t-\tau)} \left(\boldsymbol{B}_c \left( \left( 1- \frac{\tau}{\Delta t}\right) \boldsymbol{p}^{(k)} + \left( \frac{\tau}{\Delta t}\right) \boldsymbol{p}^{(k+1)} \right) + \mathtt{g}\right) \> d\tau \\ & = \left[e^{\boldsymbol{A_c}\Delta t}\right] \mathtt{x}^{(k)} + \left[\int_0^{\Delta t} e^{\boldsymbol{A_c}(\Delta t-\tau)} \boldsymbol{B}_c\right] \left( \boldsymbol{p}^{(k)} + \mathtt{g} \right) + \left[\int_0^{\Delta t} e^{\boldsymbol{A_c}(\Delta t-\tau)} \boldsymbol{B}_c \frac{\tau}{\Delta t}\right] \left( \boldsymbol{p}^{(k+1)} – \boldsymbol{p}^{(k)} \right) \\ &= \boldsymbol{A}_d \texttt{x}^{(k)} + \boldsymbol{B}_{d1} \left( \boldsymbol{p}^{(k)} + \mathtt{g} \right) + \boldsymbol{B}_{d2} \left( \boldsymbol{p}^{(k+1)} – \boldsymbol{p}^{(k)} \right) \end{split} \] Our discrete update equation is thus also linear.I personally find this part of the derivation very satisfying. We go from a continuous ODE to nice discrete update steps. The math works out quite nicely. This analysis is super useful - linear systems like the one we are working with here show up all the time. We can compute our matrices once using numerical methods for integration. We can apply our relation recursively to get our state at any discrete time point. All of these updates are linear:

\[\begin{split} \mathtt{x}^{(1)} & = \boldsymbol{A}_d \texttt{x}^{(0)} + \boldsymbol{B}_{d1} \left( \boldsymbol{p}^{(0)} + \mathtt{g} \right) + \boldsymbol{B}_{d2} \left( \boldsymbol{p}^{(1)} – \boldsymbol{p}^{(0)} \right) \\ \mathtt{x}^{(2)} & = \boldsymbol{A}_d \texttt{x}^{(1)} + \boldsymbol{B}_{d1} \left( \boldsymbol{p}^{(1)} + \mathtt{g} \right) + \boldsymbol{B}_{d2} \left( \boldsymbol{p}^{(2)} – \boldsymbol{p}^{(1)} \right) \\ &= \boldsymbol{A}_d \left(\boldsymbol{A}_d \texttt{x}^{(0)} + \boldsymbol{B}_{d1} \left( \boldsymbol{p}^{(0)} + \mathtt{g} \right) + \boldsymbol{B}_{d2} \left( \boldsymbol{p}^{(1)} – \boldsymbol{p}^{(0)} \right)\right) + \boldsymbol{B}_{d1} \left( \boldsymbol{p}^{(1)} + \mathtt{g} \right) + \boldsymbol{B}_{d2} \left( \boldsymbol{p}^{(2)} – \boldsymbol{p}^{(1)} \right) \\ &= \boldsymbol{A}_d^2 \texttt{x}^{(0)} + \left(\boldsymbol{B}_{d1} + \boldsymbol{A}_d \boldsymbol{B}_{d1}\right) \mathtt{g} + \left(\boldsymbol{A}_d \boldsymbol{B}_{d1} – \boldsymbol{A}_d \boldsymbol{B}_{d2} \right) \boldsymbol{p}^{(0)} + \left(\boldsymbol{B}_{d1} + \boldsymbol{A}_d \boldsymbol{B}_{d2} – \boldsymbol{B}_{d2} \right) \boldsymbol{p}^{(1)} + \boldsymbol{B}_{d2} \boldsymbol{p}^{(2)} \\ & \enspace \vdots \end{split}\]

Note that due to our choice of basis function, \(\boldsymbol{u}^{(k)} = [p_1^{(k)} p_2^{(k)}]^\top\) and \(\sigma^{(k)} = p_3^{(k)}\).

We can approximate the continuous version of our objective function:

\[ \int_0^{t_f} \sigma(t) \> dt \approx \sum_{k=0}^n p^{(k)}_3\]

We can pull all of this together and construct a discrete-time optimization problem. We optimize the vector \(\boldsymbol{\eta}\), which simply contains all of the \(\boldsymbol{p}\) vectors:

\[\boldsymbol{\eta} = \begin{bmatrix} \boldsymbol{p}^{(0)} \\ \boldsymbol{p}^{(1)} \\ \vdots \\ \boldsymbol{p}^{(n)} \end{bmatrix} \]

Our problem is:

\[\begin{matrix}\text{minimize}_{\boldsymbol{\eta}} & \sum_{k=0}^n p^{(k)}_3 & & \\ \text{subject to} & \mathtt{x}^{(k+1)} = \boldsymbol{A}_d \texttt{x}^{(k)} + \boldsymbol{B}_{d1} \left( \boldsymbol{p}^{(k)} + \mathtt{g} \right) + \boldsymbol{B}_{d2} \left( \boldsymbol{p}^{(k+1)} – \boldsymbol{p}^{(k)} \right) & \text{for all } k \in [0, 1, \ldots, n-1] & \text{dynamics} \\ & \|[p_1^{(k)} p_2^{(k)}]^\top\| \leq p_3^{(k)} & \text{for all } k \in [0, 1, \ldots, n] & \text{slack variable bound} \\ & \rho_1 e^{-z_o(k \Delta t)} \left(1 – (z(k \Delta t) – z_o(k \Delta t)) + \frac{1}{2}(z(k \Delta t) – z_o(k \Delta t))^2\right) \leq \sigma(k \Delta t) & \text{for all } k \in [0, 1, \ldots, n] & \text{thrust magnitude, lower} \\ & \sigma(k \Delta t) \leq \rho_2 e^{-z_o(k \Delta t)} \left(1 – (z(k \Delta t) – z_o(k \Delta t)) \right) & \text{for all } k \in [0, 1, \ldots, n] & \text{thrust magnitude, upper} \\ & p_2^{(k)} \geq p_3^{(k)} \cos(\theta) & \text{for all } k \in [0, 1, \ldots, n] & \text{thrust pointing constraint} \\ & r_y(k \Delta t) \geq \tan(\gamma_\text{gs}) r_x(k \Delta t)& \text{for all } k \in [0, 1, \ldots, n-1] & \text{glide slope constraint} \\ & \boldsymbol{r}(0) = \boldsymbol{r}_0, \dot{\boldsymbol{r}}(0) = \dot{\boldsymbol{r}}_0 & & \text{physical initial conditions} \\ & z(0) = \ln m_\text{wet} & & \text{mass initial condition} \\ & \boldsymbol{r}(t_f) = [0,0], \dot{\boldsymbol{r}}(t_f) = [0,0] & & \text{physical final conditions} \\ & z(t_f) \geq \ln m_\text{dry} & & \text{mass final condition} \end{matrix}\]

This final formulation is a finite-dimensional second-order cone program (SOCP). It can be solved by standard convex solvers to get a globally-optimal solution.

Note that we only apply our constraints at the discrete time points. This tends to be sufficient, due to the linear nature of our dynamics.

Coding it Up

Update: I originally did the implementation in Julia, but found odd behavior with the Convex.jl solvers I was using. I tried both SCS (the default recommended solver for Convex.jl) and COSMO. Neither of these solvers worked for the propulsive landing problem. Both solvers had the same behavior – they would find a solution that satisfied the terminal conditions, but would fail to satisfy the other constraints. Giving it more iterations would cause it to slowly make progress on the other constraints, but eventually it would give up and decide the problem is infeasible.

Prof. Kochenderfer noticed that Julia didn’t work for me, and recommended that I try ECOS. I swapped that in and it just worked. My Julia notebook is available here.

import cvxpy as cp
import numpy as np
import scipy

We define our parameters, which I adapted from the paper:

m_wet = 2000.0        # rocket wet mass [kg]
m_dry =  300.0        # rocket dry mass [kg]
thrust_max = 24000.0  # maximum thrust
ρ1 = 0.2 * thrust_max # lower bound on thrust
ρ2 = 0.8 * thrust_max # upper bound on thrust
α = 5e-4              # rocket specific impulse [s]
θ = np.deg2rad(30)    # max thrust angle deviation from vertical [rad]
γ = np.deg2rad(15)    # no-fly bounds angle [rad]
g = 3.71              # (martian) gravity [m/s2]
tf = 52.0             # duration of the simulation [s]
n_pts = 50            # number of discretization points
 
# initial state (x,y,x_dot,y_dot,z)
x0 = np.array([600.0, 1200.0, 50.0, -95.0, np.log(m_wet)])

We then compute some derived parameters:

t = np.linspace(0.0, tf, n_pts) # time vector [s]
N = len(t) - 2                  # number of control nodes
dt = t[1] - t[0]

We compute \(\boldsymbol{A}_c\) and \(\boldsymbol{A}_d\):

Ac = np.vstack((
        np.hstack((np.zeros((2, 2)), np.eye(2),        np.zeros((2, 1)))),
        np.hstack((np.zeros((2, 2)), np.zeros((2, 2)), np.zeros((2, 1)))),
        np.zeros((1, 5))
     ))
 
Ad = scipy.linalg.expm(Ac * dt)

We compute \(\boldsymbol{B}_c\), \(\boldsymbol{B}_{d1}\), and \(\boldsymbol{B}_{d2}\):

def integrate_trapezoidal(f, a, b, n):
    retval = f(a) * 0.5 + f(b) * 0.5
    for k in range(1, n):
        retval = retval + f(a + k*(b-a)/n)
    retval = retval * (b - a)/ n
    return retval
 
Bc = np.vstack((
        np.hstack((np.zeros((2, 2)), np.zeros((2, 1)))),
        np.hstack((np.eye(2),        np.zeros((2, 1)))),
        np.hstack((np.zeros((1, 2)), np.reshape(np.array([-α]), (1, 1))))
    ))
 
Bd1 = integrate_trapezoidal(lambda s: np.matmul(scipy.linalg.expm(Ac*(dt-s)), Bc), 0.0, dt, 10000)
Bd2 = integrate_trapezoidal(lambda s: np.matmul(scipy.linalg.expm(Ac*(dt-s))*(s/dt), Bc), 0.0, dt, 10000)

We implement our recursive calculation for the state at time index \(k\):

# gravitational 3-vector [m/s2]
g3 = np.array([0.0, -g, 0.0])
 
def get_xk(Ad, Bd1, Bd2, η, x0, g3, k):
    xk = x0
    p_km1 = np.array([0.0,0.0,0.0])
    p_k = η[0:3]
    ind_low = 0
    for j in range(0,k):
        xk = np.matmul(Ad,xk) + np.matmul(Bd1, (p_km1 + g3)) + np.matmul(Bd2, (p_k - p_km1))
        p_km1 = p_k
        ind_low = min(ind_low+3, 3*(N+1))
        p_k = η[ind_low : ind_low+3]
    return xk

and we define our problem:

# Create the CVX problem 
η = cp.Variable(3*N)
 
ω = np.tile(np.array([0,0,1]), N) # only minimize the σ's
 
# minimize fuel consumption
objective = cp.Minimize(ω.T@η)
constraints = []
 
# thrust magnitude constraints
for k in range(0, N):
    j = 3*k
    constraints.append(cp.norm(η[j:j+2]) <= η[j+2])
 
# thrust cant angle constraints
for k in range(0, N):
    j = 3*k
    uy = η[j+1]
    σ = η[j+2]
    constraints.append(uy >= np.cos(θ) * σ)
 
# thrust slack variable lower and upper bounds
z0 = lambda k: np.log(m_wet - α*ρ2*dt*k)
for k in range(0, N):
    j = 3*k
    σk = η[j+2]
    xk = get_xk(Ad, Bd1, Bd2, η, x0, g3, k)
    zk = xk[-1]
 
    a = ρ1 * np.exp(-z0(k))
    z0k = z0(k).item()
    z1k = z1(k).item()
 
    constraints.append(a * (1 - (zk-z0k) + 0.5*cp.square(zk-z0k)) <= σk)    
    constraints.append(σk <= ρ2 * np.exp(-z0k) * (1 - (zk-z0k)))
 
# glide angle constraints
for k in range(1, N):
    j = 3*k
    xk = get_xk(Ad, Bd1, Bd2, η, x0, g3, k)
    constraints.append(xk[1] >= np.tan(γ) * xk[0])
 
# terminal conditions
xf = get_xk(Ad, Bd1, Bd2, η, x0, g3, N)
constraints.append(xf[0:4] == [0,0,0,0])
constraints.append(xf[4] >= np.log(m_dry))
 
prob = cp.Problem(objective, constraints)

and we solve it!

result = prob.solve()
print(prob.status)

We can extract our solution (thrust profile and trajectory) as follows:

η_eval = η.value
 
xk = np.zeros((len(x0), len(t)))
for (i,k) in enumerate(range(0,N)):
    xk[:,i] = get_xk(Ad, Bd1, Bd2, η_eval, x0, g3, k)
 
xs = xk[0,:]
ys = xk[1,:]
us = xk[2,:]
vs = xk[3,:]
zs = xk[4,:]
 
mass = np.exp(zs)
 
u1 = η_eval[0::3]
u2 = η_eval[1::3]
σ = η_eval[2::3]
 
Tx = u1 * mass
Ty = u2 * mass
Γ = σ * mass
thrust_mag = [np.hypot(Tyi, Txi) for (Txi, Tyi) in zip(Tx, Ty)]
thrust_ang = [np.arctan2(Txi, Tyi) for (Txi, Tyi) in zip(Tx, Ty)]

The plots end up looking pretty great. Here is the landing trajectory we get for the code above:

The blue line shows the rocket’s position over time, the orange line shows the no-fly boundary, and the black line shows the ground (\(y = 0\)). The rocket has to slow down so as to avoid the no-fly zone (orange), and then nicely swoops in for the final touchdown. Here is the thrust profile:

Our thrust magnitude matches our slack variable \(\Gamma\), as expected for optimal trajectories. We also see how the solution “rides the rails”, hitting both the upper and lower bounds. This is very common for optimal control problems – we use all of the control authority available to us.

The two thrust components look like this:

We see that \(T_y\) stays positive, as it should. However, \(T_x\) starts off negative to move the rocket left, and then counters in order to stick the landing.

Our thrust angle stays within bounds:

We can of course play around with our parameters and initial conditions to get different trajectories. Here is what happens if we flip the initial horizontal velocity:

Here is what happens if we increase the minimum thrust to \(\rho_1 = 0.3 T_\text{max}\):

How to Avoid Specifying the Trajectory Duration

One remaining issue with the convex formulation above is that we have to specify our trajectory duration, \(t_f\). This, and \(n\), determine \(\Delta t\) and our terminal condition, both of which are pretty central to the problem.

We might be interested, for example, in finding the minimum-time landing trajectory. Typically, the longer the trajectory, the more fuel that is used, so being able to reduce \(t_f\) automatically can have big benefits.

The naive approach is to do a search over \(t_f\) in order to find the smallest valid trajectory duration. In fact, this is exactly what the authors suggest:

Once we have a procedure to compute the fuel optimal trajectory for a given time-of-flight, \(t_f\), we use the Golden search algorithm to determine the time-of-flight with the least fuel need, \(t^*_f\). This line search algorithm is motivated by the observation that the minimum fuel is a unimodal function of time with a global minimum

Behcet Acikmese, Lars Blackmore, Daniel P. Scharf, and Aron Wolf

I would be surprised if there were a way to optimize for \(t_f\) alongside the trajectory, but there very well could be a better approach!

The Curious Case of the Missing Constraint

As a final note, I wanted to mention that there are some additional constraints in some versions of the paper. Notably, this paper includes these constraints in its “Problem 3”:

\[ \log\left(m_\text{wet} – \alpha \rho_2 t\right) \leq z(t) \leq \log\left(m_\text{wet} – \alpha \rho_1 t\right) \]

These constraints do not appear in Lossless Convexification of Nonconvex Control Bound and Pointing Constraints of the Soft Landing Optimal Control Problem.

After thinking about it, I don’t think the constraints are needed. They are enforcing that our mass lies between an upper and lower bound. The upperbound is the mass we would have if we had minimum thrust for the entire trajectory, whereas the lowerbound is the mass we would have if we had maximum thrust for the entire trajectory. The fact that we constrain our thrusts (and constrain our final mass) mean that we don’t really need these constraints.

Conclusion

I find this problem fascinating. We go through a lot of math, so much that it can be overwhelming, but it all works out in the end. Every step ends up being understandable and justifiable. I then love seeing the final trajectories play out, and get the immediate feedback in the trajectory as we adjust the initial conditions and constraints.

I hope you enjoyed this little side project.