# 🐝 Feedback Control and the Hive Mind

When a honeybee colony must choose a new nest site, no single bee decides. Instead, thousands of individuals interact and eventually reach a collective choice. From a control-theoretic perspective, this looks like a feedback control system. But one that does not fit the classical framework of control theory.

This note explores how collective decision-making can be viewed through the lens of control theory and shows why a simple feedback design faces a fundamental limitation.


---

## 1. From Classical Control to Collective Decisions

In classical control theory, the goal is to find a control law that drives a system to a stable equilibrium.

$$
\dot{x} = f(x, u(x))
$$

The central question is:

> Does there exist a feedback control law $u:\mathbb{R}^d \rightarrow \mathbb{R}^m$ such that
> $$
> \lim_{t \to \infty} x(t) = x^{eq}
> $$
> for all $x(0) \in \mathbb{R}^d$?

In such problems, there is typically a **single equilibrium** that the controller must stabilize. In collective decision-making, however, there are **multiple competing equilibria**, and the system must decide which one to converge to.

---

## 2. A Minimal Model for a Swarm as a Mean Field Markov Process

Consider a population of bees, each choosing between two sites, $A$ and $B$.
Each bee can be in one of two states:

$$
\mathcal{S} = \{ A, B \}.
$$

Let the population’s state be described by probabilities

$$
p = [p_A \;\; p_B], \quad p_A + p_B = 1, \quad p_A, p_B \ge 0.
$$

### Transition Dynamics

Each bee may switch its preference over time.
Let $u_{AB} \ge 0$ be the rate of switching from $A$ to $B$, and $u_{BA} \ge 0$ the rate from $B$ to $A$.
The probability dynamics follow the Kolmogorov forward equations:

$$
\begin{aligned}
\dot{x}_A(t) &= -u_{AB}x_A(t) + u_{BA}x_B(t),\\
\dot{x}_B(t) &= u_{AB}x_A(t) - u_{BA}x_B(t).
\end{aligned}
$$

or equivalently,

$$
\begin{bmatrix}
\dot{x}_A \\[2pt]
\dot{x}_B
\end{bmatrix} =
\begin{bmatrix}
-u_{AB} & u_{BA} \\
u_{AB} & -u_{BA}
\end{bmatrix}
\begin{bmatrix}
x_A \\[2pt]
x_B
\end{bmatrix}.
$$

Here, $x_A(t)$ and $x_B(t)$ represent the fraction of bees supporting each option.
It can be checked that $x_A(t) + x_B(t) = 1$ for all $t \ge 0$.

---

## 3. Decision-Making as a Feedback Problem

The goal is to design feedback rates

$$
u_A(x_A, x_B, A, B), \qquad u_B(x_A, x_B, A, B)
$$

so that the population collectively chooses the better site.

A **symmetry condition** ensures that both sites are treated equivalently:

$$
u_A(x_B, x_A, A, B) = u_B(x_B, x_A, B, A).
$$

The desired stability conditions are:

$$
\begin{aligned}
(1,0) &\text{ is asymptotically stable if } A > B,\\
(0,1) &\text{ is asymptotically stable if } A < B,\\
(1,0), (0,1) &\text{ are locally asymptotically stable if } A = B.
\end{aligned}
$$

In words:
- If site $A$ is better, all bees should eventually favor $A$.
- If $B$ is better, they should favor $B$.
- If both are equal, both pure outcomes should be stable (deadlock avoidance).

---

## 4. Why This Problem Is Non-classical

In classical stabilization, the feedback law $u(x)$ is designed for a **single equilibrium**.
Here, the control law must adapt based on which site is better and still behave symmetrically when $A = B$. Most, importantly the stabilization objective is for a range of possible parameter values.

| Concept | Classical Control | Collective Decision |
|----------|------------------|---------------------|
| Equilibrium | Single target | Two competing equilibria |
| Objective | Stabilize one point | Choose between points |
| Input | Control $u(x)$ | Control $u_A, u_B$ must maintain symmetry|
| Challenge | Stability |Multiple equilibria |

---

## 5. Impossibility of Deadlock Avoidance

**Claim 1:** No smooth, symmetric feedback control law can guarantee deadlock avoidance while ensuring correct decisions for all $(A,B)$.

### Proof (Sketch)

Define the net flow toward $A$ as

$$
f(x_A, x_B, A, B) = -u_A(x_A, x_B, A, B) \, x_A + u_B(x_A, x_B, A, B) \, x_B.
$$

For asymptotic stability:

- If $A > B$, then $f(x_A, x_B, A, B) > 0$ for all interior points $(x_A, x_B) \in (0,1)^2$.
- If $A < B$, then $f(x_A, x_B, A, B) < 0$ for all interior points.

If $(1,0)$ is locally asymptotically stable when $A = B$, there must exist an interior point $(x_A^c, x_B^c)$ with

$$
f(x_A^c, x_B^c, A, A) > 0.
$$

Now consider sequences $(A_n, B_n)$ such that $A_n < B_n$ for all $n$ and
$\lim_{n \to \infty} A_n = A$, $\lim_{n \to \infty} B_n = A$.
By continuity,

$$
f(x_A^c, x_B^c, A_n, B_n) \to f(x_A^c, x_B^c, A, A) > 0.
$$

However, the condition for $A_n < B_n$ requires $f(x_A^c, x_B^c, A_n, B_n) < 0$ for all $n$, leading to a contradiction.

Therefore, no such smooth symmetric feedback law exists. ∎

---

## 6. Relaxing the Problem

The impossibility result suggests that perfect symmetry and global asymptotic stability cannot coexist.
To make progress, maybe we can relax requirements. Relaxing global stability requirement is one such possibility.

The next result demonstrates that even a simple quadratic dependence of transition rates on population proportions produces bistability between the two collective choices.

### **Claim 2**

Let $u_{AB} = A x_B^2$ and $u_{BA} = B x_A^2$. Then, if
$x_A(0) < \frac{A}{A+B}$ and $x_B(0) > \frac{B}{A+B}$, then

$$
\lim_{t \rightarrow \infty} x_A(t) = 0, \quad \lim_{t \rightarrow \infty} x_B(t) = 1
$$

Similarly, if $x_A(0) > \frac{A}{A+B}$ and $x_B(0) < \frac{B}{A+B}$, then

$$
\lim_{t \rightarrow \infty} x_A(t) = 1, \quad \lim_{t \rightarrow \infty} x_B(t) = 0
$$

---

### **Proof**

We have:

$$
x_A x_B (A x_B - B x_A) = 0
$$

This implies the equilibrium solutions of the ODE are:

$$
x_A^{eq} = 0, \quad x_B^{eq} = 1
$$

$$
x_A^{eq} = 1, \quad x_B^{eq} = 0
$$

$$
x_A^{eq} = \frac{A}{A+B}, \quad x_B^{eq} = \frac{B}{A+B}
$$

---

Let $x_A < \frac{A}{A+B}$ and hence $x_B > \frac{B}{A+B}$. Then

$$
A x_B - B x_A > 0
$$

This implies

$$
\dot{x}_A = x_A x_B (-A x_B + B x_A) < 0
$$

Thus, if $x_A(0) < \frac{A}{A+B}$ and $x_B(0) > \frac{B}{A+B}$, then

$$
\lim_{t \rightarrow \infty} x_A(t) = 0, \quad \lim_{t \rightarrow \infty} x_B(t) = 1
$$

Similarly, if $x_A(0) > \frac{A}{A+B}$ and $x_B(0) < \frac{B}{A+B}$, then

$$
\lim_{t \rightarrow \infty} x_A(t) = 1, \quad \lim_{t \rightarrow \infty} x_B(t) = 0
$$

## 7. Another Relaxation

Another way to relax the problem is to **ignore the deadlock case** ($A = B$) entirely and only consider situations where one site is strictly better than the other.
In that case, the problem becomes trivial — if every bee somehow knows which site is better, they can simply move toward it without any need for interaction or feedback from the swarm.

Formally, let $\phi : \mathbb{R} \to \mathbb{R}_{\ge 0}$ be a function satisfying

$$
\phi(x) > 0 \text{ for } x > 0, \quad \phi(x) = 0 \text{ for } x \le 0.
$$

We define the transition rates as

$$
u_{AB} = \phi(A - B), \qquad u_{BA} = \phi(B - A).
$$

### **Claim 3**

If $A > B$, then all bees eventually favor site $A$:

$$
\lim_{t \to \infty} x_A(t) = 1, \quad \lim_{t \to \infty} x_B(t) = 0.
$$

If $A < B$, then all bees eventually favor site $B$:

$$
\lim_{t \to \infty} x_A(t) = 0, \quad \lim_{t \to \infty} x_B(t) = 1.
$$

---

### **Proof**

From the population dynamics,

$$
\dot{x}_A = -u_{AB}x_A + u_{BA}x_B = -\phi(A - B) x_A + \phi(B - A)(1 - x_A).
$$

- If $A > B$, then $\phi(A - B) > 0$ and $\phi(B - A) = 0$, so
$$
\dot{x}_A = \phi(A - B)(1 - x_A) > 0.
$$
This drives $x_A(t) \to 1$ and $x_B(t) \to 0$.

- If $A < B$, then $\phi(B - A) > 0$ and $\phi(A - B) = 0$, so
$$
\dot{x}_A = -\phi(B - A)x_A < 0,
$$
which drives $x_A(t) \to 0$ and $x_B(t) \to 1$.

∎

---

This relaxation discards the symmetry and collective feedback that make the swarm problem interesting.
The result is effectively a **feedforward rule** — each bee independently switches to the better option, producing the correct decision but without any of the emergent dynamics of real swarm behavior.

This raises the question: **is there a positive role of mean-field effects or interactions between the bees?**
The first relaxation hints so, but not the second one. This will be the subject of the next post.

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## 7. Summary

- Collective decision-making can be formalized as a feedback stabilization problem on the probability simplex.
- Unlike standard control problems, there are multiple possible equilibria.
- Smooth, symmetric feedback cannot guarantee both correctness and deadlock avoidance.