# But What Do Bees Do?


In this note, we consider an odd way to address the deadlock avoidance problem. To recap, when we looked at the model in the previous note, it seemed like deadlock avoidance was fundamentally hard as a [feedback stabilization problem](https://hackmd.io/pavUBAyfQ3iOet1q_RdlwQ), unless we relaxed the problem in some way. We see how bees might have addressed this problem, partly by what I think is a form of *state-space expansion*. Let us consider this modified [model of decision making based on observations of house hunting in bees by Seeley and coauthors](https://www.science.org/doi/10.1126/science.1210361).

$$
\begin{aligned}
\dot{x}_A &= -\big(v_{AU} + v_{AB}\big)x_A + v_{UA}x_U + v_{BA}x_B,\\[4pt]
\dot{x}_B &= -\big(v_{BU} + v_{BA}\big)x_B + v_{UB}x_U + v_{AB}x_A,\\[4pt]
\dot{x}_U &= -\big(v_{UA} + v_{UB}\big)x_U + v_{AU}x_A + v_{BU}x_B.
\end{aligned}
$$

The new state $U$ denotes the *uncommitted* state. This extra state has been a source of my confusion and overthinking for 10 years since I came across the model.

The problem can now be framed again in terms of this equation. The goal is to design feedback laws $v_{i,j}(\cdot)$ so that

- $(1,0,0)$ is globally asymptotically stable if $A>B$,
- $(0,1,0)$ is globally asymptotically stable if $B>A$,
- $(1,0,0)$ and $(0,1,0)$ are locally asymptotically stable if $A = B$.

Given this objective, lets see what bees do under this context. A small modification of the [observation of bees by Seeley et al.](https://www.science.org/doi/10.1126/science.1210361) is the following choice of rates:

$$
\begin{aligned}
v_{UA} &= \gamma_A + \rho x_A, & v_{UB} &= \gamma_B + \rho x_B, \\
v_{AU} &= \sigma x_B, & v_{BU} &= \sigma x_A.
\end{aligned}
$$

A very important role is played by the *cross-inhibition* term $\sigma x_i$. In practice, bees achieve cross-inhibition by [head-butting](https://valleybees.org.au/wp-content/uploads/2017/11/HEAD-BUTTS.pdf).

The rest of the rates are set to $0$. The interpretations of the parameters are as follows:

- $\gamma_i$ is a spontaneous discovery effect proportional to the quality of site $i$.
- $\rho > 0$ captures the population recruitment effect.
- $\sigma > 0$ is the cross-inhibition effect that introduces negative feedback into the system.

Since $x_A + x_B + x_U = 1$ for all time, the dynamics live on the two-dimensional simplex, and we can eliminate $x_U$ by substituting $x_U = 1 - x_A - x_B$. The system reduces to

$$
\begin{aligned}
\dot{x}_A &= (\gamma_A + \rho x_A)(1 - x_A - x_B) - \sigma\, x_A x_B, \\[4pt]
\dot{x}_B &= (\gamma_B + \rho x_B)(1 - x_A - x_B) - \sigma\, x_A x_B.
\end{aligned}
$$

## 1. Equilibria

A direct calculation (Appendix A) shows that the equilibria of the system are organized by a single parameter,

$$
\delta := \frac{\gamma_B - \gamma_A}{\rho},
$$

which measures the quality difference between the two sites in units of recruitment strength. There are always two **vertex equilibria**, $(1,0,0)$ and $(0,1,0)$, corresponding to unanimous commitment to one site. In addition, when $|\delta| < 1$, there is a unique **interior equilibrium** $(x_A^*, x_B^*, x_U^*)$ with all three components strictly positive. At the boundary $|\delta| = 1$, the interior equilibrium collides with one of the vertices, and for $|\delta| > 1$ it leaves the simplex entirely.

So the parameter $\delta$ partitions the design space into three regimes by equilibrium count: three equilibria when the sites are close in quality, two when one is clearly better, and a transcritical bifurcation in between.

## 2. Stability

A linearization at each equilibrium (Appendix B) gives the following picture.

**Vertex equilibria.** The vertex $(1,0,0)$ has determinant $\sigma\rho(1 - \delta)$ and the vertex $(0,1,0)$ has determinant $\sigma\rho(1 + \delta)$. Both traces are negative. So:

| $\delta$ regime | $(1,0,0)$ | $(0,1,0)$ |
|---|---|---|
| $\delta < -1$ | stable | saddle |
| $-1 < \delta < 1$ | **stable** | **stable** |
| $\delta > 1$ | saddle | stable |

The regime $|\delta| < 1$ is bistable: both decisions are locally attracting. Note the structural role of $\sigma$: setting $\sigma = 0$ makes the determinant vanish at both vertices, leaving them marginally stable rather than attracting. Cross-inhibition is what gives the decision states their basin structure.

**Interior equilibrium.** Whenever it exists, the interior equilibrium is a **saddle**. The proof (Appendix B) uses the equilibrium identity $\mu x_U^* = \sigma x_A^* x_B^*$ to show that $\det J < 0$ there. The saddle has a one-dimensional stable manifold which acts as a separatrix between the basins of the two vertices.


## 3. Discussion

Putting the pieces together, the simplex dynamics has three regimes:

- **Both are kinda similar ($|\delta| < 1$).** Two stable vertices and one interior saddle. The decision choice depends on initial conditions.
- **One is kind of better ($|\delta| = 1$).** The saddle collides with the losing decision.
- **Obvious winner ($|\delta| > 1$).** Only the winning decision is stable. Almost every trajectory converges to the winner.

The state-space expansion does not satisfy the original design specification, which asked for global stability of the better decision whenever $A \neq B$. In the *Both are kinda similar ($|\delta| < 1$) regime*, the model is bistable. Therefore, the solution has created a band around the point of equal decision quality around which we achieve bistability.

The impossibility result from the previous note relied on a continuity argument at $A = B$ seems to have been gotten around by relaxing the objective, not necessarily by expanding the state space.

## 4. Could We Have Done This in Two States?

A natural question is whether the bistable band is genuinely a consequence of state-space expansion, or whether the two-state model can also exhibit it. The answer is that the two-state model can indeed reproduce a bistable band as well.

Recall the [two-state setup](https://hackmd.io/pavUBAyfQ3iOet1q_RdlwQ): $x_A + x_B = 1$, with

$$
\dot x_A = -u_{AB} x_A + u_{BA} x_B.
$$

Consider rates of the form

$$
u_{AB} = \phi(A - B) + \psi(x_A, x_B; A, B), \qquad $$$$u_{BA} = \phi(B - A) + \psi(x_B, x_A; B, A),
$$

where $\phi: \mathbb{R} \to \mathbb{R}_{\geq 0}$ is a smooth threshold function (zero for sufficiently negative argument, positive for sufficiently positive argument) and $\psi$ is a small symmetric stabilizing term that makes both vertices locally attracting. Then:

- For $|A - B|$ large, $\phi$ dominates and drives the swarm to the better site, recovering the feedforward construction from the previous note.
- For $|A - B|$ small, $\phi \approx 0$ and the $\psi$ term creates bistability at both vertices.

The width of the bistable band can be adjusted by the threshold scale of $\phi$ and can be made as small as desired. But notice the role mean-field coupling plays here. The site-discrimination work is done by $\phi(A - B)$, each bee independently reads the quality difference and switches accordingly. The mean-field term $\psi$ enters as a separate add-on to stabilize the decisions and create the band.

We test this on an example with the following choice of control:

$$
\phi(z) = \max(0,\; z - \varepsilon), \qquad \varepsilon = 0.25
$$

$$
u_{AB} = \phi(B - A) + c\, x_B^2, \qquad u_{BA} = \phi(A - B) + c\, x_A^2, \qquad c = 1
$$

With $A - B = 0.10$ and $\varepsilon = 0.25$, both $\phi$ terms vanish (we are inside the band), so the dynamics reduce to

$$
\dot x_A = -x_B^2\, x_A + x_A^2\, x_B = x_A\, x_B\,(x_A - x_B) = x_A(1 - x_A)(2x_A - 1)
$$

which is exactly the quadratic construction from the previous note with $A = B = 1$. The equilibria are at $0$, $\tfrac{1}{2}$, and $1$, with a saddle at $\tfrac{1}{2}$ and both vertices stable.

The following gif shows different swarm populations (each bee marker represents a single population) starting from different initial fractions $x_A$ and evolving under the flow. Swarms starting to the left of the saddle at $x_A = \tfrac{1}{2}$ commit to $B$; those starting to the right commit to $A$. The lower panel shows $\dot x_A$ as a function of $x_A$ — negative on the left half, positive on the right — confirming the bistable structure.


In the three-state Seeley model, by contrast, site quality enters only through the $\gamma_i$ asymmetry, and the actual decision dynamics are carried entirely by mean-field coupling. The same population interactions that produce the bistable band also do the discrimination. All of these models seem to indicate that the role of mean-field terms, that is, interaction within the population is mainly the key behind avoiding deadlock through some relaxation.

## 5. What Does This Say About Feedback Stabilizability for Collective Decision Making?

It looks like state-space expansion does not resolve the [obstruction to stabilizability](https://hackmd.io/pavUBAyfQ3iOet1q_RdlwQ). But it does relax the problem through a specific choice of control law inspired by bees. By introducing the uncommitted state $U$, we converted the rigid demand "stabilize the correct decision globally" into the softer demand "stabilize *some* vertex locally, with the correct one favored when the quality gap is large."

All the smooth symmetric feedback laws we have seen have given up something: global correctness, deadlock avoidance, or smoothness. State-space expansion gives up *global correctness in the region of near-equal quality*. The two-state construction with $\phi + \psi$ makes the same trade more crudely and less interpretably. Neither does something the other cannot.

The broader question is the following when feedback stabilizability fails, which relaxations are the most important, and which ones belong in a general theoretical framework of collective decision making as an instance of feedback stabilization?

## Reference

The model presented here is from the paper,

[Thomas D. Seeley, P. Kirk Visscher, Thomas Schlegel, Patrick M. Hogan, Nigel R. Franks, and James A. R. Marshall. Stop Signals Provide Cross Inhibition in Collective Decision-Making by Honeybee Swarms](https://www.science.org/doi/10.1126/science.1210361)

---

## Appendix A: Computing the Equilibria

At equilibrium we require $\dot{x}_A = \dot{x}_B = 0$. Subtracting the two equilibrium equations of the original three-state system yields

$$
\big[(\gamma_A+\rho x_A)-(\gamma_B+\rho x_B)\big]x_U=0,
$$

so every equilibrium satisfies either $x_U = 0$ or $x_A - x_B = (\gamma_B - \gamma_A)/\rho$.

**Case (i): $x_U = 0$.** With $x_A + x_B = 1$, the equation $\dot x_A = 0$ reduces to $\sigma x_A x_B = 0$, giving the two vertex equilibria $(1,0,0)$ and $(0,1,0)$.

**Case (ii): $x_U > 0$.** Define $\delta := (\gamma_B - \gamma_A)/\rho$. Then $x_B = x_A + \delta$ and $x_U = 1 - 2x_A - \delta$. Plugging into $\dot x_A = 0$:

$$
(\gamma_A+\rho x_A)(1-2x_A-\delta)=\sigma x_A(x_A+\delta).
$$

This is a quadratic in $x_A$, with solutions

$$
x_A^*=\frac{-B\pm\sqrt{B^2-4(2\rho+\sigma)C}}{2(2\rho+\sigma)}, \quad B := 2\gamma_A-\rho+(\rho+\sigma)\delta, \quad C := -\gamma_A(1-\delta).
$$

Once $x_A^*$ is fixed, $x_B^* = x_A^* + \delta$ and $x_U^* = 1 - 2x_A^* - \delta$.

To count interior equilibria in the simplex, define $f(x) := (\gamma_A+\rho x)(1-2x-\delta) - \sigma x(x+\delta)$. If $|\delta| < 1$, the positivity of the parameters gives $f(0) > 0$ and $f(1) < 0$, so there is exactly one root in $(0,1)$. At $\delta = 1$, $f(0) = 0$ and the root collides with the vertex $(0,1,0)$; at $\delta = -1$ it collides with $(1,0,0)$. For $|\delta| > 1$ no feasible root exists.

## Appendix B: Stability Calculations

The Jacobian of the reduced two-variable system

$$
\dot{x}_A = (\gamma_A + \rho x_A)(1 - x_A - x_B) - \sigma x_A x_B, $$$$\dot{x}_B = (\gamma_B + \rho x_B)(1 - x_A - x_B) - \sigma x_A x_B
$$

is, writing $x_U = 1 - x_A - x_B$,

$$
J =
\begin{bmatrix}
\rho x_U - (\gamma_A + \rho x_A) - \sigma x_B & -(\gamma_A + \rho x_A) - \sigma x_A \\[6pt]
-(\gamma_B + \rho x_B) - \sigma x_B & \rho x_U - (\gamma_B + \rho x_B) - \sigma x_A
\end{bmatrix}.
$$

**At $(1, 0, 0)$:** Substituting $x_A = 1$, $x_B = 0$, $x_U = 0$,

$$
J\big|_{(1,0,0)} =
\begin{bmatrix}
-(\gamma_A + \rho) & -(\gamma_A + \rho + \sigma) \\[4pt]
-\gamma_B & -(\gamma_B + \sigma)
\end{bmatrix}.
$$

The trace is $-(\gamma_A + \rho) - (\gamma_B + \sigma) < 0$. The determinant simplifies:

$$
\det J = (\gamma_A + \rho)(\gamma_B + \sigma) - \gamma_B(\gamma_A + \rho + \sigma) = \sigma(\gamma_A - \gamma_B + \rho) = \sigma\rho(1 - \delta).
$$

Stability is controlled by the sign of the determinant: $(1,0,0)$ is a stable node iff $\delta < 1$, and a saddle iff $\delta > 1$.

**At $(0, 1, 0)$:** By the $A \leftrightarrow B$ symmetry, $\det J = \sigma\rho(1 + \delta)$, so $(0,1,0)$ is stable iff $\delta > -1$.

**At the interior equilibrium.** The equilibrium identities

$$
\mu := \gamma_A + \rho x_A^* = \gamma_B + \rho x_B^*, \qquad \mu x_U^* = \sigma x_A^* x_B^*
$$

let us write the Jacobian as

$$
J\big|^* =
\begin{bmatrix}
\rho x_U^* - \mu - \sigma x_B^* & -\mu - \sigma x_A^* \\[4pt]
-\mu - \sigma x_B^* & \rho x_U^* - \mu - \sigma x_A^*
\end{bmatrix}.
$$

Expanding the determinant and cancelling, we get

$$
\det J = \rho x_U^* \cdot \big( \rho x_U^* - 2\mu - \sigma(x_A^* + x_B^*) \big).
$$

We claim the second factor is negative. From $\mu \geq \rho x_A^*$ and $\mu x_U^* = \sigma x_A^* x_B^*$,

$$
\rho x_A^* x_U^* \leq \mu x_U^* = \sigma x_A^* x_B^* \implies \rho x_U^* \leq \sigma x_B^*,
$$

and symmetrically $\rho x_U^* \leq \sigma x_A^*$. Adding these,

$$
2\rho x_U^* \leq \sigma(x_A^* + x_B^*) < 2\mu + \sigma(x_A^* + x_B^*),
$$

so $\rho x_U^* < 2\mu + \sigma(x_A^* + x_B^*)$. The second factor is therefore negative, and $\det J < 0$. The interior equilibrium is a saddle whenever it exists.