Simplifying Analyses of Chemical Reaction Networks for Approximate Majority

Abstract

Approximate Majority is a well-studied problem in the context of chemical reaction networks (CRNs) and their close relatives, population protocols: Given a mixture of two types of species with an initial gap between their counts, a CRN computation must reach consensus on the majority species. Angluin, Aspnes, and Eisenstat proposed a simple population protocol for Approximate Majority and proved correctness and $O(\log n)$ time efficiency with high probability, given an initial gap of size $\omega (\sqrt{n}\log n)$ when the total molecular count in the mixture is n. Motivated by their intriguing but complex proof, we provide simpler, and more intuitive proofs of correctness and efficiency for two bi-molecular CRNs for Approximate Majority, including that of Angluin et al. Key to our approach is to show how the bi-molecular CRNs essentially emulate a tri-molecular CRN with just two reactions and two species. Our results improve on those of Angluin et al. in that they hold even with an initial gap of $\varOmega (\sqrt{n \log n})$. Our analysis approach, which leverages the simplicity of a tri-molecular CRN to ultimately reason about bi-molecular CRNs, may be useful in analyzing other CRNs too.

A Appendix

In this appendix we (1) relate our CRN model to that of Cook et al. [3], (2) prove our lower and upper bounds on the number of B molecules in the Double-B CRN, and (3) prove the lemmas which make the analysis of Single-B parallel to that of the tri-molecular CRN.

1.1 A.1 Relationship Between Our CRN Model and that of Cook et al.

Other CRN models define reaction probabilities and computation time somewhat differently than we do, but these differences can easily be reconciled. For example, in the model of Cook et al. [3], if $k_{r}'$ is the rate constant associated with reaction $r= (s,t)$ of order $o$ and the system is in configuration $c= (x_1,x_2,\ldots ,x_m)$, then the propensity, or rate, of $r$ is

$$ \rho _{r}(c) = k'_{r} [\prod _{i=1}^m (x_i ! / (x_i - s_i)!)] / v^{o-1}. $$

If $\rho ^{tot}(c) = \sum _{r} \rho _{r}(c)$ for all reactions r of order o, then the probability that a reaction event is reaction $r$ is $\rho _{r}(c)/ \rho ^{tot}(c)$, and the expected time until a reaction event occurs is $1/ \rho ^{tot}(c)$. (In this model, reaction rate constants can be greater than 1, and may depend not only on the number of reactants of each species, but also on other properties of a species such as its shape, capturing the fact that the likelihood of different types of interactions may not all be the same.)

If in our model we set $k_{r} = k_{r}' \prod _{i=1}^m s_i!$ for each productive reaction, and normalize by $\sum _{r} k_{r}$ if necessary to ensure that $\sum _{r\in \mathcal{R}(s)} k_{r} \le 1$ (adjusting the underlying time unit accordingly), a straightforward calculation shows that, when in a given configuration $c$, the probability that a reaction event is a given reaction $r$ is the same in our model and that of Cook et al.² See the example of Fig. 4. Also, the expected time until the next reaction event differs between the models by a constant factor that is independent of $c$. Conversely, to convert from our model to that of Cook et al., divide our rate constant $k_{r}$ by $[\prod _{i=1}^m s_i!]$ (and multiply all rate constants by the same constant factor in order to adjust time units as needed).

Fig. 4.

(a) A CRN specified with respect to the Cook et al. model. The reaction rates when the system is in configuration (3, 3) are $k_{r_1}' = 18/v^2$ and $k_{r_2}' = 12/v^2$. The reaction probabilities are $\rho _{r_1}((3,3)) = 3/5$ and $\rho _{r_2}((3,3)) = 2/5$. (b) The mapping of the CRN of part (a) to our model by changing the rate constants (using Eq. 1 of footnote 2) and normalizing by $\sum k_{r}$. The probability that a reaction event is $r_1$ is $(18/14) / (30/14) = 18/30$, and the probability of $r_2$ is 12/30. Thus, reaction probabilities are preserved exactly.

1.2 A.2 Bounds on b, the Molecular Count of B, in the Double-B CRN

Here we provide a proof of Lemma 9, omitted from Sect. 3. We note that the probability that an interaction event in the interval I triggers reaction (0’) (respectively, reaction (1’), reaction (2’)) is just $x y / \left( {\begin{array}{c}n\\ 2\end{array}}\right)$ (respectively, $x b/ \left( {\begin{array}{c}n\\ 2\end{array}}\right)$, $y b/ \left( {\begin{array}{c}n\\ 2\end{array}}\right)$). In the following, we simplify calculations by replacing $\left( {\begin{array}{c}n\\ 2\end{array}}\right)$ with $n^2/2$.

Upper Bounds on b . Note that reaction (0’) has probability at most $(n/2)(n/2)/(n^2/2) = 1/2$, so at most n / 64 new B molecules are produced by reaction (0’) over interval I, in expectation, and at most n / 32 are produced, with probability $1-\exp (\varTheta (n))$. Thus, $b_\mathrm{max} \le b_\mathrm{min} + n/32$.

Given this, we can clearly assume that $b_\mathrm{min} \ge 14n/32$, since otherwise $b_\mathrm{max}$ (and, of course $b_e$) is less than 15n / 32. Thus $x +y \le 18n/32$ throughout the interval, and so reaction (0’) has probability at most $(18n/64)^2/(n^2/2)$ which is less than 1 / 6. Hence fewer than $2(n/64)/6= n/192$ new B molecules are produced by reaction (0’) over interval I, in expectation, and fewer than n / 175 are produced, with probability $1-\exp (\varTheta (n))$ (here we use a Chernoff upper tail bound). Assuming $b_0 \le 15n/32$, it follows that $b_\mathrm{max}< b_0 + n/175 < n/2$, and so $x + y > n/2$ throughout interval I.

It follows that the total probability of reactions (1’) and (2’) is at least $(n/2)(14n/32)/(n^2/2)= 14/32$ throughout interval I, which means that at least $(14/32)(n/64) > n/148$ B molecules are consumed by these reactions, in expectation, and at least n / 160 are consumed, with probability $1-\exp (\varTheta (n))$, over the course of interval I (here we use a Chernoff lower tail bound). Thus, with probability $1-\exp (\varTheta (n))$, the net change in b is less than $n/175 - n/160 < 0$, and so $b_e < b_0 \le 15n/32$. We note that this upper bound holds with probability $1-\exp (-\varTheta (n))$, which is stronger than in the statement of the lemma.

Lower Bounds on b . Note that $x-y$ is not changed by reaction (0’), and by Lemma 4, it never reaches $(x_0-y_0)/2$ through reactions (1’) and (2’). Therefore, $b_\mathrm{max} \le n/2$, it follows that $x + y \ge n/2$ and hence $x \ge n/4$. We will use this fact throughout.

We first show that even if $b_0=0$, $b_e \ge y_e/292$. Since $b_\mathrm{max} \le n/2$ it follows that reaction (2’) has probability at most $y_\mathrm{max} b_\mathrm{max} / (n^2/2) \le y_\mathrm{max}/n$. Thus reaction (2’) increases y from its minimum value $y_\mathrm{min}$ by at most $y_\mathrm{max}/64$, in expectation, and by at most $y_\mathrm{max}/32$, with high probability, over the course of interval I. Here, the high probability follows from the fact that $y_\mathrm{max} \ge y_0 \ge f_{\gamma } \lg n = \varOmega (\log n)$, and application of a Chernoff tail bound. Thus, $y_e \le y_\mathrm{max} \le y_\mathrm{min} + y_\mathrm{max}/32$ and so

Now suppose that

Thus we also have that $b_\mathrm{max} = \varOmega (\log n)$, by (*). Since $x + y \le n$, reactions (1’) and (2’) together have probability at most $n b_\mathrm{max} /(n^2/2)$, and so these reactions reduce b from its maximum value $b_\mathrm{max}$ by at most $b_\mathrm{max}/ 32$, in expectation, and by at most $b_\mathrm{max}/ 16$, with high probability, over the course of interval I. Here, the high probability follows from the fact that $b_\mathrm{max} = \varOmega (\log n)$, and application of a Chernoff tail bound. Thus, with high probability,

Then,

$$\begin{aligned} \begin{array}{lll} b_e &{} \ge (15/16) b_\mathrm{max} &{}\text {by (***)} \\ &{}> (15/16)(1/16) y_\mathrm{min} &{} \text {by (**)} \\ &{} \ge (15/16)(1/16)(31/32) y_\mathrm{max} &{}\text {by (*)} \\ &{} > y_\mathrm{max}/18 \ge y_e /18. \end{array} \end{aligned}$$

On the other hand, suppose that

Since reaction (0’) has probability at least $x_\mathrm{min} y_\mathrm{min} /(n^2/2) \ge y_\mathrm{min}/(2n)$, reaction (0’) increases b by at least $y_\mathrm{min} /64$, in expectation, and at least $y_\mathrm{min} /128$, with high probability, over the course of interval I. Since reactions (1’) and (2’) together have probability at most $n b_\mathrm{max} / (n^2/2) \le n (y_\mathrm{min}/16) / (n^2/2)$ by (****), we know that together they decrease b by at most $y_\mathrm{min}/512$, in expectation, and at most $y_\mathrm{min}/256$, with high probability, over the course of interval I. Here, the high probability follows from the fact that $y_\mathrm{min} = \varOmega (\log n)$ by (*), and application of a Chernoff tail bound. Thus the net change in b is at least $y_\mathrm{min} /128 - y_\mathrm{min}/256$, with high probability. Also,

$$\begin{aligned} \begin{array}{lll} y_\mathrm{min} /128 - y_\mathrm{min}/256 &{}= y_\mathrm{min}/256 &{} \\ &{}\ge (31/32)(1/256) y_\mathrm{max} &{} \text {by (*)} \\ &{} > y_\mathrm{max}/265. \end{array} \end{aligned}$$

So, $b_e > y_\mathrm{max}/ 265 \ge y_e /265$, even if $b_0 = 0$.

Finally, assume that $b_0 \ge y_0 / 265$. Let $b'_\mathrm{max}$ be the maximum value of b between $b_0$ and $b_\mathrm{min}$ in the course of interval I. By an argument similar to the one used for equation (***), with high probability, we get

Therefore, we have

$$\begin{aligned} \begin{array}{lll} b_\mathrm{min} &{}\ge (15/16) b_0 &{}\text {by (*****)} \\ &{}\ge (15/16) y_0 / 265 \\ &{}\ge (15/16) y_\mathrm{min} /265 \\ &{} > (15/16)(31/32) y_\mathrm{max} /265. &{}\text {by (*)} \end{array} \end{aligned}$$

and so $b > y / 292$ throughout interval I.

1.3 A.3 Adjustments Required for the Proof of Single-B

Here we describe additional adjustments to the proof of correctness and efficiency of the tri-molecular CRN that are needed to account for changes to random variables $\hat{x}$ and $\hat{y}$ due to reactions (0’x) and (0’y). Note that reactions (0’x) and (1’) increase $\hat{x}$ by 1 / 2 and decrease $\hat{y}$ by 1 / 2, while reactions (0’y) and (2’) decrease $\hat{x}$ by 1 / 2 and increase $\hat{y}$ by 1 / 2.

First, in the proof of the upper (n / 2) and lower (y / 292) bounds on b in Lemma 9, we simply adjust the probabilities of a change in $\hat{x}$ or $\hat{y}$ to account for reactions (0’x) and (0’y). (We remark that we are able to provide tighter lower and upper bounds on b with respect to variable y, i.e., $\frac{y}{2\alpha } \le b \le 2\alpha y$, where $\alpha \ge 20$, and $b = \varOmega (\log n)$, for the Single-B CRN - details omitted.) Then, utilizing the lower bound on b, Lemma 11 shows that the ratio of total probability of reactions (0’x) and (1’) to that of reactions (0’y) and (2’) is lower than the ratio of the probability of reaction (1) to that of reaction (2) in the tri-molecular CRN by at most a small constant. Therefore, the analysis of phase 1 of Single-B parallels that of the tri-molecular CRN.

Lemma 11

At any point in the computations, assuming that $\hat{x}-\hat{y} \ge \varDelta /2$, the probability that $\hat{x}-\hat{y}$ increases is at least $1/2 + \varTheta (\varDelta /n)$.

Proof

Let p denote the probability of a success ($\hat{x} - \hat{y}$ increases) and q denote the probability of a failure ($\hat{x}-\hat{y}$ increases). So, given that $x \le n$, and $y/292 < b$, we have that

$$\begin{aligned}&(1)\, \frac{q}{p} = \frac{1/2xy + yb}{1/2xy + xb} \le 1- \frac{(\hat{x}- \hat{y})b}{1/2xy + xb} \le 1 - \frac{(\varDelta /2) b}{x(1/2y + b)} \le 1- \varTheta (\varDelta /n),\\&(2)\, q + p = 1. \end{aligned}$$

It follows from Eqs. 1 and 2 that $p \ge 1/2 + \varTheta (\varDelta /4n)$.

Similarly, we can revise Lemmas 5 and 7 (and their related corollaries) to make the analysis of phases 2 and 3 of Single-B also parallel to those of the tri-molecular CRN–see Lemmas 12 and 13.

Lemma 12

At any point in the computation, if $\hat{y} = n/k$ then the probability that $\hat{y} > 2n/k$ at some subsequent point in the computation is less than $(1 - \varTheta (1))^{n/k}$.

Proof

Let p denote the probability of a success  ($\hat{y}$ decreases) and q denote the probability of a failure ( $\hat{y}$ increases). So, assuming that $x \le n$, $\hat{x}-\hat{y} \ge n-n/4k$, and $y < 292 b$, we can compute the ratio q / p on a reaction event as follows.

$$\begin{aligned} \frac{q}{p} = \frac{1/2xy + yb}{1/2xy + xb} \le 1- \frac{(\hat{x}- \hat{y})b}{1/2xy + xb} \le 1 - \frac{(n - 4n/k) b}{n(1/2y + b)} \le 1- \varTheta (1). \end{aligned}$$

By Lemma 1, we conclude that reaching a configuration where $y > 2n/k$ (which entails an excess of n / k failures to successes) is less than $(1 - \varTheta (1))^{n/k}$.

Lemma 13

At any point in the computation, if $\hat{y} = n/k$ then, assuming that $\hat{y}$ never increases to 2n / k, the probability that $\hat{y}$ decreases to $n/k - r$ within $f(n)>\varTheta (r)$ reaction events is at least $1-exp(-\varTheta (f(n))$.

Proof

The proof is completely parallel to the proof of Lemma 7. We only need to compute the probability of a success ($\hat{y}$ decrease). By Lemma 12, $q/p = 1 - \varTheta (1)$. So, considering $p+q = 1$, it’s straightforward to obtain $p \ge \frac{1}{2} + \varTheta (1)$.

Finally, we employ Lemma 8 to complete the proof of efficiency. Using the upper bound on b, which confirms that $x \ge n/4$ and the lower bound on b, which confirms $b \ge y/292$, we can conclude that the total probability of reactions (0’x), (0’y), (1’), and (2’) is at least some constant fraction of the total probability of reactions (1) and (2) in tri-molecular CRN. Therefore, the total number of interactions in Single-B is at most some constant multiple times the required number of interactions in the tri-molecular CRN.