DNA walker circuits: computational potential, design, and verification

Abstract

Unlike their traditional, silicon counterparts, DNA computers have natural interfaces with both chemical and biological systems. These can be used for a number of applications, including the precise arrangement of matter at the nanoscale and the creation of smart biosensors. Like silicon circuits, DNA strand displacement systems (DSD) can evaluate non-trivial functions. However, these systems can be slow and are susceptible to errors. It has been suggested that localised hybridization reactions could overcome some of these challenges. Localised reactions occur in DNA ‘walker’ systems which were recently shown to be capable of navigating a programmable track tethered to an origami tile. We investigate the computational potential of these systems for evaluating Boolean functions and forming composable circuits. We find that systems of multiple walkers have severely limited potential for parallel circuit evaluation. DNA walkers, like DSDs, are also susceptible to errors. We develop a discrete stochastic model of DNA walker ‘circuits’ based on experimental data, and demonstrate the merit of using probabilistic model checking techniques to analyse their reliability, performance and correctness. This analysis aids in the design of reliable and efficient DNA walker circuits.

Appendix

1.1 Proof of Lemma 2

Proof

Let $G$ be any fork gate that is reachable, otherwise it is redundant. When both the left and right output tracks are unguarded, then either track is accessible simultaneously and thus $G$ is not deterministic. Suppose only one output track of $G$ is guarded. If $G$ is only reachable when the single guard evaluates to false, then that guarded output track is never taken and $G$ is trivial (and therefore deterministic by Lemma 1). Otherwise, if $G$ is reachable only when the guard evaluates to true, then both output tracks from $G$ are accessible simultaneously and, as before, $G$ is not deterministic. Finally, suppose the left and right output tracks of $G$ have the same guard. Since $G$ is reachable, then there are only two possibilities: the common guard evaluates to true (both output tracks are accessible simultaneously), or the common guard evaluates to false, resulting in a deadlock. In either case, $G$ is not deterministic.$\square$

1.2 Proof of Thereom 1

Proof

(if—sufficiency) If $G$ is trivial it is deterministic by Lemma 1. Suppose it is not trivial. Therefore, by assumption, the output tracks have guards which are negations of each other. Thus, there is always exactly one unblocked path from $G$ and it is therefore deterministic.

(only if—necessity) Suppose, by contradiction, that $G$ is deterministic, not trivial, and does not have output track guards that are negations of each other. Since, by assumption, $G$ is not trivial, then by Lemma 2, $G$ cannot be deterministic unless it has distinct guards for its left and right output tracks. Without loss of generality, let these be $G_L$ and $G_R$, respectively ($G_L \not \equiv G_R$). Let $p$ be any path that can reach the gate.

Case 1

(p has no guards in common with an output track of G) Consider any variable assignment where $p$ reaches $G$ and set $G_L = G_R = false$. The gate $G$ is still reachable via $p$ since they do not share a guard. However, reaching $G$ results in a deadlock. Contradiction.

Case 2

( p has a guard or its negation in common with one output track of G) Suppose $p$ contains the guard $G_L$. Consider any variable assignment where $p$ reaches $G$ and therefore $G_L=true$. By assigning $G_R = true$, both output tracks of $G$ are accessible. Contradiction. The case when $p$ contains the guard $G_R$ is symmetric. Similarly, it can be shown that variable assignments exist to ensure $G$ will result in a deadlock, and thus a contradiction, when $p$ contains $\lnot G_L$ or $\lnot G_R$.

Case 3

(p has a guard or its negation in common with both output tracks of G) Since $G$ is not trivial then (i) $p$ is not guarded by both $G_L$ and $\lnot G_R$, and (ii) $p$ is not guarded by both $\lnot G_L$ and $G_R$. By condition (i) and (ii) $p$ must either contain the guards $G_L$ and $G_R$ or the guards $\lnot G_L$ and $\lnot G_R$. When $p$ reaches $G$ and the former is true, then two output tracks are accessible, and when the latter is true, a deadlock occurs. Both cases result in a contradiction.$\square$

1.3 Proof of Theorem 2

Proof

(if—sufficiency) If the non-redundant gate is a sink then any path that reaches it from its left input track cannot be extended via its right input track, and vice versa; it is therefore deterministic. Suppose the gate is not a sink and its left input track is guarded by $G$, the right by $\lnot G$ and, prior to reaching those guards, all paths that can reach the left input must traverse a track guarded by $G$ and all paths that can reach the right must traverse a track guarded by $\lnot G$. There are two cases to consider. Suppose $G$ evaluates to true. Then, no path can reach the right input since, by the assumption, those paths must traverse a track guarded by $\lnot G$ prior to reaching the gate. It follows that all paths that can reach the gate when $G$ evaluates to true must be to the left input. Furthermore, as the right input is guarded by $\lnot G$, those paths can only be extended via the output of the gate. The other case ($G$ evaluates to false) is symmetric. Furthermore, as the guards are negations of each other, they cannot simultaneously evaluate to false and cause a potential deadlock.

(only if—necessity) A non-redundant sink is deterministic, so consider the case when the gate is not a sink. By definition of a join gate that is not a sink, it must have two guarded input tracks. Let $G_L$ and $G_R$ be the guards of the left and right inputs, respectively. First, consider all paths that can reach the left input, guarded by $G_L$. It must simultaneously be true that (i) none of those paths traverse a track guarded by $\lnot G_L$ and (ii) all of those paths traverse a track guarded by $\lnot G_R$. If condition (i) is not satisfied, then there would exist a path that traverses a track guarded by $\lnot G_L$ and, to extend past the join gate, must traverse another guarded by $G_L$. As this is not possible, the path would end in a deadlock and the gate would not be deterministic. If condition (ii) is not satisfied then there would exist some path $p$ that does not traverse a track guarded by $\lnot G_R$, but may possibly traverse a track guarded by $G_R$. In this case, there exists a variable assignment where $G_R$, and all other guards on path $p$, evaluate to true. With such a variable assignment, path $p$ could be extended via the output track or the right input track. Thus, condition (ii) must also be satisfied, as otherwise the gate would not be deterministic. The conditions (and the argument that both are necessary) when considering all paths that can initially reach the right input, guarded by $G_R$, are symmetric.

The sufficiency argument shows the gate is deterministic when $G_L \equiv \lnot G_R$. It remains to show it is not deterministic otherwise. First, consider the consequence when both $G_L$ and $G_R$ evaluate to true. By condition (ii) all paths leading to the left (right) input traverse a track guarded by $\lnot G_R$ ($\lnot G_L$). In this case, no paths can reach the gate. Thus, consider when both $G_L$ and $G_R$ evaluate to false. The conditions permit that paths can reach the gate; however, if any path does it will deadlock as both inputs to the gate are blocked. Thus, for all paths that can reach the gate, it will be deterministic only when $G_L \equiv \lnot G_R$. $\square$