Cyclic Oritatami Systems Cannot Fold Infinite Fractal Curves
CCyclic Oritatami Systems Cannot FoldInfinite Fractal Curves
Yo-Sub Han Hwee KimAugust 14, 2019
Abstract
RNA cotranscriptional folding is the phenomenon in which an RNAtranscript folds upon itself while being synthesized out of a gene. The ori-tatami system (OS) is a computation model of this phenomenon, whichlets its sequence (transcript) of beads (abstract molecules) fold cotran-scriptionally by the interactions between beads according to the bindingruleset. The OS is an useful computational model for predicting and sim-ulating an RNA folding as well as constructing a biological structure. Afractal is an infinite pattern that is self-similar across different scales, andis an important structure in nature. Therefore, the fractal constructionusing self-assembly is one of the most important problems. We focus onthe problem of generating an infinite fractal instead of a partial finite frac-tal, which is much more challenging. We use a cyclic OS, which has aninfinite periodic transcript, to generate an infinite structure. We prove anegative result that it is impossible to make a Koch curve or a Minkowskicurve, both of which are fractals, using a cyclic OS. We then establishsufficient conditions of infinite aperiodic curves that a cyclic OS cannotfold.
Self-assembly is the process where smaller components—usually molecules—autonomously assemble and form a larger complex structure. Self-assemblyplays an important role in constructing biological structures and high poly-mers [16]. One well-known mathematical model of the self-assembly phenomenonis the abstract tile assembly model (aTAM) [17]. Recently, Geary et al. [2] pro-posed a new computation model called the oritatami system (OS) that simulatesthe cotranscriptional self-assembly based on the experimental RNA transcrip-tion called RNA origami [3]. In general, the OS assumes that a sequence ofmolecules is transcribed linearly, and predicts the geometric shape of the au-tonomous folding of the sequence based on the reaction rate of the folding. TheOS consists of a sequence of beads (which is the transcript) and a set of rules forpossible intermolecular reactions between beads. For each bead in the sequence,the system takes a lookahead of a few upcoming beads and determines the bestlocation of the bead that maximizes the number of possible interactions fromthe lookahead. The lookahead represents the reaction rate of the cotranscrip-tional folding and the number of interactions represents the energy level (SeeFigure 2 for the analogy between RNA origami and oritatami system).1 a r X i v : . [ c s . CC ] A ug igure 1: The motivation of the oritatami system. (a) An illustration of anRNA Origami [2], which transcribes an RNA strand that self-assembles. (b) Theproduct of an RNA Origami. (c) Abstraction of the product in the oritatamisystem. beadinteraction a bb b bbdc e ( a, d ) , ( c, e )interaction rules conformation abbcbdbbe ⇒ transcript RNA Origami Oritatami SystemNucleotides BeadsTranscript Sequence of beadsconnected by a lineh-bonds between nucleotides InteractionsCotranscriptional folding rate DelayResulting secondary structure Conformation (a) (b)
Figure 2: (a) Analogy between RNA origami and oritatami system. (b) Visu-alization of oritatami system and its terms.The OS is a computation model based on geometric aspects, and, thus, itis important to design and analyze the OS in both computational and geomet-ric perspectives. From the computational point of view, the OS is proved tobe Turing complete using a cyclic tag system simulator [2]. Designs for othercomputational problems such as binary counting [1] and Boolean formula simu-lation [5] are also established. Researchers also proved a few decision propertiesand proposed various optimization methods for the OS design [4, 6, 11]. Fromthe geometric point of view, Rogers and Seki [12] proved the decidability of geo-metric structure constructions based on the delay. Recently, Masuda et al. [10]proposed how to construct a finite Heighway dragon using a cyclic OS, whichhas a periodic transcript.A fractal is an infinite pattern that is self-similar across different scales,and is an important structure in nature. The construction of fractals is oneof the most important topics in both geometric computation models [7, 8, 9]and experiments [14, 15]. For constructing fractals by the OS, we need a fewassumptions. 2. Since the OS transcribes an RNA single strand, it is natural to considerfractal curves, which are infinite sequences of segments (and points).2. We only consider a deterministic OS (that only folds into a unique confor-mation). Note that it is trivial to make a nondeterministic OS that mayfold into several different structures including a target structure.3. We consider an infinite fractal construction. Masuda et al. [10] proposedhow to construct a finite fractal by implementing a counter and an au-tomaton periodically inside the transcript. This approach can be usedto construct an arbitrary long fractal but not an infinite one, since thecounter is finite.4. Since we need an infinite transcript with a formal finite representation, itis a natural choice to use a cyclic OS that has an infinitely repeated tran-script. Researchers already used cyclic OSs to construct conformationsthat can grow infinitely [1, 2].Although fractals have self-similarity, they do not have repeated structureswith any fixed period. In aTAM, it is known that we can assemble a certaintype of fractals infinitely even with limited tile types [9]. In OS, we claim thatunder some reasonable assumptions, it is impossible to fold some infinite fractalswith a cyclic OS. We first define the folding of the curve by an OS as mappingsegments and points of the curve to sets of points on the OS triangular lattice.Before presenting our main contributions on the impossibility of OS folding,we start with two well-known fractal curves (Koch and Minkowski curves) withrespect to the OS folding and provide a brief idea on the impossibility results.We, in particular, prove that regardless of the delay and the period, it is im-possible to fold Koch and Minkowski curves in Sections 3 and 4, respectively.Then we generalize this impossibility to infinite aperiodic curves and establishsufficient conditions together with main contributions in Section 5.We then examine two well-known fractal curves (Koch and Minkowski curves),and prove that regardless of the delay and the period, it is impossible to foldsuch curves under our assumptions. We expand the result to infinite aperiodiccurves and establish sufficient conditions to prove impossibility of the folding.
Let w = a a · · · a n be a string over Σ for some integer n and bead types a , . . . , a n ∈ Σ. The length | w | of w is n . For two indices i, j with 1 ≤ i ≤ j ≤ n , we let w [ i, j ] be the substring a i a i +1 · · · a j − a j ; we use w [ i ] to denote w [ i, i ]. We use w n to denote the concatenation of n copies of w .Oritatami systems operate on the triangular lattice Λ o with the vertex set V and the edge set E . A configuration is a triple ( P, w, H ) of a directed path P in Λ o , w ∈ Σ ∗ ∪ Σ ω , and a set H ⊆ (cid:8) ( i, j ) (cid:12)(cid:12) ≤ i, i + 2 ≤ j, { P [ i ] , P [ j ] } ∈ E (cid:9) of interactions. This is to be interpreted as the sequence w being folded whileits i -th bead w [ i ] is placed on the i -th point P [ i ] along the path and thereis an interaction between the i -th and j -th beads if and only if ( i, j ) ∈ H .Configurations ( P , w , H ) and ( P , w , H ) are congruent provided w = w , H = H , and P can be transformed into P by a combination of a translation,a reflection, and rotations by 60 degrees. The set of all configurations congruent3o a configuration ( P, w, H ) is called the conformation of the configuration anddenoted by C = [( P, w, H )]. We call w a primary structure of C .A ruleset H ⊆ Σ × Σ is a symmetric relation specifying between which beadtypes can form an interaction. An interaction ( i, j ) ∈ H is valid with respectto H , or simply H -valid , if ( w [ i ] , w [ j ]) ∈ H . We say that a conformation C is H -valid if all of its interactions are H -valid. For an integer α ≥ C is ofarity α if the maximum number of interactions per bead is α , that is, if for any k ≥ (cid:12)(cid:12) { i | ( i, k ) ∈ H } (cid:12)(cid:12) + (cid:12)(cid:12) { j | ( k, j ) ∈ H } (cid:12)(cid:12) ≤ α and this inequality holds as anequation for some k . By C ≤ α , we denote the set of all conformations of arity atmost α .Oritatami systems grow conformations by elongating them under their ownruleset. For a finite conformation C , we say that a finite conformation C isan elongation of C by a bead b ∈ Σ under a ruleset H , written as C H → b C ,if there exists a configuration ( P, w, H ) of C such that C includes a config-uration ( P · p, w · b, H ∪ H ), where p ∈ V is a point not in P and H ⊆ (cid:8) ( i, | P | +1) (cid:12)(cid:12) ≤ i ≤ | P | − , { P [ i ] , p } ∈ E, ( w [ i ] , b ) ∈ H (cid:9) . This operation is re-cursively extended to the elongation by a finite sequence of beads as follows:For any conformation C , C H → ∗ λ C ; and for a finite sequence of beads w anda bead b , a conformation C is elongated to a conformation C by w · b , writ-ten as C H → ∗ w · b C , if there is a conformation C that satisfies C H → ∗ w C and C → b C .An oritatami system (OS) is a 6-tuple Ξ = (Σ , w, H , δ, α, C σ = [( P σ , w σ , H σ )]),where H is a ruleset , δ ≥ delay , and C σ is an H -valid initial seed conforma-tion of arity at most α , upon which its transcript w ∈ Σ ∗ ∪ Σ ω is to be folded bystabilizing beads of w one at a time and minimize energy collaboratively with thesucceeding δ − C = [( P, w, H )]is U ( C ) = −| H | ; namely, the more interactions a conformation has, the morestable it becomes. The set F (Ξ) of conformations foldable by this system isrecursively defined as follows: The seed C σ is in F (Ξ); and provided that anelongation C i of C σ by the prefix w [1 : i ] be foldable (i.e., C = C σ ), its furtherelongation C i +1 by the next bead w [ i +1] is foldable if C i +1 ∈ argmin C ∈C ≤ α s.t. C i H → w [ i +1] C min n U ( C ) (cid:12)(cid:12)(cid:12) C H → ∗ w [ i +2: i + k ] C , k ≤ δ, C ∈ C ≤ α o . (1)Once we have C i +1 , we say that the bead w [ i +1] and its interactions are stabi-lized according to C i +1 . A conformation foldable by Ξ is terminal if none of itselongations is foldable by Ξ. An OS is deterministic if, for all i , there exists atmost one C i +1 that satisfies (1). Namely, a deterministic OS folds into a uniqueterminal conformation. An OS is cyclic if its transcript w = w ωo is repetition ofa string w o . We say that the OS has a period | w o | .Figure 3 illustrates an example of an OS with delay 3, arity 4, ruleset { ( a, a ) } and transcript w = aaaaaaaaa ; in (a), the system tries to stabilize the firstbead a of the transcript, and the elongation P gives 2 interactions, while theelongation P gives 4 interactions, which is the most stable one. Thus, thefirst bead a is stabilized according to the location in P . In (b) and (c), P isthe most stable elongation and a ’s are stabilized according to P . As a result,the terminal conformation is given as in (d). Note that the system grows the4 a P P P (a) (b) P P (c) P (d) ⇒ Figure 3: An example OS with delay 3 and arity 4. The seed is colored in red,elongations are colored in blue, and the stabilized beads and interactions arecolored in black.terminal conformation straight without external interactions, and we can use w = ( aaaaaa ) ω to fold an infinite periodic conformation. This example is calleda glider [1].The bead stabilization in OS is a local optimization of finding the bestposition of the bead using the next δ beads. Thus, the stabilization of a bead w [ i ]in a delay- δ OS is not affected by any bead whose distance from w [ i −
1] isgreater than δ + 1. On the triangular lattice, we can draw a hexagonal borderof radius δ + 1 from w [ i −
1] to identify the set of points that may affect thestabilization of w [ i ]. While stabilizing a bead w [ i ], we define the event horizon of w [ i ] to be a partial conformation within this hexagon. Namely, the eventhorizon is the context used to stabilize w [ i ]. Thus, if two beads w [ i ] and w [ j ]have the same event horizon, then w [ i ] and w [ j ] are stabilized congruently (SeeFigure 4). We define an event horizon for a partial transcript w [ i, j ] to be theunion of event horizons of beads in w [ i, j ] while stabilizing w [ i ]. w [ i ] w [ j ] Figure 4: Two same event horizons when δ = 2 and we have two bead types(black and white circles). The current bead, pointed by an arrow, is stabilizedin the same way in both event horizons.An L-system is a parallel rewriting system and its recursive nature makesthe system easy to describe fractal-like structures [13]. An L-system is definedas G = ( V, C, ω, P ), where • V is the set of variables that can be replaced by production rules, • C is the set of constants that do not get replaced, • ω ∈ ( V ∪ C ) ∗ is the axiom, the initial string, and5 P ⊆ V × ( V ∪ C ) ∗ is the set of production rules defining rewriting ofvariables.The system starts with ω , and as many rules as possible are applied simulta-neously for each iteration. With graphical semantics on variables and constants,the L-system is often used to represent self-similar fractals. In this paper, weassume that curves are represented by strings, whose characters represent turnsand unit segments. Then, an infinite curve is periodic if there exists a periodicstring representation with a fixed finite period, and is aperiodic otherwise. Notethat all fractal infinite curves are aperiodic. We start with one example of infinite fractal curves—the Koch curve. The Kochcurve can be constructed by starting with a segment, then recursively alteringeach line segment as follows:1. Divide the line segment into three segments of equal length.2. Draw an equilateral triangle that has the middle segment from step 1 asits base and points outward.3. Remove the line segment that is the base of the triangle from step 2.Using the L-system, the Koch curve can be encoded as follows: • Variable: F • Constants: + , −• Axiom: F • Production Rule: F → F + F − F + F ,where F denotes a segment, − denotes 120 ◦ right turn and + denotes 60 ◦ leftturn. Figure 5 illustrates the Koch curve after three iterations.Figure 5: The Koch curve after three iterationsFor this specific curve, we first make a few assumptions about the curverepresentation by a cyclic OS and the OS itself. We prove that it is impossibleto draw an infinite Koch curve using a cyclic OS under our assumptions.Let a shape be a set of points on the triangular lattice Λ o , whose grid graphis connected. A curve can be represented as a sequence of alternating points andsegments. We say that a sequence of shapes represents a curve if there existsone-to-one correspondence between shapes and alternating points and segments,and a point and a segment should be adjacent on the curve if and only if twoshapes corresponding to them are adjacent. We formally define the drawing ofthe curve by a deterministic OS as follows:6 efinition 1. Given a (possibly infinite) curve on the plane and the (possiblyinfinite) sequence ( S k ) of shapes that represents the curve, we say that a de-terministic OS draws the curve if the following condition holds: There existsa (possibly infinite) sequence ( i k ) of indices that corresponds to the sequence ofshapes, where, for all k ’s, there exists a partial configuration for w [ i k − +1 : i k ] that folds within S k . We say that the OS covers S k with the partial tran-script w [ i k − +1 : i k ] . Figure 6 shows two examples of curve drawing by an OS. Here, the tar-get curve is represented by three shapes ( S , S , S ). By Definition 1, an OSdraws the curve in Figure 6 (a) but does not in Figure 6 (b) since the partialconfiguration for w [3 : 11] is not within S . Note that shapes limit paths ofconformations, and it is not necessary to fill all points in the shape with theconformation.From the design perspective, it is crucial to assume this locality of partialconfiguration. OS designs are usually modular [1, 2, 5, 10]—a partial transcriptis folded locally under a controlled context, and we connect these partial tran-scripts to perform complex computations. Especially, for an infinite transcript,it becomes almost impossible to remove unintended interferences without thislocality. Previous cyclic OSs such as a binary counter [1] or a cyclic tag sys-tem [2] follow this assumption. Remark that the drawing in Definition 1 isgeneral in the sense that it does not restrict the OS to be infinite or cyclic, andthe curve to be infinite. ⇒ ⇒ S S S S S S b b b ⇒ ⇒ S S S S S S b b b (a) (b) Figure 6: (a) OS draws the target curve and (b) OS does not draw the targetcurve by Definition 1We assume the followings to draw the Koch curve:1. The Koch curve consists of an infinite sequence of alternating points andsegments on the triangular lattice Λ c , which is with vertical rows with unittriangles pointing left and right. We use a hexagon S p of side length d to represent a point on the curve. For a segment on the curve, we usea shape S l of d points ( l + 1 rows in total) and d + 1 points ( l rows intotal) in alternative positions which are orthogonal to the direction of thesegment (See Figure 7 (a)). The OS starts with covering the first S p , anddenote the i th S p ( S l ) by S p [ i ] ( S l [ i ]).2. We use constant numbers of beads for segments or points— p p beads in S p ,and p l beads in S l . This assumption is reasonable in the modular designof the OS.3. The period p of the OS is p = p p + p l .7 = 2 2 l = 6 p l = 9 p p = 11 (a) (b) S p S l Figure 7: (a) Shapes used to draw the Koch curve (b) An example of an OSwith p l = 9 and p p = 11Figure 7 shows an example of shapes that can be used to draw the Kochcurve, and a part of an OS that draws the curve, following the above assump-tions. In Figure 7 (a), S p ’s are drawn in red, and S l ’s are drawn in blue, where d = 2 and l = 3. In Figure 7 (b), from assumption (2), the number of beadsfor segments or points are constant, even if there are different paths in different S p ’s or S l ’s. For example, paths in S p use 11 beads and paths in S l use 9 beads.Under these assumptions, we claim the following theorem. Theorem 1.
There is no deterministic OS that can draw the Koch curve.Proof.
Assume that there exists a deterministic cyclic OS Ξ with delay δ thatdraws the Koch curve. First, we assume that δ < d + 3 l + 2, which is one lessthan the distance between two S p ’s on Λ o —apart by two unit distances on Λ c . Ifthe delay δ has an upper bound, we denote the event horizon for the maximumdelay as the “maximum” event horizon, and omit the term maximum if thecontext is clear. For convenience, we use S pl [ i ] to denote the shape resultedfrom connecting S p [ i ] and S l [ i ].We consider bead stabilization in S pl [ i ]. We have an event horizon E ( i ) thatis used to fold the conformation in S pl [ i ]. Due to the delay upper bound, all S l ’s that overlap with E ( i ) are at most three unit distances apart from S p [ i ],and all S p ’s that overlap with E ( i ) are at most two unit distances apart from S p [ i ] (See Figure 8).Since the Koch curve does not touch itself, if a point in Λ c is adjacent to twosegments of the curve, there is no segment other than two that are adjacent tothe point. Moreover, due to the self-similarity of the curve, the same statementholds for any scale of the power of 3—for a point q in Λ c , if there exist two8 p [ i ] S l [ i ] maximum event horizon ofradius 3 d + 3 l + 2(a) (b) Figure 8: (a) The boundary of a event horizon is represented by a red hexagon.(b) The boundary of the event horizon E ( i ) for beads in S pl [ i ] is represented bya red polygon. S l ’s and S p ’s that overlap with E ( i ) are depicted.points q and q on the curve that are 3 n unit distances straight away from q ,then there is no segment within 3 n unit distances from q , other than segmentson the curve from q and q (See Figure 9). q q q Figure 9: An example of q , q and q . Points q and q are 3 unit distancesstraight away from q . Aside from curves from q and q , there is no segmentwithin 3 unit distances from q (depicted by a dotted hexagon).We have nine different cases for the beads within S p ’s and S l ’s that overlapwith E ( i ) as shown in Figure 10. In each case, we first transcribe a sequence ofblue shapes (if they exist), and then a sequence of red shapes. Shapes S p [ i ] and S l [ i ] are distinguished by points within the shape. The thick hexagons represent q ’s of distance 3. We can observe that in all cases, all S p ’s and S l ’s are at mostthree unit distances apart from one of q ’s, and they are empty except the redand blue shapes. Moreover, in all cases, all beads within E ( i ) are from S p [ i − S l [ i − S l [ i ]. In other words, apartial conformation in S pl [ i ] is dependent to beads in the sequence of shapesfrom S p [ i −
4] to S l [ i − p + p , we have exactly the same partialtranscripts that fold within all S pl ’s, and, therefore, having the same path yieldsthe same partial conformation. The upper bound for the number of possiblepaths within S p ( S l ) is 5 p (5 p ). We have beads from S p [ i −
4] to S l [ i −
1] thatdetermine the partial conformation in S pl [ i ]. Thus, we have i and j such that 1 ≤ i < j ≤ p l +4 p p , and S pl [ i ] and S pl [ j ] have exactly the same conformation.Moreover, since beads from S p [ i −
4] to S l [ i −
1] are consecutively transcribed, S pl [ i ] and S pl [ j ] result in a periodic sequence of segment turns of length j − i (See Figure 11). Since the Koch curve is aperiodic, we know that the firstassumption δ < l + 3 d + 2 is wrong. 10 ) 2) 3)4) 5) 6)7) 8) 9) Figure 10: Nine cases of the beads in the union of event horizons.Now we assume that 3 l + 3 d + 2 ≤ δ < l + 12 d + 11, which is the distancebetween two S p ’s apart by 8 unit distances in the lattice for the Koch curve.Suppose that we want to stabilize beads in S pl [ i ]. The event horizon for beadsin S pl [ i ] covers only S p ’s and S l ’s which are within 8 unit distances from S p [ i ]in the lattice for the Koch curve. Similar to the previous case, we can findat most two q i ’s of distance 9, and all S p ’s and S i ’s are empty except thosewho represent the curve from q i ’s. Due to the length constraint, there can beat most 32 S p ’s and S i ’s that overlap with the event horizon. Thus, among S pl [1] to S pl [5 p l +32 p p + 1], there should exist S pl [ i ] and S pl [ j > i ] that haveexactly the same event horizon for points within, and we know that the secondassumption is also wrong. Similarly, we can expand the proof for arbitrarily large δ . Therefore, we know that for any given δ , there is no delay- δ deterministicOS that can draw the Koch curve. 11 · · · stabilizing S l [2] S l [3] S l [ i ] S l [ j ] · · · Figure 11: A series of event horizons (in dotted hexagons) for the first bead of S l [ i ]. Shapes S l ’s and S p ’s are simplified for better readability. There exists1 ≤ i < j ≤ p l +4 p p +1 such that they have exactly same event horizon forbeads within S p [ i ] ( S l [ i ]) and S p [ j ] ( S l [ j ]).Note that we can remove the third assumption about the OS period. If aperiod p is not p p + p l , beads in S pl [ i ] and S pl [ i + pj ] are exactly the same. Thus,for instance, if we assume that δ < l +3 d +2, among S pl [1] to S pl [ p · p l +4 p p +1],there should exist S pl [ i ] and S pl [ j > i ] that have exactly the same event horizonand produce the same conformation. We study another example of infinite fractal curves—the Minkowski curve. TheMinkowski curve starts from a segment, then recursively alternates each linesegment as follows:1. Divide the line segment into four segments (we call these segment 1 to 4from the start) of equal length.2. Draw a square with segment 2 as a side to the left of the original segment,and the other square with segment 3 as a side to the right.3. Remove segments 2 and 3.Using the L-system, the Minkowski curve can be encoded as follows: • Variable: F • Constants: + , −• Axiom: F • Production Rule: F → F + F − F − F F + F + F − F ,where F denotes a segment, − denotes 90 ◦ right turn and + denotes 90 ◦ leftturn. Figure 12 illustrates the Minkowski curve after three iterations.For the Minkowski curve case, since the OS works on the triangular lattice,we assume that the Minkowski curve is slanted to fit into the triangular lattice—a square in the square lattice is mapped into a rhombus in the triangular lattice.We call the lattice the rhombus lattice. We assume the followings for drawingthe Minkowski curve: 12igure 12: The Minkowski curve after three iterations1. The (tilted) Minkowski curve consists of an infinite sequence of alternatingpoints and segments on the rhombus lattice Λ r . We use a parallelogram S l of width d and length l to represent a segment on the curve, and a rhom-bus S p of side length d to represents a point on the curve (See Figure 13).The OS starts with covering the first S p , and denote the i th S p ( S l ) by S p [ i ] ( S l [ i ]).For convenience, we set up a coordinate ( x, y ) for the points of the rhombuslattice based on an arbitrary origin, where the unit x vector heads rightand the unit y vector heads upper right. Based on the coordinate, we use S l ( x, y, x , y ) to represent an S l that starts from the point ( x, y ) and thedirection is given by the vector ( x , y ). We also use S p ( x, y ) to representan S p at the point ( x, y ) (See Figure 13). d = 2 l = 5 y xP p (0 , P p (1 , P l (0 , , ,
0) = P l (1 , , − , Figure 13: Shapes used to draw the Minkowski curve, and their aliases2. We use constant numbers of beads for a connecting line and a point— p p beads for S p , and p l beads for S l .3. The OS period p is p = p p + p l .Under these assumptions, we claim the following theorem.13 heorem 2. There is no deterministic OS that can draw the tilted Minkowskicurve.Proof.
Similar to the proof for Theorem 1, we fist assume that δ < l + 3 d + 5.Suppose we want to transcribe beads in S p [ i ] = S p ( x, y ) and S l [ i ] = S l ( x, y, , x, y ). We have three casesfor the previous segment:1. S l [ i −
1] = S l ( x, y, − , S l [ i −
1] = S l ( x, y, , S l [ i −
1] = S l ( x, y, , − E ( i ) for all beadsin S pl [ i ] as in Figure 14: S p [ i ] = S p ( x, y ) S l [ i ] = S l ( x, y, , l + 3 d + 5 Figure 14: The event horizon E ( i ) for beads in S pl [ i ] is drawn in a red polygon.We observe the property of the curve based on self-similarity. For a given n ,let q be the starting point of a periodic substructure of length 8 n , and q bethe ending point. Points q and q should be 4 n distance away. Now, supposewe draw the square of size 2 · n with the center q . Then, the point q thatappears first in the square (including the boundary) is at most 7 · n segmentsaway from q (See Figure 15 (a)). Moreover, the point q that appears last inthe square is at most 8 · n segments away from q (See Figure 15 (b)).We have eleven different cases for the beads within S p ’s and S l ’s that overlapwith E ( i ) (See Figure 16). In each case, shapes S p [ i ] and S l [ i ] are distinguishedby points within the shape. The thick rhombi represent q ’s for n = 1. Wecan observe that in all cases, the union of event horizons is covered by at mosttwo squares with different q ’s, whose shapes range from S p [ i −
5] to S p [ i +6].14 q q q (a) (b) q q Figure 15: Properties of the curve when n = 1. (a) q is 56 segments away from q . (b) q is 64 segments away from q .According to the observation, we can say that a partial conformation in S pl [ i ] isdependent to beads in the sequence of shapes from S p [ i −
61] to S l [ i − i and j such that 1 ≤ i < j ≤ p l +61 p p and S pl [ i ] and S pl [ j ]have exactly the same conformation. Since this results in a periodic sequenceof segment turns, our assumption δ < l + 3 d + 5 is wrong.Now we assume that 3 l + 3 d + 5 ≤ δ < l + 15 d + 29. Suppose that we wantto stabilize beads in S pl [ i ]. The event horizon covers only S p ’s and S l ’s that arewithin the square of size 32 unit distances with the center S p [ i ]. Similar to theprevious case, we can find at most two q ’s with n = 2 where the correspondingsquares completely cover the event horizon. The earliest S p that can be such q is S p [ i − S pl [ i ] is dependent tobeads in the sequence of shapes from S p [ i − S l [ i − i and j such that 1 ≤ i < j ≤ p l +512 p p and S pl [ i ] and S pl [ j ] have exactly thesame conformation. In a similar way, we can extend the proof for arbitrarilylarge δ . Similar to the Koch curve case, we can also remove the assumptionabout the OS period. We expand the results from Sections 3 and 4 to impossibility of infinite aperiodiccurves, which includes fractals. Construction of infinite periodic curves using acyclic OS seems to be reasonable if we can design a partial OS that folds oneperiod of the curve, and we have one running example—the glider in Section 2.On the other hand, for infinite aperiodic curves, we propose sufficient conditionsthat makes curves impossible to fold. We make the following assumptions: • The curve is on an arbitrary lattice Λ, and each point (segment) in Λ ismapped to a shape S p ( S l ) in the triangular lattice Λ o . • The OS uses p p ( p l ) beads for S p ( S l ). We say p pl = p p + p l . • The OS has the period of p o . • The curve is represented by an infinite alternation of S p and S l , startingfrom S p [1]. For convenience, we refer to the union of S p [ i ] and S l [ i ] as S pl [ i ].Let δ ( n ) be an upper bound of the delay δ , dependent to a given integer n . Wepropose the condition that curves are not foldable when δ ≤ δ ( n ), and expand15 ) 2) 3)4) 5) 6)7) 8) 9)10) 11) Figure 16: Eleven cases of the beads in the union of event horizons.the result to all possible delays. Suppose we want to stabilize beads in S pl [ i ].Then, we have the maximum event horizon E ( i, n ) for beads in S pl [ i ], which isthe union of all event horizons of radius δ ( n ) + 1 whose centers are points in S pl [ i ] and points neighboring S p [ i ] in S l ’s adjacent to S p [ i ]. Now, for each i ,we have S p [ r i,n ] that appears first in E ( i, n ). Let D i,n = max ≤ j ≤ i ( j − r j,n )to be the maximum difference between j and r j,n for all j ≤ i . Then, it takes1 + gcd ( p o , p pl ) · D i,n p pl to have exactly the same conformation for the previous D i,n beads, which results in the same maximum event horizon and the sameconformation for two shapes S pl [ j ] and S pl [ k > j ]. After S pl [ j ] and S pl [ k ], it isassured that previous D i,n beads have exactly the same conformation for theconsecutively following shapes, and these shapes fold exactly the same. Since D i,n is dependent on i , we have the following theorem. Theorem 3.
If there exists i such that gcd ( p o , p pl ) · D i,n p pl ≤ i , then it is mpossible to draw a given infinite aperiodic curve with a cyclic OS whose delayis less than or equal to δ ( n ) and period is p o . In practice, the delay of the OS is bounded by the transcript length. If weconsider a cyclic OS that has an infinite transcript, the delay can be arbitrarilylarge. We extend Theorem 3 for arbitrarily large delays and obtain the followingstatement.
Theorem 4.
Suppose there exists a function δ ( n ) , where for all n ≥ , thereexists i such that gcd ( p o , p pl ) · D i,n p pl ≤ i . Then, it is impossible to draw agiven aperiodic infinite curve with a cyclic OS whose period is p o . If D i,n is dependent only on n and not i , then we may use an arbitrarilylarge i to satisfy the conditions for all n ’s regardless of gcd ( p o , p pl ). When wehave such a case, the following statement holds. Theorem 5.
If there exists a function δ ( n ) such that, for all n , D i,n is inde-pendent of i , then it is impossible to draw a given infinite aperiodic curve witha cyclic OS regardless of the delay and the period of the OS. Note that both in the proofs for Theorems 1 and 2, we have δ ( n ) that makes D i,n independent of i , which is the Theorem 5 case. The oritatami system (OS) is a computational model inspired by RNA cotran-scriptional folding, where an RNA transcript folds upon itself while synthesizedout of a gene. Since the OS is a geometric computation model, it is naturalto consider the problem of constructing fractal curves using the OS. We haveformally defined the drawing of the curve by an OS. Then we have proved thatit is impossible to draw two infinite fractal curves (Koch curve and Minkowskicurve) by a cyclic OS. Moreover, we have proposed sufficient conditions thatmake the folding of general infinite curves impossible. Our conjecture is thatall fractal curves made by edge replacements are not foldable. However, wecannot directly apply the same approach to all curves, especially to curves thattouch themselves such as Heighway dragons. Thus it is open to develop a newapproach for these self-touching curves.
Acknowledgements
Han was supported by the Basic Science Research Program through NRF fundedby MEST (2015R1D1A1A01060097). This work is partially supported by NIHR01GM109459, and by NSF’s CCF-1526485 and DMS-1800443. This researchwas also partially supported by the Southeast Center for Mathematics and Bi-ology, an NSF-Simons Research Center for Mathematics of Complex BiologicalSystems, under National Science Foundation Grant No. DMS-1764406 and Si-mons Foundation Grant No. 594594. 17 eferences [1] C. Geary, P. Meunier, N. Schabanel, and S. Seki. Efficient universal com-putation by greedy molecular folding.
CoRR , abs/1508.00510, 2015.[2] C. Geary, P. Meunier, N. Schabanel, and S. Seki. Programmingbiomolecules that fold greedily during transcription. In
Proceedings of the41st International Symposium on Mathematical Foundations of ComputerScience , pages 43:1–43:14, 2016.[3] C. Geary, P. W. K. Rothemund, and E. S. Andersen. A single-strandedarchitecture for cotranscriptional folding of RNA nanostructures.
Science ,345:799–804, 2014.[4] Y. Han and H. Kim. Ruleset optimization on isomorphic oritatami systems.In
Proceedings of the 23rd International Conference on DNA Computingand Molecular Programming , pages 33–45, 2017.[5] Y. Han, H. Kim, M. Ota, and S. Seki. Nondeterministic seedless oritatamisystems and hardness of testing their equivalence.
Natural Computing ,17(1):67–79, 2018.[6] Y. Han, H. Kim, T. A. Rogers, and S. Seki. Self-attraction removal fromoritatami systems. In
Proceedings of the 19th International Conference onDescriptional Complexity of Formal Systems , pages 164–176, 2017.[7] J. Hendricks, M. Olsen, M. J. Patitz, T. A. Rogers, and H. Thomas. Hierar-chical self-assembly of fractals with signal-passing tiles.
Natural Computing ,17(1):47–65, 2018.[8] J. Hendricks and J. Opseth. Self-assembly of 4-sided fractals in the two-handed tile assembly model. In
Proceedings of the 16th International Con-ference on Unconventional Computation and Natural Computation , pages113–128, 2017.[9] J. I. Lathrop, J. H. Lutz, and S. M. Summers. Strict self-assembly ofdiscrete sierpinski triangles.
Theoretical Computer Science , 410:384–405,2009.[10] Y. Masuda, S. Seki, and Y. Ubukata. Parametrizing string assemblingsystems. In
Proceedings of the 23rd International Conference on Imple-mentation and Applications of Automata (To appear) , 2018.[11] M. Ota and S. Seki. Rule set design problems for oritatami systems.
The-oretical Computer Science , 671:26–35, 2017.[12] T. A. Rogers and S. Seki. Oritatami system; a survey and the impossibilityof simple simulation at small delays.
Fundamenta Informaticae , 154(1-4):359–372, 2017.[13] G. Rozenburg and A. Salomaa.
The mathematical theory of L systems .Academic Pres, 1980.[14] S. Tesoro and S. E. Ahnert. Nondeterministic self-assembly of two tile typeson a lattice.
Physical Review E , 93:042412, 2016.1815] G. Tikhomirov, P. Petersen, and L. Qian. Fractal assembly of micrometre-scale dna origami arrays with arbitrary patterns.
Nature , 552(7683):67–71,2017.[16] G. M. Whitesides and M. Boncheva. Beyond molecules: Self-assemblyof mesoscopic and macroscopic components.
Proceedings of the NationalAcademy of Sciences of the United States of America , 99(8):4769–4774,2002.[17] E. Winfree.