Nine Test Tubes Generate any RE Language. C. Ferretti, G. Mauri, C. Zandron. Universita di Milano - ITALY. Abstract

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1 Nine Test Tubes Generate any RE Language C. Ferretti, G. Mauri, C. Zandron Dipartimento di Scienze dell'informazione via Comelico 39, Milano Universita di Milano - ITALY Abstract In this paper we prove that any recursively enumerable language can be generated using a distributed splicing system with a xed number of test tubes. This improves a recent result by Csuhaj-Varju, Kari, Paun, proving computational completeness only for a system with a number of tubes depending on the cardinality of the used alphabet. 1 Introduction The family of recursively enumerable languages, RE for short, marks in the usual Chomsky hierarchy for formal languages the computational power equal to that of Turing machines. That's why this family is a benchmark for studying the computational power of new models for formal languages and/or for computation itself. This is the case with the models of DNA computation recently brought to attention [1,2] as an alternative machinery to the usual silicon based computers. To compare these models to RE we rst need to map them to a suitable formal language system.one formal model apt to this domain was already suggested in 1987 by Head [4]. It is called splicing system model, and it describes one specic DNA transformation, the one operated by restriction enzymes. They cut DNA sequences at the occurrence of specic subsequences, and the thus created halves can successively rejoin with others to create new complete molecules. The model we consider here has been dened through a few modications of the original splicing model. As we will see when giving the formal denitions, it considers a set of terminals, as the original model, but also a set of nonterminals. Also, it considers the strings-molecules interacting in groups assigned to dierent test tubes, and to be from time to time redistributed among the dierent tubes according to ltering rules, as dened in [3]. So we have a system generating to ferretti@dsi.unimi.it. This work has been supported by the Italian Ministry of University (MURST), under project 40% \Algorithms and Information Structures", and by the National Research Council. 1

2 strings by interactions at two levels: inside the tubes among strings, and among the tubes in the redistribution of the sets of strings. This paper improves a result given in a preceding paper [5], where it was proved how to reach the power of RE with a xed number of test tubes, instead of a number dependent on the number of symbols contained in the alphabet as was in [3]. In [5], the number of necessary tubes was 10, while here we show that nine tubes suce. We do this by designing a kind of simple encoding/decoding sub-procedure based only on splicing. This result opens way to considerations related to the practical feasibility of this model for DNA computations and, perhaps even more important, to the eectiveness of new general purpose algorithms natively dened for splicing systems. We will diuse on these issues in the closing section of the paper. 2 Basic Denitions As usual, V is the set of all (nite length) strings over a nite alphabet V. The empty string is denoted by. The families of recursively enumerable languages and nite languages are denoted by RE and F IN, respectively. We now introduce the denitions of splicing system, and of distributed splicing system. A Head splicing system (or H system) is a triple H = (V; A; R), where V is the alphabet of H, A V is the set of axioms, and R is the set of splicing rules, with R V #V $V #V ($; # are special symbols not in V ). For x; y; z; w 2 V and r = u 1 #u 2 $u 3 #u 4 in R, we dene (x, y) `r (z, w) if and only if x = x 1 u 1 u 2 x 2, y = y 1 u 3 u 4 y 2, and z = x 1 u 1 u 4 y 2, w = y 1 u 3 u 2 x 2, for some x 1 ; x 2 ; y 1 ; y 2 2 V. For a H system H = (V; A; R) and a language L V, we write (L) = fz 2 V j(x; y) `r (z; w) or (x; y) `r (w; z); for some x; y 2 L; r 2 Rg; and dene (L) = [ i0 i (L) where 0 (L) = L i+1 (L) = i (L) [ ( i (L)) for i 0: A H system is meant to operate starting from the set of strings A, and then generate new strings iterating the splicing step `r on, them and on the strings generated during this process. The language generated in this way is (A). 2

3 A test tube system, T T for short, is a construct? = (V; (A 1 ; R 1 ; V 1 ); : : : ; (A n ; R n ; V n )) where each A i V ; R i V #V $V #V, and V i V, for 1 i n. V i is called the selector of tube i. Each triple (A i ; R i ; V i ), also called tube, operates individually in the same way as a H system (V; A i ; R i ). According to the denition of H systems, they would generate the language denoted by i (A i), but we will see that in a TT they interact among them, accepting from the others the strings belonging to V i. The set B of strings outside any language V is dened as follows B = V? Each tube i in the system starts containing only the strings of A i. One processing step (`)') of the system, moves it from the conguration (L 1 ; : : : ; L n ), where each tube i contains the strings of L i, to a conguration (L 0 1 ; : : : ; L0 n ), according to the following denition (L 1 ; : : : ; L n ) ) (L 0 1 ; : : :; L0 ) i n n[? L 0 i = [ j (L j ) \ V i ( i (L i ) \ B)) j=1 n[ i=1 V i for each i; 1 i n Finally, we state that the language generated by a T T? is the set of words appearing in the tube 1 at any processing step, when starting from the conguration (A 1 ; : : :; A n ): () is the reexive and transitive closure of the relation )) L(?) = fw 2 V jw 2 L 1 for some (A 1 ; : : : ; A n ) ) (L 1 ; : : : ; L n )g T T n (F 1 ; F 2 ) = fl(?)g such that? is a splicing system with at most n tubes, each with set of axioms from F 1 and set of rules from F 2. The set of languages generated using any number of tubes is dened by T T (F 1 ; F 2 ) = [ n1 T T n (F 1 ; F 2 ) i 3 Test Tube Systems and RE Languages In this section we prove in details our main result. Theorem 1 T T 9 (F IN; F IN) = T T (F IN; F IN) = T T (F 1 ; F 2 ) = RE for all families F 1 ; F 2 such that REG F i RE; i = 1; 2. 3

4 Proof. The inclusions T T 9 (F IN; F IN) T T (F IN; F IN) T T (F 1 ; F 2 ) are obvious. The inclusion T T 9 (F IN; F IN) RE is obvious from the Turing/Church thesis. Hence, it is sucient to prove that RE T T 9 (F IN; F IN). Take a type-0 Chomsky grammar G = (N; T; S; P ). Denote U = N [ T and construct the Test Tubes System? = (V; (A 1 ; R 1 ; V 1 ); (A 2 ; R 2 ; V 2 ); (A 3 ; R 3 ; V 3 ); (A 4 ; R 4 ; V 4 ); (A 5 ; R 5 ; V 5 ); (A 6 ; R 6 ; V 6 ); (A 7 ; R 7 ; V 7 ); (A 8 ; R 8 ; V 8 ); (A 9 ; R 9 ; V 9 )) with V = N [ T [ fx; X 0 ; Y; Y 0 ; Z; Z 0 ; H; H 0 ; R; K; g: Denote with U 1 ; : : : ; U n the symbols of the alphabet U (i.e. non terminal and terminal symbols of G) and with U n+1 the special symbol B. Dene A 1 = ; R 1 = ; V 1 = T A 2 = fxbsy; Z 0 Zg [ fzvy ju! v 2 P g [ fz@ i Y 0 j1 i n + 1g R 2 = f#uy $Z#vY ju! v 2 P g [ f#u i Y $Z#@ i Y 0 ju i 2 U [ fbgg [fz 0 #Z$XB#g V 2 = U [ fb; X; Y g A 3 = ; HH 0 g R 3 = f#@y 0 g [ f#y 0 $H#H 0 j 2 U [ fbgg V 3 = U [ fx; Y 0 g A 4 = fx R 4 = fx#$x V 4 = U [ fx; g A 5 = fzy 0 g R 5 = $Z#Y 0 g V 5 = U [ fx 0 ; g A 6 = fxzg R 6 = fx 0 #$X#Zg V 6 = U [ fx 0 ; B; Y 0 4

5 A 7 = fxu i Kj1 i n + 1g [ fry g R 7 = fx@ i #$XU i #Kj 2 U [ fbgg [ f#h 0 $R#Y g V 7 = U [ fx; H 0 g A 8 = fzzg R 8 = f#y $ZZ#g V 8 = T [ fy; Z 0 g A 9 = fzzg R 9 = f#zz$z 0 #g V 9 = T [ fz 0 g Let us examine the work of?. The rst component only selects the string produced by the others components that are terminal according to G. No such terminal string can enter a splicing, because all rules involves at least one special symbol that we add to the set of terminal and non terminal symbols. In tube 2 applications of productions of the form u! v 2 P to sentential forms Xw 1 Bw 2 uy are simulated, where w 2 uw 1 is a sentential form of G, and X; Y; B are special symbols, indicating respectively the left end and the right end of the sentential form in? and the beginning of the rotating string representing the corresponding sentential form in G. Tubes 3, 4, 5, 6 and 7 are used to rotate the symbols, so we can simulate the productions of G in the correct place. Tubes 8 and 9 are used to eliminate special symbols X and Y, so we obtain a terminal string. The construction works as follows: in the initial conguration (A 1 ; : : : ; A 9 ), only the second component can execute a splicing. There are three possibilities. We can either use a rule of the form #uy $Z#vY, for u! v 2 P (we call this a splicing of type 1), or 1a rule of the form #U i Y $Z#@ i Y 0 where U i is the i-th symbol of the alphabet U or, if i = n + 1, the symbol B (splicing of type 2), or the rule Z 0 #Z$XB# (splicing of type 3). In the following, while describing these three cases we will also denote them by `n, with n = 1; 2; 3. Consider the general case of having in tube 2 a string XwY, with w 2 U BU ; initially, w = BS. We have three possibilities for splicing: 1. (Xw 1 juy; ZjvY ) `1 (Xw 1 vy; ZuY ) for u! v 2 P and w = w1u 2. (Xw 1 ju i Y; Zj@ i Y 0 ) `2 (Xw i Y 0 ; ZU i Y ) for U i 2 U [fbg and w = w i U i 3. (Z 0 jz; XBjw 1 Y ) `3 (Z 0 w 1 Y; XBZ) for w = Bw 1 Let's examine the strings we've just created. 5

6 The string Xw 1 vy will remain in tube 2, entering new splicing of one of the three types. Clearly, the passage from Xw 1 uy to Xw 1 vy corresponds to using the rule u! v 2 P on a sux on the string bracketed by X; Y. The string ZuY will remain in tube 2, too. Such a string can enter a splicing in three cases 1. ZuY is an Axiom, then nothing new appears. 2. ZuY is used as the rst term of s splicing of the form (Zu 1 ju 0 ; Zv 0 Y ) `1 (Zu 1 v 0 Y; Zu 0 Y ), for u = u 1 u 0 and u 0! v 0 2 P ; we obtain two strings of the same form, zxy, which will remain in tube ZuY is used as the rst term in a splicing of the form (Zu 1 ju i Y; Zj@ i Y 0 ) `2 (Zu i Y 0 ; ZU i Y ), for u = u 1 U i ; U i 2 U [ fbg; the string zu i Y 0 cannot enter new splicing and cannot be transmitted to another tube. After any sequence of such splicings, the obtained strings still will be of the form ZxY, hence they will remain in tube 2 and will enter other \legal" splicing, when they are axioms, or they will enter splicing producing \useless" strings ZyY. Therefore, after a series of splicings of type 1, eventually in tube 2 a splicing of type 2 will be performed, producing strings of the form Xw i Y 0 and ZU i Y. The second string behaves exactly as we discussed above for the string ZuY. If a string Xw i Y 0 enters a new splicing in tube 2 this can only be splicing of type 3, (Z 0 jz; XBjw i Y 0 ) `3 (Z 0 w 0 ; XBZ) for w 1 = Bw 2. The string Z 0 w i Y 0 cannot enter new splicing in tube 2 and cannot be transmitted to another tube. If the string XBZ enters new splicing, this can only be of type 3, (Z 0 jz; XBjZ) `3 (Z 0 Z; XBZ) so nothing new can be created. Any string Xw i Y 0 is moved from tube 2 to tube 3 where we have to perform (Xw i?1 j@y 0 ; ) ` (Xw + 1@ i?1 ; Z@Y 0 ): The second type of rule of tube 3 will be examined below. The string Z@Y 0 cannot be transmitted to another tube and can enter only splicing of the form (Zj@Y 0 ; ) ` (Z@Y 0 ; ) hence creating nothing new. The string Xw i?1 cannot enter new splicing in tube 3, it will be transmitted to tube 4 where we have to perform (Xjw i?1 ; X ` (X i?1 ; XZ) The string XZ cannot be transmitted to another tube and can enter only splicing of the form (XjZ; X ` (XZ; X 6

7 hence creating nothing new. The string X i?1 1 cannot enter new splicing in this tube; it will be transmitted to tube 5, where the only possible splicing is (X i?1 ; ZjY 0 ) ` (X i?1 Y 0 ; ) The string cannot be transmitted to another tube and can enter only splicing of the form ; ZjY 0 ) ` ; ZY 0 ) so it can create nothing new. The string X i?1 Y 0 cannot enter new splicing in this tube; it will be transmitted to tube 6. In tube 6 we can only execute (X 0 j@w i?1 Y 0 ; XjZ 0 ) ` (X@w i?1 Y 0 ; X 0 Z 0 ) The string X 0 Z 0 cannot be transmitted to another tube and can only enter splicing of the form (X 0 jz 0 ; XjZ 0 ) ` (X 0 Z 0 ; XZ 0 ) hence producing nothing new. The string X@w i?1 Y 0 cannot enter a new splicing in this tube; it will be moved to tube 3. We started from tube 3 with the string Xw i Y 0 and now we returned to tube 3 with the string X@w i?1 Y 0. A from the right end of the string bracketed by X; Y 0 has been moved to the left end. The string X@w i?1 Y 0 cannot enter splicing of the second type in tube 3, but can enter splicing of the rst type in tube 3. By repeating the operations just described, we will return to tube 3 with the string X@@w i?2 Y 0 (i.e. we will rotate another This sequence will be repeated until all will be moved from the right end to the left end of the string bracketed by X; Y 0. It will take i steps. Then, in tube 3 we obtain a string of the form X@ i w 1 Y 0. This string cannot enter splicing of rst type in tube 3, but it can enter splicing of second type in this tube, so we can perform (X@ i w 2 jy 0 ; HjH 0 ) ` (X@ i w 2 H 0 ; HY 0 ); for w 1 = w 2 ; 2 U [ fbg: The string HY 0 cannot be transmitted to other tubes, and can only enter splicing of the form (HjY 0 ; HjH 0 ) ` (XY 0 ; HH 0 ) hence creating nothing new. The string X@ i w 1 H 0 cannot enter new splicing in this tube; it will be transmitted to tube 7. In tube 7 we have to perform one of two possible splicings. With the rst operation (X@ i jw 3 H 0 ; XU i jk) ` (XU i w 3 H 0 ; X@ i K); for w 1 = w 3 ; 2 U [ fbg: 7

8 we decode the symbol U i i (we coded U i i with the splicing of type 2 in tube 2). The string X@ i K cannot be transmitted to another tube, but it could enter the second kind of splicing in this tube, becoming X@ i Y. But this is refused by any other tube, especially tube 2 where X and Y are admitted, but The string XU i w 1 H 0 the second kind of splicing in this tube. We have to perform (XU i w 4 jh 0 ; RjY ) ` (XU i w 4 Y; RH 0 ) The string RH 0 can only become again RY, so it can't create nothing new. The string XU i w 1 Y cannot enter new splicing in this tube. It will be moved to tube 2. After this sequence of operations, we can note that having started with the string Xw 1 U i Y in tube 2 we have returned to tube 2 with the string XU i w 1 Y. A symbol from the right end of the string bracketed by X; Y has been moved to the left end. In this way, the string bracketed by X; Y can enter circular permutations as long as we want them to do that. This allows us to pass from a string Xw 1 Bw 2 Y to any string Xw 0 1 Bw0 2 Y such that w 2w 1 = w 0 2 w0 1. In this way we can \rewind" the string until its sux is the left-hand member of any rule in P that we want to simulate by a rule in R 2 of the form #uy $#vy. As the symbol B is always present (and exactly one copy of it is present as long as we do not use the rule Z 0 #Z$XB# in R 2 ), in every moment we know where the \actual beginning" of the string is placed. Consequently, using splicing of type 1 and 2 in tube 2 and splicing of tubes 3, 4, 5, 6 and 7 as described above, we can simulate every derivation in G. Conversely, exactly strings of the form Xw 1 Bw 2 Y can be obtained in this way, they correspond to strings w 2 w 1 that are sentential forms of the grammar G. We have now to consider the splicing of type 3 in tube 2 and the tubes 8 and 9. We consider rst of all the splicing of type 3 in tube 2. We already viewed what happens for the strings XBZ and Xw@ i Y 0, so we have now to consider the strings of the form XBqY for q 2 U. Using a splicing of type three we have (Z 0 jz; XBjqY ) `3 (XBZ; Z 0 qy ): If a string Z 0 qy enter a splicing in tube 2 this can be of type 1 and 2 : (Z 0 q 1 juy; ZjvY ) `1 (Z 0 q 1 vy; ZuY ); for u! v 2 P; q = q 1 u (Z 0 q1ju i Y; Zj@ i Y 0 ) `2 (Z 0 q i Y 0 ; ZU i Y ); for U i 2 U [ fbg; q = q 1 U i We have already discussed the case of ZuY; ZU i Y and Z 0 q i Y 0. The string Z 0 q 1 vy can be obtained by performing rst (XBq 1 juy; ZjvY ) `1 (XBq 1 vy; ZuY ) and then (Z 0 jz; XBjq 1 vy ) `3 (Z 0 q 1 vy; XBZ); 8

9 so it is a \legal" string. If the string Z 0 qy, obtained with a splicing of type 3 in the tube 2, have the property that q 2 T (i.e. q is a terminal string) it can be moved to tube 8. Here the only possible splicing is (Z 0 qjy; ZZj) ` (Z 0 q; ZZY ): If ZZY will enter new splicing, these are of the forms (Z 0 xjy; ZZjY ) ` (Z 0 xy; ZZY ) (ZZjY; ZZjY ) ` (ZZY; ZZY ) hence no new string is obtained. The string Z 0 q cannot enter new splicing in tube 8. It will be moved to tube 9, where we have to perform (jzz; Z 0 jq) ` (q; Z 0 ZZ): If the string Z 0 ZZ enters new splicing, these are of the forms (Z 0 jzz; Z 0 jx) ` (Z 0 x; Z 0 ZZ) (Z 0 jzz; Z 0 jzz) ` (Z 0 ZZ; Z 0 ZZ) hence nothing new can be created. The string q is terminal. It will be transmitted to all tubes, including the rst one. No splicing can be done on a terminal string. As we seen above, such a terminal string q is a string in L(G). No parasitic string can reach the rst tube, consequently L(?) = L(G). 2 The basic ideas of the construction we present here are these: We simulate the productions of the grammar using splicing operations in Tube 2. Unfortunately, using the splicing operation, we can simulate a production (in one step) only if the sub-string we are going to substitute is placed at one end of the string. The splicing operation is not able to do such a substitution (using a nite number of rules) if the sub-string is placed in the internal part of the string. To deal with this problem, we use the rotation of the characters. Look at the Example 1 below: to simulate the production U! V we rotate the characters x 3 ; x 4 and x 5 so that we can move the sub-string U to the right end of the string. This position is optimal to simulate the production using a splicing operation. After that, we rotate the string V (one character at time),and then the character x 1 and x 2. Starting with x 1 x 2 U x 3 x 4 x 5 we obtain x 1 x 2 V x 3 x 4 x 5, so we have simulate properly the production U! V of the grammar. 9

10 Example 1: x 1 x 2 U x 3 x 4 x 5 ; x 5 x 1 x 2 U x 3 x 4 ; x 4 x 5 x 1 x 2 U x 3 ; x 3 x 4 x 5 x 1 x 2 U ; x 3 x 4 x 5 x 1 x 2 U ; x 3 x 4 x 5 x 1 x 2 V ; x 3 x 4 x 5 x 1 x 2 V ; V x 3 x 4 x 5 x 1 x 2 ; x 2 V x 3 x 4 x 5 x 1 ; x 1 x 2 V x 3 x 4 x 5 2 The rotation solves the problem we've just illustrated, but it introduces a new problem. When we rotate a character, we move it from one end to the other. Using splicing rules, this operation requires more than one step, so the problem we have to deal with, is how to delete the character from the right end of the string and to put the same character in the left end of the string. Due to the multi-step process, we could delete a character from the right end and put in the left end a dierent character, generating a word which the grammar was not able to create, even if we don't want to do so. In [3] the solution to this problem was to substitute the character to rotate with a symbol that contains the information on the character substituted and then to send the string obtained to a \special" tube that rotates that specic character. In the Example 2 we show how this can be done. Example 2: We consider the string x 1 x 2 x 3 x 4 x 5 w; the character w is replaced with Y w. x 1 x 2 x 3 x 4 x 5 w ; x 1 x 2 x 3 x 4 x 5 Y w Then, the string x 1 x 2 x 3 x 4 x 5 Y w is sent to the \special" tube, the only tube able to receive this string. This special tube is the only one that contain the character Y w in the lter, and its function is to rotate the specic character w. We are sure, in this case, to put in the left end of the string the same character we have deleted from the right end of the string. 2 The problem in this solution is that we need one of these \special" tubes for every character of the language we are going to generate (because we need one tube for every type of character to rotate). Thus, the number of test tubes needed to generate a language depends on the number of dierent characters used in the language we have to generate. In the model presented here, we introduce some dierences. First of all, we number the characters of the grammar. Before rotating a character, we encode it to a number of special not present in the alphabet of the grammar. The i-th character in the order we give, is substituted with i of this symbols. This operation is done by the splicing of type 2 in the second component test tube. Then the character is moved to the left end by rotating, one at a time, the special These rotations are executed by the component tubes 2, 3, 4, 5 and 6. 10

11 When all the special symbols have been rotated, we decode i special symbols with the correspondent character U i. This is done with the component 7. In the example below we illustrate these three main phases. Example 3: We show here the three main phases of the rotation of the character x 5 of the string x 1 x 2 x 3 x 4 x 5 : Encoding : x 1 x 2 x 3 x 4 x 5 ; x 1 x 2 x 3 x Rotation 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 4 Decoding 1 x 2 x 3 x 4 ; x 5 x 1 x 2 x 3 x 4 2 This solution oer three advantages: A character is coded when it is placed in the right end of the string, thus in a suitable place for the application of splicing rules. The only character that actually rotates is the special so we need just one of the \special" tubes used in [3], about which we have discussed before. A character is decoded when it is placed in the left end of the string, thus in a suitable place for the application of splicing rules. These advantages permit us to limit the number of tubes with respect to the construction presented in [3]. Using the construction we have explained, the number of tubes doesn't depend on the number of the characters of the language we have to generate: nine Test Tubes are enough to generate any RE language. 4 Conclusions and Perspectives We proved how to build a distributed splicing system powerful enough to generate any language in RE, and using a xed number of 9 test tubes, closing one open problem from [3]: the T T hierarchy is nite. There are other some consequences to this. The technique from [3] requires jt j + 8 test tubes for languages over T. One result for T T 7 made use of closure properties of the language classes being compared to T T 7, allowing to reduce the letters to two. We can extend here that result, described below, under two respects: we move it from T T 7 to T T 6, and we need no longer the property of closure under inverse morphisms. Theorem 2 For every family F such that F RE and F is closed under intersection with regular sets and restricted morphisms, we have T T 6 (F IN; F IN)? F 6= ; 11

12 Proof. We can use the same technique as in [3], building directly for a language L 2 RE? F a system in T T 9, and then remove tubes 1, 8 and 9. The resulting T T 6 system will still generate a language in RE? F thanks to the closure properties of F. 2 Our result is still dierent from designing in detail a universal splicing system, similar to the current programmable computers, but it takes us closer to a practical implementation of a DNA computer: for each computation (language) we want, we just change the starting molecules and the restriction enzymes introduced in the test tubes, we do not change the layout on our workbench for each alphabet we need. Of course it is still a practical problem to have enough real restriction enzymes. This work has been insightful to study a case where a simple algorithm has been designed directly in terms of splicing; an algorithm not simply reproducing an usual grammatical production rule, but doing some dierent basic operation (encoding/decoding). This, of course, is not the rst case in literature, but we feel that the interest in this molecular algorithm engineering can eventually lead us to build a kind of universal grammar systems based on molecular-like operations, avoiding the uneective translations of universal Turing machines seen so far. Many interesting open problems suggested in [3] still are open, concerning the power of dierent numbers of test tubes and comparisons to dierent levels of Chomsky hierarchy. We can only add now that it is interesting to check whether it is possible to use less than 9 tubes for RE, since it is not proved that T T 8 (F IN; F IN) is properly contained in RE. Acknowledgments The authors would like to thank Gheorghe Paun for many useful discussions on these topics. References [1] Proceedings of Second DIMACS workshop on DNA computing, Princeton, 1996, to be published by ACM Press. [2] L. M. Adleman, Molecular computation of solutions of combinatorial problems, Science, 226 (1994), 1021{1024. [3] E. Csuhaj-Varju, L. Kari, G. Paun, Test Tube distributed system based on splicing, Computer and AI, 15, 2-3 (1996), 211{232. [4] T. Head, Formal language theory and DNA: an analysis of the generative capacity of specic recombinant behaviours, Bull. Math. Biology, 49 (1987), 737{759. [5] C. Zandron, C. Ferretti, G. Mauri, A reduced distributed splicing system for RE Languages, submitted for publication. 12