Tissue P systems with evolutional communication rules with two objects in the left-hand side

In the framework of Membrane Computing, several efficient solutions to computationally hard problems have been given. To find new borderlines between families of P systems that can solve them and the ones that cannot is an important task to tackle the P versus NP problem. Adding syntactic and/or semantic ingredients can mean passing from non-efficiency to presumed efficiency. Here, we try to get narrow frontiers, setting the stage to adapt efficient solutions from a family of P systems to another one. In order to do that, a solution to the SAT problem is given by means of a family of tissue P systems with evolutional symport/antiport rules and cell separation with the restriction that both the left-hand side and the right-hand side of the rules have at most two objects; that is, with recognizer P systems from TSEC(2,2)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {TSEC}}(2, 2)$$\end{document}. This result improves a previous one, when 3 objects could be used in the left-hand side of the evolutional communication rules


Introduction
Membrane computing is a bio-inspired computing paradigm based on the structure and behaviour of living cells. There are several variants of P systems, the computational models of this paradigm. It was first introduced in Pȃun (2000), defining one of the main variants, cell-like P systems that abstract the hierarchical arrangement of membranes within a single cell. In Martín-Vide et al. (2003), the idea of the interactions of networks of cells (placed in the nodes of a directed graph) between cells and between cells and their environment is used to develop tissue-like P systems, named by the ensemble of cells in living beings. Another approach with the same structure are the so-called spiking neural P systems (Ionescu et al. 2006), SN P systems for short, inspired by the way that neurons communicate with each other by means of short electrical impulses (spikes).
Within these models, several variants can be defined only by changing syntactic and/or semantic ingredients, such as kinds of rules possible, length of rules, parallelism permitted, number of objects and so on. Computational complexity theory in the framework of Membrane Computing use special variants of P systems called recognizer membrane systems, devices that, given an initial

Preliminaries
In order to do this work self-contained, we introduce some concepts that are going to be used through the paper.

Alphabets and sets
An alphabet C is a non-empty set and their elements are called symbols. A string u over C is an ordered finite sequence of symbols, that is, a mapping from a natural number n 2 N onto C. The number n is called the length of the string u and it is denoted by j u j. The empty string (with length 0) is denoted by k. The set of all strings over an alphabet C is denoted by C Ã . A language over C is a subset of C Ã .
A multiset over an alphabet C is an ordered pair ðC; f Þ where f is a mapping from C onto the set of natural numbers N. The support of a multiset m ¼ ðC; f Þ is defined as suppðmÞ ¼ fx 2 C j f ðxÞ [ 0g. A multiset is finite (resp., empty) if its support is a finite (resp., empty) set. We denote by ; the empty multiset, MðCÞ the set of all multisets over C and M þ ðCÞ ¼ MðCÞ n ; Let m 1 ¼ ðC; f 1 Þ, m 2 ¼ ðC; f 2 Þ be multisets over C, then the union of m 1 and m 2 , denoted by m 1 þ m 2 , is the multiset ðC; gÞ, when gðxÞ ¼ f 1 ðxÞ þ f 2 ðxÞ for each x 2 C.

Decision problems
A decision problem X can be informally defined as one whose solution is either yes or no. This can be formally defined by an ordered pair ðI X ; h X Þ, where I X is a language over a finite alphabet R X and h X is a total Boolean function over I X . The elements of I X are called instances of the problem X. Each decision problem X has associated a language L X over the alphabet R X as follows: L X ¼ fu 2 E X j h X ðuÞ ¼ 1g. Conversely, every language L over an alphabet R has associated a decision problem X L ¼ ðI X L ; h X L Þ as follows: I X L ¼ R Ã and h X L ðuÞ ¼ 1 if and only if u 2 L. Then, given a decision problem X we have X L X ¼ X, and given a language L over an alphabet R we have L X L ¼ L.
It is worth pointing out that any Turing machine M (with input alphabet R M ) has associated a decision problem X M ¼ ðI M ; h M Þ defined as follows: I M ¼ R Ã M , and for every u 2 R Ã M , h M ðuÞ ¼ 1 if and only if M accepts u. Obviously, the decision problem X M is solvable by the Turing machine.
A tissue P system with evolutional symport/antiport rules and cell separation of degree q ! 1 P ¼ ðC; C 0 ; C 1 ; E; M 1 ; . . .; M q ; R; i out Þ can be viewed as a set of q cells, labelled by 1; . . .; q such that (a) M 1 ; . . .; M q represent the multisets of objects initially placed in the q cells of the system; (b) E is the set of objects initially located in the environment of the system, all of them available in an arbitrary number of copies; (c) i out represents a distinguished region which will encode the output of the system. We use the term region i (0 i q) to refer to cell i in the case 1 i q and to refer to the environment in the case i ¼ 0.
A configuration at any instant of a tissue P system with evolutional symport/antiport rules and cell separation is described by the multisets of objects in each cell and the multiset of objects over C n E in the environment at that moment. The initial configuration of P ¼ ðC; C 0 ; C 1 ; E; M 1 ; . . .; M q ; R; i out Þ is ðM 1 ; . . .; M q ; ;Þ.
An evolutional symport rule ½ u i ½ j ! ½ i ½ u 0 j is applicable at a configuration C t at an instant t if there is a region i from C t which contains multiset u. By applying an evolutional symport rule, the multiset of objects in region i from C t is consumed and the multiset of objects u 0 is generated in region j from C tþ1 .
An evolutional symport rule ½ u i ½ v j ! ½ v 0 i ½ u 0 j is applicable at a configuration C t at an instant t if there is a region i from C t which contains multiset u and there is a region j which contains multiset v. By applying an evolutional symport rule, the multiset of objects u in region i and multiset of objects v in region j from C t are consumed and the multiset of objects u 0 is generated in region j and the multiset of objects v 0 in region i from C tþ1 .
A separation rule ½ a i ! ½ C 0 i ½ C 1 i is applicable at a configuration C t at an instant t if there is a cell i from C t which contains object a and i 6 ¼ i out . By applying a separation rule to such a cell i, (a) object a is consumed from such cell; (b) two new cells with label i are generated at configuration C tþ1 ; and (c) objects from C 0 from the original cell are placed in one of the new cells, while objects from C 1 from the original cell are placed in the other one.
The rules of a tissue P system with evolutional symport/ antiport rules and cell separation are applied in a maximally parallel manner, following the previous remarks, and taking into account that when a cell i is being separated at one transition step, no other rules can be applied to that cell i at that step.
A transition from a configuration C t to another configuration C tþ1 is obtained by applying rules in a maximally parallel manner following the previous remarks. A computation of the system is a (finite or infinite) sequence of transitions starting from the initial configuration, where any term of the sequence other than the first one is obtained from the previous configuration in one transition step. If the sequence is finite (called halting computation) then the last term of the sequence is a halting configuration, that is, a configuration where no rule is applicable to it. A computation gives a result only when a halting configuration is reached, and that result is encoded by the multiset of objects present in the output region i out .
A natural framework to solve decision problems is to use recognizer P systems.
Definition 2 A recognizer tissue P system with evolutional symport/antiport rules and cell separation of degree q ! 1 is a tuple where -the tuple ðC; C 0 ; C 1 ; E; M 1 ; . . .; M q ; R; i out Þ is a tissue P system with evolutional symport/antiport rules of degree q ! 1.
Tissue P systems with evolutional communication rules with two objects in the left-hand side 121 • Where C strictly contains an (input) alphabet R and two distinguished objects yes and no, and M i (1 i q) are multisets over C n R; i in 2 f1; . . .; qg is the input cell and i out is the label of the environment; -for each multiset m over the input alphabet R, any computation of the system P with input m, represented as P þ m, starts from the configuration of the form ðM 1 ; . . .; M i in þ m; . . .; M q ; ;Þ, it always halts and either object yes or object no (but not both) must appear in the environment at the last step.
For each ordered pair of natural numbers ðk 1 ; k 2 Þ; k 1 ; k 2 ! 1, the class of recognizer P systems with evolutional symport/antiport rules and cell separation with evolutional communication rules of length at most ðk 1 ; k 2 Þ is denoted by TSECðk 1 ; k 2 Þ. This means that, given an evolutional communication rule ½ u i ½ v j ! ½ v 0 i ½ u 0 j the LHS (resp., RHS) of any evolutional communication rule in a system from TSECðk 1 ; k 2 Þ involves at most k 1 ¼j u j þ j v j objects (resp., k 2 ¼j u 0 j þ j v 0 j objects).
Next, we define the concept of solving a problem in a uniform way and in polynomial time by a family of recognizer tissue P systems with evolutional symport/antiport rules and cell separation.
Definition 3 A decision problem X ¼ ðI X ; h X Þ is solvable in a uniform way and in polynomial time by a family P ¼ fPðnÞ j n 2 Ng of recognizer tissue P systems with evolutional symport/antiport rules and cell separation if the following conditions hold: 1. the family P is polynomially uniform by Turing machines; that is, there exists a Turing machine capable of constructing the elements of such a family in polynomial time; and 2. there exists a polynomial encoding (cod, s) of I X in P such that (a) for each instance u 2 I X , s(u) is a natural number and cod(u) is an input multiset of the system PðsðuÞÞ; (b) for each n 2 N, s À1 ðnÞ is a finite set; and (c) the family P is polynomially bounded 1 , sound 2 and complete 3 with regard to (X, cod, s) (Pérez-Jiménez et al. 2003).
The set of all decision problems that can be solved by recognizer tissue P systems with evolutional symport/ antiport rules and cell separation with evolutional communication rules of length at most ðk 1 ; k 2 Þ in a uniform way and polynomial time is denoted by PMC TSECðk 1 ;k 2 Þ . 4 Solution to SAT with evolutional communication rules and separation rules Pan et al. (2018) an efficient solution to the SAT problem is given by means of a family of P systems from TSECð3; 2Þ. A frontier of efficiency is given, but some open problems remain, as indicate Fig. 1 of Pan et al. (2018). It shows that the class of problems that can be solved by P systems from TSECð2; kÞ with k ! 2 is unknown. In this work we improve this borderline closing the previous open questions, giving an efficient solution of the SAT problems by means of a family of P systems from TSECð2; 2Þ. Let us briefly recall the description of the SAT problem: given a Boolean formula in conjunctive normal form (CNF), to determine whether or not there exists an assignment to its variables, called truth assignment, on which it evaluates true.
Theorem 1 SAT 2 PMC TSECð2;2Þ For each n; p 2 N, we consider the recognizer P system fyes; no; y 1 ; y 2 ; n 1 ; n 2 ; #g [ fa i;j j 1 i n; 0 j ig [ fa 0 i;j j 2 i n; 0 j i À 1g [ fa L i;j ; a R i;j j 2 i n; i;j;k j 1 i n; 1 j p; 1 k n þ j À 1g [ fx 0 i;j;k ; x 0 i;j;k ; x Ã0 i;j;k ; x 00 i;j;k ; x 00 i;j;k ; x Ã00 i;j;k ; x 000 i;j;k ; x 000 i;j;k ; x Ã000 i;j;k ; j 0 i n; 1 j p; 1 k ng l;0 j 1 l ng; 6. The set of rules R consists of the following rules: 1.1 Rules for ð4k þ 1Þ-th steps.
; for 1 i n; 1 j p; Tissue P systems with evolutional communication rules with two objects in the left-hand side 123

2.1
Rules to check satisfied clauses.
3.1 Rules to check if all clauses are satisfied by a truth assignment.
4.2 Rules to give a negative answer.
give an affirmative answer.
The input cell is the cell labelled by 1 (i in ¼ 1) and the output region is the environment (i out ¼ env).
Let u ¼ C 1^. . .^C p an instance of SAT problem consisting of p clauses C j ¼ l j;1 _ . . . _ l j;r j , 1 j p, where VarðuÞ ¼ fx 1 ; . . .; x n g, and l j;k 2 fx i ; :x i j 1 i ng, 1 j p; 1 k r j . Let us assume that the number of variables, n, and the number of clauses, p, of u, are greater than or equal to 2.
We consider the polynomial encoding (cod, s) from SAT in P defined as follows: for each u 2 I SAT with n variables and p clauses, sðuÞ ¼ hn; pi and x 4 Þ is encoded as follows: codðuÞ ¼ We use here matrix representation to make easier the view of the encoding, but let us recall that codðuÞ is a multiset of objects. We define cod k ðuÞ as the set of elements of codðuÞ when the third subscript equals k. In the same way, we define cod 0 k ðuÞ, cod 00 k ðuÞ and cod 000 k ðuÞ as the sets of elements of codðuÞ when the third subscript equals k and elements are primed, double primed and triple primed, respectively. For notation convenience, we define cod j k ðuÞ the subset of elements of cod k ðuÞ with elements of C j ; . . .; C p . For instance, cod 2 4 ðuÞ would be the following set: The Boolean formula u will be processed by the system PðsðuÞÞ þ codðuÞ. Next, we informally describe how that system works. The solution proposed follows a brute force algorithm in the framework of recognizer tissue P systems with separation and evolutional communication rules, and it consists of the following stages: • Generation stage: using separation rules each 4 steps, we produce 2 n membranes labelled by 2 containing each possible truths assignment. At the same time, we generate 2 n copies of cod n ðuÞ. This stage spends n computation steps exactly, n being the number of variables of u. • First checking stage: With rules from 2.1, we can check which clauses from the input formula u have been satisfied by a specific truth assignment. This stage takes exactly p steps. • Second checking stage: with rules from 3.1, we remove objects a j such that they are removed from a membrane if and only if the truth assignment associated to that membrane makes true clause C j . This stage takes exactly one step.
• Output stage: with rules from 4.2 and 4.3, we can give an affirmative or a negative answer depending on if the input formula is satisfiable or not. This stage spends exactly 4 steps, regardless of whether the formula is satisfiable or not.

A formal verification
In this section, an exhaustive verification of the system is given.

Generation stage
At this stage, all truth assignments for the variables associated with the Boolean formula uðx 1 ; . . .; x n Þ are going to be generated, by applying separation rules from 1.2 in membranes labelled by 2. In such manner that in the 4i þ 2-th step (1 i n À 1) of this stage, separation rule associated with an object a i;i is triggered, two new cells distributing t i and f i between them. In the last step of this stage, each membrane labelled by 2 will contain a truth assignment of the formula.
Proposition 2 Let C ¼ ðC 0 ; C 1 ; . . .; C q Þ be a computation of the system PðsðuÞÞ with input multiset codðuÞ, and let C t ðhÞ the contents of the membrane labelled by h in the configuration C t .
(B) C 4n ð1Þ ¼ fd 4n ; d 0 2 n 4n ; cod 4n ðuÞ 2 n g, and there are 2 n membranes labelled by 2 such that each of them contains objects a 1 ; . . .; a pþ1 , as well as a different subset fr 1 ; . . .; r n g, being r 2 ft; f g.
Proof ðA k Þ; 0 k 3 is going to be demonstrated by induction on k.
ðA 0 Þ The base case k ¼ 0 is trivial because at the initial configuration we have: C 0 ð1Þ ¼ fd 0 ; d 0 0 ; cod 0 ðuÞg [ fb l;0 j 1 l ng and there exists a single membranelabelled by 2 containing objects a 1 ; . . .; a pþ1 and objects a 1;0 ; . . .; a n;0 . Then, configuration C 0 yields configuration C 1 by applying the rules: Tissue P systems with evolutional communication rules with two objects in the left-hand side 125 ðA 1 Þ Thus, C 1 ð1Þ ¼ fd 1 ; d 0 1 ; cod 0 1 ðuÞg [ fb 0 l;0 j 1 l ng and in C 1 there exists one membrane labelled by 2 such that its contents is the set of objects fa 0 1;0 ; . . .; a 0 n;0 g, the object t 0 1 and objects a 0 1 ; . . .; a 0 pþ1 . Then, configuration C 1 yields configuration C 2 by applying the rules: and in C 2 there exists one membrane labelled by 2 such that its contents is the set of objects fa 1;1 ; . . .; a n;1 g, objects t L 1 and f R 1 and objects a O 1 ; . . .; a O pþ1 , for O 2 fL; Rg. Then, configuration C 2 yields configuration C 3 by applying the rules: 3 ; cod 000 1 ðuÞg, at the environment there is the multiset fb O 1;1 j O 2 fL; Rgg [ fb 2 l;1 j 2 l ng and in C 2 there exists two membranes labelled by 2 such that its contents is the set of objects fa O 2;1 ; . . .; a O n;1 g with O ¼ L (resp., O ¼ R), object t L 1 (resp., f R 1 ) and objects a O 1 ; . . .; a O pþ1 , for O ¼ L (resp., O ¼ R). Hence, the result holds for k ¼ 0 Supposing that, by induction, result is true for k (1 k n À 1); that is, ðA 0 Þ For each 4k (0 k n À 1) at configuration C 4k we have the following: -C 4k ð1Þ ¼ fd 4k ; d 0 2 k 4k ; cod k ðuÞ 2 k g [ fb 2 k l;k j k þ 1 l ng -There are 2 k membranes labelled by 2 such that each of them contains objects a kþ1;k ; . . .; a n;k ; objects r 1 ; . . .; r k , being r 2 ft; f g; and objects a 1 ; . . .; a pþ1 .
ðA 2 Þ For each 4k þ 2 (0 k n À 1) at configuration C 4kþ2 we have the following: ðA 3 Þ For each 4k þ 3 (0 k n À 1) at configuration C 4kþ3 we have the following: 4kþ3 ; cod 000 kþ1 ðuÞ 2 kþ1 g -There are 2 kþ1 membranes labelled by 2 such that each of them contains objects a kþ1;k ; . . .; a n;k ; Then, by the induction hypothesis, we want to prove the result for k þ 1.
• In order to prove (B) it is enough to notice that, on the one hand, from ðA 3 Þ configuration C 4nÀ1 4 holds: • On the other hand, configuration C 4nÀ1 yields configuration C 4n by applying the rules: • Then, we have C 4n ð1Þ ¼ fd 4n ; d 0 2 n 4n ; cod 4n ðuÞ 2 n g, and there are 2 n membranes labelled by 2 such that each of them contains objects a 1 ; . . .; a pþ1 , as well as a different subset fr 1 ; . . .; r n g, being r 2 ft; f g. h

First checking stage
Following the generation stage comes the first checking stage, where objects c j;j are created in order to know if clause C j has been satisfied by the truth assignment encoded in membranes labelled by 2. In each step, we fire rules for a single clause, therefore in p steps we can obtain objects c j;p if this clause is satisfied. This can be because of two reasons: • Literal x i appears in clause C j , and the truth value of variable x i in a truth assignment is True. Then, we can say that such truth assignment satisfies this clause; or • Literal :x i appears in clause C j , and the truth value of variable x i in a truth assignment is False. Then, we can say that such truth assignment satisfies this clause.
In any other way, variable x i has nothing to do with clause C j . At the final step of this stage, membranes labelled by 2 will have objects c j;p where C j are clauses satisfied by such truth assignment. We obtain an object a 0 pþ1 to use it in the next stage.
(a) For each k (0 k p À 1) at configuration C 4nþk we have the following: • C 4nþk ð1Þ ¼ fd 4nþk ; d 0 2 n 4nþk ; cod k n ðuÞ 2 n g • There are 2 n membranes labelled by 2 such that each of them contains -objects r 1 ; . . .; r n , being r 2 ft; f g; -objects a 1 ; . . .; a pþ1 ; and -objects c 1;k ; . . .; c k;k , where c j;k represents that clause C j has been satisfied by the truth formula encoded in such membrane.
(B) C 4nþp ð1Þ ¼ fd 4nþp ; d 0 2 n 4nþp g, and there are 2 n membranes labelled by 2 such that each of them contains objects a 1 ; . . .; a pþ1 , a different subset fr 1 ; . . .; r n g and objects c j when clause C j is satisfied in that membrane.
Proof (a) is going to be demonstrated by induction on k.
• (a) The base case k ¼ 0 is trivial because at the initial configuration we have: C 4n ð1Þ ¼ fd 4n ; d 0 2 n 4n ; cod 4n ðuÞg and there exist 2 n membranes labelled by 2 containing objects a 1 ; . . .; a pþ1 and a different subset fr 1 ; . . .; r n g, being r 2 ft; f g. Then, configuration C 4n yields configuration C 4nþ1 by applying the rules: Thus, C 4nþ1 ð1Þ ¼ fd 4nþ1 ; d 0 2 n 4nþ1 ; cod 1 4nþ1 ðuÞ 2 n g and in C 4nþ1 there exist 2 n membranes labelled by 2 such that their contents are objects a 1 ; . . .; a pþ1 , a different subset fr 1 ; . . .; r n g, being r 2 ft; f g and objects c 1;1 if some literal present in C j satisfies it. 5 Hence, the result holds for k ¼ 1.
Supposing that, by induction, result is true for k (0 k p À 1); that is, • C 4nþk ð1Þ ¼ fd 4nþk ; d 0 2 n 4nþk ; cod k n ðuÞ 2 n g • There are 2 n membranes labelled by 2 such that each of them contains -objects r 1 ; . . .; r n , being r 2 ft; f g; -objects a 1 ; . . .; a pþ1 ; and -objects c 1;k ; . . .; c k;k , where c j;k represents that clause C j has been satisfied by the truth formula encoded in such membrane.
Then, configuration C 4nþk yields configuration C 4nþkþ1 by applying the rules: for 1 i n; k þ 2 j p ½ c j;k 2 ½ c 0 ! ½ c j;kþ1 2 ½ 0 ; Thus, C 4nþkþ1 ð1Þ ¼ fd 4nþkþ1 ; d 0 2 n 4nþkþ1 ;cod kþ1 4nþkþ1 ðuÞ 2 n g and in C 4nþkþ1 there exist 2 n membranes labelled by 2 such that their contents are objects a 1 ; . . .; a pþ1 , a different subset fr 1 ; . . .; r n g, being r 2 ft; f g and objects c 1;k ; . . .; c k;k if some literal present in C j satisfies them.
In order to demonstrate (B) it is enough to notice that, on the one hand, from (a) configuration C 4nþpÀ1 holds: ; cod pÀ1 n ðuÞ 2 n g • There are 2 n membranes labelled by 2 such that each of them contains -objects r 1 ; . . .; r n , being r 2 ft; f g; -objects a 1 ; . . .; a pþ1 ; and -objects c 1;pÀ1 ; . . .; c pÀ1;pÀ1 , where c j;pÀ1 represents that clause C j has been satisfied by the truth formula encoded in such membrane.

Second checking stage
Here, when rules from 3.1 are fired at the ð4n þ p þ 1Þ-th step, objects a j within a membrane labelled by 2 are removed if and only if the truth assignment associated to that membrane makes true clause C j , that is, if there is at least one object c j in such membrane. At configuration C 4nþp we have C 4nþp ð1Þ ¼ fd 4nþp ; d 0 2 n 4nþp g and each membrane labelled by 2 contains objects a 1 ; . . .; a p and objects c j such that the corresponding truth assignment satisfies the clause C j . By applying rules from 3.1 and rule ½ d 4nþp 1 ½ c 0 ! ½ d 4nþpþ1 1 ½ 0 , object d 4nþp evolves into d 4nþpþ1 within the membrane labelled by 1, and in each membrane labelled by 2, objects a j such that their corresponding object c j;p are ''removed'' from the system, and let the next stage to check whether or not they are present, besides the object a pþ1 , that is prepared, evolving to a 0 pþ1 , to react with the remaining objects a j . This stage takes exactly one step.

Output stage
The output phase starts at the ð4n þ p þ 2Þ-th step, and takes exactly four steps, regardless of whether the input formula u is satisfied or not by some truth assignment.
• Affirmative answer: if the input formula u of SAT problem is satisfiable then at least one of the truth assignments from a membrane with label 2 has satisfied all clauses. Then, there will be a membrane labelled by 2 such that all objects a j , with 1 j p have disappeared in the previous step. At configuration C 4nþpþ1 , we have C 4nþpþ1 ð1Þ ¼ fd 4nþpþ1 g and in each membrane labelled by 2 there remain objects a j if the corresponding truth assignment does not make true clause C j and one object a 0 pþ1 . In this step, only rule ½ d 4nþpþ1 1 ½ c 0 ! ½ d 4nþpþ2 1 ½ 0 will be fired and rules ½ a j a 0 pþ1 2 ½ 0 ! ½ 2 ½ n 1 0 will be fired in membranes labelled by 2 such that at least one clause is not satisfied by the corresponding truth assignment. Then, at configuration C 4nþpþ2 , we have C 4nþpþ2 ð1Þ ¼ fd 4nþpþ2 ; n t 1 g, being t the number of truth assignments that have at least one clause not satisfied by the corresponding truth assignment, and membranes labelled by 2 contains an object a 0 pþ1 if and only if the corresponding truth assignment makes true all clauses from u, and can contain objects a j , 1 j p, if clause C j is not satisfied by the corresponding truth assignment.
In the next step, applying rules ½ 2 ½ n 1 0 ! ½ n 1 2 ½ 0 and ½ a 0 pþ1 2 ½ d 4nþpþ2 1 ! ½ y 1 2 ½ 1 , we obtain an object y 1 in a membrane labelled by 2 if and only if the corresponding truth assignment makes true the input formula. Let us remark that more than one membrane labelled by 2 can contain a truth assignment that makes true u, but in this case, we as we want to know if at least one truth assignment makes true the input formula u, we only want one object y 1 . Then, at configuration C 4nþpþ3 we have that C 4nþpþ3 ð1Þ ¼ ; and in membranes labelled by 2, we can have objects n 1 , 7 adding up to t in all membranes labelled by 2, being t the number of truth assignments that do not make true the input formula, an object a 0 pþ1 if the corresponding truth assignment makes true all clauses, excepting one membrane labelled by 2 which corresponding truth assignment makes true the input formula that will contain an object y 1 , and can contain objects a j , 1 j p, if clause C j is not satisfied by the corresponding truth assignment. In the next step the only rule that can be fired is ½ y 1 2 ½ c 0 ! ½ y 2 2 ½ 0 , that will be useful to synchronize the affirmative and the negative answer. Let us note that rule ½ n 1 2 ½ d 4nþpþ2 1 ! ½ n 2 2 ½ 1 cannot be fired because object d 4nþ3 has been consumed in the previous step by an object a 0 pþ1 . Then, at configuration C 4nþpþ4 , we have that C 4nþpþ4 ð1Þ ¼ ; and in membranes labelled by 2, we can have objects n 1 , adding up to t in all membranes labelled by 2, being t the number of truth assignments that do not make true the input formula, an object a 0 pþ1 if the corresponding truth assignment makes true all clauses, excepting one membrane labelled by 2 which corresponding truth assignment makes true the input formula that will contain an object y 2 , and can contain objects a j , 1 j p, if clause C j is not satisfied by the corresponding truth assignment. At the last step of the computation, rule ½ y 2 2 ½ 0 ! ½ 2 ½ yes 0 is fired, sending an object yes to the environment. Then, at configuration C 4nþpþ5 , we have that C 4nþpþ5 ð1Þ ¼ ; and in membranes labelled by 2, we can have objects n 1 , adding up to t in all membranes labelled by 2, being t the number of truth assignments that do not make true the input formula, an object a 0 pþ1 if the corresponding truth assignment makes true all clauses, excepting one membrane labelled by 2 which corresponding truth assignment makes true the input formula, and can contain objects a j , 1 j p, if clause C j is not satisfied by the corresponding truth assignment, and there will be an object yes in the environment. Here, the computation halts and returns an affirmative answer.
• Negative answer: If the input formula u of SAT problem is not satisfiable then none of the truth assignments encoded by a membrane labelled by 2 makes the formula u true. Thus, some object a j (1 j p) will be within all membranes labelled by 2 will not remain in such membranes. At configuration C 4nþpþ1 , we have C 4nþpþ1 ð1Þ ¼ fd 4nþpþ1 g and in each membrane labelled by 2 there remain objects a j if the corresponding truth assignment does not make true clause C j . In this step, only rules ½ a j a 0 pþ1 2 ½ 0 ! ½ 2 ½ n 1 0 , for 1 j p and rule ½ d 4nþpþ1 1 ½ c 0 ! ½ d 4nþpþ2 1 ½ 0 will be fired. Then, at configuration C 4nþpþ2 we have in the environment 2 n copies of object n 1 , C 4nþpþ2 ð1Þ ¼ fd 4nþpþ2 g and membranes labelled by 2 will contain objects a j (1 j p) when clauses C j are not satisfied by the corresponding truth assignment. In the ð4n þ p þ 3Þ-th step, rule ½ 2 ½ n 1 0 ! ½ n 1 2 ½ 0 will be fired. Here, objects n 1 will be sent to a membrane labelled by 2. Then, at configuration C 4nþpþ3 we have C 4nþpþ3 ð1Þ ¼ fd 4nþpþ2 g and membranes labelled by 2 contain objects a j (1 j p) if clause C j is not satisfied by the corresponding truth assignment, and can contain t objects n 1 (0 t 2 n ). At the ð4n þ p þ 4Þ-th step rule ½ n 1 2 ½ d 4nþpþ2 1 ! ½ n 2 2 ½ 1 is fired, since object d 4nþ3 has not been consumed by any rule from 4.3, creating an object n 2 in a membrane labelled by 2. Then, at configuration C 4nþpþ4 we have C 4nþpþ4 ð1Þ ¼ ; and membranes labelled by 2 contain objects a j (1 j p) if clause C j is not satisfied by the corresponding truth assignment, and can contain t objects n 1 (0 t 2 n ), and one of them contains an object n 2 . At the last step of the computation, rule ½ n 2 2 ½ 0 ! ½ 2 ½ no 0 is fired, sending an object no to the environment. Then, at configuration C 4nþpþ5 we have that C 4nþpþ5 ð1Þ ¼ ; and membranes labelled by 2 contain objects a j (1 j p) if clause C j is not satisfied by the corresponding truth assignment, and can contain t objects n 1 (0 t 2 n ), and there will be an object no in the environment. Here, the computation halts and returns a negative answer.

Result
Proof The family of P systems previously constructed verifies the following: • Every system of the family P is a recognizer P systems from TSECð2; 2Þ. • The family P is polynomially uniform by Turing machines because for each n; p 2 N, the rules of Pðhn; piÞ of the family are recursively defined from n; p 2 N, and the amount of resources needed to build an element of the family is of a polynomial order in n and p, as shown below: -Size of the alphabet: 9n 2 p þ 6n 2 þ 3np 2 2 À 3np þ 22n þ p 2 2 þ 13p 2 þ 14 2 H ðmaxfn 2 p; np 2 gÞ. -Initial number of cells: 2 2 Hð1Þ. -Initial number of objects in cells: n 2 þ nðp þ 2Þ þp þ 3 2 Hðn 2 Þ.
• The pair (cod, s) of polynomial-time computable functions defined fulfil the following: for each input formula u of SAT problem, sðuÞ is a natural number, codðuÞ is an input multiset of the system PðsðuÞÞ, and for each n 2 N, s À1 ðnÞ is a finite set. • The family P is polynomially bounded: indeed for each input formula u of SAT problem, the deterministic P system PðsðuÞÞ þ codðuÞ takes exactly 4n þ p þ 5 steps, being n the number of variables of u and p the number of clauses.
• The family P is sound with regard to (X, cod, s): indeed, for each formula u, if the computation of PðsðuÞÞ þ codðuÞ is an accepting computation, then u is satisfiable. • The family P is complete with regard to (X, cod, s): indeed, for each input formula u such that it is satisfiable, the computation of PðsðuÞÞ þ codðuÞ is an accepting computation.
Proof It suffices to notice that SAT problem is a NPcomplete problem, SAT 2 PMC TSECð2;2Þ , and the complexity class PMC TSECð2;2Þ is closed under polynomial-time reduction and under complement. h 6 Conclusions and future work Pan et al. (2018), a tight frontier of efficiency in the framework of tissue P systems with evolutional symport/ antiport rules and cell separation is defined by the length of the right-hand side of communication rules; that is, passing from 1 object to 2 objects is enough to pass from nonefficiency to presumable efficiency while the length of the left-hand side is at least 3. This result was demonstrated by giving a solution of the SAT problem by means of a family of P system from TSECð3; 2Þ. But an open problem was opened here: what happens with P systems from TSECðk; 2Þ (k ! 2)? Can we solve computationally hard problems restricting the length of the LHS to 2? In this paper, we focus on this topic, and we give an efficient solution to the SAT problem by means of a family of P systems from TSECð2; 2Þ, filling the gap previously open. Then, we can conclude here with a similar figure to the one presented in Pan et al. (2018), but while in the reference there are question marks in the second column from (2, 2) upwards, we have closed this problem giving demonstrating that with these types of P systems presumably hard computational problems can be efficiently solved.
Of course, after this work we can define several clear research lines to continue investigating these kinds of P systems.
-What happens when the environment ''disappears''? -Do the structure matter? By this we mean using celllike structure with this kind of rules. -In Song et al. (2017) another definition of length is given. Let k be the length of the rule defined as follows: if r ½ u i ½ v j ! ½ v 0 i ½ u 0 j , k ¼ j u j þ j v j þ j u 0 j þ j v 0 j. Then the complexity class of tissue P systems with evolutional communication rules with at most length k and cell separation is denoted by PMC TSECðkÞ . What are the borderline here? -What is the upper bound of these systems? Leporati et al. (2017) a characterization of tissue P systems with symport/antiport rules and both cell division and separation is given matching their efficiency to the class P #P , and it seems that this class of P system can reach the same complexity class. Capital Humano Línea 2. Paidi 2020, supported by the European Social Fund and Junta de Andalucía.
Funding Open Access funding provided thanks to the CRUE-CSIC agreement with Springer Nature.

Declarations
Conflict of interest The authors declare that they have no conflict of interest.
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons. org/licenses/by/4.0/.