| {"paper_meta":{"paper_id":"arxiv:0711.1177","title":"0711.1177","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"1\nConsiderations on the P NP question. \nAlfredo von Reckow. \n \nAbstract \n \nIn order to prove that the P class of problems is different to the NP class, we consider \nthe satisfability problem of propositional calculus formulae, which is an NP-complete \nproblem [1], [3]. It is shown that, for every search algorithm A, there is a set E(A) \ncontaining propositional calculus formulae, each of which requires the algorithm A to \ntake a non-polynomial time to find the truth-values of its propositional letters satisfying \nit. Moreover, E(A) βs size is an exponential function of n, which makes it impossible to \ndetect such formulae in a polynomial time. Hence, the satisfability problem does not \nhave a polynomial complexity. \n \n1. Some definitions. \n \nConsider L, the propositional calculus formal theory, as defined in Mendelson [3] and \nlet L+ be an extension of L containing the 0-letter formulas : (empty formula, or \ncontradiction) and (tautology). Let F be the countable set of formulas in L+. Let \ndenote the βlogic equivalenceβ of formulas, defined by A B iff A and B yield the same \nresult when evaluating its truth tables. \n \nIt can be easily verified that is a equivalence relationship. Let the quotient set be \nG = F / = {[A] / A F }. \n \nWe will use indistinctly the terms βatomβ or βpositive literalβ for a propositional letter. \nA βnegative literalβ is an atomβs negation. A βliteralβ is an atom or the negation of an \natom. \n \nA formula Μs cuasinorm . \n \nFor each A F , we define a quasi-norm || A || = number of different atoms in A. This \nquasi-norm has the following properties: \n \n For each A F , || A || 0. \n || A || = 0 iff A is or . \n Triangular inequality: For A and B in F ,, and any binary connective op, \n \n|| A op B || || A || + || B ||. \n \nIf A and B share atoms, the inequality is strict (including the case when A and B \n \nare the same formula). The equality occurs when A and B do not share any \n \natoms (including the case when A or B is or ). \n \nA property like || A || = | | || A || (for real) is not valid here. One way to define \nβmultiplicationβ of a formula by a positive integer is 2 A = A op A, for some binary \nconnective op. But this produces || n A || = A for any natural n. Furthermore, the \nproduct A makes no sense for any number which is not a positive integer. \n \nSince || A || | | || A || for real, the function || || is not a norm, so we call it a quasi-\nnorm.\n\n2\n \nIrreductibility \n \nTwo formulas can be equivalent, despite having different number of atoms. It is \npossible to have A B and || A || || B ||, because of the absorption laws: \n \np (p q) p \np (p q) p \n \nWe will say that a formula A is irreducible of order k iff || A || = k and it is not \nequivalent to another formula B which has less atoms than A. (i.e. if A B and || A || = \nk implies || B || k ). \n \nIn other words, A F is irreducible of order k iff \n|| A || = k and \n( B F ) (A B || B || k) \n \nA classβ cuasinorm. \n \nFor X G, letβs define \n||\n||\nmin\n||\n||\nB\nX\nX\nB \n \n, which counts the minimum number of atoms \nfiguring in a formula from the X equivalence class. \n \nHere again, || || is a quasi-norm, having the following properties: \n \n For every X G, || X || 0. \n || X || = 0 iff X is the class of all contradictions (which are equivalent to ) or \nthe class of all tautologies (which are equivalent to ). \n Triangular inequality: For X and Y in G,, and any binary connective op, \n \n|| X op Y || || X || + || Y ||, where we define X op Y = { F1 op F2 } for F1 X \nand F2 Y. \n \nThe set of order n irreducible formulas is defined by \n \nF n = { A / || A || = || [ A] || = n }. \n \nThen we have F = \n \n \nN\nn\nF\nX\nn\nX\n \n \n, \nAnd the set of all irreducible formulas is F = \nN\nn \n F n \n \nA formula is simply βirreducibleβ if it is irreducible of order n, for some natural number \nn. \n \nWe define the convertion function, \n \nf : F / F \n f( [ A ] ) = B, where B A and A is irreducible.\n\n3\nfrom each equivalence class [A], f takes an irreducible formula B. Equivalently, f can \nbe seen as an algorithm (explicity stated) which produces an irreducible formula B from \nthe class [A]. The function f converts any formula in the class [A] to the irreducible \nformula B. \n \nWe have B F n , for some natural number n, and [ B ] F n / \n \nWe define the normalization function h : F F , where h(A) is in disjunctive normal \nform (dnf). Then, for each equivalence class [A], h(f(A)) gets an irreducible and \nnormalized formula. \n \nFor any given n, let Μs consider the set of βminimalβ formulas having n propositional \nletters p1, p2, ...pn, which are the order n irreducible formulas. \n \nLet Gn = { [An ] / An is irreducible having order n }. Then G = \nN\nn \n Gn \nLet In = F n / { [An ] / B [An ] iff B is order n irreducible and B An } \n \nEach class of In contains only order n irreducible formulas. \n \nThen I = \nN\nn \nIn is the set of all equivalence classes having only irreducible formulas. \n \nIf we take one element of each equivalence class, we get the set of all irreducible \nformulas, pair wise non equivalent. \n \nNow we construct the image of the function f, taking the order n minimal formulas: \n \nHn = { Bn / Bn = f(X) for X In } where the Bn are not equivalent to each other, and \n \nH = \nN\nn \n H n. \n \nThen H is the set of all irreducible und unique formulas. In other words, each of the \nformulas in H is irreducible (because it has a minimum number of propositional \nletters, and it is impossible to eliminate a letter using absorption) and there are no two \nformulas in H being equivalent to each other (which makes each formula unique). \n \nFrom now on, we will work only with formulas from H . That is, we will assume that \nany given formula A is order n irreducible and we will ignore other formulas equivalent \nto A. \n \n2. Why this is a search problem?. \n \nWe will show that there is no algorithm to produce directly, in a straightforward way, \nthe truth values for the propositional letters in a formula, which make the formula true, \nunless there is a search involved. To this end, we show that this satisfability problem is \nequivalent to construct a βclosedβ formula for the roots of a polynomial with n \nvariables.\n\n4\n \nLet Μs define the function h, from L+ to the set of n-variable polynomials, by \n \n For a formula A of L+, having n propositional variables p1, p2, ..., pn, letβs \nassociate to each of its letters a numerical variable xi: h(pi)= xi for 1 i n. \n \n h(Β¬ A) = 1 β h(A) \n \nIf A and B are two formulas in L+, then \n \n h(A B) = h(A) h(B) \n h(A B) = h(A) + h(B) β h(A) h(B) \n h(A B) = (1 β h(A) ) (1 β h(B) ) + h(B) \n h(A B) = (1 β h(A) ) (1 β h(B) ) ( 1 + h(A) + h(B) ) + h(A) h(B) \n \n \nNote that \n \nh(A)n is equivalent to h(A) in the following sense: \nh(A)n = 0 iff h(A) = 0 and h(A)n = 1 iff h(A) = 1 \n \nNote that h is a truth function assigning to a formula A a value 0 or 1 when each of the \nvariables xi takes a value of 0 or 1. \n \nIn particular, in the resulting polynomial we can replace any factor xn by x. \n \nFor example, for the idempotency law: A A A we have: \n \nh(A A) = h(A) h(A) = h(A)2 h(A), \n \nwhere the last β β means equivalence in the sense that we get the same values for h(A)2 \nand for h(A) when each of the corresponding variables xi take a value of 0 or 1. \n \nSimilarly, with the other idempotency law, A A A, we get \nh(A A) = h(A) + h(A) β h(A)h(A) = 2h(A) β h(A)2 2h(A) β h(A) = h(A). \n \nTautologies and contradictions: \n \nIf a formula A is a tautology, then h(A) 1. \nIf a formula A is a contradiction, then h(A) 0. \n \nExample 1. \n \nFor the tautology A A we have: \nh(A A) = (1 β h(A))2 + h(A) (1 β h(A)) + h(A) = 1. \n \nThis latter equivalence can be proved in the following ways:\n\n5\na) (1 β h(A))2 (1 β h(A)), because both sides have the same value when h(A) = 0 or \nwhen h(A) = 1. \n \nb) (1 β h(A))2 + h(A) = 1 β 2h(A) + h(A)2 + h(A) = 1 β h(A) + h(A)2 \n 1 β h(A) + h(A) = 1 \n \nExample 2. \n \nFor the contradiction A Β¬A whe have: \n \nh(A Β¬A) = h(A) (1 β h(A)) = h(A) β h(A)2 h(A) β h(A) = 0. \n \nFor any formula A, the problem of finding the truth-values for its variables p1, p2, ..., pn \nsatisfaying A (i.e. making it true) is equivalent to the problem of finding the \ncorresponding numerical values for n-tuple (x1, x2, ..., xn) which are zeros of the \nfunction g, where g(A) = h(A) β 1. \n \nThis function is a polynomial with the n variables (x1, x2, ..., xn), which will be called \nthe formulaβs βcharacteristic polynomialβ, and has the form g(A) = q, where \n \nq : [0, 1]n R, takes values on the n-dimentional unit cube, and is defined by \n \n(1) \n \n \n \n \n \n)\n(\n1\n2\n1\n1\n...\n)\n,...,\n(\n2\n1\nn\nr\ni\nn\ni\nn\nin\ni\ni\nx\nx\nx\nx\nx\nq\n \n \n \n \n \n \n \nwhere each exponent ki has a value of 0 or 1, the coefficients i and are integers, \nand the variables xi have a value of 0 or 1, and r(n) = 2n. \n \nWe define 00 = 1. \n \nThe problems then is to find a value for each variable xi, either a 0 or a 1, in such a way \nthat q(x1, x2, ..., xn) = 0. Or equivalently: \n \n(2) \nFind the roots of the polynomial q \n(3) \nWith the restrictions xi in {0, 1} for i = 1, ..., n. \n \nIn other words, find q-1 (0) {0, 1}n. \n \nLemma. \n \nThere is no closed formula to calculate directly the roots of this type of polynomials. \nHence, there is no algorithm to find directly (without making any search) the truth \nvalues of the propositional letters satisfying a formula A. \n \nProof. \n \nThe equation g(A) = 0 has n variables. It can be seen as a βsystemβ of 1 equation with n \nunknowns. This is an under determined system, having infinite real roots. To\n\n6\ndetermine them, it suffices to solve g(A) = 0 for x1, which will then depend on the other \nindependent variables x2, ..., xn. \n \nSince these variables are independent (there is no relationship between them), each of \nthem can take any real value. Once we chose a value for each variable x2, ..., xn, that \nwill produce a value for x1. \n \nFrom these infinite set of values for the n-tuple (x1, ..., xn) .we have to separate those \nwhere all the variables are 0 or 1. \n \nBecause there are no additional conditions on the variables, no further relationship \nbetween them can be obtained, and this happens just because the variables x2, ..., xn are \nindependent. There is only the restriction (3 ) on the value of the solutions, but it is of \nno use for finding a closed formula for the solutions of (2). \n \nHence, we have no other way left, but to carry on an exhaustive search, and we have no \nadditional information derived from the formula which could lead us to take any \nshorcuts and reduce the number of cases (or number of rows in the truth table) to be \nconsidered. \n \nExample. \n \nFor the formula \n \nA : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3) \n \nThe associated polynomial for A is \n \nh(A) = x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 \n \nThe characteristic polynomial is \n \ng(A) = x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 \n \nand the equation g(A) β 1 = 0 is \n \nx1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 = 0. \n \nWith the restrictions xi {0, 1} for i = 1,.., 3. \n \nIn this equation, the variable x1 depends on x2 and x3. This two last variables have no \nfurther relationship to one another. If x3 takes any given value, it has no influence at all \non the values that x2 can take. The only thing we know is that we are interested in the \ncases where x2 as well as x3 take a value of 0 or 1. \n \nHence, we have no choice but to make an exhaustive search, using the 4 possible values \nfor the duple (x2, x3). The solutions will be those cases where the equation makes sense \nand yielding values of 0 or 1 for x1.\n\n7\nNote that if (x2, x3) = (1, 1) then x1 vanishes and the equation becomes 1 = 0, which \nmakes no sense. Hence (x2, x3) (1, 1). \n \nFrom other point of view, solving for x1 we get \n \n(4) \n1\n)\n(\n2\n3\n1\n2\n3\n2\n3\n2\n3\n2\n3\n2\n1\n \n \n \n \n \n \n \nx\nx\nx\nx\nx\nx\nx\nx\nx\n \n \nwhere the denominator is zero if (x2, x3) = (1, 1). It is zero for other values of these \nvariables, but it does not matter, because in none of those other cases both variables \nhave a value of 0 or 1. \n \nNote that x2 and x3 can take any real value in (4), as long as the denominator does not \nvanish. There are infinite values for x2 and x3 satisfying (4), but we are interested only \nin the 4 cases where they are 0 or 1, and there are no shorcuts for finding them. We \nmust try all the options and see which ones work. \n \nIn the general case, for a formula having n propositional letters, we can not avoid the \nneed to explore 2n-1 options for the variables x2, ...xn. \n \nOther way to search is the following: \n \nIf we put arbitrarily x1 = 0 in (4), then 2x2x3 β x2 β x3 + 1 = 0, and from here we get x2 = \n(x3 β 1) / (2x3 β 1). Here we have two cases: if si x2 = 0, then x3 = 1 and if x2 = 1, then \nx3 = 0. \n \nIf we put x1 = 0 in (4), then 2x2x3 β x2 β x3 + 1 = 3x2x3 β 2(x2 + x3) + 1, so x2 = x3 / (x3 β \n1). Here again we have two cases: if x2 = 0, then x3 = 0 and if x2 = 1, then x3 = x3 β 1, \nwhich is impossible. \n \nWe have had to analize 4 different cases, for 3 variables, corresponding to 2n-1 options, \nfor n = 2, and the variables x1, ...xn-1 \n \nWe can solve for any of the n variables as a function of the other n-1 variables. In any \nevent, we must try 2n-1 options. \n \nNotes. \n \n1. Note that the characteristic polynomial g has n variables and has 2n terms. Each term \nis a product of k variables, and there are C(n, k) = n! / (k! (n-k)!) combinations of terms \nhaving k varibles. Adding that up, there are \n \n \n \n \n \n \n \n \n \n \nn\nk\nn\nk\nn\n0\n2 \n \nterms, each one has one coefficient and n exponents. Any algorithm processing g has \nthus an input of (n+1) 2n numbers, an exponential function of n, which can not be \nprocessed in a polinomial time.\n\n8\nMoreover, if there were a βclosed formβ formula to determine the binary roots of g, then \nit would depend on the 2n coefficients of g. Since the variables x2, ..., xn are \nindependent, a closed form for each of them would be a function of those 2n \ncoefficients, and it cannot be calculated in a time t = q(n), where q is a polynomial. \n \nSuch a formula, if it existed, would behave not better than a sequential exhaustive \nsearch. \n \n2. The characteristic polynomial g has n variables x1, x2, ..., xn each one with an \nexponent 0 or 1. In any occurrence of any of this variables the exponent can be replaced \nby an odd natural number k, and the resulting polynomial has the same βbinary rootsβ \n(those being 0 or 1) of g. The exponent must be 0 or odd to avoid de case (-1)2 = 1. \n \nHence, g is equivalent to infinitely many polynomials (in the sense that all of them have \nthe same binary roots). \n \nIf there were an explicit formula to find the binary roots of g, that same formula would \nyield the binary roots of all the polinomials q equivalent to g, which can have an \narbitrarily high degree. \n \nThese comes from the idempotency laws A A A and A A A. \n \n3. It is not possible to determine, a priori, additional conditions to find the binary roots \nof g. If there were n-1 such additional conditions, then we would have a non-linear \nsystem of n equations with n variables. \n \nThe number of solutions for the characteristic polynomial can be anything from 0 (in \nthe case of a contradiction) to 2n (for a tautology). If all this options are possible, then \nno additional conditions can exist. For a formula like p1 p2 ... pn, having 2n β 1 \nsolutions, any additional condition must be redundant. For a formula having 2n β 2 \nsolutions, there can be at most one additional condition, and son on. The number of \nadditional conditions depends on the number of solutions, which is not known in \nadvance, so we donβt know beforehand how many conditions are necessary for a given \nformula. \n \nMoreover, if such conditions depend on the number of solutions, and the solutions \ndepend on the coefficients of the characteristic polynomial, then those conditions will \ndepend on those coefficients. Since there are 2n coefficients in the characteristic \npolynomial, the additional conditions will depend on 2n inputs, which can not be \nprocessed in a polynomial time. \n \nOn the other hand, if we introduce, arbitrarily, any condition g1(x1, ..., xn) = 0, then \nsome roots can be missed. \n \nFor example, if we put arbitrarily the condition x1 = 0, then we will miss all the roots \nwhere x1 = 1, in the cases where such roots exists. In the cases where g has a factor (x1 \nβ 1) then we will miss all its binary roots.\n\n9\nIf we put a condition like x1 β x2 = 0, then we might miss the roots where x1 β x2, and \nwe do not know beforehand if that is indeed the case. Simple conditions as these can \nbe, of course, verified beforehand, but more complex conditions can not. \n \nIn the example \n \nA : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3) \n \nThe characteristic polynomial is \n \nx1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 = 0. \n \nIf we add the condition x1 β x2 = 0, (which can be considere a mere supposition) the \ncharacteristic polynomial becomes \n \nx3 - x1x3 β 1 = 0. \n \nWhere x1 must be zero and x3 must be 1. x2 can be 0 or 1. \n \nThe triplet (0, 1, 0) is a root of the characteristic polynomial, but the other triplet (0, 1, \n1) is not. \n \nAdding an arbitrary condition not only leads us to miss two roots, but it introduces a \nfake root. \n \nWe have no additional information and we cannot add it arbitrarily, because we donβt \nwhere to start looking for the roots. \n \nAs have been noted in the literature (see [3]), if we know the roots (or if we a have a \ncandidate for root, somehow βguessedβ or estimated), we can replace it in the \ncharacteristic polynomial and check it in a polynomial time. \n \nBut if we donβt have a root, we have to search for it, in a sequential way. We must try \nall the options. \n \n4. Multiplying the factors of g to get the explicit form (1) requires o(mn) multiplications, \nwhere m is the number of factors and n is the number of variables. Instead of carrying \non the multiplications, we can consider the associated polynomial, when it is already \nfactored and find its roots. The remaining n-tuplets (those not being a binary root of h) \nwill be the binary root of the characteristic polynomial g. \n \nThe idea is to take advantage when h is already factored (as this might take less \noperations). \n \nIn the example, \nA : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3) \n \nThe associated polynomial is \n \nh(A) = (x1 + x2 + x3 - x1x2 β x1x3 β x2x3 + x1x2x3 )(1 - x1x2) (1 - x1x3) (1 β x2x3)\n\n10\n \nh(A) = 0 if al least one of its factors is zero. \n \nThe last thre factors give the options \n \nx1 = 1, x2 = 1, any value for x3 \nx1 = 1, x3 = 1, any value for x2 \nx2 = 1, x3 = 1, any value for x1 \n \nThe first factor gives \n \nx1 = (x2 + x3 - x2x3) / (1- x2 - x3 + x2x3) = U / ( 1 β U), for U β 0, \n \nWhere U = x2 + x3 - x2x3. All the options are: \n \nx2 \nx3 \nU \nx1 \n0 \n0 \n0 \n0 \n0 \n1 \n1 \nnot definided \n1 \n0 \n1 \nnot definided \n1 \n1 \n1 \nnot definided \n \n \n \n \nThe roots of the associated polynomial h are \n \nx1 \nx2 \nx3 \n0 \n0 \n0 \n0 \n1 \n1 \n1 \n0 \n1 \n1 \n1 \n0 \n1 \n1 \n1 \n \nSince g = h β 1, for any other triplet (x1, x2, x3), h(x1, x2, x3) is not zero, and hence it \nmust be h(x1, x2, x3) = 1. Those other three triplets must be the binary roots of g. They \nare: \n \nx1 \nx2 \nx3 \n0 \n0 \n1 \n0 \n1 \n0 \n1 \n0 \n0 \n \nBut this has required to explore all the options for the first factor. It takes an o(2n-1) \nnumber of operations, for n = 3. \n \nThe cases when two or more factors are simultaneously zero produce repeated roots, \nthat have to be eliminated from the list of roots. \n \n \n3. Exhaustive search \n \nIf there is no direct (βclosed formβ) formula to find the binary roots of the characteristic \npolynomial of a given formula, we must carry on a sequential search, checking a\n\n11\nformulaβs truth table in a sequential way, one row after another. Since there are no \nadditional conditions to guide the search, it must be made sequentially. \n \nFor the set of order n irreducible formulas, we choose an order of evaluation for the \npropositional letters and for the truth value for each atom. First, we take some \npermutation of the letters. Then, for every letter we choose if it takes the value 0 or 1 \nfirst. Then for every order of evaluation there are formulas taking necessarily an \nexponential time to find an n-tuple of truth values satisfying it. \n \n3. 1. Sequential search algorithms. \n \nIn this section we consider βuninformedβ or βblindβ search algorithms, which uses no \nadditional information (other than the given formula itself) to guide the search or to \nreduce the number of options. As was shown in section 2, such information is not \navailable beforehand, and if it were, taking it into account requires an exponential order \nalgorithm, which is no better than a sequential uninformed search. \n \nFor this reason, in this section we well consider only algorithms exploring a formulaβs \ntruth table in some order that does not depend on the formula. That is, an algorithm A \nwill explore the truth tableβs rows in the same fashion for all formulas. \n \nA βsequential search algorithmβ assigns truth values to the propositional letters p1, p2, \n..., pn occurring in a formula F in some pre-established order (out of the n! permutations \nof the n letters) and for each letter it is established if 0 or 1 is used first (out of the 2n \noptions). \n \nConsider the set of all sequential search algorithms. This set has n! 2n algorithms, \nbecause the n letters can be arranged in n! different permutations and each letter can be \nevaluated to 0 first and then to 1, or the other way around. If a formula F has n atoms ( \n|| F || = n) there are n! 2n different ways to evaluate F, depending on the order the atoms \nare evaluated and if the 0 or 1 is evaluated first for every atom. \n \nTwo sequential search algorithms A1 and A2 are βsequentially equivalentβ if and only if \nthey assign the same truth values to the same atom in the same order. The algorithms \ncan use different functions or procedures, yielding the same results in the same order, \nbut their implementations can yield different running times on a given machine, \noperating systems and programming language. \n \nTwo sequential search algorithms A1 and A2 are not equivalent, and will be considered \ndifferent, if they take the atoms in a different order or if they assign the truth values 0 \nand 1 in a different order to the same atom. \n \nEquivalently, two sequential search algorithms A1 and A2 are different if they explore \nthe truth table rows in a different order. \n \nThis sequential search is equivalent to a depth-first search in a binary tree, with a given \npriority for the order in which the atoms are evaluated and the truth value given to each \natom.\n\n12\nSequential equivalence is an equivalence relationship, so we can take an algorithm from \neach equivalence class. We will assume that we take the algorithm taking the least time \nin a particular implementation. \n \nIn order to find out if a given formula F is satisfacible, and when it is, find the \ncorresponding truth-values of its atoms, a sequential search algorithm A explores the \ntruth table rows in a certain order. The algorithm A βsolvesβ the formula F if it finds a \ntruth-value assignation for Fβs atoms, with which F is evaluated true. \n \nWe will represent the truth table for F, as evaluated by A, by A(F) = (b(1), b(2), ..., \nb(2n)), where b(k) is the value obtained in row k. \n \nWe define the function L as the function giving the first row where algorithm A finds a \nvalue of 1 when it evaluates the formula F. This is L(A(F)) = \nn\nk 2\n1min\n \n \n{ k / b(K) = 1 }. \n \nExploring with the initial order. \n \nTake F n, the set of order n irreducible formulas. Take a formula F in F n. Then || F || = \nn, and consider p1, p2, ..., pn as the n distinct atoms figuring in F, taken in the same \norder they appear in F, from left to right. We will call this the βinitial orderβ of the \natoms in F. \n \nDenote by Ai,j the algorithm which uses the i-th permutation of the atoms of F and the j-\nth way of giving the truth values to them. \n \nThe sequential search algoritm A1,1 explores Fβs truth tableβs rows in the βnaturalβ \norder, taking the atoms p1, p2, ..., pn in the same order they appear in F and for every \none of them it evaluates first the 0 and then the 1 values. \n \nThen, for the formula \n \nF: \n p1 p2 ... pn \n \nthe sequential search algoritm A1,1 must check all the rows in the truth table, to find a \nsolution only in the last row (that is, in row 2n ). Thus, L(A1,1(F)) = 2n. \n \nChanging 0s and 1s. \n \nTake now all the A1,j algorithms. All of them evaluate the atoms p1, p2, ..., pn in the \norder 1, 2, ..., n. They differ from each other because they assign the values 0 and 1 in \na different order for every letter (one of them will assign first 0 and later 1 to some atom \npi, while other one assigns first 1 and later 0 to the same letter pi ). \n \nMore precisely, if the base 2 representation of j is j = d1d2...dn, then the atom pk takes \nfirst the truth value dk and later it will take the truth value 1 β dk , for 1 β€ k β€ n. \n \nFor every j, if the A1,j algorithm evaluates first true the atom pk (that is, if dk = 1), then \nlet`s take the letter qk equal to pk. Otherwise, take qk equal to pk. Then the formula\n\n13\n(5) \n \nF: \n q1 q2 ... qn \n \nis true only when all the qj are false. To get there, the algorithm A1,j must explore first \nall the other rows in the truth table. In this case, we also have Thus, L(A1, j(F)) = 2n. \n \nThus, the algorithm A1, j(F) has an order o(2n) temporal complexity. \n \nPermuting the order of the atoms in F is irrelevant, since is has no effect on the rows \nthat have to be evaluated. In any case, the formula F will take an exponential time: \nL(Ai, j(F)) = 2n, for 1 β€ i β€ n! \n \nThe set Cn(Ai,j). \n \nFor an integer n given, we define the set Cn(Ai,j) as the set of order n irreducible and \npair-wise non equivalent formulas for whom algorithm Ai,j takes a non polynomial time. \n \nThe formula F given above in (5) is in Cn(Ai,j). It the number of formulas in this set \n(representing each one an equivalence class) where a polynomial depending on n, we \ncould detect this kind of formulas before applying the algorithm. Then we Ai,j would \nexecute in a polynomial time for the other formulas in F n. \n \nThis is not the case, as it will be shown that Cn(Ai,j) = o(g(n)), where g is an \nexponential-type function. \n \nFor every sequential search algorithm Ai,j and every formula F, it is posible to apply a \npre-process to determine if F is equivalent to a worst-posible case formula, (and in this \ncase, to evaluate it at once). Letβs call this revised algorithm Ai,j(1). \n \nBut then there is another formula F(1) forcing the revised algorithm Ai,j(1) to search all \nthe way thru the penultimate row in the truth table of F(1), because that is the only row \nwhere F(1) is true. If we repeat the process to construct algorithm Ai,j(2), which detects \nF and F(1), then the formula F(2), which is true only in the next-to-penultimate row, \nwill take an exhaustive search from row 1 to row 2n-2, and so on. \n \nMore precisely, we have two steps in the analysis: \n \nLemma 1. \n \nFor every natural number m, 1 β€ m β€ 2n, there is a formula F having n atoms, that \nis true only in row m. \n \nFor each sequential search algorithm Ai,j and every m {1, ..., 2n}, we construct a \nformula being true only on row m, by taking the formula \n \nF: \n pi1 pi2 ... pin \n \nwhich is true only in the last row of the truth table, and changing each atom for its \nnegation, depending on the order in which algorithm Ai,j assigns values to each atom.\n\n14\nTo this end, we take the base 2 representation of the index j: j = d1d2...dn. Suppose the \natom pk takes first the value dk and later on the value 1 - dk. \n \nTake the base 2 representation of m: m = e1e2...en, \n \nIf ek = dk, then qik(m) is pik, otherwise, qik(m) is Β¬ pik. \n \nThen the formula Gij(m): \nΒ¬ qi1(m) Β¬ qi2(m) .. .Β¬ qin(m) \n \nIs true only in row m and it is false in all the other rows. \n \n \n \n \n \nLemma 2. (from the end of the truth table). \n \nFor every sequential search algorithm Ai,j, there is an order n irreducible formula F, that \nis true only in the last 2 rows explored by the algorithm. \n \nProof. \n \nThis formula is constructed taking the disjunction \n \nGij(2n-1) Gij(2n) \nFor this formula, algorithm Aij(1) must explore 2n-1 rows in the truth table. \n \nOf course, we can detect this formulae and give them a different treatment, with a pre-\nprocess. \n \nCall Aij(2) be the algorithm obtained by adding such pre-process to Aij(1), to eliminate \nthe βworst possible casesβ for Aij and Aij(1). \n \nConsider now the following formula, which is true only in the last three rows, as \nexplored by algorithm Aij(2): \n \nGij(2n-2) Gij(2n-1) Gij(2n) \n \nAs well as the formulas \n \nGij(2n-2) Gij(2n-1) \nGij(2n-2) Gij(2n) \n \nWe can separate this formulas with a new pre-process, to get the algoritm Aij(3), and so \non. \n \nContinuing this way, for each algorithm Aij we construct a succession of algorithms \nAij(1), Aij(2), Aij(3), .... Every one of them has a set of worst-possible case formulas, \nand every one avoids the worst-possible case for the previous algorithms. \n \nThe worst-possible case for Aij(2n-1) is a formula whose truth table is true in all its \nsecond-half rows and false in all its first-half rows. This algorithm has exponential \ncomplexity.\n\n15\nTo find out if a formula F is true from row k, for k = 2n, 2n-1, 2n-2, ..., 2n-1 + 1, we have \nto check out 2n-1 cases. If none of them happens, then we must explore sequentially \nrows 1 to 2n-1. The worst possible case occurs with the formula which is true only in \nrow 2n-1. \n \nThe time complexity is therefore: \n \nRows to check out in the pre-process: \n2n β (2n-1 + 1) + 1 = 2n-1 \nRows to explore in the process \n \n2n-1 \n---------------------------------------------------------- \nTotal \n \n \n \n \n2n \n \nMoreover, for every k, algorithm Aij(k) has exponential complexity. \n \nWe have to check first (in the pre-process) for k worst-possible case situations of Aij(h) \nh = 1, ..., k-1. For the formulas passing this check, there are the worst-possible cases \nAij(k) demanding to explore 2n β k rows. Hence, in the worst-possible case, Aij(k) must \nexplore k + (2n - k) = 2n rows. \n \n \n \n \n \n \n \n \nConclusion: \n \nFor every way used to explore a truth-table, there is a formula requiring to explore all \nthe other rows first. If we try to avoid this worst-possible cases adding a pre-process to \nthe search algorithm, then there is another formula requiring also to explore all the other \nrows first. \n \nThis applies to any sequential search algorithm that does a βblindβ or βuninformedβ \nsearch, where no additional information is used to guide the search or to discard \noptions, which will be analized in section 3.2. \n \nAdditional notes. \n \n1. Additional complexity. \n \nThere are additional sources of complexity: each equivalence class in G = F / \ncontains formulas with an arbitrarily large number of atoms, which are equivalent to a \nformula with n atoms, due to the absorption laws. \n \nThis requires to first simplify a formula, adding extra time to the process. For an n-\natom formula F, it can be constructed infinite many other formulas with m atoms, for \nany m > n, equivalent to F by the absorption laws. Because of this, an algorithm \nprocessing formulas with n atoms will have to deal also with formulas having m > n \natoms to find out if they can be simplified to a formula with n atoms. Since m can be \narbitrarily large, the time needed will be arbitrarily large too. \n \nThatβs why we have supposed that the input formulas for any algorithm are already \norder-n irreducible. \n \n2. Extending lemma 2.\n\n16\n The Aij algorithms take into account only permutations in the order of evaluation of the \natoms and of the truth value for each atom. There are 2n n! such permutations. The \ntruth table for an n-order formula has 2n rows, having 2n! permutations. Since the proof \nof lemma 2 includes a constructive method to build a formula that is true in any given \nrow (and only in that row), that proof can be extended to cover the case of an arbitrary \npermutation of rows of the truth table. \n \n3.2. Heuristic Algorithms. \n \nAn heuristic algorithm does not explore all the rows in a formulaβs truth table. It \neliminates arbitrarily some of the rows, hoping there is a solution in the remaing rows. \nSuch algorithm will have a polynomial complexity if it explores only (n) rows for an \norder n irreductible formula F, whre is a polynomial. \n \nThis means that the algorithm eliminates a priori all the other rows. The algorithm will \nbe βacceptableβ for a formula F if some rows yielding 1 are among the explored rows. \n \nLema 3. \n \nFor every polynomial complexity heuristic algorithm H, there is a set of formulas C(H) \nfor which H is not acceptable. That is, for every formula in F in C(H), H does not \nexplore any row of F where the result is 1. \nIn other words, we can not eliminate rows arbitrarily, because that might eliminate \nprecisely those rows having a result of 1. \n \nProof. \n \nLet F be an order n irreducible formula and a polynomial depending only on n. Let Hj \nbe one of the \n \n \n \n \n \n \n)\n(\n2\nn\nn\n \nalgorithms having polynomial complexity of order (n), which \nexplores only rows 1, 2, ..., (n), evaluating the atoms p1, p2, ..., pn in the same order \nthey figure in F, from left to right. \n \nFor any n-order formula, each algorithm Hj evaluates only rows i1, i2, ..., \n)\n(n\ni \n. This \nrows are the same for all n-order formulas. The row indexes depend only on the \nalgorithm j-index and not on any given formula. \n \nTake k1 as one of the rows not explored by Hj. With the same method used in lemma 2, \nwe can construct a formula that is true only in row k1: \n \nIf atom pi is true in row k then take qi equal to pi. Otherwise take qi equal to Β¬ pi. Take \nthe conjunction of the qi . \n \nGk1: \nq1 q2 ... qn \n \nThis formula is true only in row k, and algorithm Hj missed it, since it does not explore \nthis row.\n\n17\n( Moreover, Hj will miss a solution for any disjunction of formulas Gk1 Gk2 ... Gkr, \nwhere k1, k2, ..kr are not explored by Hj) \n \nLet Hj(1) be the extensiΓ³n of Hj obtained by adding a pre-process to check row k1 \nbeforehand. Then we can construct in a similar way another formula where Hj(1) fails \nfor some formula Gk2 Μ§where row k2 is not explored by Hj(1). And then we can extend \nHj(1) to Hj(2), wich in turn will fail for some formula Gk3. \n \nThis produces a list of algorithms Hj, Hj(1), Hj(2), ..., Hj(t) The list is finite, with a \nrestriction: the last index, t, must depend polynomially on n: t = q(n) for some \npolynomial q. \n \nBut the truth table has 2n rows an hence we always can take one excluded row and build \na formula being true only on that row, causing Hj(t) to fail, no matter what polynomial \nfunction q is used. \n \n \n \n \n \n \n \n \n \n \nIn conclusion, there is a set C(Hj) having a number of equivalence classes w, where w is \nan exponential function of n, and for any formula F in a class X of C(Hj), Hj does not \nevaluate the rows where F is true, and therefore misses every solution for F. \n \n \n4. Conversion algorithms. \n \nIt is quite simple to find the truth-value assignations for the atoms satisfiying a formula \nexpressed in disjunctive normal form (dnf). There are 3 cases: \n \n1. The formula is a tautology, and is equivalent to . It is true for any assignation of \ntruth values to its atoms. \n \n2. The formula is a contradiction, and is equivalent to . It is false for any assignation \nof truth values to its atoms. In each of its βdisjointβ the formula must have the \nconjunction of a complementary pair (if some disjoint does not have any such \nconjunction, this disjoint would have a truth value assignation satisfying the formula, \nwhich is impossible). \n \n3. The formula is a contingency. It has at least one βdisjointβ and it suffices one of \nthem to be true. \n \nIn any case, it can be immediately checked out if the formula is satisfacible or not. If a \nformula is not equivalent to a contradiction, it must be satisfacible. All we need is to \ncheck sequentially the disjoints for the absence of complementary pairs. \n \nThis leads to the following algorithm: \n \nTo find the truth values satisfying a formula F \n \n1. Convert F to its disjunctive normal form. \n2. Check the disjoints sequentially from left to right until a disjoint D having no \ncomplementary pairs is found. \n3. Assign 1 to each literal in D. This will make some atoms true and some false.\n\n18\n \nApparently this is a polynomial complexity algorithm. The problem is: the conversion \nprocess (to get a formulaβs d.n.f.) is a search process in a tree, having exponential \ncomplexity. \n \nLema 4. \n \nFor every convertion algorithm there is an order n formula F requiring a time t = q(n) \nwhere q is an exponential type function. \n \nProof. \n \nA clause is a disjunction of literals, some of them positive and some negative ones. For \na given value of n, take a formula F in conjunctive normal form, having k clauses, each \nhaving m literals (obviously m β€ n). Converting F to d.n.f. we get mk disjoints, where \neach of them has k literals. \n \nIf k = n and m is close to n, the number of disjoints is in the order of nn. \n \n \nNote that evaluating the truth table requires to evaluate 2n rows, much less than nn, \nalthough both algorithms have exponential complexity. \n \n \n5. Asymptotic properties. \n \nFor a given order n, take a representative from each equivalence class F n / . We can \nidentify each class with its corresponding truth table. For any n-order formula, its truth \ntable has 2n rows. Since each row can have a resulting 0 or 1, there are \nn\n2\n2\ndifferent \ntruth tables, and therefore there are \nn\n2\n2\n equivalence classes. \n \nIn one half of all classes in F n / the formulas have true in the first row of its truth \ntable. In the other half, the formulas have false in the first row. \n \nThe formulas belonging to \n1\n2\n2\n \nn\n classes have 1 in its first row, while the formulas \nbelonging to the other \n1\n2\n2\n \nn\n classes have a 0 in its first row. \n \nIf we put a 0 in row 1, then every one of the remaining 2n-1 rows can have a 0 or 1, \ngiving a total of \n1\n2\n2\n \nn\n options. \n \nIn one fourth of all classes in F n / the formulas have 1 in row 2, knowing they have 0 \nin row 1. In a similar way, knowing rows 1, 2, ..., m-1 to have a 0, there are \nm\nn \n2\n2\n \nequivalence classes having a 1 in row m. \n \nThe number of classes whose formulas have its first 1 in row 1 or 2 or ... or m equals \nthe sum \n \nq(m) = \n1\n2\n2\n \nn\n + \n2\n2\n2\n \nn\n + \n3\n2\n2\n \nn\n+... + \nm\nn \n2\n2\n = \nn\n2\n2\n (2-1 + 2-2+ 2-3+... + 2-m)\n\n19\n= \n \n \n \n \n \n \n \n \n \n \n \n \n \nm\nn\n2\n1\n1\n22\n \n \nThe percentage of classes whose formulas are true for the first time in one of the rows 1, \n2, ..., m is then: \n \nm\nm\nm\nm\nr\n2\n1\n2\n100\n2\n1\n1\n100\n)\n(\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nLet be a s degree polynomial, for a fixed s. If m = (n), then r(m) is the percentage of \nclasses whose formulas can be solved in a polynomial time (meaning by βsolveβ to find \na n-tuple of truth values satisfying the formula). \n \nTake (x) = xs. Then m = ns and r(m) = r(ns) = \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \ns\nn\n2\n1\n1\n100\n \n \nWe get \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \ns\nn\nn\ns\nn\nn\nr\n2\n1\n1\n100\nlim\n)\n(\nlim\n= 100%. \n \nConclusion: \n \nWhen n increases, the ratio of equivalence classes having satisfactible formulas, which \nare detected by a sequential search algorithm exploring only the first (n) rows (or any \nset of pre-determined (n) rows) approaches 100%. \n \nNotes. \n \nThe ratio of order n satisfactible formulas not solved by a sequential search algorithm \nexploring (n) rows decreases to 0 as n increases. \n \nHowever, the number of equivalence classes increases very fast, as shown in the \nfollowing table: \n \nRows in table \nEquivalence classes ( = number of Truth tables) \nn u = 2n \n2u \n1\n 2 \n 4 \n2\n 4 \n 16 \n3\n 8 \n 256 \n4\n 16 \n 65.536 \n5\n 32 \n4294967296\n6\n 64 \n1,84467E+19\n7\n 128 \n3,40282E+38\n8\n 256 \n1,15792E+77\n9\n 512 \n1,3408E+154\n10\n 1.024 \n#Β‘NUM! \n11\n 2.048 \n#Β‘NUM!\n\n20\n12\n 4.096 \n#Β‘NUM! \n13\n 8.192 \n#Β‘NUM! \n14\n 16.384 \n#Β‘NUM! \n15\n 32.768 \n#Β‘NUM! \n16\n 65.536 \n#Β‘NUM! \n17\n 131.072 \n#Β‘NUM! \n18\n 262.144 \n#Β‘NUM! \n19\n 524.288 \n#Β‘NUM! \n20\n 1.048.576 \n#Β‘NUM! \n \nEven tough the ratio of equivalence classes (causing exponential complexity to an \nalgorithm) vs. total number of classes decreases, this latter number increases quite fast, \nas the following table shows: \n \nn \n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n1\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n4\n1\n3\n32\n4\n4096\n1\n5\n134217728\n128\n6\n2,8823E+17\n268435456\n7\n2,65846E+36\n6,04463E+23\n8\n4,52313E+74\n6,2771E+57\n9\n2,6187E+151\n5,5453E+129\n10\n1,7556E+305\n1,4181E+278\n16777216\n11\n#Β‘NUM! \n#Β‘NUM! \n6,8946E+215\n12\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n13\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n14\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n15\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n16\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n1\n17\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n18\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n19\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n20\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n \nIn the blank entries the calculation formula is not applicable since 2n < ns. The #Β‘NUM! \nmeans an overflow; the numeric limits of Excel on a Pentium computer has been \nexceeded. \n \nIncreasing speed for the function r. \n \nAs the following table shows, r(ns) / 100 increases quite fast. \n \ns \n \n \n \n \n \n \n \n \n \nn \n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n1\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n2\n0,75\n0,9375\n0,99609375\n0,999984741\n1\n1\n1\n1\n1\n1\n3\n0,875\n0,998046875\n0,999999993\n1\n1\n1\n1\n1\n1\n1\n4\n0,9375\n0,999984741\n1\n1\n1\n1\n1\n1\n1\n1\n5\n0,96875\n0,99999997\n1\n1\n1\n1\n1\n1\n1\n1\n6\n0,984375\n1\n1\n1\n1\n1\n1\n1\n1\n1\n\n21\n7\n0,9921875\n1\n1\n1\n1\n1\n1\n1\n1\n1\n8\n0,99609375\n1\n1\n1\n1\n1\n1\n1\n1\n1\n9\n0,998046875\n1\n1\n1\n1\n1\n1\n1\n1\n1\n10\n0,999023438\n1\n1\n1\n1\n1\n1\n1\n1\n1\n11\n0,999511719\n1\n1\n1\n1\n1\n1\n1\n1\n1\n12\n0,999755859\n1\n1\n1\n1\n1\n1\n1\n1\n1\n13\n0,99987793\n1\n1\n1\n1\n1\n1\n1\n1\n1\n14\n0,999938965\n1\n1\n1\n1\n1\n1\n1\n1\n1\n15\n0,999969482\n1\n1\n1\n1\n1\n1\n1\n1\n1\n16\n0,999984741\n1\n1\n1\n1\n1\n1\n1\n1\n1\n17\n0,999992371\n1\n1\n1\n1\n1\n1\n1\n1\n1\n18\n0,999996185\n1\n1\n1\n1\n1\n1\n1\n1\n1\n19\n0,999998093\n1\n1\n1\n1\n1\n1\n1\n1\n1\n20\n0,999999046\n1\n1\n1\n1\n1\n1\n1\n1\n1\n \nThe number 1 occurs in most places where the fraction subtracted from 1 is too small \nfor the precision handled by Excel. \n \nOther notes. \n \n1. In the preceeding anΓ‘lisis, we have not taken into account the process needed to find \nout if a formula is irreducible, since it would add processing time. A formula may be \nevaluated with all the its atoms, without trying to simplify it beforehand via the \nabsorption equivalences. \n \n2. If m < n then r(n) < 100, because the contradiction class is excluded (its table yields \nonly zeros). \n \n3. It has been shown that for every uninformed sequential search algorithm there is a set \nof formulas taking an exponential time. An βinverseβ problem can be stated as: for \nevery formula there is an uninformed sequential search algorithm giving all the \nformulaβs solutions in the first rows of the truth table, and zeros on the remaining rows. \nThe problem is to find the algorithm. This requires a search on the set of uninformed \nsequential search algorithms. This search has exponential complexity. \n \nIf a an n order formula F has m solutions (that is, there are m rows where F is true), then \nthere are m!(2n β m)! ways to permute the rows keeping the true rows first and the false \nrows last. The ratio of this number of algorithms to the total ways of exploring Fβs truth \ntable is m!(2n β m)! / (2n)! = \n \n \n \n \n \n \nm\nn\n2\n1\n, which makes it highly unlikely to find one of this \nalgorithms by chance. Using a systematic search, the vast majority of a huge number of \nalgorithms must be tested before a solution can be found. \n \nReferences: \n \n[1] Nilsson, Nils J., βArtificial Inteligence: a new synthesisβ, Morgan Kaufmann \nPublishers, San Francisco, CA, 1998 \n[2] βThe efficiency of algorithmsβ, Harry R. Lewis y Christos H. Papadimitriou, \nScientific American, January 1978.\n\n22\n[3] Mendelson, Elliott, βIntroduction to Mathematical Logic,β D. Van Nostrand \nCompany, Princeton, NJ, 1965 \n \n[4] Russell, Stuart, Norvig, Peter, βArtificial Intelligence: a modern approachβ, Prentice \nHall, NJ, 2002. \n \n[5] βMillion Dollar Minesweeperβ, Ian Stewart, Scientific American, October 2.000 \n2007.11.07\n16:36:35-05'00'","paragraphs":[{"paragraph_id":"p1","order":1,"text":"1\nConsiderations on the P NP question. \nAlfredo von Reckow."},{"paragraph_id":"p2","order":2,"text":"Abstract"},{"paragraph_id":"p3","order":3,"text":"In order to prove that the P class of problems is different to the NP class, we consider \nthe satisfability problem of propositional calculus formulae, which is an NP-complete \nproblem [1], [3]. It is shown that, for every search algorithm A, there is a set E(A) \ncontaining propositional calculus formulae, each of which requires the algorithm A to \ntake a non-polynomial time to find the truth-values of its propositional letters satisfying \nit. Moreover, E(A) βs size is an exponential function of n, which makes it impossible to \ndetect such formulae in a polynomial time. Hence, the satisfability problem does not \nhave a polynomial complexity."},{"paragraph_id":"p4","order":4,"text":"1. Some definitions."},{"paragraph_id":"p5","order":5,"text":"Consider L, the propositional calculus formal theory, as defined in Mendelson [3] and \nlet L+ be an extension of L containing the 0-letter formulas : (empty formula, or \ncontradiction) and (tautology). Let F be the countable set of formulas in L+. Let \ndenote the βlogic equivalenceβ of formulas, defined by A B iff A and B yield the same \nresult when evaluating its truth tables."},{"paragraph_id":"p6","order":6,"text":"It can be easily verified that is a equivalence relationship. Let the quotient set be \nG = F / = {[A] / A F }."},{"paragraph_id":"p7","order":7,"text":"We will use indistinctly the terms βatomβ or βpositive literalβ for a propositional letter. \nA βnegative literalβ is an atomβs negation. A βliteralβ is an atom or the negation of an \natom."},{"paragraph_id":"p8","order":8,"text":"A formula Μs cuasinorm ."},{"paragraph_id":"p9","order":9,"text":"For each A F , we define a quasi-norm || A || = number of different atoms in A. This \nquasi-norm has the following properties:"},{"paragraph_id":"p10","order":10,"text":"For each A F , || A || 0. \n || A || = 0 iff A is or . \n Triangular inequality: For A and B in F ,, and any binary connective op,"},{"paragraph_id":"p11","order":11,"text":"|| A op B || || A || + || B ||."},{"paragraph_id":"p12","order":12,"text":"If A and B share atoms, the inequality is strict (including the case when A and B"},{"paragraph_id":"p13","order":13,"text":"are the same formula). The equality occurs when A and B do not share any"},{"paragraph_id":"p14","order":14,"text":"atoms (including the case when A or B is or )."},{"paragraph_id":"p15","order":15,"text":"A property like || A || = | | || A || (for real) is not valid here. One way to define \nβmultiplicationβ of a formula by a positive integer is 2 A = A op A, for some binary \nconnective op. But this produces || n A || = A for any natural n. Furthermore, the \nproduct A makes no sense for any number which is not a positive integer."},{"paragraph_id":"p16","order":16,"text":"Since || A || | | || A || for real, the function || || is not a norm, so we call it a quasi-\nnorm."},{"paragraph_id":"p17","order":17,"text":"2"},{"paragraph_id":"p18","order":18,"text":"Irreductibility"},{"paragraph_id":"p19","order":19,"text":"Two formulas can be equivalent, despite having different number of atoms. It is \npossible to have A B and || A || || B ||, because of the absorption laws:"},{"paragraph_id":"p20","order":20,"text":"p (p q) p \np (p q) p"},{"paragraph_id":"p21","order":21,"text":"We will say that a formula A is irreducible of order k iff || A || = k and it is not \nequivalent to another formula B which has less atoms than A. (i.e. if A B and || A || = \nk implies || B || k )."},{"paragraph_id":"p22","order":22,"text":"In other words, A F is irreducible of order k iff \n|| A || = k and \n( B F ) (A B || B || k)"},{"paragraph_id":"p23","order":23,"text":"A classβ cuasinorm."},{"paragraph_id":"p24","order":24,"text":"For X G, letβs define \n||\n||\nmin\n||\n||\nB\nX\nX\nB"},{"paragraph_id":"p25","order":25,"text":", which counts the minimum number of atoms \nfiguring in a formula from the X equivalence class."},{"paragraph_id":"p26","order":26,"text":"Here again, || || is a quasi-norm, having the following properties:"},{"paragraph_id":"p27","order":27,"text":"For every X G, || X || 0. \n || X || = 0 iff X is the class of all contradictions (which are equivalent to ) or \nthe class of all tautologies (which are equivalent to ). \n Triangular inequality: For X and Y in G,, and any binary connective op,"},{"paragraph_id":"p28","order":28,"text":"|| X op Y || || X || + || Y ||, where we define X op Y = { F1 op F2 } for F1 X \nand F2 Y."},{"paragraph_id":"p29","order":29,"text":"The set of order n irreducible formulas is defined by"},{"paragraph_id":"p30","order":30,"text":"F n = { A / || A || = || [ A] || = n }."},{"paragraph_id":"p31","order":31,"text":"Then we have F ="},{"paragraph_id":"p32","order":32,"text":"N\nn\nF\nX\nn\nX"},{"paragraph_id":"p33","order":33,"text":", \nAnd the set of all irreducible formulas is F = \nN\nn \n F n"},{"paragraph_id":"p34","order":34,"text":"A formula is simply βirreducibleβ if it is irreducible of order n, for some natural number \nn."},{"paragraph_id":"p35","order":35,"text":"We define the convertion function,"},{"paragraph_id":"p36","order":36,"text":"f : F / F \n f( [ A ] ) = B, where B A and A is irreducible."},{"paragraph_id":"p37","order":37,"text":"3\nfrom each equivalence class [A], f takes an irreducible formula B. Equivalently, f can \nbe seen as an algorithm (explicity stated) which produces an irreducible formula B from \nthe class [A]. The function f converts any formula in the class [A] to the irreducible \nformula B."},{"paragraph_id":"p38","order":38,"text":"We have B F n , for some natural number n, and [ B ] F n /"},{"paragraph_id":"p39","order":39,"text":"We define the normalization function h : F F , where h(A) is in disjunctive normal \nform (dnf). Then, for each equivalence class [A], h(f(A)) gets an irreducible and \nnormalized formula."},{"paragraph_id":"p40","order":40,"text":"For any given n, let Μs consider the set of βminimalβ formulas having n propositional \nletters p1, p2, ...pn, which are the order n irreducible formulas."},{"paragraph_id":"p41","order":41,"text":"Let Gn = { [An ] / An is irreducible having order n }. Then G = \nN\nn \n Gn \nLet In = F n / { [An ] / B [An ] iff B is order n irreducible and B An }"},{"paragraph_id":"p42","order":42,"text":"Each class of In contains only order n irreducible formulas."},{"paragraph_id":"p43","order":43,"text":"Then I = \nN\nn \nIn is the set of all equivalence classes having only irreducible formulas."},{"paragraph_id":"p44","order":44,"text":"If we take one element of each equivalence class, we get the set of all irreducible \nformulas, pair wise non equivalent."},{"paragraph_id":"p45","order":45,"text":"Now we construct the image of the function f, taking the order n minimal formulas:"},{"paragraph_id":"p46","order":46,"text":"Hn = { Bn / Bn = f(X) for X In } where the Bn are not equivalent to each other, and"},{"paragraph_id":"p47","order":47,"text":"H = \nN\nn \n H n."},{"paragraph_id":"p48","order":48,"text":"Then H is the set of all irreducible und unique formulas. In other words, each of the \nformulas in H is irreducible (because it has a minimum number of propositional \nletters, and it is impossible to eliminate a letter using absorption) and there are no two \nformulas in H being equivalent to each other (which makes each formula unique)."},{"paragraph_id":"p49","order":49,"text":"From now on, we will work only with formulas from H . That is, we will assume that \nany given formula A is order n irreducible and we will ignore other formulas equivalent \nto A."},{"paragraph_id":"p50","order":50,"text":"2. Why this is a search problem?."},{"paragraph_id":"p51","order":51,"text":"We will show that there is no algorithm to produce directly, in a straightforward way, \nthe truth values for the propositional letters in a formula, which make the formula true, \nunless there is a search involved. To this end, we show that this satisfability problem is \nequivalent to construct a βclosedβ formula for the roots of a polynomial with n \nvariables."},{"paragraph_id":"p52","order":52,"text":"4"},{"paragraph_id":"p53","order":53,"text":"Let Μs define the function h, from L+ to the set of n-variable polynomials, by"},{"paragraph_id":"p54","order":54,"text":"For a formula A of L+, having n propositional variables p1, p2, ..., pn, letβs \nassociate to each of its letters a numerical variable xi: h(pi)= xi for 1 i n."},{"paragraph_id":"p55","order":55,"text":"h(Β¬ A) = 1 β h(A)"},{"paragraph_id":"p56","order":56,"text":"If A and B are two formulas in L+, then"},{"paragraph_id":"p57","order":57,"text":"h(A B) = h(A) h(B) \n h(A B) = h(A) + h(B) β h(A) h(B) \n h(A B) = (1 β h(A) ) (1 β h(B) ) + h(B) \n h(A B) = (1 β h(A) ) (1 β h(B) ) ( 1 + h(A) + h(B) ) + h(A) h(B)"},{"paragraph_id":"p58","order":58,"text":"Note that"},{"paragraph_id":"p59","order":59,"text":"h(A)n is equivalent to h(A) in the following sense: \nh(A)n = 0 iff h(A) = 0 and h(A)n = 1 iff h(A) = 1"},{"paragraph_id":"p60","order":60,"text":"Note that h is a truth function assigning to a formula A a value 0 or 1 when each of the \nvariables xi takes a value of 0 or 1."},{"paragraph_id":"p61","order":61,"text":"In particular, in the resulting polynomial we can replace any factor xn by x."},{"paragraph_id":"p62","order":62,"text":"For example, for the idempotency law: A A A we have:"},{"paragraph_id":"p63","order":63,"text":"h(A A) = h(A) h(A) = h(A)2 h(A),"},{"paragraph_id":"p64","order":64,"text":"where the last β β means equivalence in the sense that we get the same values for h(A)2 \nand for h(A) when each of the corresponding variables xi take a value of 0 or 1."},{"paragraph_id":"p65","order":65,"text":"Similarly, with the other idempotency law, A A A, we get \nh(A A) = h(A) + h(A) β h(A)h(A) = 2h(A) β h(A)2 2h(A) β h(A) = h(A)."},{"paragraph_id":"p66","order":66,"text":"Tautologies and contradictions:"},{"paragraph_id":"p67","order":67,"text":"If a formula A is a tautology, then h(A) 1. \nIf a formula A is a contradiction, then h(A) 0."},{"paragraph_id":"p68","order":68,"text":"Example 1."},{"paragraph_id":"p69","order":69,"text":"For the tautology A A we have: \nh(A A) = (1 β h(A))2 + h(A) (1 β h(A)) + h(A) = 1."},{"paragraph_id":"p70","order":70,"text":"This latter equivalence can be proved in the following ways:"},{"paragraph_id":"p71","order":71,"text":"5\na) (1 β h(A))2 (1 β h(A)), because both sides have the same value when h(A) = 0 or \nwhen h(A) = 1."},{"paragraph_id":"p72","order":72,"text":"b) (1 β h(A))2 + h(A) = 1 β 2h(A) + h(A)2 + h(A) = 1 β h(A) + h(A)2 \n 1 β h(A) + h(A) = 1"},{"paragraph_id":"p73","order":73,"text":"Example 2."},{"paragraph_id":"p74","order":74,"text":"For the contradiction A Β¬A whe have:"},{"paragraph_id":"p75","order":75,"text":"h(A Β¬A) = h(A) (1 β h(A)) = h(A) β h(A)2 h(A) β h(A) = 0."},{"paragraph_id":"p76","order":76,"text":"For any formula A, the problem of finding the truth-values for its variables p1, p2, ..., pn \nsatisfaying A (i.e. making it true) is equivalent to the problem of finding the \ncorresponding numerical values for n-tuple (x1, x2, ..., xn) which are zeros of the \nfunction g, where g(A) = h(A) β 1."},{"paragraph_id":"p77","order":77,"text":"This function is a polynomial with the n variables (x1, x2, ..., xn), which will be called \nthe formulaβs βcharacteristic polynomialβ, and has the form g(A) = q, where"},{"paragraph_id":"p78","order":78,"text":"q : [0, 1]n R, takes values on the n-dimentional unit cube, and is defined by"},{"paragraph_id":"p79","order":79,"text":"(1)"},{"paragraph_id":"p80","order":80,"text":")\n(\n1\n2\n1\n1\n...\n)\n,...,\n(\n2\n1\nn\nr\ni\nn\ni\nn\nin\ni\ni\nx\nx\nx\nx\nx\nq"},{"paragraph_id":"p81","order":81,"text":"where each exponent ki has a value of 0 or 1, the coefficients i and are integers, \nand the variables xi have a value of 0 or 1, and r(n) = 2n."},{"paragraph_id":"p82","order":82,"text":"We define 00 = 1."},{"paragraph_id":"p83","order":83,"text":"The problems then is to find a value for each variable xi, either a 0 or a 1, in such a way \nthat q(x1, x2, ..., xn) = 0. Or equivalently:"},{"paragraph_id":"p84","order":84,"text":"(2) \nFind the roots of the polynomial q \n(3) \nWith the restrictions xi in {0, 1} for i = 1, ..., n."},{"paragraph_id":"p85","order":85,"text":"In other words, find q-1 (0) {0, 1}n."},{"paragraph_id":"p86","order":86,"text":"Lemma."},{"paragraph_id":"p87","order":87,"text":"There is no closed formula to calculate directly the roots of this type of polynomials. \nHence, there is no algorithm to find directly (without making any search) the truth \nvalues of the propositional letters satisfying a formula A."},{"paragraph_id":"p88","order":88,"text":"Proof."},{"paragraph_id":"p89","order":89,"text":"The equation g(A) = 0 has n variables. It can be seen as a βsystemβ of 1 equation with n \nunknowns. This is an under determined system, having infinite real roots. To"},{"paragraph_id":"p90","order":90,"text":"6\ndetermine them, it suffices to solve g(A) = 0 for x1, which will then depend on the other \nindependent variables x2, ..., xn."},{"paragraph_id":"p91","order":91,"text":"Since these variables are independent (there is no relationship between them), each of \nthem can take any real value. Once we chose a value for each variable x2, ..., xn, that \nwill produce a value for x1."},{"paragraph_id":"p92","order":92,"text":"From these infinite set of values for the n-tuple (x1, ..., xn) .we have to separate those \nwhere all the variables are 0 or 1."},{"paragraph_id":"p93","order":93,"text":"Because there are no additional conditions on the variables, no further relationship \nbetween them can be obtained, and this happens just because the variables x2, ..., xn are \nindependent. There is only the restriction (3 ) on the value of the solutions, but it is of \nno use for finding a closed formula for the solutions of (2)."},{"paragraph_id":"p94","order":94,"text":"Hence, we have no other way left, but to carry on an exhaustive search, and we have no \nadditional information derived from the formula which could lead us to take any \nshorcuts and reduce the number of cases (or number of rows in the truth table) to be \nconsidered."},{"paragraph_id":"p95","order":95,"text":"Example."},{"paragraph_id":"p96","order":96,"text":"For the formula"},{"paragraph_id":"p97","order":97,"text":"A : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3)"},{"paragraph_id":"p98","order":98,"text":"The associated polynomial for A is"},{"paragraph_id":"p99","order":99,"text":"h(A) = x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3"},{"paragraph_id":"p100","order":100,"text":"The characteristic polynomial is"},{"paragraph_id":"p101","order":101,"text":"g(A) = x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1"},{"paragraph_id":"p102","order":102,"text":"and the equation g(A) β 1 = 0 is"},{"paragraph_id":"p103","order":103,"text":"x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 = 0."},{"paragraph_id":"p104","order":104,"text":"With the restrictions xi {0, 1} for i = 1,.., 3."},{"paragraph_id":"p105","order":105,"text":"In this equation, the variable x1 depends on x2 and x3. This two last variables have no \nfurther relationship to one another. If x3 takes any given value, it has no influence at all \non the values that x2 can take. The only thing we know is that we are interested in the \ncases where x2 as well as x3 take a value of 0 or 1."},{"paragraph_id":"p106","order":106,"text":"Hence, we have no choice but to make an exhaustive search, using the 4 possible values \nfor the duple (x2, x3). The solutions will be those cases where the equation makes sense \nand yielding values of 0 or 1 for x1."},{"paragraph_id":"p107","order":107,"text":"7\nNote that if (x2, x3) = (1, 1) then x1 vanishes and the equation becomes 1 = 0, which \nmakes no sense. Hence (x2, x3) (1, 1)."},{"paragraph_id":"p108","order":108,"text":"From other point of view, solving for x1 we get"},{"paragraph_id":"p109","order":109,"text":"(4) \n1\n)\n(\n2\n3\n1\n2\n3\n2\n3\n2\n3\n2\n3\n2\n1"},{"paragraph_id":"p110","order":110,"text":"x\nx\nx\nx\nx\nx\nx\nx\nx"},{"paragraph_id":"p111","order":111,"text":"where the denominator is zero if (x2, x3) = (1, 1). It is zero for other values of these \nvariables, but it does not matter, because in none of those other cases both variables \nhave a value of 0 or 1."},{"paragraph_id":"p112","order":112,"text":"Note that x2 and x3 can take any real value in (4), as long as the denominator does not \nvanish. There are infinite values for x2 and x3 satisfying (4), but we are interested only \nin the 4 cases where they are 0 or 1, and there are no shorcuts for finding them. We \nmust try all the options and see which ones work."},{"paragraph_id":"p113","order":113,"text":"In the general case, for a formula having n propositional letters, we can not avoid the \nneed to explore 2n-1 options for the variables x2, ...xn."},{"paragraph_id":"p114","order":114,"text":"Other way to search is the following:"},{"paragraph_id":"p115","order":115,"text":"If we put arbitrarily x1 = 0 in (4), then 2x2x3 β x2 β x3 + 1 = 0, and from here we get x2 = \n(x3 β 1) / (2x3 β 1). Here we have two cases: if si x2 = 0, then x3 = 1 and if x2 = 1, then \nx3 = 0."},{"paragraph_id":"p116","order":116,"text":"If we put x1 = 0 in (4), then 2x2x3 β x2 β x3 + 1 = 3x2x3 β 2(x2 + x3) + 1, so x2 = x3 / (x3 β \n1). Here again we have two cases: if x2 = 0, then x3 = 0 and if x2 = 1, then x3 = x3 β 1, \nwhich is impossible."},{"paragraph_id":"p117","order":117,"text":"We have had to analize 4 different cases, for 3 variables, corresponding to 2n-1 options, \nfor n = 2, and the variables x1, ...xn-1"},{"paragraph_id":"p118","order":118,"text":"We can solve for any of the n variables as a function of the other n-1 variables. In any \nevent, we must try 2n-1 options."},{"paragraph_id":"p119","order":119,"text":"Notes."},{"paragraph_id":"p120","order":120,"text":"1. Note that the characteristic polynomial g has n variables and has 2n terms. Each term \nis a product of k variables, and there are C(n, k) = n! / (k! (n-k)!) combinations of terms \nhaving k varibles. Adding that up, there are"},{"paragraph_id":"p121","order":121,"text":"n\nk\nn\nk\nn\n0\n2"},{"paragraph_id":"p122","order":122,"text":"terms, each one has one coefficient and n exponents. Any algorithm processing g has \nthus an input of (n+1) 2n numbers, an exponential function of n, which can not be \nprocessed in a polinomial time."},{"paragraph_id":"p123","order":123,"text":"8\nMoreover, if there were a βclosed formβ formula to determine the binary roots of g, then \nit would depend on the 2n coefficients of g. Since the variables x2, ..., xn are \nindependent, a closed form for each of them would be a function of those 2n \ncoefficients, and it cannot be calculated in a time t = q(n), where q is a polynomial."},{"paragraph_id":"p124","order":124,"text":"Such a formula, if it existed, would behave not better than a sequential exhaustive \nsearch."},{"paragraph_id":"p125","order":125,"text":"2. The characteristic polynomial g has n variables x1, x2, ..., xn each one with an \nexponent 0 or 1. In any occurrence of any of this variables the exponent can be replaced \nby an odd natural number k, and the resulting polynomial has the same βbinary rootsβ \n(those being 0 or 1) of g. The exponent must be 0 or odd to avoid de case (-1)2 = 1."},{"paragraph_id":"p126","order":126,"text":"Hence, g is equivalent to infinitely many polynomials (in the sense that all of them have \nthe same binary roots)."},{"paragraph_id":"p127","order":127,"text":"If there were an explicit formula to find the binary roots of g, that same formula would \nyield the binary roots of all the polinomials q equivalent to g, which can have an \narbitrarily high degree."},{"paragraph_id":"p128","order":128,"text":"These comes from the idempotency laws A A A and A A A."},{"paragraph_id":"p129","order":129,"text":"3. It is not possible to determine, a priori, additional conditions to find the binary roots \nof g. If there were n-1 such additional conditions, then we would have a non-linear \nsystem of n equations with n variables."},{"paragraph_id":"p130","order":130,"text":"The number of solutions for the characteristic polynomial can be anything from 0 (in \nthe case of a contradiction) to 2n (for a tautology). If all this options are possible, then \nno additional conditions can exist. For a formula like p1 p2 ... pn, having 2n β 1 \nsolutions, any additional condition must be redundant. For a formula having 2n β 2 \nsolutions, there can be at most one additional condition, and son on. The number of \nadditional conditions depends on the number of solutions, which is not known in \nadvance, so we donβt know beforehand how many conditions are necessary for a given \nformula."},{"paragraph_id":"p131","order":131,"text":"Moreover, if such conditions depend on the number of solutions, and the solutions \ndepend on the coefficients of the characteristic polynomial, then those conditions will \ndepend on those coefficients. Since there are 2n coefficients in the characteristic \npolynomial, the additional conditions will depend on 2n inputs, which can not be \nprocessed in a polynomial time."},{"paragraph_id":"p132","order":132,"text":"On the other hand, if we introduce, arbitrarily, any condition g1(x1, ..., xn) = 0, then \nsome roots can be missed."},{"paragraph_id":"p133","order":133,"text":"For example, if we put arbitrarily the condition x1 = 0, then we will miss all the roots \nwhere x1 = 1, in the cases where such roots exists. In the cases where g has a factor (x1 \nβ 1) then we will miss all its binary roots."},{"paragraph_id":"p134","order":134,"text":"9\nIf we put a condition like x1 β x2 = 0, then we might miss the roots where x1 β x2, and \nwe do not know beforehand if that is indeed the case. Simple conditions as these can \nbe, of course, verified beforehand, but more complex conditions can not."},{"paragraph_id":"p135","order":135,"text":"In the example"},{"paragraph_id":"p136","order":136,"text":"A : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3)"},{"paragraph_id":"p137","order":137,"text":"The characteristic polynomial is"},{"paragraph_id":"p138","order":138,"text":"x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 = 0."},{"paragraph_id":"p139","order":139,"text":"If we add the condition x1 β x2 = 0, (which can be considere a mere supposition) the \ncharacteristic polynomial becomes"},{"paragraph_id":"p140","order":140,"text":"x3 - x1x3 β 1 = 0."},{"paragraph_id":"p141","order":141,"text":"Where x1 must be zero and x3 must be 1. x2 can be 0 or 1."},{"paragraph_id":"p142","order":142,"text":"The triplet (0, 1, 0) is a root of the characteristic polynomial, but the other triplet (0, 1, \n1) is not."},{"paragraph_id":"p143","order":143,"text":"Adding an arbitrary condition not only leads us to miss two roots, but it introduces a \nfake root."},{"paragraph_id":"p144","order":144,"text":"We have no additional information and we cannot add it arbitrarily, because we donβt \nwhere to start looking for the roots."},{"paragraph_id":"p145","order":145,"text":"As have been noted in the literature (see [3]), if we know the roots (or if we a have a \ncandidate for root, somehow βguessedβ or estimated), we can replace it in the \ncharacteristic polynomial and check it in a polynomial time."},{"paragraph_id":"p146","order":146,"text":"But if we donβt have a root, we have to search for it, in a sequential way. We must try \nall the options."},{"paragraph_id":"p147","order":147,"text":"4. Multiplying the factors of g to get the explicit form (1) requires o(mn) multiplications, \nwhere m is the number of factors and n is the number of variables. Instead of carrying \non the multiplications, we can consider the associated polynomial, when it is already \nfactored and find its roots. The remaining n-tuplets (those not being a binary root of h) \nwill be the binary root of the characteristic polynomial g."},{"paragraph_id":"p148","order":148,"text":"The idea is to take advantage when h is already factored (as this might take less \noperations)."},{"paragraph_id":"p149","order":149,"text":"In the example, \nA : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3)"},{"paragraph_id":"p150","order":150,"text":"The associated polynomial is"},{"paragraph_id":"p151","order":151,"text":"h(A) = (x1 + x2 + x3 - x1x2 β x1x3 β x2x3 + x1x2x3 )(1 - x1x2) (1 - x1x3) (1 β x2x3)"},{"paragraph_id":"p152","order":152,"text":"10"},{"paragraph_id":"p153","order":153,"text":"h(A) = 0 if al least one of its factors is zero."},{"paragraph_id":"p154","order":154,"text":"The last thre factors give the options"},{"paragraph_id":"p155","order":155,"text":"x1 = 1, x2 = 1, any value for x3 \nx1 = 1, x3 = 1, any value for x2 \nx2 = 1, x3 = 1, any value for x1"},{"paragraph_id":"p156","order":156,"text":"The first factor gives"},{"paragraph_id":"p157","order":157,"text":"x1 = (x2 + x3 - x2x3) / (1- x2 - x3 + x2x3) = U / ( 1 β U), for U β 0,"},{"paragraph_id":"p158","order":158,"text":"Where U = x2 + x3 - x2x3. All the options are:"},{"paragraph_id":"p159","order":159,"text":"x2 \nx3 \nU \nx1 \n0 \n0 \n0 \n0 \n0 \n1 \n1 \nnot definided \n1 \n0 \n1 \nnot definided \n1 \n1 \n1 \nnot definided"},{"paragraph_id":"p160","order":160,"text":"The roots of the associated polynomial h are"},{"paragraph_id":"p161","order":161,"text":"x1 \nx2 \nx3 \n0 \n0 \n0 \n0 \n1 \n1 \n1 \n0 \n1 \n1 \n1 \n0 \n1 \n1 \n1"},{"paragraph_id":"p162","order":162,"text":"Since g = h β 1, for any other triplet (x1, x2, x3), h(x1, x2, x3) is not zero, and hence it \nmust be h(x1, x2, x3) = 1. Those other three triplets must be the binary roots of g. They \nare:"},{"paragraph_id":"p163","order":163,"text":"x1 \nx2 \nx3 \n0 \n0 \n1 \n0 \n1 \n0 \n1 \n0 \n0"},{"paragraph_id":"p164","order":164,"text":"But this has required to explore all the options for the first factor. It takes an o(2n-1) \nnumber of operations, for n = 3."},{"paragraph_id":"p165","order":165,"text":"The cases when two or more factors are simultaneously zero produce repeated roots, \nthat have to be eliminated from the list of roots."},{"paragraph_id":"p166","order":166,"text":"3. Exhaustive search"},{"paragraph_id":"p167","order":167,"text":"If there is no direct (βclosed formβ) formula to find the binary roots of the characteristic \npolynomial of a given formula, we must carry on a sequential search, checking a"},{"paragraph_id":"p168","order":168,"text":"11\nformulaβs truth table in a sequential way, one row after another. Since there are no \nadditional conditions to guide the search, it must be made sequentially."},{"paragraph_id":"p169","order":169,"text":"For the set of order n irreducible formulas, we choose an order of evaluation for the \npropositional letters and for the truth value for each atom. First, we take some \npermutation of the letters. Then, for every letter we choose if it takes the value 0 or 1 \nfirst. Then for every order of evaluation there are formulas taking necessarily an \nexponential time to find an n-tuple of truth values satisfying it."},{"paragraph_id":"p170","order":170,"text":"3. 1. Sequential search algorithms."},{"paragraph_id":"p171","order":171,"text":"In this section we consider βuninformedβ or βblindβ search algorithms, which uses no \nadditional information (other than the given formula itself) to guide the search or to \nreduce the number of options. As was shown in section 2, such information is not \navailable beforehand, and if it were, taking it into account requires an exponential order \nalgorithm, which is no better than a sequential uninformed search."},{"paragraph_id":"p172","order":172,"text":"For this reason, in this section we well consider only algorithms exploring a formulaβs \ntruth table in some order that does not depend on the formula. That is, an algorithm A \nwill explore the truth tableβs rows in the same fashion for all formulas."},{"paragraph_id":"p173","order":173,"text":"A βsequential search algorithmβ assigns truth values to the propositional letters p1, p2, \n..., pn occurring in a formula F in some pre-established order (out of the n! permutations \nof the n letters) and for each letter it is established if 0 or 1 is used first (out of the 2n \noptions)."},{"paragraph_id":"p174","order":174,"text":"Consider the set of all sequential search algorithms. This set has n! 2n algorithms, \nbecause the n letters can be arranged in n! different permutations and each letter can be \nevaluated to 0 first and then to 1, or the other way around. If a formula F has n atoms ( \n|| F || = n) there are n! 2n different ways to evaluate F, depending on the order the atoms \nare evaluated and if the 0 or 1 is evaluated first for every atom."},{"paragraph_id":"p175","order":175,"text":"Two sequential search algorithms A1 and A2 are βsequentially equivalentβ if and only if \nthey assign the same truth values to the same atom in the same order. The algorithms \ncan use different functions or procedures, yielding the same results in the same order, \nbut their implementations can yield different running times on a given machine, \noperating systems and programming language."},{"paragraph_id":"p176","order":176,"text":"Two sequential search algorithms A1 and A2 are not equivalent, and will be considered \ndifferent, if they take the atoms in a different order or if they assign the truth values 0 \nand 1 in a different order to the same atom."},{"paragraph_id":"p177","order":177,"text":"Equivalently, two sequential search algorithms A1 and A2 are different if they explore \nthe truth table rows in a different order."},{"paragraph_id":"p178","order":178,"text":"This sequential search is equivalent to a depth-first search in a binary tree, with a given \npriority for the order in which the atoms are evaluated and the truth value given to each \natom."},{"paragraph_id":"p179","order":179,"text":"12\nSequential equivalence is an equivalence relationship, so we can take an algorithm from \neach equivalence class. We will assume that we take the algorithm taking the least time \nin a particular implementation."},{"paragraph_id":"p180","order":180,"text":"In order to find out if a given formula F is satisfacible, and when it is, find the \ncorresponding truth-values of its atoms, a sequential search algorithm A explores the \ntruth table rows in a certain order. The algorithm A βsolvesβ the formula F if it finds a \ntruth-value assignation for Fβs atoms, with which F is evaluated true."},{"paragraph_id":"p181","order":181,"text":"We will represent the truth table for F, as evaluated by A, by A(F) = (b(1), b(2), ..., \nb(2n)), where b(k) is the value obtained in row k."},{"paragraph_id":"p182","order":182,"text":"We define the function L as the function giving the first row where algorithm A finds a \nvalue of 1 when it evaluates the formula F. This is L(A(F)) = \nn\nk 2\n1min"},{"paragraph_id":"p183","order":183,"text":"{ k / b(K) = 1 }."},{"paragraph_id":"p184","order":184,"text":"Exploring with the initial order."},{"paragraph_id":"p185","order":185,"text":"Take F n, the set of order n irreducible formulas. Take a formula F in F n. Then || F || = \nn, and consider p1, p2, ..., pn as the n distinct atoms figuring in F, taken in the same \norder they appear in F, from left to right. We will call this the βinitial orderβ of the \natoms in F."},{"paragraph_id":"p186","order":186,"text":"Denote by Ai,j the algorithm which uses the i-th permutation of the atoms of F and the j-\nth way of giving the truth values to them."},{"paragraph_id":"p187","order":187,"text":"The sequential search algoritm A1,1 explores Fβs truth tableβs rows in the βnaturalβ \norder, taking the atoms p1, p2, ..., pn in the same order they appear in F and for every \none of them it evaluates first the 0 and then the 1 values."},{"paragraph_id":"p188","order":188,"text":"Then, for the formula"},{"paragraph_id":"p189","order":189,"text":"F: \n p1 p2 ... pn"},{"paragraph_id":"p190","order":190,"text":"the sequential search algoritm A1,1 must check all the rows in the truth table, to find a \nsolution only in the last row (that is, in row 2n ). Thus, L(A1,1(F)) = 2n."},{"paragraph_id":"p191","order":191,"text":"Changing 0s and 1s."},{"paragraph_id":"p192","order":192,"text":"Take now all the A1,j algorithms. All of them evaluate the atoms p1, p2, ..., pn in the \norder 1, 2, ..., n. They differ from each other because they assign the values 0 and 1 in \na different order for every letter (one of them will assign first 0 and later 1 to some atom \npi, while other one assigns first 1 and later 0 to the same letter pi )."},{"paragraph_id":"p193","order":193,"text":"More precisely, if the base 2 representation of j is j = d1d2...dn, then the atom pk takes \nfirst the truth value dk and later it will take the truth value 1 β dk , for 1 β€ k β€ n."},{"paragraph_id":"p194","order":194,"text":"For every j, if the A1,j algorithm evaluates first true the atom pk (that is, if dk = 1), then \nlet`s take the letter qk equal to pk. Otherwise, take qk equal to pk. Then the formula"},{"paragraph_id":"p195","order":195,"text":"13\n(5)"},{"paragraph_id":"p196","order":196,"text":"F: \n q1 q2 ... qn"},{"paragraph_id":"p197","order":197,"text":"is true only when all the qj are false. To get there, the algorithm A1,j must explore first \nall the other rows in the truth table. In this case, we also have Thus, L(A1, j(F)) = 2n."},{"paragraph_id":"p198","order":198,"text":"Thus, the algorithm A1, j(F) has an order o(2n) temporal complexity."},{"paragraph_id":"p199","order":199,"text":"Permuting the order of the atoms in F is irrelevant, since is has no effect on the rows \nthat have to be evaluated. In any case, the formula F will take an exponential time: \nL(Ai, j(F)) = 2n, for 1 β€ i β€ n!"},{"paragraph_id":"p200","order":200,"text":"The set Cn(Ai,j)."},{"paragraph_id":"p201","order":201,"text":"For an integer n given, we define the set Cn(Ai,j) as the set of order n irreducible and \npair-wise non equivalent formulas for whom algorithm Ai,j takes a non polynomial time."},{"paragraph_id":"p202","order":202,"text":"The formula F given above in (5) is in Cn(Ai,j). It the number of formulas in this set \n(representing each one an equivalence class) where a polynomial depending on n, we \ncould detect this kind of formulas before applying the algorithm. Then we Ai,j would \nexecute in a polynomial time for the other formulas in F n."},{"paragraph_id":"p203","order":203,"text":"This is not the case, as it will be shown that Cn(Ai,j) = o(g(n)), where g is an \nexponential-type function."},{"paragraph_id":"p204","order":204,"text":"For every sequential search algorithm Ai,j and every formula F, it is posible to apply a \npre-process to determine if F is equivalent to a worst-posible case formula, (and in this \ncase, to evaluate it at once). Letβs call this revised algorithm Ai,j(1)."},{"paragraph_id":"p205","order":205,"text":"But then there is another formula F(1) forcing the revised algorithm Ai,j(1) to search all \nthe way thru the penultimate row in the truth table of F(1), because that is the only row \nwhere F(1) is true. If we repeat the process to construct algorithm Ai,j(2), which detects \nF and F(1), then the formula F(2), which is true only in the next-to-penultimate row, \nwill take an exhaustive search from row 1 to row 2n-2, and so on."},{"paragraph_id":"p206","order":206,"text":"More precisely, we have two steps in the analysis:"},{"paragraph_id":"p207","order":207,"text":"Lemma 1."},{"paragraph_id":"p208","order":208,"text":"For every natural number m, 1 β€ m β€ 2n, there is a formula F having n atoms, that \nis true only in row m."},{"paragraph_id":"p209","order":209,"text":"For each sequential search algorithm Ai,j and every m {1, ..., 2n}, we construct a \nformula being true only on row m, by taking the formula"},{"paragraph_id":"p210","order":210,"text":"F: \n pi1 pi2 ... pin"},{"paragraph_id":"p211","order":211,"text":"which is true only in the last row of the truth table, and changing each atom for its \nnegation, depending on the order in which algorithm Ai,j assigns values to each atom."},{"paragraph_id":"p212","order":212,"text":"14\nTo this end, we take the base 2 representation of the index j: j = d1d2...dn. Suppose the \natom pk takes first the value dk and later on the value 1 - dk."},{"paragraph_id":"p213","order":213,"text":"Take the base 2 representation of m: m = e1e2...en,"},{"paragraph_id":"p214","order":214,"text":"If ek = dk, then qik(m) is pik, otherwise, qik(m) is Β¬ pik."},{"paragraph_id":"p215","order":215,"text":"Then the formula Gij(m): \nΒ¬ qi1(m) Β¬ qi2(m) .. .Β¬ qin(m)"},{"paragraph_id":"p216","order":216,"text":"Is true only in row m and it is false in all the other rows."},{"paragraph_id":"p217","order":217,"text":"Lemma 2. (from the end of the truth table)."},{"paragraph_id":"p218","order":218,"text":"For every sequential search algorithm Ai,j, there is an order n irreducible formula F, that \nis true only in the last 2 rows explored by the algorithm."},{"paragraph_id":"p219","order":219,"text":"Proof."},{"paragraph_id":"p220","order":220,"text":"This formula is constructed taking the disjunction"},{"paragraph_id":"p221","order":221,"text":"Gij(2n-1) Gij(2n) \nFor this formula, algorithm Aij(1) must explore 2n-1 rows in the truth table."},{"paragraph_id":"p222","order":222,"text":"Of course, we can detect this formulae and give them a different treatment, with a pre-\nprocess."},{"paragraph_id":"p223","order":223,"text":"Call Aij(2) be the algorithm obtained by adding such pre-process to Aij(1), to eliminate \nthe βworst possible casesβ for Aij and Aij(1)."},{"paragraph_id":"p224","order":224,"text":"Consider now the following formula, which is true only in the last three rows, as \nexplored by algorithm Aij(2):"},{"paragraph_id":"p225","order":225,"text":"Gij(2n-2) Gij(2n-1) Gij(2n)"},{"paragraph_id":"p226","order":226,"text":"As well as the formulas"},{"paragraph_id":"p227","order":227,"text":"Gij(2n-2) Gij(2n-1) \nGij(2n-2) Gij(2n)"},{"paragraph_id":"p228","order":228,"text":"We can separate this formulas with a new pre-process, to get the algoritm Aij(3), and so \non."},{"paragraph_id":"p229","order":229,"text":"Continuing this way, for each algorithm Aij we construct a succession of algorithms \nAij(1), Aij(2), Aij(3), .... Every one of them has a set of worst-possible case formulas, \nand every one avoids the worst-possible case for the previous algorithms."},{"paragraph_id":"p230","order":230,"text":"The worst-possible case for Aij(2n-1) is a formula whose truth table is true in all its \nsecond-half rows and false in all its first-half rows. This algorithm has exponential \ncomplexity."},{"paragraph_id":"p231","order":231,"text":"15\nTo find out if a formula F is true from row k, for k = 2n, 2n-1, 2n-2, ..., 2n-1 + 1, we have \nto check out 2n-1 cases. If none of them happens, then we must explore sequentially \nrows 1 to 2n-1. The worst possible case occurs with the formula which is true only in \nrow 2n-1."},{"paragraph_id":"p232","order":232,"text":"The time complexity is therefore:"},{"paragraph_id":"p233","order":233,"text":"Rows to check out in the pre-process: \n2n β (2n-1 + 1) + 1 = 2n-1 \nRows to explore in the process"},{"paragraph_id":"p234","order":234,"text":"2n-1 \n---------------------------------------------------------- \nTotal"},{"paragraph_id":"p235","order":235,"text":"2n"},{"paragraph_id":"p236","order":236,"text":"Moreover, for every k, algorithm Aij(k) has exponential complexity."},{"paragraph_id":"p237","order":237,"text":"We have to check first (in the pre-process) for k worst-possible case situations of Aij(h) \nh = 1, ..., k-1. For the formulas passing this check, there are the worst-possible cases \nAij(k) demanding to explore 2n β k rows. Hence, in the worst-possible case, Aij(k) must \nexplore k + (2n - k) = 2n rows."},{"paragraph_id":"p238","order":238,"text":"Conclusion:"},{"paragraph_id":"p239","order":239,"text":"For every way used to explore a truth-table, there is a formula requiring to explore all \nthe other rows first. If we try to avoid this worst-possible cases adding a pre-process to \nthe search algorithm, then there is another formula requiring also to explore all the other \nrows first."},{"paragraph_id":"p240","order":240,"text":"This applies to any sequential search algorithm that does a βblindβ or βuninformedβ \nsearch, where no additional information is used to guide the search or to discard \noptions, which will be analized in section 3.2."},{"paragraph_id":"p241","order":241,"text":"Additional notes."},{"paragraph_id":"p242","order":242,"text":"1. Additional complexity."},{"paragraph_id":"p243","order":243,"text":"There are additional sources of complexity: each equivalence class in G = F / \ncontains formulas with an arbitrarily large number of atoms, which are equivalent to a \nformula with n atoms, due to the absorption laws."},{"paragraph_id":"p244","order":244,"text":"This requires to first simplify a formula, adding extra time to the process. For an n-\natom formula F, it can be constructed infinite many other formulas with m atoms, for \nany m > n, equivalent to F by the absorption laws. Because of this, an algorithm \nprocessing formulas with n atoms will have to deal also with formulas having m > n \natoms to find out if they can be simplified to a formula with n atoms. Since m can be \narbitrarily large, the time needed will be arbitrarily large too."},{"paragraph_id":"p245","order":245,"text":"Thatβs why we have supposed that the input formulas for any algorithm are already \norder-n irreducible."},{"paragraph_id":"p246","order":246,"text":"2. Extending lemma 2."},{"paragraph_id":"p247","order":247,"text":"16\n The Aij algorithms take into account only permutations in the order of evaluation of the \natoms and of the truth value for each atom. There are 2n n! such permutations. The \ntruth table for an n-order formula has 2n rows, having 2n! permutations. Since the proof \nof lemma 2 includes a constructive method to build a formula that is true in any given \nrow (and only in that row), that proof can be extended to cover the case of an arbitrary \npermutation of rows of the truth table."},{"paragraph_id":"p248","order":248,"text":"3.2. Heuristic Algorithms."},{"paragraph_id":"p249","order":249,"text":"An heuristic algorithm does not explore all the rows in a formulaβs truth table. It \neliminates arbitrarily some of the rows, hoping there is a solution in the remaing rows. \nSuch algorithm will have a polynomial complexity if it explores only (n) rows for an \norder n irreductible formula F, whre is a polynomial."},{"paragraph_id":"p250","order":250,"text":"This means that the algorithm eliminates a priori all the other rows. The algorithm will \nbe βacceptableβ for a formula F if some rows yielding 1 are among the explored rows."},{"paragraph_id":"p251","order":251,"text":"Lema 3."},{"paragraph_id":"p252","order":252,"text":"For every polynomial complexity heuristic algorithm H, there is a set of formulas C(H) \nfor which H is not acceptable. That is, for every formula in F in C(H), H does not \nexplore any row of F where the result is 1. \nIn other words, we can not eliminate rows arbitrarily, because that might eliminate \nprecisely those rows having a result of 1."},{"paragraph_id":"p253","order":253,"text":"Proof."},{"paragraph_id":"p254","order":254,"text":"Let F be an order n irreducible formula and a polynomial depending only on n. Let Hj \nbe one of the"},{"paragraph_id":"p255","order":255,"text":")\n(\n2\nn\nn"},{"paragraph_id":"p256","order":256,"text":"algorithms having polynomial complexity of order (n), which \nexplores only rows 1, 2, ..., (n), evaluating the atoms p1, p2, ..., pn in the same order \nthey figure in F, from left to right."},{"paragraph_id":"p257","order":257,"text":"For any n-order formula, each algorithm Hj evaluates only rows i1, i2, ..., \n)\n(n\ni \n. This \nrows are the same for all n-order formulas. The row indexes depend only on the \nalgorithm j-index and not on any given formula."},{"paragraph_id":"p258","order":258,"text":"Take k1 as one of the rows not explored by Hj. With the same method used in lemma 2, \nwe can construct a formula that is true only in row k1:"},{"paragraph_id":"p259","order":259,"text":"If atom pi is true in row k then take qi equal to pi. Otherwise take qi equal to Β¬ pi. Take \nthe conjunction of the qi ."},{"paragraph_id":"p260","order":260,"text":"Gk1: \nq1 q2 ... qn"},{"paragraph_id":"p261","order":261,"text":"This formula is true only in row k, and algorithm Hj missed it, since it does not explore \nthis row."},{"paragraph_id":"p262","order":262,"text":"17\n( Moreover, Hj will miss a solution for any disjunction of formulas Gk1 Gk2 ... Gkr, \nwhere k1, k2, ..kr are not explored by Hj)"},{"paragraph_id":"p263","order":263,"text":"Let Hj(1) be the extensiΓ³n of Hj obtained by adding a pre-process to check row k1 \nbeforehand. Then we can construct in a similar way another formula where Hj(1) fails \nfor some formula Gk2 Μ§where row k2 is not explored by Hj(1). And then we can extend \nHj(1) to Hj(2), wich in turn will fail for some formula Gk3."},{"paragraph_id":"p264","order":264,"text":"This produces a list of algorithms Hj, Hj(1), Hj(2), ..., Hj(t) The list is finite, with a \nrestriction: the last index, t, must depend polynomially on n: t = q(n) for some \npolynomial q."},{"paragraph_id":"p265","order":265,"text":"But the truth table has 2n rows an hence we always can take one excluded row and build \na formula being true only on that row, causing Hj(t) to fail, no matter what polynomial \nfunction q is used."},{"paragraph_id":"p266","order":266,"text":"In conclusion, there is a set C(Hj) having a number of equivalence classes w, where w is \nan exponential function of n, and for any formula F in a class X of C(Hj), Hj does not \nevaluate the rows where F is true, and therefore misses every solution for F."},{"paragraph_id":"p267","order":267,"text":"4. Conversion algorithms."},{"paragraph_id":"p268","order":268,"text":"It is quite simple to find the truth-value assignations for the atoms satisfiying a formula \nexpressed in disjunctive normal form (dnf). There are 3 cases:"},{"paragraph_id":"p269","order":269,"text":"1. The formula is a tautology, and is equivalent to . It is true for any assignation of \ntruth values to its atoms."},{"paragraph_id":"p270","order":270,"text":"2. The formula is a contradiction, and is equivalent to . It is false for any assignation \nof truth values to its atoms. In each of its βdisjointβ the formula must have the \nconjunction of a complementary pair (if some disjoint does not have any such \nconjunction, this disjoint would have a truth value assignation satisfying the formula, \nwhich is impossible)."},{"paragraph_id":"p271","order":271,"text":"3. The formula is a contingency. It has at least one βdisjointβ and it suffices one of \nthem to be true."},{"paragraph_id":"p272","order":272,"text":"In any case, it can be immediately checked out if the formula is satisfacible or not. If a \nformula is not equivalent to a contradiction, it must be satisfacible. All we need is to \ncheck sequentially the disjoints for the absence of complementary pairs."},{"paragraph_id":"p273","order":273,"text":"This leads to the following algorithm:"},{"paragraph_id":"p274","order":274,"text":"To find the truth values satisfying a formula F"},{"paragraph_id":"p275","order":275,"text":"1. Convert F to its disjunctive normal form. \n2. Check the disjoints sequentially from left to right until a disjoint D having no \ncomplementary pairs is found. \n3. Assign 1 to each literal in D. This will make some atoms true and some false."},{"paragraph_id":"p276","order":276,"text":"18"},{"paragraph_id":"p277","order":277,"text":"Apparently this is a polynomial complexity algorithm. The problem is: the conversion \nprocess (to get a formulaβs d.n.f.) is a search process in a tree, having exponential \ncomplexity."},{"paragraph_id":"p278","order":278,"text":"Lema 4."},{"paragraph_id":"p279","order":279,"text":"For every convertion algorithm there is an order n formula F requiring a time t = q(n) \nwhere q is an exponential type function."},{"paragraph_id":"p280","order":280,"text":"Proof."},{"paragraph_id":"p281","order":281,"text":"A clause is a disjunction of literals, some of them positive and some negative ones. For \na given value of n, take a formula F in conjunctive normal form, having k clauses, each \nhaving m literals (obviously m β€ n). Converting F to d.n.f. we get mk disjoints, where \neach of them has k literals."},{"paragraph_id":"p282","order":282,"text":"If k = n and m is close to n, the number of disjoints is in the order of nn."},{"paragraph_id":"p283","order":283,"text":"Note that evaluating the truth table requires to evaluate 2n rows, much less than nn, \nalthough both algorithms have exponential complexity."},{"paragraph_id":"p284","order":284,"text":"5. Asymptotic properties."},{"paragraph_id":"p285","order":285,"text":"For a given order n, take a representative from each equivalence class F n / . We can \nidentify each class with its corresponding truth table. For any n-order formula, its truth \ntable has 2n rows. Since each row can have a resulting 0 or 1, there are \nn\n2\n2\ndifferent \ntruth tables, and therefore there are \nn\n2\n2\n equivalence classes."},{"paragraph_id":"p286","order":286,"text":"In one half of all classes in F n / the formulas have true in the first row of its truth \ntable. In the other half, the formulas have false in the first row."},{"paragraph_id":"p287","order":287,"text":"The formulas belonging to \n1\n2\n2"},{"paragraph_id":"p288","order":288,"text":"n\n classes have 1 in its first row, while the formulas \nbelonging to the other \n1\n2\n2"},{"paragraph_id":"p289","order":289,"text":"n\n classes have a 0 in its first row."},{"paragraph_id":"p290","order":290,"text":"If we put a 0 in row 1, then every one of the remaining 2n-1 rows can have a 0 or 1, \ngiving a total of \n1\n2\n2"},{"paragraph_id":"p291","order":291,"text":"n\n options."},{"paragraph_id":"p292","order":292,"text":"In one fourth of all classes in F n / the formulas have 1 in row 2, knowing they have 0 \nin row 1. In a similar way, knowing rows 1, 2, ..., m-1 to have a 0, there are \nm\nn \n2\n2"},{"paragraph_id":"p293","order":293,"text":"equivalence classes having a 1 in row m."},{"paragraph_id":"p294","order":294,"text":"The number of classes whose formulas have its first 1 in row 1 or 2 or ... or m equals \nthe sum"},{"paragraph_id":"p295","order":295,"text":"q(m) = \n1\n2\n2"},{"paragraph_id":"p296","order":296,"text":"n\n + \n2\n2\n2"},{"paragraph_id":"p297","order":297,"text":"n\n + \n3\n2\n2"},{"paragraph_id":"p298","order":298,"text":"n\n+... + \nm\nn \n2\n2\n = \nn\n2\n2\n (2-1 + 2-2+ 2-3+... + 2-m)"},{"paragraph_id":"p299","order":299,"text":"19\n="},{"paragraph_id":"p300","order":300,"text":"m\nn\n2\n1\n1\n22"},{"paragraph_id":"p301","order":301,"text":"The percentage of classes whose formulas are true for the first time in one of the rows 1, \n2, ..., m is then:"},{"paragraph_id":"p302","order":302,"text":"m\nm\nm\nm\nr\n2\n1\n2\n100\n2\n1\n1\n100\n)\n("},{"paragraph_id":"p303","order":303,"text":"Let be a s degree polynomial, for a fixed s. If m = (n), then r(m) is the percentage of \nclasses whose formulas can be solved in a polynomial time (meaning by βsolveβ to find \na n-tuple of truth values satisfying the formula)."},{"paragraph_id":"p304","order":304,"text":"Take (x) = xs. Then m = ns and r(m) = r(ns) ="},{"paragraph_id":"p305","order":305,"text":"s\nn\n2\n1\n1\n100"},{"paragraph_id":"p306","order":306,"text":"We get"},{"paragraph_id":"p307","order":307,"text":"s\nn\nn\ns\nn\nn\nr\n2\n1\n1\n100\nlim\n)\n(\nlim\n= 100%."},{"paragraph_id":"p308","order":308,"text":"Conclusion:"},{"paragraph_id":"p309","order":309,"text":"When n increases, the ratio of equivalence classes having satisfactible formulas, which \nare detected by a sequential search algorithm exploring only the first (n) rows (or any \nset of pre-determined (n) rows) approaches 100%."},{"paragraph_id":"p310","order":310,"text":"Notes."},{"paragraph_id":"p311","order":311,"text":"The ratio of order n satisfactible formulas not solved by a sequential search algorithm \nexploring (n) rows decreases to 0 as n increases."},{"paragraph_id":"p312","order":312,"text":"However, the number of equivalence classes increases very fast, as shown in the \nfollowing table:"},{"paragraph_id":"p313","order":313,"text":"Rows in table \nEquivalence classes ( = number of Truth tables) \nn u = 2n \n2u \n1\n 2 \n 4 \n2\n 4 \n 16 \n3\n 8 \n 256 \n4\n 16 \n 65.536 \n5\n 32 \n4294967296\n6\n 64 \n1,84467E+19\n7\n 128 \n3,40282E+38\n8\n 256 \n1,15792E+77\n9\n 512 \n1,3408E+154\n10\n 1.024 \n#Β‘NUM! \n11\n 2.048 \n#Β‘NUM!"},{"paragraph_id":"p314","order":314,"text":"20\n12\n 4.096 \n#Β‘NUM! \n13\n 8.192 \n#Β‘NUM! \n14\n 16.384 \n#Β‘NUM! \n15\n 32.768 \n#Β‘NUM! \n16\n 65.536 \n#Β‘NUM! \n17\n 131.072 \n#Β‘NUM! \n18\n 262.144 \n#Β‘NUM! \n19\n 524.288 \n#Β‘NUM! \n20\n 1.048.576 \n#Β‘NUM!"},{"paragraph_id":"p315","order":315,"text":"Even tough the ratio of equivalence classes (causing exponential complexity to an \nalgorithm) vs. total number of classes decreases, this latter number increases quite fast, \nas the following table shows:"},{"paragraph_id":"p316","order":316,"text":"n \n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n1\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n4\n1\n3\n32\n4\n4096\n1\n5\n134217728\n128\n6\n2,8823E+17\n268435456\n7\n2,65846E+36\n6,04463E+23\n8\n4,52313E+74\n6,2771E+57\n9\n2,6187E+151\n5,5453E+129\n10\n1,7556E+305\n1,4181E+278\n16777216\n11\n#Β‘NUM! \n#Β‘NUM! \n6,8946E+215\n12\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n13\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n14\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n15\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n16\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n1\n17\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n18\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n19\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n20\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!"},{"paragraph_id":"p317","order":317,"text":"In the blank entries the calculation formula is not applicable since 2n < ns. The #Β‘NUM! \nmeans an overflow; the numeric limits of Excel on a Pentium computer has been \nexceeded."},{"paragraph_id":"p318","order":318,"text":"Increasing speed for the function r."},{"paragraph_id":"p319","order":319,"text":"As the following table shows, r(ns) / 100 increases quite fast."},{"paragraph_id":"p320","order":320,"text":"s"},{"paragraph_id":"p321","order":321,"text":"n \n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n1\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n2\n0,75\n0,9375\n0,99609375\n0,999984741\n1\n1\n1\n1\n1\n1\n3\n0,875\n0,998046875\n0,999999993\n1\n1\n1\n1\n1\n1\n1\n4\n0,9375\n0,999984741\n1\n1\n1\n1\n1\n1\n1\n1\n5\n0,96875\n0,99999997\n1\n1\n1\n1\n1\n1\n1\n1\n6\n0,984375\n1\n1\n1\n1\n1\n1\n1\n1\n1"},{"paragraph_id":"p322","order":322,"text":"21\n7\n0,9921875\n1\n1\n1\n1\n1\n1\n1\n1\n1\n8\n0,99609375\n1\n1\n1\n1\n1\n1\n1\n1\n1\n9\n0,998046875\n1\n1\n1\n1\n1\n1\n1\n1\n1\n10\n0,999023438\n1\n1\n1\n1\n1\n1\n1\n1\n1\n11\n0,999511719\n1\n1\n1\n1\n1\n1\n1\n1\n1\n12\n0,999755859\n1\n1\n1\n1\n1\n1\n1\n1\n1\n13\n0,99987793\n1\n1\n1\n1\n1\n1\n1\n1\n1\n14\n0,999938965\n1\n1\n1\n1\n1\n1\n1\n1\n1\n15\n0,999969482\n1\n1\n1\n1\n1\n1\n1\n1\n1\n16\n0,999984741\n1\n1\n1\n1\n1\n1\n1\n1\n1\n17\n0,999992371\n1\n1\n1\n1\n1\n1\n1\n1\n1\n18\n0,999996185\n1\n1\n1\n1\n1\n1\n1\n1\n1\n19\n0,999998093\n1\n1\n1\n1\n1\n1\n1\n1\n1\n20\n0,999999046\n1\n1\n1\n1\n1\n1\n1\n1\n1"},{"paragraph_id":"p323","order":323,"text":"The number 1 occurs in most places where the fraction subtracted from 1 is too small \nfor the precision handled by Excel."},{"paragraph_id":"p324","order":324,"text":"Other notes."},{"paragraph_id":"p325","order":325,"text":"1. In the preceeding anΓ‘lisis, we have not taken into account the process needed to find \nout if a formula is irreducible, since it would add processing time. A formula may be \nevaluated with all the its atoms, without trying to simplify it beforehand via the \nabsorption equivalences."},{"paragraph_id":"p326","order":326,"text":"2. If m < n then r(n) < 100, because the contradiction class is excluded (its table yields \nonly zeros)."},{"paragraph_id":"p327","order":327,"text":"3. It has been shown that for every uninformed sequential search algorithm there is a set \nof formulas taking an exponential time. An βinverseβ problem can be stated as: for \nevery formula there is an uninformed sequential search algorithm giving all the \nformulaβs solutions in the first rows of the truth table, and zeros on the remaining rows. \nThe problem is to find the algorithm. This requires a search on the set of uninformed \nsequential search algorithms. This search has exponential complexity."},{"paragraph_id":"p328","order":328,"text":"If a an n order formula F has m solutions (that is, there are m rows where F is true), then \nthere are m!(2n β m)! ways to permute the rows keeping the true rows first and the false \nrows last. The ratio of this number of algorithms to the total ways of exploring Fβs truth \ntable is m!(2n β m)! / (2n)! ="},{"paragraph_id":"p329","order":329,"text":"m\nn\n2\n1\n, which makes it highly unlikely to find one of this \nalgorithms by chance. Using a systematic search, the vast majority of a huge number of \nalgorithms must be tested before a solution can be found."},{"paragraph_id":"p330","order":330,"text":"References:"},{"paragraph_id":"p331","order":331,"text":"[1] Nilsson, Nils J., βArtificial Inteligence: a new synthesisβ, Morgan Kaufmann \nPublishers, San Francisco, CA, 1998 \n[2] βThe efficiency of algorithmsβ, Harry R. Lewis y Christos H. Papadimitriou, \nScientific American, January 1978."},{"paragraph_id":"p332","order":332,"text":"22\n[3] Mendelson, Elliott, βIntroduction to Mathematical Logic,β D. Van Nostrand \nCompany, Princeton, NJ, 1965"},{"paragraph_id":"p333","order":333,"text":"[4] Russell, Stuart, Norvig, Peter, βArtificial Intelligence: a modern approachβ, Prentice \nHall, NJ, 2002."},{"paragraph_id":"p334","order":334,"text":"[5] βMillion Dollar Minesweeperβ, Ian Stewart, Scientific American, October 2.000 \n2007.11.07\n16:36:35-05'00'"}],"pages":[{"page":1,"text":"1\nConsiderations on the P NP question. \nAlfredo von Reckow. \n \nAbstract \n \nIn order to prove that the P class of problems is different to the NP class, we consider \nthe satisfability problem of propositional calculus formulae, which is an NP-complete \nproblem [1], [3]. It is shown that, for every search algorithm A, there is a set E(A) \ncontaining propositional calculus formulae, each of which requires the algorithm A to \ntake a non-polynomial time to find the truth-values of its propositional letters satisfying \nit. Moreover, E(A) βs size is an exponential function of n, which makes it impossible to \ndetect such formulae in a polynomial time. Hence, the satisfability problem does not \nhave a polynomial complexity. \n \n1. Some definitions. \n \nConsider L, the propositional calculus formal theory, as defined in Mendelson [3] and \nlet L+ be an extension of L containing the 0-letter formulas : (empty formula, or \ncontradiction) and (tautology). Let F be the countable set of formulas in L+. Let \ndenote the βlogic equivalenceβ of formulas, defined by A B iff A and B yield the same \nresult when evaluating its truth tables. \n \nIt can be easily verified that is a equivalence relationship. Let the quotient set be \nG = F / = {[A] / A F }. \n \nWe will use indistinctly the terms βatomβ or βpositive literalβ for a propositional letter. \nA βnegative literalβ is an atomβs negation. A βliteralβ is an atom or the negation of an \natom. \n \nA formula Μs cuasinorm . \n \nFor each A F , we define a quasi-norm || A || = number of different atoms in A. This \nquasi-norm has the following properties: \n \n For each A F , || A || 0. \n || A || = 0 iff A is or . \n Triangular inequality: For A and B in F ,, and any binary connective op, \n \n|| A op B || || A || + || B ||. \n \nIf A and B share atoms, the inequality is strict (including the case when A and B \n \nare the same formula). The equality occurs when A and B do not share any \n \natoms (including the case when A or B is or ). \n \nA property like || A || = | | || A || (for real) is not valid here. One way to define \nβmultiplicationβ of a formula by a positive integer is 2 A = A op A, for some binary \nconnective op. But this produces || n A || = A for any natural n. Furthermore, the \nproduct A makes no sense for any number which is not a positive integer. \n \nSince || A || | | || A || for real, the function || || is not a norm, so we call it a quasi-\nnorm."},{"page":2,"text":"2\n \nIrreductibility \n \nTwo formulas can be equivalent, despite having different number of atoms. It is \npossible to have A B and || A || || B ||, because of the absorption laws: \n \np (p q) p \np (p q) p \n \nWe will say that a formula A is irreducible of order k iff || A || = k and it is not \nequivalent to another formula B which has less atoms than A. (i.e. if A B and || A || = \nk implies || B || k ). \n \nIn other words, A F is irreducible of order k iff \n|| A || = k and \n( B F ) (A B || B || k) \n \nA classβ cuasinorm. \n \nFor X G, letβs define \n||\n||\nmin\n||\n||\nB\nX\nX\nB \n \n, which counts the minimum number of atoms \nfiguring in a formula from the X equivalence class. \n \nHere again, || || is a quasi-norm, having the following properties: \n \n For every X G, || X || 0. \n || X || = 0 iff X is the class of all contradictions (which are equivalent to ) or \nthe class of all tautologies (which are equivalent to ). \n Triangular inequality: For X and Y in G,, and any binary connective op, \n \n|| X op Y || || X || + || Y ||, where we define X op Y = { F1 op F2 } for F1 X \nand F2 Y. \n \nThe set of order n irreducible formulas is defined by \n \nF n = { A / || A || = || [ A] || = n }. \n \nThen we have F = \n \n \nN\nn\nF\nX\nn\nX\n \n \n, \nAnd the set of all irreducible formulas is F = \nN\nn \n F n \n \nA formula is simply βirreducibleβ if it is irreducible of order n, for some natural number \nn. \n \nWe define the convertion function, \n \nf : F / F \n f( [ A ] ) = B, where B A and A is irreducible."},{"page":3,"text":"3\nfrom each equivalence class [A], f takes an irreducible formula B. Equivalently, f can \nbe seen as an algorithm (explicity stated) which produces an irreducible formula B from \nthe class [A]. The function f converts any formula in the class [A] to the irreducible \nformula B. \n \nWe have B F n , for some natural number n, and [ B ] F n / \n \nWe define the normalization function h : F F , where h(A) is in disjunctive normal \nform (dnf). Then, for each equivalence class [A], h(f(A)) gets an irreducible and \nnormalized formula. \n \nFor any given n, let Μs consider the set of βminimalβ formulas having n propositional \nletters p1, p2, ...pn, which are the order n irreducible formulas. \n \nLet Gn = { [An ] / An is irreducible having order n }. Then G = \nN\nn \n Gn \nLet In = F n / { [An ] / B [An ] iff B is order n irreducible and B An } \n \nEach class of In contains only order n irreducible formulas. \n \nThen I = \nN\nn \nIn is the set of all equivalence classes having only irreducible formulas. \n \nIf we take one element of each equivalence class, we get the set of all irreducible \nformulas, pair wise non equivalent. \n \nNow we construct the image of the function f, taking the order n minimal formulas: \n \nHn = { Bn / Bn = f(X) for X In } where the Bn are not equivalent to each other, and \n \nH = \nN\nn \n H n. \n \nThen H is the set of all irreducible und unique formulas. In other words, each of the \nformulas in H is irreducible (because it has a minimum number of propositional \nletters, and it is impossible to eliminate a letter using absorption) and there are no two \nformulas in H being equivalent to each other (which makes each formula unique). \n \nFrom now on, we will work only with formulas from H . That is, we will assume that \nany given formula A is order n irreducible and we will ignore other formulas equivalent \nto A. \n \n2. Why this is a search problem?. \n \nWe will show that there is no algorithm to produce directly, in a straightforward way, \nthe truth values for the propositional letters in a formula, which make the formula true, \nunless there is a search involved. To this end, we show that this satisfability problem is \nequivalent to construct a βclosedβ formula for the roots of a polynomial with n \nvariables."},{"page":4,"text":"4\n \nLet Μs define the function h, from L+ to the set of n-variable polynomials, by \n \n For a formula A of L+, having n propositional variables p1, p2, ..., pn, letβs \nassociate to each of its letters a numerical variable xi: h(pi)= xi for 1 i n. \n \n h(Β¬ A) = 1 β h(A) \n \nIf A and B are two formulas in L+, then \n \n h(A B) = h(A) h(B) \n h(A B) = h(A) + h(B) β h(A) h(B) \n h(A B) = (1 β h(A) ) (1 β h(B) ) + h(B) \n h(A B) = (1 β h(A) ) (1 β h(B) ) ( 1 + h(A) + h(B) ) + h(A) h(B) \n \n \nNote that \n \nh(A)n is equivalent to h(A) in the following sense: \nh(A)n = 0 iff h(A) = 0 and h(A)n = 1 iff h(A) = 1 \n \nNote that h is a truth function assigning to a formula A a value 0 or 1 when each of the \nvariables xi takes a value of 0 or 1. \n \nIn particular, in the resulting polynomial we can replace any factor xn by x. \n \nFor example, for the idempotency law: A A A we have: \n \nh(A A) = h(A) h(A) = h(A)2 h(A), \n \nwhere the last β β means equivalence in the sense that we get the same values for h(A)2 \nand for h(A) when each of the corresponding variables xi take a value of 0 or 1. \n \nSimilarly, with the other idempotency law, A A A, we get \nh(A A) = h(A) + h(A) β h(A)h(A) = 2h(A) β h(A)2 2h(A) β h(A) = h(A). \n \nTautologies and contradictions: \n \nIf a formula A is a tautology, then h(A) 1. \nIf a formula A is a contradiction, then h(A) 0. \n \nExample 1. \n \nFor the tautology A A we have: \nh(A A) = (1 β h(A))2 + h(A) (1 β h(A)) + h(A) = 1. \n \nThis latter equivalence can be proved in the following ways:"},{"page":5,"text":"5\na) (1 β h(A))2 (1 β h(A)), because both sides have the same value when h(A) = 0 or \nwhen h(A) = 1. \n \nb) (1 β h(A))2 + h(A) = 1 β 2h(A) + h(A)2 + h(A) = 1 β h(A) + h(A)2 \n 1 β h(A) + h(A) = 1 \n \nExample 2. \n \nFor the contradiction A Β¬A whe have: \n \nh(A Β¬A) = h(A) (1 β h(A)) = h(A) β h(A)2 h(A) β h(A) = 0. \n \nFor any formula A, the problem of finding the truth-values for its variables p1, p2, ..., pn \nsatisfaying A (i.e. making it true) is equivalent to the problem of finding the \ncorresponding numerical values for n-tuple (x1, x2, ..., xn) which are zeros of the \nfunction g, where g(A) = h(A) β 1. \n \nThis function is a polynomial with the n variables (x1, x2, ..., xn), which will be called \nthe formulaβs βcharacteristic polynomialβ, and has the form g(A) = q, where \n \nq : [0, 1]n R, takes values on the n-dimentional unit cube, and is defined by \n \n(1) \n \n \n \n \n \n)\n(\n1\n2\n1\n1\n...\n)\n,...,\n(\n2\n1\nn\nr\ni\nn\ni\nn\nin\ni\ni\nx\nx\nx\nx\nx\nq\n \n \n \n \n \n \n \nwhere each exponent ki has a value of 0 or 1, the coefficients i and are integers, \nand the variables xi have a value of 0 or 1, and r(n) = 2n. \n \nWe define 00 = 1. \n \nThe problems then is to find a value for each variable xi, either a 0 or a 1, in such a way \nthat q(x1, x2, ..., xn) = 0. Or equivalently: \n \n(2) \nFind the roots of the polynomial q \n(3) \nWith the restrictions xi in {0, 1} for i = 1, ..., n. \n \nIn other words, find q-1 (0) {0, 1}n. \n \nLemma. \n \nThere is no closed formula to calculate directly the roots of this type of polynomials. \nHence, there is no algorithm to find directly (without making any search) the truth \nvalues of the propositional letters satisfying a formula A. \n \nProof. \n \nThe equation g(A) = 0 has n variables. It can be seen as a βsystemβ of 1 equation with n \nunknowns. This is an under determined system, having infinite real roots. To"},{"page":6,"text":"6\ndetermine them, it suffices to solve g(A) = 0 for x1, which will then depend on the other \nindependent variables x2, ..., xn. \n \nSince these variables are independent (there is no relationship between them), each of \nthem can take any real value. Once we chose a value for each variable x2, ..., xn, that \nwill produce a value for x1. \n \nFrom these infinite set of values for the n-tuple (x1, ..., xn) .we have to separate those \nwhere all the variables are 0 or 1. \n \nBecause there are no additional conditions on the variables, no further relationship \nbetween them can be obtained, and this happens just because the variables x2, ..., xn are \nindependent. There is only the restriction (3 ) on the value of the solutions, but it is of \nno use for finding a closed formula for the solutions of (2). \n \nHence, we have no other way left, but to carry on an exhaustive search, and we have no \nadditional information derived from the formula which could lead us to take any \nshorcuts and reduce the number of cases (or number of rows in the truth table) to be \nconsidered. \n \nExample. \n \nFor the formula \n \nA : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3) \n \nThe associated polynomial for A is \n \nh(A) = x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 \n \nThe characteristic polynomial is \n \ng(A) = x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 \n \nand the equation g(A) β 1 = 0 is \n \nx1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 = 0. \n \nWith the restrictions xi {0, 1} for i = 1,.., 3. \n \nIn this equation, the variable x1 depends on x2 and x3. This two last variables have no \nfurther relationship to one another. If x3 takes any given value, it has no influence at all \non the values that x2 can take. The only thing we know is that we are interested in the \ncases where x2 as well as x3 take a value of 0 or 1. \n \nHence, we have no choice but to make an exhaustive search, using the 4 possible values \nfor the duple (x2, x3). The solutions will be those cases where the equation makes sense \nand yielding values of 0 or 1 for x1."},{"page":7,"text":"7\nNote that if (x2, x3) = (1, 1) then x1 vanishes and the equation becomes 1 = 0, which \nmakes no sense. Hence (x2, x3) (1, 1). \n \nFrom other point of view, solving for x1 we get \n \n(4) \n1\n)\n(\n2\n3\n1\n2\n3\n2\n3\n2\n3\n2\n3\n2\n1\n \n \n \n \n \n \n \nx\nx\nx\nx\nx\nx\nx\nx\nx\n \n \nwhere the denominator is zero if (x2, x3) = (1, 1). It is zero for other values of these \nvariables, but it does not matter, because in none of those other cases both variables \nhave a value of 0 or 1. \n \nNote that x2 and x3 can take any real value in (4), as long as the denominator does not \nvanish. There are infinite values for x2 and x3 satisfying (4), but we are interested only \nin the 4 cases where they are 0 or 1, and there are no shorcuts for finding them. We \nmust try all the options and see which ones work. \n \nIn the general case, for a formula having n propositional letters, we can not avoid the \nneed to explore 2n-1 options for the variables x2, ...xn. \n \nOther way to search is the following: \n \nIf we put arbitrarily x1 = 0 in (4), then 2x2x3 β x2 β x3 + 1 = 0, and from here we get x2 = \n(x3 β 1) / (2x3 β 1). Here we have two cases: if si x2 = 0, then x3 = 1 and if x2 = 1, then \nx3 = 0. \n \nIf we put x1 = 0 in (4), then 2x2x3 β x2 β x3 + 1 = 3x2x3 β 2(x2 + x3) + 1, so x2 = x3 / (x3 β \n1). Here again we have two cases: if x2 = 0, then x3 = 0 and if x2 = 1, then x3 = x3 β 1, \nwhich is impossible. \n \nWe have had to analize 4 different cases, for 3 variables, corresponding to 2n-1 options, \nfor n = 2, and the variables x1, ...xn-1 \n \nWe can solve for any of the n variables as a function of the other n-1 variables. In any \nevent, we must try 2n-1 options. \n \nNotes. \n \n1. Note that the characteristic polynomial g has n variables and has 2n terms. Each term \nis a product of k variables, and there are C(n, k) = n! / (k! (n-k)!) combinations of terms \nhaving k varibles. Adding that up, there are \n \n \n \n \n \n \n \n \n \n \nn\nk\nn\nk\nn\n0\n2 \n \nterms, each one has one coefficient and n exponents. Any algorithm processing g has \nthus an input of (n+1) 2n numbers, an exponential function of n, which can not be \nprocessed in a polinomial time."},{"page":8,"text":"8\nMoreover, if there were a βclosed formβ formula to determine the binary roots of g, then \nit would depend on the 2n coefficients of g. Since the variables x2, ..., xn are \nindependent, a closed form for each of them would be a function of those 2n \ncoefficients, and it cannot be calculated in a time t = q(n), where q is a polynomial. \n \nSuch a formula, if it existed, would behave not better than a sequential exhaustive \nsearch. \n \n2. The characteristic polynomial g has n variables x1, x2, ..., xn each one with an \nexponent 0 or 1. In any occurrence of any of this variables the exponent can be replaced \nby an odd natural number k, and the resulting polynomial has the same βbinary rootsβ \n(those being 0 or 1) of g. The exponent must be 0 or odd to avoid de case (-1)2 = 1. \n \nHence, g is equivalent to infinitely many polynomials (in the sense that all of them have \nthe same binary roots). \n \nIf there were an explicit formula to find the binary roots of g, that same formula would \nyield the binary roots of all the polinomials q equivalent to g, which can have an \narbitrarily high degree. \n \nThese comes from the idempotency laws A A A and A A A. \n \n3. It is not possible to determine, a priori, additional conditions to find the binary roots \nof g. If there were n-1 such additional conditions, then we would have a non-linear \nsystem of n equations with n variables. \n \nThe number of solutions for the characteristic polynomial can be anything from 0 (in \nthe case of a contradiction) to 2n (for a tautology). If all this options are possible, then \nno additional conditions can exist. For a formula like p1 p2 ... pn, having 2n β 1 \nsolutions, any additional condition must be redundant. For a formula having 2n β 2 \nsolutions, there can be at most one additional condition, and son on. The number of \nadditional conditions depends on the number of solutions, which is not known in \nadvance, so we donβt know beforehand how many conditions are necessary for a given \nformula. \n \nMoreover, if such conditions depend on the number of solutions, and the solutions \ndepend on the coefficients of the characteristic polynomial, then those conditions will \ndepend on those coefficients. Since there are 2n coefficients in the characteristic \npolynomial, the additional conditions will depend on 2n inputs, which can not be \nprocessed in a polynomial time. \n \nOn the other hand, if we introduce, arbitrarily, any condition g1(x1, ..., xn) = 0, then \nsome roots can be missed. \n \nFor example, if we put arbitrarily the condition x1 = 0, then we will miss all the roots \nwhere x1 = 1, in the cases where such roots exists. In the cases where g has a factor (x1 \nβ 1) then we will miss all its binary roots."},{"page":9,"text":"9\nIf we put a condition like x1 β x2 = 0, then we might miss the roots where x1 β x2, and \nwe do not know beforehand if that is indeed the case. Simple conditions as these can \nbe, of course, verified beforehand, but more complex conditions can not. \n \nIn the example \n \nA : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3) \n \nThe characteristic polynomial is \n \nx1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 = 0. \n \nIf we add the condition x1 β x2 = 0, (which can be considere a mere supposition) the \ncharacteristic polynomial becomes \n \nx3 - x1x3 β 1 = 0. \n \nWhere x1 must be zero and x3 must be 1. x2 can be 0 or 1. \n \nThe triplet (0, 1, 0) is a root of the characteristic polynomial, but the other triplet (0, 1, \n1) is not. \n \nAdding an arbitrary condition not only leads us to miss two roots, but it introduces a \nfake root. \n \nWe have no additional information and we cannot add it arbitrarily, because we donβt \nwhere to start looking for the roots. \n \nAs have been noted in the literature (see [3]), if we know the roots (or if we a have a \ncandidate for root, somehow βguessedβ or estimated), we can replace it in the \ncharacteristic polynomial and check it in a polynomial time. \n \nBut if we donβt have a root, we have to search for it, in a sequential way. We must try \nall the options. \n \n4. Multiplying the factors of g to get the explicit form (1) requires o(mn) multiplications, \nwhere m is the number of factors and n is the number of variables. Instead of carrying \non the multiplications, we can consider the associated polynomial, when it is already \nfactored and find its roots. The remaining n-tuplets (those not being a binary root of h) \nwill be the binary root of the characteristic polynomial g. \n \nThe idea is to take advantage when h is already factored (as this might take less \noperations). \n \nIn the example, \nA : (p1 p2 p3) Β¬ (p1 p2) Β¬ (p1 p3) Β¬ (p2 p3) \n \nThe associated polynomial is \n \nh(A) = (x1 + x2 + x3 - x1x2 β x1x3 β x2x3 + x1x2x3 )(1 - x1x2) (1 - x1x3) (1 β x2x3)"},{"page":10,"text":"10\n \nh(A) = 0 if al least one of its factors is zero. \n \nThe last thre factors give the options \n \nx1 = 1, x2 = 1, any value for x3 \nx1 = 1, x3 = 1, any value for x2 \nx2 = 1, x3 = 1, any value for x1 \n \nThe first factor gives \n \nx1 = (x2 + x3 - x2x3) / (1- x2 - x3 + x2x3) = U / ( 1 β U), for U β 0, \n \nWhere U = x2 + x3 - x2x3. All the options are: \n \nx2 \nx3 \nU \nx1 \n0 \n0 \n0 \n0 \n0 \n1 \n1 \nnot definided \n1 \n0 \n1 \nnot definided \n1 \n1 \n1 \nnot definided \n \n \n \n \nThe roots of the associated polynomial h are \n \nx1 \nx2 \nx3 \n0 \n0 \n0 \n0 \n1 \n1 \n1 \n0 \n1 \n1 \n1 \n0 \n1 \n1 \n1 \n \nSince g = h β 1, for any other triplet (x1, x2, x3), h(x1, x2, x3) is not zero, and hence it \nmust be h(x1, x2, x3) = 1. Those other three triplets must be the binary roots of g. They \nare: \n \nx1 \nx2 \nx3 \n0 \n0 \n1 \n0 \n1 \n0 \n1 \n0 \n0 \n \nBut this has required to explore all the options for the first factor. It takes an o(2n-1) \nnumber of operations, for n = 3. \n \nThe cases when two or more factors are simultaneously zero produce repeated roots, \nthat have to be eliminated from the list of roots. \n \n \n3. Exhaustive search \n \nIf there is no direct (βclosed formβ) formula to find the binary roots of the characteristic \npolynomial of a given formula, we must carry on a sequential search, checking a"},{"page":11,"text":"11\nformulaβs truth table in a sequential way, one row after another. Since there are no \nadditional conditions to guide the search, it must be made sequentially. \n \nFor the set of order n irreducible formulas, we choose an order of evaluation for the \npropositional letters and for the truth value for each atom. First, we take some \npermutation of the letters. Then, for every letter we choose if it takes the value 0 or 1 \nfirst. Then for every order of evaluation there are formulas taking necessarily an \nexponential time to find an n-tuple of truth values satisfying it. \n \n3. 1. Sequential search algorithms. \n \nIn this section we consider βuninformedβ or βblindβ search algorithms, which uses no \nadditional information (other than the given formula itself) to guide the search or to \nreduce the number of options. As was shown in section 2, such information is not \navailable beforehand, and if it were, taking it into account requires an exponential order \nalgorithm, which is no better than a sequential uninformed search. \n \nFor this reason, in this section we well consider only algorithms exploring a formulaβs \ntruth table in some order that does not depend on the formula. That is, an algorithm A \nwill explore the truth tableβs rows in the same fashion for all formulas. \n \nA βsequential search algorithmβ assigns truth values to the propositional letters p1, p2, \n..., pn occurring in a formula F in some pre-established order (out of the n! permutations \nof the n letters) and for each letter it is established if 0 or 1 is used first (out of the 2n \noptions). \n \nConsider the set of all sequential search algorithms. This set has n! 2n algorithms, \nbecause the n letters can be arranged in n! different permutations and each letter can be \nevaluated to 0 first and then to 1, or the other way around. If a formula F has n atoms ( \n|| F || = n) there are n! 2n different ways to evaluate F, depending on the order the atoms \nare evaluated and if the 0 or 1 is evaluated first for every atom. \n \nTwo sequential search algorithms A1 and A2 are βsequentially equivalentβ if and only if \nthey assign the same truth values to the same atom in the same order. The algorithms \ncan use different functions or procedures, yielding the same results in the same order, \nbut their implementations can yield different running times on a given machine, \noperating systems and programming language. \n \nTwo sequential search algorithms A1 and A2 are not equivalent, and will be considered \ndifferent, if they take the atoms in a different order or if they assign the truth values 0 \nand 1 in a different order to the same atom. \n \nEquivalently, two sequential search algorithms A1 and A2 are different if they explore \nthe truth table rows in a different order. \n \nThis sequential search is equivalent to a depth-first search in a binary tree, with a given \npriority for the order in which the atoms are evaluated and the truth value given to each \natom."},{"page":12,"text":"12\nSequential equivalence is an equivalence relationship, so we can take an algorithm from \neach equivalence class. We will assume that we take the algorithm taking the least time \nin a particular implementation. \n \nIn order to find out if a given formula F is satisfacible, and when it is, find the \ncorresponding truth-values of its atoms, a sequential search algorithm A explores the \ntruth table rows in a certain order. The algorithm A βsolvesβ the formula F if it finds a \ntruth-value assignation for Fβs atoms, with which F is evaluated true. \n \nWe will represent the truth table for F, as evaluated by A, by A(F) = (b(1), b(2), ..., \nb(2n)), where b(k) is the value obtained in row k. \n \nWe define the function L as the function giving the first row where algorithm A finds a \nvalue of 1 when it evaluates the formula F. This is L(A(F)) = \nn\nk 2\n1min\n \n \n{ k / b(K) = 1 }. \n \nExploring with the initial order. \n \nTake F n, the set of order n irreducible formulas. Take a formula F in F n. Then || F || = \nn, and consider p1, p2, ..., pn as the n distinct atoms figuring in F, taken in the same \norder they appear in F, from left to right. We will call this the βinitial orderβ of the \natoms in F. \n \nDenote by Ai,j the algorithm which uses the i-th permutation of the atoms of F and the j-\nth way of giving the truth values to them. \n \nThe sequential search algoritm A1,1 explores Fβs truth tableβs rows in the βnaturalβ \norder, taking the atoms p1, p2, ..., pn in the same order they appear in F and for every \none of them it evaluates first the 0 and then the 1 values. \n \nThen, for the formula \n \nF: \n p1 p2 ... pn \n \nthe sequential search algoritm A1,1 must check all the rows in the truth table, to find a \nsolution only in the last row (that is, in row 2n ). Thus, L(A1,1(F)) = 2n. \n \nChanging 0s and 1s. \n \nTake now all the A1,j algorithms. All of them evaluate the atoms p1, p2, ..., pn in the \norder 1, 2, ..., n. They differ from each other because they assign the values 0 and 1 in \na different order for every letter (one of them will assign first 0 and later 1 to some atom \npi, while other one assigns first 1 and later 0 to the same letter pi ). \n \nMore precisely, if the base 2 representation of j is j = d1d2...dn, then the atom pk takes \nfirst the truth value dk and later it will take the truth value 1 β dk , for 1 β€ k β€ n. \n \nFor every j, if the A1,j algorithm evaluates first true the atom pk (that is, if dk = 1), then \nlet`s take the letter qk equal to pk. Otherwise, take qk equal to pk. Then the formula"},{"page":13,"text":"13\n(5) \n \nF: \n q1 q2 ... qn \n \nis true only when all the qj are false. To get there, the algorithm A1,j must explore first \nall the other rows in the truth table. In this case, we also have Thus, L(A1, j(F)) = 2n. \n \nThus, the algorithm A1, j(F) has an order o(2n) temporal complexity. \n \nPermuting the order of the atoms in F is irrelevant, since is has no effect on the rows \nthat have to be evaluated. In any case, the formula F will take an exponential time: \nL(Ai, j(F)) = 2n, for 1 β€ i β€ n! \n \nThe set Cn(Ai,j). \n \nFor an integer n given, we define the set Cn(Ai,j) as the set of order n irreducible and \npair-wise non equivalent formulas for whom algorithm Ai,j takes a non polynomial time. \n \nThe formula F given above in (5) is in Cn(Ai,j). It the number of formulas in this set \n(representing each one an equivalence class) where a polynomial depending on n, we \ncould detect this kind of formulas before applying the algorithm. Then we Ai,j would \nexecute in a polynomial time for the other formulas in F n. \n \nThis is not the case, as it will be shown that Cn(Ai,j) = o(g(n)), where g is an \nexponential-type function. \n \nFor every sequential search algorithm Ai,j and every formula F, it is posible to apply a \npre-process to determine if F is equivalent to a worst-posible case formula, (and in this \ncase, to evaluate it at once). Letβs call this revised algorithm Ai,j(1). \n \nBut then there is another formula F(1) forcing the revised algorithm Ai,j(1) to search all \nthe way thru the penultimate row in the truth table of F(1), because that is the only row \nwhere F(1) is true. If we repeat the process to construct algorithm Ai,j(2), which detects \nF and F(1), then the formula F(2), which is true only in the next-to-penultimate row, \nwill take an exhaustive search from row 1 to row 2n-2, and so on. \n \nMore precisely, we have two steps in the analysis: \n \nLemma 1. \n \nFor every natural number m, 1 β€ m β€ 2n, there is a formula F having n atoms, that \nis true only in row m. \n \nFor each sequential search algorithm Ai,j and every m {1, ..., 2n}, we construct a \nformula being true only on row m, by taking the formula \n \nF: \n pi1 pi2 ... pin \n \nwhich is true only in the last row of the truth table, and changing each atom for its \nnegation, depending on the order in which algorithm Ai,j assigns values to each atom."},{"page":14,"text":"14\nTo this end, we take the base 2 representation of the index j: j = d1d2...dn. Suppose the \natom pk takes first the value dk and later on the value 1 - dk. \n \nTake the base 2 representation of m: m = e1e2...en, \n \nIf ek = dk, then qik(m) is pik, otherwise, qik(m) is Β¬ pik. \n \nThen the formula Gij(m): \nΒ¬ qi1(m) Β¬ qi2(m) .. .Β¬ qin(m) \n \nIs true only in row m and it is false in all the other rows. \n \n \n \n \n \nLemma 2. (from the end of the truth table). \n \nFor every sequential search algorithm Ai,j, there is an order n irreducible formula F, that \nis true only in the last 2 rows explored by the algorithm. \n \nProof. \n \nThis formula is constructed taking the disjunction \n \nGij(2n-1) Gij(2n) \nFor this formula, algorithm Aij(1) must explore 2n-1 rows in the truth table. \n \nOf course, we can detect this formulae and give them a different treatment, with a pre-\nprocess. \n \nCall Aij(2) be the algorithm obtained by adding such pre-process to Aij(1), to eliminate \nthe βworst possible casesβ for Aij and Aij(1). \n \nConsider now the following formula, which is true only in the last three rows, as \nexplored by algorithm Aij(2): \n \nGij(2n-2) Gij(2n-1) Gij(2n) \n \nAs well as the formulas \n \nGij(2n-2) Gij(2n-1) \nGij(2n-2) Gij(2n) \n \nWe can separate this formulas with a new pre-process, to get the algoritm Aij(3), and so \non. \n \nContinuing this way, for each algorithm Aij we construct a succession of algorithms \nAij(1), Aij(2), Aij(3), .... Every one of them has a set of worst-possible case formulas, \nand every one avoids the worst-possible case for the previous algorithms. \n \nThe worst-possible case for Aij(2n-1) is a formula whose truth table is true in all its \nsecond-half rows and false in all its first-half rows. This algorithm has exponential \ncomplexity."},{"page":15,"text":"15\nTo find out if a formula F is true from row k, for k = 2n, 2n-1, 2n-2, ..., 2n-1 + 1, we have \nto check out 2n-1 cases. If none of them happens, then we must explore sequentially \nrows 1 to 2n-1. The worst possible case occurs with the formula which is true only in \nrow 2n-1. \n \nThe time complexity is therefore: \n \nRows to check out in the pre-process: \n2n β (2n-1 + 1) + 1 = 2n-1 \nRows to explore in the process \n \n2n-1 \n---------------------------------------------------------- \nTotal \n \n \n \n \n2n \n \nMoreover, for every k, algorithm Aij(k) has exponential complexity. \n \nWe have to check first (in the pre-process) for k worst-possible case situations of Aij(h) \nh = 1, ..., k-1. For the formulas passing this check, there are the worst-possible cases \nAij(k) demanding to explore 2n β k rows. Hence, in the worst-possible case, Aij(k) must \nexplore k + (2n - k) = 2n rows. \n \n \n \n \n \n \n \n \nConclusion: \n \nFor every way used to explore a truth-table, there is a formula requiring to explore all \nthe other rows first. If we try to avoid this worst-possible cases adding a pre-process to \nthe search algorithm, then there is another formula requiring also to explore all the other \nrows first. \n \nThis applies to any sequential search algorithm that does a βblindβ or βuninformedβ \nsearch, where no additional information is used to guide the search or to discard \noptions, which will be analized in section 3.2. \n \nAdditional notes. \n \n1. Additional complexity. \n \nThere are additional sources of complexity: each equivalence class in G = F / \ncontains formulas with an arbitrarily large number of atoms, which are equivalent to a \nformula with n atoms, due to the absorption laws. \n \nThis requires to first simplify a formula, adding extra time to the process. For an n-\natom formula F, it can be constructed infinite many other formulas with m atoms, for \nany m > n, equivalent to F by the absorption laws. Because of this, an algorithm \nprocessing formulas with n atoms will have to deal also with formulas having m > n \natoms to find out if they can be simplified to a formula with n atoms. Since m can be \narbitrarily large, the time needed will be arbitrarily large too. \n \nThatβs why we have supposed that the input formulas for any algorithm are already \norder-n irreducible. \n \n2. Extending lemma 2."},{"page":16,"text":"16\n The Aij algorithms take into account only permutations in the order of evaluation of the \natoms and of the truth value for each atom. There are 2n n! such permutations. The \ntruth table for an n-order formula has 2n rows, having 2n! permutations. Since the proof \nof lemma 2 includes a constructive method to build a formula that is true in any given \nrow (and only in that row), that proof can be extended to cover the case of an arbitrary \npermutation of rows of the truth table. \n \n3.2. Heuristic Algorithms. \n \nAn heuristic algorithm does not explore all the rows in a formulaβs truth table. It \neliminates arbitrarily some of the rows, hoping there is a solution in the remaing rows. \nSuch algorithm will have a polynomial complexity if it explores only (n) rows for an \norder n irreductible formula F, whre is a polynomial. \n \nThis means that the algorithm eliminates a priori all the other rows. The algorithm will \nbe βacceptableβ for a formula F if some rows yielding 1 are among the explored rows. \n \nLema 3. \n \nFor every polynomial complexity heuristic algorithm H, there is a set of formulas C(H) \nfor which H is not acceptable. That is, for every formula in F in C(H), H does not \nexplore any row of F where the result is 1. \nIn other words, we can not eliminate rows arbitrarily, because that might eliminate \nprecisely those rows having a result of 1. \n \nProof. \n \nLet F be an order n irreducible formula and a polynomial depending only on n. Let Hj \nbe one of the \n \n \n \n \n \n \n)\n(\n2\nn\nn\n \nalgorithms having polynomial complexity of order (n), which \nexplores only rows 1, 2, ..., (n), evaluating the atoms p1, p2, ..., pn in the same order \nthey figure in F, from left to right. \n \nFor any n-order formula, each algorithm Hj evaluates only rows i1, i2, ..., \n)\n(n\ni \n. This \nrows are the same for all n-order formulas. The row indexes depend only on the \nalgorithm j-index and not on any given formula. \n \nTake k1 as one of the rows not explored by Hj. With the same method used in lemma 2, \nwe can construct a formula that is true only in row k1: \n \nIf atom pi is true in row k then take qi equal to pi. Otherwise take qi equal to Β¬ pi. Take \nthe conjunction of the qi . \n \nGk1: \nq1 q2 ... qn \n \nThis formula is true only in row k, and algorithm Hj missed it, since it does not explore \nthis row."},{"page":17,"text":"17\n( Moreover, Hj will miss a solution for any disjunction of formulas Gk1 Gk2 ... Gkr, \nwhere k1, k2, ..kr are not explored by Hj) \n \nLet Hj(1) be the extensiΓ³n of Hj obtained by adding a pre-process to check row k1 \nbeforehand. Then we can construct in a similar way another formula where Hj(1) fails \nfor some formula Gk2 Μ§where row k2 is not explored by Hj(1). And then we can extend \nHj(1) to Hj(2), wich in turn will fail for some formula Gk3. \n \nThis produces a list of algorithms Hj, Hj(1), Hj(2), ..., Hj(t) The list is finite, with a \nrestriction: the last index, t, must depend polynomially on n: t = q(n) for some \npolynomial q. \n \nBut the truth table has 2n rows an hence we always can take one excluded row and build \na formula being true only on that row, causing Hj(t) to fail, no matter what polynomial \nfunction q is used. \n \n \n \n \n \n \n \n \n \n \nIn conclusion, there is a set C(Hj) having a number of equivalence classes w, where w is \nan exponential function of n, and for any formula F in a class X of C(Hj), Hj does not \nevaluate the rows where F is true, and therefore misses every solution for F. \n \n \n4. Conversion algorithms. \n \nIt is quite simple to find the truth-value assignations for the atoms satisfiying a formula \nexpressed in disjunctive normal form (dnf). There are 3 cases: \n \n1. The formula is a tautology, and is equivalent to . It is true for any assignation of \ntruth values to its atoms. \n \n2. The formula is a contradiction, and is equivalent to . It is false for any assignation \nof truth values to its atoms. In each of its βdisjointβ the formula must have the \nconjunction of a complementary pair (if some disjoint does not have any such \nconjunction, this disjoint would have a truth value assignation satisfying the formula, \nwhich is impossible). \n \n3. The formula is a contingency. It has at least one βdisjointβ and it suffices one of \nthem to be true. \n \nIn any case, it can be immediately checked out if the formula is satisfacible or not. If a \nformula is not equivalent to a contradiction, it must be satisfacible. All we need is to \ncheck sequentially the disjoints for the absence of complementary pairs. \n \nThis leads to the following algorithm: \n \nTo find the truth values satisfying a formula F \n \n1. Convert F to its disjunctive normal form. \n2. Check the disjoints sequentially from left to right until a disjoint D having no \ncomplementary pairs is found. \n3. Assign 1 to each literal in D. This will make some atoms true and some false."},{"page":18,"text":"18\n \nApparently this is a polynomial complexity algorithm. The problem is: the conversion \nprocess (to get a formulaβs d.n.f.) is a search process in a tree, having exponential \ncomplexity. \n \nLema 4. \n \nFor every convertion algorithm there is an order n formula F requiring a time t = q(n) \nwhere q is an exponential type function. \n \nProof. \n \nA clause is a disjunction of literals, some of them positive and some negative ones. For \na given value of n, take a formula F in conjunctive normal form, having k clauses, each \nhaving m literals (obviously m β€ n). Converting F to d.n.f. we get mk disjoints, where \neach of them has k literals. \n \nIf k = n and m is close to n, the number of disjoints is in the order of nn. \n \n \nNote that evaluating the truth table requires to evaluate 2n rows, much less than nn, \nalthough both algorithms have exponential complexity. \n \n \n5. Asymptotic properties. \n \nFor a given order n, take a representative from each equivalence class F n / . We can \nidentify each class with its corresponding truth table. For any n-order formula, its truth \ntable has 2n rows. Since each row can have a resulting 0 or 1, there are \nn\n2\n2\ndifferent \ntruth tables, and therefore there are \nn\n2\n2\n equivalence classes. \n \nIn one half of all classes in F n / the formulas have true in the first row of its truth \ntable. In the other half, the formulas have false in the first row. \n \nThe formulas belonging to \n1\n2\n2\n \nn\n classes have 1 in its first row, while the formulas \nbelonging to the other \n1\n2\n2\n \nn\n classes have a 0 in its first row. \n \nIf we put a 0 in row 1, then every one of the remaining 2n-1 rows can have a 0 or 1, \ngiving a total of \n1\n2\n2\n \nn\n options. \n \nIn one fourth of all classes in F n / the formulas have 1 in row 2, knowing they have 0 \nin row 1. In a similar way, knowing rows 1, 2, ..., m-1 to have a 0, there are \nm\nn \n2\n2\n \nequivalence classes having a 1 in row m. \n \nThe number of classes whose formulas have its first 1 in row 1 or 2 or ... or m equals \nthe sum \n \nq(m) = \n1\n2\n2\n \nn\n + \n2\n2\n2\n \nn\n + \n3\n2\n2\n \nn\n+... + \nm\nn \n2\n2\n = \nn\n2\n2\n (2-1 + 2-2+ 2-3+... + 2-m)"},{"page":19,"text":"19\n= \n \n \n \n \n \n \n \n \n \n \n \n \n \nm\nn\n2\n1\n1\n22\n \n \nThe percentage of classes whose formulas are true for the first time in one of the rows 1, \n2, ..., m is then: \n \nm\nm\nm\nm\nr\n2\n1\n2\n100\n2\n1\n1\n100\n)\n(\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nLet be a s degree polynomial, for a fixed s. If m = (n), then r(m) is the percentage of \nclasses whose formulas can be solved in a polynomial time (meaning by βsolveβ to find \na n-tuple of truth values satisfying the formula). \n \nTake (x) = xs. Then m = ns and r(m) = r(ns) = \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \ns\nn\n2\n1\n1\n100\n \n \nWe get \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \ns\nn\nn\ns\nn\nn\nr\n2\n1\n1\n100\nlim\n)\n(\nlim\n= 100%. \n \nConclusion: \n \nWhen n increases, the ratio of equivalence classes having satisfactible formulas, which \nare detected by a sequential search algorithm exploring only the first (n) rows (or any \nset of pre-determined (n) rows) approaches 100%. \n \nNotes. \n \nThe ratio of order n satisfactible formulas not solved by a sequential search algorithm \nexploring (n) rows decreases to 0 as n increases. \n \nHowever, the number of equivalence classes increases very fast, as shown in the \nfollowing table: \n \nRows in table \nEquivalence classes ( = number of Truth tables) \nn u = 2n \n2u \n1\n 2 \n 4 \n2\n 4 \n 16 \n3\n 8 \n 256 \n4\n 16 \n 65.536 \n5\n 32 \n4294967296\n6\n 64 \n1,84467E+19\n7\n 128 \n3,40282E+38\n8\n 256 \n1,15792E+77\n9\n 512 \n1,3408E+154\n10\n 1.024 \n#Β‘NUM! \n11\n 2.048 \n#Β‘NUM!"},{"page":20,"text":"20\n12\n 4.096 \n#Β‘NUM! \n13\n 8.192 \n#Β‘NUM! \n14\n 16.384 \n#Β‘NUM! \n15\n 32.768 \n#Β‘NUM! \n16\n 65.536 \n#Β‘NUM! \n17\n 131.072 \n#Β‘NUM! \n18\n 262.144 \n#Β‘NUM! \n19\n 524.288 \n#Β‘NUM! \n20\n 1.048.576 \n#Β‘NUM! \n \nEven tough the ratio of equivalence classes (causing exponential complexity to an \nalgorithm) vs. total number of classes decreases, this latter number increases quite fast, \nas the following table shows: \n \nn \n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n1\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n4\n1\n3\n32\n4\n4096\n1\n5\n134217728\n128\n6\n2,8823E+17\n268435456\n7\n2,65846E+36\n6,04463E+23\n8\n4,52313E+74\n6,2771E+57\n9\n2,6187E+151\n5,5453E+129\n10\n1,7556E+305\n1,4181E+278\n16777216\n11\n#Β‘NUM! \n#Β‘NUM! \n6,8946E+215\n12\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n13\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n14\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n15\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n16\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n1\n17\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n18\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n19\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n20\n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM! \n#Β‘NUM!\n \nIn the blank entries the calculation formula is not applicable since 2n < ns. The #Β‘NUM! \nmeans an overflow; the numeric limits of Excel on a Pentium computer has been \nexceeded. \n \nIncreasing speed for the function r. \n \nAs the following table shows, r(ns) / 100 increases quite fast. \n \ns \n \n \n \n \n \n \n \n \n \nn \n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n1\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n0,5\n2\n0,75\n0,9375\n0,99609375\n0,999984741\n1\n1\n1\n1\n1\n1\n3\n0,875\n0,998046875\n0,999999993\n1\n1\n1\n1\n1\n1\n1\n4\n0,9375\n0,999984741\n1\n1\n1\n1\n1\n1\n1\n1\n5\n0,96875\n0,99999997\n1\n1\n1\n1\n1\n1\n1\n1\n6\n0,984375\n1\n1\n1\n1\n1\n1\n1\n1\n1"},{"page":21,"text":"21\n7\n0,9921875\n1\n1\n1\n1\n1\n1\n1\n1\n1\n8\n0,99609375\n1\n1\n1\n1\n1\n1\n1\n1\n1\n9\n0,998046875\n1\n1\n1\n1\n1\n1\n1\n1\n1\n10\n0,999023438\n1\n1\n1\n1\n1\n1\n1\n1\n1\n11\n0,999511719\n1\n1\n1\n1\n1\n1\n1\n1\n1\n12\n0,999755859\n1\n1\n1\n1\n1\n1\n1\n1\n1\n13\n0,99987793\n1\n1\n1\n1\n1\n1\n1\n1\n1\n14\n0,999938965\n1\n1\n1\n1\n1\n1\n1\n1\n1\n15\n0,999969482\n1\n1\n1\n1\n1\n1\n1\n1\n1\n16\n0,999984741\n1\n1\n1\n1\n1\n1\n1\n1\n1\n17\n0,999992371\n1\n1\n1\n1\n1\n1\n1\n1\n1\n18\n0,999996185\n1\n1\n1\n1\n1\n1\n1\n1\n1\n19\n0,999998093\n1\n1\n1\n1\n1\n1\n1\n1\n1\n20\n0,999999046\n1\n1\n1\n1\n1\n1\n1\n1\n1\n \nThe number 1 occurs in most places where the fraction subtracted from 1 is too small \nfor the precision handled by Excel. \n \nOther notes. \n \n1. In the preceeding anΓ‘lisis, we have not taken into account the process needed to find \nout if a formula is irreducible, since it would add processing time. A formula may be \nevaluated with all the its atoms, without trying to simplify it beforehand via the \nabsorption equivalences. \n \n2. If m < n then r(n) < 100, because the contradiction class is excluded (its table yields \nonly zeros). \n \n3. It has been shown that for every uninformed sequential search algorithm there is a set \nof formulas taking an exponential time. An βinverseβ problem can be stated as: for \nevery formula there is an uninformed sequential search algorithm giving all the \nformulaβs solutions in the first rows of the truth table, and zeros on the remaining rows. \nThe problem is to find the algorithm. This requires a search on the set of uninformed \nsequential search algorithms. This search has exponential complexity. \n \nIf a an n order formula F has m solutions (that is, there are m rows where F is true), then \nthere are m!(2n β m)! ways to permute the rows keeping the true rows first and the false \nrows last. The ratio of this number of algorithms to the total ways of exploring Fβs truth \ntable is m!(2n β m)! / (2n)! = \n \n \n \n \n \n \nm\nn\n2\n1\n, which makes it highly unlikely to find one of this \nalgorithms by chance. Using a systematic search, the vast majority of a huge number of \nalgorithms must be tested before a solution can be found. \n \nReferences: \n \n[1] Nilsson, Nils J., βArtificial Inteligence: a new synthesisβ, Morgan Kaufmann \nPublishers, San Francisco, CA, 1998 \n[2] βThe efficiency of algorithmsβ, Harry R. Lewis y Christos H. Papadimitriou, \nScientific American, January 1978."},{"page":22,"text":"22\n[3] Mendelson, Elliott, βIntroduction to Mathematical Logic,β D. Van Nostrand \nCompany, Princeton, NJ, 1965 \n \n[4] Russell, Stuart, Norvig, Peter, βArtificial Intelligence: a modern approachβ, Prentice \nHall, NJ, 2002. \n \n[5] βMillion Dollar Minesweeperβ, Ian Stewart, Scientific American, October 2.000 \n2007.11.07\n16:36:35-05'00'"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"G = F / = {[A] / A F }.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"For each A F , we define a quasi-norm || A || = number of different atoms in A. This","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"|| A || = 0 iff A is or .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"A property like || A || = | | || A || (for real) is not valid here. One way to define","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"βmultiplicationβ of a formula by a positive integer is 2 A = A op A, for some binary","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"connective op. But this produces || n A || = A for any natural n. Furthermore, the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"We will say that a formula A is irreducible of order k iff || A || = k and it is not","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"equivalent to another formula B which has less atoms than A. (i.e. if A B and || A || =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"|| A || = k and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"|| X || = 0 iff X is the class of all contradictions (which are equivalent to ) or","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"|| X op Y || || X || + || Y ||, where we define X op Y = { F1 op F2 } for F1 X","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"F n = { A / || A || = || [ A] || = n }.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"Then we have F =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"And the set of all irreducible formulas is F =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"f( [ A ] ) = B, where B A and A is irreducible.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"Let Gn = { [An ] / An is irreducible having order n }. Then G =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"Let In = F n / { [An ] / B [An ] iff B is order n irreducible and B An }","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"Then I =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"Hn = { Bn / Bn = f(X) for X In } where the Bn are not equivalent to each other, and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"H =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"associate to each of its letters a numerical variable xi: h(pi)= xi for 1 i n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"h(Β¬ A) = 1 β h(A)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"h(A B) = h(A) h(B)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"h(A B) = h(A) + h(B) β h(A) h(B)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"h(A B) = (1 β h(A) ) (1 β h(B) ) + h(B)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"h(A B) = (1 β h(A) ) (1 β h(B) ) ( 1 + h(A) + h(B) ) + h(A) h(B)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"h(A)n = 0 iff h(A) = 0 and h(A)n = 1 iff h(A) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"h(A A) = h(A) h(A) = h(A)2 h(A),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"h(A A) = h(A) + h(A) β h(A)h(A) = 2h(A) β h(A)2 2h(A) β h(A) = h(A).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"h(A A) = (1 β h(A))2 + h(A) (1 β h(A)) + h(A) = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"a) (1 β h(A))2 (1 β h(A)), because both sides have the same value when h(A) = 0 or","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"when h(A) = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"b) (1 β h(A))2 + h(A) = 1 β 2h(A) + h(A)2 + h(A) = 1 β h(A) + h(A)2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"1 β h(A) + h(A) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"h(A Β¬A) = h(A) (1 β h(A)) = h(A) β h(A)2 h(A) β h(A) = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"function g, where g(A) = h(A) β 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"the formulaβs βcharacteristic polynomialβ, and has the form g(A) = q, where","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"and the variables xi have a value of 0 or 1, and r(n) = 2n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"We define 00 = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"that q(x1, x2, ..., xn) = 0. Or equivalently:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"With the restrictions xi in {0, 1} for i = 1, ..., n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"The equation g(A) = 0 has n variables. It can be seen as a βsystemβ of 1 equation with n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"determine them, it suffices to solve g(A) = 0 for x1, which will then depend on the other","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"h(A) = x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"g(A) = x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"and the equation g(A) β 1 = 0 is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"With the restrictions xi {0, 1} for i = 1,.., 3.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"Note that if (x2, x3) = (1, 1) then x1 vanishes and the equation becomes 1 = 0, which","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"where the denominator is zero if (x2, x3) = (1, 1). It is zero for other values of these","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"If we put arbitrarily x1 = 0 in (4), then 2x2x3 β x2 β x3 + 1 = 0, and from here we get x2 =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"(x3 β 1) / (2x3 β 1). Here we have two cases: if si x2 = 0, then x3 = 1 and if x2 = 1, then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"x3 = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"If we put x1 = 0 in (4), then 2x2x3 β x2 β x3 + 1 = 3x2x3 β 2(x2 + x3) + 1, so x2 = x3 / (x3 β","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"1). Here again we have two cases: if x2 = 0, then x3 = 0 and if x2 = 1, then x3 = x3 β 1,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"for n = 2, and the variables x1, ...xn-1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"is a product of k variables, and there are C(n, k) = n! / (k! (n-k)!) combinations of terms","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"coefficients, and it cannot be calculated in a time t = q(n), where q is a polynomial.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"(those being 0 or 1) of g. The exponent must be 0 or odd to avoid de case (-1)2 = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"On the other hand, if we introduce, arbitrarily, any condition g1(x1, ..., xn) = 0, then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"For example, if we put arbitrarily the condition x1 = 0, then we will miss all the roots","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"where x1 = 1, in the cases where such roots exists. In the cases where g has a factor (x1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"If we put a condition like x1 β x2 = 0, then we might miss the roots where x1 β x2, and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"x1 + x2 + x3 β 2x1x2 β 2x1x3 β 2x2x3 + 3x1x2x3 β 1 = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"If we add the condition x1 β x2 = 0, (which can be considere a mere supposition) the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"x3 - x1x3 β 1 = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"h(A) = (x1 + x2 + x3 - x1x2 β x1x3 β x2x3 + x1x2x3 )(1 - x1x2) (1 - x1x3) (1 β x2x3)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"h(A) = 0 if al least one of its factors is zero.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"x1 = 1, x2 = 1, any value for x3","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"x1 = 1, x3 = 1, any value for x2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"x2 = 1, x3 = 1, any value for x1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"x1 = (x2 + x3 - x2x3) / (1- x2 - x3 + x2x3) = U / ( 1 β U), for U β 0,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"Where U = x2 + x3 - x2x3. All the options are:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"Since g = h β 1, for any other triplet (x1, x2, x3), h(x1, x2, x3) is not zero, and hence it","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"must be h(x1, x2, x3) = 1. Those other three triplets must be the binary roots of g. They","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"number of operations, for n = 3.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"|| F || = n) there are n! 2n different ways to evaluate F, depending on the order the atoms","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"We will represent the truth table for F, as evaluated by A, by A(F) = (b(1), b(2), ...,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"value of 1 when it evaluates the formula F. This is L(A(F)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"{ k / b(K) = 1 }.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"Take F n, the set of order n irreducible formulas. Take a formula F in F n. Then || F || =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"solution only in the last row (that is, in row 2n ). Thus, L(A1,1(F)) = 2n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"More precisely, if the base 2 representation of j is j = d1d2...dn, then the atom pk takes","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"For every j, if the A1,j algorithm evaluates first true the atom pk (that is, if dk = 1), then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"all the other rows in the truth table. In this case, we also have Thus, L(A1, j(F)) = 2n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"L(Ai, j(F)) = 2n, for 1 β€ i β€ n!","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"This is not the case, as it will be shown that Cn(Ai,j) = o(g(n)), where g is an","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq88","equation_number":null,"raw_text":"To this end, we take the base 2 representation of the index j: j = d1d2...dn. Suppose the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq89","equation_number":null,"raw_text":"Take the base 2 representation of m: m = e1e2...en,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq90","equation_number":null,"raw_text":"If ek = dk, then qik(m) is pik, otherwise, qik(m) is Β¬ pik.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq91","equation_number":null,"raw_text":"To find out if a formula F is true from row k, for k = 2n, 2n-1, 2n-2, ..., 2n-1 + 1, we have","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq92","equation_number":null,"raw_text":"2n β (2n-1 + 1) + 1 = 2n-1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq93","equation_number":null,"raw_text":"h = 1, ..., k-1. For the formulas passing this check, there are the worst-possible cases","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq94","equation_number":null,"raw_text":"explore k + (2n - k) = 2n rows.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq95","equation_number":null,"raw_text":"There are additional sources of complexity: each equivalence class in G = F /","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq96","equation_number":null,"raw_text":"restriction: the last index, t, must depend polynomially on n: t = q(n) for some","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq97","equation_number":null,"raw_text":"For every convertion algorithm there is an order n formula F requiring a time t = q(n)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq98","equation_number":null,"raw_text":"If k = n and m is close to n, the number of disjoints is in the order of nn.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq99","equation_number":null,"raw_text":"q(m) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq100","equation_number":null,"raw_text":"Let be a s degree polynomial, for a fixed s. If m = (n), then r(m) is the percentage of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq101","equation_number":null,"raw_text":"Take (x) = xs. Then m = ns and r(m) = r(ns) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq102","equation_number":null,"raw_text":"Equivalence classes ( = number of Truth tables)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq103","equation_number":null,"raw_text":"n u = 2n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq104","equation_number":null,"raw_text":"table is m!(2n β m)! / (2n)! =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":44007,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}} |