Indistinguishable binomial decision tree of 3-SAT: Proof of class P is a proper subset of class NP

This paper solves a long standing open problem of whether NP-complete problems could be solved in polynomial time on a deterministic Turing machine by showing that the indistinguishable binomial decision tree can be formed in a 3-SAT instance. This paper describes how to construct the decision tree and explains why 3-SAT has no polynomial-time algorithm when the decision tree is formed in the 3-SAT instance. The indistinguishable binomial decision tree consists of polynomial numbers of nodes containing an indistinguishable variable pair but generates exponentially many paths connecting the clauses to be used for sequences of resolution steps. The number of paths starting from the root node and arriving at a child node forms a binomial coefficient. In addition, each path has an indistinguishable property from one another. Due to the exponential number of paths and their indistinguishability, if an indistinguishable binomial decision tree is constructed in which there exist one or more paths generating an empty clause, the number of calculation steps needed to extract the empty clause is not polynomially bounded. This result leads to the conclusion that class P is a proper subset of class NP. One Sentence Summary: This paper solves the P versus NP problem by showing that the indistinguishable binomial decision tree can be formed in a 3-SAT instance. INTRODUCTION The Boolean satisfiability (SAT) problem is to determine whether there exists a feasible set to satisfy a given Boolean formula. SAT is the first known example of a NP-complete problem 2 (Cook–Levin theorem) and thousands of NP-compete problems have been identified by reducing the SAT to the problems. The SAT problem is divided by tractable SAT such as 2-SAT and Horn-SAT and intractable SAT such as 3-SAT 3 . In the above tractable instances, 2-SAT is NLcomplete 4 and Horn-SAT is P-complete 5 . In addition, 3-SAT is NP-complete 2 . Hence, SAT is a good research object to search for the relationship of the classes, NL, P, and NP. It is known that NL⊆P⊆NP, but unknown whether NL=P and whether P=NP. These two questions have been open problems for several decades. Most complexity theorists expect that NL⊊ P ⊊ NP. Especially, the P versus NP problem 6 , as one of the famous unsolved problems in mathematics and computer science, is to clarify the relationship for the inclusion of the classes P and NP 7 . The obvious way to prove P = NP is to show that one or more NP-complete problems have a polynomial-time algorithm. Researchers have found thousands of NP-complete problems since Karp's research 8,9 . However, although there are so many NP-complete problems, researchers have failed to find a polynomial-time algorithm for any of the problems. Hence, with the belief that P ≠ NP, various proof techniques have been studied to distinguish between P and NP. However, all known proof techniques such as relativizing 10 , natural 11~16 , and algebrizing 17~20 proofs were insufficient to prove that P ≠ NP. Resolving the question whether 3-SAT has a polynomial-time algorithm is equivalent to the P versus NP problem because 3-SAT is NPcomplete. This paper solves the P versus NP problem by showing that 3-SAT has no polynomialtime algorithm. The SAT problem is organized by clauses. The simplest clause is the unit clause containing only one variable. This paper shows that the simplest clause can be converted to a logically equivalent, highly complicated clause group, which is termed as a binomial decision tree. The decision tree has polynomial number of nodes but generates exponentially many paths arriving at the polynomial number of nodes. One of the distinctive features of the indistinguishable binomial decision tree is the indistinguishability of paths in that the paths cannot be divided into groups according to the arrival node. As the paths cannot be grouped, every algorithm must search for all exponential number of paths in order to verify whether there exists a node containing a specific variable. The P versus NP problem is explained as to whether every problem whose solution can be quickly (in polynomial time) verified, can also be solved quickly. We will show that every algorithm must investigate all nodes to decide the satisfiability of an instance. An instance containing the binomial decision tree is quickly verifiable because the number of nodes is polynomially bounded. However, to solve the problem, that is, to investigate all nodes cannot be executed quickly because exponentially many paths must be investigated to decide the satisfiability. Therefore, the indistinguishable binomial decision tree clearly explains the relationship of verifiability and solvability of NP-complete problems. This paper describes how to construct the indistinguishable binomial decision tree. RESULTS Decision chain and generalized unit clause In order to verify the satisfiability of an instance represented by conjunctive normal form (CNF), we investigate whether we can generate an empty clause from some set of clauses. In a logical approach, the only way to generate a logically equivalent new clause is to apply the resolution rule. For instance, there is a resolution step to apply the resolution rule: , (1) a c b c a b    The dividing line stands for entails, which means that (a˅c)˄(b˅¬c) is logically equivalent to (a˅c)˄(b˅¬c)˄(a˅b). The input clauses must have a variable and its complement, which is called as resolved variable. We use the concept of the decision chain 1 to represent a sequence of resolution steps. The decision chain is a linked list of the resolved variables while sequentially applying the resolution rule. Figure 1 shows an example of a sequence of resolution steps and its representation by a decision chain. Original clauses Newly generated clause A ˅ B ˅ C A ˅ ¬D, B ˅ D ˅ ¬E A ˅ B ˅ ¬E A ˅ B ˅ ¬E, C ˅ E A ˅ B ˅ C A ˅ ¬D B ˅ D ˅ ¬E C ˅ E ¬D