Note for the P Versus NP Problem (II)

1. Introduction

The P versus $NP$ problem is a fundamental question in computer science that asks whether problems whose solutions can be easily checked can also be easily solved [1]. “Easily” here means solvable in polynomial time, where the computation time grows proportionally to the input size [1,2]. Problems solvable in polynomial time belong to the class P, while $NP$ includes problems whose solutions can be verified efficiently given a suitable “certificate” [2].

The central question is whether P and $NP$ are the same. Most researchers believe that P is a strict subset of $NP$, meaning that some problems are inherently harder to solve than to verify. Resolving this problem has profound implications for fields like cryptography and artificial intelligence [3,4]. The P versus $NP$ problem is widely considered one of the most challenging open questions in computer science. Techniques like relativization and natural proofs have yielded inconclusive results, suggesting the problem’s difficulty [5,6]. Similar problems, such as the $VP$ versus $VNP$ problem in algebraic complexity, remain unsolved [7].

Resolving the P versus $NP$ problem is often described as a “holy grail” of computer science. A positive resolution could revolutionize our understanding of computation and potentially lead to groundbreaking algorithms for critical problems. The problem is listed among the Millennium Prize Problems. While recent years have seen progress in related areas, such as finding efficient solutions to specific instances of $NP$-complete problems, the core question of P versus $NP$ remains unanswered [8]. A polynomial-time algorithm for any $NP$-complete problem would directly imply P equals $NP$ [9]. Our work focuses on presenting such an algorithm for a well-known $NP$-complete problem.

2. Background and Ancillary Results

$NP$-complete problems are the Everest of computational challenges. Despite the ease of verifying proposed solutions with a succinct certificate, finding these solutions efficiently remains an elusive goal. A problem is classified as $NP$-complete if it satisfies two stringent criteria within computational complexity theory:

• Efficient Verifiability: Solutions can be quickly checked using a concise proof [9].
• Universal Hardness: Every problem in the class $NP$ can be reduced to this problem without significant computational overhead [9].

The implications of finding an efficient algorithm for a single $NP$-complete problem are profound. Such a breakthrough would serve as a master key, unlocking efficient solutions for all problems in NP, with transformative consequences for fields like cryptography, artificial intelligence, and planning [3,4].

Illustrative examples of $NP$-complete problems include:

• Boolean Satisfiability Problem (SAT): Given a logical expression, determine if there exists an assignment of truth values to its variables that makes the entire expression true [10].
• Exact k-Coloring Problem: Given a graph G and a positive integer k, determine if there exists a valid coloring of G such that exactly k vertices have the same color and no adjacent vertices have the same color. This problem is equivalent to finding an independent set of size k, an $NP$-complete problem [11].
• Partition into Triangles (PT) Problem: Given a graph $G=(V,E)$, where V is the set of vertices and E is the set of edges, the $PT$ problem asks whether the vertices of G can be partitioned into disjoint sets $V_1,V_2,\dots ,V_q$, each containing exactly 3 vertices, such that each $V_i$ induces a triangle in G [11]. In other words, we want to know if we can divide the graph into q disjoint triangles.

The provided examples represent a small subset of the extensively studied $NP$-complete problems relevant to our current work. A bipartite graph, denoted as $(U,V,E)$, is an undirected graph characterized by the existence of two node sets $U,V$ and edges in E that only connect nodes from opposite sets. An independent set $V^{\prime }$ is a subset of vertices in a graph G where no two vertices in the set are connected by an edge. In addition, a vertex cover (sometimes called a node cover) of a graph G is a subset of its vertices, denoted by $V^{\prime }$, such that every edge in G has at least one endpoint in $V^{\prime }$.

Definition 1. 

Exact Independent Vertex Cover (EIVC) Problem

INSTANCE: An undirected graph $G=(V,E)$ and a positive integer k.

QUESTION: Is there set $V^{\prime }$ of exactly k vertices such that $V^{\prime }$ is both a vertex cover and an independent set in G?

REMARKS: Solving the $EIVC$ problem is akin to determining a 2-coloring of a bipartite graph such that one of the partitions contains exactly k vertices. This task can be accomplished in polynomial time, given the efficient solvability of the 2-coloring problem in bipartite graphs.

Formally, an instance of Boolean Satisfiability Problem (SAT) is a Boolean formula $\varphi$ which is composed of:

• Boolean variables: $x_1,x_2,\dots ,x_n$;
• Boolean connectives: Any Boolean function with one or two inputs and one output, such as ∧(AND), ∨(OR), ¬(NOT), ⇒(implication), ⇔(if and only if);
• and parentheses.

A truth assignment for a Boolean formula $\varphi$ is a set of values for the variables in $\varphi$. A satisfying truth assignment is a truth assignment that causes $\varphi$ to be evaluated as true. A Boolean formula with a satisfying truth assignment is satisfiable. The problem $SAT$ asks whether a given Boolean formula is satisfiable [11].

We define a $CNF$ Boolean formula using the following terms: A literal in a Boolean formula is an occurrence of a variable or its negation [9]. A Boolean formula is in conjunctive normal form, or $CNF$, if it is expressed as an AND of clauses, each of which is the OR of one or more literals [9]. A Boolean formula is in 2-conjunctive normal form or $2CNF$, if each clause has exactly two distinct literals [9].

For example, the Boolean formula:

$$(x_1\vee \neg x_2)\wedge (x_3\vee x_2)\wedge (\neg x_1\vee \neg x_3)$$

is in $2CNF$. The first of its three clauses is $(x_1\vee \neg x_2)$, which contains the two literals $x_1$ and $\neg x_2$. Finally, we introduce the last problem:

Definition 2. 

Exact Monotone Xor 2-Satisfiability (EMX2SAT) Problem

INSTANCE: An n-variable $2CNF$ formula with monotone clauses (meaning the variables are never negated) using logic operators ⊕ (instead of using the operator ∨) and a positive integer k.

QUESTION: Is there exists a satisfying truth assignment in which exactly k of the variables are true?

By presenting the $NP$-completeness and an efficient solution to Partition into Triangles, we would establish a proof that P equals $NP$.

3. Main Result

This is a main insight.

Theorem 1. 

The problem $PT$ can be reduced to $EMX2SAT$ in polynomial time [12].

Proof. 

We can build an equivalent $EMX2SAT$ instance for any given instance $G=(V,E)$ of the $PT$ problem:

• Variables:
  • Create a variable for each vertex v in the original graph G. Denote this variable as v itself.
  • For each pair of disconnected vertices u and v in G (i.e., $(u,v)\notin E$), introduce a new variable denoted by $x_{uv}$.
• Clauses:
  • For every two disconnected vertices $u,v\in V$ (i.e., $(u,v)\notin E$), construct two clauses using the new variables:

    –

    $(u\oplus x_{uv})$: This enforces that either vertex u is true or the new variable $x_{uv}$ is true (XOR).

    –

    $(x_{uv}\oplus v)$: This enforces that either the new variable $x_{uv}$ is true or vertex v is true (XOR).

Key Points about the Construction:

• The two clauses for each pair of disconnected vertices u and v in G ensures that both variables u and v have the same truth value.
• Moreover, the two clauses together imply that $x_{uv}$ is true exactly when both u and v are false. This means that for any pair of vertices u and v that are not directly connected in the graph, at least one of the three variables u, v, or $x_{uv}$ must be assigned the value true.

Mapping between PT solutions and EMX2SAT assignments:

• A satisfying truth assignment in the $EMX2SAT$ formula corresponds to a valid partition into triangles in the original graph G if:

  –

  Variables assigned false represent the vertices in the graph G.

  –

  If the new variable $x_{uv}$ is assigned the value true, it means that the disconnected vertices u and v must belong to different triangles in the partition.

Why this construction works:

• The clauses guarantee that, in a satisfying truth assignment, a vertex and the corresponding new variables must have compatible truth values.
• If a graph can be partitioned into q triangles, then there exists a corresponding truth assignment where the original variables (representing vertices) are false and exactly one of the new variables $x_{uv}$ is true for each pair of disconnected vertices u and v.

Equivalence and Complexity:

• A q-triangle partition of G can be mapped to a satisfying truth assignment of the $EMX2SAT$ formula with exactly $k=9·{\displaystyle \frac{q·(q-1)}{2}}$ true variables, and vice versa.
• If we partition the graph into q disjoint sets, each vertex in a set $V_i$ will be disconnected from $3·(q-i)$ vertices that belong to the sets $V_{i+1},V_{i+2},\dots ,V_q$. By summing the number of disconnected pairs for all vertices in all sets, we arrive at a total of $9·{\displaystyle \frac{q·(q-1)}{2}}$ disconnected pairs due to $9·(q-1)+9·(q-2)+\dots +9·\left(1\right)=9·{\displaystyle \frac{q·(q-1)}{2}}$. For any pair of disconnected vertices u and v, the variable $x_{uv}$ is assigned the value true if and only if both u and v are assigned the value false. As a result, there will be a total of $9·{\displaystyle \frac{q·(q-1)}{2}}$ true variables in the satisfying truth assignment, since all the original variables corresponding to vertices are set to false. It’s important to note that a satisfying truth assignment with exactly $9·{\displaystyle \frac{q·(q-1)}{2}}$ true variables directly corresponds to a partition of the graph G into q triangles.
• Given that $PT$ is an $NP$-complete problem and we have demonstrated a polynomial-time reduction from $PT$ to $EMX2SAT$, we can conclude that $EMX2SAT$ is also $NP$-complete.

In essence, the proof demonstrates that solving $EMX2SAT$ is equivalent to finding a closure of exactly k vertices in the $PT$ problem. This implies that $EMX2SAT$ inherits the $NP$-completeness property from $PT$. □

These are the main theorems.

Theorem 2. 

$EIVC\in P$.

Proof. 

Given the efficient solvability of the 2-coloring problem in bipartite graphs, we claim that the $EIVC$ problem can be accomplished within polynomial time. This is a straight-forward dynamic programming algorithm similar to solve subset sum: Let $(A_1,B_1),(A_2,B_2)\dots ,(A_p,B_p)$ be the sides of partitions $A_i$ and $B_i$ in a connected component i of the bipartite graph $(U,V,E)$, such that every vertex in a single partition has the same color.

Now, we create a dynamic programming table $DP[i,t]$ that stores whether it is possible to have a bipartite graph with exactly t vertices on one color using the i first components. The bi-dimensional array $DP$ of dimensions p by k is initialized such that all elements are $\mathit{false}$, with the exception of $DP[0,0]$ which is set to $\mathit{true}$. Using the recurrence

$$DP[i,t]=DP[i-1,t-\mid A_i\mid ]\vee DP[i-1,t-\mid B_i\mid ],$$

we correctly decide whether there exists an entire partitioning of exactly k vertices with the same color after by examining $DP[p,k]$. This is a polynomial time algorithm since the running time is bounded by $O(\mid U\mid +\mid V\mid +\mid E\mid +p·k)$. Identifying 2-color partitions takes $O\left(\right|U|+|V|+|E\left|\right)$ time using breadth-first search algorithm (BFS), while finding k vertices of the same color requires $O(p·k)$ iterations. □

Theorem 3. 

$EMX2SAT\in P$ [12].

Proof. 

There is a connection between finding a satisfying truth assignment in $EMX2SAT$ with exactly k true variables and finding a set of exactly k vertices that is both a vertex cover and an independent set in a specific graph construction.

Here’s a breakdown of the equivalence:

• Graph Construction:
  • Each vertex in the new graph represents a variable in the $EMX2SAT$ formula.
  • Edges are created between variables based on the structure of the $2CNF$ clauses: If two variables appear in a clause (e.g., $(x\oplus y)$), then an edge is drawn between the corresponding vertices in the graph.
• $EMX2SAT$ and the Graph:
  • A truth assignment in $EMX2SAT$ where exactly k variables are true directly translates to a set of exactly k vertices in the constructed graph where true variables correspond to the vertices included in the set.
  • The properties of $EMX2SAT$ clauses ensure that:

    –

    Vertex Cover: The chosen vertices cover all the edges (due to the structure of the clauses and the way edges are formed). This satisfies the vertex cover condition.

    –

    Independent Set: The chosen vertices don’t have any edges connecting them (because the variables are connected in the graph, and only one variable from each clause can be true). This satisfies the independent set condition.

Therefore, finding a satisfying truth assignment with exactly k true variables in $EMX2SAT$ is indeed equivalent to finding a set of exactly k vertices that fulfills both vertex cover and independent set requirements in the corresponding graph. However, we know the problem of finding a set of exactly k vertices that is both a vertex cover and an independent set can be easily solved in polynomial time. Consequently, the instances of the problem $EMX2SAT$ can be solved in polynomial time as well. □

This is a key finding.

Theorem 4. 

$P=NP$.

Proof. 

A polynomial-time solution to any $NP$-complete problem would establish the equivalence of P and $NP$ [9]. Given that Partition into Triangles is an $NP$-complete problem, a polynomial-time solution for it, as presented here, would directly imply P equals $NP$. □

4. Conclusions

A definitive proof that P equals $NP$ would fundamentally reshape our computational landscape. The implications of such a discovery are profound and far-reaching:

• Algorithmic Revolution.

  –

  The most immediate impact would be a dramatic acceleration of problem-solving capabilities. Complex challenges currently deemed intractable, such as protein folding, logistics optimization, and certain cryptographic problems, could become efficiently solvable [3]. This breakthrough would revolutionize fields from medicine to cybersecurity. Moreover, everyday optimization tasks, from scheduling to financial modeling, would benefit from exponentially faster algorithms, leading to improved efficiency and decision-making across industries [3].

• Scientific Advancements.

  –

  Scientific research would undergo a paradigm shift. Complex simulations in fields like physics, chemistry, and biology could be executed at unprecedented speeds, accelerating discoveries in materials science, drug development, and climate modeling [3]. The ability to efficiently analyze massive datasets would provide unparalleled insights in social sciences, economics, and healthcare, unlocking hidden patterns and correlations [3].

• Technological Transformation.

  –

  Artificial intelligence would be profoundly impacted. The development of more powerful AI algorithms would be significantly accelerated, leading to breakthroughs in machine learning, natural language processing, and robotics [8]. While the cryptographic landscape would face challenges, it would also present opportunities to develop new, provably secure encryption methods [8].

• Economic and Societal Benefits.

  –

  The broader economic and societal implications are equally significant. A surge in innovation across various sectors would be fueled by the ability to efficiently solve complex problems. Resource optimization, from energy to transportation, would become more feasible, contributing to a sustainable future [3].

In conclusion, a proof of $P=NP$ would usher in a new era of computational power with transformative effects on science, technology, and society. While challenges and uncertainties exist, the potential benefits are immense, making this a compelling area of continued research.