Dense Complete Set For NP

A sparse language is a formal language such that the number of strings of length n is bounded by a polynomial function of n . We create a class with the opposite definition, that is a class of languages that are dense instead of sparse. We define a dense language on m as a formal language (a set of binary strings) where there exists a positive integer n 0 such that the counting of the number of strings of length n ≥ n 0 in the language is greater than or equal to 2 n − m where m is a real number and 0 < m ≤ 1. We call the complexity class of all dense languages on m as DENSE ( m ). We prove that there exists an NP–complete problem that belongs to DENSE ( m ) for every possible value of 0 < m ≤ 1.


Summary
In computational complexity theory, a sparse language is a formal language (a set of strings) such that the complexity function, counting the number of strings of length n in the language, is bounded by a polynomial function of n. The complexity class of all sparse languages is called SP ARSE. SP ARSE contains T ALLY , the class of unary languages, since these have at most one string of any one length. Fortune showed in 1979 that if any sparse language is coNP-complete, then P = N P (this is Fortune's theorem) [5]. Mahaney used this to show in 1982 that if any sparse language is NP-complete, then P = N P [6]. A simpler proof of this based on left-sets was given by Ogihara and Watanabe in 1991 [7]. Mahaney's argument does not actually require the sparse language to be in N P , so there is a sparse NP-hard set if and only if P = N P [6].
We create a class with the opposite definition, that is a class of languages that are dense instead of sparse. We show there is a sequence of languages that are in NP-complete, but their density grows as much as we go forward into the iteration of the sequence. The first element of the sequence is a variation of the NP-complete problem known as HAM-CYCLE [8]. The next element in the sequence is constructed from this new version of HAM-CYCLE. Indeed, each language is created from its previous one in the sequence. Since the density grows according we move forward into the sequence, then there exists a language so much dense such that its density tends to 0 when the bit-length n of the binary strings tends to infinity. However, this incredible dense language is still NP-complete.

Basic Definitions
Let Σ be a finite alphabet with at least two elements, and let Σ * be the set of finite strings over Σ The language accepted by a Turing machine M , denoted L(M ), has an associated alphabet Σ and is defined by We denote by t M (w) the number of steps in the computation of M on input w [1]. For n ∈ N we denote by T M (n) the worst case run time of M ; that is where Σ n is the set of all strings over Σ of length n [1]. We say that M runs in polynomial time if there is a constant k such that for all n, T M (n) ≤ n k + k [1]. In other words, this means the language L(M ) can be accepted by the Turing machine M in polynomial time. Therefore, P is the complexity class of languages that can be accepted in polynomial time by deterministic Turing machines [4]. A verifier for a language L is a deterministic Turing machine M , where L = {w : M (w, c) = "yes" for some string c}.
We measure the time of a verifier only in terms of the length of w, so a polynomial time verifier runs in polynomial time in the length of w [1]. A verifier uses additional information, represented by the symbol c, to verify that a string w is a member of L. This information is called certificate. N P is also the complexity class of languages defined by polynomial time verifiers [8]. If N P is the class of problems that have succinct certificates, then the complexity class coN P must contain those problems that have succinct disqualifications [8]. That is, a "no" instance of a problem in coN P possesses a short proof of its being a "no" instance [8].
A function f : Σ * → Σ * is a polynomial time computable function if some deterministic Turing machine M , on every input w, halts in polynomial time with just f (w) on its tape [9]. Let {0, 1} * be the infinite set of binary strings, we say that a language L 1 ⊆ {0, 1} * is polynomial time reducible to a language L 2 ⊆ {0, 1} * , written A simple graph is an undirected graph without multiple edges or loops [4]. An instance of the language HAM-CYCLE is a simple graph G = (V, E) where V is the set of vertices and E is the set of edges, each edge being an unordered pair of vertices [4]. We say (u, v) ∈ E is an edge in a simple graph G = (V, E) where u and v are vertices. For a simple graph 2, ..., k [4]. A Hamiltonian cycle is a simple cycle of the simple graph which contains all the vertices of the graph. A simple graph that contains a hamiltonian cycle is said to be hamiltonian; otherwise, it is nonhamiltonian [4]. The problem HAM-CYCLE asks whether a simple graph is hamiltonian [4]. In this work, we are going to represent the simple graphs with an adjacency-matrix [4]. For the adjacency-matrix representation of a simple graph G = (V, E), we assume that the vertices are numbered 1, 2, . . . , |V | in some arbitrary manner. The adjacency-matrix representation of a simple graph G consists of a |V | × |V | matrix A = (a i,j ) such that a i,j = 1 when (i, j) ∈ E and a i,j = 0 otherwise [4]. In this way, every simple graph of k vertices could be represented by a binary string of k 2 bits.
Observe the symmetry along the main diagonal of the adjacency matrix in this kind of graph that is called simple. We define the transpose of a matrix A = (a i,j ) to be the matrix A T = (a T i,j ) given by a T i,j = a j,i . Hence the adjacency matrix A of a simple graph is its own transpose A = A T .

▶ Definition 3. The language NON-SIMPLE contains all the graph that are represented by an adjacency-matrix A such that
Proof. Given a binary string x, we can check whether x is an adjacency-matrix which is not equal to its own transpose in time O(|x| 2 ) just iterating each bit a i,j in x and checking whether a i,j ̸ = a j,i or a i,j = 1 when i = j where | . . . | represents the bit-length function [4]. Proof. Given a binary string z such that z = xy and the bit-length of x is equal to (⌊ |z|⌋) 2 , we can decide in polynomial time whether x / ∈ NON-SIMPLE just verifying when x = x T and a i,i = 0 for all vertex i. In this way, we can reduce in polynomial time a simple graph G = (V, E) of k vertices encoded as the binary string x such that when x has k 2 bits and x / ∈ NON-SIMPLE then

x ∈ HAM-CYCLE if and only if xy ∈ HAM-CYCLE'
where y could be the empty string. In this way, we can reduce in polynomial time each element of HAM-CYCLE to some element of HAM-CYCLE'. Therefore, HAM-CYCLE' is in NP-hard. Moreover, we can check in polynomial time over a binary string z such that z = xy and the bit-length of x is equal to (⌊ |z|⌋) 2 whether x ∈ HAM-CYCLE or x ∈ NON-SIMPLE since HAM-CYCLE ∈ N P and NON-SIMPLE ∈ N P because of P ⊆ N P Choosing a graph on n vertices at random is the same as including each edge in the graph with probability 1 2 , independently of the other edges [2]. You get a more general model of random graphs if you choose each edge with probability p [2]. This model is known as G n,p [2]. It turns out that for any constant p > 0, the probability that G n,p contains a Hamiltonian cycle tends to 1 when n tends to infinity [2]. In fact, this is true whenever p > c×log n n for some constant c. In particular this is true for p = 1 2 , which is our case [2]. For all the binary strings z such that z = xy and the bit-length of x is equal to (⌊ |z|⌋) 2 , the amount of elements of size |z| in HAM-CYCLE' is equal to the number of binary strings . Since the number of Hamiltonian graphs increases as much as we go further on n, it does not seem surprising either that once n gets large most binary strings belong to HAM-CYCLE'. Moreover, the amount of binary strings which have some bit-length k 2 and belongs to NON-SIMPLE is considerably superior to the amount of strings with the same bit-length which are valid simple graphs. Actually, we can affirm for a sufficiently large positive integer n ′ 0 , all the binary strings of length n ≥ n ′ 0 which belong to HAM-CYCLE' are indeed more than or equal to 2 n−1 elements. In this way, we show that HAM-CYCLE' ∈ DEN SE(1). ◀ ▶ Definition 8. We will define a sequence of languages HAM-CYCLE' k for every possible integer 1 ≤ k. We state HAM-CYCLE' 1 as the language HAM-CYCLE'. Recursively, from a language HAM-CYCLE' k , we define HAM-CYCLE' k+1 as follows: A binary string xy complies with xy ∈ HAM-CYCLE' k+1 if and only if x and y are binary strings, x ∈ HAM-CYCLE' k or y ∈ HAM-CYCLE' k such that |x| = ⌊ |xy| 2 ⌋ where | . . . | represents the bit-length function and ⌊. . .⌋ is the floor function.

F. Vega
not belong to any HAM-CYCLE' k ′ , because its substrings of a quadratic length are also adjacency-matrix of only zeros. Suppose, we have an instance y of HAM-CYCLE' k ′ . We can reduce y in HAM-CYCLE' k ′ to zy in HAM-CYCLE' k ′ +1 such that where the binary string z is exactly a sequence of ⌊ |zy| 2 ⌋ zeros. We can do this since we already know z / ∈ HAM-CYCLE' k ′ . Certainly, if the membership zy ∈ HAM-CYCLE' k ′ +1 is true, z / ∈ HAM-CYCLE' k ′ and |z| = ⌊ |zy| 2 ⌋, then y ∈ HAM-CYCLE' k ′ also holds according to the Definition 8. Since this reduction remains in polynomial time for every positive integer 1 ≤ k ′ , then we show that HAM-CYCLE' k ′ +1 is in NP-hard. Moreover, HAM-CYCLE' k ′ +1 is also in NP-complete, because of the Lemma 9. ◀ ▶ Theorem 11. For every integer 1 ≤ k, if the language HAM-CYCLE' k is in DEN SE(k ′ ) for every instance of bit-length n ′ ≥ n 0 , then HAM-CYCLE' k+1 is in DEN SE( k ′ 2 ) for every instance of bit-length n ′ ≥ 2 × n 0 .
Proof. If the language HAM-CYCLE' k is in DEN SE(k ′ ) for every instance of bit-length n ′ ≥ n 0 , then for every integer n ≥ n 0 the amount of elements of size n+i in HAM-CYCLE' k+1 (where i ≥ n 0 and i = ⌊ n+i 2 ⌋) is greater than or equal to This is because there must be more than or equal to 2 i−k ′ elements of size i in HAM-CYCLE' k which are prefixes of the binary strings of size n + i in the language HAM-CYCLE' k+1 . We multiply that amount by 2 n since this is the number of different combinations of suffixes with length n in the binary strings of size n + i. Moreover, there must be more than or equal to 2 n−k ′ elements of size n in HAM-CYCLE' k which are suffixes of the binary strings of size n + i in HAM-CYCLE' k+1 . We multiply that amount by (2 i − 2 i−k ′ ) since this is the number of different combinations of prefixes with length i in the binary strings of size n + i just avoiding to count the previous prefixes twice. If we join both properties, we obtain the sum described by the formula above. Indeed, this formula can be simplified to and extracting a common factor we obtain which is equal to Nevertheless, for every real number 0 < k ′ ≤ 1 we have that Plot the function f(x) on the interval [-3, 3] which is equivalent to that it is true for every real number 0 < k ′ ≤ 1. We can check in the Figure 1 that the function f (x) = 2 x × (2 − 2 x 2 ) is greater than or equal to 1 over the interval [0, 1]. Thus Since there are more than or equal to 2 n ′ −( k ′ 2 ) elements of the language HAM-CYCLE' k+1 with length n ′ ≥ 2 × n 0 therefore, we show that HAM-CYCLE' k+1 is in DEN SE( k ′ 2 ) for every instance of bit-length n ′ ≥ 2 × n 0 . ◀ ▶ Lemma 12. HAM-CYCLE' k ∈ DEN SE( 1 2 k−1 ) for every instance of bit-length n ≥ 2 k−1 × n ′ 0 , where the constant n ′ 0 is the positive integer used in the Definition 1 and Lemma 7 for HAM-CYCLE'.
Proof. According to the Lemma 7, HAM-CYCLE' 1 is in DEN SE(1) for every instance of bit-length n ≥ 2 0 × n ′ 0 = n ′ 0 . Consequently, due to Theorem 11, HAM-CYCLE' 2 is in DEN SE( 1 2 ) for every instance of bit-length n ≥ 2 1 × n ′ 0 . Moreover, HAM-CYCLE' 3 is in DEN SE( 1 4 ) for every instance of bit-length n ≥ 2 2 × n ′ 0 and so forth . . . and thus, for every language HAM-CYCLE' k , we have that HAM-CYCLE' k ∈ DEN SE( 1 2 k−1 ) for every instance of bit-length n ≥ 2 k−1 × n ′ 0 . ◀ ▶ Definition 13. We will define a language HAM-CYCLE' ∞ as follows: A binary string x complies with x ∈ HAM-CYCLE' ∞ if and only if we obtain that x ∈ HAM-CYCLE' k and 2 k−1 × n ′ 0 ≤ |x| < 2 k × n ′ 0 where | . . . | represents the bit-length function and the constant n ′ 0 is the positive integer used in the Definition 1 and Lemma 7 for HAM-CYCLE'.
Proof. We can calculate the value of k from some binary string x that is approximately ⌈log 2 ( |x| n ′ 0 )⌉, where ⌈. . .⌉ is the ceiling function. In this way, we should know if Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 26 October 2021 doi:10.20944/preprints201908.0037.v8