The discrete Hopfield neural network 3-satisfiability (DHNN-3SAT) model represents an innovative application of deep learning techniques to the Boolean SAT problem. Existing research indicated that the DHNN-3SAT model demonstrated significant advantages in handling 3SAT problem instances of varying scales and complexities. Compared to traditional heuristic algorithms, this model converged to local minima more rapidly and exhibited enhanced exploration capabilities within the global search space. However, the model faced several challenges and limitations. As constraints in SAT problems dynamically increased, decreased, or changed, and as problem scales expanded, the model's computational complexity and storage requirements may increase dramatically, leading to reduced performance in handling large-scale SAT problems. To address these challenges, this paper first introduced a method for designing network synaptic weights based on fundamental logical clauses. This method effectively utilized the synaptic weight information from the original SAT problem within the DHNN network, thereby significantly reducing redundant computations. Concrete examples illustrated the design process of network synaptic weights when constraints were added, removed, or updated, offering new approaches for managing the evolving constraints in SAT problems. Subsequently, the paper presented a DHNN-3SAT model optimized by genetic algorithms combined with K-modes clustering. This model employed genetic algorithm-optimized K-modes clustering to effectively cluster the initial space, significantly reducing the search space. This approach minimized the likelihood of redundant searches and reduced the risk of getting trapped in local minima, thus improving search efficiency. Experimental tests on benchmark datasets showed that the proposed model outperformed traditional DHNN-3SAT models, DHNN-3SAT models combined with genetic algorithms, and DHNN-3SAT models combined with imperialist competitive algorithms across four evaluation metrics. This study not only broadened the application of DHNN in solving 3SAT problems but also provided valuable insights and guidance for future research.
Citation: Xiaojun Xie, Saratha Sathasivam, Hong Ma. Modeling of 3 SAT discrete Hopfield neural network optimization using genetic algorithm optimized K-modes clustering[J]. AIMS Mathematics, 2024, 9(10): 28100-28129. doi: 10.3934/math.20241363
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The discrete Hopfield neural network 3-satisfiability (DHNN-3SAT) model represents an innovative application of deep learning techniques to the Boolean SAT problem. Existing research indicated that the DHNN-3SAT model demonstrated significant advantages in handling 3SAT problem instances of varying scales and complexities. Compared to traditional heuristic algorithms, this model converged to local minima more rapidly and exhibited enhanced exploration capabilities within the global search space. However, the model faced several challenges and limitations. As constraints in SAT problems dynamically increased, decreased, or changed, and as problem scales expanded, the model's computational complexity and storage requirements may increase dramatically, leading to reduced performance in handling large-scale SAT problems. To address these challenges, this paper first introduced a method for designing network synaptic weights based on fundamental logical clauses. This method effectively utilized the synaptic weight information from the original SAT problem within the DHNN network, thereby significantly reducing redundant computations. Concrete examples illustrated the design process of network synaptic weights when constraints were added, removed, or updated, offering new approaches for managing the evolving constraints in SAT problems. Subsequently, the paper presented a DHNN-3SAT model optimized by genetic algorithms combined with K-modes clustering. This model employed genetic algorithm-optimized K-modes clustering to effectively cluster the initial space, significantly reducing the search space. This approach minimized the likelihood of redundant searches and reduced the risk of getting trapped in local minima, thus improving search efficiency. Experimental tests on benchmark datasets showed that the proposed model outperformed traditional DHNN-3SAT models, DHNN-3SAT models combined with genetic algorithms, and DHNN-3SAT models combined with imperialist competitive algorithms across four evaluation metrics. This study not only broadened the application of DHNN in solving 3SAT problems but also provided valuable insights and guidance for future research.
Let A denote the class of functions f which are analytic in the open unit disk Δ={z∈C:|z|<1}, normalized by the conditions f(0)=f′(0)−1=0. So each f∈A has series representation of the form
f(z)=z+∞∑n=2anzn. | (1.1) |
For two analytic functions f and g, f is said to be subordinated to g (written as f≺g) if there exists an analytic function ω with ω(0)=0 and |ω(z)|<1 for z∈Δ such that f(z)=(g∘ω)(z).
A function f∈A is said to be in the class S if f is univalent in Δ. A function f∈S is in class C of normalized convex functions if f(Δ) is a convex domain. For 0≤α≤1, Mocanu [23] introduced the class Mα of functions f∈A such that f(z)f′(z)z≠0 for all z∈Δ and
ℜ((1−α)zf′(z)f(z)+α(zf′(z))′f′(z))>0(z∈Δ). | (1.2) |
Geometrically, f∈Mα maps the circle centred at origin onto α-convex arcs which leads to the condition (1.2). The class Mα was studied extensively by several researchers, see [1,10,11,12,24,25,26,27] and the references cited therein.
A function f∈S is uniformly starlike if f maps every circular arc Γ contained in Δ with center at ζ ∈Δ onto a starlike arc with respect to f(ζ). A function f∈C is uniformly convex if f maps every circular arc Γ contained in Δ with center ζ ∈Δ onto a convex arc. We denote the classes of uniformly starlike and uniformly convex functions by UST and UCV, respectively. For recent study on these function classes, one can refer to [7,9,13,19,20,31].
In 1999, Kanas and Wisniowska [15] introduced the class k-UCV (k≥0) of k-uniformly convex functions. A function f∈A is said to be in the class k-UCV if it satisfies the condition
ℜ(1+zf″(z)f′(z))>k|zf′(z)f′(z)|(z∈Δ). | (1.3) |
In recent years, many researchers investigated interesting properties of this class and its generalizations. For more details, see [2,3,4,14,15,16,17,18,30,32,35] and references cited therein.
In 2015, Sokół and Nunokawa [33] introduced the class MN, a function f∈MN if it satisfies the condition
ℜ(1+zf″(z)f′(z))>|zf′(z)f(z)−1|(z∈Δ). |
In [28], it is proved that if ℜ(f′)>0 in Δ, then f is univalent in Δ. In 1972, MacGregor [21] studied the class B of functions with bounded turning, a function f∈B if it satisfies the condition ℜ(f′)>0 for z∈Δ. A natural generalization of the class B is B(δ1) (0≤δ1<1), a function f∈B(δ1) if it satisfies the condition
ℜ(f′(z))>δ1(z∈Δ;0≤δ1<1), | (1.4) |
for details associated with the class B(δ1) (see [5,6,34]).
Motivated essentially by the above work, we now introduce the following class k-Q(α) of analytic functions.
Definition 1. Let k≥0 and 0≤α≤1. A function f∈A is said to be in the class k-Q(α) if it satisfies the condition
ℜ((zf′(z))′f′(z))>k|(1−α)f′(z)+α(zf′(z))′f′(z)−1|(z∈Δ). | (1.5) |
It is worth mentioning that, for special values of parameters, one can obtain a number of well-known function classes, some of them are listed below:
1. k-Q(1)=k-UCV;
2. 0-Q(α)=C.
In what follows, we give an example for the class k-Q(α).
Example 1. The function f(z)=z1−Az(A≠0) is in the class k-Q(α) with
k≤1−b2b√b(1+α)[b(1+α)+2]+4(b=|A|). | (1.6) |
The main purpose of this paper is to establish several interesting relationships between k-Q(α) and the class B(δ) of functions with bounded turning.
To prove our main results, we need the following lemmas.
Lemma 1. ([8]) Let h be analytic in Δ with h(0)=1, β>0 and 0≤γ1<1. If
h(z)+βzh′(z)h(z)≺1+(1−2γ1)z1−z, |
then
h(z)≺1+(1−2δ)z1−z, |
where
δ=(2γ1−β)+√(2γ1−β)2+8β4. | (2.1) |
Lemma 2. Let h be analytic in Δ and of the form
h(z)=1+∞∑n=mbnzn(bm≠0) |
with h(z)≠0 in Δ. If there exists a point z0(|z0|<1) such that |argh(z)|<πρ2(|z|<|z0|) and |argh(z0)|=πρ2 for some ρ>0, then z0h′(z0)h(z0)=iℓρ, where
ℓ:{ℓ≥n2(c+1c)(argh(z0)=πρ2),ℓ≤−n2(c+1c)(argh(z0)=−πρ2), |
and (h(z0))1/ρ=±ic(c>0).
This result is a generalization of the Nunokawa's lemma [29].
Lemma 3. ([37]) Let ε be a positive measure on [0,1]. Let ϝ be a complex-valued function defined on Δ×[0,1] such that ϝ(.,t) is analytic in Δ for each t∈[0,1] and ϝ(z,.) is ε-integrable on [0,1] for all z∈Δ. In addition, suppose that ℜ(ϝ(z,t))>0, ϝ(−r,t) is real and ℜ(1/ϝ(z,t))≥1/ϝ(−r,t) for |z|≤r<1 and t∈[0,1]. If ϝ(z)=∫10ϝ(z,t)dε(t), then ℜ(1/ϝ(z))≥1/ϝ(−r).
Lemma 4. ([22]) If −1≤D<C≤1, λ1>0 and ℜ(γ2)≥−λ1(1−C)/(1−D), then the differential equation
s(z)+zs′(z)λ1s(z)+γ2=1+Cz1+Dz(z∈Δ) |
has a univalent solution in Δ given by
s(z)={zλ1+γ2(1+Dz)λ1(C−D)/Dλ1∫z0tλ1+γ2−1(1+Dt)λ1(C−D)/Ddt−γ2λ1(D≠0),zλ1+γ2eλ1Czλ1∫z0tλ1+γ2−1eλ1Ctdt−γ2λ1(D=0). |
If r(z)=1+c1z+c2z2+⋯ satisfies the condition
r(z)+zr′(z)λ1r(z)+γ2≺1+Cz1+Dz(z∈Δ), |
then
r(z)≺s(z)≺1+Cz1+Dz, |
and s(z) is the best dominant.
Lemma 5. ([36,Chapter 14]) Let w, x and\ y≠0,−1,−2,… be complex numbers. Then, for ℜ(y)>ℜ(x)>0, one has
1. 2G1(w,x,y;z)=Γ(y)Γ(y−x)Γ(x)∫10sx−1(1−s)y−x−1(1−sz)−wds;
2. 2G1(w,x,y;z)= 2G1(x,w,y;z);
3. 2G1(w,x,y;z)=(1−z)−w2G1(w,y−x,y;zz−1).
Firstly, we derive the following result.
Theorem 1. Let 0≤α<1 and k≥11−α. If f∈k-Q(α), then f∈B(δ), where
δ=(2μ−λ)+√(2μ−λ)2+8λ4(λ=1+αkk(1−α);μ=k−αk−1k(1−α)). | (3.1) |
Proof. Let f′=ℏ, where ℏ is analytic in Δ with ℏ(0)=1. From inequality (1.5) which takes the form
ℜ(1+zℏ′(z)ℏ(z))>k|(1−α)ℏ(z)+α(1+zℏ′(z)ℏ(z))−1|=k|1−α−ℏ(z)+αℏ(z)−αzℏ′(z)ℏ(z)|, |
we find that
ℜ(ℏ(z)+1+αkk(1−α)zℏ(z)ℏ(z))>k−αk−1k(1−α), |
which can be rewritten as
ℜ(ℏ(z)+λzℏ(z)ℏ(z))>μ(λ=1+αkk(1−α);μ=k−αk−1k(1−α)). |
The above relationship can be written as the following Briot-Bouquet differential subordination
ℏ(z)+λzℏ′(z)ℏ(z)≺1+(1−2μ)z1−z. |
Thus, by Lemma 1, we obtain
ℏ≺1+(1−2δ)z1−z, | (3.2) |
where δ is given by (3.1). The relationship (3.2) implies that f∈B(δ). We thus complete the proof of Theorem 3.1.
Theorem 2. Let 0<α≤1, 0<β<1, c>0, k≥1, n≥m+1(m∈ N ), |ℓ|≥n2(c+1c) and
|αβℓ±(1−α)cβsinβπ2|≥1. | (3.3) |
If
f(z)=z+∞∑n=m+1anzn(am+1≠0) |
and f∈k-Q(α), then f∈B(β0), where
β0=min{β:β∈(0,1)} |
such that (3.3) holds.
Proof. By the assumption, we have
f′(z)=ℏ(z)=1+∞∑n=mcnzn(cm≠0). | (3.4) |
In view of (1.5) and (3.4), we get
ℜ(1+zℏ′(z)ℏ(z))>k|(1−α)ℏ(z)+α(1+zℏ′(z)ℏ(z))−1|. |
If there exists a point z0∈Δ such that
|argℏ(z)|<βπ2(|z|<|z0|;0<β<1) |
and
|argℏ(z0)|=βπ2(0<β<1), |
then from Lemma 2, we know that
z0ℏ′(z0)ℏ(z0)=iℓβ, |
where
(ℏ(z0))1/β=±ic(c>0) |
and
ℓ:{ℓ≥n2(c+1c)(argℏ(z0)=βπ2),ℓ≤−n2(c+1c)(argℏ(z0)=−βπ2). |
For the case
argℏ(z0)=βπ2, |
we get
ℜ(1+z0ℏ′(z0)ℏ(z0))=ℜ(1+iℓβ)=1. | (3.5) |
Moreover, we find from (3.3) that
k|(1−α)ℏ(z0)+α(1+z0ℏ′(z0)ℏ(z0))−1|=k|(1−α)(ℏ(z0)−1)+αz0ℏ′(z0)ℏ(z0)|=k|(1−α)[(±ic)β−1]+iαβℓ|=k√(1−α)2(cβcosβπ2−1)2+[αβℓ±(1−α)cβsinβπ2]2≥1. | (3.6) |
By virtue of (3.5) and (3.6), we have
ℜ(1+zℏ′(z0)ℏ(z0))≤k|(1−α)ℏ(z0)+α(1+z0ℏ(z0)ℏ(z0))−1|, |
which is a contradiction to the definition of k-Q(α). Since β0=min{β:β∈(0,1)} such that (3.3) holds, we can deduce that f∈B(β0).
By using the similar method as given above, we can prove the case
argℏ(z0)=−βπ2 |
is true. The proof of Theorem 2 is thus completed.
Theorem 3. If 0<β<1 and 0≤ν<1. If f∈k-Q(α), then
ℜ(f′)>[2G1(2β(1−ν),1;1β+1;12)]−1, |
or equivalently, k-Q(α)⊂B(ν0), where
ν0=[2G1(2β(1−μ),1;1β+1;12)]−1. |
Proof. For
w=2β(1−ν), x=1β, y=1β+1, |
we define
ϝ(z)=(1+Dz)w∫10tx−1(1+Dtz)−wdt=Γ(x)Γ(y) 2G1(1,w,y;zz−1). | (3.7) |
To prove k-Q(α)⊂B(ν0), it suffices to prove that
inf|z|<1{ℜ(q(z))}=q(−1), |
which need to show that
ℜ(1/ϝ(z))≥1/ϝ(−1). |
By Lemma 3 and (3.7), it follows that
ϝ(z)=∫10ϝ(z,t)dε(t), |
where
ϝ(z,t)=1−z1−(1−t)z(0≤t≤1), |
and
dε(t)=Γ(x)Γ(w)Γ(y−w)tw−1(1−t)y−w−1dt, |
which is a positive measure on [0,1].
It is clear that ℜ(ϝ(z,t))>0 and ϝ(−r,t) is real for |z|≤r<1 and t∈[0,1]. Also
ℜ(1ϝ(z,t))=ℜ(1−(1−t)z1−z)≥1+(1−t)r1+r=1ϝ(−r,t) |
for |z|≤r<1. Therefore, by Lemma 3, we get
ℜ(1/ϝ(z))≥1/ϝ(−r). |
If we let r→1−, it follows that
ℜ(1/ϝ(z))≥1/ϝ(−1). |
Thus, we deduce that k-Q(α)⊂B(ν0).
Theorem 4. Let 0≤α<1 and k≥11−α. If f∈k-Q(α), then
f′(z)≺s(z)=1g(z), |
where
g(z)=2G1(2λ,1,1λ+1;zz−1)(λ=1+αkk(1−α)). |
Proof. Suppose that f′=ℏ. From the proof of Theorem 1, we see that
ℏ(z)+zℏ′(z)1λℏ(z)≺1+(1−2μ)z1−z≺1+z1−z(λ=1+αkk(1−α);μ=k−αk−1k(1−α)). |
If we set λ1=1λ, γ2=0, C=1 and D=−1 in Lemma 4, then
ℏ(z)≺s(z)=1g(z)=z1λ(1−z)−2λ1/λ∫z0t(1/λ)−1(1−t)−2/λdt. |
By putting t=uz, and using Lemma 5, we obtain
ℏ(z)≺s(z)=1g(z)=11λ(1−z)2λ∫10u(1/λ)−1(1−uz)−2/λdu=[2G1(2λ,1,1λ+1;zz−1)]−1, |
which is the desired result of Theorem 4.
The present investigation was supported by the Key Project of Education Department of Hunan Province under Grant no. 19A097 of the P. R. China. The authors would like to thank the referees for their valuable comments and suggestions, which was essential to improve the quality of this paper.
The authors declare no conflict of interest.
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