1. Introduction and Background Results

(1)

We examine the application of NL in determining the parameter t in the Dai-Liao CG method (5).

(2)

A theoretical analysis is accomplished to confirm the global convergence of the proposed method.

(3)

A numerical comparison is given between the proposed FDL algorithm and other known DL algorithms.

2. Fuzzy Neutrosophic Dai-Liao Conjugate Gradient Method

(1)

Neutrosophication maps the input ${\mathsf{\Delta }}_k:=f\left({\mathbf{x}}_k\right)-f\left({\mathbf{x}}_{k+1}\right)$ into neutrosophic ordered triplets $(\mathrm{T}\left({\mathsf{\Delta }}_k\right),\mathrm{I}\left({\mathsf{\Delta }}_k\right),\mathrm{F}\left({\mathsf{\Delta }}_k\right))$. The MFs are defined with the aim to improve the CG iterative rule exploiting numerical experience. The sigmoid function with the slope defined by ${\varsigma }_1$ at the crossover point $\mathsf{\Delta }={\varsigma }_2$ is a proper choice for $\mathrm{T}$:

$$\mathrm{T}\left(\mathsf{\Delta }\right)=1/\left(1+{\mathrm{e}}^{-{\varsigma }_1(\mathsf{\Delta }-{\varsigma }_2)}\right).$$

(24)

A proper choice for $\mathrm{F}$ is the following sigmoid function:

$$\mathrm{F}\left(\mathsf{\Delta }\right)=1/(1+{\mathrm{e}}^{{\varsigma }_1(\mathsf{\Delta }-{\varsigma }_2)}).$$

(25)

The Gaussian function with the standard deviation ${\varsigma }_1$ and the mean ${\varsigma }_2$ defines the indeterminacy $\mathrm{I}$:

$$\mathrm{I}\left(\mathsf{\Delta }\right)={\mathrm{e}}^{-\frac{{(\mathsf{\Delta }-{\varsigma }_2)}^2}{2{\varsigma }_1^2}}.$$

(26)

Then the neutrosophication of $\mathsf{\Delta }\in \mathbb{R}$ is defined as the transition $\mathsf{\Delta }\to \langle \mathrm{T}\left(\mathsf{\Delta }\right),\mathrm{I}\left(\mathsf{\Delta }\right),\mathrm{F}\left(\mathsf{\Delta }\right)\rangle$, where the MFs are determined in (24)–(26).

(2)

Neutrosophic inference between an input fuzzy set $\mathfrak{I}$ and an output fuzzy set is based on the subsequent “IF–THEN” regulations:

$$\begin{gathered} {\Re }_1:\mathrm{If}\mathfrak{I}=S_P⟹\mathfrak{O}=\{\mathrm{T},\mathrm{I},\mathrm{F}\} \\ {\Re }_2:\mathrm{If}\mathfrak{I}=S_N⟹\mathfrak{O}=\{\mathrm{T},\mathrm{I},\mathrm{F}\}. \end{gathered}$$

Fuzzy sets $S_P$ and $S_N$ point, respectively, to positive or negative errors. Applying the unification $\Re ={\Re }_1\cup {\Re }_2$, we define ${\mathfrak{O}}_i=\mathfrak{I}\circ {\Re }_i$, $i=1,2$, where ∘ denotes the fuzzy transformation. In addition, for a fuzzy vector $\zeta =\{\mathrm{T}\left({\mathsf{\Delta }}_k\right),\mathrm{I}\left({\mathsf{\Delta }}_k\right),\mathrm{F}\left({\mathsf{\Delta }}_k\right)\}$, it follows ${\kappa }_{\mathfrak{I}\circ \Re }\left(\zeta \right)={\kappa }_{\mathfrak{I}\circ {\Re }_1}\bigvee {\kappa }_{\mathfrak{I}\circ {\Re }_2}=sup({\kappa }_{\mathfrak{I}}\bigwedge {\kappa }_{{\mathfrak{O}}_i})$, $i=1,2$, where ⋀ and ⋁ denote the $(min,max,max)$ and $(max,min,min)$ operator, respectively. In this research, the centroid defuzzification method is utilized to generate a vector of crisp outputs ${\zeta }^{*}\in {\mathbb{R}}^3$:

$${\zeta }^{*}=\frac{{\int }_{\mathfrak{O}}\zeta {\kappa }_{\mathfrak{I}\circ \Re }\left(\zeta \right)\mathrm{d}\zeta }{{\int }_{\mathfrak{O}}{\kappa }_{\mathfrak{I}\circ \Re }\left(\zeta \right)\mathrm{d}\zeta }.$$

(3)

De-neutrosophication is based on the transformation $\langle \mathrm{T}\left({\mathsf{\Delta }}_k\right),\mathrm{I}\left({\mathsf{\Delta }}_k\right),\mathrm{F}\left({\mathsf{\Delta }}_k\right)\rangle \to {\nu }_k\in \mathbb{R}$ resulting in a crisp value ${\nu }_k$ and suggested as:

$${\nu }_k=2-\left(\mathrm{T}\left({\mathsf{\Delta }}_k\right)+\mathrm{I}\left({\mathsf{\Delta }}_k\right)+\mathrm{F}\left({\mathsf{\Delta }}_k\right)\right).$$

(27)

The backtracking line search from [29] begins from $\alpha =1$ and generates further step sizes which ensure decrease of the goal function in each iteration. Algorithm 1, restated from [30], is used to define the primary step size ${\alpha }_k$.

Algorithm 1 The backtracking line search.

Algorithm 2 of the FDL method is described as follows:

Algorithm 2 Fuzzy neutrosophic Dai-Liao (FDL) conjugate gradient method.

3. Convergence Examination

Proof. 

In the initial stage it follows ${\mathbf{g}}_0^{\mathrm{T}}{\mathbf{d}}_0=-{\parallel {\mathbf{g}}_0\parallel }^2$. Since $c=1$, obviously (40) is satisfied in the initial stage $k=0$. Assume (40) for some $k\ge 1$. Applying the inner product of both the left and right hand side in (4) with ${\mathbf{g}}_k^{\mathrm{T}}$, it is concluded

$$\begin{array}{cc}\hfill {\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_k& =-\parallel {\mathbf{g}}_k{\parallel }^2+{\beta }_k^{\mathrm{FDL}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\hfill \\ \hfill & =-\parallel {\mathbf{g}}_k{\parallel }^2+\left(\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}-{\nu }_k\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{s}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}\right){\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\hfill \\ \hfill & =-\parallel {\mathbf{g}}_k{\parallel }^2+\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}-{\nu }_k\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{s}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\hfill \\ \hfill & =-\parallel {\mathbf{g}}_k{\parallel }^2+\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}-{\nu }_k\frac{{\alpha }_{k-1}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\hfill \\ \hfill & =-{\parallel {\mathbf{g}}_k\parallel }^2+\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}-{\nu }_k\frac{{\alpha }_{k-1}{\left({\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\right)}^2}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}.\hfill \end{array}$$

(41)

Using (21) in conjunction with (38) and ${\alpha }_{k-1}>0$, we conclude

$${\nu }_k\frac{{\alpha }_{k-1}{\left({\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\right)}^2}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}\ge 0.$$

(42)

Now from (41) and (42) it follows

$$\begin{array}{cc}\hfill {\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_k& \le -{\parallel {\mathbf{g}}_k\parallel }^2+\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\hfill \\ \hfill & =\frac{-\parallel {\mathbf{g}}_k{\parallel }^2{\left({\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}\right)}^2+\left({\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}\right)\left({\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\right)\left({\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}\right)}{{\left({\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}\right)}^2}.\hfill \end{array}$$

(43)

Applying the inequality $P^{\mathrm{T}}Q\le \frac{1}{2}{(\parallel P\parallel }^2+{\parallel Q\parallel }^2)$ to the equality (43) with $P=\frac{1}{\sqrt{2}}\left({\mathbf{d}}_{k-1}^T{\mathbf{y}}_{k-1}\right){\mathbf{g}}_k$ and $Q=\sqrt{2}\left({\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\right){\mathbf{y}}_{k-1}$, it is obtained

(44)

The requirement (40) is satisfied for $c=\left(\frac{3}{4}-\frac{L^2}{{\theta }^2}\right)$ in (44) and an arbitrary $k\ge 0$. □

Proof. 

Suppose the opposite. Since (45) is not valid, we conclude the existence of $c_1>0$ satisfying

$$\parallel {\mathbf{g}}_k\parallel \ge c_1,\mathrm{for}\mathrm{all}k.$$

(46)

Squaring both sides of (19) one derives

$$\parallel {\mathbf{d}}_k{\parallel }^2=\parallel {\mathbf{g}}_k{\parallel }^2-2{\beta }_k^{\mathrm{FDL}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}+{\left({\beta }_k^{\mathrm{FDL}}\right)}^2{\parallel {\mathbf{d}}_{k-1}\parallel }^2.$$

(47)

Taking into account (20), we obtain

$$\begin{array}{cc}\hfill -2{\beta }_k^{\mathrm{FDL}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}& =-2\left(\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}-{\nu }_k\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{s}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}\right){\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\hfill \\ \hfill & =-2\left(\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}-{\nu }_k\frac{{\alpha }_{k-1}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\right)\hfill \\ \hfill & =-2\left(\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}-{\nu }_k\frac{{\alpha }_{k-1}{\left({\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}\right)}^2}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}\right).\hfill \end{array}$$

(48)

Now from (42), it follows

$$\begin{array}{cc}\hfill -2{\beta }_k^{\mathrm{EDL}}{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}& \le 2\left|\frac{{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{y}}_{k-1}}{{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}\right||{\mathbf{g}}_k^{\mathrm{T}}{\mathbf{d}}_{k-1}|\le 2\frac{\parallel {\mathbf{g}}_k\parallel \parallel {\mathbf{y}}_{k-1}\parallel }{\left|{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}\right|}\parallel {\mathbf{g}}_k\parallel \parallel {\mathbf{d}}_{k-1}\parallel \hfill \\ \hfill & =2\frac{{\alpha }_{k-1}\parallel {\mathbf{g}}_k{\parallel }^2\parallel {\mathbf{y}}_{k-1}\parallel \parallel {\mathbf{d}}_{k-1}\parallel }{{\alpha }_{k-1}{\mathbf{d}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}\hfill \\ \hfill & =2\frac{\parallel {\mathbf{g}}_k{\parallel }^2\parallel {\mathbf{y}}_{k-1}\parallel \parallel {\mathbf{s}}_{k-1}\parallel }{{\mathbf{s}}_{k-1}^{\mathrm{T}}{\mathbf{y}}_{k-1}}\hfill \\ \hfill & \le 2\frac{\parallel {\mathbf{g}}_k{\parallel }^{2}L\parallel {\mathbf{s}}_{k-1}\parallel \parallel {\mathbf{s}}_{k-1}\parallel }{\theta \parallel {\mathbf{s}}_{k-1}{\parallel }^2}\hfill \\ \hfill & =\frac{2L}{\theta }{\parallel {\mathbf{g}}_k\parallel }^2.\hfill \end{array}$$

(49)

Now, an application of (20) initiates

(50)

Further (21) and (50) lead to

$${\beta }_k^{\mathrm{FDL}}\le \frac{\left(L+1\right)\parallel {\mathbf{g}}_k\parallel }{\theta \parallel {\mathbf{d}}_{k-1}\parallel }.$$

(51)

Using (49) and (51) in (47), we get

$$\begin{array}{cc}\hfill \parallel {\mathbf{d}}_k{\parallel }^2& \le \parallel {\mathbf{g}}_k{\parallel }^2+\frac{2L}{\theta }\parallel {\mathbf{g}}_k{\parallel }^2+\frac{{(L+1)}^2{\parallel {\mathbf{g}}_k\parallel }^2}{{\theta }^2{\parallel {\mathbf{d}}_{k-1}\parallel }^2}\parallel {\mathbf{d}}_{k-1}{\parallel }^2\hfill \\ \hfill & =\parallel {\mathbf{g}}_k{\parallel }^2+\frac{2L}{\theta }\parallel {\mathbf{g}}_k{\parallel }^2+\frac{{(L+1)}^2{\parallel {\mathbf{g}}_k\parallel }^2}{{\theta }^2}\hfill \\ \hfill & =\left(1+\frac{2L}{\theta }+\frac{{(L+1)}^2}{{\theta }^2}\right)\parallel {\mathbf{g}}_k{\parallel }^2\hfill \\ \hfill & =\left(\frac{\theta +2L}{\theta }+\frac{{(L+1)}^2}{{\theta }^2}\right)\parallel {\mathbf{g}}_k{\parallel }^2\hfill \\ \hfill & =\frac{(\theta +2L)\theta +{(L+1)}^2}{{\theta }^2}{\parallel {\mathbf{g}}_k\parallel }^2.\hfill \end{array}$$

(52)

Dividing both sides of (52) by ${∥{\mathbf{g}}_k∥}^4$ and utilizing (46), it can be decided

$$\begin{gathered} \frac{\parallel {\mathbf{d}}_k{\parallel }^2}{\parallel {\mathbf{g}}_k{\parallel }^4}\le \frac{(\theta +2L)\theta +{(L+1)}^2}{{\theta }^2}·\frac{1}{c_1^2}, \\ \frac{\parallel {\mathbf{g}}_k{\parallel }^4}{\parallel {\mathbf{d}}_k{\parallel }^2}\ge \frac{{\theta }^2·c_1^2}{(\theta +2L)\theta +{(L+1)}^2}. \end{gathered}$$

(53)

The inequalities in (53) imply

$$\sum _{k=0}^{\infty }\frac{\parallel {\mathbf{g}}_k{\parallel }^4}{\parallel {\mathbf{d}}_k{\parallel }^2}\ge \sum _{k=0}^{\infty }\frac{{\theta }^2·c_1^2}{(\theta +2L)\theta +{(L+1)}^2}=\infty .$$

(54)

Therefore, $\parallel {\mathbf{g}}_k\parallel \ge c_1$ causes a contradiction with Lemma 2. □

4. Numerical Results

5. Conclusions

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