Research article Special Issues

Fatigue detection method for UAV remote pilot based on multi feature fusion

  • In recent years, UAV industry is developing rapidly and vigorously. However, so far, there is no relevant research on the fatigue detection method for UAV remote pilot, which is the core technology to ensure the flight safety of UAV. Aiming at this problem, a fatigue detection method for UAV remote pilot is proposed in this paper. Specifically, we first build a UAV operator fatigue detection database (OFDD). By analyzing the fatigue features in the database, we find that multiple facial features are highly correlated to the fatigue state, especially the head posture, and the temporal information is essential for distinguish between yawn and speaking in the study of UAV remote pilot fatigue detection. Based on these findings, a fatigue detection method for UAV remote pilots was proposed by efficiently locating the related facial regions, a multiple features extraction module to extract the eye, mouth and head posture features, and an efficient temporal fatigue decision module based on SVM. The experimental results show that this method not only performs well on the traditional driver dataset, but also achieves an accuracy rate of 97.05%; and it achieves the highest detection accuracy rate of 97.32% on the UAV remote pilots fatigue detection dataset OFDD.

    Citation: Lei Pan, Chongyao Yan, Yuan Zheng, Qiang Fu, Yangjie Zhang, Zhiwei Lu, Zhiqing Zhao, Jun Tian. Fatigue detection method for UAV remote pilot based on multi feature fusion[J]. Electronic Research Archive, 2023, 31(1): 442-466. doi: 10.3934/era.2023022

    Related Papers:

    [1] Doston Jumaniyozov, Ivan Kaygorodov, Abror Khudoyberdiyev . The algebraic classification of nilpotent commutative algebras. Electronic Research Archive, 2021, 29(6): 3909-3993. doi: 10.3934/era.2021068
    [2] Fei Ma, Min Yin, Yanhui Teng, Ganglian Ren . Nonlinear generalized semi-Jordan triple derivable mappings on completely distributive commutative subspace lattice algebras. Electronic Research Archive, 2023, 31(8): 4807-4817. doi: 10.3934/era.2023246
    [3] Xiuhai Fei, Cuixian Lu, Haifang Zhang . Nonlinear Jordan triple derivable mapping on $ * $-type trivial extension algebras. Electronic Research Archive, 2024, 32(3): 1425-1438. doi: 10.3934/era.2024066
    [4] Quanguo Chen, Yong Deng . Hopf algebra structures on generalized quaternion algebras. Electronic Research Archive, 2024, 32(5): 3334-3362. doi: 10.3934/era.2024154
    [5] Kengo Matsumoto . $ C^* $-algebras associated with asymptotic equivalence relations defined by hyperbolic toral automorphisms. Electronic Research Archive, 2021, 29(4): 2645-2656. doi: 10.3934/era.2021006
    [6] Kailash C. Misra, Sutida Patlertsin, Suchada Pongprasert, Thitarie Rungratgasame . On derivations of Leibniz algebras. Electronic Research Archive, 2024, 32(7): 4715-4722. doi: 10.3934/era.2024214
    [7] Peigen Cao, Fang Li, Siyang Liu, Jie Pan . A conjecture on cluster automorphisms of cluster algebras. Electronic Research Archive, 2019, 27(0): 1-6. doi: 10.3934/era.2019006
    [8] Yizheng Li, Dingguo Wang . Lie algebras with differential operators of any weights. Electronic Research Archive, 2023, 31(3): 1195-1211. doi: 10.3934/era.2023061
    [9] Juxiang Sun, Guoqiang Zhao . Gorenstein invariants under right Quasi-Frobenius extensions. Electronic Research Archive, 2025, 33(6): 3561-3570. doi: 10.3934/era.2025158
    [10] Ming Ding, Zhiqi Chen, Jifu Li . The properties on F-manifold color algebras and pre-F-manifold color algebras. Electronic Research Archive, 2025, 33(1): 87-101. doi: 10.3934/era.2025005
  • In recent years, UAV industry is developing rapidly and vigorously. However, so far, there is no relevant research on the fatigue detection method for UAV remote pilot, which is the core technology to ensure the flight safety of UAV. Aiming at this problem, a fatigue detection method for UAV remote pilot is proposed in this paper. Specifically, we first build a UAV operator fatigue detection database (OFDD). By analyzing the fatigue features in the database, we find that multiple facial features are highly correlated to the fatigue state, especially the head posture, and the temporal information is essential for distinguish between yawn and speaking in the study of UAV remote pilot fatigue detection. Based on these findings, a fatigue detection method for UAV remote pilots was proposed by efficiently locating the related facial regions, a multiple features extraction module to extract the eye, mouth and head posture features, and an efficient temporal fatigue decision module based on SVM. The experimental results show that this method not only performs well on the traditional driver dataset, but also achieves an accuracy rate of 97.05%; and it achieves the highest detection accuracy rate of 97.32% on the UAV remote pilots fatigue detection dataset OFDD.



    The algebraic classification (up to isomorphism) of algebras of dimension n from a certain variety defined by a certain family of polynomial identities is a classic problem in the theory of non-associative algebras. There are many results related to the algebraic classification of small-dimensional algebras in many varieties of non-associative algebras [11,12,2,3,4,6,13,9,16]. So, algebraic classifications of 2-dimensional algebras [16,19], 3-dimensional evolution algebras [1], 3-dimensional anticommutative algebras [17], 4-dimensional division algebras [5,7], 4-dimensional nilpotent algebras [13] and 6-dimensional anticommutative nilpotent algebras [12] have been given. In the present paper, we give the algebraic classification of 5-dimensional nilpotent commutative algebras. The variety of commutative algebras is defined by the following identity: xy=yx. It contains commutative CD-algebras, Jordan algebras, mock-Lie algebras and commutative associative algebras as subvarieties. On the other hand, it is a principal part in the varieties of weakly associative algebras and flexible algebras.

    The algebraic study of central extensions of associative and non-associative algebras has been an important topic for years (see, for example, [10,20] and references therein). Our method for classifying nilpotent commutative algebras is based on the calculation of central extensions of nilpotent algebras of smaller dimensions from the same variety (first, this method has been developed by Skjelbred and Sund for Lie algebra case in [20]) and the classifications of all complex 5-dimensional nilpotent commutative (non-Jordan) CD-algebras [11]; nilpotent Jordan (non-associative) algebras [9]; and nilpotent associative commutative algebras [18].

    Throughout this paper, we use the notations and methods well written in [10], which we have adapted for the commutative case with some modifications. Further in this section we give some important definitions.

    Let (A,) be a complex commutative algebra and V be a complex vector space. The C-linear space Z2(A,V) is defined as the set of all bilinear maps θ:A×AV such that θ(x,y)=θ(y,x). These elements will be called cocycles. For a linear map f from A to V, if we define δf:A×AV by δf(x,y)=f(xy), then δfZ2(A,V). We define B2(A,V)={θ=δf :fHom(A,V)}. We define the second cohomology space H2(A,V) as the quotient space Z2(A,V)/B2(A,V).

    Let Aut(A) be the automorphism group of A and let ϕAut(A). For θZ2(A,V) define the action of the group Aut(A) on Z2(A,V) by

    ϕθ(x,y)=θ(ϕ(x),ϕ(y)).

    It is easy to verify that B2(A,V) is invariant under the action of Aut(A). So, we have an induced action of Aut(A) on H2(A,V).

    Let A be a commutative algebra of dimension m over C and V be a C-vector space of dimension k. For the bilinear map θ, define on the linear space Aθ=AV the bilinear product " [,]Aθ" by [x+x,y+y]Aθ=xy+θ(x,y) for all x,yA,x,yV. The algebra Aθ is called a k-dimensional central extension of A by V. One can easily check that Aθ is a commutative algebra if and only if θZ2(A,V).

    Call the set Ann(θ)={xA:θ(x,A)=0} the annihilator of θ. We recall that the annihilator of an algebra A is defined as the ideal Ann(A)={xA:xA=0}. Observe that Ann(Aθ)=(Ann(θ)Ann(A))V.

    The following result shows that every algebra with a non-zero annihilator is a central extension of a smaller-dimensional algebra.

    Lemma 1.1. Let A be an n-dimensional commutative algebra such that

    dim(Ann(A))=m0.

    Then there exists, up to isomorphism, a unique (nm)-dimensional commutative algebra A and a bilinear map θZ2(A,V) with Ann(A)Ann(θ)=0, where V is a vector space of dimension m, such that AAθ and A/Ann(A)A.

    Proof. Let A be a linear complement of Ann(A) in A. Define a linear map P:AA by P(x+v)=x for xA and vAnn(A), and define a multiplication on A by [x,y]A=P(xy) for x,yA. For x,yA, we have

    P(xy)=P((xP(x)+P(x))(yP(y)+P(y)))=P(P(x)P(y))=[P(x),P(y)]A.

    Since P is a homomorphism, P(A)=A and A is a commutative algebra and also A/Ann(A)A, which gives us the uniqueness. Now, define the map θ:A×AAnn(A) by θ(x,y)=xy[x,y]A. Thus, Aθ is A and therefore θZ2(A,V) and Ann(A)Ann(θ)=0.

    Definition 1.2. Let A be an algebra and I be a subspace of Ann(A). If A=A0I then I is called an annihilator component of A. A central extension of an algebra A without annihilator component is called a non-split central extension.

    Our task is to find all central extensions of an algebra A by a space V. In order to solve the isomorphism problem we need to study the action of Aut(A) on H2(A,V). To do that, let us fix a basis e1,,es of V, and θZ2(A,V). Then θ can be uniquely written as θ(x,y)=si=1θi(x,y)ei, where θiZ2(A,C). Moreover, Ann(θ)=Ann(θ1)Ann(θ2)Ann(θs). Furthermore, θB2(A,V) if and only if all θiB2(A,C). It is not difficult to prove (see [10,Lemma 13]) that given a commutative algebra Aθ, if we write as above θ(x,y)=si=1θi(x,y)eiZ2(A,V) and Ann(θ)Ann(A)=0, then Aθ has an annihilator component if and only if [θ1],[θ2],,[θs] are linearly dependent in H2(A,C).

    Let V be a finite-dimensional vector space over C. The Grassmannian Gk(V) is the set of all k-dimensional linear subspaces of V. Let Gs(H2(A,C)) be the Grassmannian of subspaces of dimension s in H2(A,C). There is a natural action of Aut(A) on Gs(H2(A,C)). Let ϕAut(A). For W=[θ1],[θ2],,[θs]Gs(H2(A,C)) define ϕW=[ϕθ1],[ϕθ2],,[ϕθs]. We denote the orbit of WGs(H2(A,C)) under the action of Aut(A) by Orb(W). Given

    W1=[θ1],[θ2],,[θs],W2=[ϑ1],[ϑ2],,[ϑs]Gs(H2(A,C)),

    we easily have that if W1=W2, then si=1Ann(θi)Ann(A)=si=1Ann(ϑi)Ann(A), and therefore we can introduce the set

    Ts(A)={W=[θ1],,[θs]Gs(H2(A,C)):si=1Ann(θi)Ann(A)=0},

    which is stable under the action of Aut(A).

    Now, let V be an s-dimensional linear space and let us denote by E(A,V) the set of all non-split s-dimensional central extensions of A by V. By above, we can write

    E(A,V)={Aθ:θ(x,y)=si=1θi(x,y)ei  and  [θ1],[θ2],,[θs]Ts(A)}.

    We also have the following result, which can be proved as in [10,Lemma 17].

    Lemma 1.3. Let Aθ,AϑE(A,V). Suppose that θ(x,y)=si=1θi(x,y)ei and ϑ(x,y)=si=1ϑi(x,y)ei. Then the commutative algebras Aθ and Aϑ are isomorphic if and only if

    Orb[θ1],[θ2],,[θs]=Orb[ϑ1],[ϑ2],,[ϑs].

    This shows that there exists a one-to-one correspondence between the set of Aut(A)-orbits on Ts(A) and the set of isomorphism classes of E(A,V). Consequently we have a procedure that allows us, given a commutative algebra A of dimension ns, to construct all non-split central extensions of A. This procedure is:

    1. For a given commutative algebra A of dimension ns, determine H2(A,C), Ann(A) and Aut(A).

    2. Determine the set of Aut(A)-orbits on Ts(A).

    3. For each orbit, construct the commutative algebra associated with a representative of it.

    The idea of the definition of a CD-algebra comes from the following property of Jordan and Lie algebras: the commutator of any pair of multiplication operators is a derivation. This gives three identities of degree four, which reduce to only one identity of degree four in the commutative or anticommutative case. Namely, a commutative algebra is a commutative CD-algebra (CCD-algebra) if it satisfies the following identity:

    ((xy)a)b+((xb)a)y+x((yb)a)=((xy)b)a+((xa)b)y+x((ya)b).

    The above described method gives all commutative (CCD- and non-CCD-) algebras. But we are interested in developing this method in such a way that it only gives non-CCD commutative algebras, because the classification of all CCD-algebras is done in [11]. Clearly, any central extension of a commutative non-CCD-algebra is a non-CCD-algebra. But a CCD-algebra may have extensions which are not CCD-algebras. More precisely, let D be a CCD-algebra and θZ2C(D,C). Then Dθ is a CCD-algebra if and only if

    θ(x,y)=θ(y,x),
    θ((xy)a,b)+θ((xb)a,y)+θ(x,(yb)a)=θ((xy)b,a)+θ((xa)b,y)+θ(x,(ya)b).

    for all x,y,a,bD. Define the subspace Z2D(D,C) of Z2C(D,C) by

    Z2D(D,C)={θZ2C(D,C):θ(x,y)=θ(y,x),θ((xy)a,b)+θ((xb)a,y)+θ(x,(yb)a)=θ((xy)b,a)+θ((xa)b,y)+θ(x,(ya)b) for all x,y,a,bD}.

    Observe that B2(D,C)Z2D(D,C). Let H2D(D,C)=Z2D(D,C)/B2(D,C). Then H2D(D,C) is a subspace of H2C(D,C). Define

    Rs(D)={WTs(D):WGs(H2D(D,C))},
    Us(D)={WTs(D):WGs(H2D(D,C))}.

    Then Ts(D)=Rs(D)Us(D). The sets Rs(D) and Us(D) are stable under the action of Aut(D). Thus, the commutative algebras corresponding to the representatives of Aut(D) -orbits on Rs(D) are CCD-algebras, while those corresponding to the representatives of Aut(D)-orbits on Us(D) are not CCD-algebras. Hence, we may construct all non-split commutative non-CCD-algebras A of dimension n with s-dimensional annihilator from a given commutative algebra A of dimension ns in the following way:

    1. If A is non-CCD, then apply the procedure.

    2. Otherwise, do the following:

    (a) Determine Us(A) and Aut(A).

    (b) Determine the set of Aut(A)-orbits on Us(A).

    (c) For each orbit, construct the commutative algebra corresponding to one of its representatives.

    Let us introduce the following notations. Let A be a nilpotent algebra with a basis e1,e2,,en. Then by Δij we will denote the bilinear form Δij:A×AC with Δij(el,em)=δilδjm, if ij and lm. The set {Δij:1ijn} is a basis for the linear space of bilinear forms on A, so every θZ2(A,V) can be uniquely written as θ=1ijncijΔij, where cijC. Let us fix complex number ηk (ηkk=1, ηlk1 for 0<l<k). For denote our algebras, we will use the following notations:

    NΞjjth5dimensional family ofcommutative nonCCDalgebras with parametrs Ξ.Nijjth idimensional nonCCDalgebra.Nijjth idimensional CCDalgebra.

    Remark 1. All families of algebras from our final list do not have intersections, but inside some families of algebras there are isomorphic algebras. All isomorphisms between algebras from a certain family of algebras constucted from the representative (Σ) are given in the list of distinct orbit representations. The notation (Ξ)O(Ξ1)=O(Ξ2) represents that the elements (Ξ1) and (Ξ2) have the same orbit.

    Thanks to [8] we have the complete classification of complex 4-dimensional nilpotent commutative algebras. It will be re-written by some different way for separating CCD- and non-CCD-algebras.

    N301,N401:e1e1=e2H2C=H2DN302,N402:e1e1=e2e1e2=e3H2CH2DN303,N403:e1e2=e3H2C=H2DN304,N404:e1e1=e2e2e2=e3H2CH2DN405:e1e1=e2e1e3=e4H2C=H2DN406:e1e1=e2e3e3=e4H2C=H2DN407:e1e1=e4e2e3=e4H2C=H2DN408:e1e1=e2e1e2=e3e2e2=e4H2CH2DN409:e1e1=e2e2e3=e4H2CH2DN410:e1e1=e2e1e2=e4e3e3=e4H2CH2DN411:e1e1=e2e1e3=e4e2e2=e4H2CH2DN412:e1e1=e2e2e2=e4e3e3=e4H2CH2DN413(λ):e1e1=e2e1e2=e3e1e3=e4e2e2=λe4H2CH2DN414:e1e2=e3e1e3=e4H2CH2DN415:e1e2=e3e1e3=e4e2e2=e4H2CH2DN416:e1e2=e3e1e3=e4e2e3=e4H2CH2DN417:e1e2=e3e3e3=e4H2CH2DN418:e1e1=e4e1e2=e3e3e3=e4H2CH2DN419:e1e1=e4e1e2=e3e2e2=e4e3e3=e4H2CH2DN401:e1e1=e2e1e2=e3e2e3=e4N402:e1e1=e2e1e2=e3e1e3=e4e2e3=e4N403:e1e1=e2e1e2=e3e3e3=e4N404:e1e1=e2e1e2=e3e2e2=e4e3e3=e4N405:e1e1=e2e1e3=e4e2e2=e3N406:e1e1=e2e1e2=e4e1e3=e4e2e2=e3N407:e1e1=e2e2e2=e3e2e3=e4N408:e1e1=e2e1e3=e4e2e2=e3e2e3=e4N409:e1e1=e2e2e2=e3e3e3=e4N410:e1e1=e2e2e2=e3e1e2=e4e3e3=e4N411(λ):e1e1=e2e1e2=λe4e2e2=e3e2e3=e4e3e3=e4

    Here we will collect all information about N302:

    CohomologyAutomorphismsN302e1e1=e2e1e2=e3H2D(N302)=[Δ13],[Δ22],H2C(N302)=H2D(N302)[Δ23],[Δ33]ϕ=(x00yx20z2xyx3)

    Let us use the following notations:

    1=[Δ13],2=[Δ22],3=[Δ23],4=[Δ33].

    Take θ=4i=1αiiH2C(N302). Since

    ϕT(00α10α2α3α1α3α4)ϕ=(ααα1αα2α3α1α3α4),

    we have

    α1=(α1x+α3y+α4z)x3,α2=(α2x2+4α3xy+4α4y2)x2,α3=(α3x+2α4y)x4,α4=α4x6.

    We are interested only in (α3,α4)(0,0) and consider the vector space generated by the following two cocycles:

    θ1=α11+α22+α33+α44  and  θ2=β11+β22+β33.

    Thus, we have

    α1=(α1x+α3y+α4z)x3,β1=(β1x+β3y)x3,α2=(α2x2+4α3xy+4α4y2)x2,β2=(β2x+4β3y)x3,α3=(α3x+2α4y)x4,β3=β3x5.α4=α4x6.

    Consider the following cases.

    1. α40, then:

    (a) β3=0,β20,β1=0, then by choosing x=2α24, y=α3α4, z=α232α1α4, we have the representatives 4,2;

    (b)  β3=0,β20,β10, then by choosing

    x=2α24β2, y=α3α4β2, z=α23(2β1+β2)+2α4(α2β1α1β2),

    we have the representatives 4,1+α2α0;

    (c) β3=0,β2=0,β10, then by choosing y=xα32α4, we have two representatives 4,1 and 2+4,1, depending on α23=α2α4 or not. The first representative will be joint with the family from the case (1b);

    (d) β30,4α2β23=4β2α3β3β22α4,β2=4β1, then by choosing

    x=4β3α4,y=β2α4,z=β2α34α1β3,

    we have the representative 4,3;

    (e) β30,4α2β23=4β2α3β3β22α4,β24β1, then by choosing

    x=4β1β24β3,y=β224β1β216β23,z=(4β1β2)(8β1α3β34β1β2α48α1β33+β22α4)32β33α4,

    we have the representative 4,1+3;

    (f) β30,4α2β234β2α3β3β22α4, then by choosing

    x=4α2β234β2α3β3+β22α44β23α4, y=β2α4β224α3β2β3+4α2β238β23α4, z=(8β1α3β34β1β2α48α1β33+β22α4)4α2β234β2α3β3+β22α416β33α4α4,

    we have the family of representatives 2+4,α1+3.

    2. α4=0,α30, then we may suppose that β3=0 and

    (a)  if β10,β2=4β1,α2=4α1, then by choosing x=α3,y=α1,z=0, we have the representative 3,1+42;

    (b) if β10,β2=4β1,α24α1, then by choosing x=α24α1α3,y=4α21α1α2α23,z=0, we have the representative 24(2+3),1+42;

    (c) if β10,β24β1, then by choosing x=α3(β24β1),y=β1α2α1β2,z=0, we have the family of representatives 3,1+α2α4, which will be jointed with the case (2a);

    (d) if β1=0, then we have the representative 33,2.

    Summarizing, we have the following distinct orbits:

    1,2+4,1+42,24(2+3),1+λ2,3,1+λ2,4,α1+3,2+4,1+3,4,2,33,2,4,3,4.

    Note that the algebras constructed from the orbits 1+42,24(2+3), 1+λ2,3, 1+α2,4, 2,33 and 2,4 are parts of some families of algebras which found below. Hence, we have the following new algebras:

    N12:e1e1=e2e1e2=e3e1e3=e4e2e2=e5e3e3=e5N4168:e1e1=e2e1e2=e3e1e3=e4e2e2=4e424e5e2e3=24e5Nλ,0170:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e3=e5Nλ,0184:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e3e3=e5Nα13:e1e1=e2e1e2=e3e1e3=αe4e2e2=e5e2e3=e4e3e3=e5N14:e1e1=e2e1e2=e3e1e3=e4e2e3=e4e3e3=e5N176:e1e1=e2e1e2=e3e2e2=e4e2e3=3e5N080:e1e1=e2e1e2=e3e2e2=e4e3e3=e5N15:e1e1=e2e1e2=e3e2e3=e4e3e3=e5

    Here we will collect all information about N304:

    N304e1e1=e2e2e2=e3H2D(N304)=[Δ12],H2C(N304)=H2D(N304)[Δ13],[Δ23],[Δ33]ϕ=(x000x20z0x4)

    Let us use the following notations:

    1=[Δ12],2=[Δ13],3=[Δ23],4=[Δ33].

    Take θ=4i=1αiiH2C(N304). Since

    ϕT(0α1α2α10α3α2α3α4)ϕ=(αα1α2α1αα3α2α3α4),

    we have

    α1=(α1x+α3z)x2,α2=(α2x+α4z)x4,α3=α3x5,α4=α4x8.

    Consider the following cases:

    1. α40, then consider the vector space generated by the following two cocycles:

    θ1=α11+α22+α33+α44  and  θ2=β11+β22+β33.

    Thus, we have

    α1=(α1x+α3z)x2,β1=(β1x+β3z)x2,α2=(α2x+α4z)x4,β2=β2x5,α3=α3x6,β3=β3x6.α4=α4x8.

    Then we consider the following subcases:

    (a) β3=0,α3=0, then we have:

    (i) if β1=0,α1=0, then we have the representative 4,2;

    (ii) if β1=0,α10, then by choosing x=5α1α41, we have the representative 1+4,2;

    (iii) if β10,β2=0, then by choosing x=1 and z=α2α14, we have the representative 4,1;

    (iv)  if β10,β20, then by choosing x=β1β21 and z=α1β2β1α2α4β1β2, we have the representative 4,1+2.

    (b) β3=0,α30, then we have:

    (i) if β2=0, then by choosing x=α3α41 and z=α2α3α34, we have the representative 3+4,1;

    (ii) if β20,β1=0, then by choosing

    x=α3α41 and z=α1α31α14,

    we have the representative 3+4,2;

    (iii) if β20,β10,β2α3=β1α4, then by choosing x=α3α41 and z=0, we have the family of representatives α1+3+4,1+2;

    (iv) if β20,β10,β2α3β1α4, then by choosing

    x=β1β2 and z=(α1β2β1α2)β1(β1α4β2α3)β2,

    we have the family of representatives α3+4,1+2α0,1, which will be jointed with the case (1(a)iv).

    (c) β30,α3=0, then we have:

    (i) if β2=0,α10, then by choosing x=5α1α14 and z=α25α1α64, we have the family of representatives 1+4,α1+3;

    (ii) if β2=0,α1=0, then by choosing z=α2xα4, we have two representatives 4,3 or 4,1+3 depending on whether β1α4=α2β3 or not;

    (iii) if β20, then by choosing x=β2β3 and z=α2β2β3α4, we have the family of representatives α1+4,β1+2+3.

    2. α4=0,α30, then we may suppose that β3=0. Thus, we have

    α1=(α1x+α3z)x2,β1=β1x3,α2=α2x5,β2=β2x5,α3=α3x6,

    and consider the following subcases:

    (a) β2=0, then we have two representatives 3,1 or 2+3,1, depending on whether α2=0 or not;

    (b) β20,α2=0, then by choosing z=α1xα3, we have two representatives 3,2 or 3,1+2, depending on whether β1=0 or not.

    3. α4=0,α3=0,β3=0,β2=0,α20, then we have the representative 2,1.

    Summarizing, we have the following distinct orbits:

    1,2, 1,2+3, 1,3, 1,3+4, 1,4, 1+2,α1+3+4O(α)=O(α), 1+2,3, 1+2,α3+4α1, β1+2+3,α1+4, α1+3,1+4O(α)=O(η3α)=O(η23α), 1+3,4, 1+4,2, 2,3, 2,3+4, 2,4, 3,4.

    Note that, the orbit 1,2 after a change of the basis of the constructed algebra gives a part of the family Nα79, which will be found below. Hence, we have the following new algebras:

    N076:e1e1=e2e1e2=e3e1e4=e5e2e2=e4N16:e1e1=e2e1e2=e4e1e3=e5e2e2=e3e2e3=e5N17:e1e1=e2e1e2=e4e2e2=e3e2e3=e5N18:e1e1=e2e1e2=e4e2e2=e3e2e3=e5e3e3=e5N19:e1e1=e2e1e2=e4e2e2=e3e3e3=e5Nα20:e1e1=e2e1e2=e4+αe5e1e3=e4e2e2=e3e2e3=e5e3e3=e5N21:e1e1=e2e1e2=e4e1e3=e4e2e2=e3e2e3=e5Nα122:e1e1=e2e1e2=e4e1e3=e4e2e2=e3e2e3=αe5e3e3=e5Nα,β23:e1e1=e2e1e2=βe4+αe5e1e3=e4e2e2=e3e2e3=e4e3e3=e5Nα24:e1e1=e2e1e2=αe4+e5e2e2=e3e2e3=e4e3e3=e5N25:e1e1=e2e1e3=e4e2e2=e3e2e3=e4e3e3=e5N26:e1e1=e2e1e2=e4e1e3=e5e2e2=e3e3e3=e4N27:e1e1=e2e1e3=e4e2e2=e3e2e3=e5N28:e1e1=e2e1e3=e4e2e2=e3e2e3=e5e3e3=e5N29:e1e1=e2e1e3=e4e2e2=e3e3e3=e5N30:e1e1=e2e2e2=e3e2e3=e4e3e3=e5

    Here we will collect all information about N402:

    N402e1e1=e2e1e2=e3H2D(N402)=[Δ13],[Δ22],[Δ14],[Δ24],[Δ44]H2C(N402)=H2D(N402)[Δ23],[Δ33],[Δ34]ϕ=(x000qx200w2xqx3re00t)

    Let us use the following notations:

    1=[Δ13],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N402). Since

    ϕT(00α1α20α3α4α5α1α4α6α7α2α5α7α8)ϕ=(ααα1α2αα3α4α5α1α4α6α7α2α5α7α8),

    we have

    α1=(α1x+α4q+α6w+α7e)x3,α2=(α1x+α4q+α6w+α7e)r+(α2x+α5q+α7w+α8e)t,α3=(α3x2+4α4xq+4α6q2)x2,α4=(α4x+2α6q)x4,α5=(α4r+α5t)x2+2(α6r+α7t)xq,α6=α6x6,α7=(α6r+α7t)x3,α8=α6r2+2α7rt+α8t2.

    We interested in (α2,α5,α7,α8)(0,0,0,0) and (α4,α6,α7)(0,0,0). Let us consider the following cases:

    1. α6=0,α7=0, then α40 and we have the following subcases:

    (a) α8=0,α2α4α1α5=0, then we have a split extension;

    (b) α8=0,α2α4α1α50,α3=4α1, then by choosing

    x=4α2α4α1α5,t=α24,r=α4α5,q=α14α2α4α1α5α4,

    we have the representative 2+4;

    (c) α8=0,α2α4α1α50,α34α1, then by choosing

    x=α34α1α4,t=(α34α1)4α24(α2α4α1α5),r=α5(α34α1)4α34(α1α5α2α4),q=4α21α1α3α24,

    we have the representative 2+3+4;

    (d) α80,α3=4α1, then by choosing

    x=α4α8,t=α34α28,q=α1α8,r=α24α5α28,e=α1α5α2α4,

    we have the representative 4+8;

    (e) α80,α34α1, then by choosing

    x=α34α1α4,t=(α34α1)5α24α8,q=4α21α1α3α24,r=α5(α34α1)5α34α8,e=(4α1α3)(α2α4α1α5)α24α8,

    we have the representative 3+4+8.

    2. α6=0,α70, then we have the following subcases:

    (a) α4=0,α3=0, then by choosing

    x=2α27,q=α5α7,e=2α1α7,w=α25+2α1α82α2α7,t=2α7,r=α8,

    we have the representative 7;

    (b) α4=0,α30, then by choosing

    x=1, q=α52α7, e=α1α7, w=α25+2α1α82α2α72α27, t=α3α7, r=α3α82α27,

    we have the representative 3+7;

    (c) α40,α3α272α4α5α7+α24α8=0, then by choosing

    x=α7,t=α4,e=α34α14α7,r=α4α82α7,q=α3α74α4,w=4α1α4α84α2α4α7+α3(α5α7α4α8)4α4α37,

    we have the representative 4+7;

    (d) α40,α3α272α4α5α7+α24α80, then by choosing

    x=α32α4+α5α7α4α82α27,q=α3(α3α272α4α5α7+α24α8)8α24α27,w=(α3α272α4α5α7+α24α8)(4α2α4α74α1α4α8+α3(α5α7+α4α8))8α24α47,e=(4α1α3)(α3α272α4α5α7+α24α8)8α4α37,t=(α3α272α4α5α7+α24α8)24α4α57,r=α8(α3α272α4α5α7+α24α8)28α4α67,

    we have the representative 4+5+7.

    3. α60, then we have the following subcases:

    (a) α6α8α27=0,α5α6α4α7=0,α2α6α1α7=0, then we have a split extension;

    (b) α6α8α27=0,α5α6α4α7=0,α2α6α1α70,α3α6α24=0, then by choosing

    x=1,t=α26α2α6α1α7,q=α42α6,r=α6α7α1α7α2α6,e=0,w=α242α1α6α6,

    we have the representative 2+6;

    (c) α6α8α27=0,α5α6α4α7=0,α2α6α1α70,α3α6α240, then by choosing

    x=α3α6α24α26,t=(α3α6α24)5α36(α2α6α1α7),q=α4α3α6α242α26,r=(α3α6α24)5α7α46(α2α6α1α7),e=0,

    and w=(α242α1α6)α3α6α242α36, we have the representative 2+3+6;

    (d) α6α8α27=0,α5α6α4α70,2α6(α2α6α1α7)=α4(α5α6α4α7), then by choosing

    t=α26α5α6α4α7x4, q=α42α6x, r=α6α7α4α7α5α6x4, e=0, w=α242α1α6α6x,

    we have the representatives 5+6 and 3+5+6 depending on whether α3α6=α24 or not;

    (e) α6α8α27=0,α5α6α4α70,2α6(α2α6α1α7)α4(α5α6α4α7), then by choosing x=2α6(α2α6α1α7)α4(α5α6α4α7)2α26(α5α6α4α7), t=α26α5α6α4α7x4, q=α42α6x, r=α6α7α4α7α5α6x4, e=0, w=α242α1α6α6x, we have the representative 2+α3+5+6;

    (f) α6α8α270,α5α6α4α7=0, then by choosing

    t=α6x3α6α8α27,q=α4x2α6,r=α7x3α6α8α27,e=(α1α7α2α6)xα6α8α27,w=(α242α26+α1α8α2α7α27α6α8)x,

    we have the representatives 6+8 and 3+6+8 depending on whether α3α6=α24 or not.

    (g) α6α8α270,α5α6α4α70, then by choosing

    x=α5α6α4α7α26(α6α8α27),t=(α5α6α4α7)3α26(α27α6α8)2,q=α4(α4α7α5α6)2α6α26(α6α8α27),r=α7(α4α7α5α6)3α36(α27α6α8)2,e=α6(α5α6α4α7)(α4α5α6α24α7+2α6(α2α6+α1α7))2α36(α6α8α27)3,w=α6(α5α6α4α7)(α24α8α4α5α7+2α6(α2α7α1α8))2α36(α6α8α27)3,

    we have the representative α3+5+6+8.

    Summarizing, we have the following distinct orbits

    2+3+4,2+α3+5+6,2+3+6,2+4,2+6,3+4+8,3+5+6,α3+5+6+8,3+6+8,3+7,4+5+7,4+7,4+8,5+6,6+8,7,

    which gives the following new algebras:

    N31:e1e1=e2e1e2=e3e1e4=e5e2e2=e5e2e3=e5Nα32:e1e1=e2e1e2=e3e1e4=e5e2e2=αe5e2e4=e5e3e3=e5N33:e1e1=e2e1e2=e3e1e4=e5e2e2=e5e3e3=e5N34:e1e1=e2e1e2=e3e1e4=e5e2e3=e5N35:e1e1=e2e1e2=e3e1e4=e5e3e3=e5N36:e1e1=e2e1e2=e3e2e2=e5e2e3=e5e4e4=e5N37:e1e1=e2e1e2=e3e2e2=e5e2e4=e5e3e3=e5Nα38:e1e1=e2e1e2=e3e2e2=αe5e2e4=e5e3e3=e5e4e4=e5N39:e1e1=e2e1e2=e3e2e2=e5e3e3=e5e4e4=e5N40:e1e1=e2e1e2=e3e2e2=e5e3e4=e5N41:e1e1=e2e1e2=e3e2e3=e5e2e4=e5e3e4=e5N42:e1e1=e2e1e2=e3e2e3=e5e3e4=e5N43:e1e1=e2e1e2=e3e2e3=e5e4e4=e5N44:e1e1=e2e1e2=e3e2e4=e5e3e3=e5N45:e1e1=e2e1e2=e3e3e3=e5e4e4=e5N46:e1e1=e2e1e2=e3e3e4=e5

    Here we will collect all information about N404:

    N404e1e1=e2e2e2=e3H2D(N404)=[Δ12],[Δ14],[Δ24],[Δ44],H2C(N404)=H2D(N404)[Δ13],[Δ23],[Δ33],[Δ34]ϕ=(x0000x200y0x4rz00t)

    Let us use the following notations:

    1=[Δ12],2=[Δ13],3=[Δ14],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N404). Since

    ϕT(0α1α2α3α10α4α5α2α4α6α7α3α5α7α8)ϕ=(αα1α2α3α1αα4α5α2α4α6α7α3α5α7α8),

    we have

    α1=(α1x+α4y+α5z)x2,α2=(α2x+α6y+α7z)x4,α3=(α2x+α6y+α7z)r+(α3x+α7y+α8z)t,α4=α4x6,α5=(α4r+α5t)x2,α6=α6x8,α7=(α6r+α7t)x4,α8=α6r2+2α7rt+α8t2.

    We interested in (α3,α5,α7,α8)(0,0,0,0) and (α2,α4,α6,α7)(0,0,0,0). Let us consider the following cases:

    1. α6=0,α7=0,α4=0, then α20 and we have the following cases:

    (a) if α8=0,α5=0, then by choosing t=1 and r=α3α2, we have a split extension;

    (b) if α8=0,α50, then by choosing

    x=α2α5, t=α42α25, z=α1α2, r=α32α3α25, y=0,

    we have the representative 2+5;

    (c) if α80,α5=0,α1=0, then by choosing

    x=α3, t=α2α28, z=α3, r=0, y=0,

    we have the representative 2+8;

    (d) if α80,α5=0,α10, then by choosing

    x=α1α21, t=4α51α32α18, z=α1α12α3α18, r=0, y=0,

    we have the representative 1+2+8;

    (e) if α80,α50, then by choosing

    x=α25α2α8,t=α55α22α38,z=α1α5α2α8,r=α45(α1α8α3α5)α32α38,y=0,

    we have the representative 2+5+8.

    2. α6=0,α7=0,α40, then by choosing r=α5α4t,y=α1x+α5zα4, we have α1=α5=0. Now we can suppose that α1=0,α5=0, and we have the following subcases:

    (a) if α8=0,α3=0, then we have a split extension;

    (b) if α8=0,α30,α2=0, then by choosing x=α3, y=0, z=0, r=0, t=α43, we have the representative 3+4;

    (c) if α8=0,α30,α20, then by choosing x=α2α4, y=0, z=0, r=0,t=α52α3α44, we have the representative 2+3+4;

    (d) if α80,α2=0, then by choosing x=1,y=0,z=α3α8,r=0,t=α4α8, we have the representative 4+8;

    (e) if α80,α20, then by choosing x=α3α4,y=0,z=α2α3α4α8,r=0,t=α32α54α8, we have the representative 2+4+8.

    3. α6=0,α70, then by choosing r=α8t2α7,y=(α2α8α3α7)xα27,z=α2xα7, we have α2=α3=α8=0. Now we can suppose that α2=0, α3=0, α8=0, and consider the following cases:

    (a) if α4=0,α5=0,α1=0, then we have the representative 7;

    (b) if α4=0,α5=0,α10, then by choosing x=1α7,t=α1,y=0,z=0,r=0, we have the representative 1+7;

    (c) if α4=0,α50,α1=0, then by choosing x=α5α7,t=1,y=0,z=0,r=0, we have the representative 5+7;

    (d) if α4=0,α50,α10, then by choosing x=α5α7,t=α21α5α7,y=0,z=0,r=0, we have the representative 1+5+7;

    (e) if α40,α5=0,α1=0, then by choosing x=α7,t=α4,y=0,z=0,r=0, we have the representative 4+7;

    (f) if α40,α5=0,α10, then by choosing x=3α1α4,t=3α1α24α37,y=0,z=0,r=0, we have the representative 1+4+7;

    (g) if α40,α50, then by choosing x=α5α7,t=α4α5α27,y=0,z=0,r=0, we have the representative α1+4+5+7.

    4. α60, then by choosing r=α7tα6,y=α2x+α7zα6, we have α2=α7=0. Now we can suppose that α2=0,α7=0, and we have:

    (a) if α8=0,α5=0, then α30 and we have the following subcases:

    (i) α4=0,α1=0, then by choosing x=α3,t=α63α6,y=0,z=0,r=0, we have the representative 3+6;

    (ii) α4=0,α10, then by choosing x=5α1α6,t=5α71α53α26,y=0,z=0,r=0, we have the representative 1+3+6;

    (iii) α40, then by choosing x=α4α6,t=5α74α23α56,y=0,z=0,r=0, we have the representative α1+3+4+6.

    (b) α8=0,α50, then we have the following subcases:

    (i) α4=0,α3=0, then by choosing x=6α5α6,t=1,z=α16α55α6,y=0,r=0, we have the representative 5+6;

    (ii) α4=0,α30, then by choosing x=α3α5,t=α63α75α6,z=α1α3α25,y=0,r=0, we have the representative 3+5+6;

    (iii) α40, then by choosing x=α4α6,t=α34α5α26,z=α1α15α4α16,y=0,r=0, we have the representative α3+4+5+6.

    (c) α80, then we have the following subcases:

    (i) α5=0,α4=0,α1=0, then by choosing x=1,t=α6α8,z=α3α8,y=0,r=0, we have the representative 6+8;

    (ii) α5=0,α4=0,α10, then by choosing

    x=5α1α16,t=10α81α36α58,z=α3α185α1α16,y=0,r=0,

    we have the representative 1+6+8;

    (iii) α5=0,α40, then by choosing

    x=α4α16, t=α24α36α18, z=α3α18α4α16, y=0, r=0,

    we have the representative α1+4+6+8;

    (iv) α50, then by choosing x=4α25α6α8, t=α25α6α38, z=α3α54α6α58, y=0, r=0, we have the representative α1+β4+5+6+8.

    Summarizing, we have the following distinct orbits:

    1+2+8,1+3+6,α1+3+4+6O(α)=O(α),α1+β4+5+6+8O(α,β)=O(α,β)=O(±iα,β),α1+4+6+8O(α)=O(α),1+4+7,α1+4+5+7O(α)=O(α),1+5+7,1+6+8,1+7,2+3+4,2+4+8,2+5,2+5+8,2+8,3+4,α3+4+5+6O(α)=O(α),3+5+6,3+6,4+7,4+8,5+6,5+7,6+8,7.

    Hence, we have the following new algebras:

    N47:e1e1=e2e1e2=e5e1e3=e5e2e2=e3e4e4=e5N48:e1e1=e2e1e2=e5e1e4=e5e2e2=e3e3e3=e5Nα49:e1e1=e2e1e2=αe5e1e4=e5e2e2=e3e2e3=e5e3e3=e5Nα,β50:e1e1=e2e1e2=αe5e2e2=e3e2e3=βe5e2e4=e5e3e3=e5e4e4=e5Nα51:e1e1=e2e1e2=αe5e2e2=e3e2e3=e5e3e3=e5e4e4=e5N52:e1e1=e2e1e2=e5e2e2=e3e2e3=e5e3e4=e5Nα53:e1e1=e2e1e2=αe5e2e2=e3e2e3=e5e2e4=e5e3e4=e5N54:e1e1=e2e1e2=e5e2e2=e3e2e4=e5e3e4=e5N55:e1e1=e2e1e2=e5e2e2=e3e3e3=e5e4e4=e5N56:e1e1=e2e1e2=e5e2e2=e3e3e4=e5N57:e1e1=e2e1e3=e5e1e4=e5e2e2=e3e2e3=e5N58:e1e1=e2e1e3=e5e2e2=e3e2e3=e5e4e4=e5N59:e1e1=e2e1e3=e5e2e2=e3e2e4=e5N60:e1e1=e2e1e3=e5e2e2=e3e2e4=e5e4e4=e5N61:e1e1=e2e1e3=e5e2e2=e3e4e4=e5N62:e1e1=e2e1e4=e5e2e2=e3e2e3=e5Nα63:e1e1=e2e1e4=αe5e2e2=e3e2e3=e5e2e4=e5e3e3=e5N64:e1e1=e2e1e4=e5e2e2=e3e2e4=e5e3e3=e5N65:e1e1=e2e1e4=e5e2e2=e3e3e3=e5N66:e1e1=e2e2e2=e3e2e3=e5e3e4=e5N67:e1e1=e2e2e2=e3e2e3=e5e4e4=e5N68:e1e1=e2e2e2=e3e2e4=e5e3e3=e5N69:e1e1=e2e2e2=e3e2e4=e5e3e4=e5N70:e1e1=e2e2e2=e3e3e3=e5e4e4=e5N71:e1e1=e2e2e2=e3e3e4=e5

    Here we will collect all information about N408:

    N408e1e1=e2e1e2=e3e2e2=e4H2D(N408)=[Δ13],[Δ14]+3[Δ23]H2C(N408)=H2D(N408)[Δ14],[Δ24],[Δ33],[Δ34],[Δ44]ϕ=(x000yx200z2xyx30ty2x2yx4)

    Let us use the following notations:

    1=[Δ13],2=[Δ14]+3[Δ23],3=[Δ14],4=[Δ24],5=[Δ33],6=[Δ34],7=[Δ44].

    Take θ=7i=1αiiH2C(N408). Since

    ϕT(00α1α2+α3003α2α4α13α2α5α6α2+α3α4α6α7)ϕ=(ααα1α2+α3αα3α2α4α13α2α5α6α2+α3α4α6α7),

    we have

    α1=(α1x+3α2y+α5z+α6t)x3+((α2+α3)x+α4y+α6z+α7t)x2y,α2=13(3α2x3+(α4+2α5)x2y+3α6xy2+α7y3)x2,α3=((α2+α3)x+α4y+α6z+α7t)x413(3α2x3+(α4+2α5)x2y+3α6xy2+α7y3)x2,α4=(α4x2+2α6xy+α7y2)x4,α5=(α5x2+2α6xy+α7y2)x4,α6=(α6x+α7y)x6,α7=α7x8.

    We are interested in (α3,α4,α5,α6,α7)(0,0,0,0,0), (α2+α3,α4,α6,α7)(0,0,0,0) and (α1,α2,α5,α6)(0,0,0,0). Let us consider the following cases:

    1. α7=0,α6=0,α5=0,α4=0, then α30, α2+α30 and (α1,α2)(0,0).

    (a)  if α2α34, then by choosing x=4α2+α3,y=α1, we have the representative α2+3α0,14,1;

    (b) if α2=α34, then we have the representatives

    142+3 and 1142+3

    depending on α1=0 or not.

    2. α7=0,α6=0,α5=0,α40, then by choosing y=3α2α4x, we have α2=0. This we can suppose α2=0, which implies α10 and choosing x=α1α14, we have the representative 1+α3+4.

    3. α7=0,α6=0,α50.

    (a)  if α4=0, then α2α3 and choosing

    x=α2+α3α5, y=3α2α3+3α232α25, z=(α2+α3)(2α1α5+12α2α3+3α23)4α35,

    we have the representative 2+5;

    (b) if α40,α4α5,2(α2α5α2α4+α3α5)+α3α4=0, then by choosing

    x=2(α4α5),y=3α3,z=0,t=0,

    we have the representative α4+5α0,1;

    (c)  if α40,α4α5,2(α2α5α2α4+α3α5)+α3α40, then by choosing

    x=2(α2α5α2α4+α3α5)+α3α42(α25α4α5),y=3α3(2(α2α5α2α4+α3α5)+α3α4)2α5(α5α4)2,z=(2α2(α4α5)α3(α4+2α5))(4α1(α4α5)224α2α3(α4α5)+3α23(α4+2α5))8(α4α5)3α25,t=0,

    we have the family of representatives 2+α4+5α0,1;

    (d) if α40,α4=α5, then by choosing y=α2xα5 and z=(α3α5α1α5+3α22)xα25, we have the representatives 4+5 and 3+4+5 depending on whether α3=0 or not. Note that 4+5=2+4+5 and it will be jointed with the family from the case (3c).

    4. if α7=0,α60, then by choosing x=1, y=(α4+2α5)236α2α6α42α56α6,

    z=y2α3α6+2y(α5α4)3α6 and t=x2α1+xy(4α2+α3)+xzα5+y(yα4+zα6)α6),

    we have α1=α2=α3=0. Now we can suppose that α1=0,α2=0,α3=0, and we have the following cases:

    (a) if α4=0,α5=0, then by choosing x=1,y=0,z=0,t=0, we have the representative 6;

    (b)  if α4=0,α50, then by choosing x=4α53α6,y=8α259α26,z=0,t=0, we have the representative 4+145+6;

    (c) if α40, then by choosing x=α4α6,y=0,z=0,t=0, we have the family of representatives 4+α5+6, which will be jointed with the representative from the case (4b).

    5. if α70, then by choosing x=1, y=α6α7, t=2α36+2(α4α5)α6α73α3α273α37 and z=0, we have α3=0,α6=0. Now we can suppose that α3=0,α6=0, and we have the following cases:

    (a) if α50, then by choosing x=α5α17,y=0,z=α21α15α17,t=0, we have the family of representatives α2+β4+5+7;

    (b) if α5=0,α2=0, then α10 and we have the family of representatives 1+α4+7;

    (c) if α5=0,α20, then by choosing x=3α2α17,y=0,z=0,t=0, we have the family of representatives α1+2+β4+7.

    Summarizing all cases we have the following distinct orbits

    1142+3,α1+2+β4+7O(α,β)=O(η3α,η23β)=O(η23α,η3β),1+α3+4O(α)=O(α),1+α4+7O(α)=O(α),α2+3α0,1,2+α4+5,α2+β4+5+7O(α,β)=O(α,β),3+4+5,α4+5α0,1,4+α5+6,6,

    which gives the following new algebras:

    N72:e1e1=e2e1e2=e3e1e3=e5e1e4=34e5e2e2=e4e2e3=34e5Nα,β73:e1e1=e2e1e2=e3e1e3=αe5e1e4=e5e2e2=e4e2e3=3e5e2e4=βe5e4e4=e5Nα74:e1e1=e2e1e2=e3e1e3=e5e1e4=αe5e2e2=e4e2e4=e5Nα75:e1e1=e2e1e2=e3e1e3=e5e2e2=e4e2e4=αe5e4e4=e5Nα0,176:e1e1=e2e1e2=e3e1e4=(1+α)e5e2e2=e4e2e3=3αe5Nα77:e1e1=e2e1e2=e3e1e4=e5e2e2=e4e2e3=3e5e2e4=αe5e3e3=e5Nα,β78:e1e1=e2e1e2=e3e1e4=αe5e2e2=e4e2e3=3αe5e2e4=βe5e3e3=e5e4e4=e5N79:e1e1=e2e1e2=e3e1e4=e5e2e2=e4e2e4=e5e3e3=e5Nα0,180:e1e1=e2e1e2=e3e2e2=e4e2e4=αe5e3e3=e5Nα81:e1e1=e2e1e2=e3e2e2=e4e2e4=e5e3e3=αe5e3e4=e5N82:e1e1=e2e1e2=e3e2e2=e4e3e4=e5

    Here we will collect all information about N409:

    N409e1e1=e2e2e3=e4H2D(N409)=[Δ12],[Δ13],[Δ22],[Δ33]H2C(N409)=H2D(N409)[Δ14],[Δ24],[Δ34],[Δ44]ϕ=(x0000x20000r0t0sx2r)

    Let us use the following notations:

    1=[Δ12],2=[Δ13],3=[Δ14],4=[Δ22],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N409). Since

    ϕT(0α1α2α3α1α40α5α20α6α7α3α5α7α8)ϕ=(αα1α2α3α1α4αα5α1αα6α7α3α5α7α8),

    we have

    α1=(α1x+α5t)x2,α2=(α2x+α7t)r+(α3x+α8t)s,α3=(α3x+α8t)x2r,α4=α4x4,α5=α5x4r,α6=(α6r+α7s)r+(α7r+α8s)s,α7=(α7r+α8s)x2r,α8=α8r2x4.

    We are interested in (α3,α5,α7,α8)(0,0,0,0). Let us consider the following cases:

    1. α8=0,α7=0,α5=0, then α30 and we have

    (a) if α1=0,α4=0,α6=0, then by choosing x=1,r=α3,s=α2,t=0, we have the representative 3;

    (b) if α1=0,α4=0,α60, then by choosing x=α6,r=α3α26,s=α2α26,t=0, we have the representative 3+6;

    (c) if α1=0,α40,α6=0, then by choosing x=α23,r=α3α4,s=α2α4,t=0, we have the representative 3+4;

    (d) if α1=0,α40,α60, then by choosing x=α31α4α6, r=α32α34α6, s=α2α33α34α6, t=0, we have the representative 3+4+6;

    (e) if α10,α4=0,α6=0, then by choosing x=1, r=α1α31, s=α1α2α23, t=0, we have the representative 1+3;

    (f) if α10,α4=0,α60, then by choosing x=3α1α6α23, r=α1α13, s=α1α2α23, t=0, we have the representative 1+3+6;

    (g) if α10,α40, then by choosing

    x=α1α14, r=α1α13, s=α1α2α23, t=0,

    we have the family of representatives 1+3+4+α6.

    2. α8=0,α7=0,α50 and we have

    (a) if α3=0,α2=0,α4=0,α6=0, then by choosing r=1,x=α5,t=α1,s=0, we have the representative 5;

    (b) if α3=0,α2=0,α4=0,α60, then by choosing x=α5α6, r=α55α36,s=0,t=α1α6, we have the representative 5+6;

    (c) if α3=0,α2=0,α40,α6=0, then by choosing x=1,r=α4α51,t=α1α51,s=0, we have the representative 4+5;

    (d) if α3=0,α2=0,α40,α60, then by choosing

    x=4α4α6α25, r=α4α51, t=α14α4α6α65,s=0,

    we have the representative 4+5+6;

    (e) if α3=0,α20,α4=0,α6=0, then by choosing

    r=1,x=3α2α51,t=α13α2α45,s=0,

    we have the representative 2+5;

    (f) if α3=0,α20,α4=0,α60, then by choosing

    x=3α2α51, r=α163α42α15, t=α13α2α45,s=0,

    we have the representative 2+5+6;

    (g) if α3=0,α20,α40, then by choosing

    x=3α2α15,r=α4α51,t=α13α2α45,s=0,

    we have the family of representatives 2+4+5+α6;

    (h) if α30,α4=0,α6=0, then by choosing x=α3α51,r=α3,s=α2,t=α1α3α25, we have the representative 3+5;

    (i) if α30,α4=0,α60, then by choosing

    x=α3α51,r=α43α35α16,s=α2α33α35α16,t=α1α3α25,

    we have the representative 3+5+6;

    (j) if α30,α40, then by choosing x=α3α51,r=α4α51,s=α2α4α13α15,t=α1α3α25, we have the family of representatives 3+4+5+α6.

    3. α8=0,α70, then by choosing x=2α27, t=α3α62α2α7, s=α6, r=2α7, we have α2=α6=0. Now we can suppose that α2=0 and α6=0, then for s=0 and t=0, we have:

    (a) if α1=0,α3=0,α4=0,α5=0, then by choosing r=1,x=1, we have the representative 7;

    (b) if α1=0,α3=0,α4=0,α50, then by choosing x=α7,r=α5α7, we have the representative 5+7;

    (c) if α1=0,α3=0,α40,α5=0, then by choosing x=α7,r=α4, we have the representative 4+7;

    (d) if α1=0,α3=0,α40,α50, then by choosing x=α4α7α35,r=α4α51, we have the representative 4+5+7;

    (e) if α1=0,α30,α5=0, then by choosing r=α3,x=α5, we have the family of representatives 3+α4+7;

    (f) if α1=0,α30,α50, then by choosing x=α3α51,r=α23α15α17, we have the family of representatives 3+α4+5+7;

    (g) if α10,α3=0,α4=0,α5=0, then by choosing x=α1α7,r=α1, we have the representative 1+7;

    (h) if α10,α3=0,α4=0,α50, then by choosing x=3α1α7α25,r=3α21α15α17, we have the representative 1+5+7;

    (i) if α10,α3=0,α40, then by choosing x=α1α41,r=α1α14α17, we have the family of representatives 1+4+α5+7;

    (j) if α10,α30, then by choosing x=α1α7α23,r=α1α31, we have the family of representatives 1+3+α4+β5+7.

    4. α80, then by choosing x=α8,t=α3,s=α7,r=α8, we have α3=α7=0. Now we can suppose that α3=0 and α7=0, then for s=0 and t=0, we have

    (a) if α1=0,α2=0,α4=0,α6=0, then we have the representatives 8 and 5+8, depending on whether α5=0 or not;

    (b) if α1=0,α2=0,α4=0,α60, then we have the representatives 6+8 and 5+6+8, depending on whether α5=0 or not;

    (c) if α1=0,α2=0,α40,α6=0, then we have the representatives 4+8 and 4+5+8, depending on whether α5=0 or not;

    (d) if α1=0,α2=0,α40,α60, then by choosing x=4α6α81,r=α4α81, we have the representative 4+α5+6+8;

    (e) if α1=0,α20,α4=0,α5=0, then we have the representatives 2+8 and 2+6+8, depending on whether α6=0 or not;

    (f) if α1=0,α20,α4=0,α50, then by choosing x=3α2α51,r=α5α81, we have the representative 2+5+α6+8;

    (g) if α1=0,α20,α40, then by choosing x=6α22α14α18, r=α4α81, we have the representative 2+4+α5+β6+8;

    (h) if α10,α2=0,α4=0,α5=0, then we have the representatives 1+8 and 1+6+8 depending on whether α6=0 or not;

    (i) if α10,α2=0,α4=0,α50, then by choosing x=α1α8α25,r=α5α81, we have the representative 1+5+α6+8;

    (j) if α10,α2=0,α40, then by choosing x=α1α41,r=α4α81, we have the representative 1+4+α5+β6+8;

    (k) if α10,α20, then by choosing x=5α22α11α18,r=5α41α32α18, we have the representative 1+2+α4+β5+γ6+8.

    Summarizing, we have the following distinct orbits:

    1+2+α4+β5+γ6+8O(α,β,γ)=O(η5α,η35β,η5γ)=O(η25α,η5β,η25γ)=O(η35α,η45β,η35γ)=O(η45α,η25β,η45γ),1+3,1+3+α4+β5+7,1+3+4+α6,1+3+6,1+4+α5+β6+8O(α,β)=O(α,β),1+4+α5+7O(α,β)=O(α,β),1+5+α6+8,1+5+7,1+6+8,1+7,1+8,2+4+5+α6O(α)=O(η3α)=O(η23α),2+4+α5+β6+8,2+5,2+5+6,2+5+α6+8O(α,β)=O(α,β)=O(α,η23β)=O(α,η23β)=O(α,η3β)=O(α,η3β),2+6+8,2+8,3,3+4,3+4+5+α6,3+α4+5+7,3+4+6,3+5,3+5+6,3+6,4+5,4+5+6,4+α5+6+8O(α)=O(α),4+5+7,4+5+8,4+7,4+8,5,5+6,5+6+8,5+7,5+8,6+8,7,8,

    which gives the following new algebras:

    Nα,β,γ83:e1e1=e2e1e2=e5e1e3=e5e2e2=αe5e2e3=e4e2e4=βe5e3e3=γe5e4e4=e5N84:e1e1=e2e1e2=e5e1e4=e5e2e3=e4Nα,β85:e1e1=e2e1e2=e5e1e4=e5e2e2=αe5e2e3=e4e2e4=βe5e3e4=e5Nα86:e1e1=e2e1e2=e5e1e4=e5e2e2=e5e2e3=e4e3e3=αe5N87:e1e1=e2e1e2=e5e1e4=e5e2e3=e4e3e3=e5Nα,β88:e1e1=e2e1e2=e5e2e2=e5e2e3=e4e2e4=αe5e3e3=βe5e4e4=e5Nα89:e1e1=e2e1e2=e5e2e2=e5e2e3=e4e2e4=αe5e3e4=e5Nα90:e1e1=e2e1e2=e5e2e3=e4e2e4=e5e3e3=αe5e4e4=e5N91:e1e1=e2e1e2=e5e2e3=e4e2e4=e5e3e4=e5N92:e1e1=e2e1e2=e5e2e3=e4e3e3=e5e4e4=e5N93:e1e1=e2e1e2=e5e2e3=e4e3e4=e5N94:e1e1=e2e1e2=e5e2e3=e4e4e4=e5Nα95:e1e1=e2e1e3=e5e2e2=e5e2e3=e4e2e4=e5e3e3=αe5Nα,β96:e1e1=e2e1e3=e5e2e2=e5e2e3=e4e2e4=αe5e3e3=βe5e4e4=e5N97:e1e1=e2e1e3=e5e2e3=e4e2e4=e5N98:e1e1=e2e1e3=e5e2e3=e4e2e4=e5e3e3=e5Nα99:e1e1=e2e1e3=e5e2e3=e4e2e4=e5e3e3=αe5e4e4=e5N100:e1e1=e2e1e3=e5e2e3=e4e3e3=e5e4e4=e5N101:e1e1=e2e1e3=e5e2e3=e4e4e4=e5N102:e1e1=e2e1e4=e5e2e3=e4N103:e1e1=e2e1e4=e5e2e2=e5e2e3=e4Nα104:e1e1=e2e1e4=e5e2e2=e5e2e3=e4e2e4=e5e3e3=αe5Nα105:e1e1=e2e1e4=e5e2e2=αe5e2e3=e4e2e4=e5e3e4=e5N106:e1e1=e2e1e4=e5e2e2=e5e2e3=e4e3e3=e5N107:e1e1=e2e1e4=e5e2e3=e4e2e4=e5N108:e1e1=e2e1e4=e5e2e3=e4e2e4=e5e3e3=e5N109:e1e1=e2e1e4=e5e2e3=e4e3e3=e5N110:e1e1=e2e2e2=e5e2e3=e4e2e4=e5N111:e1e1=e2e2e2=e5e2e3=e4e2e4=e5e3e3=e5Nα112:e1e1=e2e2e2=e5e2e3=e4e2e4=αe5e3e3=e5e4e4=e5N113:e1e1=e2e2e2=e5e2e3=e4e2e4=e5e3e4=e5N114:e1e1=e2e2e2=e5e2e3=e4e2e4=e5e4e4=e5N115:e1e1=e2e2e2=e5e2e3=e4e3e4=e5N116:e1e1=e2e2e2=e5e2e3=e4e4e4=e5N117:e1e1=e2e2e3=e4e2e4=e5N118:e1e1=e2e2e3=e4e2e4=e5e3e3=e5N119:e1e1=e2e2e3=e4e2e4=e5e3e3=e5e4e4=e5N120:e1e1=e2e2e3=e4e2e4=e5e3e4=e5N121:e1e1=e2e2e3=e4e2e4=e5e4e4=e5N122:e1e1=e2e2e3=e4e3e3=e5e4e4=e5N123:e1e1=e2e2e3=e4e3e4=e5N124:e1e1=e2e2e3=e4e4e4=e5

    Here we will collect all information about N410:

    N410e1e1=e2e1e2=e4e3e3=e4H2D(N410)=[Δ13],[Δ14],[Δ22],[Δ23],[Δ33]H2C(N410)=H2D(N410)[Δ24],[Δ34],[Δ44]ϕ=(x000yx2zrx0z0r0tz2+2xysx3),r2=x3

    Let us use the following notations:

    1=[Δ13],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N410). Since

    ϕT(00α1α20α3α4α5α1α4α6α7α2α5α7α8)ϕ=(ααα1α2αα3α4α5α1α4α6+αα7α2α5α7α8),

    we have

    α1=(α3y+α4z+α5t)zrx+(α1x+α4y+α6z+α7t)r+(α2x+α5y+α7z+α8t)s,α2=(α2x+α5y+α7z+α8t)x3,α3=α3x4+2α5x2(z2+2xy)+α8(z2+2xy)2,α4=(α3x2+α5(z2+2xy))zrx+(α4x2+α7(z2+2xy))r+(α5x2+α8(z2+2xy))s,α5=(α5x2+α8(z2+2xy))x3,α6=(α4rα3zrx+α5s)zrx+(α6rα4zrx+α7s)r+(α7rα5zrx+α8s)s(α3y+α4z+α5t)x2(α2x+α5y+α7z+α8t)(z2+2xy),α7=(α7rα5zrx+α8s)x3,α8=α8x6.

    We are interested in (α5,α7,α8)(0,0,0). Let us consider the following cases:

    1. α8=0,α5=0, then α70. Now by choosing

    y=α22+α2α3+α4α72α27x,z=α2α7x,s=3α22α3+α2(α23+6α4α7)+α7(α3α4+2α6α7)4α37x3,t=α27(α242α1α7)+α32α3+α22(α23+3α4α7)+2α2α7(α3α4+α6α7)2α47x,

    we have α1=0,α2=0,α4=0,α6=0. Then we have the representatives 7 or 3+7 depending on whether α3=0 or not.

    2. α8=0,α50, then by choosing

    y=α2α5+α27α25x,z=α7α5x,s=α3α7α4α5α25x3,t=α2α3α5+α25α6+3α4α5α72α3α27α35x,

    we have α2=α4=α6=0 and α7=0. Therefore, we can suppose that α2=0,α4=0,α6=0,α7=0, and have the following cases:

    (a) if α1=0,α3=0, then we have the representative 5;

    (b) if α1=0,α30, then by choosing x=α3α5,r2=x3, we have the representative 3+5;

    (c) if α10, then by choosing x=5α21α25,r2=x3, we have the family of representatives 1+α3+5.

    3. α80, then by choosing y=α5x2+α8z22α8x,s=x(α5zα7x)α8,t=α2x+α5y+α7zα8, we have α2=α5=α7=0. Therefore, we can suppose that α2=0,α5=0,α7=0, and have the following cases:

    (a) if α3=0,α4=0,α6=0, then we have the representative 8 and 1+8 depending on whether α1=0 or not.

    (b) if α3=0,α4=0,α60, then by choosing x=3α6α8,r2=x3,z=α13α26α8, we have the representative 6+8;

    (c) if α3=0,α40, then by choosing x=5α24α28,r2=x3,z=α635α34α28, we have the representative α1+4+8;

    (d) if α30, then by choosing x=α3α8,r2=x3,z=α4α3α8, we have the representative α1+3+β6+8.

    Summarizing, we have the following distinct orbits:

    1+α3+5O(α)=O(η45α)=O(η35α)=O(η25α)=O(η5α),α1+3+β6+8O(α,β)=O(α,β)=O(η3α,η23β)=O(η3α,η23β)=O(η23α,η3β)=O(η23α,η3β),α1+4+8O(α)=O(α)=O(η45α)=O(η45α)=O(η35α)=O(η35α)=O(η25α)=O(η25α)=O(η5α)=O(η5α),1+8,3+5,3+7,5,6+8,7,8,

    which gives the following new algebras:

    Nα125:e1e1=e2e1e2=e4e1e3=e5e2e2=αe5e2e4=e5e3e3=e4Nα,β126:e1e1=e2e1e2=e4e1e3=αe5e2e2=e5e3e3=e4+βe5e4e4=e5Nα127:e1e1=e2e1e2=e4e1e3=αe5e2e3=e5e3e3=e4e4e4=e5N128:e1e1=e2e1e2=e4e1e3=e5e3e3=e4e4e4=e5N129:e1e1=e2e1e2=e4e2e2=e5e2e4=e5e3e3=e4N130:e1e1=e2e1e2=e4e2e2=e5e3e3=e4e3e4=e5N131:e1e1=e2e1e2=e4e2e4=e5e3e3=e4N132:e1e1=e2e1e2=e4e3e3=e4+e5e4e4=e5N133:e1e1=e2e1e2=e4e3e3=e4e3e4=e5N134:e1e1=e2e1e2=e4e3e3=e4e4e4=e5

    Here we will collect all information about N411:

    N411e1e1=e2e1e3=e4e2e2=e4H2D(N411)=[Δ12],[Δ22],[Δ23],[Δ33]H2C(N411)=H2D(N411)[Δ14],[Δ24],[Δ34],[Δ44]ϕ=(x0000x200z0x30t2xzsx4)

    Let us use the following notations:

    1=[Δ12],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N411). Since

    ϕT(0α10α2α1α3α4α50α4α6α7α2α5α7α8)ϕ=(αα1αα2α1α3+αα4α5αα4α6α7α2α5α7α8),

    we have

    α1=(α1x+α4z+α5t)x2+2(α2x+α7z+α8t)xz,α2=(α2x+α7z+α8t)x4,α3=(α3x2+4α5xz+4α8z2)x2(α6z+α7t)x3(α2x+α7z+α8t)s,α4=(α4x+2α7z)x4+(α5x+2α8z)xs,α5=(α5x+2α8z)x5,α6=α6x6+2α7x3s+α8s2,α7=(α7x3+α8s)x4,α8=α8x8.

    We are interested in (α2,α5,α7,α8)(0,0,0,0). Let us consider the following cases:

    1. α8=0,α7=0,α5=0, then α20 and we have

    (a)  if α4=0,α6=0, then by choosing x=2α2,z=α1,s=8α22α3,t=0, we have the representative 2;

    (b)  if α4=0,α60, then by choosing x=α2α6,z=α12α6,s=α1α2α6+2α22α32α36,t=0, we have the representative 2+6;

    (c) if α40,α4=2α2,α10, then by choosing

    x=α1α2,z=0,s=α1α3α1α22α2,t=0,

    we have the family of representatives 1+224+α6;

    (d)  if α40,α4=2α2,α1=0,α60, then by choosing x=α2α16, z=0, s=α22α3α36, t=0, we have the representative 224+6;

    (e)  if α40,α4=2α2,α1=0,α6=0, then by choosing x=α2, z=0, s=α22α3, t=0, we have the representative 224;

    (f) if α40,α42α2,α6=0, then by choosing

    x=α4+2α2,z=α1,s=α3(α4+2α2)3α2,t=0,

    we have the the family of representatives 2+α4α0,2, which will be jointed with the cases (1a) and (1b);

    (g) if α40,α42α2,α60, then by choosing

    x=α2α6,z=α1α2α6(α4+2α2),s=α1α22α6+2α32α3+α22α3α4α36(α4+2α2),t=0,

    we have the family of representatives 2+α4+6α0,2, which will be jointed with the cases (1b) and (1d).

    2. α8=0,α7=0,α50, then we have

    (a)  if α6=0,α2=0, then by choosing

    x=4α5,z=α3,s=64α4α25,t=α3α44α1α5α5

    we have the representative 5;

    (b)  if α6=0,α20, then by choosing

    x=α2α5,z=α22α4+α2α3α54α35,s=α32α4α45,t=(α2α4+2α22)(α2α4+α3α5)4α1α2α254α45,

    we have the representative 2+5;

    (c)  if α60,α6=4α5,α3=0,α2=0, then by choosing x=α5,z=0,s=α4α25,t=α1, we have the representative 5+46;

    (d)  if α60,α6=4α5,α2α4+α3α5=0,α20, then by choosing

    x=α2α5,z=0,s=α32α4α45,t=α1α2α25,

    we have the representative 2+5+46;

    (e) if α60,α6=4α5,α2α4+α3α50, then by choosing

    x=α2α4+α3α5α5,z=0, s=α4(α2α4+α3α5)3α45, t=α1α2α4+α3α5α25,

    we have the family of representatives α2+3+5+46;

    (f) if α60,α64α5,α2=0, then by choosing

    x=α64α5,z=α3, s=α4(4α5α6)3α5, t=4α1α5α1α6α3α4α5,

    we have the family of representatives 5+α6α0,4, which will be jointed with the cases (2a) and (2c);

    (g) if α60,α64α5,α20, then by choosing

    x=α2α5,z=α2(α2α4+α3α5)α25α64α25,s=α4α32α45,t=α2(2α22α4+α3α4α5+α2(α24+2α3α5)α1α5(4α5α6))α35(4α5α6),

    we have the family of representatives 2+5+α6(α0,4), which will be jointed with the cases (2a) and (2c).

    3. α8=0,α70, then by choosing z=α2α7x, s=α62α7x3, t=α2(α64α5)+α3α7α27x, we have α2=α3=α6=0. Now we can suppose that α2=0,α3=0,α6=0 and have the following subcases:

    (a) if α1=0,α4=0,α5=0, then we have the representative 7;

    (b) if α1=0,α4=0,α50, then by choosing x=α5α7,z=0,s=0,t=0, we have the representative 5+7;

    (c) if α1=0,α40, then by choosing x=α4α7,z=0,s=0,t=0, we have the family of representatives 4+α5+7;

    (d) if α10, then by choosing x=4α1α7,z=0,s=0,t=0, we have the family of representatives 1+α4+β5+7.

    4. α80, then by choosing z=α52α8x,s=α7α8x3,t=α5α72α2α82α28x, we have α2=α5=α7=0. Now we can suppose that α2=0,α5=0,α7=0 and have the following subcases:

    (a) if α1=0,α3=0,α4=0,α6=0, then we have the representative 8;

    (b) if α1=0,α3=0,α4=0,α60, then by choosing x=α6α8,z=0,s=0,t=0, we have the representative 6+8;

    (c) if α1=0,α3=0,α40, then by choosing x=3α4α8,z=0,s=0,t=0, we have the family of representatives 4+α6+8;

    (d) if α1=0,α30, then by choosing x=4α3α8,z=0,s=0,t=0, we have the family of representatives 3+α4+β6+8;

    (e) if α10, then by choosing x=5α1α8,z=0,s=0,t=0, we have the family of representatives 1+α3+β4+γ6+8.

    Summarizing, we have the following distinct orbits:

    1+224+α6O(α)=O(α),1+α3+β4+γ6+8O(α,β,γ)=O(η5α,η25β,η35γ)=O(η25α,η45β,η5γ)=O(η35α,η5β,η45γ)=O(η45α,η35β,η25γ),1+α4+β5+7O(α,β)=O(α,β)=O(α,iβ)=O(α,iβ),α2+3+5+46O(α)=O(α),2+α4,2+α4+6,2+5+α6,3+α4+β6+8,4+α5+7,4+α6+8,5+α6,5+7,6+8,7,8,

    which gives the following new algebras:

    Nα135:e1e1=e2e1e2=e5e1e3=e4e1e4=e5e2e2=e4e2e3=2e5e3e3=αe5Nα,β,γ136:e1e1=e2e1e2=e5e1e3=e4e2e2=e4+αe5e2e3=βe5e3e3=γe5e4e4=e5Nα,β137:e1e1=e2e1e2=e5e1e3=e4e2e2=e4e2e3=αe5e2e4=βe5e3e4=e5Nα138:e1e1=e2e1e3=e4e1e4=αe5e2e2=e4+e5e2e4=e5e3e3=4e5Nα139:e1e1=e2e1e3=e4e1e4=e5e2e2=e4e2e3=αe5Nα140:e1e1=e2e1e3=e4e1e4=e5e2e2=e4e2e3=αe5e3e3=e5Nα141:e1e1=e2e1e3=e4e1e4=e5e2e2=e4e2e4=e5e3e3=αe5Nα,β142:e1e1=e2e1e3=e4e2e2=e4+e5e2e3=αe5e3e3=βe5e4e4=e5Nα143:e1e1=e2e1e3=e4e2e2=e4e2e3=e5e2e4=αe5e3e4=e5Nα144:e1e1=e2e1e3=e4e2e2=e4e2e3=e5e3e3=αe5e4e4=e5Nα145:e1e1=e2e1e3=e4e2e2=e4e2e4=e5e3e3=αe5N146:e1e1=e2e1e3=e4e2e2=e4e2e4=e5e3e4=e5N147:e1e1=e2e1e3=e4e2e2=e4e3e3=e5e4e4=e5N148:e1e1=e2e1e3=e4e2e2=e4e3e4=e5N149:e1e1=e2e1e3=e4e2e2=e4e4e4=e5

    Here we will collect all information about N412:

    N412e1e1=e2e2e2=e4e3e3=e4H2D(N412)=[Δ12],[Δ13],[Δ23],[Δ33]H2C(N412)=H2D(N412)[Δ14],[Δ24],[Δ34],[Δ44]ϕ±=(x0000x20000±x20t0sx4)

    Let us use the following notations:

    1=[Δ12],2=[Δ13],3=[Δ14],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N412). Since

    ϕT±(0α1α2α3α10α4α5α2α4α6α7α3α5α7α8)ϕ±=(αα1α2α3α10α4α5α2α4α6α7α3α5α7α8),

    we have

    α1=(α1x+α5t)x2,α2=(α3x+α8t)s±(α2x+α7t)x2,α3=(α3x+α8t)x4,α4=(α5s±α4x2)x2,α5=α5x6,α6=α6x4±2α7sx2+α8s2,α7=(α8s±α7x2)x4,α8=α8x8.

    We will consider only the action of ϕ+ for find representatives and after that we will see that the set of our representatives gives distinct orbits under action of ϕ+ and ϕ. We are interested in (α3,α5,α7,α8)(0,0,0,0). Let us consider the following cases:

    1. α8=0,α5=0,α7=0, then α30 and we have the following subcases:

    (a) if α1=0,α4=0,α6=0, then by choosing x=α3,s=α2α3,t=0, we have the representative 3;

    (b) if α1=0,α4=0,α60, then by choosing x=α6α3,s=α2α26α33,t=0, we have the representative 3+6;

    (c) if α1=0,α40, then by choosing x=α4α3,s=α2α24α33,t=0, we have the representative 3+4+α6;

    (d) if α10, then by choosing x=α1α3,s=α1α2α23,t=0, we have the representative 1+3+α4+β6.

    2. α8=0,α5=0,α70, then we have the following subcases:

    (a) if α1=0,α3=0,α4=0, then by choosing x=1,s=α62α7,t=α3α62α2α72α27, we have the representative 7;

    (b) if α1=0,α3=0,α40, then by choosing x=α4α7,s=α4α62α27,t=α4(α3α62α2α4)2α27α7, we have the representative 4+7;

    (c) if α1=0,α30, then by choosing x=α3α7,s=α23α62α37,t=α23α62α2α3α72α37, we have the representative 3+α4+7;

    (d) if α10, then by choosing x=3α1α7,s=3α21α368α57,t=3α1(α3α62α2α7)2α273α7, we have the representative 1+α3+β4+7.

    3. α8=0,α50, then by choosing t=α1α5x,s=α4α5x2, we have α1=α4=0. Now we can suppose that α1=0,α4=0 and have the following subcases:

    (a) if α2=0,α3=0,α6=0, then we have the family of representatives 5+α7;

    (b) if α2=0,α3=0,α60, then by choosing x=α6α5,s=0,t=0, we have the family of representatives 5+6+α7;

    (c) if α2=0,α30, then by choosing x=α3α5,s=0,t=0, we have the family of representatives 3+5+α6+β7;

    (d) if α20, then by choosing x=3α2α5,s=0,t=0, we have the family of representatives 2+α3+5+β6+γ7.

    4. α80, then by choosing t=α3α8x,s=α7α8x2, we have α3=α7=0. Now we can suppose that α3=0,α7=0 and have the following subcases:

    (a) if α1=0,α2=0,α4=0,α5=0,α6=0, then we have the representative 8;

    (b) if α1=0,α2=0,α4=0,α5=0,α60, then by choosing x=4α6α8,s=0,t=0, we have the representative 6+8;

    (c) if α1=0,α2=0,α4=0,α50, then by choosing x=α5α8,s=0,t=0, we have the family of representatives 5+α6+8;

    (d) if α1=0,α2=0,α40, then by choosing x=4α4α8,s=0,t=0, we have the family of representatives 4+α5+β6+8;

    (e) if α1=0,α20, then by choosing x=5α2α8,s=0,t=0, we have the family of representatives

    2+α4+β5+γ6+8;

    (f) if α10, then by choosing x=5α1α8,s=0,t=0, we have the family of representatives

    1+α2+β4+γ5+μ6+8.

    Summarizing all cases we have the following distinct orbits:

    1+α2+β4+γ5+μ6+8O(α,β,γ,μ)=O(±α,±η45β,η25γ,η45μ)=O(±α,η35β,η45γ,η35μ)=O(±α,±η25β,η5γ,η25μ)=O(±α,η5β,η35γ,η5μ),1+3+α4+β6O(α,β)=O(α,β)=O(α,β)=O(α,β),1+α3+β4+7O(α,β)=O(η3α,η23β)=O(η23α,η3β),2+α3+5+β6+γ7O(α,β,γ)=O(α,β,γ)=O(η3α,η23β,γ)=O(η3α,η23β,γ)=O(η23α,η3β,γ)=O(η23α,η3β,γ),2+α4+β5+γ6+8O(α,β,γ)=O(α,β,γ)=O(η45α,η25β,η45γ)=O(η45α,η25β,η45γ)=O(η35α,η45β,η35γ)=O(η35α,η45β,η35γ)=O(η25α,η5β,η25γ)=O(η25α,η5β,η25γ)=O(η5α,η35β,η5γ)=O(η5α,η35β,η5γ),3,3+4+α6,3+α4+7O(α)=O(α),3+5+α6+β7O(α,β)=O(α,β),3+6,4+α5+β6+8O(α,β)=O(iα,β)=O(iα,β)=O(α,β),4+7,5+6+α7O(α,β)=O(α,β),5+α6+8,5+α7O(α)=O(α),6+8,7,8,

    which gives the following new algebras:

    Nα,β,γ,μ150:e1e1=e2e1e2=e5e1e3=αe5e2e2=e4e2e3=βe5e2e4=γe5e3e3=e4+μe5e4e4=e5Nα,β151:e1e1=e2e1e2=e5e1e4=e5e2e2=e4e2e3=αe5e3e3=e4+βe5Nα,β152:e1e1=e2e1e2=e5e1e4=αe5e2e2=e4e2e3=βe5e3e3=e4e3e4=e5Nα,β,γ153:e1e1=e2e1e3=e5e1e4=αe5e2e2=e4e2e4=e5e3e3=e4+βe5e3e4=γe5Nα,β,γ154:e1e1=e2e1e3=e5e2e2=e4e2e3=αe5e2e4=βe5e3e3=e4+γe5e4e4=e5N155:e1e1=e2e1e4=e5e2e2=e4e3e3=e4Nα156:e1e1=e2e1e4=e5e2e2=e4e2e3=e5e3e3=e4+αe5Nα157:e1e1=e2e1e4=e5e2e2=e4e2e3=αe5e3e3=e4e3e4=e5Nα,β158:e1e1=e2e1e4=e5e2e2=e4e2e4=e5e3e3=e4+αe5e3e4=βe5N159:e1e1=e2e1e4=e5e2e2=e4e3e3=e4+e5Nα,β160:e1e1=e2e2e2=e4e2e3=e5e2e4=αe5e3e3=e4+βe5e4e4=e5N161:e1e1=e2e2e2=e4e2e3=e5e3e3=e4e3e4=e5Nα162:e1e1=e2e2e2=e4e2e4=e5e3e3=e4+e5e3e4=αe5Nα163:e1e1=e2e2e2=e4e2e4=e5e3e3=e4+αe5e4e4=e5Nα164:e1e1=e2e2e2=e4e2e4=e5e3e3=e4e3e4=αe5N165:e1e1=e2e2e2=e4e3e3=e4+e5e4e4=e5N166:e1e1=e2e2e2=e4e3e3=e4e3e4=e5N167:e1e1=e2e2e2=e4e3e3=e4e4e4=e5

    Here we will collect all information about N413(λ):

    N413(λ)e1e1=e2e1e2=e3e1e3=e4e2e2=λe4H2D(N413(2))=[Δ22],4[Δ23]+[Δ14],[Δ24],H2C(N413(2))=H2D(N413(2))[Δ23],[Δ33],[Δ34],[Δ44]H2D(N413(λ)λ2)=[Δ22],(3λ2)[Δ23]+[Δ14],H2C(N413(λ)λ2)=H2D(N413(λ)[Δ23],[Δ24],[Δ33],[Δ34],[Δ44]ϕ=(x000yx200z2xyx30tλy2+2xz(λ+2)x2yx4)

    Let us use the following notations:

    1=[Δ14]+(3λ2)[Δ23],2=[Δ22],3=[Δ23],4=[Δ24],5=[Δ33],6=[Δ34],7=[Δ44].

    Take θ=7i=1αiiH2C(N413(λ)). Since

    ϕT(000α10α2(3λ2)α1+α3α40(3λ2)α1+α3α5α6α1α4α6α7)ϕ=(αααα1αα2+λα(3λ2)α1+α3α4α(3λ2)α1+α3α5α6α1α4α6α7),

    we have

    α1=(α1x+α4y+α6z+α7t)x4,α2=α2x4+4λ(α6y+α7z)xy2+λ2α7y4+4(α4z+(α3+(3λ2)α1)y)x3+2(4α6yz+2α7z2+(2α5+λα4)y2)x2λ((λ+2)(α4y+α6z+α7t)y+((α3+4λα1)y+α5z+α6t)x)x2,α3=[(λ+2)(α4x2+2α6xy+2α7xz+λα7y2)y+((α3+(3λ2)α1)x2+2α5xy+2α6xz+λα6y2)x]x2(3λ2)(α1x+α4y+α6z+α7t)x4,α4=(α4x2+2α6xy+2α7xz+λα7y2)x4,α5=(α5x2+2(λ+2)α6xy+(λ+2)2α7y2)x4,α6=(α6x+(λ+2)α7y)x6,α7=α7x8.

    We are interested in

    (α3,α4,α5,α6,α7)(0,0,0,0,0) and (α1,α4,α6,α7)(0,0,0,0).

    Let us consider the following cases:

    1. α7=0,α6=0,α5=0,α4=0, then α10, α30 and

    (a)  if λ{1,2,4}, (λ4)α34(1λ)(λ2)α1, then by choosing y=α2x(λ4)α3+4(λ1)(λ2)α1, we have the family of representatives

    α1+3α{0,(λ4)4(1λ)(λ2)};λ1,2,4;

    (b) if λ{1,2,4}, (λ4)α3=4(1λ)(λ2)α1, α2=0, then we have the family of representatives λ44(1λ)(λ2)1+3λ1,2,4, which we will be jointed with the family from the case (1a);

    (c) if λ{1,2,4}, (λ4)α3=4(1λ)(λ2)α1, α20, then by choosing x=α2α3,y=0,z=0,t=0, we have the family of representatives (λ4)1+4(1λ)(λ2)(2+3)λ1,2,4;

    (d) if λ{1,2,4}, then by choosing some suitable x and y we have the family of representatives α1+3α0,λ{1,2,4}, which will be jointed with the family from the case (1a).

    2. α7=0,α6=0,α5=0,α40, then we have

    (a) if α3=2(2λ)α1, then by choosing

    x=4α24,y=4α1α4,z=α1α3(4λ)α2α4α21(812λ+3λ2),t=0,

    we have the representative 4;

    (b) if α32(2λ)α1, then by choosing

    x=α3+2(λ2)α1α4,y=α1(α3+2(λ2)α1)α24,z=(2(2λ)α1α3)(α2α4+(λ4)α1α3+(3λ212λ+8)α21)4α34,t=0,

    we have the representative 3+4.

    3. α7=0,α6=0,α50, then

    (a) if α4=0, then α10 and

    (i)  if λ0, then by choosing

    x=α1α5,y=α1α32α25,z=α1(2α2α5+(λ2)α23+4(λ23λ+2)α1α3)2λα35,t=0,

    we have the family of representatives 1+5λ0;

    (ii) if λ=0, then by choosing x=α1α5,y=α1α32α25,z=0,t=0, we have the family representative 1+α2+5α0,λ=0 and the representative 1+5λ=0, which will be jointed with the family from the case (3(a)i).

    (b) if α40 and λ=0, then we have the followings:

    (i) if α3α4=2α1(2α4+α5), then by choosing

    x=4α34, y=4α1α24, z=4α1α3α4α2α244α21(2α4+α5), t=0,

    we have the family of representatives α4+5α0,λ=0;

    (ii) if α3α42α1(2α4+α5), then by choosing

    x=α3α42α1(2α4+α5)α4α5),y=α1(2α1(2α4+α5)α3α4)α24α5),z=(2α1(2α4+α5)α3α4)(α2α244α1α3α4+4α21(2α4+α5))4α44α5,t=0,

    we have the family of representatives 3+α4+5α0,λ=0;

    (c) if α40 and λ0, then we have the followings:

    (i)  if 4α4=λα5, 4λ(λ4)α1α3+λ2α2α5+4(3λ312λ2+8λ+16)α21=0, λα3+2(λ22λ4)α1=0, then by choosing x=1,y=α1α4,z=0,t=0, we have the family of representatives λ44+5λ0;

    (ii)  if 4α4=λα5, 4λ(λ4)α1α3+λ2α2α5+4(3λ312λ2+8λ+16)α21=0, λα3+2(λ22λ4)α10, then by choosing

    x=λα3+2(λ22λ4)α1λα5,y=4α1(λα3+2(λ22λ4)α1)λ2α25,z=0,t=0,

    we have the family of representatives 3+λ44+5λ0;

    (iii) if 4α4=λα5, 4λ(λ4)α1α3+λ2α2α5+4(3λ312λ2+8λ+16)α210, then by choosing

    x=4λ(λ4)α1α3+λ2α2α5+4(3λ312λ2+8λ+16)α21λα5,y=4α14λ(λ4)α1α3+λ2α2α5+4(3λ312λ2+8λ+16)α21λ2α25,z=0,t=0,

    we have the family of representatives 2+α3+λ44+5λ0;

    (iv) if λ0,4α4λα5, then by choosing

    y=α1α4x,z=α2α24+(λ4)α1α3α4+α21(4α5+(3λ212λ+8)α4)α24(4α4λα5)x,t=0,

    we have two families of representatives

    α4+5αλ4 and 3+α4+5αλ4

    depending on α3α42α1α5+2(λ2)α1α4=0 or not. These families will be jointed with representatives from cases (3(c)i) and (3(c)ii).

    4. α7=0,α60, then by choosing y=α42α6x,z=α242α1α62α26x, we have α1=0,α4=0. Since we can suppose that α1=0,α4=0 and

    (a)  if λ0,α3=0, then by choosing t=α2λα6x, we have the representatives 6λ0 and 5+6λ0 depending on α5=0 or not;

    (b)  if λ0,α30, then by choosing x=α3α6,t=α2α3λα6α6, we have the family of representatives 3+α5+6λ0;

    (c) if λ=0,α2=0,α3=0, then we have the representatives 6λ=0 and 5+6λ=0 depending on α5=0 or not, which will be jointed with representatives from the case (4a);

    (d) if λ=0,α2=0,α30, then by choosing x=α3α6,t=0, we have the family of representatives 3+α5+6(λ=0), which will be jointed with the family of representatives from the case (4b);

    (e) if λ=0,α20, then by choosing x=3α2α6,t=0, we have the family of representatives 2+α3+β5+6λ=0.

    5. α70,λ2, then by choosing

    y=α6α7(λ+2)x,z=2(λ+2)2α4α7(λ+4)α262(λ+2)2α27x,t=(λ2+6λ+8)α4α6α72(λ+2)2α1α27(λ+4)α362(λ+2)2α37x,

    we have α1=0,α4=0,α6=0. Now we can suppose that α1=0,α4=0,α6=0 then we have

    (a)  if α3=0,α5=0,α2=0, then we have the representative 7λ2;

    (b)  if α3=0,α5=0,α20, then by choosing x=4α2α17, we have the representative 2+7λ2;

    (c)  if α50, then by choosing x=α5α17, y=α32α5α7, z=α234α35α7, t=0, we have the family of representatives α2+5+7λ2.

    (d)  if α30,α5=0 then by choosing x=3α3α17, we have the family of representatives α2+3+7λ2.

    6. α70,λ=2, then

    (a) if α6=0,α5=0, then by choosing z=y2xα4x2α7,t=xα1+yα4α7, we have α1=0 and α4=0. Now consider the followings:

    (i) if α3=8α1,α2α7α24=0, then we have the representative 7λ=2, which will be jointed with the representative from the case (5a);

    (ii) if α3=8α1,α2α7α240, then by choosing x=4α2α7α24α27,y=0, we have the representative 2+7λ=2, which will be jointed with the representative from the case (5b);

    (iii) if α38α1, then by choosing x=3α38α1α7,y=α2α7α2448α1α76α3α7x, we have the representative 3+7λ=2.

    (b) if α6=0,α50, then by choosing

    x=α5α7,y=8α1α32α5α7,z=α7(α38α1)22α4α254α5α7α5α7,t=α3α42α1(4α4+α5)2α7α5α7,

    we have the family of representatives α2+5+7λ=2, which will be jointed with the representative from the case (5c).

    (c) if α60, then we have the following cases:

    (i)  if α5α7=α26,8α1α7+α4α6=α3α7, then by choosing

    x=α6α7,y=0,z=α4α62α27,t=α6(α4α62α1α7)2α37,

    we have the family of representatives α2+5+6+7λ=2;

    (ii) if α5α7=α26,8α1α7+α4α6α3α7, then by choosing

    x=α6α7,y=α6(α2α7α242α1α6)6α7(α4α6+8α1α7α3α7)),z=y2xα4x2α7α6yα7,t=xα1+yα4+zα6α7,

    we have the family of representatives

    α3+5+6+7α0,λ=2;

    (iii) if α5α7α260, then by choosing

    x=α6α7,y=α6(α4α6+8α1α7α3α7)2α7(α26+α5α7),z=y2xα42α7xα6α7y,t=(α4α62α1α7)x22α6α7y2+2(α26α4α7)xy2α27x,

    we have the family of representatives

    α2+β5+6+7β1,λ=2,

    which will be jointed with the family from the case (6(c)i).

    Summarizing all cases we have the following distinct orbits:

    (λ4)1+4(1λ)(λ2)(2+3)λ{1;2;4},1+α2+5λ=0,α0,α1+3α0,1+5,2+α3+λ44+5O(α)=O(α)λ0,2+α3+β5+6O(α,β)=O(η23α,ηβ)=O(η3α,η23β)λ=0,α2+3+7O(α)=O(η3α)=O(η23α)λ2,α2+β5+6+7λ=2,α2+5+7,2+7,3+4,3+α4+5α0,3+α5+6,α3+5+6+7α0,λ=2,3+7λ=2,4λ2,α4+5α0,5+6,6,7.

    Now we have the following new algebras

    Nλ1;2;4168:e1e1=e2e1e2=e3e1e3=e4e1e4=(λ4)e5e2e2=λe4+4(1λ)(λ2)e5e2e3=λ(λ+2)e5Nα0169:e1e1=e2e1e2=e3e1e3=e4e1e4=e5e2e2=αe5e2e3=2e5e3e3=e5Nλ,α0170:e1e1=e2e1e2=e3e1e3=e4e1e4=αe5e2e2=λe4e2e3=(1+α(3λ2))e5Nλ171:e1e1=e2e1e2=e3e1e3=e4e1e4=e5e2e2=λe4e2e3=(3λ2)e5e3e3=e5Nλ0,α172:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4+e5e2e3=αe5e2e4=λ4e5e3e3=e5Nα,β173:e1e1=e2e1e2=e3e1e3=e4e2e2=e5e2e3=αe5e3e3=βe5e3e4=e5Nλ2,α174:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4+αe5e2e3=e5e4e4=e5Nα,β175:e1e1=e2e1e2=e3e1e3=e4e2e2=2e4+αe5e3e3=βe5e3e4=e5e4e4=e5Nλ,α176:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4+αe5e3e3=e5e4e4=e5Nλ177:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4+e5e4e4=e5Nλ178:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e3=e5e2e4=e5Nλ,α0179:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e3=e5e2e4=αe5e3e3=e5Nλ,α180:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e3=e5e3e3=αe5e3e4=e5Nα0181:e1e1=e2e1e2=e3e1e3=e4e2e2=2e4e2e3=αe5e3e3=e5e3e4=e5e4e4=e5N182:e1e1=e2e1e2=e3e1e3=e4e2e2=2e4e2e3=e5e4e4=e5Nλ2183:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e4=e5Nλ,α0184:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e2e4=αe5e3e3=e5Nλ185:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e3e3=e5e3e4=e5Nλ186:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e3e4=e5Nλ187:e1e1=e2e1e2=e3e1e3=e4e2e2=λe4e4e4=e5

    Here we will collect all information about N414:

    N414e1e2=e3e1e3=e4H2D(N414)=[Δ11],[Δ22],[Δ23],[Δ33]H2C(N414)=H2D(N414)[Δ14],[Δ24],[Δ34],[Δ44]ϕ=(x0000q000rxq0tsxrx2q)

    Let us use the following notations:

    1=[Δ11],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N414). Since

    ϕT(α100α20α3α4α50α4α6α7α2α5α7α8)ϕ=(α1ααα2αα3α4α5αα4α6α7α2α5α7α8),

    we have

    α1=α1x2+2α2xt+α8t2,α2=(α2x+α8t)x2q,α3=(α3q+α4r+α5s)q+(α4q+α6r+α7s)r+(α5q+α7r+α8s)s,α4=(α4q+α6r+α7s)xq+(α5q+α7r+α8s)xr,α5=(α5q+α7r+α8s)x2q,α6=(α6q2+2α7qr+α8r2)x2,α7=(α7q+α8r)x3q,α8=α8x4q2.

    We are interested in (α2,α5,α7,α8)(0,0,0,0). Let us consider the following cases:

    1. α8=0,α7=0,α5=0, then α20 and we have

    (a) if α6=0,α4=0,α3=0, then by choosing x=2α2,q=1,r=0,s=0,t=α1, we have the representative 2;

    (b) if α6=0,α4=0,α30, then by choosing x=α3,q=α2α23,r=0,s=0,t=α1α32α2, we have the representative 2+3;

    (c) if α6=0,α40, then by choosing x=α4,q=α2α4,r=α2α32,s=0,t=α1α42α2, we have the representative 2+4;

    (d) if α60,α3α6α24=0, then by choosing x=α6,q=α2,r=α2α4α6,s=0,t=α1α62α2, we have the representative 2+6;

    (e) if α60,α3α6α240, then by choosing

    x=α3α6α24α6,q=α2α3α6α24α26,r=α2α4α3α6α24α36,s=0,t=α1α3α6α242α2α6,

    we have the representative 2+3+6.

    2. α8=0,α7=0,α50, then we have

    (a)  if α6=0,α2=0, then by choosing

    x=1,r=α4α5q,s=2α24α3α52α25q,t=0,

    we have the representatives 5 and 1+5 depending on whether α1=0 or not;

    (b)  if α6=0,α20, then by choosing

    x=α5,q=α2,r=α2α4α5,s=α2(2α24α3α5)2α25,t=α1α52α2,

    we have the representatives 2+5;

    (c) if α60,α5=α6, then we have the following subcases:

    (i)  if α2=0,α4=0,α1=0, then we have the representative 56;

    (ii)  if α2=0,α4=0,α10, then by choosing

    x=1,q=α1α5,r=0,s=α3α12α5α5,t=0,

    we have the representative 1+56;

    (iii) if α2=0,α40,α1=0, then by choosing

    x=α4α5,q=1,r=0,s=α32α5,t=0,

    we have the representative 4+56;

    (iv) if α2=0,α40,α10, then by choosing

    x=α4α5,q=α1α5,r=0,s=α3α12α5α5,t=0,

    we have the representative 1+4+56;

    (v)  if α20,α4=0, then by choosing

    x=α5,q=α2,r=0,s=α2α32α5,t=α1α52α2,

    we have the representative 2+56;

    (vi) if α20,α40, then by choosing

    x=α4α5,q=α2α4α25,r=0,s=α2α3α42α35,t=α1α42α2α5,

    we have the representative 2+4+56.

    (d) if α60,α5α6, then we have the following subcases:

    (i) if α2=0,α1=0, then by choosing

    x=1,q=1,s=α24(2α5+α6)α3(α5+α6)22α5(α5+α6)2,r=α4α5+α6,t=0,

    we have the family of representatives 5+α6α0,1, which will be jointed with the cases (2a) and (2(c)i);

    (ii) if α2=0,α10, then by choosing

    x=1,q=α1α5,r=α4α1(α5+α6)α5,s=(α24(2α5+α6)α3(α5+α6)2)α12α5(α5+α6)2α5,t=0,

    we have the family of representatives 1+5+α6α0,1, which will be jointed with the cases (2a) and (2(c)ii);

    (iii) if α20, then by choosing

    x=α5,q=α2, r=α2α4α5+α6, s=α2(α24(2α5+α6)α3(α5+α6)2)2α5(α5+α6)2,t=α1α52α2,

    we have the family of representatives 2+5+α6α0,1, which will be jointed with the cases (2b) and (2(c)v).

    3. α8=0,α70 then by choosing r=α5α7q,s=α5α6α4α7α27q, we have α4=α5=0. Therefore, we can suppose that α4=0,α5=0, thus we have

    (a) if α2=0,α1=0,α3=0, then we have the representatives 7 and 6+7 depending on whether α6=0 or not;

    (b) if α2=0,α1=0,α30, then by choosing x=3α3α7,q=1,r=0,s=0,t=0, we have the family of representatives 3+α6+7;

    (c) if α2=0,α10,α3=0, then we have the representatives 1+7 and 1+6+7 depending on whether α6=0 or not;

    (d) if α2=0,α10,α30, then by choosing

    x=3α3α7,q=6α31α3α27,r=0,s=0,t=0,

    we have the family of representatives 1+3+α6+7;

    (e) if α20,α3=0, then by choosing q=α2α7,r=0,s=0,t=α12α2x, we have the representatives 2+7 and 2+6+7 depending on whether α6=0 or not;

    (f) if α20,α30, then by choosing x=3α3α7,q=α2α7,r=0,s=0,t=α13α32α23α7 we have the family of representatives 2+3+α6+7.

    4. α80 then by choosing r=α7α8q,s=α27α5α8α28q,t=α2α8x, we have α2=α5=α7=0. Therefore, we can suppose that α2=0,α5=0,α7=0, then we have

    (a) if α1=0,α3=0,α4=0, then we have the representatives 8 and 6+8 depending on whether α6=0 or not;

    (b) if α1=0,α3=0,α40, then by choosing x=3α4α8,q=1,r=0,s=0,t=0, we have the family of representatives 4+α6+8;

    (c) if α1=0,α30, then by choosing x=4α3α8,q=1,r=0,s=0,t=0, we have the family of representatives 3+α4+β6+8;

    (d) if α10,α3=0,α4=0, then we have the representatives 1+8 and 1+6+8 depending on whether α6=0 or not;

    (e) if α10,α3=0,α40, then by choosing x=3α4α8,q=6α31α24α8,r=0,s=0,t=0, we have the family of representative 1+4+α6+8;

    (f) if α10,α30, then by choosing x=4α3α8,q=α14α3α8,r=0,s=0,t=0, we have the family of representative 1+3+α4+β6+8.

    Summarizing, we have the following distinct orbits:

    1+3+α4+β6+8O(α,β)=O(iα,β)=O(iα,β)=O(α,β),1+3+α6+7O(α)=O(η3α)=O(η23α),1+4+56,1+4+α6+8O(α)=O(η3α)=O(η23α),1+5+α6,1+6+7,1+6+8,1+7,1+8,2,2+3,2+3+6,2+3+α6+7O(α)=O(η3α)=O(η23α),2+4,2+4+56,2+5+α6,2+6,2+6+7,2+7,3+α4+β6+8O(α,β)=O(iα,β)=O(iα,β)=O(α,β),3+α6+7O(α)=O(η3α)=O(η23α),4+56,4+α6+8O(α)=O(η3α)=O(η23α),5+α6,6+7,6+8,7,8,

    which gives the following new algebras:

    Nα,β188:e1e1=e5e1e2=e3e1e3=e4e2e2=e5e2e3=αe5e3e3=βe5e4e4=e5Nα189:e1e1=e5e1e2=e3e1e3=e4e2e2=e5e3e3=αe5e3e4=e5N190:e1e1=e5e1e2=e3e1e3=e4e2e3=e5e2e4=e5e3e3=e5Nα191:e1e1=e5e1e2=e3e1e3=e4e2e3=e5e3e3=αe5e4e4=e5Nα191:e1e1=e5e1e2=e3e1e3=e4e2e4=e5e3e3=αe5N192:e1e1=e5e1e2=e3e1e3=e4e3e3=e5e3e4=e5N193:e1e1=e5e1e2=e3e1e3=e4e3e3=e5e4e4=e5N194:e1e1=e5e1e2=e3e1e3=e4e3e4=e5N195:e1e1=e5e1e2=e3e1e3=e4e4e4=e5N196:e1e2=e3e1e3=e4e1e4=e5N197:e1e2=e3e1e3=e4e1e4=e5e2e2=e5N198:e1e2=e3e1e3=e4e1e4=e5e2e2=e5e3e3=e5Nα199:e1e2=e3e1e3=e4e1e4=e5e2e2=e5e3e3=αe5e3e4=e5N200:e1e2=e3e1e3=e4e1e4=e5e2e3=e5N201:e1e2=e3e1e3=e4e1e4=e5e2e3=e5e2e4=e5e3e3=e5Nα202:e1e2=e3e1e3=e4e1e4=e5e2e4=e5e3e3=αe5N203:e1e2=e3e1e3=e4e1e4=e5e3e3=e5N204:e1e2=e3e1e3=e4e1e4=e5e3e3=e5e3e4=e5N205:e1e2=e3e1e3=e4e1e4=e5e3e4=e5Nα,β206:e1e2=e3e1e3=e4e2e2=e5e2e3=αe5e3e3=βe5e4e4=e5Nα207:e1e2=e3e1e3=e4e2e2=e5e3e3=αe5e3e4=e5N208:e1e2=e3e1e3=e4e2e3=e5e2e4=e5e3e3=e5Nα209:e1e2=e3e1e3=e4e2e3=e5e3e3=αe5e4e4=e5Nα210:e1e2=e3e1e3=e4e2e4=e5e3e3=αe5N211:e1e2=e3e1e3=e4e3e3=e5e3e4=e5N212:e1e2=e3e1e3=e4e3e3=e5e4e4=e5N213:e1e2=e3e1e3=e4e3e4=e5N214:e1e2=e3e1e3=e4e4e4=e5

    Here we will collect all information about N415:

    N415e1e2=e3e1e3=e4e2e2=e4H2D(N415)=[Δ11],[Δ22],[Δ23],[Δ33]H2C(N415)=H2D(N415)[Δ14],[Δ24],[Δ34],[Δ44]ϕ=(x0000x2000rx30tsxrx4)

    Let us use the following notations:

    1=[Δ11],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N415). Since

    ϕT(α100α20α3α4α50α4α6α7α2α5α7α8)ϕ=(α1ααα2αα3+αα4α5αα4α6α7α2α5α7α8),

    we have

    α1=α1x2+2α2xt+α8t2,α2=(α2x+α8t)x4,α3=x4α3+2rx2α4+2sx2α5+r2α6+2rsα7+s2α8x(rxα2+tx2α7+rtα8),α4=(α4x2+α6r+α7s)x3+(α5x2+α7r+α8s)xr,α5=(α5x2+α7r+α8s)x4,α6=(α6x4+2α7x2r+α8r2)x2,α7=(α7x2+α8r)x5,α8=α8x8.

    We are interested in (α2,α5,α7,α8)(0,0,0,0). Let us consider the following cases:

    1. α8=0,α7=0,α5=0, then α20 and we have

    (a)  if α6=0,α4=0, then by choosing x=2α2,r=4α2α3,s=0,t=α1, we have the representative 2;

    (b)  if α6=0,α40,α2=2α4, then we have the representatives 22+4 and 22+3+4 depending on whether α3=0 or not;

    (c) if α6=0,α40,α22α4, then by choosing x=α22α4,r=α3(α22α4),s=0,t=α1(2α4α2)2α2, we have the family of representatives 2+α4α0,12, which will be jointed with representatives from the cases (1a) and (1b);

    (d) if α60, then by choosing x=α2α6,r=α22α4α36,s=0,t=α12α6, we have the representative 2+α3+6.

    2. α8=0,α7=0,α50 then we have

    (a) if α5α6, then we have the following subcases:

    (i)  if α2=0,α1=0, then by choosing

    x=2α5(α5+α6),s=2α5(α24(2α5+α6)α3(α5+α6)2),r=4α4α25(α5+α6),

    we have the family of representatives 5+α6α1;

    (ii)  if α2=0,α10, then by choosing

    x=4α1α5, r=α4α1(α5+α6)α5, s=((α5+α6)(2α24α2α4)α3(α5+α6)2α24α6)α12α5(α5+α6)2α5,t=0,

    we have the family of representatives 1+5+α6α1;

    (iii)  if α20, then by choosing

    x=α2α5,r=α22α4α25(α5+α6),s=α22((α5+α6)(2α24α2α4)α3(α5+α6)2α24α6)2α35(α5+α6)2,t=α12α5,

    we have the family of representatives 2+5+α6α1.

    (b) if α6=α5, then we have the following subcases:

    (i) if α4=0,α2=0,α1=0, then we have the representative 56, which will be jointed with the family from the case (2(a)i);

    (ii) if α4=0,α2=0,α10, then by choosing x=4α1α5,r=0,s=α3α12α5α4,t=0, we have the representative 1+56, which will be jointed with the family from the case (2(a)ii);

    (iii) if α4=0,α20, then by choosing x=α2α5,r=0,s=α22α32α35,t=α12α5, we have the representative 2+56, which will be jointed with the family from the case (2(a)iii);

    (iv) if α40, then by choosing x=α4α5, s=α3α242α35, r=0, we have the families of representatives

    α1+4+56 and α2+4+56α0

    depending on α2=0 or not.

    3. α8=0,α70, then by choosing

    r=α5α7x2,s=α5α6α4α7α27x2,t=α3α272α4α5α7+α25α6+α2α5α7α37x,

    we have α3=α4=α5=0. Therefore, we can suppose that α3=0,α4=0,α5=0, and we have

    (a) if α1=0,α2=0, then we have the representatives 7 and 6+7 depending on whether α6=0 or not;

    (b) if α1=0,α20, then by choosing x=α2α17,r=0,s=0,t=0, we have the family of representative 2+α6+7;

    (c) if α10, then by choosing x=5α1α7,r=0,s=0,t=0, we have the family of representative 1+α2+β6+7.

    4. α80, then by choosing r=α7α8x2,t=α2α8x,s=α5x2+α7rα8 we have α2=α5=α7=0. Therefore, we can suppose that α2=0,α5=0,α7=0, then we have

    (a) if α1=0,α3=0,α4=0, then we have the representatives 8 and 6+8 depending on whether α6=0 or not;

    (b) if α1=0,α3=0,α40, then by choosing x=3α4α18,r=0,s=0,t=0, we have the family of representative 4+α6+8;

    (c) if α1=0,α30, then by choosing x=4α3α18,r=0,s=0,t=0, we have the family of representative 3+α4+β6+8;

    (d) if α10, then by choosing x=6α1α18,r=0,s=0,t=0, we have the family of representative 1+α3+β4+γ6+8.

    Summarizing all cases we have the following distinct orbits:

    1+α2+β6+7O(α,β)=O(η25α,η5β)=O(η45α,η25β)=O(η5α,η35β)=O(η35α,η45β),1+α3+β4+γ6+8O(α,β,γ)=O(η3α,β,η23γ)=O(η3α,β,η23γ)=O(η23α,β,η3γ)=O(η23α,β,η3γ)=O(α,β,γ),α1+4+56,1+5+α6,22+3+4,2+α3+6,2+α4,α2+4+56α0,2+5+α6,2+α6+7O(α)=O(α),3+α4+β6+8O(α,β)=O(iα,β)=O(iα,β)=O(α,β),4+α6+8O(α)=O(η3α)=O(η23α),5+α6,6+7,6+8,7,8,

    which gives the following new algebras:

    Nα,β215:e1e1=e5e1e2=e3e1e3=e4e1e4=αe5e2e2=e4e3e3=βe5e3e4=e5Nα,β,γ216:e1e1=e5e1e2=e3e1e3=e4e2e2=e4+αe5e2e3=βe5e3e3=γe5e4e4=e5Nα217:e1e1=αe5e1e2=e3e1e3=e4e2e2=e4e2e3=e5e2e4=e5e3e3=e5Nα218:e1e1=e5e1e2=e3e1e3=e4e2e2=e4e2e4=e5e3e3=αe5N219:e1e2=e3e1e3=e4e1e4=2e5e2e2=e4+e5e2e3=e5Nα220:e1e2=e3e1e3=e4e1e4=e5e2e2=e4+αe5e3e3=e5Nα221:e1e2=e3e1e3=e4e1e4=e5e2e2=e4e2e3=αe5Nα0222:e1e2=e3e1e3=e4e1e4=αe5e2e2=e4e2e3=e5e2e4=e5e3e3=e5Nα223:e1e2=e3e1e3=e4e1e4=e5e2e2=e4e2e4=e5e3e3=αe5Nα224:e1e2=e3e1e3=e4e1e4=e5e2e2=e4e3e3=αe5e3e4=e5Nα,β225:e1e2=e3e1e3=e4e2e2=e4+e5e2e3=αe5e3e3=βe5e4e4=e5Nα226:e1e2=e3e1e3=e4e2e2=e4e2e3=e5e3e3=αe5e4e4=e5Nα227:e1e2=e3e1e3=e4e2e2=e4e2e4=e5e3e3=αe5N228:e1e2=e3e1e3=e4e2e2=e4e3e3=e5e3e4=e5N229:e1e2=e3e1e3=e4e2e2=e4e3e3=e5e4e4=e5N230:e1e2=e3e1e3=e4e2e2=e4e3e4=e5N231:e1e2=e3e1e3=e4e2e2=e4e4e4=e5

    Here we will collect all information about N416:

    N419 e1e1=e4e1e2=e3e2e2=e4e3e3=e4 H2D(N419)=[Δ11],[Δ13],[Δ22],[Δ23]H2C(N419)=H2D(N419)[Δ14],[Δ24],[Δ34],[Δ44]
    ϕ1=(x0000q0000xq0ts01),x2=1,q2=1ϕ2=(0p00y00000yp0ts01),y2=1,p2=1

     | Show Table
    DownLoad: CSV

    Let us use the following notations:

    1=[Δ11],2=[Δ14],3=[Δ22],4=[Δ23],5=[Δ24],6=[Δ33],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N416). Since

    ϕT(α100α20α3α4α50α4α6α7α2α5α7α8)ϕ=(α1ααα2αα3α4+αα5αα4+αα6α7α2α5α7α8),

    in the case ϕ=ϕ1, we have

    α1=α1x2+2α2xt+α8t2,α2=(α2x+α8t)x3,α3=α3x2+2α5xs+α8s2,α4=(α4x+α7s)x2α7x2t,α5=(α5x+α8s)x3,α6=α6x4,α7=α7x5,α8=α8x6;

    and on the opposite case, for ϕ=ϕ2, we have

    α1=α3y2+2α5ty+α8t2,α2=(α5y+α8t)y3,α3=α1y2+2α2sy+α8s2,α4=((st)α7yα4)y2,α5=(yα2+sα8)y3,α6=α6y4,α7=α7y5,α8=α8y6.

    We are interested in (α2,α5,α7,α8)(0,0,0,0). Let us consider the following cases:

    1.  α8=0,α7=0,α5=0, then α20 and

    (a) if α40, then by choosing ϕ=ϕ1, x=α4α12, t=α1α42α22, we have the family of representatives 2+α3+4+β6;

    (b) if α4=0,α30, then by choosing ϕ=ϕ1, x=α3α12, t=α1α32α32, we have the family of representatives 2+3+α6;

    (c) if α4=0,α3=0, then by choosing ϕ=ϕ1, x=2α2, t=α1, s=0, we have the family of representatives 2+α6.

    2. α8=0,α7=0,α50 and

    (a) if α20,α40, then by choosing

    ϕ=ϕ1, x=α4α5, t=α1α42α2α5, s=α3α42α25,

    we have the following family of representatives

    α2+4+5+β6α0;

    (b) if α20,α4=0, then by choosing

    ϕ=ϕ1, x=2α2α5, t=α1α5, s=α2α3,

    we have the following family of representatives α2+5+β6α0;

    (c) if α2=0, then by choosing ϕ=ϕ2, y=1, t=0, s=0, we have the representative with α5=0 and α20, which was considered above.

    3. α8=0,α70, then we have

    (a) if α2=0,α5=0,α1=0,α3=0, then we have the representatives 7 and 6+7 depending on whether α6=0 or not;

    (b) if α2=0,α5=0,α10, then by choosing

    ϕ=ϕ1, x=3α1α17,s=0,t=α43α1α17,

    we have the family of representatives 1+α3+β6+7;

    (c) if α20, then by choosing

    ϕ=ϕ1, x=α2α17,s=(α1α7+2α2α4)/(2α27),t=α1/(2α7),

    we have the family of representatives 2+α3+β5+γ6+7.

    4. α80, then by choosing ϕ=ϕ1, t=α2α8x,s=α5α8x, we have α2=α5=0. Now we can suppose that α2=0,α5=0 and we have

    (a) if α1=0,α3=0,α4=0,α6=0, then we have the representatives 8 and 7+8 depending on whether α7=0 or not;

    (b) if α1=0,α3=0,α4=0,α60, then by choosing ϕ=ϕ1, x=α6α18, s=0, t=0, we have the family of representative 6+α7+8;

    (c) if α1=0,α3=0,α40, then by choosing ϕ=ϕ1, x=3α4α18,s=0,t=0, we have the family of representatives 4+α6+β7+8;

    (d) if α10, then by choosing ϕ=ϕ1, x=4α1α18,s=0,t=0, we have the family of representatives 1+α3+β4+γ6+μ7+8.

    Summarizing, we have the following distinct orbits:

    1+α3+β4+γ6+μ7+8O(α,β,γ,μ)=O(α,iβ,γ,iμ)=O(α,iβ,γ,iμ)=O(α,β,γ,μ)=O(1α,β4α3,γα,μ4α)=O(1α,iβ4α3,γα,iμ4α)=O(1α,β4α3,γα,μ4α)=O(1α,β4α3,γα,μ4α),1+α3+β6+7O(α,β)=O(α,η3β)=O(α,η23β)=O(α1,η3β3α1)=O(α1,η23β3α1)=O(α1,β3α1),2+α3+4+β6,2+α3+β5+γ6+7O(α,β,γ)=O(αβ4,1β,γβ),2+3+α6,α2+4+5+β6O(α,β)=O(α1,βα1)α0,α2+5+β6O(α,β)=O(α1,βα1)α0,2+α6,4+α6+β7+8O(α,β)=O(η23α,η3β)=O(η3α,η23β)=O(η23α,η3β)=O(η3α,η23β)=O(α,β),6+7,6+α7+8O(α)=O(α),7,7+8,8,

    which gives the following new algebras:

    Nα,β,γ,μ232:e1e1=e5e1e2=e3e1e3=e4e2e2=αe5e2e3=e4+βe5e3e3=γe5e3e4=μe5e4e4=e5Nα,β233:e1e1=e5e1e2=e3e1e3=e4e2e2=αe5e2e3=e4e3e3=βe5e3e4=e5Nα,β234:e1e2=e3e1e3=e4e1e4=e5e2e2=αe5e2e3=e4+e5e3e3=βe5Nα,β,γ235:e1e2=e3e1e3=e4e1e4=e5e2e2=αe5e2e3=e4e2e4=βe5e3e3=γe5e3e4=e5Nα236:e1e2=e3e1e3=e4e1e4=e5e2e2=e5e2e3=e4e3e3=αe5Nα0,β237:e1e2=e3e1e3=e4e1e4=αe5e2e3=e4+e5e2e4=e5e3e3=βe5Nα0,β238:e1e2=e3e1e3=e4e1e4=αe5e2e3=e4e2e4=e5e3e3=βe5Nα239:e1e2=e3e1e3=e4e1e4=e5e2e3=e4e3e3=αe5Nα,β240:e1e2=e3e1e3=e4e2e3=e4+e5e3e3=αe5e3e4=βe5e4e4=e5N241:e1e2=e3e1e3=e4e2e3=e4e3e3=e5e3e4=e5Nα242:e1e2=e3e1e3=e4e2e3=e4e3e3=e5e3e4=αe5e4e4=e5N243:e1e2=e3e1e3=e4e2e3=e4e3e4=e5N244:e1e2=e3e1e3=e4e2e3=e4e3e4=e5e4e4=e5N245:e1e2=e3e1e3=e4e2e3=e4e4e4=e5

    Here we will collect all information about N417:

    N419 e1e1=e4e1e2=e3e2e2=e4e3e3=e4 H2D(N419)=[Δ11],[Δ13],[Δ22],[Δ23]H2C(N419)=H2D(N419)[Δ14],[Δ24],[Δ34],[Δ44]
    ϕ1=(x0000q0000xq0ts01),x2=1,q2=1ϕ2=(0p00y00000yp0ts01),y2=1,p2=1

     | Show Table
    DownLoad: CSV

    Let us use the following notations:

    1=[Δ11],2=[Δ13],3=[Δ14],4=[Δ22],5=[Δ23],6=[Δ24],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N417). Since

    ϕT(α10α2α30α4α5α6α2α50α7α3α6α7α8)ϕ=(α1αα2α3αα4α5α6α2α50α7α3α6α7α8),

    then in the case ϕ=ϕ1, we have

    α1=α1x2+2α3xt+α8t2,α2=(α2x+α7t)xq,α3=(α3x+α8t)x2q2,α4=α4q2+2α6qs+α8s2,α5=(α5q+α7s)xq,α6=(α6q+α8s)x2q2,α7=α7x3q3,α8=α8x4q4;

    and in the opposite case ϕ=ϕ2, we have

    α1=α4p2+2α6pt+α8t2,α2=(α5p+α7t)py,α3=(α6p+α8t)p2y2,α4=α1y2+2α3sy+α8s2,α5=(α2y+α7s)py,α6=(α3y+α8s)p2y2,α7=α7p3y3,α8=α8p4y4.

    We are interested in (α3,α6,α7,α8)(0,0,0,0). Let us consider the following cases:

    1. α8=0,α7=0,α6=0, then α30 and choosing ϕ=ϕ1, t=α12α3x, we get α1=0. Now consider the following subcases:

    (a) if α2=0,α4=0,α5=0, then we have the representative 3;

    (b) if α2=0,α4=0,α50, then by choosing ϕ=ϕ1, x=α5α3,q=1,s=0,t=α1α52α3α3, we have the representative 3+5;

    (c) if α2=0,α40, then by choosing ϕ=ϕ1, x=3α4α3,q=1,s=0,t=α13α52α33α3, we have the representative 3+4+α5;

    (d) if α20,α4=0,α5=0, then by choosing ϕ=ϕ1, x=α2,q=1α3,s=0,t=α1α22α3, we have the representative 2+3;

    (e) if α20,α4=0,α50, then by choosing ϕ=ϕ1, x=α5α3,q=α22α3α5,s=0,t=α1α52α3α3, we have the representative 2+3+5;

    (f) if α20,α40, then by choosing ϕ=ϕ1, x=3α4α33,q=3α32α23α14, we have the family of representative 2+3+4+α5.

    2. α8=0,α7=0,α60, and α3=0, then by choosing some suitable automorphism ϕ2 we have α30 which is the case considered above. Now we can suppose that α30, and choosing t=α12α3x,s=α42α6x, we have α1=0,α4=0. Therefore, we can suppose that α1=0,α4=0. Consider the following subcases:

    (a) α2=0,α5=0, then by choosing ϕ=ϕ1, x=α6,q=α3,s=0,t=0, we have the representative 3+6;

    (b) α20, then by choosing ϕ=ϕ1, x=α13α2α6,q=α2α16,s=0,t=0, we have the family of representatives 2+3+α5+6.

    3. α8=0,α70, then by choosing ϕ=ϕ1, t=α2α17x,s=α5α17q, we have α2=0,α5=0. Therefore, we can suppose that α2=0,α5=0. Consider the following subcases:

    (a) if α1=0,α4=0,α3=0,α6=0, then we have the representative 7;

    (b) if α1=0,α4=0,α3=0,α60, then by choosing ϕ=ϕ1, x=α6α17,q=1,s=0,t=0, we have the representative 6+7;

    (c) if α1=0,α4=0,α30, and α6=0, then by choosing some suitable automorphism ϕ2, we have α60. Thus we can consider the case α60 and choosing ϕ=ϕ1, x=α6α17,q=α3α17,s=0,t=0, we have the representative 3+6+7;

    (d) if α1=0,α40,α3=0,α6=0, then by choosing ϕ=ϕ1, x=1,q=α4α17,s=0,t=0, we have the representative 4+7;

    (e) if α1=0,α40,α3=0,α60, then by choosing ϕ=ϕ1, x=α6α71,q=α4α27α36, we have the representative 4+6+7;

    (f) if α1=0,α40,α30, then by choosing

    ϕ=ϕ1, x=3α4α13,q=α3α17,s=0,t=0,

    we have the family of representatives 3+4+α6+7;

    (g) if α10. In case of α4=0, choosing some suitable automorphism ϕ2, we have α40. Thus, we can suppose α40, and choosing

    ϕ=ϕ1, x=8α34α11α27,q=8α31α14α27,s=0,t=0,

    we have the family of representatives 1+α3+4+β6+7.

    4. α80, then by choosing ϕ=ϕ1, t=α3α8x,s=α6α8q, we have α3=0,α6=0. Consider the following cases:

    (a) if α1=0,α4=0,α2=0,α5=0, then we have the representatives 8 and 7+8 depending on whether α7=0 or not;

    (b) if α1=0,α4=0,α2=0,α50, then we have the representatives 5+8 and 5+7+8 depending on whether α7=0 or not;

    (c) if α1=0,α4=0,α20. In case of α5=0, choosing some suitable automorphism ϕ2, we have α50. Thus, we can suppose α50, and choosing

    ϕ=ϕ1, x=5α35α22α18,q=5α32α25α18,s=0,t=0,

    we have the family of representatives 2+5+α7+8;

    (d) if α1=0,α40,α2=0,α5=0, then we have the representatives 4+8 and 4+7+8 depending on whether α7=0 or not;

    (e) if α1=0,α40,α2=0,α50, then by choosing

    ϕ=ϕ1, x=α4α15,q=α25α14α14α18,s=0,t=0,

    we have the family of representatives 4+5+α7+8;

    (f) if α1=0,α40,α20, then by choosing

    ϕ=ϕ1, x=8α34α22α18,q=4α22α14α18,s=0,t=0,

    we have the family of representatives 2+4+α5+β7+8;

    (g) if α10 then by choosing some suitable automorphism ϕ2, we have α40. Thus, we can suppose α40, and choosing

    ϕ=ϕ1, x=6α24α11α18,q=6α21α14α18,s=0,t=0,

    we have the family of representatives

    1+α2+4+β5+γ7+8.

    Summarizing, we have the following distinct orbits:

    1+α2+4+β5+γ7+8O(α,β,γ)=O(η23α,η23β,η23γ)=O(η23α,η23β,η23γ)=O(η23α,η23β,η23γ)=O(η23α,η23β,η23γ)=O(η3α,η3β,η3γ)=O(η3α,η3β,η3γ)=O(η3α,η3β,η3γ)=O(η3α,η3β,η3γ)=O(α,β,γ)=O(α,β,γ)=O(α,β,γ)=O(β,α,γ)=O(η23β,η23α,η23γ)=O(η23β,η23α,η23γ)=O(η23β,η23α,η23γ)=O(η23β,η23α,η23γ)=O(η3β,η3α,η3γ)=O(η3β,η3α,η3γ)=O(η3β,η3α,η3γ)=O(η3β,η3α,η3γ)=O(β,α,γ)=O(β,α,γ)=O(β,α,γ),1+α3+4+β6+7O(α,β)=O(η4α,η4β)=O(η4α,η4β)=O(η34α,η34β)=O(η34α,η34β)=O(iα,iβ)=O(iα,iβ)=O(α,β)=O(β,α)=O(η4β,η4α)=O(η4β,η4α)=O(η34β,η34α)=O(η34β,η34α)=O(iβ,iα)=O(iβ,iα)=O(β,α),2+3,2+3+4+α5O(α)=O(η3α)=O(η23α),2+3+5,2+3+α5+6O(α)=O(α1),2+4+α5+β7+8O(α,β)=O(η34α,η34β)=O(η34α,η34β)=O(η4α,η4β)=O(η4α,η4β)=O(iα,iβ)=O(iα,iβ)=O(α,β),2+5+α7+8O(α)=O(η25α)=O(η45α)=O(η5α)=O(η35α),3,3+4+α5O(α)=O(η3α)=O(η23α),3+4+α6+7O(α)=O(η3α)=O(η23α),3+5,3+6,3+6+7,4+5+α7+8O(α)=O(α),4+6+7,4+7,4+7+8,4+8,5+7+8,5+8,6+7,7,7+8,8,

    which gives the following new algebras:

    Nα,β,γ246:e1e1=e5e1e2=e3e1e3=αe5e2e2=e5e2e3=βe5e3e3=e4e3e4=γe5e4e4=e5Nα,β247:e1e1=e5e1e2=e3e1e4=αe5e2e2=e5e2e4=βe5e3e3=e4e3e4=e5N248:e1e2=e3e1e3=e5e1e4=e5e3e3=e4Nα249:e1e2=e3e1e3=e5e1e4=e5e2e2=e5e2e3=αe5e3e3=e4N250:e1e2=e3e1e3=e5e1e4=e5e2e3=e5e3e3=e4Nα251:e1e2=e3e1e3=e5e1e4=e5e2e3=αe5e2e4=e5e3e3=e4Nα,β252:e1e2=e3e1e3=e5e2e2=e5e2e3=αe5e3e3=e4e3e4=βe5e4e4=e5Nα253:e1e2=e3e1e3=e5e2e3=e5e3e3=e4e3e4=αe5e4e4=e5N254:e1e2=e3e1e4=e5e3e3=e4Nα255:e1e2=e3e1e4=e5e2e2=e5e2e3=αe5e3e3=e4Nα256:e1e2=e3e1e4=e5e2e2=e5e2e4=αe5e3e3=e4e3e4=e5N257:e1e2=e3e1e4=e5e2e3=e5e3e3=e4N258:e1e2=e3e1e4=e5e2e4=e5e3e3=e4N259:e1e2=e3e1e4=e5e2e4=e5e3e3=e4e3e4=e5Nα260:e1e2=e3e2e2=e5e2e3=e5e3e3=e4e3e4=αe5e4e4=e5N261:e1e2=e3e2e2=e5e2e4=e5e3e3=e4e3e4=e5N262:e1e2=e3e2e2=e5e3e3=e4e3e4=e5N263:e1e2=e3e2e2=e5e3e3=e4e3e4=e5e4e4=e5N264:e1e2=e3e2e2=e5e3e3=e4e4e4=e5N265:e1e2=e3e2e3=e5e3e3=e4e3e4=e5e4e4=e5N266:e1e2=e3e2e3=e5e3e3=e4e4e4=e5N267:e1e2=e3e2e4=e5e3e3=e4e3e4=e5N268:e1e2=e3e3e3=e4e3e4=e5N269:e1e2=e3e3e3=e4e3e4=e5e4e4=e5N270:e1e2=e3e3e3=e4e4e4=e5

    Here we will collect all information about N418:

    N418e1e1=e4e1e2=e3e3e3=e4H2D(N418)=[Δ11],[Δ13],[Δ22],[Δ23]H2C(N418)=H2D(N418)[Δ14],[Δ24],[Δ34],[Δ44]ϕ±=(x0000±10000±x0ts0x2)

    Let us use the following notations:

    1=[Δ11],2=[Δ13],3=[Δ14],4=[Δ22],5=[Δ23],6=[Δ24],7=[Δ34],8=[Δ44].

    Take θ=8i=1αiiH2C(N418). Since

    ϕT±(α10α2α30α4α5α6α2α50α7α3α6α7α8)ϕ±=(α1αα2α3αα4α5α6α2α50α7α3α6α7α8),

    we have

    α1=α1x2+2α3xt+α8t2,α2=±(α2x+α7t)x,α3=(α3x+α8t)x2,α4=α4±2α6s+α8s2,α5=(α5±α7s)x,α6=(±α6+α8s)x2,α7=±α7x3,α8=α8x4.

    We are interested in (α3,α6,α7,α8)(0,0,0,0). Let us consider ϕ=ϕ+ and the following cases:

    1. α8=0,α7=0,α6=0, then α30 and choosing t=α12α3x, we get α1=0. Now consider the following subcases:

    (a) if α2=0,α4=0,α5=0, then we have the representative 3;

    (b) if α2=0,α4=0,α50, then by choosing x=α5α3,s=0,t=α1α52α3α3, we have the representative 3+5;

    (c) if α2=0,α40, then by choosing x=3α4α3,s=0,t=α13α42α33α3, we have the family of representatives 3+4+α5;

    (d) if α20, then by choosing x=α2α13,s=0,t=α1α22α23, we have the family of representatives 2+3+α4+β5.

    2. α8=0,α7=0,α60, then by choosing s=α42α6x, we have α4=0. Consider the following cases:

    (a) α3=0, then we have two families of representatives α1+β2+6 and α1+β2+5+6 depending on whether α5=0 or not;

    (b) α30 then by choosing x=α6α3,s=α42α6,t=α1α62α23, we have the family of representatives α2+3+β5+6.

    3. α8=0,α70, then by choosing t=α2α17x,s=α5α17, we have α2=0,α5=0. Thus, we can suppose that α2=0,α5=0 and now consider the following cases:

    (a) if α1=0,α4=0,α6=0, then we have the family of representatives α3+7;

    (b) if α1=0,α4=0,α60, then by choosing x=α6α17,s=0,t=0, we have the family of representatives α3+6+7;

    (c) if α1=0,α40, then by choosing x=3α4α17,s=0,t=0, we have the family of representatives α3+4+β6+7;

    (d) if α10, then by choosing x=α1α17,s=0,t=0, we have the family of representatives 1+α3+β4+γ6+7.

    4. α80, then by choosing t=α3α8x,s=α6α8, we have α3=0,α6=0. Thus, we can suppose that α3=0,α6=0. Consider the following cases:

    (a) if α1=0,α2=0,α4=0,α5=0, then we have the representatives 8 and 7+8 depending on whether α7=0 or not;

    (b) if α1=0,α2=0,α4=0,α50, then by choosing x=3α5α18,s=0,t=0, we have the family of representatives 5+α7+8;

    (c) if α1=0,α2=0,α40 then by choosing x=4α4α18,s=0,t=0, we have the family of representatives 4+α5+β7+8;

    (d) if α1=0,α20, then by choosing x=α2α18,s=0,t=0, we have the family of representatives 2+α4+β5+γ7+8;

    (e) if α10, then by choosing x=α1α81,s=0,t=0, we have the family of representatives 1+α2+β4+γ5+μ7+8.

    Summarizing all cases, we have the following distinct orbits:

    1+α2+β4+γ5+μ7+8O(α,β,γ,μ)=O(α,β,γ,μ)=O(α,β,γ,μ)=O(α,β,γ,μ),α1+β2+5+6O(α,β)=O(α,β),α1+β2+6O(α,β)=O(α,β),1+α3+β4+γ6+7O(α,β,γ)=O(α,β,γ),2+3+α4+β5O(α,β)=O(α,β),α2+3+β5+6,2+α4+β5+γ7+8O(α,β,γ)=O(α,iβ,iγ)=O(α,iβ,iγ),3,3+4+α5O(α)=O(η3α)=O(η23α),α3+4+β6+7O(α,β)=O(α,β)=O(α,η3β)=O(α,η23β)=O(α,η3β)=O(α,η23β),3+5,α3+6+7O(α,β)=O(α,β),α3+7O(α)=O(α),4+α5+β7+8O(α,β)=O(iα,iβ)=O(iα,iβ)=O(α,β)=O(α,β)=O(iα,iβ)=O(iα,iβ)=O(α,β),5+α7+8O(α)=O(η3α)=O(η23α)=O(α)=O(η3α)=O(η23α),7+8,8,

    which gives the following new algebras:

    Here we will collect all information about


     | Show Table
    DownLoad: CSV

    Let us use the following notations:

    Take Since

    then, in the case we have

    For define the main families of representatives, we will use and for find equal orbits we will use other automorphisms. We are interested in

    Let us consider the following cases:

    if then and choosing we have the family of representatives

    if and then by choosing some suitable automorphism we have which is the case considered above;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    We are interested in and consider following cases:

    then and we have the following subcases:

    if then we have

    if then we have the representative

    if then by choosing we have the representative

    if then by choosing we have the family of representatives

    (b) if then by choosing we have the family of representatives and depending on whether or not, which will be jointed with the cases (1(a)i) and (1(a)ii).

    2. then we have the following subcases:

    if

    if then we have the representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    (b) if then by choosing we have the family of representatives

    (c) if then we have

    if then by choosing we have the representative

    if then by choosing we have the family of representatives which will be jointed with a representative from the case (2(a)i);

    if then by choosing we have the family of representatives which will be jointed with the family from the case (2(a)i).

    3. then we consider the following subcases:

    if then choosing we have representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    4. then by choosing we have Thus, we can suppose that and now consider following subcases:

    then we have representatives and depending on whether or not;

    then by choosing we have the family of representatives

    then by choosing we have the family of representatives

    then by choosing we have the family of representatives

    Summarizing all cases, we have the following distinct orbits:

    Hence, we have the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    We are interested in and consider following cases:

    if then and we have

    if then we have the family of representatives

    if then by choosing we have the family of representatives

    2. if then by choosing we have the family of representatives

    3. if then by choosing

    we have the family of representatives

    4. if then by choosing we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    We are interested in and consider following cases:

    then and we have the following subcases:

    if then we have the representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    2. then we have the following subcases:

    if then we have the representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    3. then we have the following subcases:

    if then we have representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    4. then we have the following subcases:

    then we have representatives and depending on whether or not;

    then by choosing we have the family of representatives

    then by choosing we have the family of representatives

    then by choosing we have the family of representatives

    then by choosing

    we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    We are interested in and consider following cases:

    if then and we have the family of representatives

    if then we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    We are interested in and consider following cases:

    then and we have the following subcases:

    if then we have the representative

    if then by choosing we have the representative

    if then by choosing we have the family of representatives

    (b)

    if then choosing we have the family of representatives which will be jointed with the case (1(a)i);

    if then by choosing we have the family of representatives which will be jointed with the case (1(a)ii).

    2. then we have the following subcases:

    if then we have representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    3. then by choosing we have Thus, we can suppose that and consider following subcases:

    if then we have representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    (b) then by choosing we have the family of representatives

    4.

    if then then by choosing we have representatives and depending on whether or not;

    if then by choosing

    we have the family of representatives

    if then by choosing

    we have the family of representatives

    if then by choosing

    we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    Since and for we have the following cases:

    if then and we have the following subcase:

    if then we have the family of representatives

    if then by choosing we have the family of representative

    2. if then by choosing we have the family of representatives

    3. if then we have the following subcases:

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    4. if then by choosing we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    We are interested in and consider following cases:

    then and we have the following subcases:

    if then we have representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    2. then we have the following subcases:

    if then we have the familty of representative

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    3. then we have the following subcases:

    if then we have representatives and depending on whether or not;

    if then choosing we have the family of representatives

    if then by choosing we have the family of representatives

    if then choosing we have the family of representatives

    4. then we have the following subcases:

    then we have representatives and depending on whether or not;

    then by choosing we have family of representatives

    then by choosing we have family of representatives

    then by choosing we have the family of representatives

    then by choosing

    we have family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    Since we have the following cases:

    if then and we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    Since we have the following cases:

    then and we have the following subcases:

    if then we have representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    2. then we have the following cases:

    if then we have representatives and depending on whether or not;

    if then choosing we have the family of representatives

    if then by choosing we have the family of representatives

    3. then we have the following cases:

    if then we have representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing

    we have the family of representatives

    4. then we have the following cases:

    if then choosing we have representatives and depending on whether or not;

    if then by choosing we have the family of representatives

    if then by choosing

    we have the family of representatives

    then by choosing

    we have the family of representatives

    then by choosing

    we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    Since we have the following cases:

    if then and we have the family of representatives

    if then we have the family of representatives

    if then we have the family of representatives

    if then we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Here we will collect all information about

    Let us use the following notations:

    Take Since

    we have

    Since we have the following cases:

    if then and we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    if then by choosing we have the family of representatives

    Summarizing, we have the following distinct orbits:

    which gives the following new algebras:

    Remark 2. Note that the algebras and are isomorphic. Hence, there are some additional isomorphism relations for algebras from the present subsection

    Theorem 5.1. Let be a complex -dimensional nilpotent commutative algebra. Then we have one of the following situations.

    If is associative, then is isomorphic to one algebra listed in [15].

    If is a non-associative Jordan algebra, then is isomorphic to one algebra listed in [9].

    If is a non-Jordan -algebra, then is isomorphic to one algebra listed in [11].

    If is a non--algebra, then is isomorphic to one algebra listed in the following list.



    [1] Drones by the Numbers, Federal Aviation Administration, UAS quarterly activity reports, 2022. Available from: https://www.faa.gov/uas/resources/by_the_numbers.
    [2] Civil Aviation Administration of China, Annual report of Chinese civil aviation pilot development 2021, Dalian maritime university press, Dalian, China, 2022. Available from: https://pilot.caac.gov.cn/jsp/phone/airPhoneNewsDetail.jsp?uuid=abbc4b8e-1d42-4aaa-a0b4-6ab4ef2eb1df&code=Statistical_info#down.
    [3] N. Aung, P. Tewogbola, The impact of emotional labor on the health in the workplace: a narrative review of literature from 2013–2018, AIMS Public Health, 6 (2019), 268–275. https://doi.org/10.3934/publichealth.2019.3.268 doi: 10.3934/publichealth.2019.3.268
    [4] M. Hoang, E. Hillier, C. Conger, D. N. Gengler, C. W. Welty, C. Mayer, et al., Evaluation of call volume and negative emotions in emergency response system telecommunicators: a prospective, intensive longitudinal investigation, AIMS Public Health, 9 (2022), 403–414. https://doi.org/10.3934/publichealth.2022027
    [5] R. Parasuraman, D. R. Davies, Decision theory analysis of response latencies in vigilance, J. Exp. Psychol. Hum. Percept. Perform., 2 (1976), 578–590. https://doi.org/10.1037/0096-1523.2.4.578 doi: 10.1037/0096-1523.2.4.578
    [6] G. P. Krueger, Sustained military performance in continuous operations: combatant fatigue, rest and sleep needs, in Handbook of Military Psychology, John Wiley & Sons press, (1991), 255–277.
    [7] C. D. Wickens, W. S. Helton, J. G. Hollands, S. Banbury, Engineering Psychology and Human Performance, Routledge Press, New York, USA, 2021. https://doi.org/10.4324/9781003177616
    [8] A. P. Tvaryanas, Human factors considerations in migration of unmanned aircraft system (UAS) operator control, Defense Technical Information Center press, Brooks, USA, 2006.
    [9] X. Qiu, F. Tian, Q. Shi, Q. Zhao. B. Hu, Designing and application of wearable fatigue detection system based on multimodal physiological signals, in 2020 IEEE International Conference on Bioinformatics and Biomedicine (BIBM), IEEE, Seoul, Korea (South), (2020), 716–722. https://doi.org/10.1109/BIBM49941.2020.9313129
    [10] T. Iida, Y. Ito, M. Kanazashi, S. Murayama, T. Miyake, Y. Yoshimaru, et al., Effects of psychological and physical stress on oxidative stress, serotonin, and fatigue in young females induced by objective structured clinical examination: pilot study of u-8-OHdG, u-5HT, and s-HHV-6, Int. J. Tryptophan Res., 14 (2021). https://doi.org/10.1177/11786469211048443
    [11] X. Li, G. Li, L. Peng, L. Yan, C. Zhang, Driver fatigue detection based on speech feature transfer learning, J. China Railw. Soc., 42 (2020), 74–81.
    [12] G. Zhao, Y. He, H. Yang, Y. Tao, Research on fatigue detection based on visual features, IET Image Proc., 16 (2022), 1044–1053. https://doi.org/10.1049/ipr2.12207 doi: 10.1049/ipr2.12207
    [13] L. Wang, C. Zhang, X. Yin, R. Fu, H. Wang, A non-contact driving fatigue detection technique based on driver's physiological signals, Automot. Eng., 40 (2018), 333–341. https://doi.org/10.19562/j.chinasae.qcgc.2018.03.014 doi: 10.19562/j.chinasae.qcgc.2018.03.014
    [14] X. Li, G. Li, J. Shi, L. Peng, Fatigue driving detection based on speech psychoacoustic analysis, Chin. J. Sci. Instrum., 39 (2018), 166–175. https://doi.org/10.19650/j.cnki.cjsi.J1702539 doi: 10.19650/j.cnki.cjsi.J1702539
    [15] W. Zheng, K. Zheng, G. Li, W. Liu, C. Liu, J. Liu, et al., Vigilance estimation using a wearable EOG device in real driving environment, IEEE Trans. Intell. Transp. Syst., 21 (2020), 170–184. https://doi.org/10.1109/TITS.2018.2889962
    [16] X. Wang, C. Xu, Driver drowsiness detection based on non-intrusive metrics considering individual difference, Accid. Anal. Prev., 95 (2016), 350–357. https://doi.org/10.1016/j.aap.2015.09.002 doi: 10.1016/j.aap.2015.09.002
    [17] W. Feng, Y. Cao, X. Li, W. Hu, Face fatigue detection based on improved deep convolutional neural network, Comput. Intell. Neurosci., 20 (2020), 5680–5687.
    [18] Z. Li, F. Liu, W. Yang, S. Peng, J. Zhou, A survey of convolutional neural networks: analysis, applications, and prospects, IEEE Trans. Neural Networks Learn. Syst., 2021 (2021), 1–21. https://doi.org/10.1109/TNNLS.2021.3084827 doi: 10.1109/TNNLS.2021.3084827
    [19] D. Chen, F. Liu, Z. Li, Deep learning based single sample per person face recognition: A survey, preprint, arXiv: 2006.11395.
    [20] Y. Ed-doughmi, N. Idrissi, Driver fatigue detection using recurrent neural networks, in Proceedings of the 2nd International Conference on Networking, Information Systems & Security (NISS19), ACM, Rabat, Morocco, (2019), 1–6. https://doi.org/10.1145/3320326.3320376
    [21] C. Zheng, B. Xiaojuan, W. Yu, Fatigue driving detection based on Haar feature and extreme learning machine, J. China Univ. Posts Telecommun., 23 (2016), 91–100. https://doi.org/10.1016/S1005-8885(16)60050-X doi: 10.1016/S1005-8885(16)60050-X
    [22] R. Huang, Y. Wang, Z. Li, Z. Lei, Y. Xu, RF-DCM: multi-granularity deep convolutional model based on feature recalibration and fusion for driver fatigue detection, IEEE Trans. Intell. Transp. Syst., 23 (2022), 630–640.https://doi.org/10.1109/TITS.2020.3017513 doi: 10.1109/TITS.2020.3017513
    [23] W. Gu, Y. Zhu, X. Chen, L. He, B. Zheng, Hierarchical CNN-based real-time fatigue detection system by visual-based technologies using MSP model, IET Image Proc., 12 (2018), 2319–2329. https://doi.org/10.1049/iet-ipr.2018.5245 doi: 10.1049/iet-ipr.2018.5245
    [24] S. Dey, S. A. Chowdhury, S. Sultana, M. A. Hossain, M. Dey, S. K. Das, Real time driver fatigue detection based on facial behaviour along with machine learning approaches, in 2019 IEEE International Conference on Signal Processing, Information, Communication & Systems (SPICSCON), IEEE, Dhaka, Bangladesh, (2019), 135–140. https://doi.org/10.1109/SPICSCON48833.2019.9065120
    [25] Z. Xiao, Z. Hu, L. Geng, F. Zhang, J. Wu, Y. Li, Fatigue driving recognition network: fatigue driving recognition via convolutional neural network and long short-term memory units, IET Intel. Transport Syst., 13 (2019), 1410–1416. https://doi.org/10.1049/iet-its.2018.5392 doi: 10.1049/iet-its.2018.5392
    [26] W. Liu, J. Qian, Z. Yao, X. Jiao, J. Pan, Convolutional two-stream network using multi-facial feature fusion for driver fatigue detection, Future Internet, 11 (2019), 115. https://doi.org/10.3390/fi11050115 doi: 10.3390/fi11050115
    [27] L. Geng, X. Liang, Z. Xiao, Y. Li, Real-time driver fatigue detection based on morphology infrared features and deep learning, Infrared Laser Eng., 47 (2018), 203009. https://doi.org/10.3788/IRLA201847.0203009 doi: 10.3788/IRLA201847.0203009
    [28] F. Liu, D. Chen, J. Zhou, F. Xu, A review of driver fatigue detection and its advances on the use of RGB-D camera and deep learning, Eng. Appl. Artif. Intell., 116 (2022), 105399. https://doi.org/10.1016/j.engappai.2022.105399 doi: 10.1016/j.engappai.2022.105399
    [29] S. Abtahi, M. Omidyeganeh, S. Shirmohammadi, B. Hariri, YawDD: A yawning detection dataset, in Proceedings of the 5th ACM Multimedia Systems Conference, ACM, Singapore, (2014), 24–28. https://doi.org/10.1145/2557642.2563678
    [30] K. Diaz-Chito, A. Hernández-Sabaté, A. M. López, A reduced feature set for driver head pose estimation, Appl. Soft Comput., 45 (2016), 98–107. https://doi.org/10.1016/j.asoc.2016.04.027
    [31] F. Song, X. Tan, X. Liu, S. Chen, Eyes closeness detection from still images with multi-scale histograms of principal oriented gradients, Pattern Recognit., 47 (2014), 2825–2838. https://doi.org/10.1016/j.patcog.2014.03.024 doi: 10.1016/j.patcog.2014.03.024
    [32] C. Weng, Y. Lai, S. Lai, Driver drowsiness detection via a hierarchical temporal deep belief network, in Asian Conference on Computer Vision (ACCV), Springer Cham, (2016), 117–133. https://doi.org/10.1007/978-3-319-54526-4_9
    [33] R. Gross, Face databases, in Handbook of Face Recognition, Springer, (2005), 301–327. https://doi.org/10.1007/0-387-27257-7_14
    [34] G. Jocher, A. Stoken, J. Borovec, A. Chaurasia, T. Xie, C. Y. Liu, Ultralytics/yolov5: v5.0-yolov5-p6 1280 models, Zenodo, 2021 (2021). https://doi.org/10.5281/zenodo.4679653
    [35] J. F. Cohn, A. J. Zlochower, J. Lien, T. Kanade, Automated face analysis by feature point tracking has high concurrent validity with manual FACS coding, Psychophysiology, 36 (1999), 35–43. https://doi.org/10.1017/s0048577299971184 doi: 10.1017/s0048577299971184
    [36] S. Dubuisson, F. Davoine, M. Masson, A solution for facial expression representation and recognition, Signal Process. Image Commun., 17 (2002), 657–673. https://doi.org/10.1016/S0923-5965(02)00076-0 doi: 10.1016/S0923-5965(02)00076-0
    [37] N. A. Rahman, K. C. Wei, J. See, RGB-H-CbCr skin colour model for human face detection, Fac. Inf. Technol., 4 (2007), 1–6.
    [38] M. Sun, L. Liang, H. Wang, W. He, L. Zhao, Facial landmark detection based on cascade convolutional neural network, J. Univ. Chin. Acad. Sci., 37 (2020), 562–569. https://doi.org/10.7523/j.issn.2095-6134.2020.04.017 doi: 10.7523/j.issn.2095-6134.2020.04.017
    [39] X. Guo, S. Li, J. Yu, J. Zhang, J. Ma, L. Ma, et al., PFLD: A practical facial landmark detector, preprint, arXiv: 1902.10859.
    [40] P. Yan, D. Yan, C. Du, Design and implementation of a driver's eye state recognition algorithm based on PERCLOS, Chin. J. Electron., 4 (2014), 669–672.
    [41] V. F. Ferrario, C. Sforza, G. Serrao, G. Grassi, E. Mossi, Active range of motion of the head and cervical spine: a three-dimensional investigation in healthy young adults, J. Orthop. Res., 20 (2002), 122–129. https://doi.org/10.1016/S0736-0266(01)00079-1 doi: 10.1016/S0736-0266(01)00079-1
    [42] H. Wang, Z. Li, X. Ji, Y. Wang, Face R-CNN, preprint, arXiv: 1706.01061.
    [43] B. Yan, C. Yang, F. Chen, K. Takeda, C. Wang, FDNet: a deep learning approach with two parallel cross encoding pathways for precipitation nowcasting, J. Comput. Sci. Technol., 1 (2021). https://jcst.ict.ac.cn/EN/10.1007/s11390-021-1103-8
    [44] J. Li, Y. Wang, C. Wang, Y. Tai, J. Qian, J. Yang, et al., DSFD: dual shot face detector, in Proceedings of the IEEE/CVF Conference on Computer Vision and Pattern Recognition (CVPR), IEEE, Long Beach, USA, (2019), 5060–5069. https://doi.org/10.1109/CVPR.2019.00520
    [45] J. Deng, J. Guo, E. Ververas, I. Kotsia, S. Zafeiriou, RetinaFace: single-shot multi-level face localisation in the wild, in Proceedings of the IEEE/CVF Conference on Computer Vision and Pattern Recognition (CVPR), IEEE, Seattle, USA, (2020), 5203–5212. https://doi.org/10.1109/cvpr42600.2020.00525
    [46] W. Tian, Z. Wang, H. Shen, W. Deng, Y. Meng, B. Chen, et al., Learning better features for face detection with feature fusion and segmentation supervision, preprint, arXiv: 1811.08557.
    [47] S. Zhang, X. Zhu, Z. Lei, H. Shi, X. Wang, S. Z. Li, S3fd: single shot scale-invariant face detector, in Proceedings of the IEEE International Conference on Computer Vision (ICCV), IEEE, Venice, Italy, (2017), 192–201. https://doi.org/10.1109/iccv.2017.30
    [48] J. Wang, Y. Yuan, G. Yu, Face attention network: an effective face detector for the occluded faces, preprint, arXiv: 1711.07246.
    [49] H. Zhang, C. Wu, Z. Zhang, Y. Zhu, H. Lin, Z. Zhang, et al., ResNeSt: Split-attention networks, in Proceedings of the IEEE/CVF Conference on Computer Vision and Pattern Recognition, IEEE, New Orleans, USA, (2022), 2736–2746. https://doi.org/10.1109/CVPRW56347.2022.00309
    [50] M. Sandler, A. Howard, M. Zhu, A. Zhmoginov, L. Chen, Mobilenetv2: Inverted residuals and linear bottlenecks, in Proceedings of the IEEE Conference on Computer Vision and Pattern Recognition (CVPR), IEEE, Salt Lake, USA, (2018), 4510–4520. https://doi.org/10.1109/CVPR.2018.00474
    [51] K. Sun, Y. Zhao, B. Jiang, T. Cheng, B. Xiao, D. Liu, et al., High-resolution representations for labeling pixels and regions, preprint, arXiv: 1904.04514.
    [52] M. Tan, Q. Le, Efficientnet: Rethinking model scaling for convolutional neural networks, in International Conference on Machine Learning (ICML), PMLR, Long Beach, USA, (2019), 6105–6114.
  • This article has been cited by:

    1. Patrícia Damas Beites, Amir Fernández Ouaridi, Ivan Kaygorodov, The algebraic and geometric classification of transposed Poisson algebras, 2023, 117, 1578-7303, 10.1007/s13398-022-01385-4
  • Reader Comments
  • © 2023 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)
通讯作者: 陈斌, bchen63@163.com
  • 1. 

    沈阳化工大学材料科学与工程学院 沈阳 110142

  1. 本站搜索
  2. 百度学术搜索
  3. 万方数据库搜索
  4. CNKI搜索

Metrics

Article views(13284) PDF downloads(206) Cited by(5)

Article outline

Figures and Tables

Figures(10)  /  Tables(7)

/

DownLoad:  Full-Size Img  PowerPoint
Return
Return

Catalog