Research article

Fuzzy permutation entropy derived from a novel distance between segments of time series

  • Received: 11 June 2020 Accepted: 27 July 2020 Published: 05 August 2020
  • MSC : 94A17

  • As effective tools for time series analyzing, a variety of information entropies have been widely applied in engineering, economic, biomedicine and other fields. In this paper, we define a new distance between finite sequence based on inversion and derive a new entropy, Fuzzy Permutation Entropy(FPE). A comparison of recognition performance for WGN, 1/f noise, periodical, or chaotic sequence and sine waves shows that FPE is valid on distinguishing deterministic signals from stochastic signals. Further contrast studying versus FE and PE shows that FPE is more sensitive to the alteration of the complexity of time series and more effective on separating different signals. Moreover, FPE is used to explore the distinction of the complexity of various traffic flows via the time headways which is simulated from the improved brake light rule model. We hope it is conducive to design Intelligent Transportation Management System.

    Citation: Zelin Zhang, Zhengtao Xiang, Yufeng Chen, Jinyu Xu. Fuzzy permutation entropy derived from a novel distance between segments of time series[J]. AIMS Mathematics, 2020, 5(6): 6244-6260. doi: 10.3934/math.2020402

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  • As effective tools for time series analyzing, a variety of information entropies have been widely applied in engineering, economic, biomedicine and other fields. In this paper, we define a new distance between finite sequence based on inversion and derive a new entropy, Fuzzy Permutation Entropy(FPE). A comparison of recognition performance for WGN, 1/f noise, periodical, or chaotic sequence and sine waves shows that FPE is valid on distinguishing deterministic signals from stochastic signals. Further contrast studying versus FE and PE shows that FPE is more sensitive to the alteration of the complexity of time series and more effective on separating different signals. Moreover, FPE is used to explore the distinction of the complexity of various traffic flows via the time headways which is simulated from the improved brake light rule model. We hope it is conducive to design Intelligent Transportation Management System.


    Consider a non trivial group A and take an element t not in A. Consider an equation,

    s(t)=a1tp1a2tp2...antpn=1,

    where, ajϵA, pj = ±1 and aj 1 if it appears between two cyclically adjacent t having exponents with different signs. The equation s(t)=1 is solvable over the group A if it has a solution in some group extension of A. Precisely, let G be a group containing A, if there exist an injective homomorphism, say λ:AG, such that, g1gp1g2gp2gngpn=1 in G, where gj=λ(aj) and gG then the equation s(t)=1 is solvable over A. A conjecture stated in [2] by Levin asserts that if A is torsion- free, then every equation over A is solvable. Bibi and Edjvet [1], using in particular results of Prishchepov [3], has shown that the conjecture is true for equations of length at most seven. Here we prove the following: Theorem: The singular equation,

    s(t)=atbt1ctdtet1ftgt1ht1=1,

    of length eight is solvable over any torsion-free group.

    Levin showed that any equation over a torsion-free group is solvable if the exponent sum p1+p2+pn is same as length of the equation. Stallings proved that any equation of the form a1ta2tamtb1t1b2t1bnt1=1, is solvable over torsion-free group. Klyachko showed that any equation with exponent sum ±1 is solvable over torsion free group.

    All necessary definitions concerning relative presentations and pictures can be found in [4]. A relative group presentation P is a triplet <A,t|r> where A is a group, t is disjoint from A and r is a set of cyclically reduced words in A<t>. If the relative presentation P is orientable and aspherical then the natural homomorphism AG is injective. P is orientable and aspherical implies s(t)=1 is solvable. We use weighttest to establish asphericity. The star graph Γ of a relative presentation P=<A,t|r> is a directed graph with vertex set {t,t1} and edges set r, the set of all cyclic permutations of elements of {r,r1} which begin with t or t1. Sr write S=Tαj where αjA and T begins and ends with t or t1. The initial function i(S) is the inverse of last symbol of T and terminal function τ(S) is the first symbol of T. The labeling function on the edges is defined by π(S)=αj and is extended to paths in the usual way. A weight function θ on the star graph Γ is a real valued function given to the edges of Γ. If d is an edge of Γ, then θ(d) = θ(ˉd). A weight function θ is aspherical if the following three conditions are satisfied:

    (1) Let R ϵr be cyclically reduced word, say R=tϵ11a1...tϵnnan, where ϵi = ±1 and ai A. Then,

    nι=1(1θ(tϵιιaι...tϵnnantϵ11a1...tϵι1ι1aι1))2.

    (2) Each admissible cycle in Γ has weight at least 2.

    (3) Each edge of Γ has a non-negative weight.

    The relative presentation P is aspherical if Γ admits an aspherical weight function. For convenience we write s(t)=1 as,

    s(t)=atbt1ctdtet1ftgt1ht1=1,

    where a,b,c,e,f,gA/{1} and d,hA. Moreover, by applying the transformation y=td, if necessary, it can be assumed without any loss that d=1 in A.

    (1) Since c(2, 2, 3, 3, 3, 3, 3, 3) = 0, so at least 3 admissible cycles must be of order 2.

    (2) ac and ac1 implies c2 = 1, a contradiction.

    (3) af and af1 implies f2 = 1, a contradiction.

    (4) be and be1 implies e2 = 1, a contradiction.

    (5) bg and bg1 implies g2 = 1, a contradiction.

    (6) If two of ac, af and cf1 are admissible then so is the third.

    (7) If two of ac1, af1 and cf1 are admissible then so is the third.

    (8) If two of ac,cf and af1 are admissible then so is the third.

    (9) If two of ac1, af and cf are admissible then so is the third.

    (10) If two of be,eg, bg1 are admissible then so is the third.

    (11) If two of be1, eg and bg are admissible then so is the third.

    (1) a=c1,b=e1

    (2) a=c1,b=g1

    (3) a=c1,e=g1

    (4) a=c1,d=h1

    (5) a=c1,b=e

    (6) a=c1,b=g

    (7) a=c,b=g1

    (8) a=c,e=g1

    (9) a=c,d=h1

    (10) a=c,b=e

    (11) a=c,b=g

    (12) a=c,e=g

    (13) a=f1,b=g1

    (14) a=f1,d=h1

    (15) a=f1,b=g

    (16) c=f1,b=e1

    (17) c=f1,b=g1

    (18) c=f1,d=h1

    (19) c=f1,b=e

    (20) c=f1,b=g

    (21) c=f1,e=g

    (22) a=f,d=h1

    (23) a=f,b=e

    (24) a=f,b=g

    (25) c=f,b=g1

    (26) c=f,e=g1

    (27) c=f,d=h1

    (28) c=f,b=e

    (29) b=e,d=h1

    (30) a=c1,a=f1,d=h1

    (31) a=c1,a=f1,c=f

    (32) a=c1,a=f,c=f1

    (33) a=c1,c=f1,d=h1

    (34) a=c1,c=f,d=h1

    (35) a=c,a=f,c=f

    (36) a=c,a=f1,c=f1

    (37) a=f1,c=f1,d=h1

    (38) a=f1,c=f,d=h1

    (39) a=f,c=f1,d=h1

    (40) a=f,c=f,d=h1

    (41) b=e1,b=g1,d=h1

    (42) b=e1,b=g1,e=g

    (43) b=e,b=g1,e=g1

    (44) b=e1,b=g,e=g1

    (45) b=e1,b=g,d=h1

    (46) b=g1,e=g1,d=h1

    (47) b=g1,e=g,d=h1

    (48) b=g,e=g1,d=h1

    (49) b=g,e=g,d=h1

    (50) a=c1,a=f,c=f1,d=h1

    (51) a=c1,a=f1,c=f,d=h1

    (52) a=c1,b=e,b=g1,d=h1

    (53) a=c,a=f1,c=f1,d=h1

    (54) a=c,a=f,c=f,d=h1

    (55) b=e1,b=g1,e=g,d=h1

    (56) a=c1,b=e,b=g1,e=g1,d=h1

    (57) a=c,a=f,c=f,b=e1,d=h1

    (58) a=c,a=f,c=f,b=e,d=h1

    (59) c=f,b=e,b=g1,e=g1,d=h1

    (60) a=c,a=f1,c=f1,b=e1,b=g1,d=h1

    (61) a=c,a=f,c=f,b=e1,b=g1,d=h1

    (62) a=f1,c=f,b=e1,b=g,e=g1,d=h1

    (63) a=f,c=f1,b=e,b=g1,e=g1,d=h1

    (64) a=f,c=f1,b=e1,b=g,e=g1,d=h1

    (65) a=c1,a=f,c=f1,b=e,b=g1,e=g1,d=h1

    (66) a=c1,a=f1,c=f,b=e1,b=g,e=g1,d=h1

    (67) a=c,a=f,c=f,b=e1,b=g,e=g1,d=h1

    Recall that P=<A,t|s(t)>=1 where s(t)=atbt1ctdtet1ftgt1ht1=1 and a,b,c,e,f,gA/{1} and d,hA. We will show that some cases of the given equation are solvable by applying the weight test. Also the transformation y=td, leads to the assumption that d=1 in A.

    In this case the relative presentation P is given as :

    P=<A,t|s(t)=atbt1ctdtet1ftgt1ht1>

    = <A,t|s(t)=c1te1t1ct2et1ftgt1ht1=1>.

    Suppose, x=tet1 which implies that x1=te1t1.

    Using above substitutions in relative presentation we have new presentation

    Q=<A,t,x|t1c1x1ctxftgt1h=1=x1tet1>. Observe that in new presentation Q we have two relators R1=t1c1x1ctxftgt1h and R2=x1tet1. The star graph Γ1 for new presentation Q is given by the Figure 1.

    Figure 1.   .

    Let γ1c1,γ2c,γ31,γ4f,γ5g,γ6h and η11,η2e,η31.

    Assign weights to the edges of the star graph Γ1 as follows:

    θ(γ1)=θ(γ2)=θ(η2)=1 and θ(γ3)=θ(γ4)=θ(γ5)=θ(γ5)=θ(γ6)=θ(η1)=θ(η3)=12. The given weight function θ is aspherical and all three conditions of weight test are satisfied.

    (1) Observe that, (1θ(γj))=6(1+1+12+12+12)+12)=2 and (1θ(ηj))=3(0+1+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contraction. For example, γ3γ6η1 is a cycle of length 2 having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1atbt1a1t2et1ftb1t1h and R2=ta1t1. The star graph Γ2 for new presentation Q is given by the Figure 1.

    (1) Observe that, (1θ(γj))=6(0+1+1+12+1+12)=2 and (1θ(ηj))=3(0+1+1)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example γ6γ3η1 is a cycle of length 2, having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbx1tg1t1ftgt1h and R2=x1t1at. The star graph Γ2 for new presentation Q is given by the Figure 1.

    (1) Observe that, (1θ(γj))=6(0+1+1+12+1+12)=2 and (1θ(ηj))=3(0+1+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ5η3γ1η3 is a cycles of length 2 having weight less than 2, which implies b=g1, a contradiction.

    (3) Each admissible cycle has non-negative weight.

    In this case we have two relators R1=xbx1et1ftg and R2=t1a1t2. The star graph Γ3 for the relative presentation is given by the Figure 1.

    (1) Observe that (1θ(γj))=4(0+1+12+12)=2 and (1θ(ηj))=4(0+12+1+12)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbx1tbt1ftgt1ht1 and R2=x1t1at. The star graph Γ4 for the relative presentation is given by the Figure 1.

    (1) Observe that (1θ(γj))=7(0+1+1+1+12+12+1)=2 and (1θ(ηj))=3(0+12+1+12)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ5η3γ1η3 is a cycle of length 2 having weight less than 2, which implies g=b1 a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axa1t2et1fxh and R2=x1tb1t1. The star graph Γ5 for the relative presentation Q is given by the Figure 2.

    Figure 2.   .

    (1) Observe that, (1θ(γj))=6(12+1+12+1+12+12)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6γ3η3 is a cycle of length 2 having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight,

    In this case we have two relators R1=xg1xtet1ftgt1h and R2=x1t1ct. The star graph Γ6 for the relative presentation is given by the Figure 2.

    (1) Observe that, (1θ(γj))=6(32+12+1+12+12+0)=2 and (1θ(ηj))=3(0+1+0)=2

    (2) Each admissible cycle having weight less than 2 leads to a contradiction, For example, the cycle γ1η1η3, γ6η1η3γ2 and γ6η1η3γ1 are the cycles of length 2 having weight less than 2, which implies g=1, h=1 and h=g, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xht1atbt1atx1f and R2=x1tgt1. The star graph Γ7 for the relative presentation Q is given by the Figure 2.

    (1) Observe that, (1θ(γj))=6(12+1+1+1+12+0)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each possible admissible cycle having weight less than 2 leads to a contradiction. For Example η2γ5γ1 is a cycle of length 2 having weight less than 2, which implies g=h1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbxtet1ftg and R2=x1t1at. The star graph Γ8 for the relative presentation is given by the Figure 2.

    (1) Observe that, (1θ(γj))=6(12+32+0+12+12+1)=2 and (1θ(ηj))=3(0+1+0)=2.

    (2) Each edge having weight less than 2 leads to a contradiction. For example, γ2η1η3 is a cycle of length 2 having weight less than 2, which implies b=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axatxftgt1h and R2=x1tb1t1. The star graph Γ9 for the relative presentation is given by the Figure 3.

    Figure 3.   .

    (1) Observe that, (1θ(γj))=6(1+12+12+12+1+12)=2 and (1θ(ηj))=3(0+12+12)=2.

    (2) Each edge having weight less than 2 leads to a contradiction. For example, γ6γ5η1 is a cycles of length 2 having weight less than 2, which implies that h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axat2et1fxh and R2=x1tgt1. The star graph Γ10 for the relative presentation is given by the Figure 3.

    (1) Observe that, (1θ(γj))=6(12+1+12+1+12+12)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each edge having weight less than 2 leads to a contradiction. For example, γ6γ3η3 is a cycles of length 2 having weight less than 2, which implies that h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1ctbt1axfx1th and R2=x1t2et1. The star graph Γ11 for the relative presentation is given by the Figure 3.

    (1) Observe that (1θ(γj))=6(1+12+0+12+1+1)=2 and (1θ(ηj))=4(1+0+12+12)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbt1ct2ex1b1t1h and R2=x1t1at. The star graph Γ12 for the relative presentation is given by the Figure 3.

    (1) Observe that, (1θ(γj))=6(12+12+0+1+1+1)=2 and (1θ(ηj))=3(0+1+0)=2.

    (2) Each edge having weight less than 2 leads to a contradiction. For example, γ6γ3η1 is a cycles of length 2 having weight less than 2, which implies that h=1, a contradiction.

    (3) Each edge has non negative weight.

    In this case we have two relators R1=t1xbt1ct2exg and R2=x1t1ft. The star graph Γ13 for the relative presentation is given by the Figure 4.

    Figure 4.   .

    (1) (1θ(γj))=6(12+12+1+1+1+0)=2 and (1θ(ηj))=3(0+12+12)=2.

    (2) Each edge having weight less than 2 leads to a contradiction. For example, γ6γ4γ1 amd γ6γ5η1η3 are the cycles of length 2 having weight less than 2, which implies g=1 and g=e1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbt1ct2ex1bt1h and R2=x1t1at. The star graph Γ13 for the relative presentation Q is given by the Figure 4.

    (1) Observe that, (1θ(γj))=6(12+1+1+12+1+0)=2 and (1θ(ηj))=3(0+12+12)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6η1γ3 is a cycle of length 2 having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1ac1tx1ctxgt1h and R2=x1tb1t1c1t. The star graph Γ14 for the relative presentation Q is given by the Figure 4.

    (1) Observe that, (1θ(γj))=6(1+12+1+1+12+0)=2 and (1θ(ηj))=4(0+1+12+12)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6η1γ4 is a cycles of length 2, having weight less than 2, which implies h=1 a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axtex1h and R2=x1tbt1ct. The star graph Γ15 for the relative presentation Q is given by the Figure 4.

    (1) Observe that, (1θ(γj))=4(1+0+1+0)=2 and (1θ(ηj))=4(1+0+0+1)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ4η4γ2γ1 is a cycle of length 2 having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t2atbx1texg and R2=x1t1et1c1t. The star graph Γ16 for the relative presentation Q is given by the Figure 4.

    (1) (1θ(γj))=4(1+1+12+12+12)=2 and (1θ(ηj))=4(12+12+12+12)=2.

    (2) Each admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1atbxtbx1gt1h and R2=x1t1ct. The star graph Γ17 for the relative presentation Q is given by the Figure 5.

    Figure 5.   .

    (1) (1θ(γj))=6(1+12+1+12+1+12+12)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6η3γ3 is a cycle of length 2 having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axct2et1c1tbt1h and R2=x1tbt1. The star graph Γ18 for the relative presentation Q is given by the Figure 5.

    (1) Observe that, (1θ(γj))=6(12+1+12+1+12+12)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6γ3η3 is a cycle of length 2 having weight less than 2, which implies h=1 a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1atbxtex1et1h and R2=x1t1ct. The star graph Γ19 for the relative presentation Q is given by the Figure 5.

    (1) Observe that, (1θ(γj))=6(12+1+12+1+12+12)=2 and (1θ(ηj))=3(0+1+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6η3γ3 is a cycle of length 2 having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbt1ct2exg and R2=x1t1at. The star graph Γ20 for the Q relative presentation is given by the Figure 5.

    (1) Observe that, (1θ(γj))=6(1+12+1+1+0+12)=2 and (1θ(ηj))=3(0+12+12)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axctxatget1h and R2=x1tbt1.

    The star graph Γ21 for the relative presentation Q is given by the Figure 6.

    Figure 6.   .

    (1) Observe that, (1θ(γj))=6(12+12+1+12+12+1)=2 and (1θ(ηj))=3(0+12+12)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6γ3η1, is a cycle of length 2 having weight less than 2, which implies h=1 a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xct2exh and R2=x1t1atbt1. The star graph Γ22 for the relative presentation Q is given by the Figure 6.

    (1) Observe that, (1θ(γj))=4(12+12+12+12)=2 and (1θ(ηj))=4(12+12+12+12)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ23η1 is a cycle of length 2, having weight less than 2, which implies e2=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1atg1xtexgt1h and R2=x1t1ft>. The star graph Γ23 for the relative presentation is given by the Figure 6.

    (1) Observe that, (1θ(γj))=6(1+12+12+12+1+12)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6η3γ3 is a cycle of length 2 having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1atbxtexe1t1h and R2=x1t1ct. The star graph Γ23 for the relative presentation Q is given by the Figure 6.

    (1) Observe that, (1θ(γj))=6(1+12+12+12+1+12)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ6η3γ3 has weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t2atbxtexg and R2=x1t1ct. The star graph Γ24 for the relative presentation Q is given by the Figure 6.

    (1) Observe that, (1θ(γj))=6(1+12+1+1+12+0)=2 and (1θ(ηj))=3(0+1+0)=2.

    (2) Each possible admissible cycle has weight at least 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axt1cxctgt1h and R2=x1t2bt1. The star graph Γ25 for the relative presentation Q is given by the Figure 7.

    Figure 7.   .

    (1) (1θ(γj))=4(12+12+12+12)=2 and (1θ(ηj))=4(12+12+12+12)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xcta1t2xftg and R2=x1t2atbt1. The star graph Γ26 for the relative presentation Q is given by the Figure 7.

    (1) (1θ(γj))=6(1+12+1+12+12+12)=2 and (1θ(ηj))=5(12+1+12+12+12)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbx1tex1g and R2=x1t1at. The star graph Γ27 for the relative presentation Q is given by the Figure 7.

    (1) (1θ(γj))=5(1+0+1+1+0)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axctx1fte1t1h and R2=x1tbt1. The star graph Γ27 for the new presentation Q is given by the Figure 7.

    (1) Observe that, (1θ(γj))=5(1+0+1+1+0)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbx1tex1gt1h and R2=x1t1at. The star graph Γ27 for the relative presentation Q is given by the Figure 7.

    (1) Observe that, (1θ(γj))=5(0+1+1+0+1)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative watch.

    In this case we have two relators: R1=xbx1tet1c1tgt1h and R2=x1t1at. The star graph Γ28 for the relative presentation Q is given by the Figure 7.

    (1) Observe that, (1θ(γj))=5(0+1+1+1+12+12)=2 and (1θ(ηj))=3(0+12+12)=2.

    (2) Each possible admissible cycle having weight less than 2 leads to a contrdiction. For example, γ5η3γ1 is a cycle of length 2 having weight less than 2, which implies g=b1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbx1texg and R2=x1t1at. The star graph Γ29 for the relative presentation Q is given by the Figure 8.

    Figure 8.   .

    (1) Observe that, (1θ(γj))=5(1+0+1+0+1)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbx1tex1g and R2=x1t1at. The star graph Γ30 for the relative presentation Q is given by the Figure 8.

    (1) Observe that, (1θ(γj))=5(1+0+1+1+0)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbxtexgt1h and R2=x1t1at. The star graph Γ31 for the relative presentation Q is given by the Figure 8.

    (1) Observe that, (1θ(γj))=5(1+0+12+12+1)=2 and (1θ(ηj))=3(12+0+12)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbxtex1gt1h and R2=x1t1at. The star graph Γ32 for the relative presentation Q is given by the Figure 8.

    (1) Observe that, (1θ(γj))=5(0+1+0+1+1)=2 and (1θ(ηj))=3(1+0+0)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbxtex1g and R2=x1t1at. The star graph Γ33 for the relative presentation Q is given by the Figure 9.

    Figure 9.   .

    (1) Observe that (1θ(γj))=5(1+1+0+12+12)=2 and (1θ(ηj))=3(12+0+12)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbt1ct2ex1g and R2=x1t1at. The star graph Γ38 for the relative presentation Q is given by the Figure 9.

    (1) Observe that, (1θ(γj))=6(1+12+1+1+12+0)=2 and (1θ(ηj))=3(0+12+12)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ2γ16η1η3γ5 is the cycle of length 2 having weight less than 2, which leads to a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbx1texg and R2=x1t1at. The star graph Γ35 for the relative presentation Q is given by the Figure 9.

    (1) Observe that, (1θ(γj))=5(0+1+1+1+0)=2 and (1θ(ηj))=3(1+0+0)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbx1tbxb1 and R2=x1t1at. The star graph Γ35 for the relative presentation Q is given by the Figure 9.

    (1) Observe that, (1θ(γj))=5(1+0+1+0+1)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbxtexg and R2=x1t1at. The star graph Γ36 for the relative presentation Q is given by the Figure 9.

    (1) Observe that, (1θ(γj))=5(0+1+1+1+0)=2 and (1θ(ηj))=3(1+0+0)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axctx1fx1 and R2=x1tbt1. The star graph Γ37 for the relative presentation Q is given by the Figure 10.

    Figure 10.   .

    (1) Observe that, (1θ(γj))=5(12+12+1+1+0)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axctx1fx1h and R2=x1tbt1. The star graph Γ37 for the relative presentation Q is given by the Figure 9.

    (1) Observe that, (1θ(γj))=5(12+12+1+1+0)=2 and (1θ(ηj))=3(12+0+12)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axctxfx1h and R2=x1tbt1. The star graph Γ38 for the relative presentation Q is given by the Figure 10.

    (1) Observe that, (1θ(γj))=5(0+1+1+0+1)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2, so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axctx1fte1t1h and R2=x1tbt1. The star graph Γ39 for the relative presentation Q is given by the Figure 10.

    (1) Observe that, (1θ(γj))=6(12+1+0+12+1+1)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example, γ5η1η3 is the cycle of length 2 having weight less than 2, which implies h=1, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axctx1fx and R2=x1tbt1. The star graph Γ40 for the relative presentation Q is given by the Figure 10.

    (1) Observe that, (1θ(γj))=6(12+1+0+12+1+1)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=|t1axctxfx1 and R2=x1tbt1. The star graph Γ41 for the relative presentation Q is given by the Figure 11.

    Figure 11.   .

    (1) Observe that, (1θ(γj))=6(12+1+0+12+1+1)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axa1txfx1 and R2=x1tbt1. The star graph Γ41 for the relative presentation Q is given by the Figure 11.

    (1) Observe that, (1θ(γj))=5(0+1+1+0+1)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=x1t1axctx1f and R2=x1tbt1. The star graph Γ42 for the relative presentation Q is given by the Figure 11.

    (1) Observe that, (1θ(γj))=5(0+12+12+1+1)=2 and (1θ(ηj))=3(12+0+12)=2.

    (2) Each admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) each edge has non-negative weight.

    In this case we have two relators R1=xatbt1cx1f and R2=x1tbt2. The star graph Γ43 for the relative presentation Q is given by the Figure 11.

    (1) Observe that, (1θ(γj))=6(12+1+0+12+1+1)=2 and (1θ(ηj))=3(12+12+0)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xt1axctxf and R2=x1tbt1. The star graph Γ44 for the relative presentation Q is given by the Figure 11.

    (1) Observe that, (1θ(γj))=5(12+12+12+12+1)=2 and (1θ(ηj))=3(12+0+12)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbx1etxg and R2=x1t2c1t. The star graph Γ45 for the relative presentation Q is given by the Figure 12.

    Figure 12.   .

    (1) Observe that, (1θ(γj))=4(0+1+1+0)=2 and (1θ(ηj))=4(1+0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbxtex1g and R2=x1t1at. The star graph Γ46 for the relative presentation Q is given by the Figure 12.

    (1) Observe that, (1θ(γj))=5(1+1+0+0+1)=2 and (1θ(ηj))=3(1+0+0)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbxtexg and R2=x1t1at. The star graph Γ47 for the relative presentation Q is given by the Figure 12.

    (1) Observe that, (1θ(γj))=5(1+1+0+1+0)=2 and (1θ(ηj))=3(1+0+0)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=x1t1axctx1f and R2=x1tbt1. The star graph Γ48 for the relative presentation Q is given by the Figure 12.

    (1) Observe that, (1θ(γj))=5(0+0+1+1+1)=2 and (1θ(ηj))=3(0+0+0)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xc1x1tcx1f and R2=x1tb1t2. The star graph Γ49 for the relative presentation Q is given by the Figure 13.

    Figure 13.   .

    (1) Observe that, (1θ(γj))=4(0+1+1+0)=2 and (1θ(ηj))=4(1+0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbxtb1xg and R2=x1t1at. The star graph Γ50 for the relative presentation Q is given by the Figure 13.

    (1) Observe that, (1θ(γj))=5(12+1+12+1+0)=2 and (1θ(ηj))=3(12+0+12=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example γ2η1γ15 is a cycle of length 2 having weight less than 2 which implies b=g, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=|t1xbxtbxg and R2=x1t1at. The star graph Γ50 for the relative presentation Q is given by the Figure 12.

    (1) Observe that, (1θ(γj))=5(12+1+12+1+0)=2 and (1θ(ηj))=3(12+0+12)=2.

    (2) Each admissible cycle having weight less than 2 leads to a contradiction. For example γ2η1γ15 is a cycle of length 2 having weight less than 2 which implies b=g, a contradiction.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=x1t1axctxc and R2=x1tbt1. The star graph Γ51 for the relative presentation Q is given by the Figure 13.

    (1) Observe that, (1θ(γj))=5(1+0+0+1+1)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xat1x1atxta1 and R2=x1tbt1t2. The star graph Γ52 for the relative presentation Q is given by the Figure 13.

    (1) Observe that, (1θ(γj))=6(0+1+0+1+1+1)=2 and (1θ(ηj))=4(0+0+1+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbxtb1xb1 and R2=x1t1at. The star graph Γ53 for the relative presentation Q is given by the Figure 14.

    Figure 14.   .

    (1) Observe that, (1θ(γj))=5(12+1+12+1+0)=2 and (1θ(ηj))=3(12+0+12)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=xbt1x1bxtb1tb1 and R2=x1t1a1tbt1. The star graph Γ54 for the relative presentation Q is given by the Figure 14.

    (1) Observe that, (1θ(γj))=6(12+1+0+1+12+1)=2 and (1θ(ηj))=4(1+0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axa1txax1 and R2=x1tbt1. The star graph Γ55 for the relative presentation Q is given by the Figure 14.

    (1) Observe that, (1θ(γj))=5(0+1+1+0+1)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbx1tb1xb and R2=x1t1at. The star graph Γ56 for the relative presentation Q is given by the Figure 14.

    (1) Observe that, (1θ(γj))=5(1+0+1+0+1)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1xbx1tb1x1b and R2=x1t1at. The star graph Γ57 for the relative presentation Q is given by the Figure 14.

    (1) Observe that, (1θ(γj))=5(1+0+1+1+0)=2 and (1θ(ηj))=3(0+0+1)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In this case we have two relators R1=t1axatx1ax and R2=x1tbt1. The star graph Γ58 for the relative presentation Q is given by the Figure 15.

    Figure 15.  Γ58.

    (1) Observe that, (1θ(γj))=5(1+0+1+0+1)=2 and (1θ(ηj))=3(1+0+0)=2.

    (2) Each possible admissible cycle has weight atleast 2 so θ admits an aspherical weight function.

    (3) Each edge has non-negative weight.

    In all the cases, 3 conditions of weight test are satisfied.

    Hence the equation s(t)=atbt1ctdtet1ftgt1ht1=1 where a,b,c,e,f,gA/{1} and d,hA is solvable.

    In this article, we reviewed some basic concepts of combinatorial group theory (like torsion-free group, equations over groups, relative presentation, weight test) and discussed the two main conjectures in equations over torsion-free groups. We investigated all possible cases and solved the singular equation of length eight over torsion free group by using weight test. This result will be useful in dealing with equations over torsion-free groups.

    The authors wish to express their gratitude to Prince Sultan University for facilitating the publication of this article through the Theoretical and Applied Sciences Lab.

    All authors declare no conflicts of interest in this paper.



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