We remember here some definitions, notations and results which will be the basis of what follows 1 MULTIVALUED FUNCTIONS, FUZZY SUBSETS AND JOIN SPACES Piergiulio CORSINI,* Razieh MAHJOOB** * Dept. of Biology and Agro-Industrial Economy, Via delle Scienze 208, 33100 UDINE (ITALY) e-mail: corsini2002@yahoo.com web site: http://ijpam.uniud.it/journal/curriculum_corsini.htm ** Dept. of Mathematics - Faculty of Basic Science, University of Semnan, SEMNAN, (IRAN) ra_mahjoob@yahoo.com ABSTRACT One has considered the Hypergroupoid ΗΓ = < H;οΓ > associated with a multivalued function Γ from H to a set D, defined as follows: ∀ x ∈ H, x οΓ x = ⎨y⏐ Γ(y) ∩ Γ(x) ≠ ∅⎬ , ∀ (y,z) ∈ H2 , y οΓ z = y οΓ y ∪ z οΓ z , and one has calculated the fuzzy grade ∂(ΗΓ) for several functions Γ defined on sets H, such that ⎮H⎮ ∈ ⎨3, 4, 5, 6, 8, 9, 16⎬. INTRODUCTION The analysis of the connections between Hyperstructures and Fuzzy Sets dates since 1993 when Corsini defined and studied the join spaces Hμ obtained from the fuzzy set < Η, μ > , and a little later Zahedi and Ameri considered fuzzy hypergroups. These subjects were studied in the following years by several scientists in Romania, Iran, Greece, Italy, Canada. In 1993 Corsini associated a hypergroupoid with every fuzzy subset, and he proved that this hypergroupoid is a join space [8]. In 2003 Corsini [14] associated a fuzzy set μΗ with every hypergroupoid < Η, o > and considered the sequence of the fuzzy subsets μΗ and of the join spaces Hμ constructed from a hypergroup. This sequence has been studied in depth for several classes of hypergroups by Corsini [14], Corsini–Cristea [16], [17], [18], Corsini–Leoreanu-Fotea [22], Corsini– Leoreanu–Iranmanesh [23], Cristea [25], [26], Stefanescu–Cristea [70], Leoreanu-Fotea V. – Leoreanu L. [ 53] . In this paper one has considered the hypergroupoid < Η, oΓ > associated with a multivalued function Γ from a set H to a set D, defined as follows ∀ x ∈ H, x oΓ x = {y⏐Γ(y) ∩ Γ(x) ≠ ∅ }, Ratio Mathematica, 20, 2010 2 ∀ (y,z) ∈ H2 , y oΓ z = y oΓ y ∪ z oΓ z and one has calculated the fuzzy grade ∂(HΓ) , for several functions Γ defined on sets H such that ⏐Η⏐ ∈ {3, 4, 5, 6, 8, 9, 16}. We can remark that we have ∂(Η) = s+1, for all the examinated cases with the exception of (136 ), (236), (336), (129), if n = 2sq , where m.c.d. (q,2) = 1. We remember here some definitions, notations and results which will be the basis of what follows. With every fuzzy subset (H; μA ) of a set Η , it is possible to associate a hypergroupoid <Η; ομ>, where the hyperoperation < ομ> is defined by: ∀ (x,y) ∈ H2 , (I) x ομ y = ⎨ z ⏐ min ⎨ μA(x), μA(y)⎬ ≤ μA(z) ≤ max ⎨ μA(x), μA(y)⎬ ⎬ One proved [8] that <Η; ομ> is a join space. With every hypergroupoid < Η; ο >, it is possible to associate a fuzzy subset, as follows: Set ∀ (x,y) ∈ H2 , ∀ u ∈ H , μx,y (u) = 0 ⇔ u ∉ x ο y if u ∈ x ο y , μx,y (u) = 1/⏐x ο y⏐ , set ∀ u ∈ H , A(u) = ∑(x,y)∈H2 μx,y (u) , Q(u) = ⎨ (x,y) ⏐u ∈ x ο y ⎬ , q(u) = ⏐ Q(u) ⏐ , (II) μH(u) = A(u) / q(u) , see [14]. So it is clear that, given a hypergroupoid < Η; ο >, a sequence of fuzzy subsets and of join spaces is determined μH = μ1 , μ2 ,…. μm+1 ,…, < Η; ο > = 0H , 1H, … mH…, such that ∀j ≥ 1, jμ = Hj 1−μ , and jH is the join space associated, after (I), with jμ . We call “fuzzy grade of H”, if it exists, the number ∂(H) (or f.g.(H)) = min ⎨s⏐ mH ≈ m+1H ⎬ and “strong fuzzy grade of H” , if it exists, the number s.f.g.(H) = min ⎨s⏐ mH = m+1H ⎬ , see [17]. In this paper one has determined 6 hypergroupoids of 3 elements such that ∂(H) = 0, 4 hypergroupoids of 3 elements such that ∂(H) = 1, 5 hypergroupoids of 4 elements such that ∂(H) = 0, 8 hypergroupoids of 4 elements such that ∂(H) = 1, 12 hypergroupoids of 4 elements such that ∂(H) = 2, 5 hypergroupoids of 4 elements such that ∂(H) = 3, 2 hypergroupoids of 5 elements such that ∂(H) = 1, 2 hypergroupoids of 6 elements such that ∂(H)= 1, 8 hypergroupoids of 6 elements such that ∂(H) = 2, 3 hypergroupoids of 6 elements such that ∂(H) = 3, 1 hypergroupoid of 8 elements such that ∂(H)= 4, 1 hypergroupoid of 9 elements such that ∂(H)= 2, 1 hypergroupoid of 16 elements such that ∂(H)= 5. Ratio Mathematica, 20, 2010 3 $ 1. Let Γ be a multivalued function from a set H = {u1, u2,..., un} to a set D, i.e. Γ : H → P*(D). Then we have the following THEOREM 1 If there exists d ∈ D, such that ∀i, Γ(ui) ∋ d, then ∂(HΓ) = 0. Indeed, we have ∀i, xi oΓ xi = {uj⎮Γ(uj) ∩ Γ(uj) ≠ Ø} = H, therefore ∀(i, j), ui oΓ uj = H. Whence oH = T, from which ∀s, sH = oH, so ∂(HΓ) = 0. THEOREM 2 Let Γ be a multivalued function from a set H to a set D, that is Γ : H → P*(D), and let < oΓ > be the hyperoperation defined in H : ∀x ∈ H, xoΓ x = {z⎮Γ(z) ∩ Γ( x) ≠ Ø}, ∀(y, z), y oΓ z = y oΓ y ∪ z oΓ z. Then the hypergroupoid < H; oΓ > is a commutative quasi-join space, that is ∀(a, b, c, d) ∈ H4, (j) a / b ∩ c / d ≠ Ø ⇒ a oΓ d ∩ b oΓ c ≠ Ø. Let’s suppose a / b ∩ c / d ∋ v, that is a ∈ b oΓ v, c ∈ d oΓ v. Then, since b o v = b oΓ b ∪ v oΓ v, d oΓ v = d oΓ d ∪ v oΓ v, and ∀(x, y) ∈ H2, y ∈ x oΓ x ⇒ x ∈ y oΓ y, at least one of the following cases is verified (I) a ∈ b oΓ b, c ∈ d oΓ d, (II) a ∈ b oΓ b, c ∈ v oΓ v (III) a ∈ v oΓ v, c ∈ d oΓ d, (IV) a ∈ v oΓ v, c ∈ v oΓ v (I) implies b ∈ a oΓ a, whence b ∈ a oΓ d, and we have also b ∈ b oΓ b ⊂ b oΓ c (II) We find b ∈ a oΓ d ∩ b oΓ c as in (I). (III) We obtain c ∈ d oΓ d ⊂ a oΓ d, and also c ∈ c oΓ c ⊂ b oΓ c. (IV) implies v ∈ a oΓ a ⊂ a oΓ d and also v ∈ c oΓ c ⊂ b oΓ c. Therefore the implication (j) is always satisfied whence < H; oΓ > is a quasi-join space. Ratio Mathematica, 20, 2010 4 $ 2. Set H = {u1, u2, u3}. Then there are functions Γ : H → P*(D) such that the fuzzy grade of the associated sequence is respectively 0, 1. (103) Set Γ(u1) = {d1}, Γ(u2) = Γ(u3) = {d2, d3}. We have clearly So μ1(u1) = 0.467, μ1(u2) = μ1(u3) = 0.417. It follows 1H = 0H. By consequence ∂(103) = 0. (203) Set Γ(u1) = {d1, d2}, Γ(u2) = Γ(u3) = {d3}. We have One obtains μ1(u1) = 0.467, μ1(u2) = μ1(u3) = 0.417. So 1H = 0H, then ∂(20 3) = 0. (303) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d1}. We have 1H = 0H = T, ∂(303) = 0. (403) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2 }, Γ(u3) = {d3}. We have We obtain μ(1) = 0.417 = μ(2) , μ(3) = 0.467 , So ∂(403) = 0. (503) Set Γ(u1) = Γ(u2) = {d1}, Γ(u3) = {d3}. We have 0H u1 u2 u3 u1 u1 H H u2 u2 u3 u2 u3 u3 u2 u3 0H u1 u2 u3 u1 u1 H H u2 u2 u3 u2 u3 u3 u2 u3 0H u1 u2 u3 u1 H H H u2 H H u3 H 0H u1 u2 u3 u1 u1 u2 u1 u2 H u2 u1 u2 H u3 u3 Ratio Mathematica, 20, 2010 5 As in (403), we obtain ∂(503) = 0. (113) Let⎮H⎮= 3 = ⎮D⎮. Set Γ(u1) = {d1}, Γ(u2) = {d2}, Γ(u3) = {d3}. So we have We have clearly μ1(u1) = μ1(u2) = μ1(u3) = 0.6. Therefore we obtain 1H = T, whence ∂(11 3) = 1. (213) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2}, Γ(u3) = {d3}. We have We obtain : μ1(u1) = 0.37, μ1(u2) = μ1(u3) = 0.354. By consequence, So we have : μ2(u1) = 0.467, μ2(u2) = μ2(u3) = 0.417. From this, we obtain 2H = 1H, whence ∂(21 3) = 1. (313) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = { d1, d2}, Γ(u3) = {d3}. We have 0H u1 u2 u3 u1 u1 u2 u1 u2 H u2 u1 u2 H u3 u3 0H u1 u2 u3 u1 u1 u1 u2 u1 u3 u2 u2 u2 u3 u3 u3 0H u1 u2 u3 u1 H H H u2 u1 u2 H u3 u1 u3 1H u1 u2 u3 u1 u1 H H u2 u2 u3 u2 u3 u3 u2 u3 Ratio Mathematica, 20, 2010 6 See (213). So we obtain again ∂(313) = 1. (413) Set H = {u1, u2, u3}, Γ(u1) = {d1}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d1}. So we have By consequence μ1(u1) = 0.354 = μ1(u2), μ1(u3) = 0.370. Therefore we obtain Hence μ2(u1) = μ2(u2) = 0.4167, μ2(u3) = 0.467. It follows 2H = 1H. Therefore ∂(413) = 1. $ 3. Set H = {u1, u2, u3, u4}. Then there are functions Γ : H → P*(D) such that the fuzzy grade of the associated sequence is respectively 0, 1, 2, 3. (104) Set Γ(u1) = {d1, d2}, Γ(u2) = Γ(u3) = Γ(u4) = {d3, d4}. Then we have We obtain μ1(u1) = 0.357, μ1(u2) = μ1(u3) = μ1(u4) = 0.300. By consequence 1H = 0H and therefore ∂(104 )= 0. 0H u1 u2 u3 u1 H H H u2 u1 u2 H u3 u1 u3 0H u1 u2 u3 u1 u1 u3 H H u2 u2 u3 H u3 H 1H u1 u2 u3 u1 u1 u2 u1 u2 H u2 u1 u2 H u3 u3 0H u1 u2 u3 u4 u1 u1 H H H u2 u2 u3 u4 u2 u3 u4 u2 u3 u4 u3 u2 u3 u4 u2 u3 u4 u4 u2 u3 u4 Ratio Mathematica, 20, 2010 7 (204 ) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = Γ(u3) = Γ(u4) = {d4} . Then We have as in (1) 0H = 1H so ∂(20 4) = 0. (304) Set Γ(u1) = {d1, d2}, Γ(u2) = {d3, d4}, Γ(u3) = Γ(u4) = {d4}. Also in this case By consequence 0H = 1H from which ∂(304) = 0. (404 ) Set Γ(u1) = {d1, d2, d3, d4}, Γ(u2) = {d2, d3, d4}, Γ(u3) = Γ(u4) = {d4}. We have Clearly, 1H = 0H = T. So ∂(404) = 0. (114) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d3}, Γ(u4) = {d4}. We have Whence μ1(u1) = μ1(u3) = 0.333 μ1(u2) = 0.344, μ1(u4) = 0.405. from which we obtain 1H : 0H u1 u2 u3 u4 u1 u1 H H H u2 u2 u3 u4 u2 u3 u4 u2 u3 u4 u3 u2 u3 u4 u2 u3 u4 u4 u2 u3 u4 0H u1 u2 u3 u4 u1 u1 H H H u2 u2 u3 u4 u2 u3 u4 u2 u3 u4 u3 u2 u3 u4 u2 u3 u4 u4 u2 u3 u4 0H u1 u2 u3 u4 u1 H H H H u2 H H H u3 H H u4 H 0H u1 u2 u3 u4 u1 u1 u2 u1 u2 u3 u1 u2 u3 u1 u2 u4 u2 u1 u2 u3 u1 u2 u3 H u3 u2 u3 u2 u3 u4 u4 u4 Ratio Mathematica, 20, 2010 8 Hence μ2(u1) = μ2(u3) = 0.36, μ2(u2) = 0.394, μ2(u4) = 0.429. By consequence 2H = 1H , then ∂(11 4 )= 1. (214) Set Γ(u1) = {d1, d2}, Γ(u2) = {d3, d4}, Γ(u3) = {d3}, Γ(u4) = {d4}. So we have Whence μ1(u1) = 0.405 μ1(u2) = 0.344, μ(u3) = μ(u4) = 0.3. We obtain 1H : So we have : μ2(u1) = 0.429, μ2(u2) = 0.394, μ2(u3) = μ2(u4) = 0.361 whence one finds that 2H = 1H. It follows ∂(21 4) = 1. (314) Set Γ(u1) = {d1}, Γ(u2) = {d2}, Γ(u3) = {d3}, Γ(u4) = {d4}. So Then ∀ i, μ1(ui) = 0.571. By consequence 1H = T and therefore ∂(314) = 1. 1H u1 u2 u3 u4 u1 u1 u3 u1 u3 u1 u3 u2 H u2 u1 u3 u1 u3 u2 H u3 u2 u2 u4 u4 u4 0H u1 u2 u3 u4 u1 u1 H u1 u2 u3 u1 u2 u4 u2 u2 u3 u4 u2 u3 u4 u2 u3 u4 u3 u2 u3 u2 u3 u4 u4 u2 u4 1H u1 u2 u3 u4 u1 u1 u1 u2 H H u2 u2 u2 u3 u4 u2 u3 u4 u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u1 u2 u1 u3 u1 u4 u2 u2 u2 u3 u2 u4 u3 u3 u3 u4 u4 u4 Ratio Mathematica, 20, 2010 9 (414) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d1}, Γ(u3) = {d2}, Γ(u4) = {d3}. We have So μ1(u1) = 0.328, μ1(u2) = μ1(u3) = μ1(u4) = 0.299. Hence So μ2(u1) = 0.357 μ2(u2) = μ2(u3) = μ2(u4) = 0.3 from which 2H = 1H. Therefore ∂(41 4 )= 1. (514) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = Γ(u4) = {d4}. We have μ1(u1) = μ1(u2) = μ1(u3) = μ1(u4) = 0.333. So 1H = T, whence ∂(51 4) = 1. (614) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2}, Γ(u3) = {d3, d4}, Γ(u4) = {d4}. We have clearly ∀ i, μ1(u2) = μ1(u1). Therefore 1H = T and by consequence ∂(61 4 )= 1. (714) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d3}, Γ(u3) = {d4}, Γ(u4) = {d4}. 0H u1 u2 u3 u4 u1 H H H H u2 u1 u2 u1 u2 u3 u1 u2 u4 u3 u1 u3 u1 u3 u4 u4 u1 u4 1H u1 u2 u3 u4 u1 u1 H H H u2 u2 u3 u4 u2 u3 u4 u2 u3 u4 u3 u2 u3 u4 u2 u3 u4 u4 u2 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 Ratio Mathematica, 20, 2010 10 See (514) and (614). We have 1H = T, whence ∂(714) = 1. (604) Set Γ(u1) = {d1, d2}, Γ(u2) = {d3, d4}, Γ(u3) = Γ(u4) = {d4}. We have So ∂(604 )= 0. (504) Set Γ(u1) = {d1, d2}, Γ(u2) = Γ(u3) = Γ(u4) = {d4}. We have So μ(1) = 0.357 , μ(2) = μ(3) = μ(4)=0.3 . It follows ∂(504 )=0. (124 ) Set Γ(u1) = {d1, d2}, Γ(u2) = Γ(u3) = {d2, d3}, Γ(u4) = {d3, d4}. One obtains μ1(u1) = 0.256 = μ1(u4), μ1(u2) = μ1(u3) = 0.260, whence we obtain 1H: 0H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 H H H u2 u2 u3 u4 u2 u3 u4 u2 u3 u4 u3 u2 u3 u4 u2 u3 u4 u4 u2 u3 u4 0H u1 u2 u3 u4 u1 u1 H H H u2 u2 u3 u4 u2 u3 u4 u2 u3 u4 u3 u2 u3 u4 u2 u3 u4 u4 u2 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 u3 H H H u2 H H H u3 H H u4 u4 u2 u3 Ratio Mathematica, 20, 2010 11 Therefore 2H = T (the total hypergroup). Then ∂(12 4 )= 2. (224) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d4}, Γ(u4) = {d4}. Then whence μ1(u1) = 0.292 = μ1(u4), μ1(u2) = μ1(u3) = 0.3. So, we have Then μ1(u1) = μ1(u4) = μ1(u2) = μ1(u3) , whence 2H = T, and by consequence ∂(224 )= 2. (324) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d3, d4}, Γ(u3) = { d2, d4}, Γ(u4) = {d3}. We have μ1(u1) = μ1(u2) = 0.260, μ1(u3) = μ1(u4) = 0.256, whence we obtain 1H u1 u4 u2 u3 u1 u1 u4 u1 u4 H H u4 u1 u4 H H u2 u2 u3 u2 u3 u3 u2 u3 0H u1 u2 u3 u4 u1 u1 u2 u1 u2 u3 H H u2 u1 u2 u3 H H u3 u2 u3 u4 u2 u3 u4 u4 u3 u4 1H u1 u4 u2 u3 u1 u1 u4 u1 u4 H H u4 u1 u4 H H u2 u2 u3 u2 u3 u3 u2 u3 0H u1 u2 u3 u4 u1 H H H H u2 H H H u3 u1 u2 u3 H u4 u1 u2 u4 Ratio Mathematica, 20, 2010 12 Then 2H = T, from which ∂(32 4)= 2. (424 ) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d4} , Γ(u3) = { d3}, Γ(u4) = {d4}. We have: Then μ1(u1) = 0.3 = μ1(u2), μ1(u3) = μ1(u4) = 0.292. It follows Therefore 2H = T, whence ∂(42 4 )= 2. (524 ) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = Γ(u3) = {d3, d4}, Γ(u4) = {d4}. We have So μ1(u1) = μ1(u4) = 0.23, μ1(u2) = μ1(u3) = 0.260. By consequence, we obtain Therefore we have 2H = T, whence ∂(52 4 )= 2. 1H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 u3 H u1 u2 u3 H u2 u1 u2 u4 H u1 u2 u4 u3 u1 u3 H u4 u2 u4 1H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 u3 H H H u2 H H H u3 H H u4 u2 u3 u4 1H u1 u4 u2 u3 u1 u1 u4 u1 u4 H H u4 u1 u4 H H u2 u2 u3 u2 u3 u3 u2 u3 Ratio Mathematica, 20, 2010 13 (624) Set Γ(u1) = {d1}, Γ(u2) = {d1, d2}, Γ(u3) = {d2, d3, d4}, Γ(u4) = {d4}. We have Hence μ1(u1) = 0.292 = μ1(u4), μ1(u2) = μ1(u3) = 0.3. We obtain whence ∀ i, μ1(ui) = μ1(u1). Then 2H = T. Therefore ∂(62 4 )= 2. (724) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d4}, Γ(u3) = {d3, d4}, Γ(u4) = {d4}. We have μ1(u1) = μ1(u4) = 0.256, μ1(u2) = μ1(u3) = 0.260. So, we obtain 1H: Then ∀ i, μ2(ui) = 0.389, so 2H = T, and ∂(72 4 )= 2. 0H u1 u2 u3 u4 u1 u1 u2 u1 u2 u3 H H u2 u1 u2 u3 H H u3 u2 u3 u4 u2 u3 u4 u4 u3 u4 1H u1 u4 u2 u3 u1 u1 u4 u1 u4 H H u4 u1 u4 H H u2 u2 u3 u2 u3 u3 u2 u3 0H u1 u2 u3 u4 u1 u1 u2 u3 H H H u2 H H H u3 H H u4 u2 u3 u4 1H u1 u4 u2 u3 u1 u1 u4 u1 u4 H H u4 u1 u4 H H u2 u2 u3 u2 u3 u3 u2 u3 Ratio Mathematica, 20, 2010 14 (824) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2}, Γ(u3) = {d3}, Γ(u4) = {d4}. We have We obtain μ1(u1) = 0.389 = μ1(u2), μ1(u3) = 0.476 = μ1(u4). By consequence, Therefore 2H = T and ∂(82 4 )= 2. (924) Set Γ(u1) = Γ(u2) = {d1, d2}, Γ(u3) = {d3}, Γ(u4) = {d4}. We have See (824). Therefore ∂(924 )= 2. (1024 ) Set Γ(u1) = {d1, d2}, Γ(u2) = {d3}, Γ(u3) = Γ(u4) = {d4}. We obtain whence μ1(u1) = μ1(u2) = 0.476, μ(u3) = μ(u4) = 0.389. So, we obtain 0H u1 u2 u3 u4 u1 u1 u2 u1 u2 u1 u2 u3 u1 u2 u4 u2 u1 u2 u1 u2 u3 u1 u2 u4 u3 u3 u3 u4 u4 u4 1H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 u1 u2 u1 u2 u3 u1 u2 u4 u2 u1 u2 u1 u2 u3 u1 u2 u4 u3 u3 u3 u4 u4 u4 0H u1 u2 u3 u4 u1 u1 u1 u2 u1 u3 u4 u1 u3 u4 u2 u2 u2 u3 u4 u2 u3 u4 u3 u3 u4 u3 u4 u4 u3 u4 Ratio Mathematica, 20, 2010 15 Hence 2H = T, from which ∂(102 4 )= 2. (1124 ) Set Γ(u1) = {d1}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d4}, Γ(u4) = {d4, d1}. We have Hence μ1(u1) = μ1(u3) = 0.291, μ1(u3) = μ1(u4) = 0.3. One obtains 1H as follows: From 1H, one finds μ2(ui) = μ2(uj), ∀(i, j). Therefore 2H = T, so ∂(112 4 )= 2. (1224) Let Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d3}, Γ(u4) = {d1}. We have Then μ1(u1) = 0.3, μ1(u2) = 0.3, μ1(u3) = 0.2917, μ1(u4) = 0.2917. By consequence, 1H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u4 H H u1 u3 u4 u2 u2 u3 u2 u3 u4 H u3 u2 u3 u4 H u4 u1 u3 u4 1H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 u4 H H u1 u2 u4 u2 u1 u2 u3 u1 u2 u3 H u3 u2 u3 H u4 u1 u4 Ratio Mathematica, 20, 2010 16 Therefore we have 2H = T (the total hypergroup) whence ∂(1224 )= 2. (134) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d2}, Γ(u4) = {d3}. We have μ1(u1) = 0.272 = μ1(u3), μ1(u2) = 0.286, μ1(u4) = 0.271, whence So we have μ2(u2) = 0.405 = μ2(u4), μ2(u1) = μ2(u3) = 0.34. We obtain We have clearly μ3(u1) = μ3(u3) = μ3(u2) = μ3(u4). Therefore 3H is the total hypergroup of order 4 and ∂(134 )= 3. 1H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 u3 H u1 u2 u3 H u2 H H H u3 u1 u2 u3 H u4 u2 u4 1H u2 u1 u3 u4 u2 u2 u2 u1 u3 u2 u1 u3 H u1 u1 u3 u1 u3 u1 u3 u4 u3 u1 u3 u1 u3 u4 u4 u4 2H u1 u3 u2 u4 u1 u1 u3 u1 u3 H H u3 u1 u3 H H u2 u2 u4 u2 u4 u4 u2 u4 Ratio Mathematica, 20, 2010 17 (234) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = Γ(u4) = {d3, d4}. We obtain the following We have μ1(u1) = 0.27083, μ1(u2) = 0.286, μ1(u3) = μ1(u4) = 0.272. Therefore the second hypergroupoid is Hence μ2(u2) = μ2(u1) = 0.405, μ2(u3) = μ2(u4) = 0.369. By consequence we have again From 2H we obtain 3H = T. Then ∂(23 4 )= 3. (334) If Γ(u1) = {d1, d2, d3}, Γ(u2) = {d3, d4}, Γ(u3) = Γ(u4) = {d4}. One finds the same sequence as in (234). Therefore ∂(334 )= 3 0H u1 u2 u3 u4 u1 u1 u2 H H H u2 H H H u3 u2 u3 u4 u2 u3 u4 u4 u2 u3 u4 1H u2 u3 u4 u1 u2 u2 u2 u3 u4 u2 u3 u4 H u3 u3 u4 u3 u4 u3 u4 u1 u4 u3 u4 u3 u4 u1 u1 u1 2H u2 u1 u3 u4 u2 u2 u1 H H H u1 u2 u1 H H u3 u3 u4 H u4 u3 u4 0H u1 u2 u3 u4 u1 u1 u2 H H H u2 H H H u3 u2 u3 u4 u2 u3 u4 u4 u2 u3 u4 Ratio Mathematica, 20, 2010 18 (434) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d4}, Γ(u4) = {d4}. We have So μ1(u1) = 0.272 = μ1(u2), μ1(u3) = 0.286, μ1(u4) = 0.271. Hence Then μ2(u3) = μ2(u4) = 0.405, μ2(u1) = μ2(u2) = 0.369, from which we obtain Hence 3H = T and ∂(43 4 )= 3. $ 4. Set H = {u1, u2, u3, u4, u5}. Then there are functions Γ : H → P*(D) such that the fuzzy grade of the associated sequence is respectively 1, 2. (115) Let ⎮H⎮= 5 = ⎮D⎮, Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d4}, Γ(u4) = {d4}, Γ(u5) = {d5}. We have So μ1(u1) = 0.29167= μ1(u4), μ1(u2) = μ1(u3) = 0.2936, μ1(u5) = 0.370. We obtain 0H u1 u2 u3 u4 u1 u1 u2 u3 u1 u2 u3 H H u2 u1 u2 u3 H H u3 H H u4 u3 u4 1H u3 u1 u2 u4 u3 u3 u3 u1 u2 u3 u1 u2 H u1 u1 u2 u1 u2 u1 u2 u4 u2 u1 u2 u1 u2 u4 u4 u4 2H u1 u2 u3 u4 u1 u1 u2 u1 u2 H H u2 u1 u2 H H u3 u3 u4 u3 u4 u4 u3 u4 0H u1 u2 u3 u4 u5 u1 u1 u2 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u1 u2 u5 u2 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u1 u2 u3 u5 u3 u2 u3 u4 u2 u3 u4 u2 u3 u4 u5 u4 u3 u4 u3 u4 u5 u5 u5 Ratio Mathematica, 20, 2010 19 From this we have μ2(u5) = 0.348, μ2(u2) = μ2(u3) = 0.3067, μ2(u1) = μ2(u4) = 0.3. So μ2(u5) > μ2(u2) = μ(u3) > μ(u1) = μ(u4). It follows that 2H = 1H. Therefore ∂(11 5) = 1. (225) Set⎮H⎮= 5 = ⎮D⎮, Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d3, d4}, Γ(u3) = {d3, d4, d5}, Γ(u4) = {d4}, Γ(u5) = {d5}. So we have Whence we obtain μ1(u1) = 0.228, μ1(u2) = 0.234, μ1(u3) = 0.2447, μ1(u4) = μ1(u1) μ1(u5) = 0.231. We have μ2(u3) = 0.3852, μ2(u2) = 0.3644, μ2(u5) = 0.3412, μ2(u1) = μ2(u4) = 0.3208. So 2H = 1H and by consequence ∂(22 5) = 1. $ 5. Set H = {u1, u2, u3, u4, u5, u6}. Then there are functions Γ : H → P*(D) such that the fuzzy grade of the associated sequence is respectively 1, 2, 3. 1H u5 u2 u3 u1 u4 u5 u5 u5 u2 u3 u5 u2 u3 H H u2 u2 u3 u2 u3 u2 u3 u1 u4 u2 u3 u1 u4 u3 u2 u3 u2 u3 u1 u4 u2 u3 u1 u4 u1 u1 u4 u1 u4 u4 u1 u4 0H u1 u2 u3 u4 u5 u1 u1 u2 u3 u1 u2 u3 u4 H u1 u2 u3 u4 u1 u2 u3 u5 u2 u1 u2 u3 u4 H u1 u2 u3 u4 H u3 H H H u4 u2 u3 u4 u2 u3 u4 u5 u5 u3 u5 1H u3 u2 u5 u1 u4 u3 u3 u3 u2 u3 u2 u5 H H u2 u2 u2 u5 u2 u5 u1 u4 u2 u5 u1 u4 u5 u5 u5 u1 u4 u5 u1 u4 u1 u1 u4 u1 u4 u4 u1 u4 Ratio Mathematica, 20, 2010 20 (116) Set ⎮H⎮= 6 = ⎮D⎮, Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d4}, Γ(u4) = {d4, d5}, Γ(u5) = Γ(u6) = { d5, d6}. We have So μ1(u1) = 0.231667, μ1(u2) = 0.2284, μ1(u3) = 0.22654, μ1(u4) = 0.22656, μ1(u5) = μ1(u6) = 0.219. Hence we obtain Therefore μ2(u1) = 0.348, μ2(u2) = 0.3315, μ2(u4) = 0.317, μ2(u3) = 0.303, μ2(u5) = μ2(u6) = 0.29. By consequence 2H = 1H, hence ∂(11 6) = 1. (216) Set Γ(u1) = {d1}, Γ(u2) = {d2, d3, d4}, Γ(u3) = {d3, d4, d5}, Γ(u4) = {d4}, Γ(u5) = {d5}, Γ(u6) = {d5, d6}. We have 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u1 u2 u3 u1 u2 u3 u4 H u1 u2 u4 u5 u6 u1 u2 u4 u5 u6 u2 u1 u2 u3 u1 u2 u3 u4 H H H u3 u2 u3 u4 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u4 u3 u4 u5 u6 u3 u4 u5 u6 u3 u4 u5 u6 u5 u4 u5 u6 u4 u5 u6 u6 u4 u5 u6 1H u1 u2 u4 u3 u5 u6 u1 u1 u1 u2 u1 u2 u4 u1 u2 u3 u4 H H u2 u2 u2 u4 u2 u4 u3 u2 u4 u3 u5 u6 u2 u4 u3 u5 u6 u4 u4 u4 u3 u4 u3 u5 u6 u4 u3 u5 u6 u3 u3 u3 u5 u6 u3 u5 u6 u5 u5 u6 u5 u6 u6 u5 u6 Ratio Mathematica, 20, 2010 21 We obtain μ1(u1) = 0.303 μ1(u2) = 0.22469 μ1(u3) = 0.24 μ1(u4) = μ1(u2) = μ1(u5) = μ1(u6). Setting {u2, u4, u5, u6} = P, we have One finds 2H = 1H. So ∂(21 6) = 1. (126) Set ⎮H⎮= 6 = ⎮D⎮, Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d4}, Γ(u4) = {d4, d5}, Γ(u5) = {d5}, Γ(u6) = {d6}. So we have μ1(u1) = 0.268, μ1(u2) = 0.267, μ1(u3) = 0.260, μ1(u4) = μ1(u2), μ1(u5) = μ1(u1), μ1(u6) = 0.348. 0H u1 u2 u3 u4 u5 u6 u1 u1 u1 u2 u3 u4 H u1 u2 u3 u4 u1 u3 u5 u6 u1 u3 u5 u6 u2 u2 u3 u4 u2 u3 u4 u5 u6 u2 u3 u4 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u3 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u4 u2 u3 u4 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u5 u3 u5 u6 u3 u5 u6 u6 u3 u5 u6 1H u1 u3 u2 u4 u5 u6 u1 u1 u1 u3 H H H H u3 u3 H H H H u2 P P P P u4 P P P u5 P P u6 P 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 u1 u2 u4 u5 u1 u2 u6 u2 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 u1 u2 u3 u4 u5 u1 u2 u3 u6 u3 u2 u3 u4 u2 u3 u4 u5 u2 u3 u4 u5 u2 u3 u4 u6 u4 u3 u4 u5 u3 u4 u5 u3 u4 u5 u6 u5 u4 u5 u4 u5 u6 u6 u6 Ratio Mathematica, 20, 2010 22 Therefore we obtain So μ2(u6) = μ2(u3) = 0.315 μ2(u1) = μ2(u5) = μ2(u2) = μ2(u4) = 0.279. Setting Γ = {u1, u5, u2, u4}, Q = {u6, u3} we have So we have μ1(u1) = μ1(u5) = μ1(u2) = μ1(u4) = 0.208 μ1(u6) = μ1(u3) = 0.233 It follows 3H = 2H, by consequence ∂(12 6) = 2. (226) Set⎮D⎮= 6 = ⎮H⎮, Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d4}, Γ(u4) = {d4, d5}, Γ(u5) = {d5, d6}, Γ(u6) = {d6}. We have So we obtain μ1(u1) = 0.2467 = μ1(u6), μ1(u2) = 0.243 = μ1(u5), μ1(u3) = μ1(u4) = 0.2407 Whence 1H u6 u1 u5 u2 u4 u3 u6 u6 u1 u5 u6 u1 u5 u6 u1 u5 u6 u2 u4 u1 u5 u6 u2 u4 H u1 u1 u5 u1 u5 u1 u5 u2 u4 u1 u5 u2 u4 u1 u5 u2 u4 u3 u5 u1 u5 u1 u5 u2 u4 u1 u5 u2 u4 u1 u5 u2 u4 u3 u2 u2 u4 u2 u4 u2 u4 u3 u4 u2 u4 u2 u4 u3 u3 u3 2H u1 u5 u2 u4 u6 u3 u1 P P P P H H u5 P P P H H u2 P P H H u4 P H H u6 Q Q u3 Q 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 u1 u2 u4 u5 u6 u1 u2 u5 u6 u2 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 H u1 u2 u3 u5 u6 u3 u2 u3 u4 u2 u3 u4 u5 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u4 u3 u4 u5 u3 u4 u5 u6 u3 u4 u5 u6 u5 u4 u5 u6 u4 u5 u6 u6 u5 u6 Ratio Mathematica, 20, 2010 23 Hence μ2(u1) = μ2(u6) = μ2(u3) = μ2(u4) = 0.2667, μ2(u2) = μ2(u5) = 0.2619. Therefore we set P = {u1, u6, u3, u4}. We obtain We have clearly 3H = 2H, whence ∂(22 6) = 2. (326) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = { d3, d4}, Γ(u3) = { d4, d5}, Γ(u4) = {d5}, Γ(u5) = {d5, d6}, Γ(u6) = {d6}. So we have μ1(u1) = 0.233, μ1(u2) = 0.230, μ1(u3) = 0.228, μ1(u4) = 0.218, μ1(u5) = μ1(u3) = 0.228, μ1(u6) = 0.231. We obtain 1H u1 u6 u2 u5 u3 u4 u1 u1 u6 u1 u6 u1 u6 u2 u5 u1 u6 u2 u5 H H u6 u1 u6 u1 u6 u2 u5 u1 u6 u2 u5 H H u2 u2 u5 u2 u5 u2 u5 u3 u4 u2 u5 u3 u4 u5 u2 u5 u2 u5 u3 u4 u2 u5 u3 u4 u3 u3 u4 u3 u4 u4 u3 u4 2H u1 u6 u3 u4 u2 u5 u1 P P P P H H u6 P P P H H u3 P P H H u4 P H H u2 u2 u5 u2 u5 u5 u2 u5 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u1 u2 u3 u1 u2 u3 u4 u5 u1 u2 u3 u4 u5 H u1 u2 u5 u6 u2 u1 u2 u3 u1 u2 u3 u4 u5 u1 u2 u3 u4 u5 H u1 u2 u3 u5 u6 u3 u2 u3 u4 u5 u2 u3 u4 u5 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u4 u3 u4 u5 u3 u4 u5 u6 u3 u4 u5 u6 u5 u3 u4 u5 u6 u3 u4 u5 u6 u6 u5 u6 Ratio Mathematica, 20, 2010 24 We have μ2(u1) = 0.345 > μ2(u6) = 0.326 > μ2(u4) = 0.324> μ2(u2) = 0.306 > μ2(u3) = μ2(u5) = 0.296. Therefore we have We obtain now μ3(u1) = 0.348 > μ3(u6) = 0.33158> μ3(u4) = 0.317 > μ3(u2) = 0.303 > μ3(u3) = μ3(u5) = 0.29. Therefore 3H = 2H and it follows that ∂(32 6) = 2. (426) Set Γ(u1) = {d1, d2}, Γ(u2) = { d2, d3}, Γ(u3) = { d3, d4}, Γ(u4) = {d2}, Γ(u5) = {d3}, Γ(u6) = {d4}. We have 1H u1 u6 u2 u3 u5 u4 u1 u1 u1 u6 u1 u6 u2 u1 u6 u2 u3 u5 u1 u6 u2 u3 u5 H u6 u6 u6 u2 u6 u2 u3 u5 u6 u2 u3 u5 u6 u2 u3 u5 u4 u2 u2 u2 u3 u5 u2 u3 u5 u2 u3 u5 u4 u3 u3 u5 u3 u5 u3 u5 u4 u5 u3 u5 u3 u5 u4 u4 u4 2H u1 u6 u4 u2 u3 u5 u1 u1 u1 u6 u1 u6 u4 u1 u6 u4 u2 H H u6 u6 u6 u4 u6 u4 u2 u6 u4 u2 u3 u5 u6 u4 u2 u3 u5 u4 u4 u4 u2 u2 u3 u5 u2 u3 u5 u2 u2 u2 u3 u5 u2 u3 u5 u3 u3 u5 u3 u5 u5 u3 u5 Ratio Mathematica, 20, 2010 25 We find : μ1(u1) = 0.210 = μ1(u4) μ1(u2) = 0.221 μ1(u3) = 0.216 μ1(u5) = 0.208 μ1(u6) = 0.219. Hence we obtain We have : μ2(u5) = 0.324 μ2(u2) = 0.345 μ2(u6) = 0.326 μ2(u3) = 0.3058 μ2(u1) = μ2(u4) = 0.296. Therefore μ2(u2) > μ2(u6) > μ2(u5) > μ2(u3) > μ2(u1) = μ2(u4). Therefore we have We can see that 3H = 2H and it follows that ∂(426) = 2. 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u4 u1 u2 u3 u4 u5 H u1 u2 u4 u1 u2 u3 u4 u5 u1 u2 u3 u4 u6 u2 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 u1 u2 u3 u4 u5 H u3 u2 u3 u5 u6 H u2 u3 u5 u6 u2 u3 u5 u6 u4 u1 u2 u4 u1 u2 u3 u4 u5 u1 u2 u3 u4 u6 u5 u2 u3 u5 u2 u3 u5 u6 u6 u3 u6 1H u5 u4 u1 u3 u6 u2 u5 u5 u5 u4 u1 u5 u4 u1 u5 u4 u1 u3 u5 u4 u1 u3 u6 H u4 u4 u1 u4 u1 u3 u1 u4 u4 u1 u3 u6 u4 u1 u3 u6 u2 u1 u4 u1 u3 u1 u4 u4 u1 u3 u6 u4 u1 u3 u6 u2 u3 u3 u3 u6 u3 u6 u2 u6 u6 u6 u2 u2 u2 2H u2 u6 u5 u3 u1 u4 u2 u2 u2 u6 u2 u6 u5 u2 u6 u5 u3 H H u6 u6 u6 u5 u6 u5 u3 u6 u5 u3 u1 u4 u6 u5 u3 u1 u4 u5 u5 u5 u3 u5 u3 u1 u4 u5 u3 u1 u4 u3 u3 u3 u1 u4 u3 u1 u4 u1 u1 u4 u1 u4 u4 u1 u4 Ratio Mathematica, 20, 2010 26 (526) Set Γ(u1) = {d1}, Γ(u2) = {d2, d3}, Γ(u3) = {d3, d4}, Γ(u4) = {d4, d5}, Γ(u5) = {d5, d6} Γ(u6) = {d6}. We have We obtain μ1(u1) = 0.348, μ1(u2) = 0.268=μ1(u6), μ1(u3) = 0.2667=μ1(u5), μ1(u4) = 0.260. So, we have Now we obtain μ2(u1) = 0.324 μ2(u4)= 0.315, μ2(u5) = μ2(u3) =μ2(u2) = μ2(u6) =0.279. Setting P = { u2, u6 ,u3, u5 }, we find 2H We have clearly 3H = 2H whence ∂(52 6) = 2. 0H u1 u2 u3 u4 u5 u6 u1 u1 u1 u2 u3 u1 u2 u3 u4 u1 u3 u4 u5 u1 u4 u5 u6 u1 u5 u6 u2 u2 u3 u2 u3 u4 u2 u3 u4 u5 u2 u3 u4 u5 u6 u2 u3 u5 u6 u3 u2 u3 u4 u2 u3 u4 u5 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u4 u3 u4 u5 u3 u4 u5 u6 u3 u4 u5 u6 u5 u4 u5 u6 u4 u5 u6 u6 u5 u6 1H u1 u2 u6 u3 u5 u4 u1 u1 u1 u2 u6 u1 u2 u6 u1 u2 u6 u3 u5 u1 u2 u6 u3 u5 H u2 u2 u6 u2 u6 u2 u6 u3 u5 u2 u6 u3 u5 u2 u6 u5 u4 u3 u6 u2 u6 u2 u6 u3 u5 u2 u6 u3 u5 u2 u6 u5 u4 u3 u3 u3 u5 u3 u5 u5 u4 u3 u5 u3 u5 u5 u4 u3 u4 u4 2H u1 u4 u2 u6 u3 u5 u1 u1 u4 u1 u4 H H H H u4 u1 u4 H H H H u2 P P P P u6 P P P u3 P P u5 P Ratio Mathematica, 20, 2010 27 (626) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d3, d4}, Γ(u3) = {d4, d5}, Γ(u4) = {d5, d6}, Γ(u5) = {d5}, Γ(u6) = {d6}. So, denoting {ui, ui+1,…, uj-1, uj} by uij, we have We have μ1(u1) = 0.233, μ1(u2) = 0.2302, μ1(u3) = μ1(u4) =0.228, μ1(u5) = 0.218, μ1(u6) = 0.2308. Hence We have μ2(u1) = 0.345454, μ2(u6) = 0.326316, μ2(u2) = 0.305797, μ2(u3) = μ2(u4) = 0.29615, μ2(u5) = 0.32424. From this, we have 2H as follows One can see that 3H = 2H, therefore ∂(626) = 2. 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u13 u15 H u15 u1 u2 u4 u6 u2 u13 u15 H u15 u14 u6 u3 u25 u26 u25 u26 u4 u36 u36 u36 u5 u35 u36 u6 u4 u6 1H u1 u6 u2 u3 u4 u5 u1 u1 u1 u6 u1 u6 u2 u1 u6 u2 u3 u4 u1 u6 u2 u3 u4 H u6 u6 u6 u2 u6 u2 u3 u4 u6 u2 u3 u4 u6 u2 u3 u4 u5 u2 u2 u2 u3 u4 u2 u3 u4 u2 u5 u3 u4 u3 u3 u4 u3 u4 u3 u4 u5 u4 u3 u4 u3 u4 u5 u5 u5 2H u1 u6 u5 u2 u3 u4 u1 u1 u1 u6 u1 u6 u5 u1 u6 u5 u2 H H u6 u6 u6 u5 u6 u5 u2 u6 u5 u2 u3 u4 u6 u5 u2 u3 u4 u5 u5 u5 u2 u5 u2 u3 u4 u5 u2 u3 u4 u2 u2 u2 u3 u4 u2 u3 u4 u3 u3 u4 u3 u4 u4 u3 u4 Ratio Mathematica, 20, 2010 28 (726) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d4}, Γ(u3) = { d3, d4, d5}, Γ(u4) = {d4, d5, d6}, Γ(u5) = {d5}, Γ(u6) = {d6}. We have μ1(u1) = 0.22, μ1(u2) = 0.20864, μ1(u3) = 0.22762, μ1(u4) = μ1(u3) μ1(u5) = μ1(u2) = 0.20864, μ1(u6) = μ1(u1)= 0.22. Hence, μ1(u3) = μ1(u4) = 0.22762 > μ1(u1) = μ1(u6) = 0.22 > μ1(u2) = μ1(u5) = 0.20864. We obtain We have μ2(u3) = μ2(u4) = 0.2667, μ2(u1) = 0.2619 = μ2(u6), μ2(u2) = μ2(u5) = μ2(u3) = μ2(u4) = 0.2667. Set P = {u3, u4, u2, u5}, Q= {u1, u6}. We obtain It follows that μ3(u3) = μ3(u4) = μ3(u2) = μ3(u5) =0.208, μ3(u1) = μ3(u6) = 0.233. We have clearly 3H = 2H, so ∂(72 6) = 2. 0H u1 u2 u3 u4 u5 u6 u1 u1 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 H u1 u3 u4 u5 u1 u3 u4 u6 u2 u2 u3 u4 u1 u2 u3 u4 u5 u2 u3 u4 u5 u6 u2 u3 u4 u5 u2 u3 u4 u6 u3 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 H u4 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u5 u3 u4 u5 u3 u4 u5 u6 u6 u4 u6 1H u3 u4 u1 u6 u2 u5 u3 u3 u4 u3 u4 u3 u4 u1 u6 u3 u4 u1 u6 H H u4 u3 u4 u3 u4 u1 u6 u3 u4 u1 u6 H H u1 u1 u6 u1 u6 u1 u6 u2 u5 u1 u6 u2 u5 u6 u1 u6 u1 u6 u2 u5 u1 u6 u2 u5 u2 u2 u5 u2 u5 u5 u2 u5 2H u3 u4 u2 u5 u1 u6 u3 P P P P H H u4 P P P H H u2 P P H H u5 P H H u1 Q Q u6 Q Ratio Mathematica, 20, 2010 29 (826) Set Γ(u1) = {d1, d2}, Γ(u2) = {d2, d3, d4}, Γ(u3) = {d3, d4, d5} Γ(u4) = {d5, d6}, Γ(u5) = {d5}, Γ(u6) = {d6}. We obtain μ1(u1) = 0.233, μ1(u2) = 0.230247, μ1(u3) = 0.228125, μ1(u4) = μ1(u3) μ1(u5) = 0.218518, μ1(u6) = 0.230833. From this, we have 1H. From 1H we obtain : μ2(u1) = 0.34545, μ2(u6) = 0.3263, μ2(u5) = 0.324242, μ2(u2) = 0.305797, μ2(u3) = μ2(u4) = 0.296. Therefore we find 2H as follows From 2H we obtain : μ3(u1) = 0.34848, μ3(u6) = 0.331579, μ3(u5) = 0.31739 μ3(u2) = 0.302898, μ3(u3) = μ3(u4) = 0.29 We have clearly 3H = 2H, by consequence ∂(82 6) = 2. 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u1 u2 u3 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 u1 u2 u4 u6 u2 u1 u2 u3 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 u1 u2 u3 u4 u6 u3 u2 u3 u4 u5 u2 u3 u4 u5 u6 u2 u3 u4 u5 u2 u3 u4 u5 u6 u4 u3 u4 u5 u6 u3 u4 u5 u6 u3 u4 u5 u6 u5 u3 u4 u5 u3 u4 u5 u6 u6 u4 u6 1H u1 u6 u2 u3 u4 u5 u1 u1 u1 u6 u1 u2 u6 u1 u6 u2 u3 u4 u1 u6 u2 u3 u4 H u6 u6 u2 u6 u2 u6 u3 u4 u2 u6 u3 u4 u2 u6 u3 u4 u5 u2 u2 u2 u3 u4 u2 u3 u4 u2 u3 u4 u5 u3 u3 u4 u3 u4 u3 u4 u5 u4 u3 u4 u3 u4 u5 u5 u5 2H u1 u6 u5 u2 u3 u4 u1 u1 u1 u6 u1 u6 u5 u1 u6 u5 u2 H H u6 u6 u6 u5 u6 u5 u2 u6 u5 u2 u3 u4 u6 u5 u2 u3 u4 u5 u5 u5 u2 u5 u2 u3 u4 u5 u2 u3 u4 u2 u2 u2 u3 u4 u2 u3 u4 u3 u3 u4 u3 u4 u4 u3 u4 Ratio Mathematica, 20, 2010 30 (136) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d3, d4}, Γ(u3) = {d3, d4, d5}, Γ(u4) = {d4, d5, d6} Γ(u5) = {d5}, Γ(u6) = {d6}. We obtain μ1(u1) = 0.2006173, μ1(u2) = 0.2005208, μ1(u3) = 0.20714, μ1(u4) = 0.211905, μ1(u5) = 0.198765, μ1(u6) = 0.206667. By consequence we have 1H. Hence we have μ2(u4) = 0.354545 = μ2(u5), μ2(u3) = 0.34035 = μ2(u2) μ2(u6) = 0.33188 = μ2(u1) from which we obtain 2H. From 2H it follows μ3 (u4) = μ3(u5) = μ3(u6) =μ3(u1) =0.26667 μ3 (u2) = μ3(u3) = 0.26190. 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 u1 u2 u3 u4 u6 u2 u1 u2 u3 u4 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 u1 u2 u3 u4 u6 u3 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 H u4 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u5 u3 u4 u5 u3 u4 u5 u6 u6 u4 u6 1H u4 u3 u6 u1 u2 u5 u4 u4 u4 u3 u4 u3 u6 u4 u3 u6 u1 u4 u3 u6 u1 u2 H u3 u3 u3 u6 u3 u6 u1 u3 u6 u1 u2 u3 u6 u1 u2 u5 u6 u6 u6 u1 u6 u1 u2 u6 u1 u2 u5 u1 u1 u1 u2 u1 u2 u5 u2 u2 u2 u5 u5 u5 2H u4 u5 u3 u2 u6 u1 u4 u4 u5 u4 u5 u4 u5 u3 u2 u4 u5 u3 u2 H H u5 u4 u5 u4 u5 u3 u2 u4 u5 u3 u2 H H u3 u3 u2 u3 u2 u3 u2 u6 u1 u3 u2 u6 u1 u2 u3 u2 u3 u2 u6 u1 u3 u2 u6 u1 u6 u6 u1 u6 u1 u1 u6 u1 Ratio Mathematica, 20, 2010 31 Set P = {u4, u5, u6, u1}, Q = {u3, u2}. Then we obtain 3H as follows From 3H, it follows that 4H = 3H and we have finally ∂(136) = 3. (236) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d3, d4}, Γ(u3) = {d3, d4, d5}, Γ(u4) = {d4, d5, d6}, Γ(u5) = {d5}, Γ(u6) = {d6}. whence ∂(13 6)= ∂(236)=3. (336) Set Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d4}, Γ(u3) = {d3, d4, d5}, Γ(u4) = {d4, d5, d6} Γ(u5) = {d5}, Γ(u6) = {d6}. We have See (13 6). We have ∂(33 6)= ∂(136)=3. 3H u4 u5 u6 u1 u3 u2 u4 P P P P H H u5 P P P H H u6 P P H H u1 P H H u3 Q Q u2 Q 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 u1 u2 u3 u4 u6 u2 u1 u2 u3 u4 u1 u2 u3 u4 u5 H u1 u2 u3,u4,u5 u1,u2 u3 u4 u6 u3 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 H u4 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u2 u3 u4,u5 u6 u5 u3 u4 u5 u3 u4 u5 u6 u6 u4 u6 0H u1 u2 u3 u4 u5 u6 u1 u1 u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 u1 u2 u3 u4 u6 u2 u1 u2 u3 u4 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 u1 u2 u3 u4 u6 u3 u1 u2 u3 u4 u5 H u1 u2 u3 u4 u5 H u4 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u5 u3 u4 u5 u3 u4 u5 u6 u6 u4 u6 Ratio Mathematica, 20, 2010 32 $ 6. (148) Set H = {u1, u2, u3, u4, u5, u6, u7, u8} and Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d3, d4}, Γ(u3) = {d3, d4, d5}, Γ(u4) = { d4, d5, d6}, Γ(u5) = {d5, d6, d7}, Γ(u6) = {d7, d8}, Γ(u7) = {d7}, Γ(u8) = {d8}. So, denoting {ui, ui+1,…, uj-1, uj} by uij, we have We have μ1(u1) = 0.1756, μ1(u2) = 0.17470, μ1(u3) = 0.1754978, μ1(u4) = 0.1729, μ1(u5) = 0.1803, μ1(u6) = 0.1813, μ1(u7) = 0.175641 = μ1(u1), μ1(u8) = 0.19073. So μ1(u8) > μ1(u6) > μ1(u5) > μ(u7) = μ(u1) > μ1(u3) > μ1(u2) > μ(u4). One obtains 1H as follows from which μ2(u4) = μ2(u8) = 0.2890, μ2(u2) = μ2(u6) = 0.2724, μ2(u1) = μ2(u7) = 0.247567, μ2(u3) = μ2(u5) = 0.2549. So we obtain 2H. 0H u1 u2 u3 u4 u5 u6 u7 u8 u1 u13 u14 u15 u15 u17 u13 u58 u13 u57 u13 u6 u8 u2 u14 u15 u15 u17 H u17 u14 u6 u8 u3 u15 u15 u17 H u17 u16 u8 u4 u25 u27 u28 u27 u26 u8 u5 u37 u38 u37 u38 u6 u58 u58 u58 u7 u57 u58 u8 u6 u8 1H u4 u2 u3 u1 u7 u5 u6 u8 u4 u4 u4 u2 u4 u2 u3 u4 u2 u3 u1 u7 u4 u2 u3 u1 u7 u4 u2 u3 u1 u7 u5 u4 u2 u3 u1 u7 u5 u6 H u2 u2 u2 u3 u2 u3 u1 u7 u2 u3 u1 u7 u2 u3 u1 u7 u5 u2 u3 u1 u7 u5 u6 u2 u3 u1 u7 u5 u6 u8 u3 u3 u3 u1 u7 u3 u1 u7 u3 u1 u5 u7 u3 u1 u7 u5 u6 u3 u1 u7 u5 u6 u8 u1 u1 u7 u1 u7 u1 u7 u5 u1 u7 u5 u6 u1 u7 u5 u6 u8 u7 u1 u7 u1 u7 u5 u1 u7 u5 u6 u1 u7 u5 u6 u8 u5 u5 u5 u6 u5 u6 u8 u6 u6 u6 u8 u8 u8 Ratio Mathematica, 20, 2010 33 Hence we have : μ3(u4) = μ3(u8) = μ3(u1) = μ3(u7) = 0.22619, μ3(u2) = μ3(u6) = μ3(u3) = μ3(u5) = 0.2197. Setting P= {u4, u8, u1, u7}, Q = {u3, u5, u2, u6}, we find 3H From this, we obtain : ∀i, μ4(ui) = 0.166. It follows 4H = T, whence ∂(14 8) = 4. $7. (12 9) Let H={ ui | 1 ≤ i ≤ 9} and for i ≤ 7 , set Γ(ui) = { di, di+1, di+2}, Γ(u8) = {d8}, Γ(u9) = {d9}. We obtain 2H u4 u8 u2 u6 u3 u5 u1 u7 u4 u4 u8 u4 u8 u4 u8 u2 u6 u4 u8 u2 u6 u4 u8 u2 u6 u3 u5 u4 u8 u2 u6 u3 u5 H H u8 u4 u8 u4 u8 u2 u6 u4 u8 u2 u6 u4 u8 u2 u6 u3 u5 u4 u8 u2 u6 u3 u5 H H u2 u2 u6 u2 u6 u2 u6 u3 u5 u2 u6 u3 u5 u2 u6 u3 u5 u1 u7 u2 u6 u3 u5 u1 u7 u6 u2 u6 u2 u6 u3 u5 u2 u6 u3 u5 u2 u6 u3 u5 u1 u7 u2 u6 u3 u5 u1 u7 u3 u3 u5 u3 u5 u3 u5 u1 u7 u3 u5 u1 u7 u5 u3 u5 u3 u5 u1 u7 u3 u5 u1 u7 u1 u1 u7 u1 u7 u7 u1 u7 3H u4 u8 u1 u7 u2 u6 u3 u5 u4 P P P P H H T H u8 P P P H H H H u1 P P H H H H u7 P H H H H u2 Q Q Q Q u6 Q Q Q u3 Q Q u5 Q Ratio Mathematica, 20, 2010 34 From 0H we have μ1(u9) = 0.1729 > μ1(u7) = 0.1648 > μ1(u1) = 0.1616 > μ1(u3) = 0.160 > μ1(u6) = 0.1599 > μ1(u8) = 0.1597 > μ1(u2) = 0.159169 > μ1(u5) = 0.157387 > μ1(u4) = 0.159218. From these data, we obtain 1H as follows 0H u1 u2 u3 u4 u5 u6 u7 u8 u9 u1 u1u2 u3 u1 u2 u3 u4 u1 u2 u3 u4 u5 u1 u2 u3 u4 u5 u6 u1 u2 u3 u4 u5 u6 u7 u1 u2 u3 u4 u5 u6 u7 u8 u1 u2 u3 u5 u6 u7 u8 u9 u1 u2 u3 u6 u7 u8 u1 u2 u3 u7 u9 u2 u1 u2 u3 u4 u1 u2 u3 u4 u5 u1 u2 u3 u4 u5 u6 u1 u2 u3 u4 u5 u6 u7 u1 u2 u3 u4 u5 u6 u7 u8 H u1 u2 u3 u4 u6 u7 u8 u1 u2 u3 u4 u7 u9 u3 u1 u2 u3 u4 u5 u1 u2 u3 u4 u5 u6 u1u2 u3 u4 u5 u6 u7 u1 u2 u3 u4 u5 u6 u7 u8 H u1 u2 u3 u4 u5 u6 u7 u8 u1 u2 u3 u4 u5 u7 u9 u4 u2 u3 u4 u5 u6 u2 u3 u4 u5 u6 u7 u2 u3 u4 u5 u6 u7 u8 u2 u3 u4 u5 u6 u7 u8 u9 u2 u3 u4 u5 u6 u7 u8 u2 u3 u4 u5 u6 u7 u9 u5 u3 u4 u5 u6 u7 u3 u4 u5 u6 u7 u8 u3 u4 u5 u6 u7 u8 u9 u3 u4 u5 u6 u7 u8 u3 u4 u5 u6 u7 u9 u6 u4 u5 u6 u7 u8 u4 u5 u6 u7 u8 u9 u4 u5 u6 u7 u8 u4 u5 u6 u7 u8 u9 u7 u5 u6 u7 u8 u9 u5 u6 u7 u8 u9 u5 u6 u7 u8 u9 u8 u6 u7 u8 u6 u7 u8 u9 u9 u7 u9 1H u9 u7 u1 u3 u6 u8 u4 u2 u5 u9 u9 u9 u7 u9 u7 u1 u9 u7 u1 u3 u9 u7 u1 u3 u6 u9 u7 u1 u3 u6 u8 u9 u7 u1 u3 u6 u8 u4 u9 u7 u1 u3 u6 u8 u2 u4 H u7 u7 u7 u1 u7 u1 u3 u7 u1 u3 u6 u7 u1 u3 u6 u8 u7 u1 u3 u6 u8 u4 u7 u1 u3 u6 u8 u2 u4 u7 u1 u3 u6 u8 u2 u5 u4 u1 u1 u1 u3 u1 u3 u6 u1 u3 u6 u8 u1 u3 u6 u8 u4 u1 u3 u6 u8 u2 u4 u1 u3 u6 u8 u2 u5 u4 u3 u3 u3 u6 u3 u6 u8 u3 u6 u8 u4 u3 u6 u8 u2 u4 u3 u6 u8 u2 u5 u4 u6 u6 u6 u8 u6 u8 u4 u6 u8 u2 u4 u6 u8 u2 u5 u4 u8 u8 u8 u4 u8 u2 u4 u8 u2 u5 u4 u4 u4 u2 u4 u2 u5 u4 u2 u2 u5 u2 u5 u5 Ratio Mathematica, 20, 2010 35 From 1H we obtain as follows, μ2 and then 2H. μ2(u9) = μ2(u5) = 0.2740 > μ2(u7) = μ2(u2) = 0.261085 > μ2(u1) = μ2(u4) = = 0.250716 > μ2(u3) = μ2(u8) = 0.24495 > μ2(u6) = 0.243116. From 2H we obtain μ3(u9) = μ3(u5) = 0.211805 , μ3(u7) = μ3(u2) = 0.205433, μ3(u1) = μ3 (u4) = 0.20504, μ3 (u3) = μ3(u8) = 0.21155, μ3(u6) = 0.24407. Then we have 3H as follows 2H u9 u5 u7 u2 u1 u4 u3 u8 u6 u9 u5 u9 u5 u9 u5 u9 u7 u2 u5 u9 u7 u2 u5 u9 u7 u2 u1 u4 u5 u9 u7 u2 u1 u4 u5 u9 u7 u2 u3 u8 u1 u4 u5 u9 u7 u3 u8 u2 u1 u4 H u5 u5 u9 u5 u9 u7 u2 u5 u9 u7 u2 u5 u9 u7 u2 u1 u4 u5 u9 u7 u2 u1 u4 u5 u9 u7u2 u1 u4 u3 u8 u5 u9 u7 u2 u1 u4 u3u8 H u7 u7 u2 u7 u2 u7 u2 u1 u4 u7 u2 u1 u4 u7 u2 u1 u4 u3 u8 u7 u2 u1 u4 u3 u8 u7 u2 u6 u1 u4 u3 u8 u2 u7 u2 u7 u2 u1 u4 u7 u2 u1 u4 u7 u2 u1 u4 u3 u8 u7 u2 u1 u4 u3 u8 u7 u2 u6 u1 u4 u3 u8 u1 u1 u4 u1 u4 u1 u4 u3 u8 u1 u4 u3 u8 u1 u4 u6 u3 u8 u4 u1 u4 u1 u4 u3 u8 u1 u4 u3 u8 u1 u4 u6 u3 u8 u3 u3 u8 u3 u8 u3 u8 u6 u8 u3 u8 u3 u8 u6 u6 u6 3H u6 u9 u5 u3 u8 u2 u7 u1 u4 u6 u6 u6 u9 u5 u6 u9 u5 u6 u9 u5 u3 u8 u6 u9 u5 u3 u8 u6 u9 u5 u3 u8 u2u7 u6 u9 u5 u3 u8 u2 u7 H H u9 u9 u5 u9 u5 u9 u5u3 u8 u9 u5 u3 u8 u9 u5 u3 u8 u2 u7 u9 u5 u3 u8 u2 u7 u9 u5 u3u8 u2 u7 u1 u4 u9 u5 u3 u8 u2 u7 u1 u4 u5 u9 u5 u9 u5 u3 u8 u9 u5 u3 u8 u9 u5 u3 u8 u2 u7 u9 u5 u3 u8 u2 u7 u9 u5 u3 u8 u2 u7 u1 u4 u9 u5 u3u8 u2 u7 u1 u4 u3 u3 u8 u3 u8 u3 u8 u2 u7 u3 u8 u2 u7 u3 u8 u2 u7 u1 u4 u3 u8 u2 u7 u1 u4 u8 u3 u8 u3 u8 u2 u7 u3 u8 u2 u7 u3 u8 u2 u7 u1 u4 u3 u8 u2 u7 u1 u4 u2 u2 u7 u2 u7 u2 u7 u1 u4 u2 u7 u1 u4 u7 u2 u7 u2 u7 u1 u4 u2 u7 u1 u4 u1 u1 u4 u1 u4 u4 u1 u4 Ratio Mathematica, 20, 2010 36 It is possible to verify that the function φ : 2H → 3H defined as follows φ(u3) = u9, φ(u8) = u5, φ(u1) = u3, φ(u4) = u8, φ(u9) = u1, φ( u5) = u4, φ(u7) = u7, φ(u2) = u2, φ(u6) = u6, is a hypergroup isomorphism. It follows that the fuzzy grade of (12 9) is 2. $ 8. (1516) Set H = {ui⎮1 ≤ i ≤ 16}, D = {di⎮1 ≤ i ≤ 16}, Γ(u1) = {d1, d2, d3}, Γ(u2) = {d2, d3, d4}, and ∀ i : i ≤ 13, Γ(ui) = {di, di+1, di+2}, Γ(u14) = {d15, d16}, Γ(u15) = {d15}, Γ(u16) = {d16}. Since ∀ i, we have ui ο ui = {uj⎮ Γ(uj) ∩ Γ(ui) ≠ ∅}, it follows that we have u1 ο u1 = { u1, u2, u3}, u2 ο u2 = { u1, u2, u3, u4}, u3 ο u3 = { u1, u2, u3, u4, u5}, ∀ i : 4 ≤ i ≤ 13, ui ο ui = { ui-2, u i-1, ui, ui+1, ui+2}, u14 ο u14 = { u13, u14, u15, u16}, u15 ο u15 = { u13, u14, u15}, u16 ο u16 = { u14, u16}. For 0H we have the following table : Ratio Mathematica, 20, 2010 37 0H u1 u2 u3 u4 u5 u6 u7 u8 u9 u10 u11 u12 u13 u14 u15 u16 u1 u13 u14 u15 u16 u17 u18 u13 u59 u13 u610 u13 u711 u13 u812 u13 u913 u13 u1014 u13 u1115 u13 u1316 u13 u1315 u13 u14 u16 u2 u14 u15 u16 u17 u18 u19 u14 u610 u14 u711 u14 u812 u14u913 u14u1014 u14 u1115 u14u1316 u14u1315 u14u14 u16 u3 u15 u16 u17 u18 u19 u110 u15 u711 u15u812 u15u913 u15u1014 u15u1115 u15u1316 u15u1315 u15u14 u16 u4 u26 u27 u28 u29 u210 u211 u26u812 u26u913 u26u1014 u26u1115 u26u1316 u26u1315 u26u14 u16 u5 u37 u38 u39 u310 u311 u312 u37u913 u37u1014 u37u1115 u37u1316 u37u1315 u37u14 u16 u6 u48 u49 u410 u411 u412 u413 u48u1014 u48u1115 u48u1316 u48u1315 u48u14 u16 u7 u59 u510 u511 u512 u513 u514 u59u1115 u59u1316 u59u1315 u59u14 u16 u8 u610 u611 u612 u613 u614 u615 u610u1316 u610u1315 u610u14 u16 u9 u711 u712 u713 u714 u715 u711u1316 u715 u711u14 u16 u10 u812 u813 u814 u815 u816 u815 u812u14 u16 u11 u913 u914 u915 u916 u915 u914u16 u12 u1014 u1015 u1016 u1015 u1014u16 u13 u1115 u1116 u1115 u1116 u14 u1316 u1316 u1316 u15 u1315 u1316 u16 u14 u16 Ratio Mathematica, 20, 2010 38 From 0H we obtain μ1(u16) = 0.15673, μ1(u15) = 0.13992, μ1(u14) = 0.141293, μ1(u1) = 0.13867, μ1(u2) = 0.134942, μ1(u13) = 0.134215, μ1(u3) = 0.132574, μ1(u4) = 0.129700, μ1(u5) = 0.128076, μ1(u6) = 0.127554, μ1(u7) = 0.126581, μ1(u12) = 0.1283654, μ1(u11) = 0.126441, μ1(u10) = 0.126878, μ1(u8) = 0.12671, μ1(u9) = 0.126608. For 1H, set v1 = u16, v2 = u14, v3 = u15, v4 = u1, v5 = u2, v6 = u13, v7 = u3, v8 = u4, v9 = u12, v10 = u5, v11 = u6, v12 = u10, v13 = u8, v14 = u9, v15 = u7, v16 = u11. ∀(i, j), such that i ≤ j set vij = {vi, vi+1,...vj}. So we have vi ο1 vj = vij. For 2H we have v1 ο2 v1 = v1 ο2 v16 = v16 ο2 v16 = {v1, v16}, v2 ο2 v2 = v2 ο2 v15 = v15 ο2 v15 = {v2, v15}. Generally, vi ο2 vi = vi ο2 v16-(i-1) = v16-(i-1) ο2 v16-(i-1) = {vi, v16-(i-1) }. For i < j, vi ο2 vj = U jsi ≤≤ vs ο2 vs. Set P1 = {v1, v16, v8, v9}, P2 = {v2, v15, v7, v10}, P3 = {v3, v14, v6, v11}, P4 = {v4, v13, v5, v12}. Then for 3H, ∀k: 1 ≤ k ≤ 14, we have ∀( vi, vj) ∈ Pk x Pk, vi ο3 vj = Pk. If s < t, ∀( vi, vj) ∈ Ps x Pt, we have vi ο3 vj = U tus ≤≤ Pu. For 4H, setting Q1 = P1 U P4, Q2 = P2 U P3, we have ∀( vi, vj) ∈ Qi x Qj, vi ο4 vj = Qi U Qj. By consequence, if i ≠ j, vi ο4 vj = H and vi ο4 vi = Qi. Since ⎮Q1⎮=⎮Q2⎮, we have ∀ vi ∈ Q1, ∀ vj∈ Q2, μ4(vi) = μ4(vj). It follows that 5H = T (total hypergroup) and by consequence ∂(15 16) = 5. 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