






摘要: 以直覺相似度為擾動參數, 對直覺模糊取式(IFMP)、 直覺模糊拒取式(IFMT)問題的反向三I支持算法和反向三I約束算法進行魯棒性估計, 結果表明, 輸出的結果不會因為輸入的微小改變而產生顯著變化, 即兩種算法均具有良好的魯棒性.
關鍵詞: 直覺模糊推理; 相似度; 反向三I支持算法; 反向三I約束算法; 魯棒性
中圖分類號: O159文獻標志碼: A文章編號: 1671-5489(2025)02-0391-08
Robustness of Reverse Triple I AlgorithmBased on Intuitionistic Similarity
YUAN Yidan, HUI Xiaojing, WANG Qian
(School of Mathematics and Computer Science, Yan’an University, Yan’an 716000, Shaanxi Province, China)
Abstract: By using the intuitionistic similarity as a perturbation parameter, weestimated the robustness of the reverse triple I sustaining algorithm and
reverse triple I restriction algorithm for IFMP and IFMT problems. The results show that the output results will not significantly change due to small changes in the input,
indicating that both algorithms have good robustness.
Keywords: intuitionistic fuzzy inference; similarity; reverse triple I sustaining algorithm; reverse triple I restriction algorithm; robustness
收稿日期: 2024-05-16. 網絡首發日期: 2024-11-25.
第一作者簡介: 袁一丹(2000—), 女, 漢族, 碩士研究生, 從事數理邏輯與不確定性推理的研究, E-mail: 1757628434@qq.com. 通信作者簡介: 惠小靜(1973—), 女, 漢族, 博士, 教授, 從事數理邏輯與不確定性推理的研究, E-mail: xhmxiaojing@163.com.
基金項目: 國家自然科學基金(批準號: 12261090; 12301456).
網絡首發地址: https://link.cnki.net/urlid/22.1340.O.20241122.1535.002.
模糊推理在模糊集理論[1]中占有重要地位, 其基本模型是模糊取式(FMP)和模糊拒取式(FMT). 針對這兩種模型, Zadeh[2]提出了CRI(compositional rule of inference)算法, 此后, 該算法在模糊控制領域得到廣泛應用. 但鑒于CRI算法缺乏嚴格的邏輯依據, 王國俊[3]提出了全蘊涵三I算法, 將模糊推理納入到邏輯體系內, 對經典的CRI算法進行了有效改進. 文獻[4-8]進一步推廣了三I算法, 討論了算法的還原性及魯棒性.
為更確切地考慮事物的模糊性, Atanassov[9]推廣了模糊集, 提出了直覺模糊集的概念, 引入了猶豫度, 降低了模糊信息的損失, 在群決策、 模式識別、 醫療診斷等領域有重要應用. 但由于直覺模糊蘊涵算子的復雜性, 因此直覺模糊推理還未形成完整的理論體系. 文獻[10-11]初步提出了直覺三角模和直覺模糊蘊涵的概念, 為直覺模糊推理的邏輯系統構建了合理的理論框架; 文獻[12]基于直覺三角模給出了剩余型直覺模糊蘊涵算子的統一形式, 確立了該算子與模糊蘊涵算子的關系; 文獻[13-15]針對剩余型直覺模糊蘊涵算子, 研究了直覺模糊推理三I算法、 三I約束算法和反向三I算法; 文獻[16]根據模糊連接詞的靈敏度討論了直覺模糊推理系統的魯棒性; 文獻[17-18]借助直覺模糊距離證明了基于Lukasiewicz蘊涵算子的直覺模糊推理三I算法和SIS(subsethood infer subsethood)算法的魯棒性; 文獻[19]利用雙剩余定義了直覺模糊集的相似度, 并探討了直覺模糊推理α-反向三I支持算法的魯棒性, 但給出的相似度減弱了其程度化意義; 文獻[20]構造了新的直覺相似度, 分析了直覺模糊取式(IFMP)問題的三I算法和CRI算法的魯棒性. 本文針對IFMP和直覺模糊拒取式(IFMT)問題, 基于直覺相似度分別討論反向三I支持算法和反向三I約束算法的魯棒性.
1 預備知識
定義1[9]設X≠, A={(x,At(x),Af(x))x∈X}稱為X上的直覺模糊集, 其中At: X→[0,1]和Af: X→[0
,1]分別是X上的隸屬函數和非隸屬函數, 且0≤At+Af≤1; 若x∈X, At(x)+Af(x)=1, 則A退化為模糊集. 用IFS(X)表示X上的全體直覺模糊集.
定義2[13]令IFS={(m,n)∈[0,1]2m+n≤1}, 在IFS上定義偏序關系≤: α,β∈IFS, α=(a1,a2), β=(b1,b2), α≤
β.a1≤b1, a2≥b2, 其中α∧β=(a1∧b1,a2∨b2), α∨β=(a1∨b1,a2∧b2), 最小元0=(0,1), 最大元1=(1,0), 顯然(IFS,≤)為完備格.
定義3[21]設是[0,1]上的三角模, →:[0,1]2→[0,1]是[0,1]上的二元函數. 若ab≤c.a≤b→c, a,b,c∈[0,1], 則(
,→)稱為伴隨對, ([0,1],,→)稱為剩余格.
定義4[21]設([0,1],→)是剩余格, a,b∈[0,1], 規定ab=(a→b)(b→a).
定義5[21]若二元運算→滿足下列條件:
1) a→b=1.a≤b;
2) a≤b→c.b≤a→c;
3) a→(b→c)=b→(a→c);
4) 1→a=a;
5) a→∧i∈Ibi=∧i∈I(a→bi), (∨i∈Iai)→b=∧i∈I(ai→b);
6) a→b關于a單調不增, 關于b單調不減.
則稱→是[0,1]上的正則蘊涵算子.
定義6[13]α,β∈IFS, α=(a1,a2), β=(b1,b2), 在直覺模糊集上定義,: αβ=
(a1b1,a2b2), αβ=(a1b1,a2b2), 其中是與對偶的三角余模.
定理1[13]若直覺三角模由左連續三角模生成, 則在直覺模糊集上有二元運算→, 使得αβ≤γ
.α≤β→γ且β→γ=∨{λ∈IFSλβ ≤γ}.
定義7[20]設X={x1,x2,…xn}, A,B∈IFS(X), A={(x,At(x),Af(x))}, B={(x,Bt(x),Bf(x))}. 令S(A,B)=∧
ni=1[(At(xi)Bt(xi))(Af(xi)Bf(xi))], 則S稱為A,B的相似度.
引理1[20]對于剩余格([0,1],,→), 如果a,b,c,d∈[0,1], ∈{∧,∨,,→,}, 則有:
1) (ab)∧(cd)≤(a∧c)(b∧d);
2) (ab)∧(cd)≤(a∨c)(b∨d);
3) (ab)(cd)≤(ac)(bd);
4) ab≤ab;
5) a→b≤b→a.
定理2[20]設A(x),A′(x),B(x),B′(x)∈IFS(X), 如果S(A,A′)≥δ1, S(B,B′)≥δ2, 則
S(AB,A′B′)≥δ21δ22.
定理3[20]設A(x),A′(x),B(x),B′(x)∈IFS(X), 如果S(A,A′)≥δ1, S(B,B′)≥δ2, 則
S(A→B,A′→B′)≥δ31δ32.
2 反向三I支持算法的魯棒性
下面基于直覺相似度研究IFMP和IFMT問題的反向三I支持算法的魯棒性.
定理4[15]設(,→)是IFS上的直覺伴隨對, 則:
1) IFMP問題的反向三I支持解B(y)為B(y)=∧x∈X{A(x)(A(x)→B(y))}, y∈Y, 其中B(y)=(Bt(y),Bf(y))可分解為
Bt(y)=∧x∈X{At(x)((At(x)→Bt(y))∧(A-f(x)→B-f(y)))},
Bf(y)=∨x∈X{Af(x)(1-A-f(x)→B-f(y))},
式中A-f(x)=1-Af(x).
2) IFMT問題的反向三I支持解A(x)為A(x)=∨y∈Y{(A(x)→B(y))→B(y)}, x∈X, 其中A(x)=(At(x),Af(x))可分解為
At(x)=∨y∈Y{(((At(x)→Bt(y))∧(A-f(x)→B-f(y)))→Bt(y))∧((A-f(x)→B-f(y))→B-f(y))},
Af(x)=∧y∈Y{1-(A-f(x)→B-f(y))→B-f(y)}.
定理5 設B,B′分別為定理4中IFMP(A,B,A)和IFMP(A′,B′,A′)問題的直覺模糊推理反向三I支持解, 若S(A,A
′)≥δ1, S(B,B′)≥δ2, S(A,A′)≥δ3, 則S(B,B′)≥δ31δ32δ23.
證明: 設B(y)=(Bt(y),Bf(y)), B′(y)=(Bt′(y),Bf′(y)).
首先, 分析Bt(y)Bt′(y)的取值范圍:
Bt(y)Bt′(y)=∧x∈X{At(x)((At(x)→Bt(y))∧(A-f(x)→B-f(y)))}
∧x∈X{At′(x)((A′t(x)→B′t(y))∧(
A′-f(x)→B′-f(y)))}≥∧x∈X{At(x)((At(x)→Bt(y))∧(A-f(x)→B-f(y)))
At′(x)((A′t(x)→B′t(y))∧(A′-f(x)→B′
-f(y)))}≥∧x∈X{(At(x)At′(x))((At(x)→Bt(y))∧
(A-f(x)→B-f(y))(A′t(x)→B′t(y))∧(A′-f(x)→B′-f(y)))}≥
∧x∈Xδ3((At(x)→Bt(y))(A′t(x)→B′t(y)))
((A-f(x)→B-f(y))(A′-f(x)→B′-f(y)))≥
∧x∈Xδ3((At(x)A′t(x))(Bt(y)B′t(y)))
((A-f(x)A′-f(x))(B-f(y)B′-f(y)))≥
∧x∈Xδ3((At(x)A′t(x))(Bt(y)B′t(y)))((Af(x)A′f(x))(Bf(y)B′
f(y)))≥δ3(δ1δ2)(δ1δ2),
由于和是對偶的, 因此ab=1-(1-a)(1-b)=(ab), 則
Bf(y)=∨x∈X{Af(x)(1-A-f(x)→B-f(y))}=∨x∈X{(A-f(x)(A-f(x)→B-f(y)))}.
其次, 分析Bf(y)Bf′(y)的取值范圍:
Bf(y)Bf′(y)=∨x∈X{(A-f(x)(A-f(x)→B-f(y)))}∨x∈X
{(A-f(x)(A′-f(x)→B′-f(y)))}≥
∧x∈X{(A-f(x)(A-f(x)→B-f(y)))(A-f′(x)(A′-f(x)→B′-f(y)))}≥∧x∈X{(A-f(x)(A-f(x)→B-f(y)))(A-f′
(x)(A′-f(x)→B′-f(y)))}≥∧x∈X{(A-f(x)A-f′(x))((A
-f(x)→B-f(y))(A′-f(x)→B′-f(y)))}≥∧x∈X{(Af(x)Af′(x))((A-f(x)→B
-f(y))(A′-f(x)→B′-f(y)))}≥∧x∈Xδ3(A-f(x)A′-f(x))(B
-f(y)B′-f(y))≥∧x∈Xδ3(Af(x)A′f(x))(Bf(y)B′f(y))≥δ3δ1δ2.
最后, 可得B與B′之間的相似度為
S(B(y),B′(y))=∧y∈Y((Bt(y)Bt′(y))(Bf(y)Bf′(y)))≥
δ3(δ1δ2)(δ1δ2)δ3δ1δ2=δ31δ32δ23.
例1 設X={x1,x2,x3,x4,x5}(xj∈X, j=1,2,…,5), Y={y1,y2,y3,y4,y5}(yi∈Y,i
=1,2,…,5)是非空論域, A=(At,Af), B=(Bt,Bf), A=(At,Af)是3個直覺模糊集, A,A∈IFS(X), B∈IFS(Y). 基于Lukasiewicz蘊涵算子考慮, 其中
At={0.125,0.361,0.752,0.663,0.725},Af={0.790,0.610,0.206,0.291,0.227},
A′t={0.123,0.357,0.752,0.663,0.723},A′f={0.790,0.608,0.201,0.291,0.228},
Bt={0.876,0.564,0.173,0.131,0.072},Bf={0.113,0.422,0.811,0.370,0.912},B′t={0.881,0.564,0.173,0.126,0.072}
,B′f={0.110,0.422,0.811,0.368,0.908},
At={0.481,0.142,0.246,0.065,0.864},Af={0.312,0.628,0.728,0.899,0.027},At′={0.478,0.137,0.249,0.065,0.864},
Af′={0.312,0.628,0.731,0.901,0.022}.
由定義7得
S(A,A′)=∧ni=1[(At(xi)A′t(xi))(Af(xi)A′f(xi))]=0.994,S(B,B′)=∧
ni=1[(Bt(yi)B′t(yi))(Bf(yi)B′f(yi))]=0.992,S(A,A′)=∧ni=1
[(At(xi)At′(xi))(Af(xi)Af′(xi))]=0.994.
由定理5得
S(B,B′)≥δ31δ32δ23=0.99430.99230.9942=0.946.
定理6 設A,A′分別是定理4中IFMT(A,B,B)和IFMT(A′,B′,B′)問題的直覺模糊推理反向三I支持解, 若S(A,A′
)≥δ1, S(B,B′)≥δ2, S(B,B′)≥δ3, 則S(A,A′)≥δ41δ42δ33.
證明: 設A(x)=(At(x),Af(x)), A′(x)=(At′(x),Af′(x)).
首先, 分析At(x)At′(x)的取值范圍:
At(x)At′(x)=∨y∈Y{(((At(x)→Bt(y))∧(A-f(x)→B-f(y)))→Bt(y))∧
((A-f(x)→B-f(y))→B-f(y))}∨y∈Y{(((A′t(x)→B′t(y))∧(A′-f(x)→B′-f(y)))→Bt′(y))∧((A′-f(x)→B′-f(y))→B
-f′(y))}≥∧y∈Y{(((At(x)→Bt(y))∧(A-f(x)→B-f(y)))→Bt(y))∧((A-f(x)→B-f(y))→
B-f(y))(((A′t(x)→B′t(y))∧(A′-f(x)
→B′-f(y)))→Bt′(y))∧((A′-f(x)→B′-f(y))→B-f′(y))}≥
∧y∈Y{((((At(x)→Bt(y))∧(A-f(x)→B-f(y)))→Bt(y))(((A′t(x)→B′t(y))∧
(A′-f(x)→B′-f(y)))→Bt′(y)))(((A-f(x)→B-f(y))→B-f(y
))((A′-f(x)→B′-f(y))→B-f′(y)))}≥∧y∈Y(((At(x)→Bt(y))∧(A-f(x)→B
-f(y)))((A′t(x)→B′t(y))∧(A′-f(x)→B′-f(y))))(Bt(y)B
t′(y))((A-f(x)→B-f(y))(A′-f(x)→B′-f(y)))
(B-f(y)B-f′(y))≥∧y∈Y((At(x)→Bt(y))(A′t(x)→B′t(y)))((A-f(x)→B
-f(y))(A′-f(x)→B′-f(y)))δ3(A-f(x)A′-f(x))(B-f(y)B′-f(y))(Bf(y)Bf′(y))≥
∧y∈Y((At(x)A′t(x))(Bt(y)B′t(y)))((A-f(x)A′-f(x))(B-f(y)
B′-f(y)))δ3(Af(x)A′f(x))(Bf(y)B′f(y))δ3≥
∧y∈Y(δ1δ2)((Af(x)A′f(x))(Bf(y)B′f(y)))δ3(δ1δ2)δ3≥
(δ1δ2)(δ1δ2)δ3(δ1δ2)δ3.
其次, 分析Af(x)Af′(x)的取值范圍:
Af(x)Af′(x)=∧y∈Y{1-(A-f(x)→B-f(y))→B-f(y)}
∧y∈Y{1-(A′-f(x)→B′-f(y))→B-f′(y)}≥∧y∈Y{(1-(A-f(x)→B
-f(y))→B-f(y))(1-(A′-f(x)→B′-f(y))→B-f′(y))}≥
∧y∈Y{((A-f(x)→B-f(y))→B-f(y))
((A′-f(x)→B′-f(y))→B-f′(y))}≥
∧y∈Y{((A-f(x)→B-f(y))→B-f(y))((A′-f(x)→B′-f(y))→B-f′
(y))}≥∧y∈Y((A-f(x)→B-f(y))(A′-f(x)→B′-f(y)))(B-f(y)B-f′(y))≥
∧y∈Y((A-f(x)A′-f(x))(B
-f(y)B′-f(y)))(Bf(y)Bf′(y))≥∧y∈Y((Af(x)A′f
(x))(Bf(y)B′f(y)))δ3≥(δ1δ2)δ3.
最后, 可得A與A′之間的相似度為
S(A(x),A′(x))=∧x∈X((At(x)At′(x))(Af(x)Af′(x)))≥
(δ1δ2)(δ1δ2)δ3(δ1δ2)δ3(δ1δ2)δ3=δ41δ42δ33.
注1 由定理5和定理6知, 如果δi→1(i=1,2,3), 則可得S(B,B′)→1, 表明輸入A,B和A產生微小的擾動不會導致IFMP(
A,B,A)的反向三I支持解B有較大偏差, 因此IFMP問題的反向三I支持算法是魯棒的; 同理分析可知IFMT問題的反向三I支持算法也是魯棒的.
3 反向三I約束算法的魯棒性
下面基于直覺相似度研究IFMP和IFMT問題的反向三I約束算法的魯棒性.
定理7[15]設(,→)是IFS上的直覺伴隨對, 則:
1) IFMP問題的反向三I約束解B(y)為B(y)=∨x∈X{A(x)(A(x)→B(y))}, y∈Y, 其中B(y)=(Bt(y),Bf(y))可分解為
Bt(y)=∨x∈X{At(x)((At(x)→Bt(y))∧(A-f(x)→B-f(y)))},Bf(y)=∧x∈X{Af(x)(1-A-f(x)→B-f(y))}.
2) IFMT問題的反向三I約束解A(x)為A(x)=∧y∈Y{(A(x)→B(y))→B(y)}, x∈X, 其中A(x)=(At(x),Af(x))可分解為
At(x)=∧y∈Y{(((At(x)→Bt(y))∧(A-f(x)→B-f(y)))→Bt(y))∧((A-f(x)→B-f(y))→B-f(y))},
Af(x)=∨y∈Y{1-(A-f(x)→B-f(y))→B-f(y)}.
定理8 設B,B′分別是定理7中IFMP(A,B,A)和IFMP(A′,B′,A′)問題的直覺模糊推理反向三I約束解, 若S(A,A′)≥δ1,
S(B,B′)≥δ2, S(A,A′)≥δ3, 則S(B,B′)≥δ31δ32δ23.
證明: 設B(y)=(Bt(y),Bf(y)), B′(y)=(Bt′(y), Bf′(y)).
首先, 分析Bt(y)Bt′(y)的取值范圍:
Bt(y)Bt′(y)=∨x∈X{At(x)((At(x)→Bt(y))∧(A-f(x)→B-f(y)))}
∨x∈X{At′(x)((A′t(x)→B′t(y))∧(A′-f(x)→B′-f(y)))}≥
∧x∈X{At(x)((At(x)→Bt(y))∧(A-f(x)→B-f(y)))
At′(x)((A′t(x)→B′t(y))∧(A′-f(x)→B′-f(y)))}≥
∧x∈X{(At(x)At′(x))((At(x)→Bt(y))∧
(A-f(x)→B-f(y))(A′t(x)→B′t(y))∧(A′-f(x)→B′-f(y)))}≥
∧x∈Xδ3((At(x)→Bt(y))(A′t(x)→B′t(y)))
((A-f(x)→B-f(y))(A′-f(x)→B′-f(y)))≥
∧x∈Xδ3((At(x)A′t(x))(Bt(y)B′t(y)))
((A-f(x)A′-f(x))(B-f(y)B′-f(y)))≥
∧x∈Xδ3((At(x)A′t(x))(Bt(y)B′t(y)))
((Af(x)A′f(x))(Bf(y)B′f(y)))≥δ3(δ1δ2)(δ1δ2).
其次, 由于Bf(y)=∧x∈X{Af(x)(1-A-f(x)→B-f(y))}=∧x∈X{(A-f(x)
(A-f(x)→B-f(y)))}, 進而分析Bf(y)Bf′(y)的取值范圍:
Bf(y)Bf′(y)=∧x∈X{(A-f(x)(A-f(x)→B-f(y)))}
∧x∈X{(A-f′(x)(A′-f(x)→B′-f(y)))}≥
∧x∈X{(A-f(x)(A-f(x)→B-f(y)))
(A-f′(x)(A′-f(x)→B′-f(y)))}≥
∧x∈X{(A-f(x)(A-f(x)→B-f(y)))(A-f′(x)(A′-f(x)→B′-f(y)))}≥
∧x∈X{(A-f(x)A-f′(x))((A-f(x)→B-f(y))(A′-f(x)→B′-f(y)))}≥
∧x∈X{(Af(x)Af′(x))((A-f(x)→B-f(y))(A′-f(x)→B′-f(y)))}≥
∧x∈Xδ3(A-f(x)A′-f(x))(B-f(y)B′-f(y))≥
∧x∈Xδ3(Af(x)A′f(x))(Bf(y)B′f(y))≥δ3δ1δ2.
最后, 可得B與B′之間的相似度為
S(B(y),B′(y))=∧y∈Y((Bt(y)Bt′(y))(Bf(y)Bf′(y)))≥
δ3(δ1δ2)(δ1δ2)δ3δ1δ2=δ31δ32δ23.
定理9 設A,A′分別是定理7中IFMT(A,B,B)和IFMT(A′,B′,B′)問題的直覺模糊推理反向三I約束解, 若S(A,A′
)≥δ1, S(B,B′)≥δ2, S(B,B′)≥δ3, 則S(A,A′)≥δ41δ42δ33.
證明: 設A(x)=(At(x),Af(x)), A′(x)=(At′(x),Af′(x)).
首先, 分析At(x)At′(x)的取值范圍:
At(x)At′(x)=∧y∈Y{(((At(x)→Bt(y))∧(A-f(x)→B-f(y)))→Bt(y))∧
((A-f(x)→B-f(y))→B-f(y))}∧y∈Y{(((A′t(x)→B′t(y))∧
(A′-f(x)→B′-f(y)))→Bt′(y))∧
((A′-f(x)→B′-f(y))→B-f′(y))}≥
∧y∈Y{(((At(x)→Bt(y))∧(A-f(x)→B-f(y)))→Bt(y))∧((A-f(x)→B-f(y))→B-f(y))(((A
′t(x)→B′t(y))∧(A′-f(x)→B′-f(y)))→Bt′(y))∧
((A′-f(x)→B′-f(y))→B-f′(y))}≥
∧y∈Y{((((At(x)→Bt(y))∧(A-f(x)→B-f(y)))→Bt(y))
(((A′t(x)→B′t(y))∧(A′-f(x)→B′-f(y)))→Bt′(y)))
(((A-f(x)→B-f(y))→B-f(y))((A′-f(x)→B′-f(y))→B-f′(y)))}≥
∧y∈Y(((At(x)→Bt(y))∧(A-f(x)→B-f(y)))
((A′t(x)→B′t(y))∧(A′-f(x)→B′-f(y))))
(Bt(y)Bt′(y))((A-f(x)→B-f(y))
(A′-f(x)→B′-f(y)))(B-f(y)B-f′(y))≥
∧y∈Y((At(x)→Bt(y))(A′t(x)→B′t(y)))((A-f(x)→B-f(y))
(A′-f(x)→B′-f(y)))δ3(A-f(x)A′-f(x))
(B-f(y)B′-f(y))(Bf(y)Bf′(y))≥
∧y∈Y((At(x)A′t(x))(Bt(y)B′t(y)))((A-f(x)A′-f(x))(B-f(y)B′-f(y)))
δ3(Af(x)A′f(x))(Bf(y)B′f(y))δ3≥
∧y∈Y(δ1δ2)((Af(x)A′f(x))(Bf(y)B′f(y)))δ3(δ1δ2)δ
3≥(δ1δ2)(δ1δ2)δ3(δ1δ2)δ3.
其次, 分析Af(x)Af′(x)的取值范圍:
Af(x)Af′(x)=∨y∈Y{1-(A-f(x)→B-f(y))→B-f(y)}
∨y∈Y{1-(A′-f(x)→B′-f(y))→B-f′(y)}≥
∧y∈Y{(1-(A-f(x)→B-f(y))→B-f(y))(1-(A′-f(x)→B′-f(y))→B-f′(y))}≥
∧y∈Y{((A-f(x)→B-f(y))→B-f(y))
((A′-f(x)→B′-f(y))→B-f′(y))}≥
∧y∈Y{((A-f(x)→B-f(y))→B-f(y))((A′-f(x)→B′-f(y))→B-f′(y))}≥
∧y∈Y((A-f(x)→B-f(y))(A′-f(x)→B′-f(y)))(B-f(y)B-f′(y))≥
∧y∈Y((A-f(x)A′-f(x))(B-f(y)B′-f(y)))(Bf(y)Bf′(y))≥
∧y∈Y((Af(x)A′f(x))(Bf(y)B′f(y)))δ3≥(δ1δ2)δ3.
最后, 可得A與A′之間的相似度為
S(A(x),A′(x))=∧x∈X((At(x)At′(x))(Af(x)Af′(x)))≥
(δ1δ2)(δ1δ2)δ3(δ1δ2)δ3(δ1δ2)δ3=δ41δ42δ33.
注2 由定理8和定理9知, 如果δi→1(i=1,2,3), 則可得S(B,B′)→1, 表明輸入A,B和A產生微小的擾動不會導致IFMP(
A,B,A)的反向三I約束解B有較大偏差, 因此IFMP問題的反向三I約束算法是魯棒的; 同理分析可知IFMT問題的反向三I約束算法也是魯棒的.
綜上, 本文基于直覺相似度分析了直覺模糊推理反向三I支持算法和反向三I約束算法的魯棒性, 結果表明, IFMP(IFMT)問題的兩種算法有相同的魯棒性.
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