版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领
文档简介
1、T28 1202102$Vol. 28 No. 1Feb. 2021CHINESE JOURNAL OF ENGINEERING MATHEMATICS$:1005-3085(2021)01-0055-06Th$%$%fl1,21 , $1 , %1(1- ff$,: 541004; 2- ;fi, ; 551700) : $fR%$%,$fR%$%,fR% $ M 0 $ y% M %R% $ 1.: $: AMS(2000) 11B85$: $: A1$%yfi%,fl$fi%,yI$f%.$fi,$R%$f$% y,%f$%.%$f$%K%1-7 .K,$R%$%Q, 8 KfR% ,
2、9 fR%$%$%$%,%L%fR%$%$.$f R%$%,$fR%$% R%fi,R% $ y% .$% $%$%.f$%$ 9,10.2y$ 1 $ M =(X, Y, S, , ) R%$, s0 S,K s S, X , S = (s0 , ),% S = (s0 , ) | X .% s0 M %R.%,fi$R%$.y 1 M =(X, Y, S, , ) %$,s0 S,% M s0 R, M % s0 %. 1 M = (X, Y, S, , ) %$,B S %, M % B $%.$: 2007-08-07. $: fl (19818$R),Q,T. :%. : (6047
3、3005):ff (0832103):ffR (2007106020701M48).y$ 2 M = (X, Y, S, , ),M r = (Y, X, Sr, r, r) %$, s S,Q sr Sr,K X ,$ r(sr, (s, ) = 0 ,$ | = , sr s % . sr Sr,Q s S, sr % s % , sr % . 1 M = (X, Y, S, , ),M r = (Y, X, Sr, r, r) %$,sr Sr % s0 S %0 ,K% X ,r(sr , (s0 , ) (s0 , ) % .0 2 M = (X, Y, S, , ) s0 R, M
4、 %R%$ M % . 10,M = (X, Y, S, , ) %$ M % .f s0 S,% s0 ,% s0 % , s0 R% M %$ ,%f%. 3 M = (X, Y, S, , ) $,s0 S, s0 % s0 R% M %$ M1 .fR%$.y 2 M = (X, Y, S, , ) M r = (Y, X, Sr, r, r) %R%$,M r % M % , sr Sr % M % % Gr(z)sr = 0,$ Gr(z) M r %,M $ M %. M M r % H (z) H r(z). 3, $ H r(z)H (z) =zr I ,I . sr % M
5、 % ,K% X ,$ r(sr, (M , ) = 0 ,$ 0 X . Gr(z)sr + H r(z)H (z)X (z) = X (z) + z X (z),$ X (z), X (z) 00 , 0 %R,f M r % M % , H r(z)H (z) = z I ,f G (z)s =rrrkX0 (z).,M r M r R, sr = r(M r , ), X , k 0,r(M r , )r(sr, (M , ) = r(M r , )r(r(M r , ), (M , )= r(M r , (M , ) = r(M r , )0 ,%Gr(z)M r + H r(z)(
6、X (z) + zk H (z)X (z)= Gr(z)M r + H r(z)X (z) + zk (X (z) + z X (z),0H r(z)X (z) + zk H r(z)H (z)X (z) = H r(z)X (z) + zk X (z) + zk+ X (z),0$ X (z) % %R,f zk X (z) = 0, Gr(z)sr = X (z) = 0.00 sr Sr fl Gr(z)sr = 0,K% X ,r(sr, (M , ) %RGr(z)sr + H r(z)(G(z)M + H (z)X (z) = Gr(z)sr + H r(z)H (z)X (z)
7、= z X (z), r(sr, (M , ) = 0 ,$ 0 $fifi. sr % M % .fl,% M r % M r % M % ,% M % % M r . 2 M = (X, Y, S, , ) %R%$, M r = (Y, X, Sr, r, r)% M % % sr Sr,K% X ,r(sr, (M , ) = 0 ,$ M % M %,|0 | = . .K% s S, X , M R, X , s = (M , ), (s, ) = (M , ), ),f(M , )(s, ) = (M , )(M , ), ) = (M , ), sr Sr, r(sr, (M
8、, ) = 0 , |0 | = ,%r(sr, (M , ) = r(sr, (M , )(s, )= r(sr, (M , )r(r(sr, (M , ), (s, ) = 0 = rr ,0$ |r| = |, |r | = ,f r(sr, (M , ),0r(r(sr, (M , ), (s, ) = r , |r | = ,00 s %K M r % M % . 3 M = (X, Y, S, , ) %R%$,M r = (Y, X, Sr, r, r) %$,%flK% sr Sr, X , sr Sr,0sr = r(sr , (M , ),0 M r % M % %K% s
9、r Sr, X , 0 X ,r(sr, (M , ) = 0 .3 M1 = (X, Y, S1 , 1 , 1 ),M2 = (Y, Z, S2 , 2 , 2 ) %$,M1 M2 % (K) $ C (M1 , M2 ) = (X, Z, S1 S2 , , ),$(s1 , s2 ), x) = (1 (s1 , x), 2 (s2 , 1 (s1 , x),(s1 , s2 ), x) = 2 (s2 , 1 (s1 , x), (s1 , s2 ) S1 S2 ,x X.h M1 M2 .$ M WIFA M . WIFA M %,fl WIFAM1 M2 , M M1 M2 .
10、y$ 31$ M = (X, X, X, , ) y,$(x, xr) = xr,(x, xr) = x, x, xr X, Md . Md % M d ,Q$ y. 1,2 %,$ M = (X, Y, S, , ), M %K s $ r,fW Mr,s = (s, x0 , , xr1 ) | x0 , , xr1 X , s %$ r $.|W M | r,M = min |W M |r,sr,ssS s %$ r $ M %$ r $. M = (X, Y, S, , ) %$,s S, y Y ,$Out(s) = (s, x) | x X ,d(s) = |Out(s)|,St
11、(s, y) = t S | x X, s.t. t = (s, x) and (s, x) = y. 4 M = (X, Y, S, , ) s0 S R%fl 1 WIFA,|X | = |Y |,%K s S,d(s0 ) d(s), M %$% 1. |X | = |Y | = n, M %fl 1 %,K% s S, delay(s) 1 (delay(s) $ s % 2). d(s0 ) , d(s0 ) n. 4 % 2,K% sr S, d(sr) = d(s0 ) = n, |X | = |Y | = n,K% sr S,Q (sr, x) = (s, y),x, y X
12、, x = y,f sr 0 . M fl 1 z.f d(s) = d(s0 ) 1, M 0 WIFA y% |W M | = 1.r,s0 4 M = (X, Y, S, , ) R% $,%K s S = (M , ) | X k , k ,(s, 0) = 0, M 0 WIFA y. M = (X, Y, S, , ) %R% $,Q M 0 WIFA y, 4,$ |W M | = 1.%K 1 , 2 r,MX ,1 = 2 ,(M , 1 ) = (M , 2 ), (M , 1 0 ) = (M , 2 0 ). 1 0 , 2 0%R X1 (z), X2 (z),M %
13、 H (z), H (z)X1 (z) = H (z)X2 (z),% H (z)X1 (z) X2 (z) = 0, 11 % 1 X1 (z) X2 (z) = 0,% X1 (z) = X2 (z), 1 0 = 2 0 ,% 1 = 2 . 1 = 2 z. 5 M = (X, Y, S, , ) s0 S R%fl 1 WIFA,|X | = |Y | = n % 1,M = 1. 1 M = (X, Y, S, , ),$ X = Y = a, b, c, d, e, f ,S = s1 , s2 , $f:1,|W M | = 1,M , M 0 WIFA 1 y% n 1,s0
14、(s1 , a) = s1 ,(s1 , b) = s2 , (s1 , c) = s1 , (s1 , d) = s2 , (s1 , e) = s1 , (s1 , f ) = s2 ,(s1 , a) = a;(s1 , b) = a; (s1 , c) = b; (s1 , d) = b; (s1 , e) = c; (s1 , f ) = c;(s2 , a) = s1 ,(s2 , b) = s2 , (s2 , c) = s1 , (s2 , d) = s2 , (s2 , e) = s1 , (s2 , f ) = s2 ,(s2 , a) = d;(s2 , b) = d;
15、(s2 , c) = e; (s2 , d) = e; (s2 , e) = f ; (s2 , f ) = f.%, M R%$,s1 , s2 % M %R.% M $y.f 1 %$%fl:fl % % n y WIFA M %$y?$:1 $. $%KJ. $:A ,1993, 23(7): 759-765Bao F. Composition and decomposition of weakly invertible finite automataJ. Science in China, Series A,1993, 23(7): 759-7652 I. $%flJ. Q,2005,
16、 42(4): 690-696Wang H J. Two results of decomposing weakly invertible finite automataJ. Journal of Computer Research and Development, 2005, 42(4): 690-6963 ,$,%. $%J. ,2005, 28(9): 1501-1507Cao F, Deng P M, Yi Z. Decomposition of weakly invertible finite automataJ. Chinese Journal of Com- puters, 20
17、05, 28(9): 1501-15074 y,$fl,$. $%J. ff$:,2006, 24(1):30-33Yan H Y, Xie Z W, Deng P M, et al. Functions of zero state of linear finite automataJ. Journal of GuangxiNormal University (Natural Science Edition), 2006, 24(1): 30-335 ,$. y$%J. ,1994, 17(5): 330-337Gao X, Bao F. Decomposition of binary wea
18、kly invertible finite automataJ. Chinese Journal of Computers,1994, 17(5): 330-3376 . $%flJ. ,1993, 16(8): 629-632Gao X. Two results about the decomposition of delay step of weakly invertible finite automataJ. ChineseJournal of Computers, 1993, 16(8): 629-6327 ,$,%. Moore %J. ff$:,2003, 21(4):44-47C
19、ao F, Deng P M, Yi Z. Some results of invertibility of moore automataJ. Journal of Guangxi NormalUniversity (Natural Science Edition), 2003, 21(4): 44-47 8 %. %J. ,1987, 30(5): 679-687Shen H. Semigroup structure of automataJ. Acta Mathematica Sinica, 1987, 30(5): 679-687 9 . $%M. fl:$,1979Tao R J. I
20、nvertibility of Finite AutomataM. Beijing: Science Press, 1979 10 . M. fl:$,1986Tao R J. Introduction to Finite AutomataM. Beijing: Science Press, 198611 ,. $%-%J. $ (A),1996, 26(5): 429-436Dai Z D, Ye D F. Invertibility of linear finite automata-classification and enumeration of transfer functionJ.
21、 Science in China, Series A, 1996, 26(5): 429-436Some Characteristics of Finite Automata Generated by Single StateHUANG Fei-dan1,2 , MENG Chun-feng1 , DENG Pei-min1 , YI Zhong1(1- College of Mathematical Science, Guangxi Normal University, Guilin 541004;2- Department of Mathematics, Bijie College, Bijie, Guizhou 551700)Abstract: This paper investigates the weakly invertible and decomposable properties of the finiteautomata generated by a singl
温馨提示
- 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
- 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
- 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
- 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
- 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
- 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
- 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。
最新文档
- 江苏南京江浦高级中学2025届高考英语一模试卷含解析
- 2025届黑龙江省哈尔滨市阿城区二中高三冲刺模拟数学试卷含解析
- 心脏血管外科实习报告
- 江苏省清江市清江中学2025届高三一诊考试数学试卷含解析
- 2025届山东省莱芜一中高三六校第一次联考英语试卷含解析
- 2023年水煤浆燃烧器资金需求报告
- 2024年气象、水文仪器及装置项目立项申请报告
- 2024年作物收获机械项目申请报告模板
- 小区广告制作合同范例
- 卖票合同范例
- 大学《传播学概论》试卷及答案
- 工程设计费收费标准
- -坚定目标赢在执行 主题班会课件
- 英语语法基础知识教学讲义课件
- 青岛版二年级上册数学《乘法的初步认识》单元整体备课设计
- 房屋买卖协议书电子版模板
- 食品分析习题(有答案)
- 研究思路图模板
- DB14T 1722-2018 桥梁加固用碳纤维复合材料板材的耐久性应用技术要求
- 装修工程验收单
- 骨折诊疗与护理考核试题与答案
评论
0/150
提交评论