T^p chi Tin hqc vk Diiu kliiSn hpc, T.27, S.3 (2011), 241-252
VAN DE KET NHAP THONG TIN BIEU DIEN BANG BO 4
vdl NGur
NGHTA DLTA TREN DAI S6 GIA TCTNGUYfiN VAN LONG, H O A N G VAN THONG Dai hoc Giao thong van tdi
Tdm tat. Bai toan k^t nh$,p thong tin ndi chung ddng vai tro quan trgng trong qu4 trinh lay quylt dinh va do do no cd y nghia ufng dung r0ng ldn, d^c biet la bki todn ket nh?Lp thong tin md vi con ngudi thudng quyet dinh thong qua thong tin md ngon ngii. Cho den nay cdc phudng phdp gidi bai todn nay chu yeu d\ta tren cdc tap md. Trong [8] va [7] cdc tdc gia da nghien ciiu giai bai todn nay khi ling dung ngii nghia ngon ngii dvra tren d^\ so gia tijf (DSGT) vdi bieu diin ngfl nghia tUcJng iing bang do; lieu bo 2 va bp 3. Trong bai bdo nay, thong tin md cua thang ddnh gid duoc bieu diin bang bo 4 dl khac phuc mot so thigu sdt trong [7]. No cho phep biiu thi thong tin md day dii hdn va cho phep tinh todn tren cdc thong tin ngon ngii trong thang ddnh gid triic tiep hdn va do do kgt qua ket nhap tu nhign va chinh xdc hdn.
Abstract. The role of information aggregation problems is in general very important in the decision process and hence they can find applications in many fields, particularly, for the fuzzy informa- tion aggregation problem since human makes decisions by means of vague linguistic terms. So far there are many methods have been exploited to solve these problems but mainly based on fuzzy sets.
The works [8] and [7] were devpted to solve this problem, applying hedge-algebra-based semantics of linguistic terms to representing linguistic data by 2-tuples or 3-tuples. In this paper we introduce 4-tuple representation exploiting more deeply hedge-algebra-based semantics of linguistic terms for solving this problem. This allows represent the semantics of linguistic symbols in rating scale more completely and hence it produces output results more exactly and naturaly.
1.
M 6 DAURat nhieu van de cua viec lay quy^t dinh hang ngay cua con ngiidi deu dua tren cac thdng tin khdng chinh xac cd ban chat md, khdng chac chdn. Ly thuyet tap md da tao ra cac md hinh toan hoc va cac cdng cu cho phep diia ra cac each tiep can khac nhau de giai cac bai toan lay quyet dinh trong nhiing Hnh viic khac nhau. Nhiing thanh tilu ilng dung cua ly thuyet tap md lai thdc day sii phat trien ciia chinh ly thuyet dd va do vay da ra ddi cac chuyen nganh nhii cdng nghe logic md (Fuzzy logics technology), cdng nghg ket nhap (Aggregation technology), tinh toan vdi cac txt ngdn ngii (Computing with Words-CW) ...
Tinh toan vdi cac tii ngdn ngii la mot ly thuyet cho phep thao tac tren cac tii de giai quyet cac bai toan ma dii lieu dau vao va dau ra can bieu thi bang cac tii ngdn ngii, dac biet ddi vdi cac bai toan lay quyet dinh. Mdt bai toan quan trgng cua qua trinh lay quyet dinh la tdng hop, ket gdp cac thong tin, dii heu ngdn ngii tii cac ngudn khac nhau (tii cac chuyen gia chang han), de cd mot thong tin danh gia dai dien chung, diidc goi la bdi todn kit nhdp tren cac tii ngon ngii. Bai toan cd t h i diidc md ta nhii sau:
242 NGUY&N VAN LONG, H O A N G VAN T H O N G
Gia si'r ngiidi (luyet djnh phai lay (luyet djnh chgn mOt phiidng dn "tdt nhat" trong m phudng 4n Igfa cliMii Ai, i = 1, . .,m, tren cd sd lay ^ kien d4nh gid cua n chuyen gia Cj, j = I, 11. Trong mdi triidng tiidng tin ngdn ngii, eae chuyen gia bilu thj danh gia cua minh bang eae tfr ngdn ngii (thang danh gia, ngdn ngii) liy trong t9.p S = SQ,- • - ,Sg. Ky hi?u Xij la y kien danh gux eiia eliuyen gia j ve phiidng An Ai-
Mgt >eii can tt.r nhien la can djnh gia y kien tong hgp cua cdc chuyen gia ddi vdi tiing phiTdng an. nghia la ta can sir di^ng mgt phep toan ket nh^p f(« tfch hgp cac f kien {xij : j = 1,. ., 7)} eua eae elmyen gia. Mgt each hinh thiic todn hgc, toan tuf ket nhdp Id mgt dnh x?i (J* : {so Sg}" —> {soi • • • ^ Sg}. Anh x^ ndy phai dirge xdc djnh sao cho ket qua ciia phep todn @ ( s t i , . . . . .s,„) eo the'' xem Id bien thj y kien tdp the ciia n chuyen gia.
Nhir vdy, l)ai todn ket nhdp Id mgt bdi t.oaii cdt yeu trong qud trinh lay quyet dinh, va vi gid tri ciia todn ti'r cung la mgt tir ngdn ngfi, nen ngiidi ta can sii dting cdng cu tfnh todn tren cdc tfr ngdn ngfr. Cd nhifMi phiidng phdp ti^p can tfnh todn khdc nhau [1-5, 7, 8, 10] de giai quyet van de nay. De ldm rd ngi dung nghign cutu cua bdi bdo, ta se phan tfch tong quan mgt vdi phudng phdp.
1.1. Phi:fdng p h a p t i n h t o a n n g o n ngU diia t r e n n g u y e n ly m d r o n g c u a t a p md Y tudng chinh cua phUdng phdp Id cdc phep tfnh ket nhdp kinh dign nhu phep trung binh so hgc, trung binh cd trgng so . . . , cd the chuyen thanh cdc phep tfnh tUdng iing tren cac tap md, ching ban phep lay trung binh cgng md, trung binh cdng md cd trgng so tren cac tap md . . . Khi dd, cdc tii ngdn ngii trong tap S dUdc xem la cdc nhdn cua cac tdp md. Cac phep ket nhap md thuc hien trgn cdc tap md ciia cdc nhdn trong tap S se cho ket qua la tap md. Ndi chung tap md ket qua khdc vdi cdc tap md cua cdc nhan, hay nd khdng bigu thi cho mdt nhan ngdn ngii nao trong S. Dieu nay dan den sii cdn thiet phai phdt triln cdc phuong phdp xdp xi ngon ngit, tiic la tim mgt nhan ngdn ngii trong S cd tdp md xap xi tdp md ket qua nhat.
1.2. Phi/cTng p h a p t i n h t o a n t r e n cac k y h i e u n g o n ngi?
Gia sut y kien ddnh gid theo mgt tigu chf dugc bilu thi bdng cdc tii ngdn ngii trong tap S = {SQ,. - .,Sg} dugc sdp tuyen tfnh theo ngii nghia cua chung sao cho: Sj < Sj neu va chi nlu i < j.Vi khdng t h i tfnh true tiep trgn cdc tii ngn ngUdi ta mugn ckn triic tfnh todn cua doan [0, g] bao hdm cac chi so di thi;c hi?n vi?c ket nhdp so hgc. Y tudng nay the hien nhU sau [3-5].
Gia sut ta lay ket nhdp tdp cdc tii ngdn ngii trong A = {ai,. ., Op}, Oi E S.Ta thiic hign mgt hodn vi cdc chi sd cua tdp A, A = { a ^ i , . . . , a,rp}, sao cho a,ri > a-irj neu i < j - Xet mdt phep kit nhap so hgc g nao dd, g se cdm sinh mOt phep ket nhdp 9* trgn tap S dUdc dinh nghia nhu sau: Tfnh g{'Ki,..., TTp) E [0, p], vdi TFI, . . . , Tfp la cdc chi so cua cdc phan tut trong A. Ddt i* = round(5'(7ri, . . , TTp)), trong dd round la phep lam tron so hgc. Khi dd phan tut Si*
dugc xem la kit qua ket nhdp g*{aT,^,. ., a^rp). Phep kit nhap g* dudc dinh nghia nhu tren la mdt han che, yi nd chiu mgt rang bugc ft t u nhien va mat mat nhilu thong tin. Hiln nhign, thay vi dung chi so cdc tii nhu tren, neu ta sut dung ngii nghia dinh lugng cua cdc tut ngdn ngii trong thang ddnh gid thi dinh nghia phep ket nhap g* cd cd sd hdp ly hdn (xem [10]).
VAN D ^ KET NHAP THONG TIN Bifiu DI6N BANG BO 4 ... 243 1.3. P h t f d n g p h a p t i n h t o a n n g o n ngiJ di^a t r e n bieu d i i n ddf lieu b o 2
^Trong phUdng phdp tren ta can lam trdn bdng bilu thiic i* = round(p(7ri, • •, TTp)) de kit qua la mgt tii ngdn ngii Oi* trong tdp S. Tuy nhien vi?c ldm trdn ldm m4t mdt thdng tin vd cac tdc gia [4] da dUa ra cdch bieu diln dii li?u bO 2 de khdc phuc si; mat mdt thdng tin ndy
k-1 .k
\—h-\ I tj I I
0 b, g
Hinh 1.1. Bilu diln bg 2
Y tudng ciia phUdng phdp nhu sau: gid tri g{ni, . . , TTp) chUa ldm trdn se Id mgt so thuc b E[0,g]. k-0.5 <b < k + 0.5. De bieu thi rd cac thdng tin cdc tdc gid trong [5] de xuat dang bilu dien bg 2, {sk, bk). trong do bk = b - k E [-0.5, -f-0.5). Tii Sk dugc ggi Id tam cua thdng tin cdn Ok bilu thi gid tri chuyin dich tCr gid tri goe b ve tit ngdn ngii Sk- Bieu dien nay ciing xdc dinh mdt dnh x^ sau:
^•[0,g]-^Sx[-0.5,0.5);A{b) = {sk,bk), vdibk = b-k,k = vound{b) (1.1) vd, do dd, ta cd
A-':Sx [-0.5,0.5) ^ [0, g]; A-\{sk, bk)) = k + bkE[0, g].
Vdi bilu dign nhu vdy, cac phep kit nhdp trgn cdc tut ngdn ngii dUdc dinh nghia thdng qua cdc phep kit nhap so hgc nhd viec chuyin doi cua dnh xa A" ' . - 1
1.4. Phufdng p h a p t i n h t o a n n g o n ngiJ dtfa t r e n bieu d i i n diJ lieu b o 3
Trong [8,7], cdc tdc gia da nghign ciiu phdt triln viec tfnh todn trgn cdc tut diia trgn dai so gia tut (DSGT) vdi viec tan dung cdc khoang do do tinh md cua cdc tut ngdn ngii. Theo cdch tilp can ciia DSGT, ngii nghia cdc tut ngdn ngii cd t h i bieu thi qua cdc khoang tfnh md mute /, vdi I chi do ddi cua sdu bilu diln cdc tii ngdn ngii. Cho trudc I vd xet mdt tut s bat ky do dai I. Bilu diln ngii nghia bg 3 [7] cua tut s la
{s;v{s);n{s)) (1.2) trong dd v{s) la gid tri dinh lUdng cua tut s, il{s) la khoang tfnh md cua tut s. Ta bilt rang
v{s) E Q{s).
Lgi fch cua phUdng phdp nay Id thay cho viec phai xet tap chi so cua cdc tii, mdt dai ludng mang it thdng tin ve ngii nghia ciia tur, ta sut dung gid tri thiic v{s) va khoang tfnh md Cl{s) chiia nd mang ngii nghia dinh lUdng cua tut s. Rd rang bieu diln bd 3 mang nhieu thdng tin hdn va tu nhien hdn so vdi bilu diln bd 2.
Nhugc dilm cua cdch tilp can ndy la gia tri thiic v{s) dugc xdc dinh bdi ham dinh lUdng V la mdt rang bugc chat, vi chang han kit qua cua ph6p kit nhdp so hgc khdng nhi.t thilt triing vdi gid tri thuc cua tut ngdn ngii bieu thi kit qua kit nhdp trgn cdc bg 3 dii heu. Ngoai ra trong [7] ciing dUa ra mdt rang budc chat la cdc tut kit nhdp dugc bieu thi bang cdc bd 3 dii Ueu tren cumg dd dai I.
De khac phuc nhiing thilu sdt ngu tren, trong bdi bdo nay chiing tdi md rdng bieu diln bg 3 thanh mgt bilu diln bd 4 bieu thi nhieu thdng tin hdn vd cho phep hnh hoat hdn trong
244 NGUVftN VAN LONC, HOANG VAN THONG
vi?c giai eae bdi todn ket nhdp 'i*''! <''»" '"f "K"ii ngfi vd trong gidi quylt bdi todn quylt dinh diia tren y kien nhieu ehuyen gia,.
Ngi (iuiifi, edn I9J eiia bdi bao diigc td eh tie nhU sau. Mt;c 2 trinh bay phUdng phdp hinh thiic hda l)i(Mi dien ngii nghia bdng bg 4 cac tir ngdn ngfi trong thang ddnh gid vd cdc phep ket nhdp tien dilt li(Mi bieu dien bg 4. Mtie 3 ino ta ea,(li gidi bdi todn liy quylt djnh bdng cac phUdng phap bieu dien bg 1 do bdi bdo de xuat vd so sdch kit qua vdi cdc kit qua cua cac phUdng phap khac. Viee phdn tieli so sdnh ear ket qud ciia cdc phiidng phdp se chiing td lgi fch va hicu qua ciia phUdng phap mdi.
2. X A Y D U N G T H A N G D I ^ M N G O N NGIJf V<3l NGLf N G H I A Bi^u DIIJN BANG BO 4
2 . 1 . Lan c a n ngu: n g h i a ciia cdc tit n g o n nguf
Ddt S = X^k) = XiU. . U Xfc, nhd rdng Xi Id tdp cdc tir cd dg ddi /.
Nhu dd trinh bdy trong Myc 1, doi vdi bdi todn ddt ra ta can xet thang ddnh gia bdng cac tut ngdn ngii S = X(^k) = SQ, .. .,Sg.De dua ra cdch bieu dien ngii nghia bdng bg 4, trudc hit chiing ta dinh nghia ldn can ngU nghia cua cdc tir ngdn ngii. Ext. DSGT AX— {X, G, C. H, E, $, <).
Dl dinh nghia ldn can ngii nghia mute fc, ta xet tap cdc khoang tfnh md miic fc -I-1, h+i- Ta bilt rdng Z^+i tao thdnh mgt phdn ho^ich cua doan [0,1].
Xet tdp cdc khoang tfnh md miic fc -I-1, Ik+\- Ta se phdn cum cdc khoang tfnh md trong Ik+i nhu sau. Vi cdc khoang tinh md trong Ik+i rdi nhau vd kl tilp nhau, ta sdp xep chiing theo thut tu tdng dan. Vdi 2 <p,q, ddt
Hi = {hi, h.j:l<i< [p/2]&I < j < [q/2]}. Hi = {hi, h-j : [p/2] < i < pk[q/2] < j < q}
Viec phdn cum se dUdc thiic hi?n nhu sau. Ta duy§t tut trdi sang phai cdc khoang tfnh md
^k+i{hiyj), hi E H va yj E Xk, trong h+i- Chiing se dugc xep vao cung mgt cum nlu cac khoang ke tilp cd cdc gia tut hi cimg thuge tdp H,,, vdi n = 1 hodc n = 2. Cd t h i kiem tra thay rdng tdp
Cy,i = {^k+i{hiy) : hi E Hi}.yE Xk, (2.3) chfnh la mdt cum theo cdch phdn cym ndy va cdc khoang trong Ik-^\ ndm giuta hai cum dang
(2.3) kl tilp Id cum chiia mgt so khodng 3^fc+i(/ity) : hi E Hi-
Ky hifu c la tap tat ca cdc cum theo cdch phan cuni trgn va moi cuna cua nd dUdc ky hieu la C. Cd the de ddng kiem tra thay rdng hg C la mgt phdn cum cua tdp h+i-
(i) Cdc cum C Id rdi nhau;
(ii) Mgi khodng tfnh md trong Ik+i diu thuge vdo mgt cum C ndo dd.
Ddt X(fe) = A:iU...UA'fc va 4^
Sk{C) = Y{^k+i : ^k+i e C} (2.4) M e n h d e 2 . 1 . (i) Sk{C) la mgt khoang md d ddu mut phdi vd ddng d ddu miit trdi, trii cum
sinh ra khodng tan cUng phdi Id doq/n ddng cd hai ddu miit, vd ho Sk{C) : C E \ la mdt phan hoach cua doan [0,1];
(ii) Vdi moi y E -X^(fc)> ton tQ,i m0t khodng Sk{C) sao cho gid tri dinh lucfng v{y) nam trong Sk{C) cimg vdi hai khodng tinh md mite k + 1 ki vdi no.
Viec chiing minh menh de trgn la rat ddn gidn. -".hiing tdi khdng trinh bay d ddy.
VAN D E K E T NHAP THONG TIN B I ^ U DifiN BANG BO 4 ... 245 D i n h n g h i a 2 . 1 . Vdi mgi y E X^k), khodng (2.4) cd tfnh chit v{y) E Sk{C), va nd Id dilm
trong cua khoang Sk{C) vdi tdpd t u nhign trgn mien so th^c, dUdc ggi Id ldn cdn ngii nghia miic fc cua tut ngdn ngii y.
2.2. B i e u d i i n tuf n g o n ngu? b a n g b o 4 ngi? n g h i a
Cho mgt DSGT tuyin tfnh vd diy du A X = {X, G, C, H, <) va xet mdt tdp hiiu han cac tii ngdn ngii 5 e X . T^p S dugc ggi la ddy dii ngii nghia, hay d l ddn gidn ggi la diy du, ngu
hx E S =^ kxE S. vi\/k E H (2.5) Khdi niem day dii bao dam rdng thang ddnh gid XQ d moi miic sinh tii ciia nd cd du mdt
cdc tut ngdn ngii sinh bang gia tut. Rd rdng rdng Xf^k) la diy dii ngii nghia.
Vf du, ta xet mgt DSGT A X vdi G = {bad, good}, H = {R, V}, trong do cac chii cdi la sU riit ggn tUdng iing ciia cdc tur Rather vd Very. Gia sut 5 = {F_ bad, bad, R_ bad, medium, R^good, good, V-good). Thang ddnh gid nay la diy du va dugc bilu thi bdng dang cay nhu trong Hinh 2.2. Tinh day dii ciia tap S dugc kilm chiing bdng cdch xet cdc ndt con cua mdt ndt cha, nlu chiing suih tut cha bdng gia tut thi cd dii mdt cdc gia t\l RvaV
Tap XQ Id tap sdp tuyen tinh vdi quan he thii tu Id: V^bad < bad < R^bad < medium <
RR^good < R^good < VR^good < good < V-good.
medium
good
V_bad R_bad Rjgcod V_good Hinh 2.1. Cdy bilu diln tap S
Dudi ddy chung ta dua ra cdch bieu diln thdng tin bd 4 cua cdc tii ngdn ngii trong thang ddnh gid ngdn ngii.
D i n h n g h i a 2.2. Xet mgt DSGT A X = {X, G, C, H, <) dUdc trang bi mgt do do tinh md fm. Gia sii I Id so nguyen dUdng. Bilu diln bg 4 miic I cua mdt tut ngdn ngii s E X \a bd 4 ngii nghia (s, v{s),r, Si{s)), trong dd v{s) Id gid tri dinh lUdng cua tut s, Si{s) la khoang lan can ngii nghia miic I ciia s, ggi Id miic ldn can, vd r, v{s) E Si{s). Hieu so r — v{s) dUdc ggi la do lech ngii nghia cua gid tri dugc chgn r
Khdc vdi bilu diln bg 3 trong [7], cdc tii s khdng bi rang budc phai cd ciing do ddi vd thgm thanh phin mdi r Thanh phin nay bao dam cdc phep tfnh kit nhdp trgn bd 4 la ddng.
Ngoai ra, ve thuc tiln nd cho phep cdc chuygn gia cd the cho diem ddnh gid bang so thiic trong mien tham chilu ciia biln ngdn ngii.
Vi ta khdng t h i tfnh todn true tiep tren cdc tur s, cdch bilu diln ngU nghia cua tut nhu vay cho phep ta tinh todn trgn gid tri thuc r. Y nghia cua dnh xa ngii nghia dinh lUdng v{s) la gid tri thuc mang nhieu thdng tin cua tur s nhat vd cd the xem nd tUdng ting vdi tam diem (core) cua tap md bilu thi ngii nghia cua 5. Vdi y nghia dd, hieu sd r — v{s) mang thdng tin ve do phii hdp cua r doi vdi tur 5. Nhu vdy thay vi tfnh todn tren cdc chi so cua tap S, trong each tilp can nay chung ta tinh todn trgn sd thiic r ndm trong ldn can «S/(s)-khoang ldn can ngii nghia cua s.
246 NGUYEN VAN LONG, H O A N G VAN T H O N G
2.3. X a y duTng bi4u d i i n bd 4 ngtir n g h i a cho thang ddnh gid n g o n nguT
Trong nhieu iiio hinh iing dung ngudi ta mong mudn cho ph6p cdc chuygn gia ddnh gid cdc tieu chf doi vdi cac ddi tugng can ddnh gid d l ra quyet djnh theo thang dilm ngdn ngii.
X6t mgt thang ddnh gia ngdn ngfr S cd thii tU tuyen tfnh d\fa tren ngii nghia cua chiing.
Gia silt tdp thang ddnh gia S eo tfnh chat "ke tiep", bdo ddm tfnh da d9,ng cua thang danh gid ngdn ngfr, sau:
{'is E S){3s'E S){3h E H){s = hs') (2.6) Y nghia eua (2.6) Id thang ddnh gid ngdn ngii lien tuc ve miic md td ngut nghia sinh ra bdi
cdc gia tir: khdng co tii ndo trong thang ddnh gid md phdi sit dy/ng hai ldn tdc dgng gia tU ki tiip vdo mot tH: ndo dd dd cd trong thang ddnh gid mdi .sinh dU0c ra nd. Ndi khdc di, neu can hai lin tac dgng gia tut thi cd nghia trong tdp S cdn thieu mgt so t ir ngdn ngii ndo dd.
Vf du, thang ddnh gid ngdn ngii S = bad, medium, RR-good, VR-good, good, RV^good, VV-good Id khdng diy vi ''RR-good" chi co the sinh dugc tii "good" trong 5^ bdng vi?c siJt dung hai lin tdc dgng lign tiep gia ti'r R.
Thang ddnh gid ngdn ngut bdng bO 4 ngfr nghia dugc dinh nghia nhu sau.
D i n h n g h i a 2 . 3 . Cho mgt thang ddnh gid ngdn ngii 5 C X gom cac tur ngdn ngd:, va mgt so nguygn dUdng I, mgt bilu dien bO 4 ngu: nghia cua S, {{s, v{s), r, S{s)) . s E S,r E S{s)}, r Id kf hieu bien thUc, dugc ggi Id mgt thang ddnh gid ngdn ngii di^a trgn bg 4 ngii nghia nlu tap S{s) : s E S ldp thanh mgt phdn ho9,ch ciia do^n [0,1], tiic la ta cd:
(ij S{s) E S{s') = 0 , Vs, s' e 5 vd s 7^ s';
{ii)S{s) = [0,l].
Gid tri thuc r trong bg 4 ngii nghia ddng vai trd nhU dilm thuc trong thang danh gid thiic trong trudng hgp kinh dien. 0 ddy doan [0,1] Id thang dilm thUc dugc chuan hda d l tien trinh bdy ve phUdng phdp ludn. Mdi quan h? dinh lugng giiia s va r t h i hi?n quan he ngii nghia trong bilu dien bg 4. Dilu ki?n (i) bao dam rdng mgt gid tri thuc r khdng t h i Id bieu thi ciuig liic cho hai tut ngdn ngii khdc nhau. (ii) ndi rdng bat ky gid tri ndo ciia thang dilm thuc ciing cd t h i dugc chgn d l bieu thi mgt tii ndo dd trong S vdi dg sai l?ch dugc do bdng r — v{s)-
Vdi cdch bilu diln bd 4 ngii nghia vi?c so sdnh hai bO dugc xdc dinh bdi thut tU cua cac tut s trong S vd thii ty: cua gid trf thifc trong thdnh phdn thU 3 ciia bg 4. Cu t h i Id:
{si,v{si),ri, S{si)) <df {sj,v{sj),rj, S{sj)) neu Si < Sj hodc Si = Sj k r^ < rj.
va {si, v{si),ri, S{si)) =df {sj, v{sj)^rj, S{sj)) nlu Si = Sj k ri = rj.
Vi {S{s) : s E S} ldp thdnh mgt phdn h o ^ h , cd t h i thay rdng {si,v{si),ri, S{si)) <
{sj,v{sj),rj, S{sj)) <^ ri < rj. Mgt bdi todn ddt ra Id, cho thang ddnh gid ngdn ngii S, co the hay khdng xdy dung dugc mgt thang ddnh gid ngdn ngii dvfa trdn bg 4 ngii nghia cho 5.
Gia sut ks la dg ddi ldn nhat cua cdc tut trong S, nghia la ks la cdn tren diing ciia cac |s|
vdi mgi s trong S. Ta se dua trgn cdc khodng tinh md ciia S dk xdy dung cac khoang S{s), vdi / Id miic ldn cdn. Trudc hit ta dUa ra khdi nifm "Khodng ldn cdn kl" cua s.
Nhu trgn, ta kf hifu X^ Id tdp cdc tut trong A" cd dg ddi /. X6t tdp cdc khoang tfnh md mute I, ll = {Ji{x) •.XEXI} = \J{{Ji{hiu) : i E [-q,p]} : u E X j - i } . Vdi I > ks, nhu chiing ta bilt cac gid tri v{s), s E S, diu Id cdc dau miit cua cdc khoang tinh md miic /. Hdp ciia hai i:hoang tinh md mute / cd cimg chung dau miit v{s) dugc ggi la khodng ldn cdn ki cua s,
sEjS.
Mfnh de 2.1 chiu sU rang budc \a2 <p,q dugc md rgng bdng mgnh d l sau.
M e n h d l 2.2. Gid st? tap S cd tinh chdt ki tiip. Khi dd mUc I nho nhdt de cdc khodng l&n cdn ki cua v{s), cUng ki hi$u la Si{s), vdi s E S, rdi nhau la:
VAN DE KET NHAP THONG TIN BIEU DIEN BANG BO 4 ... 247 (i) l = ks + l, vdip,q> 2;
(ii) l = ks + 2 vdip=l hoac q=l.
Menh de 2.2 bao dam rdng doi vdi tdp S cd tfnh ke tiep ta ludn ludn xdy dung dugc he khoang ldn can miic / rdi nhau. Vigc md rgng mgt he khodng rdi nhau nhu vdy dl nd trd thanh phdn hoach cua [0,1] Id khdng khd khdn ve mdt thudt todn.
Sau day ta dua ra thudt todn xdy dung bieu diln bd 4 ngii nghia ciia tdp S cdc tii cd tfnh kg tilp.
Thuat toan 2.1. Xdy dUng biiu diin bo 4 ngit nghia cua S.
Xet danh sach cac tut ngdn ngii 5 = SQ, s i , . . . . Sm thda dieu kifn (2.4) vdi thii tu thda man Si < Sj nlu vd chi neu i < j .
Vdi I la mgt so nguygn dudng thda dieu kifn trong Mfnh de 2.2, xet danh sach Ii = {Ii{xj) : Xj E Xi,j = 0,1,...,n} gom cac khodng tfnh md miic / thda dieu kien Ii{xj) < Ii{x'j) khi va chi khi j < j ' Kf hifu khoang lan cdn liln kl miic / Id Si{s), s E S.
Xet vi tut Incl{Ii{xj).Si{si)) ndi rdng khoang Ii{xj) chiia trong khoang Si{si).
Bvtdc 1: Xuat phdt tut so(i := 0) va tur khoang tinh md diu tien Ji{xj){j := 0) trong danh sach / / . Luu gid tri chi so 0 vdo biln Index;
Kilm tra dilu kifn Incl{Ji{xj),Si{sQ)) lin lugt doi vdi cdc Ji{xj) cho den khi tim dugc j d i u tign thda dilu kign Connect{Ji{xj),Si{sQ)) = "true" Dat Tempi{so) :=
Si{so) U Yindex<k<jli{xk)- Thay j tim dugc vao bien Index va tang chi so j them lj:=j + l-
BtCdc 2: T ^ g i:=i + l.
Kilm tra dilu kifn Incl{Ji{xj),Si{si)) lan lugt tiep cdc Ji{xj) cho din khi tim dugc j thda dilu kifn Incl{Ji{xj), Si{si)) = "true" Dat Center = round{{j + Index)/2) va dat S{si-i) := Tempi{si-i)UYjndex<k<CenterIi{xk) vd dua S{si-i) vao danh sach NEIGH- SYS. Dat Tempi{si) := «S/(sj) ^Ycenter<k<jJi{xk)- Thay j tim dugc vao biln Index va tang chi so j them 1 : j := j + 1.
- Nlu i < m, quay vl Budc 2.
Si/dfc 3: Dat S{sm) := Si{sm)Yindex<k<nli{xk) vd dua vdo danh sach NEIGH-SYS.
Vdi mdi Si, tinh v{si) va thdnh lap tap bg 4 ngii nghia {si,v{si),r, S{si)), vdi r E S{si) E NEIGH-SYS da dugc xac dinh trong Budc 2, va dua vao tap OUTPUT.
- Xuat danh sdch kit qua OUTPUT vd kit thiic thuat todn.
Dinh ly 2.1. . Vdi I thoa mdn Menh di 2.2, thuat todn luon luon dicng, dd phitc tap tuyin tinh vd kit sudt (ouput) cua Thudt todn 2.1 la mot phdn hoach cua doan [0,1]. Do do, tap biiu diin bo 4 {{s,v{s),r, S{s)) : s E S,r E S{s)} cua S la thang ddnh gid ngon ngit dUa tren bd 4 ngU nghia.
Chimg minh. Theo gia thiet ciia I, cdc gid tri v{si), i = 0 , 1 , . . . , m, diu la cdc d i u niiit cua cdc khoang tfnh md mute I trong Ii = {Ji{xj) : Xj E Xi,j = 0 , 1 , . . . , n}. Chii y rang hg 11 la mgt phdn hoach cha doan [0,1] va, do vdy, ludn tdn tai khoang Ji{xj) thda dilu kifn Connect{Ji{xj), Si{si)) = "true" Theo Dinh ly 2.2, giiia hai gia tri lien tilp b i t ky, v{si) va
|(si+i), diu cd it n h i t 2 khoang tfnh md cua Ii- Mgt nuta so cdc khoang dd dUdc sii dung
248 NGUVfeN VAN LONG, H O A N G V A N T H O N G
d l xdy dung khodng lan can ^(.Sj) vd niia cdn 1^,1 d l xay dUng S{si+i). Lum y rdng tat ca eae khodng tfnh md miic / mdm ben trai v{so) dugc silt dung d l xdy di/ng S{so)) va t i t ca cac khoang tfnh md iniie / mdm ben phdi v{sm) dugc sut dung d l xdy di^ng S{sm)- Tur dd suy ra taj) {5(.s,) : / ^ 0 , 1 , . ., m} t9,o thdnh mOt phdn ho9,ch ciia do^in [0,1].
Theo each xay di,mg, ludn ludn cd 2 khodng tfnh md miic I ke bfn trdi vd ke bgn phai cua v{.Sj) ndm trong 5/(.s,) vd do dd v{si) E S{.si), i = 0 , 1 , . . . , m.
Ta thay rdng thudt toan chi duyft mgt lan danh sdch Ii vd do dd dc ddng chiing minh
cdc khdng dmh edn l^ii trong phdt bilu djnh 15^' trfn. • 3. C A C P H E P K E T N H A P T R f i N C A C B O 4 N G 0 N G H I A
3.1. Bai todn ket nhdp vd ph^p ket nhap
Trong thuc te ed rat nhilu van de dan din bdi todn tim f kiln ddnh gid ldm dgi difn chung cho y kien eua eae elmyen gia ve ingt van de ndo dd. Vice tong hgp va kit hgp cac y kiln rifng ciia timg chuyen gia dugc ggi Id vi$c kit nhdp (aggregation) vd phep tfnh thuc hifn vifc ket nliA}) eae y kiln rieng biet dugc ggi Id ph4p kit nhdp (aggregation operator). Trong cdch tiep cdn tinh todn trgn tir, thang ddnh gid la thang cac- tur ngdn ngii vd phep tfnh kit nhdp ciing phai cho kit qua la mgt tii trong thang diem.
Tudng tu nhu trong [4,5,7], chung ta se md rgng cdc phep kit nhdp thdng thudng trgn so thuc "dai difn" ngii nghia dinh lugng trong bieu diln bg 4 ngii nghia ciia cdc tut trong S.
Trong phdn nay ta xet mdt thang diem ngdn ngii dua trdn bg 4 ngii nghia cua S = So, Sl,. ., Sm, Ts = {{si, v{si), ri, S{si)) : r^ 6 S{si), z = 0 , 1 , . . . , m}.
Tudng tu nhu trong [5], ta dinh nghia dnh xa sau:
A : [0,1] ^ Ts, A(r) = {si, v{si), r, S{si)), r E S{si), vdi r € [0,1] (3.7) Dinh nghia dnh xa (3.1) la chinh vi hg {S{si) : i = 0 , 1 , . ., m} Id mdt phdn hoach cua
[0,1] ngn chi ton tai duy nhat mgt tii Si thda man (3.1).
Anh xa ngugc A ^ dugc xdc dinh bdi cdng thiic
A - n ( s i , v{si),r, S{si))) = r,rE [0,1] (3.8) D i n h nghia 3.1. Cho g la mgt phep kit nhdp p - n g d i trgn cdc so thi/c cua doan [0,1],
g : [0,1]P -> [0,1]. Phep p-ngdi g*,g* : {Ts)^ -^ Ts, dugc ggi la phdp kit nhdp md rdng cua phep kit nhdp thdng thudng g sang miln cdc bg 4 chiia cdc tut ngdn ngii nlu nd dUdc dinh nghia nhu sau:
Vdi vectd dii lifu {{sik, v{sik), rik, S{sik)) : fc = 1,. . ,p),
g*{{sii, v{sii), m, S{sii)),..., {sip, v{sip), rip, S{sip))) = A{g{rii, .., g{rip)) (3.9) Dinh nghia phep kit nhdp trfn tut dua trgn bilu diln bg 4 ngii nghia ciia chiing mang nhilu thdng tin ngii nghia ciia tut hdn so vdi vifc thUc hifn phep kit nhdp tren cdc chi so vi nhiing ly do sau:
(i) Cac khoang ldn can S{si) cua cdc tur mang ngii nghia dinh lUdng cua cdc tut vdi miic tinh md I. Chiing dugc xdc dinh dua tren dg dg tinh md cua cdc tut ngdn ngii. Vi vdy, gid tri thuc rj cung mang nhilu thdng tin ngii nghia cua tut nhilu hdn cdc chi so ciia chiing;
VAN Dfe K^T NHAP THONG TIN Bifiu DifiN BANG BO 4 ... 249 (ii) Theo dinh nghia cua bg 4 ngii nghia, gid tri ngii nghia dinh lugng v{si) cua Si cd the
dugc xem Id gid tri thuc mang thdng tin ngii nghia phii hgp vdi si nhat va hieu so {ri* - v{si*)) cho ta thdng tin vl df lech cua vifc chuyen doi gid tri thuc vo gid tri ngdn ngii.
Sau day ta chi ra mgt so phep kit nhdp thdng dung.
1) Phep trung binh cong so hgc
Cho cac bg 4 ngii nghia tren thang dilm Ts, {sik, v{sik), rik, S{sik)), k=l,...,p. Theo Dinh nghia 3.1, phep ket nhdp trung binh so hgc
9anth{0'l^ --,ap) = - y ak p ^-^
i<fc<p
cam sinh mgt phdp kit nhdp tron cdc tut vdi bieu diln bg 4 ngii nghia sau:
9*ariiii{si\,v{six),rix,S{sii)),...,{sip,v{sip),rip,S{sip))) = A{- ^ nk) (3.10)
i<fc<p
nghia la gid tri cua ga^th Id bg 4 {si*,v{si*), \ Y.i<k<p^k), S{si*)) iing vdi i* sao cho - X ! ^ifc e S{si,).
l<A;<p
2) Phep trung binh cgng cd trong so
Cho cac bd 4 ngii nghia trgn thang dilm Ts, {sik, v{sik),rik, S{sik)), fc = 1 , . . . ,p. Phep trung binh cd trgng so thdng thudng
9weight{0'l, . . . , ttp) = 2Z '^kO'k, Wk >Ova ^ tWfc = 1.
l<fc<p 1<A:<P
cam sinh mdt phep kit nhdp trung binh cd trgng so ^^gjo^jj sau:
^weig/it((5il) ^ ( ^ i l ) ) ^ i l . S{Sii)), ---, {Sip, v{sip),rip, S{Sip))) = A ( ^ tOft'^ifc) l<fc<p
nghia la gid tri cua gl^ight la bd 4 {si*,v{si*), J2i<k<p ^k^k, S{si*)) iing vdi i* sao cho
^ WkTik e S{si»).
l<k<p
3) Phep kit nhdp trong sd cd thit tU (ordered weighted aggregation operator)
Cho cdc bg 4 ngii nghia trgn thang dilm Ts, (sijfc, v{sik), rik, S{sik)), fc = 1,...,p. Phep kit nhdp trgng so cd thii tu thdng thudng
gOr-weight{ai, - •, Op) = ^ WkO^, Wk > 0 va } Wk = 1,
l<k<p l<fc<P
trong dd vectd (of,. •, a^") thu dugc tut vifc sap xep ddy o i , . ., Op theo thii tu giam din, se cam sinh mgt phep kit nhdp trgng so cd thd: tu ^^gigAt ^^^•
9ch-weightiiSil,v{Sii),rii, S{Sii)), . . . , {Sip, v{Sip), rip, S{Sip))) = A ( ^ WkO,'^) l<k<p
250 NGUVfiN VAN IVONG, H O A N G VAN T H O N G
nghia la gia tr} eiia g^^^^^t la bg 4 (sj*, f (.Sj-). Ei<fc<p ^^fe^^. S{8i»)) ling vdi i* sao cho y ^ WkO^ € ^(.S,;*).
3.2. Vf d u m i n h h o a m o t d cdc phifdng p h d p t i ^ p cfin
T^ hay lay vf du vi(>c gi;ii bai toan quyet djnh ngon ngfi (Linguistic decision problem) dl so sdnh gifra hai phUdng phdp tiep edn.
Trong 11.5] dUa ra \ i du ve bai todn ling dung nhU sau. Gia silt mOt cdng ty phdn phoi muon ky hgp ddng \'()i mgt cong ty tU vin ldm mgt bdo cdo phdn tfch tong quan ve cac kha ndng cung cip eae giai phap xay dung lio mdy tfnh hifn en trfn thi trUdng de quylt dinh phUdng an phfi hgp nhat cho nhu can ciia cdng ty. Nhutng phUdng an lua chgn bao gom:
Xi
Unix
^2
Windows-NT
•-^.3
AS/400
X4
VMS
Cdng ty tu vin lap mgt nhdm gom 4 phong thUc hifn tu vin ndy, bao gom Pi
Cost analysis
P2
System analysis
P3 Risk analysis
P4
Technology analysis
Mdi phdng cung cap y kiln ddnh gid cua minh ddi vdi tuttig phUdng an (Alternatives) bdng mgt vectd ddnh gia dUa trgn y kiln chuygn gia (Expert) bilu thi qua thang dilm ngdn ngii S = X(2) = {EL, VL, L, M, H, VH, EH}, trong dd cac ky higu trong 5 Id cac chii vilt tdt cua "Extremely Low", "Very Low", "Low", "Medium", "High", "Very High" vd -'Extremely High"
Cdc y kiln dUdc cho trong bang sau:
Al E x Pi P2 P3 P4
X i
Vl m
h h
Xi
m I Vl h
X3
m Vl m
I
X4
I h m I
Chiing tdi thuc hifn giai bai todn ndy vdi phUdng cdc phUdng phdp trinh biy d trfn va so sdnh kit qua trong bdng 3.1. Dudi ddy chiing toi md td chi tilt phUdng phdp gidi dua trfn bilu diln cdc tut bdng bO 4 ngii nghia.
So vdi vf du trong [4], d ddy chiing ta da thay tii N (Negative) bdng tii EL (Extremely Low) vd P (Positive) bdng tut EH (Extremely High). Doi vdi 3 phUdng phdp trdn, viec thay dii ndy khdng anh hudng din bdn chit cua phUdng phdp ludn vi thang dilm ngdn ngii chi la mgt ddy cdc ky hifu ngdn ngii. Y nghia cdc tut chi cd tinh gdi nhd va thdn thifn vdi ngudi diing. Tuy nhifn, theg each tilp cdn dua trfn ngii nghia ciia DSGT, ngii nghia ciia cac ky hifu lai quan trgng. Vifc thay doi nhu tren de bao dam cdc tur se ndm trong tap n l n cua mgt DSGT vd, t i t nhien, vifc chgn ndy se anh hudng din kit qud tfnh todn.
Gia sut cdc tur trong S = {El, VI, I, m, h, Vh, Eh}, trong dd I := Low, h := High va m := Medium, cd ngii nghia dinh lUdng trfn khdng gian tham chilu [0, Ij. 5 la mdt tap con
VAN DE K 6 T NHAP THONG TIN BIEU DIEN BANG BO 4 251 trong mgt DSGT A X = {X, G, C, H, <), vdi G = {/, h}, H = {R, Lt}, trong dd R la vilt tat cua tu Rather" va Lt Id viet tdt cua tut "Little", H+ = {V,E}.
Gia sut do do tfnh md cua cdc tii dugc cho nhu sau: fm{l) = 0.5 = fm{h), n{R) = fi{E) = 0.24; fx{V) = fj,{Lt) = 0.26. Theo Mfnh d l 2.1, ta cd bilu diln bg 4 vdi miic ngii nghia / = 3, nghia Id vifc tinh cdc ldn can ngii nghia ciia S dua tren cdc khoang tfnh md ciia cdc tii X3 ' Kgt qua tfnh todn thu dugc nhu sau:
v{El) = 0.06 1^(1^0 = 0.185 v{l) = 0.25 v{m) = 0.5 v{h) = 0.75 v{Vh) = 0.8488 v{Eh) = 0.94 Ss{El) = [0, (0.0288+0.0312) * 2) = [0,0.12);
S3{Vl) = [0.12,0.0338 + 0.0312*2) = [0.12,0.2162);
5 3 ( 0 = [0.2162,0.0338 + 0.12) = [0.2114,0.37);
S3{m) = [0.37,0.13*2) = [0.37,0.63);
S3{h) = [0.63,0.12 + 0.0338) = [0.63,0.7838);
S3{Vh) = [0.7838.0.0312* 2 + 0.0338) = {0.7838,0.88);
S3{Eh) = [0.88, (0.0312 + 0.0288) * 2) = [0.88,1.0];
Khi dd ta hodn todn chi ra dugc cdc gid tri bd 4 cua thang ddnh gia cua S. Chang han ta cd cac gid tri: {El, 0.06, r, [0.0,0.12)), r E [0.0,0.12); {VI, 0.185, r, [0.12,0.2162)), r E
[0.12,0.2162); (V/i, 0.8488, r, [0.7838,0.88)), r € [0.7838,0.88) ...
Dl so sdnh, tUdng tu nhu doi vdi trudng hgp bilu dien bdng bg 2 trong [5,8], ta sii dung phep kit nhap trung binh so hgc:
Doi vdi phUdng an xi : (0.185 + 0.5 + 0.75 + 0.75)/4 = 2.185/4 = 0.54625 E S3{m) Doi vdi phudng dn Xi : (0.5 + 0.25 + 0.185 + 0.75)/4 = 1.685/4 = 0.42125 E Sslm);
Ddi vdi phudng an X3 : (0.5 + 0.185 + 0.5 + 0.25)/4 = 1.435/4 = 0.35875 E Sslm);
Doi vdi phUdng an X4 : (0.25 + 0.75 + 0.5 + 0.25)/4 = 1.75/4 = 0.4375 E S3{m);
Theo dinh nghia thii tu trgn cdc bd 4 ngii nghia, rci la phUdng an dugc lua chgn. Nhu vay, phudng an xi cd dilm ngdn ngii bilu thi bdng bd 4 ngii nghia (m, 0.5,0.54625, [0.37,0.63)) la cao nhat va trung vdi kit qua lua chgn cua phudng phap bilu diln bang bd 2 dugc trinh bay d dilm 3).
Bang 3.1 So sach cdc kit qua ldi giai cua 4 phuong phdp, cac d chii dam la cac phUdng an dugc lua chon.
Phirong an
X i X2 XS X4
PhUdng phap biiu diin ngii nghia cac tH bing tap md appi{Cxi) = m appi{Cx2) = m
appi{Cx3) — 1 o p p i ( C x 4 ) = na
PhUdng phdp dua tren ky hi^u
ngon ngii trong thang diem
m m 1 m
PhUdng phap dua tren bieu dign
dien cac tiit biing bp 2
( m , 0.0) (m, -0.5) (1, 0.25) (m, -0.25)
PhUdng phap giai dila tren bilu diin cac tii b ^ g
b6 4 ngii nghia.
(m, 0.5, 0.54625, [0.37, 0.63)) (m, 0.5, 0.42125, [0.37, 0.63)) (1, 0.5, 0.54625, [0.2114, 0.37))
(m, 0.5, 0.4375, [0.37, 0.63))
Phudng phdp bieu dien bd 4 vd bd 2 cho phep chgn mdt phUdng dn tdt nhat trong khi dd phUdng phdp bilu diln ngii nghia cdc tur bang tap md vd bdng ki hifu ngdn ngii trong thang diem lai cho phep chgn ca 3 phUdng an.
4. K E T L U A N
Bai todn liy quylt dinh vdi thdng tin md ddn den cdc cdch tilp can khdc nhau trong viec giai bdi todn kit nhdp cac y kiln ddnh gid dua trgn thang diem danh gia ngdn ngii. Cdc
252 NGUvfeN VAN LONG, HOANG VAN THONG
phUdng phdp dUa tren bieu dien ngii nghia hAng tdp md thudng m i t mdt thdng tin trong vifc bieu dien ngfr nghia eiie tir ngon ngii va, trong tfnh todn tren cdc tii. Chdng h9.n vifc tfnh todn tren eae chi sd oua thang diem ngon ngfr da ditgc sdp tuyin tinh Id khdng tU nhifn, vi cdc chi sd l)i(Mi tin qua ft thong tin ve ngfr nghia ciia, cue tir ngdn ngii. DSGT cung cap cho ta cdng cu bii'^u dien thong tin ngdn ngfr vdi nhieu Uu diem ddng ke: (i) Cau triic DSGT phdn dnh kha trung thanh ngfr nghia eae tir ngon ngii; (ii) Cd t6n tg,! nidi quan he chdt che vd phong phii giiia ngfr nghia djnh tinh va dinh lugng; (iii) Vifc bieu dien ngii nghia nhd bd 4 ngut nghia mang nhieu thdng tin. Vdi nhfrng Uu diem nhU vdy phiidng phdp mdi de bilu diln vd ket nhap Ciie thdng tin ngdn ngii dugc de xuat trong l)ai bax) se bao ham dugc nhiing thdng tin ngii nghia hfiu fch trong viec giiii cac bai toan ling dung.
Nhu ta da phan tfch d tren, phUdng phap mdi chdt, lgc dugc nhieu thdng tin ngii nghia cua tir ngdn ngfr. Ngodi ra, phg^m vi iing dung ciia phiidng phdp rgng vd linh boat hdn vi nd t^o ra khd ndng xay dung phUdng phap ddnh gid cho phfp viia cho diem ngdn ngii, vira cho diem thUc D(^ so sdnh chiing to hieu ndng ciia phUdng phdp, sut dung phUdng phdp dugc de xuat de giai bdi toan da dugc gidi bdng ear phiidng phdp khdc. Kit qua chi ra rdng phUdng phdp mdi cho ket qua tot vd do sut dung do tfnh mem d^o ciia nd (cho phep ddnh gid bdng dilm thuc hodc ngdn ngfr, tfnh todn ket nhdp tren cdc gid tri ngfr nghia dinh lugng, nghia la gid tri thire. ...).
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Ngdy nhdn bdi 26 7 2010 Nhdn lai sau sita I4 11 2011
