Một cách tiếp cận để xấp xỉ dữ liệu trong cơ sở dữ liệu mờ

haˆn hoa.ch cu’a I(x
′). Do do´ ∃∆(n+1) = I(h1x′) ∈ P (n+1) : I(h1x′) = I(x) ⊆ ∆(n+1).
Vaˆ.y ≈k du´ng vo´
.i k = n+ 1, hay x ≈n+1 x.
T´ınh doˆ´i xu´.ng: ∀x, y ∈ Dom(Ai), neˆ´u x ≈k y th`ı theo di.nh ngh˜ıa ∃∆
k ∈ P k : I(x) ⊆ ∆k
va` I(y) ⊆ ∆k hay ∃∆k ∈ P k : I(y) ⊆ ∆k va` I(x) ⊆ ∆k. Vaˆ.y y ≈k x th`ı y ≈k x.
T´ınh ba˘´t ca`ˆu: Ta chu´.ng minh ba`˘ng phu.o.ng pha´p qui na.p.
Tru.`o.ng ho.. p k = 1:
Ta co´ P 1 = {I(c+), I(c−)}, neˆ´u x ≈1 y va` y ≈1 z th`ı ∃∆
1 = I(c+) ∈ P 1 : I(x) ⊆ ∆1 va`
I(y) ⊆ ∆1 va` I(z) ⊆ ∆1 hoa˘. c ∃∆
1 = I(c−) ∈ P 1 : I(x) ⊆ ∆1 va` I(y) ⊆ ∆1 va` I(z) ⊆ ∆1, co´
ngh˜ıa la` ∃∆1 ∈ P 1 : I(x) ⊆ ∆1 va` I(z) ⊆ ∆1 hay x ≈1 z. Vaˆ.y ≈k du´ng vo´
.i k = 1.
Gia’ su.’ quan heˆ. ≈k du´ng vo´
.i tru.`o.ng ho.. p k = n co´ ngh˜ıa la` ta co´ ∀x, y, z ∈ Dom(Ai) neˆ´u
x ≈n y va` y ≈n z th`ı x ≈n z.
Ta ca`ˆn chu´.ng minh quan heˆ. ≈k du´ng vo´
.i tru.`o.ng ho..p k = n+1. Tu´
.c la` ∀x, y, z ∈ Dom(Ai)
neˆ´u x ≈n+1 y va` y ≈n+1 z th`ı x ≈n+1 z.
Theo gia’ thieˆ´t neˆ´u x ≈n+1 y va` y ≈n+1 z th`ı ∃∆
(n+1) ∈ P (n+1) : I(x) ⊆ ∆(n+1) va` I(y) ⊆
∆(n+1) va` I(z) ⊆ ∆(n+1), co´ ngh˜ıa la` ∃∆(n+1) ∈ P (n+1) : I(x) ⊆ ∆(n+1) va` I(z) ⊆ ∆(n+1).
Vaˆ.y x ≈n+1 z.
Boˆ’ de`ˆ 3.2. Cho u = hn..h1x va` v = h
′
m...h
′
1x la` bieˆ’u dieˆ˜n ch´ınh ta˘´c cu’a u va` v doˆ´i vo´
.i x.
(1) Neˆ´u u = v th`ı u ≈k v vo´.i mo. i k.
(2) Neˆ´u h1 6= h
′
1 th`ı u ≈|x| v.
Chu´.ng minh:
(1) Theo Boˆ’ de`ˆ 3.1, v`ı u = v neˆn ta co´ u ≈k u hay v ≈k v , vo´.i mo.i k.
(2) Neˆ´u u| = |v| = 2, tu´.c la` u = h1x va` v = h
′
1x, do h1 6= h
′
1 neˆn u 6= v. Ta co´
I(h1x) ⊆ I(x), I(h′1x) ⊆ I(x) va` I(h1x) 6⊂ I(h
′
1x) neˆn ∃∆
1 = I(x) ∈ P 1 : I(h1x) ⊆ ∆1 va`
I(h′1x) ⊆ ∆
1 hay h1x ≈1 h′1x. Vaˆ.y u ≈|x| v.
Neˆ´u |u| 6= |v|, do h1 6= h′1 neˆn I(h1x) 6⊂ I(h
′
1x) (1’). Gia’ su
.’ ∃k > 1 sao cho u ≈k v th`ı
MOˆ. T CA´CH TIE´ˆP CAˆ. N DEˆ
’ XA´ˆP XI’ DU˜
.
LIEˆ. U 115
∃∆k ∈ P k = {I(hk−1...h1x), I(h
′
k−1...h
′
1x)}, vo´
.i P k la` moˆ.t phaˆn hoa.ch cu’a I(x) : I(u) ⊆ ∆
k
va` I(v) ⊆ ∆k.
Neˆ´u cho.n ∆
k = I(hk−1...h1x) th`ı I(u) ⊆ I(hk−1...h1x) va` I(v) ⊆ I(hk−1...h1x) hay
I(hn...h1x) ⊆ I(hk−1...h1x) va` I(h′m...h
′
1x) ⊆ I(hk−1...h1x) die`ˆu na`y maˆu thuaˆ’n v`ı I(h
′
m...h
′
1x) 6⊂
I(hk−1...h1x) do (1’).
Neˆ´u cho.n∆
k = I(h′k−1...h
′
1x) th`ı I(hn...h1x) ⊆ I(h
′
k−1...h
′
1x) va` I(h
′
m...h
′
1x) ⊆ I(h
′
k−1...h
′
1x),
die`ˆu na`y maˆu thuaˆ’n v`ı I(hn...h1x) 6⊂ I(h′k−1...h
′
1x) do (1’). Vaˆ.y khoˆng toˆ`n ta. i k > 1 sao cho
u ≈k v hay k = 1. Vaˆ.y u ≈|x| v.
Di.nh ly´ 3.1. Xe´t P
k = {I(x) : x ∈ Xk} vo´.i Xk = {x ∈ X : |x| = k} la` moˆ. t phaˆn hoa. ch,
u = hn...h1x va` v = h
′
m...h
′
1x la` bieˆ’u dieˆ˜n ch´ınh ta˘´c cu’a u va` v doˆ´i vo´
.i x.
(1) Neˆ´u u ≈k v th`ı u ≈k′ v, ∀0 < k
′ < k.
(2) Neˆ´u toˆ`n ta. i moˆ. t chı’ soˆ´ j 6 min(m, n) lo´
.n nhaˆ´t sao cho vo´.i mo. i s = 1...j, ta co´
hs = h
′
s th`ı u ≈j+|x| v.
Chu´.ng minh: (1) Ta co´ P k = {I(hk−1...h1x), I(h′k−1...h1x)}. Vı` u ≈k v neˆn theo di.nh ngh˜ıa
∃∆k ∈ P k : I(u) ⊆ ∆k va` I(v) ⊆ ∆k (1’).
Ta la. i co´ P
1 = {I(x)}, P 2 = {I(h1x), I(h′1x)}, ..., P
k = {I(hk−1...h1x), I(h′k−1...h1x)}.
Ma˘. t kha´c ta co´ I(hk−1...h1x) ⊆ I(hk−2...h1x) ⊆ ... ⊆ I(h1x) ⊆ I(x) va` I(h
′
k−1...h
′
1x) ⊆
I(h′k−2...h
′
1x) ⊆ ... ⊆ I(h
′1x) ⊆ I(x) neˆn ∃∆k = I(hk−1...h1x) ∈ P k hoa˘.c ∃∆
k = I(h′k−1...h
′
1x) ∈
P k va` ∃∆k−1 = I(hk−2...h1x) ∈ P
k−1 hoa˘. c ∃∆
k−1 = I(h′k−2...h
′
1x) ∈ P
k − 1... va` ∃∆2 =
I(h1x) ∈ P
2 hoa˘. c ∃∆
2 = I(h′1x) ∈ P
2 va` ∃∆1 = I(x) ∈ P1 sao cho: ∆
k ⊆ ∆k−1 ⊆ ... ⊆
∆2 ⊆ ∆1 (2’).
Tu`. (1’) va` (2’) ta co´ I(u) ⊆ ∆k ⊆ ∆k−1 ⊆ ... ⊆ ∆2 ⊆ ∆1 va` I(v) ⊆ ∆k ⊆ ∆k−1 ⊆
... ⊆ ∆2 ⊆ ∆1, co´ ngh˜ıa la` ∀0 < k′ < k luoˆn ∃∆k
′
∈ P k
′
: I(u) ⊆ ∆k
′
va` I(v) ⊆ ∆k
′
. Vaˆ.y
∀0 < k′ < k neˆ´u u ≈k v th`ı u ≈k′ v.
(2): Neˆ´u j = 1 ta co´ h1 = h
′
1, khi do´ u = hn...h2h1x va` v = h
′
m...h
′
2h
′
1x hay u = hn...h2h1x
va` v = h′m...h
′
2h1x. Da˘.t x
′ = h1x ta co´ u = hn...h2x
′ va` v = h′m...h
′
2x
′. Vı` h2 6= h
′2 neˆn theo
Boˆ’ de`ˆ 2.3 ta co´ u ≈|x′| v (do |x
′| = 2, |x| = 1) hay u ≈2 v. Vaˆ.y u ≈j+|x| v.
Neˆ´u j 6= 1, da˘.t k = j, ta ca`ˆn chu´
.ng minh u ≈k+|x| v. Vı` u ≈k v neˆn theo gia’ thieˆ´t ta co´
∀s = 1...k ta co´ hs = h
′
s. Khi do´ u = hn...h2h1x va` v = h
′
m...h
′
2h
′
1x hay u = hn.hkhk−1...h1x
va` v = h′m...hkhk−1...h1x.
Da˘.t x
′ = hkhk−1...h1x ta co´ u = hn...hk+1x
′ va` v = h′m...h
′
k+1x
′. Vı` hk+1 6= h
′
k+1 neˆn
theo Boˆ’ de`ˆ 2.2 ta co´ u ≈|x′| v hay u ≈k+|x| v (do |x
′| = k, |x| = 1).
Heˆ. qua’ 3.1. Neˆ´u u ∈ H(v) th`ı u ≈|v| v.
Di.nh ly´ 3.2. Xe´t P
k = {I(x) : x ∈ Xk} vo´.i X k = {x ∈ X : |x| = k}, u = hn...h1x va`
v = h′m...h
′
1x la` bieˆ’u dieˆ˜n ch´ınh ta˘´c cu’a u va` v doˆ´i vo´
.i x. Neˆ´u toˆ`n ta. i chı’ soˆ´ k 6 min(m, n)
lo´.n nhaˆ´t sao cho u ≈k v th`ı u 6=k+1 v.
Heˆ. qua’ 3.2. (1) Neˆ´u u ∈ H(v) th`ı u 6=|v|+1 v
(2) Neˆ´u u 6=k v th`ı u 6=k′ v ∀0 < k < k
′
Di.nh ly´ 3.3. Xe´t P
k = {I(x) : x ∈ Xk} vo´.i X k = {x ∈ X : |x| = k}, u = hn...h1x va`
v = h′m...h
′
1x la` bieˆ’u dieˆ˜n ch´ınh ta˘´c cu’a u va` v doˆ´i vo´
.i x. Neˆ´u u k v th`ı vo´
.i
116 NGUYE˜ˆN CA´T HOˆ`, NGUYE˜ˆN COˆNG HA`O
mo. i a ∈ H(u), vo´
.i mo. i b ∈ H(v) ta co´ a k b.
3.2. Mie`ˆn tri. cu’a thuoˆ.c t´ınh trong quan heˆ. co´ chu´
.a gia´ tri. soˆ´
Tru.`o.ng ho.. p mie`ˆn tri. cu’a thuoˆ.c t´ınh co´ chu´
.a gia´ tri. soˆ´, chu´ng ta se˜ bieˆ´n doˆ’i ca´c gia´ tri. soˆ´
tha`nh ca´c gia´ tri. ngoˆn ngu˜
. tu.o.ng u´.ng theo moˆ. t ngu˜
. ngh˜ıa xa´c di.nh. Tru
.´o.c tieˆn, ta di xaˆy
du.. ng moˆ. t ha`m IC chuyeˆ’n moˆ. t soˆ´ ve`ˆ moˆ. t gia´ tri. thuoˆ.c [0, 1] va` ha`m Φk deˆ’ chuyeˆ’n moˆ. t gia´
tri. trong [0, 1] tha`nh moˆ.t gia´ tri. ngoˆn ngu˜
. x tu.o.ng u´.ng trong da. i soˆ´ gia tu
.’ X.
Di.nh ngh˜ıa 3.5. Cho Dom(Ai) = Num(Ai)∪LV (Ai), v la` ha`m di.nh lu
.o.. ng ngu˜
. ngh˜ıa cu’a
Ai. Ha`m IC : Dom(Ai)→ [0, 1] du.o.. c xa´c di.nh nhu
. sau:
Neˆ´u LV (Ai) = ∅ va` Num(Ai) 6= ∅ th`ı ∀ω ∈ Dom(Ai) ta co´ IC(ω) =
ω − ψmin
ψmax − ψmin
vo´.i
Dom(Ai) = [ψmin, ψmax] la` mie`ˆn tri. kinh dieˆ’n cu’a Ai.
Neˆ´uNum(Ai) 6= ∅, LV (Ai) 6= ∅ th`ı ∀ω ∈ Dom(Ai) ta co´ IC(ω) = {ω
∗v(ψmaxLV )}/ψmax,
vo´.i LV (Ai) = [ψminLV , ψmaxLV ] la` mie`ˆn tri. ngoˆn ngu˜
. cu’a Ai.
Vı´ du. 3.5. Cho Dom(Tuoi) = {0...100, ... raˆ´t raˆ´t tre’ ,......., raˆ´t raˆ´t gia`}.
Num(Tuoi) = {20, 25, 27, 30, 45, 60, 75, 66, 80}.
LV (Tuoi) = {tre’ , raˆ´t tre’ , gia`, kha´ tre’ , kha´ gia`, ı´t gia`, raˆ´t gia`, raˆ´t raˆ´t tre’}, Dom(Tuoi) =
Num(Tuoi) ∪ LV (Tuoi).
Neˆ´u LV (Tuoi) = ∅ khi do´ Dom(Tuoi) = Num(Tuoi) = {20, 25, 27, 30, 45, 60, 75, 66, 80}.
Do do´ ∀ω ∈ Dom(Tuoi), ta co´ Dom(Tuoi) = {0,2, 0,25, 0,27, 0,3, 0,45, 0,6, 0,75, 0,66,
0,8}.
Neˆ´u Num(Ai) 6= ∅ va` LV (Ai) 6= ∅ ta co´ Dom(Tuoi) = Num(Tuoi) ∪ LV (Tuoi) = {tre’ ,
raˆ´t tre’ , gia`, kha´ tre’ , kha´ gia`, ı´t gia´, raˆ´t gia`, raˆ´t raˆ´t tre’ , 20, 25, 27, 30, 45, 60, 75, 66,
80}. Gia’ su.’ t´ınh du.o.. c v(ψmaxLV ) = v(raˆ´t raˆ´t gia`) = 0,98. Khi do´ ∀ω ∈ Num(Ai) ta co´
IC(ω) = {ω.v(ψmaxLV )}/ψmax = (ω × 0, 98)/100, hay ∀ω ∈ Num(Ai) su.’ du. ng IC(ω), ta co´
Num(Ai) = {0,196, 0,245, 0,264, 0,294, 0,441, 0,588, 0,735, 0,646, 0,784}.
Neˆ´u ta cho.n ca´c tham soˆ´ W va` doˆ. do t´ınh mo`
. cho ca´c gia tu.’ sao cho v(ψmaxLV ) ≈ 1, 0
th`ı ({ω × v(ψmaxLV )}/ψmax) ≈ 1−
ψmax − ω
ψmax − ψmin
.
Di.nh ngh˜ıa 3.6. Cho da. i soˆ´ gia tu
.’ X = (X,G,H,6), v la` ha`m di.nh lu
.o.. ng ngu˜
. ngh˜ıa cu’a
X . φk : [0, 1]→X go. i la` ha`m ngu
.o.. c cu’a ha`m v theo mu´
.c k du.o.. c xa´c di.nh:
∀a ∈ [0, 1], Φk(a) = x
k khi va` chı’ khi a ∈ I(xk), vo´.i xk ∈X k.
Vı´ du. 3.6. Cho da. i soˆ´ gia tu
.’ X = (X,G,H,6), trong do´ H+ = {ho.n, raˆ´t} vo´.i ho.n < raˆ´t
va` H− = {´ıt, kha’ na˘ng} vo´.i ı´t > kha’ na˘ng, G = {nho’, lo´.n}. Gia’ su.’ cho W = 0, 6, fm(ho.n)
= 0, 2, fm(raˆ´t) = 0, 3, fm(´ıt) = 0, 3, fm(kha’ na˘ng) = 0, 2.
Ta co´ P 2 = {I(ho.n lo´.n), I(raˆ´t lo´.n), I (´ıt lo´.n), I(kha’ na˘ng lo´.n), I(ho.n nho’), I(raˆ´t nho’),
I (´ıt nho’), I(kha’ na˘ng nho’)} la` phaˆn hoa.ch cu’a [0, 1]. fm(nho’) = 0, 6, fm(lo´
.n) =0, 4, fm(raˆ´t
lo´.n) = 0, 12, fm(kha’ na˘ng lo´.n) = 0, 08. Ta co´ |I(raˆ´t lo´.n)| = fm(raˆ´t lo´.n) = 0, 12, hay I(raˆ´t
lo´.n) = [0, 88, 1]. Do do´ theo di.nh ngh˜ıa Φ2(0, 9) = raˆ´t lo´
.n v`ı 0, 9 ∈ I(raˆ´t lo´.n).
Tu.o.ng tu.. ta co´ |I(kha’ na˘ng lo´
.n)| = fm(kha’ na˘ng lo´.n) = 0, 08, hay I(kha’ na˘ng lo´.n) =
[0, 72, 0, 8]. Do do´ theo di.nh ngh˜ıa Φ2(0, 75) = kha’ na˘ng lo´
.n v`ı 0, 75 ∈ I(kha’ na˘ng lo´.n).
MOˆ. T CA´CH TIE´ˆP CAˆ. N DEˆ
’ XA´ˆP XI’ DU˜
.
LIEˆ. U 117
Trong pha`ˆn na`y, gia’ su.’ chu´ng toˆi chı’ xe´t ca´c pha`ˆn tu.’ du.o.. c sinh tu`
. pha`ˆn tu.’ lo´.n.
H`ınh 3.1. T´ınh mo`. cu’a pha`ˆn tu.’ sinh lo´.n
Di.nh ly´ 3.4. Cho da. i soˆ´ gia tu
.’ X = (X,G,H,6), v la` ha`m di.nh lu
.o.. ng ngu˜
. ngh˜ıa cu’a
X, Φk la` ha`m ngu
.o.. c cu’a v, ta co´
(1) ∀xk ∈Xk, Φk(v(xk)) = xk
(2) ∀a ∈ I(xk), ∀b ∈ I(yk), xk 6=k y
k, neˆ´u a < b th`ı Φk(a) <k Φk(b).
Chu´.ng minh.
(1) Da˘. t a = v(x
k) ∈ [0, 1]. Vı` v(xk) ∈ I(xk) neˆn a ∈ I(xk). Theo di.nh ngh˜ıa ta co´
Φk(v(x
k)) = xk.
(2) Vı` xk 6=k yk neˆn theo di.nh ngh˜ıa ta co´ x
k <k y
k hoa˘. c y
k <k x
k, suy ra v(xk) < v(yk)
hoa˘. c v(y
k) < v(xk). Ma˘.t kha´c ta co´ v(x
k) ∈ I(xk) va` v(yk) ∈ I(yk), theo gia’ thieˆ´t a < b do
do´ xk <k y
k. Hay Φk(a) <k Φk(b).
3.3. Thuaˆ.t toa´n xa´c di.nh gia´ tri. chaˆn ly´ cu’a die`ˆu kieˆ.n mo`
.
Nhu. trong Mu˜c 3 da˜ tr`ınh ba`y, mie`ˆn tri. cu’a thuoˆ.c t´ınh mo`
. trong quan heˆ. cu’a lu
.o.. c doˆ` co
.
so.’ du˜. lieˆ.u phu´
.c ta.p va` co´ theˆ’ nhaˆ.n gia´ tri. nhu
. soˆ´, gia´ tri. ngoˆn ngu˜
. hoa˘. c vu`
.a gia´ tri. soˆ´ vu`
.a
gia´ tri. ngoˆn ngu˜
.. Vı` vaˆ.y, ta di xaˆy du
.
. ng thuaˆ.t toa´n da´nh gia´ die`ˆu kieˆ.n mo`
. deˆ’ la`m co. so.’ cho
vieˆ.c thao ta´c va` t`ım kieˆ´m du˜
. lieˆ.u sau na`y.
Go.i Dom(Ai) = Num(Ai)∪LV (Ai) la` mie`ˆn tri. cu’a thuoˆ.c t´ınh mo`
. Ai trong moˆ. t quan heˆ.
cu’a lu.o.. c doˆ` co
. so.’ du˜. lieˆ.u. Khi do´, thuaˆ.t toa´n du
.o.. c xaˆy du
.
. ng nhu
. sau.
Thuaˆ.t toa´n 3.1
Va`o: Cho r la` moˆ.t quan heˆ. xa´c di.nh treˆn taˆ.p vu˜ tru. ca´c thuoˆ.c t´ınh U.
Die`ˆu kieˆ.n t[Ai] ≈k u, vo´
.i u la` moˆ.t gia´ tri. soˆ´ hoa˘. c gia´ tri. ngoˆn ngu˜
..
Ra: Vo´.i mo. i t ∈ r sao cho (t[Ai] ≈k u) = true.
Phu.o.ng pha´p
// Di xaˆy du.. ng ca´c P
k = {I(t[Ai]) : |t[Ai]| = k, ∀t ∈ r}, theo [2], moˆ. t gio´
.i ha.n ho
.
. p ly´ deˆ’ phu`
ho.. p trong thu
.
. c teˆ´ ta cho k 6 4. Tru
.´o.c tieˆn, ta chuyeˆ’n ca´c gia´ tri. soˆ´ tha`nh gia´ tri. ngoˆn ngu˜
..
(1) for moˆ˜i t ∈ r do
(2) if t[Ai] ∈ Num(Ai) then t[Ai] = Φk(IC(t[Ai]))
//Xaˆy du.. ng ca´c P
k du.. a va`o doˆ. da`i ca´c tu`
..
(3) k = 1
(4) While k 6 4 do
(5) P k = ∅
118 NGUYE˜ˆN CA´T HOˆ`, NGUYE˜ˆN COˆNG HA`O
(6) for moˆ˜i t ∈ r do
(7) if |t[Ai]| = k then P
k = P k ∪ {I(t[Ai])}
(8) k = k + 1
// Xa´c di.nh gia´ tri. chaˆn ly´ cu’a (t[Ai] ≈k u).
(9) if u ∈ Num(Ai) then u
′ = Φk(IC(u))
(10) k = 4 // Phaˆn hoa.ch tu
.o.ng u´.ng vo´.i mu´.c lo´.n nhaˆ´t.
(11) While k > 0 do
(12) for moˆ˜i ∆k ∈ P k do
(13) if (I(t[Ai]) ⊆ ∆k and I(u) ⊆ ∆k) or (I(t[Ai]) ⊆ ∆k and I(u′) ⊆ ∆k) then
{(t[Ai] ≈k u) = true} or {(t[Ai] ≈k u′) = true}
(14) exit
(15) k = k − 1
Thuaˆ.t toa´n 3.2
Va`o: Cho r la` moˆ.t quan heˆ. xa´c di.nh treˆn taˆ.p vu˜ tru. ca´c thuoˆ.c t´ınh U.
Die`ˆu kieˆ.n t[Ai]θu, vo´
.i u la` moˆ. t gia´ tri. soˆ´ hoa˘.c gia´ tri. ngoˆn ngu˜
., θ ∈ {6=k, k}.
Ra: Vo´.i mo. i t ∈ r sao cho (t[Ai]θu) = true
Phu.o.ng pha´p
(1) Su.’ du.ng ca´c bu
.´o.c tu`. (1)-(8) trong Thuaˆ.t toa´n 3.1
(2) if u ∈ Num(Ai) then u
′ = Φk(IC(u))
(3) k = 1
(4) While k 6 4 do
(5) for vo´.i mo. i ∆
k ∈ P k do
(6) if {I(t[Ai]) 6⊂ ∆
k or I(u) 6⊂ ∆k} then (t[Ai] 6=k u) = true
(7) if {v(t[Ai]) > v(u)} then (t[Ai] >k u) = true
(8) else if (t[Ai] <k u) = true
(7) if {I(t[Ai]) 6⊂ ∆k or I(u′) 6⊂ ∆k} then (t[Ai] 6=k u′) = true
(9) if {v(t[Ai]) > v(u′)} then or (t[Ai] >k u′) = true
(10) else if (t[Ai] <k u
′) = true
(11) k = k + 1
3.4. Vı´ du. . Cho lu
.o.. c doˆ` quan heˆ. U = {SOCM,HOTEN, SUCKHOE,TUOI,LUONG}
va` quan heˆ. Luong Tuoi du
.o.. c xa´c di.nh nhu
. sau:
Ba’ng 3.1. Quan heˆ. Lu
.o.ng tuoˆ’i
MOˆ. T CA´CH TIE´ˆP CAˆ. N DEˆ
’ XA´ˆP XI’ DU˜
.
LIEˆ. U 119
Socm Hoten Suckhoe Tuoi Luong
11111 Pha.m Tro.ng Ca`ˆu raˆ´t raˆ´t toˆ´t 31 2.800.000
22222 Nguye˜ˆn Va˘n Ty´ raˆ´t toˆ´t 85 cao
33333 Tra`ˆn Tieˆ´n xaˆ´u 32 2.000.000
44444 Vu˜ Hoa`ng ho.n xaˆ´u 45 500.000
55555 An Thuyeˆn raˆ´t xaˆ´u 41 raˆ´t cao
66666 Thuaˆ.n Yeˆ´n kha’ na˘ng xaˆ´u 61 thaˆ´p
77777 Va˘n Cao ho.n toˆ´t 59 ı´t cao
88888 Thanh Tu`ng kha’ na˘ng toˆ´t 75 1.500.000
99999 Nguye˜ˆn Cu.`o.ng ı´t toˆ´t 25 kha´ thaˆ´p
(a) T`ım nhu˜.ng ca´n boˆ. co´ TUOI ≈2 ho
.n gia` va` SUCKHOE ≈2 kha’ na˘ng toˆ´t.
(b) T`ım nhu˜.ng ca´n boˆ. co´ TUOI ≈1 tre’ hoa˘.c co´ LUONG 6=1 cao.
Tru.´o.c heˆ´t chu´ng ta se˜ xem mie`ˆn tri. cu’a SUCKHOE, TUOI va` LUONG la` ba da. i soˆ´ gia
tu.’ va` du.o.. c xa´c di.nh nhu
. sau:
XSuckhoe = X suckhoe, Gsuckhoe, Hsuckhoe,6), vo´
.i Gsuckhoe = {toˆ´t, xaˆ´u}, H
+
suckhoe = {raˆ´t,
ho.n}, H−suckhoe = {kha’ na˘ng, ı´t}, raˆ´t > ho
.n va` ı´t > kha’ na˘ng.
Wsuckhoe = 0, 6, fm(xaˆ´u) = 0, 6, fm(toˆ´t) = 0, 4, fm(raˆ´t) = 0, 3, fm(kha´) = 0, 2, fm(kha’
na˘ng) = 0, 2, fm(´ıt) = 0, 3.
XTuoi = (X tuoi, Gtuoi, Htuoi,6), vo´
.i Gtuoi = {tre’, gia`}, H
+
t uoi = {raˆ´t, ho
.n}, H−tuoi =
{kha’ na˘ng, ı´t}, raˆ´t > ho.n va` ı´t > kha’ na˘ng. Wtuoi = 0, 4, fm(tre’) = 0, .4, fm(gia`) =
0, 6, fm(raˆ´t) = 0, 3, fm(kha´) =0, 15, fm(kha’ na˘ng) = 0, 25, fm(´ıt) = 0, 3.
XLuong = (X luong , Gluong, Hluong,6), vo´
.i Gluong = {cao, thaˆ´p}, H
+
luong = {raˆ´t, ho
.n},
H−luong = {kha’ na˘ng, ı´t}, raˆ´t > ho
.n va` ı´t > kha’ na˘ng. Wluong = 0, 6, fm(thaˆ´p) =
0, 6, fm (cao) = 0, 4, fm(raˆ´t) = 0, 25, fm(kha´) = 0, 25, fm(kha’ na˘ng) = 0, 25, fm(´ıt) =
0, 25.
Doˆ´i vo´.i thuoˆ.c t´ınh TUOI: Ta co´ fm(raˆ´t tre’) = 0, 12, fm(ho
.n tre’) = 0, 06, fm(´ıt tre’) =
0, 12, fm(kha’ na˘ng tre’) = 0, 1.
Vı` raˆ´t tre’ < ho.n tre’ < tre’ < kha’ na˘ng tre’ < ı´t tre’ neˆn I(raˆ´t tre’) = [0, 0, 12], I(ho.n tre’)
= [0, 12, 0, 18], I(kha’ na˘ng tre’) = [0, 18, 0, 3], I (´ıt tre’) = [0, 3, 0, 4].
Ta co´ fm(raˆ´t gia`) = 0, 18, fm(ho.n gia`) = 0, 09, fm(´ıt gia`) = 0, 18, fm(kha’ na˘ng gia`)
= 0, 15.
Vı` ı´t gia` < kha’ na˘ng gia` < gia` < ho.n gia` < raˆ´t gia` neˆn I (´ıt gia`) = [0, 4, 0, 58], I(kha’
na˘ng gia`) = [0, 58, 0, 73], I(ho.n gia`) = [0, 73, 0, 82], I(raˆ´t gia`) = [0, 82, 1].
Neˆ´u cho.n ψ1 = 100 ∈X tuoi khi do´ ∀ω ∈ Num(TUOI), su
.’ du. ng IC(ω) ta co´Num(TUOI) =
{0, 31, 0, 85, 0, 32, 0, 45, 0, 41, 0, 61, 0, 59, 0, 75, 0, 25}.
Do do´ Φ2(0, 31) = ı´t tre’ v`ı 0, 31 ∈ I (´ıt tre’), tu.o.ng tu.. Φ2(0, 85) = raˆ´t gia`, Φ2(0, 32) =
ı´t tre’, Φ2(0, 45) = ı´t gia`, Φ2(0, 41) = ı´t gia`, Φ2(0, 61) = kha’ na˘ng gia`, Φ2(0, 59) = kha’ na˘ng
gia`, Φ2(0, 75) = ho
.n gia`, Φ2(0, 25) = kha’ na˘ng tre’.
Doˆ´i vo´.i thuoˆ.c t´ınh LUONG: Ta co´fm(raˆ´t thaˆ´p) = 0, 15, fm(kha´ thaˆ´p) = 0, 15, fm(´ıt
thaˆ´p) = 0, 15, fm(kha’ na˘ng thaˆ´p) = 0, 15.
Vı` raˆ´t thaˆ´p < ho.n thaˆ´p< thaˆ´p< kha’ na˘ng thaˆ´p < ı´t thaˆ´p neˆn I(raˆ´t thaˆ´p) = [0, 0, 15], I(ho.n
thaˆ´p) = [0, 15, 0, 3], I(kha’ na˘ng thaˆ´p) = [0, 3, 0, 45], I (´ıt thaˆ´p) = [0, 45, 0, 6].
120 NGUYE˜ˆN CA´T HOˆ`, NGUYE˜ˆN COˆNG HA`O
Ta co´ fm(raˆ´t cao) = 0, 1, fm(ho.n cao) = 0, 1, fm(´ıt cao) = 0, 1, fm(kha’ na˘ng cao) =
0, 1.
Vı` ı´t cao < kha’ na˘ng cao < cao < ho.n cao < raˆ´t cao neˆn I (´ıt cao) = [0, 6, 0, 7], I(kha’
na˘ng cao) = [0, 7, 0, 8], I(ho.n cao) = [0, 8, 0, 9], I(raˆ´t cao) = [0, 9, 1].
Neˆ´u cho.n ψ2 =raˆ´t raˆ´t cao ∈ X luong va` ψ1 = 3.000.000, ta co´ v(raˆ´t raˆ´t cao) = 0, 985
khi do´ ∀ω ∈ Num(LUONG) = {2.800.000, 2.000.000, 500.000, 1.500.000}, su.’ du. ng IC(ω) =
{ω × v(ψ2)}/ψ1, ta co´ Num(LUONG) = {0, 92, 0, 65, 0, 16, 0, 49}.
Do do´ Φ2(0, 92)= raˆ´t cao, Φ2(0, 65) = ı´t cao, Φ2(0, 16) = ho.n thaˆ´p, Φ2(0, 49) = ı´t cao.
Vaˆ.y, nhu˜
.ng ca´n boˆ. co´ TUOI ≈2 ho
.n gia` va` SUCKHOE ≈2 kha’ na˘ng toˆ´t la`:
Ba’ng 3.2. Keˆ´t qua’ t`ım kieˆ´m cu’a v´ı du. (a)
Socm Hoten Suckhoe Tuoi Luong
88888 Thanh Tu`ng kha’ na˘ng toˆ´t 75 1.500.000
va` nhu˜.ng ca´n boˆ. co´ TUOI ≈1 tre’ hoa˘. c co´ LUONG 6=1cao.
Ba’ng 3.2. Keˆ´t qua’ t`ım kieˆ´m cu’a v´ı du. (b)
Socm Hoten Suckhoe Tuoi Luong
11111 Pha.m Tro.ng Ca`ˆu raˆ´t raˆ´t toˆ´t 31 2.800.000
33333 Tra`ˆn Tieˆ´n xaˆ´u 32 2.000.000
44444 Vu˜ Hoa`ng kha´ xaˆ´u 45 500.000
66666 Thuaˆ.n Yeˆ´n kha’ na˘ng xaˆ´u 61 thaˆ´p
99999 Nguye˜ˆn Cu.`o.ng ı´t toˆ´t 25 kha´ thaˆ´p
4. KEˆ´T LUAˆ. N
Ba`i ba´o xem xe´t moˆ. t ca´ch tro.n ve.n vieˆ.c da´nh gia´ deˆ’ doˆ´i sa´nh ca´c gia´ tri. khi mie`ˆn tri. thuoˆ.c
t´ınh cu’a moˆ. t quan heˆ. trong co
. so.’ du˜. lieˆ.u mo`
. nhaˆ.n gia´ tri. da da.ng.Vieˆ.c da´nh gia´ na`y la` phu`
ho.. p vo´
.i thu.. c teˆ´, bo
.’ i v`ı gia´ tri. cu’a ngoˆn ngu˜
. la` tu.o.ng doˆ´i phu´.c ta.p. Treˆn co
. so.’ na`y, ba`i ba´o
da˜ phaˆn t´ıch ca´c quan heˆ. doˆ´i sa´nh giu˜
.a hai gia´ tri. theo ngu˜
. ngh˜ıa mo´.i. Tu`. do´ du.a ra moˆ. t
soˆ´ v´ı du. ve`ˆ ca´c thao ta´c du˜
. lieˆ.u theo ca´ch tieˆ´p caˆ.n mo´
.i. Vaˆ´n de`ˆ xaˆy du.. ng ca´c phu. thuoˆ.c du˜
.
lieˆ.u treˆn moˆ h`ınh co
. so.’ du˜. lieˆ.u mo`
. theo ca´ch tieˆ´p caˆ.n da. i soˆ´ gia tu
.’ se˜ du.o.. c gio´
.i thieˆ.u trong
nhu˜.ng ba`i ba´o tieˆ´p theo.
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Nhaˆ. n ba`i nga`y 6 - 1 - 2006