Transcript 知識推論法
Slide 1
Chapter 4
Methods of Inference
知識推論法
Slide 2
4.1 Deductive and Induction （演繹與歸納）
• Deduction(演繹): Logical reasoning in which
conclusions must follow from their premises.
• Induction(歸納): Inference from the specific case to
the general.
• Intuition(直觀): No proven theory.
• Heuristics(啟發): Rules of thumb (觀測法) based upon
experience.
• Generate and test: Trial and error.
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Slide 3
• Abduction(反推): Reasoning back from a true
conclusion to the premises that may have caused the
conclusion.
• Autoepitemic(自覺、本能): Self-knowledge
• Nonmonotonic(應變知識): previous knowledge
may be incorrect when new evidence is obtained
• Analogy(類推): based on the similarities to
another situation
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Slide 4
Syllogism （三段論）
• Syllogism（三段論）is simple, well-understood
branch of logic that can be completely proven.
– Premise(前提): Anyone who can program is
intelligent
– Premise(前提): John can program
– Conclusion(結論): Therefore, John is intelligent.
• In general, a syllogism is any valid deductive
argument having two premises and a conclusion.
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Slide 5
Categorical Syllogism
(定言三段論)
定言命題的型態
形態
概要
意思
A
E
I
O
所有S為P
沒有S為P
某些S為P
某些S不為P
完全肯定
完全否定
部分肯定
部分否定
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Slide 6
•三段論的標準形態（Standard form）
大前提：所有M為P
小前提：所有S為M
結論：所有S為P
- P代表結論的「謂詞」(Predicate)，又稱為「大詞」（Major term）
- S代表結論的「主詞」(Subject)，又稱作「小詞」（Minor term）。
- 含有大詞的前提稱為「大前提」（Major premise）；
- 含有小詞的前提稱為「小前提」（Minor premise）。
- M稱為「中詞」（Middle term）
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Slide 7
Mood(模式)
• Patterns of Categorical Statement
Figure-1
Figure-2
Figure-3
Figure-4
MP
SM
PM
SM
MP
MS
PM
MS
AAA-1
AAA-2
AAA-3
AAA-4
大前提
MP
PM
MP
PM
小前提
SM
SM
MS
MS
大前提
小前提
• 4種AAA模式
形態
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Slide 8
• ex: AAA-1
– 所有M為P
所有S為M
∴所有S為P
• We use decision procedure(決策程序) to prove
the validity of syllogistic argument
• The decision procedure for syllogisms can be done
using Venn Diagrams(維思圖)
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Slide 9
• ex: Decision procedure for Syllogism AEE-1
所有M為P
沒有S為M
∴沒有S為P
S
P
S
P
S
P
M
M
M
(a)維 思 圖
(b)大 前 提 後
(c)小 前 提 後
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Slide 10
General Rule under “some” quantifiers
1. If a class is empty, it is shaded.
2. Universal statement, A and E, are always drawn
before particular ones.
3. If a class has at least one member, mark it with a *.
4. If a statement does not specify in which of two
adjacent classed an object exists, place a * on the
line between the classes.
5. If an area has been shaded, no * can be put in it.
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Slide 11
ex: Decision procedure for Syllogism IAI-1
某些P為M
所有M為S
∴某些S為P
S
P
S
P
＊
M
(a) 所 有 M 是 S
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M
(b) 一 些 P 是 M
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4.2 State and problem spaces
(狀態與問題空間)
• Tree（樹狀結構）: nodes, edges
• Directed or undirected
• Digraph（雙向圖）: a graph with directed
edges
• Lattice（晶格）: a directed acyclic graph
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Slide 13
• A useful method of describing the behavior of an
object is to define a graph called the state space.
[state(狀態) and action(行動)]
–
–
–
–
–
–
Initial state
Operator
State space
Path
Goal test
Path cost
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Slide 14
Finite State Machine
（有限狀態機器 ）
• Determining valid strings WHILE,WRITE, and BEGIN
H
I
L
E
W
R
T
I
開始
B
N
G
E
not G
I
成功
not I
not E
not N
錯誤
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Slide 15
Finding solution in problem space
• State space（狀態空間） can be thought as a
problem space（問題空間）.
• Finding the solution to a problem in a problem
space involves finding a valid path from start to
success( answer).
• The state space for the Monkey and Bananas
Problem
• Traveling salesman problem（旅行推銷員問題）
• Graph algorithms, AND-OR Trees, etc.
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Slide 16
Ex: Monkey and
猴子在躺
椅上
跳下躺椅
Bananas Problem
– 房子裡有一懸掛的香蕉
– 房子裡只有一張躺椅跟一把梯子
– 猴子無法直接拿到香蕉
• 指示：
跳下躺椅
移動梯子
把梯子移到香蕉下的位置
爬上梯子
摘下香蕉
• 初始狀態：
觀察到躺椅
在香蕉下
觀察到躺椅
不在香蕉下
躺椅不位
於香蕉下
• 假設：
–
–
–
–
–
猴子在地
板上
移動躺椅
觀察到猴子
不在梯子上
觀察到猴子
不在梯子上
猴子在梯
子上
躺椅位於
香蕉下
猴子不在
梯子上
移動猴子
觀察到梯子
不在香蕉下
觀察到梯子
在香蕉下
梯子不位
於香蕉下
移動梯子
梯子在香
蕉下
爬上梯子
– 猴子在躺椅上
4. 知識推論法
猴子在梯
子上
S.S. Tseng & G.J. Hwang
摘下香蕉
猴子成功
得到香蕉
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Slide 17
D
B
Ex: Travel Salesman Problem
（旅行推銷員問題）
A
C
A
B
A
C
C
A
(a) 旅 行 推 銷 員 的 問 題 描 述
D
B
D
B
B
C
A
A
C
A B
D
D
B
D
C
B
B
C
D
C A C
(b) 搜 尋 路 徑 (粗 線 是 解 答 路 徑 )
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Slide 18
Ill-structured problem
(非結構化問題)
• Ill-structured problems（非結構化問題）
have uncertainties associated with it.
–
–
–
–
–
–
Goal not explicit
Problem space unbounded
Problem space not discrete
Intermediate states difficult to achieve
State operators unknown
Time constraint
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Slide 19
Ex:旅遊代理人
特徵
客戶的反應
目標不明顯
我在想到底要去哪裡
問題空間範圍未被介定
我不確定要去哪裡
問題狀態不是離散的
我只是想去旅遊，目的地並不
重要
中間的狀態不易實行
我沒有足夠的錢去
狀態的可用運算元未知
我不知道怎麼可以籌到錢
時間限制
我必須儘快出發
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Slide 20
4.3 Rules of Inference(規則式推論)
• Syllogism（三段論） addresses only a small
portion of the possible logic statements.
• Propositional logic
pq
p______
q
Inference is called direct reasoning (直接推論),
modus ponens (離斷率), law of detachment (分離
律) , and assuming the antecedent (假設前提).
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Slide 21
Truth table for Modus Ponense（離斷率）
p
T
T
F
F
4. 知識推論法
q
T
F
T
F
p→q
T
F
T
T
(p→q)p
T
F
F
F
S.S. Tseng & G.J. Hwang
(p→q) p→q
T
T
T
T
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Slide 22
Law of Inference
Schemata
1.Law of Detachment
p→q
p
∴q
2.Law of the Contrapositive
p→q
∴~q→~p
3. Law of Modus Tollens
p→q
~q
∴~p
4.Chain Rules(Law of the Syllogism)
P→q
q→r
∴p→r
5.Law of Disjunctive Inference
pq
~p
∴q
~(~p)
∴p
6.Law of the Double Negation
4. 知識推論法
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pq
~q
∴p
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Slide 23
7.De Morgan’s Law
~(pq)
∴~p ~q
~(pq)
∴~p ~q
8.Law of Simplification
pq
∴p
~(pq)
∴~q
9.Law of Conjunction
p
q
∴pq
10.Law of Disjunctive Addition
p
∴pq
11. Law of Conjunctive Argument
~(pq)
p
∴~q
~(pq)
q
∴~p
Table 3.8 Some Rules of Inference for Propositional Logic
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Slide 24
Resolution in propositional Logic
（命題邏輯分解）
F: Rules or facts known to be TRUE
S: A conclusion to be Proved
• Convert all the propositions of F to clause form.
2. Negate S and convert the result to clause form. Add it
to the set of clauses obtained in step 1.
3. Repeat until either a contradiction is found or no
progress can be made:
(1) Select two clauses.
Call these the Parent clauses.
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Slide 25
(2) Resolve them together.
The resulting clause, called the resolvent, will be the
disjunction of all of the literals of both of the parent
clauses with the following exception：If there are any
pairs of literals L and ~L. Such that one of the parent
clauses contains L and the other contaions ~L, then
eliminate both L and ~L from the resolvent.
(3) If the resolvent is the empty clause, then a
contradiction has been found. If it is not, then add it
to the set of clauses available to the procedure.
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Slide 26
Given Axioms
Converted to Clause Form
p
p
(p q) r
1.
～p ～q r
(s t) q
2.
～s q
3.
～t q
4.
t
t
5.
p =下雨
q = 騎車
s = 路線熟悉
t = 路途遠
r = 穿雨衣
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Slide 27
～r
～ｐ ～ｑｒ
p
～p ～ｑ
～q
～t ｑ
～t
t
Resolution in Propositional Logic
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Slide 28
Resolution with quantifiers
Example（from Nilsson）：
Whoever can read (R) is literate (L).
Dolphins (D) aren’t literate (~L).
Some dolphins (D) are intelligent (I).
To prove：Some who are intelligent (I) can’t read (~R).
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Slide 29
Translating：
x[R(x)→ L(x)]
x [ D ( x ) → ~L ( x ) ]
x[D(x) ＆ I(x)]
To prove:
4. 知識推論法
x[I(x)&~R(x)]
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Slide 30
(1) － (4)：
x [~ R ( x ) OR L ( x ) ] & y [ ~ D ( y )
OR ~ L ( y ) ] & D ( A ) & I ( A ) &
z [~ I ( z ) OR R ( z ) ]
(5) － (9)：
C1=~R(x)
OR L(x)
C2=~D(y)
OR ~L(y)
C3=D(A)
C4=I(A)
C5= ~I(z)
4. 知識推論法
OR R(z)
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Slide 31
• The second order logic can have quantifiers
that range over function and predicate
symbols
• If P is any predicate of one document
– Then
– x =y (for every P [P(x) P(y) ]
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Slide 32
4.4 Inference Chain (推斷鏈)
D3
Initial facts
A2
A1
D2
B
C
D1
E
Solution
inference + inference +… + inference
Chain
backward
chaining
forward
chaining
Infer from initial
facts to solutions
4. 知識推論法
Assume that some solution
is true, and try to prove
the assumption by finding
the required facts
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Slide 33
•
Forward Chaining（前向鏈結）:
Rule1: elephant(x) mammal(x)
Rule2: mammal(x) animal(x)
Fact：John is an elephant.
elephant (John) is true
X=John (Unification)
elephant(x) mammal(x)
Mammal(John) is true
X’=X=John
mammal(x’) animal(x’)
• Unification（變數替代）
The process of finding substitutions for
variables to make arguments match.
4. 知識推論法
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animal(John) is true
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Slide 34
Forward Chaining（前向推論）
Rule1：A1 and B1 C1
Rule2：A2 and C1 D2
Rule3： A3 and B2 D3
Rule4：C1 and D3 G
Facts：A1 is true, A2 is true , A3 is true, B1 is true, B2 is true
{A1, A2, A3, B1, B2} match {r1, r3}
fire r1 {A1, A2, A3, B1, B2, C1} match {r1, r2, r3}
fire r2 {A1, A2, A3, B1, B2, C1, D2} match{r1, r2, r3}
fire r3 {A1, A2, A3, B1, B2, C1 D2, D3} match{r1, r2, r3, r4}
fire r4 {A1, A2, A3, B1, B2, C1 D2, D3, G }
4. 知識推論法
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GOAL
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Slide 35
Backward Chaining (反向推論)
rule1 : A1 and B1
rule2 : A2 and C1
rule3 : A3 and B2
rule4 : C1 and D3
rule5 : C1 and D4
1. Assume G’ is true
C1
D2
D3
G
G’
facts : A1, A2, B1, B2, A3
2. Assume G is true
R5
R4
Verify C1 and D4
Verify C1 and D3
R1
R3
Verify A1 and B1
OK
Verify A3 and B2
OK
D4 is unknown, ask user.
If D4 is FALSE, give up.
Verify A1 and B1
OK
4. 知識推論法
S.S. Tseng & G.J. Hwang
OK
OK
OK
35
Slide 36
A1
A2
B1
A3
B2
C1
D2
D3
?
G
GOAL
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Slide 37
• Good application of forward chaining（前向鏈結）
Goal
Facts
Broad and Not Deep
or
too many possible goals
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Slide 38
• Good application of backward chaining（後向鏈結）
Narrow and
Deep
GOALS
Facts
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Slide 39
• Forward Chaining（前向鏈結）
Planning
Monitoring
Control
Data-driven
Explanation not facilitated
• Backward chaining（後向鏈結）
Diagnosis
Goal-driven
Explanation facilitated
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Slide 40
Analogy
• Try to relate old situations as guides to new
ones
• Consider tic-tac-toe with values as a magic
square (15 game)
»6
»7
»2
1
5
9
8
3
4
• 18 game from set {2,3,4,5,6,7,8,9,10}
• 21 game from set {3,4,5,6,7,8,9,10,11}
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Slide 41
Nonmonotonic reasoning
• In nonmonotonic system, the theorems do
not necessarily increase as the number of
axioms increases.
• As a very simple example, suppose there is
a fact that asserts the time. As soon as time
changes by a second, the old fact is no
longer valid.
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Slide 42
4.5 Reasoning Under Uncertainty
(不確定性推論)
• Uncertainty can be considered as the lack of adequate
information to make a decision.
• Classical probability, Bayescian probability, DempsterShafer theory, and Zadeh’s fuzzy theory.
• In the MYCIN and PROSPECTOR systems conclusion are
arrived at even when all the evidence needed to absolutely
prove the conclusion is not known.
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Slide 43
Many different types of error can contribute to uncertainty
Example
Turn the value off
Turn value-1
Turn value-1 off
Value is stuck
Value is not stuck
Turn value-1 to 5
Turn value-1 to 5.4
Turn value-1 to 5.4 or 6 or 0
Value-1 setting is 5.4 or 5.5 or 5.1
Value-1 setting is 7.5
Value-1 is not stuck because it’s
never been stuck before
Output is normal and so value-1
is in good condition
4. 知識推論法
Error
Ambiguous
Incomplete
Incorrect
False positive (接受錯誤值)
False negative (拒絕正確值)
Imprecise
Inaccurate
Unreliable
Random error
Systematic error
Invalid induction
Reason
What value?
Which way?
Correct is on
Value is not stuck
Value is stuck
Correct is 5.4
Correct is 9.2
Equipment error
Statistical fluctuation (波動)
Mis-calibration (刻度)
Invalid deduction
Value is stuck in open position
S.S. Tseng & G.J. Hwang
Value is stuck
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Slide 44
• A hypothesis is an assumption to be tested.
• Type 1 error (false positive) means acceptance of a
hypothesis when it is not true.
• Type 2 error (false negative) means rejection of a hypothesis
when it is true.
• Error of measurement
– Precision
• The millimeter(公釐) ruler is more precise than
centimeter ruler.
– accuracy
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Slide 45
Error & Induction
The process of induction is the opposite of deduction
The fire alarm goes off (響起)
∴ There is a fire.
An even stronger argument is
The fire alarm goes off & I smell smoke
∴ There is a fire.
Although this is a strong argument, it is not proof that there is
a fire.
My clothes are burning
4. 知識推論法
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Slide 46
Deductive errors
p→q
q
∴p
If John is a father, than John is a man
John is a man
∴ John is a father
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Slide 47
Baye’s Theorem （貝氏定理）
• Conditional probability（條件機率）, P(A | B) , states the
probability of event B occurred. Crash= Brand X(0.6)+
Not X(0.1)=0.7
• P( X|C) =
P( C | X) P(X) =
P(C)
(0.75)(0.8) =
0.7
6
7
• Suppose you have a drive and don’t know its brand, what
is the probability that if it crashes, it is Brand X? nonBrand X?
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Slide 48
Decision Tree for the Disk drive Crashes
Act
Choose
Brand X
P(X)=0.8
Don’t Choose
Brand X
P(X’)=0.2
No Crash
P(C’ | X’)=0.5
P(C’∩X’)=0.1
Crash
P(C | X’)=0.5
P(C ∩X’)=0.1
Prior
P(Hi)
Crash
P(C | X)=0.75
No Crash
P(C’ | X)=0.25
P(C’ ∩ X)=0.2
P(C ∩X)=0.6
Conditional
P(E | Hi )
Joint -P(E ∩ Hi )
=P(E | Hi ) P( Hi )
0.1
P(X’ | C’) =
0.1+0.2
0.1
P(X’ | C) =
=1/3
4. 知識推論法
0.1+0.6
=1/7
0.2
P(X | C’) =
0.2+0.1
=2/3
S.S. Tseng & G.J. Hwang
0.6
P(X | C) =
0.6+0.1
=6/7
Posterior
P(H i | E) = P (E ∩ Hi)
iP(E∩Hi)
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Slide 49
P(Hi | E) =
P(E∩Hi)
P(E ∩Hj)
j
=
P(E | H i) P(Hi)
P(E | Hj) P(Hj)
j
P(E | Hi)P(Hi)
P(E)
• Bayes’ Theorem（貝氏定理） is commonly used for
decision tree analysis of business and the social science.
=
• Used in Prospector expert system to decide
favorite sites of mineral exploration
4. 知識推論法
S.S. Tseng & G.J. Hwang
49
Slide 50
Hypothetical Reasoning and Backward Induction.
Probabilities
No Oil
Oil
P(O’)=0.4
Prior
P(O)=0.6
Subjective Opinion
of Site - P (Hi)
-Test
+Test
-Test
+Test
P(- | O’)
P(+ | O’)
P(- | O)
P(+ | O)
=0.9
=0.1
=0.2
=0.8
P(-∩O’)
P(+∩O’)
P(-∩O)
P(+∩O)
=0.36
=0.04
=0.12
=0.48
P(+)=P(+∩O)+P(+∩O’)=0.48+0.04=0.52
Conditional
Seismic Test
Result
Joint P(E∩H)
=P(E | Hi)
P(Hi)
P(-)=P(-∩O)+P(-∩O’)=0.12+0.36=0.48
4. 知識推論法
S.S. Tseng & G.J. Hwang
50
Slide 51
-Test
+Test
Probabilities
P(-)=0.48
P(+)=0.52
Unconditional
P (E)
No Oil
Oil
No Oil
Oil
P(O’|-)
P(O|-)
P(O’|+)
P(O|+)
= (9)(4)
= (2)(6)
= (1)(4)
= (8)(6)
0.48
0.48
= 3/4
= 1/4
0.52
0.52
= 1/13
Posterior
of Site - P(Hi | E)
P ( E| Hi) P (Hi)
=
P(E)
= 12/13
P(-∩O)
P(+∩O)
P(-∩O)
P(+∩O)
Joint -P(E∩H)
=0.36
=0.04
=0.12
=0.48
=P(Hi | E) P(E)
4. 知識推論法
S.S. Tseng & G.J. Hwang
51
Slide 52
•
•
•
•
Oil release , if successful
Drilling expense
Seismic survey
Expected payoff (success)
$1250000
-$200000
-$50000
• 846153=1000000 *12/13 – 1000000*1/13
• Fail
• -500000= 1000000*1/4- 1000000*3/4
• Expected payoff (total)
• 416000= 846153*0.52 – 50000 * 0.48
4. 知識推論法
S.S. Tseng & G.J. Hwang
52
Slide 53
Temporal reasoning and
Markov chain
• Temporal reasoning: reasoning about events
that depend on time
• Temporal logic
• The system’s progression through a
sequence of status is called a Stochastic
process if it is probabilistic.
4. 知識推論法
S.S. Tseng & G.J. Hwang
53
Slide 54
– P11
P12
– P21
P22
– Where Pmn is the probability of a transition
from state m to state n.
S = { P1,P2, …, Pn} where P1+P2+…+Pn= 1
S2 = S1 T
0.1
0.9
S2 = [0.8,0.2]
= [0.2,0.8]
0.6
4. 知識推論法
0.4
S.S. Tseng & G.J. Hwang
54
Slide 55
• Assume 10 percent of all people who now use
Brand X drive will buy another Brand X when
needed. 60 percent of people who don’t use Brand
X will buy Brand X when they need a new drive.
Over a period of time, how many people will use
Brand X?
S3 = [0.5,0.5], S4 = [0.35,0.65],
S5 = [0.425,0.575], S6 = [0.3875,0.6125]
S7 = [0.40625,0.59375], S8 = [0.396875,0.602125]
Steady state matrix
4. 知識推論法
S.S. Tseng & G.J. Hwang
55
Slide 56
The odds of belief
• “The patient is covered with red spots”
• Proposition A: “The patient has measles”
• P(A|B) :(degree of belief that A is true,
given B)is not necessarily a probability if
the events and propositions can not be
repeated or has a math basis.
4. 知識推論法
S.S. Tseng & G.J. Hwang
56
Slide 57
• The odds on A against B given some event
C are odds =P(A|C)/ P(B|C)
• If B = A’
– odds =P(A|C)/ P(A’|C) =P(A|C)/ (1-P(A|C) )
• Likelihood of P = 0.95
– Odds = .95/(1-.95) = 19 to 1
4. 知識推論法
S.S. Tseng & G.J. Hwang
57
Slide 58
Sufficiency and necessity
Bayes’ Theorem is
P(E | H)P(H)
P(H|E) =
P(E)
Negation P(H’|E) = P(E | H’)P(H’)
P(E)
4. 知識推論法
S.S. Tseng & G.J. Hwang
58
Slide 59
P(H | E)
＝
P(H’ | E)
P(E | H) P(H)
P(E | H’) P(H’)
Defining the prior odds on H as
O(H) =
P(H)
P(H’)
P(H | E)
O(H | E) =
4. 知識推論法
P(H’ | E)
S.S. Tseng & G.J. Hwang
59
Slide 60
Likelihood ratio
P(E | H)
LS=
P(E | H’)
O(H | E) = LS O(H)
odds-likelihood form of Bayes’ Theorem.
The factor LS is also called likelihood of sufficiency
because if LS =∞ then the evidence E is logically
sufficient for concluding that H is true.
4. 知識推論法
S.S. Tseng & G.J. Hwang
60
Slide 61
Likelihood of necessity, LN, is defined similarly to LS as
LN=
O(H | E’)
O(H)
P(H | E’)
P(H’ | E’)
P(E’ | H)
=
P(E’ | H’)
=
P(H)
P(H’)
O(H | E’) = LN O(H)
LN=0,then P(H | E’) = 0. This means that H must be false
when E’ true. Thus if E is not present then H is false,
which means that E is necessary for H.
4. 知識推論法
S.S. Tseng & G.J. Hwang
61
Slide 62
LS
0
Small(0
Large(1<
4. 知識推論法
Effect on Hypothesis
H is false when E is true or
E’ is necessary for concluding H
E is unfavorable for concluding H
E has no effect on belief of H
E is favorable for concluding H
E is logically sufficient for H or
Observing E means H must be true
S.S. Tseng & G.J. Hwang
62
Slide 63
LN
0
small(0
large(1<
4. 知識推論法
Effect on Hypothesis
H is false when E is true or E is necessary for H
Absence of E is unfavorable for concluding H
Absence of E has no effect on H
Absence of E is favorable of H
Absence of E is logically sufficient for H
S.S. Tseng & G.J. Hwang
63
Slide 64
Uncertainty in inference chains
• Uncertainty may be present in rules,
evidence used by the rules, or both.
4. 知識推論法
S.S. Tseng & G.J. Hwang
64
Slide 65
Expert Inconsistency
If LS > 1 then P(E | H’) < P(E | H)
1 – P(E | H’) > 1 – P(E | H)
1-P(E | H)
LN=
1-P(E | H’)
<1
Case 1：LS>1 and LN <1
Case 2：LS<1 and LN >1
Case 3：LS= LN = 1
4. 知識推論法
S.S. Tseng & G.J. Hwang
65
Slide 66
Exercise
• 考慮以下的事實與規則，試以前向鏈結和後向鏈結描述其推論過程。
事實: A1, A2, A3, A4, B1, B2
規則: R1: A1 and A3 --> C2
R2: A1 and B1 --> C1
R3: A2 and C2 --> D2
R4: A3 and B2 --> D3
R5: C1 and D2 --> G1
R6: B1 and B2 --> D4
R7: A1 and A2 and A3 --> D2
R8: C1 and D3 --> G2
R9: C2 and A4 --> G3
目標: G1, G2 and G3
4. 知識推論法
S.S. Tseng & G.J. Hwang
66
Chapter 4
Methods of Inference
知識推論法
Slide 2
4.1 Deductive and Induction （演繹與歸納）
• Deduction(演繹): Logical reasoning in which
conclusions must follow from their premises.
• Induction(歸納): Inference from the specific case to
the general.
• Intuition(直觀): No proven theory.
• Heuristics(啟發): Rules of thumb (觀測法) based upon
experience.
• Generate and test: Trial and error.
4. 知識推論法
S.S. Tseng & G.J. Hwang
2
Slide 3
• Abduction(反推): Reasoning back from a true
conclusion to the premises that may have caused the
conclusion.
• Autoepitemic(自覺、本能): Self-knowledge
• Nonmonotonic(應變知識): previous knowledge
may be incorrect when new evidence is obtained
• Analogy(類推): based on the similarities to
another situation
4. 知識推論法
S.S. Tseng & G.J. Hwang
3
Slide 4
Syllogism （三段論）
• Syllogism（三段論）is simple, well-understood
branch of logic that can be completely proven.
– Premise(前提): Anyone who can program is
intelligent
– Premise(前提): John can program
– Conclusion(結論): Therefore, John is intelligent.
• In general, a syllogism is any valid deductive
argument having two premises and a conclusion.
4. 知識推論法
S.S. Tseng & G.J. Hwang
4
Slide 5
Categorical Syllogism
(定言三段論)
定言命題的型態
形態
概要
意思
A
E
I
O
所有S為P
沒有S為P
某些S為P
某些S不為P
完全肯定
完全否定
部分肯定
部分否定
4. 知識推論法
S.S. Tseng & G.J. Hwang
5
Slide 6
•三段論的標準形態（Standard form）
大前提：所有M為P
小前提：所有S為M
結論：所有S為P
- P代表結論的「謂詞」(Predicate)，又稱為「大詞」（Major term）
- S代表結論的「主詞」(Subject)，又稱作「小詞」（Minor term）。
- 含有大詞的前提稱為「大前提」（Major premise）；
- 含有小詞的前提稱為「小前提」（Minor premise）。
- M稱為「中詞」（Middle term）
4. 知識推論法
S.S. Tseng & G.J. Hwang
6
Slide 7
Mood(模式)
• Patterns of Categorical Statement
Figure-1
Figure-2
Figure-3
Figure-4
MP
SM
PM
SM
MP
MS
PM
MS
AAA-1
AAA-2
AAA-3
AAA-4
大前提
MP
PM
MP
PM
小前提
SM
SM
MS
MS
大前提
小前提
• 4種AAA模式
形態
4. 知識推論法
S.S. Tseng & G.J. Hwang
7
Slide 8
• ex: AAA-1
– 所有M為P
所有S為M
∴所有S為P
• We use decision procedure(決策程序) to prove
the validity of syllogistic argument
• The decision procedure for syllogisms can be done
using Venn Diagrams(維思圖)
4. 知識推論法
S.S. Tseng & G.J. Hwang
8
Slide 9
• ex: Decision procedure for Syllogism AEE-1
所有M為P
沒有S為M
∴沒有S為P
S
P
S
P
S
P
M
M
M
(a)維 思 圖
(b)大 前 提 後
(c)小 前 提 後
4. 知識推論法
S.S. Tseng & G.J. Hwang
9
Slide 10
General Rule under “some” quantifiers
1. If a class is empty, it is shaded.
2. Universal statement, A and E, are always drawn
before particular ones.
3. If a class has at least one member, mark it with a *.
4. If a statement does not specify in which of two
adjacent classed an object exists, place a * on the
line between the classes.
5. If an area has been shaded, no * can be put in it.
4. 知識推論法
S.S. Tseng & G.J. Hwang
10
Slide 11
ex: Decision procedure for Syllogism IAI-1
某些P為M
所有M為S
∴某些S為P
S
P
S
P
＊
M
(a) 所 有 M 是 S
4. 知識推論法
S.S. Tseng & G.J. Hwang
M
(b) 一 些 P 是 M
11
Slide 12
4.2 State and problem spaces
(狀態與問題空間)
• Tree（樹狀結構）: nodes, edges
• Directed or undirected
• Digraph（雙向圖）: a graph with directed
edges
• Lattice（晶格）: a directed acyclic graph
4. 知識推論法
S.S. Tseng & G.J. Hwang
12
Slide 13
• A useful method of describing the behavior of an
object is to define a graph called the state space.
[state(狀態) and action(行動)]
–
–
–
–
–
–
Initial state
Operator
State space
Path
Goal test
Path cost
4. 知識推論法
S.S. Tseng & G.J. Hwang
13
Slide 14
Finite State Machine
（有限狀態機器 ）
• Determining valid strings WHILE,WRITE, and BEGIN
H
I
L
E
W
R
T
I
開始
B
N
G
E
not G
I
成功
not I
not E
not N
錯誤
4. 知識推論法
S.S. Tseng & G.J. Hwang
14
Slide 15
Finding solution in problem space
• State space（狀態空間） can be thought as a
problem space（問題空間）.
• Finding the solution to a problem in a problem
space involves finding a valid path from start to
success( answer).
• The state space for the Monkey and Bananas
Problem
• Traveling salesman problem（旅行推銷員問題）
• Graph algorithms, AND-OR Trees, etc.
4. 知識推論法
S.S. Tseng & G.J. Hwang
15
Slide 16
Ex: Monkey and
猴子在躺
椅上
跳下躺椅
Bananas Problem
– 房子裡有一懸掛的香蕉
– 房子裡只有一張躺椅跟一把梯子
– 猴子無法直接拿到香蕉
• 指示：
跳下躺椅
移動梯子
把梯子移到香蕉下的位置
爬上梯子
摘下香蕉
• 初始狀態：
觀察到躺椅
在香蕉下
觀察到躺椅
不在香蕉下
躺椅不位
於香蕉下
• 假設：
–
–
–
–
–
猴子在地
板上
移動躺椅
觀察到猴子
不在梯子上
觀察到猴子
不在梯子上
猴子在梯
子上
躺椅位於
香蕉下
猴子不在
梯子上
移動猴子
觀察到梯子
不在香蕉下
觀察到梯子
在香蕉下
梯子不位
於香蕉下
移動梯子
梯子在香
蕉下
爬上梯子
– 猴子在躺椅上
4. 知識推論法
猴子在梯
子上
S.S. Tseng & G.J. Hwang
摘下香蕉
猴子成功
得到香蕉
16
Slide 17
D
B
Ex: Travel Salesman Problem
（旅行推銷員問題）
A
C
A
B
A
C
C
A
(a) 旅 行 推 銷 員 的 問 題 描 述
D
B
D
B
B
C
A
A
C
A B
D
D
B
D
C
B
B
C
D
C A C
(b) 搜 尋 路 徑 (粗 線 是 解 答 路 徑 )
4. 知識推論法
S.S. Tseng & G.J. Hwang
17
Slide 18
Ill-structured problem
(非結構化問題)
• Ill-structured problems（非結構化問題）
have uncertainties associated with it.
–
–
–
–
–
–
Goal not explicit
Problem space unbounded
Problem space not discrete
Intermediate states difficult to achieve
State operators unknown
Time constraint
4. 知識推論法
S.S. Tseng & G.J. Hwang
18
Slide 19
Ex:旅遊代理人
特徵
客戶的反應
目標不明顯
我在想到底要去哪裡
問題空間範圍未被介定
我不確定要去哪裡
問題狀態不是離散的
我只是想去旅遊，目的地並不
重要
中間的狀態不易實行
我沒有足夠的錢去
狀態的可用運算元未知
我不知道怎麼可以籌到錢
時間限制
我必須儘快出發
4. 知識推論法
S.S. Tseng & G.J. Hwang
19
Slide 20
4.3 Rules of Inference(規則式推論)
• Syllogism（三段論） addresses only a small
portion of the possible logic statements.
• Propositional logic
pq
p______
q
Inference is called direct reasoning (直接推論),
modus ponens (離斷率), law of detachment (分離
律) , and assuming the antecedent (假設前提).
4. 知識推論法
S.S. Tseng & G.J. Hwang
20
Slide 21
Truth table for Modus Ponense（離斷率）
p
T
T
F
F
4. 知識推論法
q
T
F
T
F
p→q
T
F
T
T
(p→q)p
T
F
F
F
S.S. Tseng & G.J. Hwang
(p→q) p→q
T
T
T
T
21
Slide 22
Law of Inference
Schemata
1.Law of Detachment
p→q
p
∴q
2.Law of the Contrapositive
p→q
∴~q→~p
3. Law of Modus Tollens
p→q
~q
∴~p
4.Chain Rules(Law of the Syllogism)
P→q
q→r
∴p→r
5.Law of Disjunctive Inference
pq
~p
∴q
~(~p)
∴p
6.Law of the Double Negation
4. 知識推論法
S.S. Tseng & G.J. Hwang
pq
~q
∴p
22
Slide 23
7.De Morgan’s Law
~(pq)
∴~p ~q
~(pq)
∴~p ~q
8.Law of Simplification
pq
∴p
~(pq)
∴~q
9.Law of Conjunction
p
q
∴pq
10.Law of Disjunctive Addition
p
∴pq
11. Law of Conjunctive Argument
~(pq)
p
∴~q
~(pq)
q
∴~p
Table 3.8 Some Rules of Inference for Propositional Logic
4. 知識推論法
S.S. Tseng & G.J. Hwang
23
Slide 24
Resolution in propositional Logic
（命題邏輯分解）
F: Rules or facts known to be TRUE
S: A conclusion to be Proved
• Convert all the propositions of F to clause form.
2. Negate S and convert the result to clause form. Add it
to the set of clauses obtained in step 1.
3. Repeat until either a contradiction is found or no
progress can be made:
(1) Select two clauses.
Call these the Parent clauses.
4. 知識推論法
S.S. Tseng & G.J. Hwang
24
Slide 25
(2) Resolve them together.
The resulting clause, called the resolvent, will be the
disjunction of all of the literals of both of the parent
clauses with the following exception：If there are any
pairs of literals L and ~L. Such that one of the parent
clauses contains L and the other contaions ~L, then
eliminate both L and ~L from the resolvent.
(3) If the resolvent is the empty clause, then a
contradiction has been found. If it is not, then add it
to the set of clauses available to the procedure.
4. 知識推論法
S.S. Tseng & G.J. Hwang
25
Slide 26
Given Axioms
Converted to Clause Form
p
p
(p q) r
1.
～p ～q r
(s t) q
2.
～s q
3.
～t q
4.
t
t
5.
p =下雨
q = 騎車
s = 路線熟悉
t = 路途遠
r = 穿雨衣
4. 知識推論法
S.S. Tseng & G.J. Hwang
26
Slide 27
～r
～ｐ ～ｑｒ
p
～p ～ｑ
～q
～t ｑ
～t
t
Resolution in Propositional Logic
4. 知識推論法
S.S. Tseng & G.J. Hwang
27
Slide 28
Resolution with quantifiers
Example（from Nilsson）：
Whoever can read (R) is literate (L).
Dolphins (D) aren’t literate (~L).
Some dolphins (D) are intelligent (I).
To prove：Some who are intelligent (I) can’t read (~R).
4. 知識推論法
S.S. Tseng & G.J. Hwang
28
Slide 29
Translating：
x[R(x)→ L(x)]
x [ D ( x ) → ~L ( x ) ]
x[D(x) ＆ I(x)]
To prove:
4. 知識推論法
x[I(x)&~R(x)]
S.S. Tseng & G.J. Hwang
29
Slide 30
(1) － (4)：
x [~ R ( x ) OR L ( x ) ] & y [ ~ D ( y )
OR ~ L ( y ) ] & D ( A ) & I ( A ) &
z [~ I ( z ) OR R ( z ) ]
(5) － (9)：
C1=~R(x)
OR L(x)
C2=~D(y)
OR ~L(y)
C3=D(A)
C4=I(A)
C5= ~I(z)
4. 知識推論法
OR R(z)
S.S. Tseng & G.J. Hwang
30
Slide 31
• The second order logic can have quantifiers
that range over function and predicate
symbols
• If P is any predicate of one document
– Then
– x =y (for every P [P(x) P(y) ]
4. 知識推論法
S.S. Tseng & G.J. Hwang
31
Slide 32
4.4 Inference Chain (推斷鏈)
D3
Initial facts
A2
A1
D2
B
C
D1
E
Solution
inference + inference +… + inference
Chain
backward
chaining
forward
chaining
Infer from initial
facts to solutions
4. 知識推論法
Assume that some solution
is true, and try to prove
the assumption by finding
the required facts
S.S. Tseng & G.J. Hwang
32
Slide 33
•
Forward Chaining（前向鏈結）:
Rule1: elephant(x) mammal(x)
Rule2: mammal(x) animal(x)
Fact：John is an elephant.
elephant (John) is true
X=John (Unification)
elephant(x) mammal(x)
Mammal(John) is true
X’=X=John
mammal(x’) animal(x’)
• Unification（變數替代）
The process of finding substitutions for
variables to make arguments match.
4. 知識推論法
S.S. Tseng & G.J. Hwang
animal(John) is true
33
Slide 34
Forward Chaining（前向推論）
Rule1：A1 and B1 C1
Rule2：A2 and C1 D2
Rule3： A3 and B2 D3
Rule4：C1 and D3 G
Facts：A1 is true, A2 is true , A3 is true, B1 is true, B2 is true
{A1, A2, A3, B1, B2} match {r1, r3}
fire r1 {A1, A2, A3, B1, B2, C1} match {r1, r2, r3}
fire r2 {A1, A2, A3, B1, B2, C1, D2} match{r1, r2, r3}
fire r3 {A1, A2, A3, B1, B2, C1 D2, D3} match{r1, r2, r3, r4}
fire r4 {A1, A2, A3, B1, B2, C1 D2, D3, G }
4. 知識推論法
S.S. Tseng & G.J. Hwang
GOAL
34
Slide 35
Backward Chaining (反向推論)
rule1 : A1 and B1
rule2 : A2 and C1
rule3 : A3 and B2
rule4 : C1 and D3
rule5 : C1 and D4
1. Assume G’ is true
C1
D2
D3
G
G’
facts : A1, A2, B1, B2, A3
2. Assume G is true
R5
R4
Verify C1 and D4
Verify C1 and D3
R1
R3
Verify A1 and B1
OK
Verify A3 and B2
OK
D4 is unknown, ask user.
If D4 is FALSE, give up.
Verify A1 and B1
OK
4. 知識推論法
S.S. Tseng & G.J. Hwang
OK
OK
OK
35
Slide 36
A1
A2
B1
A3
B2
C1
D2
D3
?
G
GOAL
4. 知識推論法
S.S. Tseng & G.J. Hwang
36
Slide 37
• Good application of forward chaining（前向鏈結）
Goal
Facts
Broad and Not Deep
or
too many possible goals
4. 知識推論法
S.S. Tseng & G.J. Hwang
37
Slide 38
• Good application of backward chaining（後向鏈結）
Narrow and
Deep
GOALS
Facts
4. 知識推論法
S.S. Tseng & G.J. Hwang
38
Slide 39
• Forward Chaining（前向鏈結）
Planning
Monitoring
Control
Data-driven
Explanation not facilitated
• Backward chaining（後向鏈結）
Diagnosis
Goal-driven
Explanation facilitated
4. 知識推論法
S.S. Tseng & G.J. Hwang
39
Slide 40
Analogy
• Try to relate old situations as guides to new
ones
• Consider tic-tac-toe with values as a magic
square (15 game)
»6
»7
»2
1
5
9
8
3
4
• 18 game from set {2,3,4,5,6,7,8,9,10}
• 21 game from set {3,4,5,6,7,8,9,10,11}
4. 知識推論法
S.S. Tseng & G.J. Hwang
40
Slide 41
Nonmonotonic reasoning
• In nonmonotonic system, the theorems do
not necessarily increase as the number of
axioms increases.
• As a very simple example, suppose there is
a fact that asserts the time. As soon as time
changes by a second, the old fact is no
longer valid.
4. 知識推論法
S.S. Tseng & G.J. Hwang
41
Slide 42
4.5 Reasoning Under Uncertainty
(不確定性推論)
• Uncertainty can be considered as the lack of adequate
information to make a decision.
• Classical probability, Bayescian probability, DempsterShafer theory, and Zadeh’s fuzzy theory.
• In the MYCIN and PROSPECTOR systems conclusion are
arrived at even when all the evidence needed to absolutely
prove the conclusion is not known.
4. 知識推論法
S.S. Tseng & G.J. Hwang
42
Slide 43
Many different types of error can contribute to uncertainty
Example
Turn the value off
Turn value-1
Turn value-1 off
Value is stuck
Value is not stuck
Turn value-1 to 5
Turn value-1 to 5.4
Turn value-1 to 5.4 or 6 or 0
Value-1 setting is 5.4 or 5.5 or 5.1
Value-1 setting is 7.5
Value-1 is not stuck because it’s
never been stuck before
Output is normal and so value-1
is in good condition
4. 知識推論法
Error
Ambiguous
Incomplete
Incorrect
False positive (接受錯誤值)
False negative (拒絕正確值)
Imprecise
Inaccurate
Unreliable
Random error
Systematic error
Invalid induction
Reason
What value?
Which way?
Correct is on
Value is not stuck
Value is stuck
Correct is 5.4
Correct is 9.2
Equipment error
Statistical fluctuation (波動)
Mis-calibration (刻度)
Invalid deduction
Value is stuck in open position
S.S. Tseng & G.J. Hwang
Value is stuck
43
Slide 44
• A hypothesis is an assumption to be tested.
• Type 1 error (false positive) means acceptance of a
hypothesis when it is not true.
• Type 2 error (false negative) means rejection of a hypothesis
when it is true.
• Error of measurement
– Precision
• The millimeter(公釐) ruler is more precise than
centimeter ruler.
– accuracy
4. 知識推論法
S.S. Tseng & G.J. Hwang
44
Slide 45
Error & Induction
The process of induction is the opposite of deduction
The fire alarm goes off (響起)
∴ There is a fire.
An even stronger argument is
The fire alarm goes off & I smell smoke
∴ There is a fire.
Although this is a strong argument, it is not proof that there is
a fire.
My clothes are burning
4. 知識推論法
S.S. Tseng & G.J. Hwang
45
Slide 46
Deductive errors
p→q
q
∴p
If John is a father, than John is a man
John is a man
∴ John is a father
4. 知識推論法
S.S. Tseng & G.J. Hwang
46
Slide 47
Baye’s Theorem （貝氏定理）
• Conditional probability（條件機率）, P(A | B) , states the
probability of event B occurred. Crash= Brand X(0.6)+
Not X(0.1)=0.7
• P( X|C) =
P( C | X) P(X) =
P(C)
(0.75)(0.8) =
0.7
6
7
• Suppose you have a drive and don’t know its brand, what
is the probability that if it crashes, it is Brand X? nonBrand X?
4. 知識推論法
S.S. Tseng & G.J. Hwang
47
Slide 48
Decision Tree for the Disk drive Crashes
Act
Choose
Brand X
P(X)=0.8
Don’t Choose
Brand X
P(X’)=0.2
No Crash
P(C’ | X’)=0.5
P(C’∩X’)=0.1
Crash
P(C | X’)=0.5
P(C ∩X’)=0.1
Prior
P(Hi)
Crash
P(C | X)=0.75
No Crash
P(C’ | X)=0.25
P(C’ ∩ X)=0.2
P(C ∩X)=0.6
Conditional
P(E | Hi )
Joint -P(E ∩ Hi )
=P(E | Hi ) P( Hi )
0.1
P(X’ | C’) =
0.1+0.2
0.1
P(X’ | C) =
=1/3
4. 知識推論法
0.1+0.6
=1/7
0.2
P(X | C’) =
0.2+0.1
=2/3
S.S. Tseng & G.J. Hwang
0.6
P(X | C) =
0.6+0.1
=6/7
Posterior
P(H i | E) = P (E ∩ Hi)
iP(E∩Hi)
48
Slide 49
P(Hi | E) =
P(E∩Hi)
P(E ∩Hj)
j
=
P(E | H i) P(Hi)
P(E | Hj) P(Hj)
j
P(E | Hi)P(Hi)
P(E)
• Bayes’ Theorem（貝氏定理） is commonly used for
decision tree analysis of business and the social science.
=
• Used in Prospector expert system to decide
favorite sites of mineral exploration
4. 知識推論法
S.S. Tseng & G.J. Hwang
49
Slide 50
Hypothetical Reasoning and Backward Induction.
Probabilities
No Oil
Oil
P(O’)=0.4
Prior
P(O)=0.6
Subjective Opinion
of Site - P (Hi)
-Test
+Test
-Test
+Test
P(- | O’)
P(+ | O’)
P(- | O)
P(+ | O)
=0.9
=0.1
=0.2
=0.8
P(-∩O’)
P(+∩O’)
P(-∩O)
P(+∩O)
=0.36
=0.04
=0.12
=0.48
P(+)=P(+∩O)+P(+∩O’)=0.48+0.04=0.52
Conditional
Seismic Test
Result
Joint P(E∩H)
=P(E | Hi)
P(Hi)
P(-)=P(-∩O)+P(-∩O’)=0.12+0.36=0.48
4. 知識推論法
S.S. Tseng & G.J. Hwang
50
Slide 51
-Test
+Test
Probabilities
P(-)=0.48
P(+)=0.52
Unconditional
P (E)
No Oil
Oil
No Oil
Oil
P(O’|-)
P(O|-)
P(O’|+)
P(O|+)
= (9)(4)
= (2)(6)
= (1)(4)
= (8)(6)
0.48
0.48
= 3/4
= 1/4
0.52
0.52
= 1/13
Posterior
of Site - P(Hi | E)
P ( E| Hi) P (Hi)
=
P(E)
= 12/13
P(-∩O)
P(+∩O)
P(-∩O)
P(+∩O)
Joint -P(E∩H)
=0.36
=0.04
=0.12
=0.48
=P(Hi | E) P(E)
4. 知識推論法
S.S. Tseng & G.J. Hwang
51
Slide 52
•
•
•
•
Oil release , if successful
Drilling expense
Seismic survey
Expected payoff (success)
$1250000
-$200000
-$50000
• 846153=1000000 *12/13 – 1000000*1/13
• Fail
• -500000= 1000000*1/4- 1000000*3/4
• Expected payoff (total)
• 416000= 846153*0.52 – 50000 * 0.48
4. 知識推論法
S.S. Tseng & G.J. Hwang
52
Slide 53
Temporal reasoning and
Markov chain
• Temporal reasoning: reasoning about events
that depend on time
• Temporal logic
• The system’s progression through a
sequence of status is called a Stochastic
process if it is probabilistic.
4. 知識推論法
S.S. Tseng & G.J. Hwang
53
Slide 54
– P11
P12
– P21
P22
– Where Pmn is the probability of a transition
from state m to state n.
S = { P1,P2, …, Pn} where P1+P2+…+Pn= 1
S2 = S1 T
0.1
0.9
S2 = [0.8,0.2]
= [0.2,0.8]
0.6
4. 知識推論法
0.4
S.S. Tseng & G.J. Hwang
54
Slide 55
• Assume 10 percent of all people who now use
Brand X drive will buy another Brand X when
needed. 60 percent of people who don’t use Brand
X will buy Brand X when they need a new drive.
Over a period of time, how many people will use
Brand X?
S3 = [0.5,0.5], S4 = [0.35,0.65],
S5 = [0.425,0.575], S6 = [0.3875,0.6125]
S7 = [0.40625,0.59375], S8 = [0.396875,0.602125]
Steady state matrix
4. 知識推論法
S.S. Tseng & G.J. Hwang
55
Slide 56
The odds of belief
• “The patient is covered with red spots”
• Proposition A: “The patient has measles”
• P(A|B) :(degree of belief that A is true,
given B)is not necessarily a probability if
the events and propositions can not be
repeated or has a math basis.
4. 知識推論法
S.S. Tseng & G.J. Hwang
56
Slide 57
• The odds on A against B given some event
C are odds =P(A|C)/ P(B|C)
• If B = A’
– odds =P(A|C)/ P(A’|C) =P(A|C)/ (1-P(A|C) )
• Likelihood of P = 0.95
– Odds = .95/(1-.95) = 19 to 1
4. 知識推論法
S.S. Tseng & G.J. Hwang
57
Slide 58
Sufficiency and necessity
Bayes’ Theorem is
P(E | H)P(H)
P(H|E) =
P(E)
Negation P(H’|E) = P(E | H’)P(H’)
P(E)
4. 知識推論法
S.S. Tseng & G.J. Hwang
58
Slide 59
P(H | E)
＝
P(H’ | E)
P(E | H) P(H)
P(E | H’) P(H’)
Defining the prior odds on H as
O(H) =
P(H)
P(H’)
P(H | E)
O(H | E) =
4. 知識推論法
P(H’ | E)
S.S. Tseng & G.J. Hwang
59
Slide 60
Likelihood ratio
P(E | H)
LS=
P(E | H’)
O(H | E) = LS O(H)
odds-likelihood form of Bayes’ Theorem.
The factor LS is also called likelihood of sufficiency
because if LS =∞ then the evidence E is logically
sufficient for concluding that H is true.
4. 知識推論法
S.S. Tseng & G.J. Hwang
60
Slide 61
Likelihood of necessity, LN, is defined similarly to LS as
LN=
O(H | E’)
O(H)
P(H | E’)
P(H’ | E’)
P(E’ | H)
=
P(E’ | H’)
=
P(H)
P(H’)
O(H | E’) = LN O(H)
LN=0,then P(H | E’) = 0. This means that H must be false
when E’ true. Thus if E is not present then H is false,
which means that E is necessary for H.
4. 知識推論法
S.S. Tseng & G.J. Hwang
61
Slide 62
LS
0
Small(0
Large(1<
4. 知識推論法
Effect on Hypothesis
H is false when E is true or
E’ is necessary for concluding H
E is unfavorable for concluding H
E has no effect on belief of H
E is favorable for concluding H
E is logically sufficient for H or
Observing E means H must be true
S.S. Tseng & G.J. Hwang
62
Slide 63
LN
0
small(0
large(1<
4. 知識推論法
Effect on Hypothesis
H is false when E is true or E is necessary for H
Absence of E is unfavorable for concluding H
Absence of E has no effect on H
Absence of E is favorable of H
Absence of E is logically sufficient for H
S.S. Tseng & G.J. Hwang
63
Slide 64
Uncertainty in inference chains
• Uncertainty may be present in rules,
evidence used by the rules, or both.
4. 知識推論法
S.S. Tseng & G.J. Hwang
64
Slide 65
Expert Inconsistency
If LS > 1 then P(E | H’) < P(E | H)
1 – P(E | H’) > 1 – P(E | H)
1-P(E | H)
LN=
1-P(E | H’)
<1
Case 1：LS>1 and LN <1
Case 2：LS<1 and LN >1
Case 3：LS= LN = 1
4. 知識推論法
S.S. Tseng & G.J. Hwang
65
Slide 66
Exercise
• 考慮以下的事實與規則，試以前向鏈結和後向鏈結描述其推論過程。
事實: A1, A2, A3, A4, B1, B2
規則: R1: A1 and A3 --> C2
R2: A1 and B1 --> C1
R3: A2 and C2 --> D2
R4: A3 and B2 --> D3
R5: C1 and D2 --> G1
R6: B1 and B2 --> D4
R7: A1 and A2 and A3 --> D2
R8: C1 and D3 --> G2
R9: C2 and A4 --> G3
目標: G1, G2 and G3
4. 知識推論法
S.S. Tseng & G.J. Hwang
66