Chapter 7
METHODS OF INFERENCE
知識推論法
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7.1演繹與歸納（ Deductive and Induction ）
• 演繹( Deduction): 藉由前提假設而推論出結論
• 歸納(Induction): 一葉知秋,見微知柱; 從小處歸
納出一個大觀.
• 直觀(Intuition): 尚未被證實的理論.
• 啟發(Heuristics): 從既有經驗所獲得的規則.
• Generate and test: 不斷的嘗試, 從錯誤中學習.
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• Abduction: 從已成立的結論往回推論, 以
得導致此結論的前提.
• Autoepitemic: Self-knowledge
• Nonmonotonic: 舊有的知識可能被新的
知識更新,取代
• 類比(Analogy): 藉由比較方式, 類似的前
提會導致相同的結論(ex:醫療診斷)
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三段論（ Syllogism ）
• 三段論（ Syllogism ）是一種邏輯論證,
容易理解且已被為整證明過.
– 前提(Premise): 會寫程式的人都很聰明
– 前提(Premise): John會寫程式
– 結論(Conclusion): 因此, John很聰明.
• 一般而言,三段論根據演繹的論證包含了
兩個前提和一個結論.
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定言三段論(Categorical Syllogism)
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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•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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模式 (Mood)
• 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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• ex: AAA-1
– 所有M為P
所有S為M
∴所有S為P
• 我們利用 決策程序(decision procedure)來證明
論證是否有效
• 三段論運用維思圖 (Venn Diagrams)來驗證
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• ex: AAA-1的決策程序
所有M為P
所有S為M
∴所有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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• ex: AEE-1的決策程序
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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“某些”的表示規則
• 1. 如果該類別是空的,則填上較暗色彩.
• 2. 完全肯定或否定的陳述句(Universal
statement): A 和 E, 會先畫
• 3. 若該類別至少有一個物件,則以 *標示.
• 4. 若兩個相鄰的類別沒有一個物件存在,
則用*標在線上.
• 5. 已經塗上暗色的的部分不需要標上*.
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ex: IAI-1的決策程序
某些P為M
所有M為S
∴某些P為S
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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7.2
狀態與問題空間(State and
problem spaces)
• 樹狀結構（ Tree ）: 點(nodes), 線(edges)
• 有向性(Directed)/無向性(Indirected)
• 雙向圖（Digraph）: 圖中所有的線都是
有方向性的
• 晶格（Lattice）: 有方向性但沒有循環性
的圖
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• 藉由描述物件在圖中的行為來定義該圖,
此方法我們稱做“問題狀態空間”
–
–
–
–
–
–
初始狀態(Initial state)
可用的運算元(Operator)
狀態空間(State space)
路徑(Path)
目標測試(Goal test)
路徑成本(Path cost)
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有限狀態機器
（ Finite State Machine ）
• 用以辨識 WHILE和WRITE的有限狀態機器
H
I
L
E
W
R
T
I
開始
B
N
G
E
not G
not E
I
成功
not I
not N
錯誤
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在問題空間中尋解
(Finding solution in problem space)
• 狀態空間（ State space ）可以視為一個
問題空間（ problem space ）.
• 在問題空間裡求解需要找出一條有效的
路徑(開始->答案).
• “猴子與香蕉問題”的問題空間
• 旅行推銷員問題（Traveling salesman
problem）
• 圖形演算法, AND-OR Trees…等等.
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猴子在躺
椅上
Ex: 猴子與香蕉
跳下躺椅
觀察到躺椅
不在香蕉下
問題
• 假設：
– 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：
– 猴子在躺椅上
猴子在地
躺椅位於
板上
觀察到躺椅 香蕉下
在香蕉下
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移動躺椅
躺椅不位
猴子不在
於香蕉下 觀察到猴子 梯子上
不在梯子上
觀察到猴子
不在梯子上
移動猴子
猴子在梯
梯子不位
子上
觀察到梯子 於香蕉下
不在香蕉下
觀察到梯子
在香蕉下
移動梯子
梯子在香
蕉下
爬上梯子
猴子在梯
子上
摘下香蕉
猴子成功
得到香蕉
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Ex:旅行推銷員問題
（ Travel Salesman Problem ）
D
B
A
C
A
(a) 旅行推銷員的問題描述
B
A
C
C
D
A B
D
B
B
C
A
C
AB D
D
B
A C D
B
B
C
D
C AC
(b) 搜尋路徑(粗線是解答路徑)
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非結構化問題
(Ill-structured problem)
• 非結構化問題（ Ill-structured problems ）
是指很多不確定因素的結合.
–
–
–
–
–
–
目標不明顯
問題空間範圍尚未被介定
問題狀態空間非離散的
中間狀態的不易實行
可用運算元未知
時間限制
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Ex:旅遊代理人
特徵
客戶的反應
目標不明顯
我在想要到底要去哪裡
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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7.3 規則式推論(Rules of Inference)
• 三段論（ Syllogism ） 只是邏輯陳述的其中一
種方式.
• 命題邏輯(Propositional logic)
pq
p______
q
這種命題邏輯的推論型態有很多種名子: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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離斷率的真值表（ Truth table for Modus Ponense ）
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
(p→q)p
T
F
F
F
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(p→q)  p→q
T
T
T
T
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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
pq
~p
∴q
~(~p)
∴p
6.Law of the Double Negation
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pq
~q
∴p
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7.De Morgan’s Law
~(pq)
∴~p ~q
~(pq)
∴~p  ~q
8.Law of Simplification
pq
∴p
~(pq)
∴~q
9.Law of Conjunction
p
q
∴pq
10.Law of Disjunctive Addition
p
∴pq
11. Law of Conjunctive Argument
~(pq)
p
∴~q
~(pq)
q
∴~p
Table 3.8 Some Rules of Inference for Propositional Logic
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命題邏輯分解
（ Resolution in propositional Logic ）
1. 將命題(proposition)轉換成clause form形式.
2. 重複以下動作直到產生矛盾或是無法再繼續
(1) 選擇兩個clauses, 我們稱之為parent clauses
(2) 將parent clauses作disjunction,產生出來的
clause稱為resolvent, 再把有L和~L這樣形式的
元素移除
(3) 若最後resolvent是空的(empty),則矛盾產生;
若否則再加入另一個clause作disjunction,同(2)
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Given Axioms
Converted to Clause Form
p
p
1.
(p
q)
r
～p
～q
(s
t)
q
～s
q
3.
～t
q
4.
t
Figure.
t
r
2.
5.
命題邏輯的facts
(A Few Facts in Propositional Logic)
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～ｐˇ～ｑˇｒ
～r
p
～p ˇ～ｑ
～q
～t ˇｑ
～t
Figure.
t
命題邏輯之解
(Resolution in Propositional
Logic)
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加入量詞的推論
(Resolution with quantifiers)
EX（form Nilsson）：
任何會閱讀(Read)的皆具有讀寫能力(Literate).
海豚(Dolphin)沒有讀寫能力
一些海豚很聰明(Intelligent).
試證明：存在一些很聰明但沒有閱讀能力的海豚.
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轉換：


x [ R ( x ) －→
L(x)]
x [ D ( x ) －→ －L ( x ) ]
 x[D(x) ＆
To prove:
I(x)]
x[I(x)&-R(x)]
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(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) OR R(z)
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• 第二敘述邏輯 (The second order logic) 的
量詞 (quantifiers)範圍含括函數符號
(function symbols)與敘述符號( predicate
symbols)
• 如果P是一文件中的任一敘述(predicate
–則
– x =y = ( for every P [P(x)  P(y) ]
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7.4 推斷鏈
D3
Initial facts
A2
A1
D2
B
C
D1
E
Solution
inference + inference +… + inference
Chain
backward
Chaining(向後鏈結)
forward
Chaining(向前鏈結)
從事實來推論結果
假設可能的解答是成立的,再
尋求相關的證據證明之
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•
前向鏈結（ Forward Chaining ）：
rule1：
大象(x)
哺乳類(x)
rule2：
哺乳類(x)
動物(x)
face：John 是一知大象.
大象(John) 成立
X=John (變數替代-Unification)
大象(x)
哺乳類(x)
哺乳類(John) 成立
X’=X=John
哺乳類(x)
動物(x’)
動物(John) 成立
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• 變數替代（ Unification ）
將變數(variable)用事實(fact)取代,反覆的進行變
數替代最後即可得出結論.
rule1：A1 and B1
C1
rule2：A2 and C1
D2
rule3： A3 and B2
D3
rule4：C1 and D3
G
facts：A1 is true
B1 is true
A2 is true
A3 is true
B2 is true
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• 前向推論（ Forward reasoning ）：
{A1, A2, A3, B1, B2, B3}
fire r1 {A1, A2, A3, B1, B2, B3,
C1 }
match
{r1, r3}
match
{r1, r2, r3}
match
fire r2 {A1, A2, A3, B1, B2, B3, C1,
D2}
fire r3 {A1, A2, A3, B1, B2, B3, C1 D2,
{r1, r2, r3}
D3 }
fire r4 {A1, A2, A3, B1, B2, B3, C1 D2, D3,
{r1, r2, r3, r4}
G}
GOAL
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rule1：A1 and B1
C1
rule2：A2 and C1
D2
rule3： A3 and B2
D3
rule4：C1 and D3
D4
rule5： C1 and D4
G’
facts： A1, A2, B1, B2, A3
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•後向推論（ Backward reasoning ）：
1. 假設 G’ 成立
R5
Verify C1 and D4
訊問使用者D4之值, 若為
False則G’不成立
R1
Verify A1 and B1
OK
OK
2. 假設 G 成立
R4
Verify C1 and D3
R3
Verify A1 and B1
OK
OK
Verify A3 and B2
OK
OK
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A1
A2
B1
A3
B2
C1
D2
D3
?
G
GOAL
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•前向鏈結（ Good application of forward chaining ）
Goal
Facts
Broad and Not Deep
or
too many possible goals
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•後向鏈結（ Good application of backward chaining ）
Narrow and
Deep
GOALS
Facts
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•前向鏈結（ Forward Chaining ）
Planning
Monitoring
Control
Data-driven
Explanation not facilitated
•後向鏈結（ Backward chaining ）
Diagnosis
Goal-driven
Explanation facilitated
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類比 (Analogy)
• 將舊有相關的情境作為推論的參考
• 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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Nonmonotonic reasoning
• 在nonmonotonic system中,隨著元素的增
加,並不一定會增加定理的數量.
• 以一個簡單的例子來說,假設某一事實是
和時間相關的,若今時間改變了,則該事實
就不再是有效(令人信服)的了.
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7.5 Reasoning Under Uncertainty
(不確定性推論)
• 不確定性 (Uncertainty)是指在作決策時缺少足夠的資訊.
• 處理不確定性的理論有:Classical probability, Bayescian
probability, Dempster-Shafer theory, and Zadeh’s fuzzy
theory.
• 在 MYCIN 和 PROSPECTOR的系統中,即使所有前提確
實證明了結論是未知的,也會產生很多結論.
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導致不確定的錯誤
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
Error
Ambiguous
Incomplete
Incorrect
False positive
False negative
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
Miscalibration
Value is stuck
Invalid deduction
Value is stuck in open
position
• 假說是需被驗證的.
• Type 1 error (false positive) 是指接受了一個不會成立(F)的假設.
• Type 2 error (false negative) 是指否決了一個會被成立(T)的假設.
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• 度量的誤差(Error of measurement)
– 精確度(Precision)
• 公釐量尺的精確度比公分量尺高.
– 正確性(accuracy)
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錯誤與歸納(Error & Induction)
“歸納”的過程和“演繹”是相反的
火警警報聲響起
∴ 有火災發生了.
加入一個更強烈的參數:
火警警報聲響起
& 聞道煙味
∴有火災發生了.
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雖然加入了一個強烈的參數, 但依然無法證明確
實發生火災.
衣服著火了
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演繹錯誤
(Deductive errors)
p→q
q
∴p
If the value is in good condition, than the output is normal
The output is normal
∴ The value is in good condition.
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Baye’s Theorem （貝氏定理）
• Conditional probability（條件機率）, P(A | B) , 是指在B
事件成立的情況之下A事件發生的機率. 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
• 假設你擁有某不知名廠牌的硬碟壞掉了,那麼此硬碟是
X牌和其他非X牌的機率各是多少?
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硬碟損壞的決策樹
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
=1/3
0.1
P(X’ | C) =
0.1+0.6
=1/7
0.2
P(X | C’) =
0.2+0.1
=2/3
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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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P(E∩Hi) =
P(E | H i) P(Hi)
P(E ∩Hj)
P(E | Hj) P(Hj)
j
j
P(E | Hi)P(Hi)
=
P(E)
• Bayes’ Theorem（貝氏定理）利用決策樹作分
析,最常用於商業事務或社會科學.
P(Hi | E) =
• 探礦專家系統可決定何處是值得探勘的
Expert Systems sstseng
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假設推論與向後歸納
(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
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-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)
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•
•
•
•
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
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時態推論 (Temporal reasoning)
和Markov chain
• 時態推論(Temporal reasoning): 推論的事
件和時間相關
• 時態邏輯(Temporal logic)
• The system’s progression through a
sequence of status is called a Stochastic
process(隨機過程) if it is probabilistic.
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– P11
P12
– P21
P22
– Pmn 是指從狀態m轉換到狀態n的機率.
S = { P1,P2, …, Pn} where P1+P2+…+Pn= 1
S2 = S1 T
S2 = [0.8,0.2]
= [0.2,0.8]
0.1
0.9
0.6
0.4
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• 假設有10%的人擁有X牌的硬碟會再買該牌的
硬碟.60%的人不使用X牌硬碟但下次會買.問再
一段時間之後會有多少人使用X牌硬碟?
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
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The odds of belief
• “這個病人渾身佈滿紅疹”
• 提議A: “這個病人罹患麻疹”
• P(A|B) :(在B事件下A成立的可信度)is
not necessarily a probability if the events
and propositions can not be repeated or has
a math basis.
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• 在C事件下A,B相斥的機率 =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
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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)
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P(H | E)
P(E | H) P(H)
＝
P(H’ | E) P(E | H’) P(H’)
Defining the prior odds on H as
P(H)
O(H) =
P(H’)
P(H | E)
O(H | E) =
P(H’ | E)
Likelihood ratio
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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.
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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.
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LS
0
Small(0<LS<<1)
1
Large(1<<LS)

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
Expert Systems sstseng
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LN
0
small(0<LN<<1)
1
large(1<<LN)

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
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推論鏈中不確定性
(Uncertainty in inference chains)
• Uncertainty may be present in rules,
evidence used by the rules, or both.
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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
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Exercise
1.
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
Expert Systems sstseng
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