Matakuliah
Tahun
Versi
: T0264/Inteligensia Semu
: Juli 2006
: 2/2
Pertemuan 4
Heuristic Search Techniques
Problem Reduction
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Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• << TIK-99 >>
• << TIK-99>>
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Outline Materi
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•
•
•
•
Materi 1
Materi 2
Materi 3
Materi 4
Materi 5
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3.4 Problem Reduction
A Simple AND - OR Graph
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3.4 Problem Reduction
AND - OR Graphs
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Algoritm : Problem Reduction
1. Initialize the graph to the starting node
2. Loop until the starting node is labeled SOLVED
or until its cost goes above FUTILITY :
a. Traverse the graph, starting at the initial node and
following the current best path, and accumulate
the set of nodes that are on that path and have
not yet been expanded or labeled as solved
b. Pick one of these unecpanded nodes and expand
it. If there are no successors, assign FUTILITY as
the value of this node. Otherwise, add its
successors to the graph abd for each of them
computer f’. If f’ of any node is O, mark that node
as SOLVED
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Algoritm : Problem Reduction
c. Change the f’ estimate of the newly expanded
node to reflect the new information provided by
its successors. Propagate this change
backward through the graph. If any node
contains a successors arc whose descendants
are all solved, label the node that is visited
while going up the graph, decide which of its
successor arcs is the most promising and mark
it as part of the current best path
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The Operation of Problem Reduction
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3.5 Constraint Satisfaction
Algorithm : Constrain Satisfaction
1. Propagate available constraints : First set
OPEN to the set of all objects that must have
values assigned to them in a complete solution.
Then do until an inconsistency is detected or
until OPEN is empty :
a.
b.
c.
Select an object OB from OPEN. Strengthen the
set of constraints that apply to OB.
If this set is different from the set that was
assigned the last time OB was examined or if this
is the first OB has been examined, then add
to
OPEN all objects that share anyconstraints with
OB.
Remove OB from OPEN.
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3.5 Constraint Satisfaction
2. If solution, then quit and report the solution.
3. If contradiction, then return failure.
4. If neither, then guess : Loop until a solution is
found or all possible solution have been
eliminated :
a. Select an object whose value is not yet
determined & select a way of strengthening the
constraints on that object.
b. Recursively invoke constraint satisfaction with the
current set of constraints augmented by the
strengthening constraint just selected.
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A Criptarithmetic Problem
Problem :
SEND
+ MORE
MONEY
Initial State :
• No two letters have the same value
• The sums of the digits must be as shown
in the problem
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Initial State
M = 1
S = 8 or 9
O = 0 or 1
 O = 0
N = E or E+1  N = E+1
C2 = 1
N+R > 8
E <> 9
Solving
a Criptarithmetic
Problem
Send
+ More
Money
E=2
N = 3
R = 8 or 9
2+D = Y or 2+D =
10+Y
C1=0
2+D
N+R
R =
S =
= Y
= 10+E
9
8
C1=1
2+D = 10 + Y
D = 8+Y
D = 8 or 9
D=8
D=9
Y=0
conflict
Y=1
conflict
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A Criptarithmetic Problem
• Ulasan :
• Solusi Cryptarithmetic Problem dengan Constraint
satisfaction
•
S E N D
M Ø R E +
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
MØ N E Y
C3 C2 C1
S E N D
M Ø R E +
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
MØ N E Y
• C3 / C2 / C1 ( C ) = Carry yang bernilai 1 atau 0.
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A Criptarithmetic Problem
• Initial state :
• [if !supportLists] 1. [endif]  No two letters have the
same value.
( masing-masing huruf hanya memiliki 1 nilai dengan range
[0-9] )
• [if !supportLists]2. [endif] The sums of the digit must be
as shown in the problem.
• Jumlah digit antar huruf musti sama dengan soal.
• Mis : S+M = MØ, maka S+M harus bernilai belasan agar
tercapai 2 digit)
• Steps for solving the problem
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A Criptarithmetic Problem
• S + M = MØ nilai S dan M harus berkisar diantara [0-9],
karena hanya 1 digit. Maka nilai M adalah 1 (M = 1)
karena S+M+C3 tidak bisa lebih besar dari 19.
• Karena M = 1, maka (S + 1(M) + C3 = 1(M)Ø) nilai S
adalah 8(+1) atau 9 agar didapat 2 digit nilai (10 atau 11). 8
(+1) maksudnya 8 ditambah Carry(C3).
• Nilai Ø adalah 0 (Ø = 0) karena S + 1(M) + C3 harus
bernilai 10 supaya didapat 2 digit. Nilai 11 bisa didapat
kalau menggunakan S = 9 dan C3 = 1, tapi nilai 1 sudah
buat M, karena itu nilai Ø tidak boleh 1.
• Kemungkinan 1 : S = 8 , C3 = 1 , Ø = 0
• E + Ø = (1) N nilai E + Ø harus bernilai min 10 agar bisa
didapat nilai C3 = 1
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A Criptarithmetic Problem
• C2 + E + Ø = N , (1+) 9(E) + 0(Ø) = (1)N , maka 10 = (1)N
maka N = 0,hal ini menciptakan kontradiksi karena nilai N =
Ø. Jadi kemungkinan ini salah
• Kemungkinan 2 : S = 9 , C3 = 0 , Ø = 0
E + Ø = N, C2 + E + Ø = N , maka nilai C2 harus 1 karena
nilai E tidak boleh sama dengan N.
• Didapat persamaan 1 + E + 0 = N, maka N = E+1.
• Karena Nilai pasti tidak didapat, maka kita jalankan
langkah 1.
• N + R = E (hasil disini harus bernilai belasan agar C2 =
1),D + E = Y.
• Langkah 1 sudah dijalankan , tapi tidak ketemu nilai
variabel, karena itu kita
• jalankan langkah 4.
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A Criptarithmetic Problem
• Kita mulai dengan tebakan kalo E = 2, maka N = 3.
Pemilihan nilai E, karena variabel E paling banyak muncul
dan paling banyak berinteraksi dengan huruf lain. C1 + N +
R = (1)E , C1 + 3 + R = (1)2, berapa nilai C1 + R + 3 agar
didapat nilai 12? .
• Jawabannya : C1 = 1, R = 8. Lalu kita masukan ke
persamaan lanjutan, D + E = Y (Hasil disini harus belasan
agar didapat C1 = 1), D + 2 = Y, dari sini didapatkan
kesimpulan bahwa kemungkinan ini tidak mungkin, karena
sebesar apapun nilai D + 2 tidak akan mencapai nilai
belasan.
• Kita lanjutkan tebakan kalo E = 3, maka N = 4. C1 + N + R,
C1 + 4 + R = (1)3, berapa C1 + R + 4 agar didapat 13? R
= 8 dan C1 = 1.
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A Criptarithmetic Problem
• Kita masukan ke D + E = (1)Y, kemungkinan ini tidak
mungkin karena D + 3 hanya bernilai paling besar 10(D =
7), dan menyebabkan Y = 0. (Ø = 0)
• Sekarang jika E = 4 ,maka N = 5. C1 + N + R, C1 + 5 + R =
(1)4, berapa C1 + 5 + R agar didapat 14 ? R = 8 dan C1 =
1. Kita masukan ke D + E = (1)Y kemungkinan ini tidak
mungkin karena D(7) + 4 hanya bernilai paling besar 11,
dan menyebabkan Y = 1. (M = 1)
• Sekarang jika E = 5 ,maka N = 6. C1 + N + R, C1 + 6 + R =
(1)5, berapa C1 + 6 + R agar didapat 15 ? R = 8 dan C1 =
1.
• Kita masukan ke D + E = (1)Y kemungkinan ini mungkin
karena D(7) + 5 bernilai paling besar 12, dan
menyebabkan Y = 2. Didapatkan jawaban, maka kita
jalankan langkah 2.
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A Criptarithmetic Problem
Jawaban nilai-nilai variabel agar didapat hasil yang
sesuai dengan soal :
•
•
•
•
•
M=1
Ø=0
S = 9 —>
E = 5
N=6
•
•
•
•
•
•
•
R =8
D=7
Y=2
C1 = 1
C2 = 1
C3 = 0
C3 C2 C1
0 1 1
S E N D
9 5 6 7
M Ø R E + —>
1 0 8 5 +
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
M Ø N E Y
1 0 6 5 2
• Semoga Bisa Dimengerti
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3.6 Means-ends analysis
• Collection of search strategy that can reason
either forward or backward, but for given
problem, one direction or the other must be
chosen.
• A mixed strategy would make it possible to solve
the major part of problem first and then go back
and solve the small problem that arise “gluing”
the big pieces together.
• The mean-ends analysis process centers
around the detection of differences between the
current state and the goal state.
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The Robot ’s Operators
Operator
PUSH (obj,loc)
Preconditions
Results
at(robot,obj) 
at(obj,loc) 
large(obj) 
at(robot,loc)
clear(obj) 
armempty
at(robot,obj) 
at(obj,loc) 
small(obj)
at(robot,loc)
WALK(loc)
none
at(robot,loc)
PICKUP(obj)
at(robot,obj)
holding(obj)
PUTDOWN(obj)
holding(obj)
 holding(obj)
PLACE(obj1,obj2)
at(robot,obj2) 
On(obj1,obj2)
CARRY(obj,loc)
holding(obj1)
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A Difference Table
Push
Move object
Move robot
Clear object
*
Carry
Walk
Pick
up
Put Place
down
*
*
*
Get object on object
*
Get arm empty
Be holding object
*
*
*
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Means-ends analysis
The Progress of The Means-Ends Analysis Method
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Algorithm : Means-Ends Analysis
1. Compare CURRENT to GOAL.If there are no
differences between them, then return.
2. Otherwise, select the most important difference
and reduce it by doing the following until
success or failure is signaled :
a. Select an as yet untried operator O that is
applicable to the current difference. If there are no
such operators, then signal failure.
b. Attempt to apply O to CURRENT. Generate
descriptions of two states : O-START, a state in
which O ‘s preconditions are satisfied and ORESULT, the state that would result if O were
applied in O-START.
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Algorithm : Means-Ends Analysis
c.
If
(FIRST-PART  MEA(CURRENT,O-START))
And
(LAST-PART  MEA(O-RESULT,GOAL))
are succesful, the signal success and return
the result of concatenating FIRST-PART,O,
and LAST-PART.
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<< CLOSING>>
End of Pertemuan 4
Good Luck
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