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Bootaufgabe

Zustandsmodellierung u. Ubergangsfunktion

Zustandsmodellierung und Nachfolgerfunktion

Einfuhrung in die Methoden der Kunstlichen Intelligenz - 1 -

Zustand = (# E links, # K links , Boot (l / r))

Startzustand = (2,2,l)

Zielzustand = (0,0,r)

alternativ:

Zustand = ({Personen links},{Personen rechts},Boot (l/r)),

z.B. ({K, K, E, E}, ∅, r)



nf((0,0, l)) = ∅nf((0,1, l)) = {(0,0, r)}nf((0,2, l)) = {(0,1, r), (0,0, r)}nf((1,0, l)) = {(0,0, r)}nf((1,1, l)) = {(1,0, r), (0,1, r)}nf((1,2, l)) = {(0,2, r), (1,1, r), (1,0, r)}nf((2,0, l)) = {(1,0, r)}nf((2,1, l)) = {(1,1, r), (2,0, r)}nf((2,2, l)) = {(1,2, r), (2,1, r), (2,0, r)}



nf((0,0, r)) = {(1,0, l), (0,1, l), (0,2, l)}nf((0,1, r)) = {(1,1, l), (0,2, l)}nf((0,2, r)) = {(1,2, r)}nf((1,0, r)) = {(2,0, l), (1,1, l), (1,2, l)}nf((1,1, r)) = {(2,1, l), (1,2, l)}nf((1,2, r)) = {(2,2, r)}nf((2,0, r)) = {(2,1, l), (2,2, l)}nf((2,1, r)) = {(2,2, l)}nf((2,2, r)) = ∅

Breitensuche

ohne Zyklenerkennung

Breitensuche ohne Zyklenerkennung


(K, K, E, E | − − | l)44

ttjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj jj

**TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTOO

��

(K, K, E |E | r) (E, E |K, K | r)OO

��

(K, E, E |K | r)

(K, E, E |K | l)OO

��

(K, E, E |K, E | r)OO

��. . .



L = (K, K, E, E | − − | l), enthalt keinen Zielknoten

(K, K, E, E | − − | l)44



��

(K, K, E |E | r) (E, E |K, K | r)OO

��

(K, E, E |K | r)

(K, E, E |K | l)OO

��

(K, E, E |K, E | r)OO

��. . .



N(L) Menge der direkten Nachfolger

(K, K, E, E | − − | l)44



��

(K, K, E |E | r) (E, E |K, K | r)OO

��

(K, E, E |K | r)

(K, E, E |K | l)OO

��

(K, E, E |K, E | r)OO

��. . .



L := N(L) , L enthalt keinen Zielknoten

(K, K, E, E | − − | l)44



��

(K, K, E |E | r) (E, E |K, K | r)OO

��

(K, E, E |K | r)

(K, E, E |K | l)OO

��

(K, E, E |K, E | r)OO

��. . .



N(L) Menge der direkten Nachfolger von L.

(K, K, E, E | − − | l)44



��

(K, K, E |E | r) (E, E |K, K | r)OO

��

(K, E, E |K | r)

(K, E, E |K | l)OO

��

(K, E, E |K, E | r)OO

��. . .




(K, K, E, E | − − | l)44



��

(K, K, E |E | r) (E, E |K, K | r)OO

��

(K, E, E |K | r)

(K, E, E |K | l)OO

��

(K, E, E |K, E | r)OO

��. . .



N(L) Menge der direkten Nachfolger von L.

(K, K, E, E | − − | l)44



��

(K, K, E |E | r) (E, E |K, K | r)OO

��

(K, E, E |K | r)

(K, E, E |K | l)OO

��

(K, E, E |K, E | r)OO

��. . .




(K, K, E, E | − − | l)44



��

(K, K, E |E | r) (E, E |K, K | r)OO

��

(K, E, E |K | r)

(K, E, E |K | l)OO

��

(K, E, E |K, E | r)OO

��. . .

Aufgabe 2

Teil a)

Tiefensuche


(A1)

ssgggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg@@

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(B1)

||yyyy

yyyy

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�� &&MMMMMMMMMMMMMMMMMMMMMMMMMMMMMM(B2) . . . (B∞)

(C11)

��

(C12) . . . (C1∞)

(D111) . . .

Tiefensuche


(A1)


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(B1)

||yyyy

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(C11)

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(C12) . . . (C1∞)

(D111) . . .

Tiefensuche


⇒(B2)wird nie erreicht!

(A1)


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(B1)

||yyyy

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(C11)

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(C12) . . . (C1∞)

(D111) . . .

Breitensuche


(A1)


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(B1)

||yyyy

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(C11)

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(C12) . . . (C1∞)

(D111) . . .

Breitensuche


⇒(C11)wird nie erreicht!

(A1)


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(B1)

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(C11)

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(C12) . . . (C1∞)

(D111) . . .

Aufgabe 2

Teil b

Vollstandige Suchstrategie


Wir nummerieren die Knoten mit Sequenzen s(N)durch:

s(Wurzel) = [1]

Sei Ni das i-te Kind von Knoten N , dann

s(Ni) = s(N)++[i]



[1]

ttjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

�� **UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU

''OOOOOOOOOOOOOOOOOOOOOOOOOOOOO

[1,1]

��

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��3

3333

3333

3333

333

xxqqqqqqqqqqqqqqqqqqqqqqqq

""FFFFFFFFFFFFFFFFFFF[1,2]

��

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��3

3333

3333

3333

333

""FFFFFFFFFFFFFFFFFFF[1,3] . . . [1,∞]

[1,1,1]

��~~}}}}

}}}}

}}}}

}}}}

}[1,1,2] [1,1,3] . . . [1,1,∞] [1,2,1] [1,2,2] . . . [1,2,∞]

[1,1,1,1] [1,1,1,2] . . .



k := 1

while true do

L := Menge der Knoten mit Sequenzsumme k;

Wenn L einen Zielknoten enthalt,dann gebe Knoten aus und stoppe;

k := k + 1



k=1

[1]


�� **TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT


[1,1]

��

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��3

3333

3333

3333

333


""FFFFFFFFFFFFFFFFFFF[1,2]

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��3

3333

3333

3333

333

""FFFFFFFFFFFFFFFFFFF[1,3] . . . [1,∞]

[1,1,1]

��~~}}}}

}}}}

}}}}

}}}}

}[1,1,2] [1,1,3] . . . [1,1,∞] [1,2,1] [1,2,2] . . . [1,2,∞]

[1,1,1,1] [1,1,1,2] . . .



k=2

[1]




[1,1]

��

��

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��2

2222

2222

2222

22


""EEEE

EEEE

EEEE

EEEE

EEE

[1,2]

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2222

2222

2222

222

""EEEEEEEEEEEEEEEEEEE[1,3] . . . [1,∞]

[1,1,1]

��~~}}}}

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}[1,1,2] [1,1,3] . . . [1,1,∞] [1,2,1] [1,2,2] . . . [1,2,∞]

[1,1,1,1] [1,1,1,2] . . .



k=3

[1]




[1,1]

��

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��2

2222

2222

2222

222


""EEEEEEEEEEEEEEEEEEE[1,2]

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2222

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2222

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""EEEE

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EEEE

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[1,3] . . . [1,∞]

[1,1,1]

��~~}}}}

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}[1,1,2] [1,1,3] . . . [1,1,∞] [1,2,1] [1,2,2] . . . [1,2,∞]

[1,1,1,1] [1,1,1,2] . . .



k=4

[1]



''OOOOOOOOOOOOOOOOOOOOOOOOOOOO

[1,1]

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��2

2222

2222

2222

222

xxqqqqqqqqqqqqqqqqqqqqqqqqq

""EEEEEEEEEEEEEEEEEEE[1,2]

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2222

2222

2222

222

""EEEEEEEEEEEEEEEEEEE[1,3] . . . [1,∞]

[1,1,1]

��~~~~~~

~~~~

~~~~

~~~~

~[1,1,2] [1,1,3] . . . [1,1,∞] [1,2,1] [1,2,2] . . . [1,2,∞]

[1,1,1,1] [1,1,1,2] . . .



Korrektheit:

• Pro Iteration sind nur endlich viele Knoten zutesten.

• Wenn N ein Zielknoten ist, dann seis(N) = [j1, . . . , jn], wobei j, n ∈ N.

• In (n∑

i=1ji)-ter Iteration wird Knoten N getestet.

• Sogar der erfolgreiche Pfad lasst sich anhand vons(N) bestimmen.

Blatt 3

A-Stern


A1

''PPPPPPPPPPPPPPPPPPP

S

377oooooooooooooooooooo

2 ''OOOOOOOOOOOOOOOOOOOO C 1 // G

B4

77nnnnnnnnnnnnnnnnnnn

A-Stern


h(X) = 0, fur alle x ∈ {S, A, B, C, G}h is unterschatzendh ist monoton (h(N) ≤ g(N, N ′) + h(N ′)fur alle N mit Nachfolger N ′)

A1


S



B4


A-Stern


h(X) = 0, fur alle x ∈ {S, A, B, C, G}

Open = {(S,0, [])}Closed = ∅N = S

A1


S



B4


A-Stern



Open = {}Closed = {(S,0, [])}N = S

A1


S



B4


A-Stern



Open = {(A,3, [S]), (B,2, [S])}Closed = {(S,0, [])}N = S

A1


S



B4


A-Stern



Open = {(A,3, [S]), (B,2, [S])}Closed = {(S,0, [])}N := min{g(N) + h(N)|N ∈ Open} = B

A1


S



B4


A-Stern



Open = {(A,3, [S])}Closed = {(S,0, []), (B,2, [S])}N = B

A1


S



B4


A-Stern



Open = {(A,3, [S]), (C,6, [B, S])}Closed = {(S,0, []), (B,2, [S])}N = B

A1

''OOOOOOOOOOOOOOOOOOOO

S



B4


A-Stern



Open = {(A,3, [S]), (C,6, [B, S])}Closed = {(S,0, []), (B,2, [S])}N := min{g(N) + h(N)|N ∈ Open} = A

A1

''PPPPPPPPPPPPPPPPPPPP

S



B4


A-Stern



Open = {(C,6, [B, S])}Closed = {(S,0, []), (B,2, [S]), (A,3, [S])}N = A

A1


S



B4


A-Stern



Open = {(C,4, [A, S])}Closed = {(S,0, []), (B,2, [S]), (A,3, [S])}N = A

A1


S



B4


A-Stern



Open = {(C,4, [A, S])}Closed = {(S,0, []), (B,2, [S]), (A,3, [S])}N = min{g(N) + h(N)|N ∈ Open} = C

A1


S



B4


A-Stern



Open = {}Closed = {(S,0, []), (B,2, [S]), (A,3, [S]), (C,4, [A, S])}N = C

A1


S



B4


A-Stern



Open = {(G,5, [C, A, S])}Closed = {(S,0, []), (B,2, [S]), (A,3, [S]), (C,4, [A, S])}N = C

A1


S



B4


A-Stern



Open = {(G,5, [C, A, S])}Closed = {(S,0, []), (B,2, [S]), (A,3, [S]), (C,4, [A, S])}N = min{g(N) + h(N)|N ∈ Open} = G

A1


S



B4


A-Stern


Fertig da G Zielknoten

Open = {(G,5, [C, A, S])}Closed = {(S,0, []), (B,2, [S]), (A,3, [S]), (C,4, [A, S])}N = G

A1


S

377nnnnnnnnnnnnnnnnnnnn


B4


A-Stern


Mit besserer Heuristik kann man dasBesuchen von B sparen!

A1


S

377nnnnnnnnnnnnnnnnnnnn


B4


Springpuzzle


H H V V V HMogliche Zuge?

H H V V V H

H H V V V H

H H V V H V

H H V V V H

H V V H V H

Bootaufgabe - ki.informatik.uni-frankfurt.de€¦ · Zustandsmodellierung und Nachfolgerfunktion...

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