TU Wien:Einführung in wissensbasierte Systeme VU (Egly)/Übungen SS12/Blatt 3 - Beispiel 4

You have the following information:

1. When there is a cash transport, usually there are valuable things to steal.
2. There is a cash transport.
3. When there are valuable things to steal, usually criminals show up.
4. When there are valuable things to steal, usually the police shows up.
5. When there are valuable things to steal and the police shows up, criminals definitely do not show up.

• (a) Formalise this situation in terms of a default theory. It should have an extension indicating that criminals show up, as well as an extension that states that the police is there. Use
  • cash for "there is a cash transport",
  • valuable for "there are valuable things to steal",
  • criminals for "criminals show up", and
  • police for "the police shows up".

• (b) Replace information 5. from above by a formalisation of

5'. If there are valuable things to steal and the police shows up, then criminals usually do not come.

How does the default theory change? Which extensions emerge after this replacement? Are all of them plausible?

Lösungsvorschlag

a

1. $\delta _{1}={\frac {{\text{cash}}:{\text{valuable}}}{\text{valuable}}}$
   
   
2. ${\text{cash}}\,$
   
   
3. $\delta _{2}={\frac {{\text{valuable}}:{\text{criminals}}}{\text{criminals}}}$
   
   
4. $\delta _{3}={\frac {{\text{valuable}}:{\text{police}}}{\text{police}}}$
   
   
5. $({\text{valuable}}\land {\text{police}})\rightarrow \neg {\text{criminals}}$
   
   

${\begin{aligned}T&=(W,\Delta )\\W&=\{{\text{cash}},({\text{valuable}}\land {\text{police}})\rightarrow \neg {\text{criminals}}\}\\\Delta &=\{\delta _{1},\delta _{2},\delta _{3}\}\end{aligned}}$

$E_{1}={\text{Cn}}(W\cup \{{\text{valuable}},{\text{criminals}}\})$

$E_{2}={\text{Cn}}(W\cup \{{\text{valuable}},{\text{police}}\})$

b

5'. $\delta _{4}={\frac {{\text{valuable}},{\text{police}}:\neg {\text{criminals}}}{\neg {\text{criminals}}}}$

${\begin{aligned}T&=(W,\Delta )\\W&=\{{\text{cash}}\}\\\Delta &=\{\delta _{1},\delta _{2},\delta _{3}\,\delta _{4}\}\end{aligned}}$

Mit $\delta _{2}$ ${\text{criminals}}$ inferieren, blockiert $\delta _{4}$, daher gibt es zwei Extensions.

$E_{1}={\text{Cn}}(W\cup \{{\text{valuable}},{\text{criminals}},{\text{police}}\})$

$E_{2}={\text{Cn}}(W\cup \{{\text{valuable}},{\text{police}},\neg {\text{criminals}}\})$