Monkey and Bananas Exercise

1
Monkey and Bananas Exercise

• Notes on Exercise 3.10 of Bratko
• For CSCE 580 Sp03
• Marco Valtorta

2
Bratkos Solution

• let canget(State,Actions) be the relation that
  holds if the monkey can get from State to a state
  in which it has the bananas by carrying out the
  moves described in the list Actions.
• If the monkey already has the bananas, there is
  nothing that it needs to do.
• canget( state(_,_,_,has), ).
• Definition continues on the next slide

3
Bratkos Solution (ctd.)

• The monkey can get from State to a state in which
  it has the bananas by doing Action followed by
  Actions if
• (a) the monkey can move from State to NewState by
  doing Action, and
• (b) the monkey can move from NewState to a state
  in which it has the bananas by doing Actions.
• canget( State, ActionActions) -
• move( State, Action, NewState),
• canget( NewState, Actions).

4
Alternate Solution

• canget1(State, Actions, Path) if the monkey can
  get the bananas from State by doing the actions
  in the difference list Path Actions
• If the monkey is already in a state in which it
  can get the bananas, then there is nothing that
  it needs to do
• canget1( state(_,_,_,has), Actions, Actions).
• Definition continues on the next slide

5
Alternate Solution (Ctd.)

• The monkey can get the bananas from State1 by
  doing the Actions in Path - Actions if
• (a) the monkey can move to State2 by doing Move,
  and
• (b) the monkey can get the bananas from State2 by
  doing the actions in Path - (Actions Move)
• canget1( State1, Actions, Path) -
• move( State1, Move, State2),
• canget1( State2, Move Actions, Path).

6
Alternate Solution (Ctd.)

• There is a procedural reading to the alternate
  solution accumulate the solution path into the
  Actions list going down the goal tree, then save
  it at the bottom of the goal tree
• The solution Path is built backwards! We need
• canget( State, Actions) -
• canget1( State, , ReverseActions),
• rev(Actions, ReverseActions).
• Actions can be thought of as an accumulator.
• Accumulators and difference lists are good for
  efficiency, but in this exercise the simpler
  solution is faster too!