CS533: Homework 1 Solutions:

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Exercise 11.4:

There are many possible answers. I generally accepted any answer that
did not violate the intended semantics of the domain. Here is one.

a) The initial state is:

   At(Monkey,A), At(Bananas,B), At(Box,C), Height(Monkey,Low),
   Height(Box,Low), Height(Bananas,High), Pushable(Box), Climbable(Box)

b) The actions are:

   Go(x,y)
     Pre: At(Monkey,x)
     Add: At(Monkey,y)
     Del: At(Monkey,x)

   Push(b,x,y)
     Pre: At(Monkey,x),Pushable(b)
     Add: At(b,y),At(Monkey,y)
     Del: At(b,x),At(Monkey,x)

   ClimbUp(b,x)
     Pre: At(Monkey,x),At(b,x),Climbable(b),Height(Monkey,Low)
     Add: On(Monkey,b),Height(Monkey,High)
     Del: Height(Monkey,Low)

   Grasp(b)
     Pre: Height(Monkey,High), Height(b,High)
     Add: Have(Monkey,b)
     Del:

   ClimbDown(b)
     Pre: On(Monkey,b), Height(Monkey,High)
     Add: Height(Monkey,Low)
     Del: On(Monkey,b), Height(Monkey,High)

   UnGrasp(b)
     Pre: Have(Monkey,b)
     Add: 
     Del: Have(Monkey,b)

c) In situation calculus, the goal is a state s such that:

        Have(Monkey,Bananas,s) & (exists x At(Box,x,s0) & At(Box,x,s))

   where s0 is the initial state situation. 

   In STRIPS, we can only talk about the goal state; there is no way of 
   representing the fact that there must be some relation (such as       
   equality  of locations of an object) between two states within the 
   plan. So there is no way to represent this goal via a standard
   STRIPS planning problem. 

d) Actually, we did include the Pushable precondition. This is an
   example of the qualification problem.

Exercise 11.10:

A forward state-space planner maintains a partial plan that is a
strict linear sequence of actions. Each search node corresponds to an
action sequence. The initial search node is the empty sequence. The
search, or refinement, operators add "applicable" actions to the end
of a sequence. An action is applicable to an action sequence if it can
be taken in the state that results from executing the action sequence
from the initial state. The goal nodes of the search space are any
nodes where the result of executing the action sequence from the
initial state satisfies the goal condition of the planning problem.

Similarly, a backward state-space planner maintains a partial plan
that is a reversed sequence of actions. The refinement operator is to
add an action to the beginning of the sequence as long as the actions
effects are compatible with the state at the beginning of the
sequence. The goal nodes are defined in the same way as for forward
state-space search.

Exercise 11.11:

The initial state is:

   On(B,Table), On(C,A), On(A,Table), Clear(B), Clear(C)

The goal is:
   
   On(A,B), On(B,C)

First we'll explain why it is an anomaly for a noninterleaved
planner. There are two subgoals. Suppose we decide to work on On(A,B)
first.  We can clear C off of A and then move A on to B. But then
there is no way to achieve On(B,C) without undoing the work we have
done. Similarly, if we work on the subgoal On(B,C) first we can
immediately achieve it in one step, but then we have to undo it to get
A on B.

Exercises 11.13

There are many possible answers. I generally accepted any answer that
did not violate the intended semantics of the domain. Here is one.

go(x,y)
  PRE: at(x),on(FLOOR),same-room(x,y)
  ADD: at(y)
  DEL: at(x)

push(b,x,y)
  PRE: at(x),on(FLOOR),box-at(b,x),same-room(x,y)
  ADD: at(y),box-at(b,y)
  DEL: at(x),box-at(b,x)

climb-up(b,x)
  PRE: at(x),box-at(b,x),on(FLOOR)
  ADD: on(b)
  DEL: on(FLOOR)

climb-down(b,x)
  PRE: on(b),at(x)
  ADD: on(FLOOR)
  DEL: on(b)

turn-on(s,b)
  PRE: on(b),box-at(b,s),switch(s)
  ADD: switch-on(s)
  DEL: 

turn-off(s,b)
  PRE: on(b),box-at(b,s),switch(s)
  ADD: 
  DEL: switch-on(s)

Note that I treated switches as locations.

To encode Figure 11.17 as a planning problem we will use the following
constants:

boxes: b1,b2,b3,b4
rooms: r1,r2,r3,r4  
switches: s1,s2,s3,s4
doors: d1,d2,d3,d4    

Doors, rooms, and switches will be viewed as locations.

We will have the following facts describing the initial state:

same-room(di,ri),same-room(di,ri)  for i=1,2,3,4
same-room(ri,si),same-room(si,ri)  for i=1,2,3,4
same-room(di,dj)  for i=1,2,3,4; j=1,2,3,4
box-at(bi,r1) for i=1,2,3,4
at(r3)

The plan for Shakey to get b2 to r2 is as follows:

go(r3,d3),go(d3,d1),go(d1,r1),push(b2,r1,d1),push(b2,d1,d2),push(b2,d2,r2)

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Exercise A:

   A standard propositional STRIPS planning problem (as described in
   class) specifies: 1) a goal condition, 2) an initial state, and 3)
   a set of propositional STRIPS actions with positive
   preconditions. We will refer to the set of all standard STRIPS
   problems as STANDARD.

   i) Let NEG-PRE be the class of problems that are specified as
      above, except that actions are allowed to have negative
      preconditions. (Note that NEG-PRE is a strict superset of
      STANDARD.) Show how to convert any problem p1 in NEG-PRE to a
      problem p2 in STANDARD such that p1 has a solution iff p2 has a
      solution. In other words, give a reduction from Plan-SAT for
      NEG-PRE to Plan-SAT for STANDARD. Prove your answer for full
      credit.

I gave 6 points for the correct construction and 4 points for the
correct proof. The correctness of the construction may seem obvious,
but it is important to be able to set up a formal argument. Sometimes
these obvious constructions have holes that only show up in the
proof (though this is not one of those cases).

Construction:

Given a problem P in NEG-PRE we will create a new problem P' in
STANDARD as follows:

Let I be the initial state of P, G be the goal, and A be the set of
propositional strips actions. Also let PROPS be the set of
propositions involved in P.

For each proposition p in P introduce a new proposition not-p. 

The initial state I' of P' is equal to 

      I' = I \union {not-p | p \in PROPS - I}

The goal G' of P' will just be G. 

For each action a in P we will add the following action a' to P':

   PRE(a') = {p | p is in PRE(a)} \union 
             {not-p | p is a negative precondtion of a}

   ADD(a') = ADD(a) \union {not-p | p in DEL(a)}

   DEL(a') = DEL(a) \union {not-p | p in ADD(a)}

These action definition will effectively maintain the not-p
propositions so that they are part of the state whenever p is false. 

Proof of correctness:

Let P be a problem in NET-PRE and P' be the corresponding standard
problem as obtained from the above construction.

We want to show that P is satisfiable iff P' is satisfiable.

For action a in P we will denote the corresponding action in P' as
a'. 

Also for any sates s (i.e. a set of positive propositions) let 
s'=s \union {not-p | p \in PROPS - s}. That is s' explicitly lists the
negative facts that are not in s'. 

The trickiest part of the proof is to select a good inductive
hypothesis. After that, it is straightforward. 

To form the inductive hypothesis we introduce the concept of
state-action sequence. A state action sequence is simply an
alternating sequence of states and actions. We say that a
state-action sequence s1,a1,s2,a2,...,a_{n-1},sn  is "legal" iff for
each i, the preconditions of ai are satisfied in si and s_{i+1} is the
result of applying ai in si. 

Given a state-action sequence S involving actions and proposition in P
we define S' to be the corresponding action sequence that results by
replacing each action a by a' and each state s by s'. So S' is a
state-action sequence involving actions and proposition in P'. We will
prove the following statement about state-action sequences:

   For any state-action sequence S involving actions and proposition
   in P, if S is a legal sequence then so is S'. 

   Proof: Use induction on the number of actions n in the
          state-action sequence.

          Base Case: n=0 is trivially true

          Inductive Case: Assume that the statement holds for all
          state-action sequences with fewer than n actions. Show that
          it holds for sequences with n actions. 

          Let A=s1,a1,...,a_n,s_{n+1}. We have the following,

          s1,a1,...,a_n,s_{n+1} is legal   
 
                     iff
           
          both 1) s1,a1,...,a_{n-1},s_n is legal, and 
               2) a_n is applicable in s_n and results in state s_{n+1}

                     iff

          both 1) (s1)',(a1)',...,(a_{n_1})',(s_n)' is legal 
                  [by the inductive hypothesis], and 
               2) (a_n)' is applicable in (s_n)' and yields (s_{n+1})'
                  [this is a simple case analysis that follow from the
                   way we have defined (s_n)' and (s_{n+1})']
 
                     iff

          (s1)',(a1)',...,(a_n)',(s_{n+1})' is legal
                
          That completes the proof.

From this fact we can prove what we want.

   P is satisfiable   

          iff   

   P has a legal state-action sequence that begins with I and ends
   with s such that G is a subset of s. 

          iff

   P' has a legal state-action sequence that begins with I' and ends
   with a state s' where G is a subset of s'

          iff

   P' is satisfiable

   The last fact follows because s' is by definition a superset of s.

That completes the proof.

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   ii) Let DISJ-GOAL be the class of problems that are standard STRIPS
       problems only they have disjunctive goals. A disjunctive goal
       is a disjunction of standard goals, where a standard goal is a
       set of propositions. For example, the following is a
       disjunctive goal:

            {on(a,b),on(b,c)} or {on(c,b),on(b,a)}

      We say that a problem in DISJ-GOAL is solvable iff there is an
      action sequence from the initial state to some state that
      satisfies at least one of the goal disjuncts. So in our example
      above we must reach a state with either the tower (a,b,c) or
      (c,b,a).

      Repeat part (a) for DISJ-GOAL.

Construction:

Intuitively we will add a new action for each disjunct of the
goal. Each of these actions will have the corresponding goal as a
precondition and will add a new proposition GOAL. The goal of our new
problem is simply GOAL.

Given a problem P in DISJ-GOAL we will create a new problem P' in
STANDARD as follows:

Let I be the initial state of P, G=g1 or g2 or ... or gn be the goal, and
A be the set of propositional strips actions.

Create a new proposition GOAL that we will use to define P'.
Define the action ACTi as follows,

ACTi
  PRE: gi
  ADD: GOAL
  DEL

The initial state I' of P' is equal to I 
The goal of P' is GOAL
The action set of P' is A \union {ACTi | i=1,...,n}

We now prove that P has a solution iff P' has a solution. 

Proof:
 
To prove the forward direction, assume that P has a solution. This
means that there is a sequence of actions a1,...,an from A that lead
from I to a state s that satisfies one of the gi. Consider the action
sequence a1,...,an,ACTi where i is the index of the satisfied goal. It
is clear that ACTi can be executed in s and that it leads to a goal
state of P'. Thus P' has a solution.

To prove the backward direction, assume that P is not solvable, and
thus that there is no action sequence involving actions in A that
achieve one of the gi. For the sake of contradition assume that P' is
solvable. This means that the is some action sequence that achieves
GOAL. But his can only be done by executing one of the ACTi actions,
which depends on having one of the gi achievable by actions in A. Thus
we get a contradiction.