Assigned: Feb. 23
Due: Mar. 30
In this assignment, you are to implement the Davis-Putnam algorithm and use it to solve a simple form of an adventure game.
The puzzle is as follows: You are given (1) a maze with treasures and tolls at particular nodes; (2) a maximum number of steps. The object is to find a path through the maze from START to GOAL in at most the given number of steps.
Example: Suppose that the maze is as follows:
Then the path "START, D, E, G, F, GOAL" succeeds in 5 steps. The path "START, D, A, B, A, D, C,GOAL" succeeds in 7 steps.
Assume that there is only one instance of each treasure in the maze. Therefore, each treasure is in one of three states: Either at its original location, or in the possession of the player, or spent as a toll. There may be more than one node that requires a particular treasure as a toll; if so, it is only possible to visit one of them.
A node may have more than one treasure as toll or may supply more than one treasure.
You may assume that START and GOAL have neither tolls nor treasures.
You will write three programs.
There are 12 kinds of propositions.
1. The player is only at one place at a time.
For any time I, for any two distinct nodes M and N, ~(At(M,I) ^ At(N,I)).
In CNF this becomes ~At(M,I) V ~At(N,I).
For example, ~At(C,2) V ~At(F,2).
2. If the player has treasure T at time I then T is not available at time I.
Thus Has(T,I) => ~Available(T,I).
In CNF, ~Has(T,I) V ~Available(T,I).
For example, ~Has(Gold,1) V ~Available(Gold,1).
3. The player must move on edges. Suppose that node N is connected to
M1 ... Mk. For any time I, if the player is
at node N at time I, then the
player moves to M1 or to M2 ... at time I+1.
Thus At(N,I) => At(M1,I+1) V ... V At(Mk,I+1).
In CNF, ~At(N,I) V At(M1,I+1) V ... V At(Mk,I+1).
For example, ~At(C,2) V At(START,3) V At(D,3) V At(F,3)
Assume that there is a self-loop from GOAL to itself, so that once the player has reached GOAL he can stay there.
4. If node N has toll T and the player is at N at time I+1, then the player
must have T at time I.
Thus, At(N,I+1) => Has(T,I).
In CNF ~At(N,I+1) V Has(T,Ii).
For example, ~At(B,3) V Has(Wand,2)
5. If treasure T is initially at node N and is available at time I
and the player is at N at time I+1, then at time I+1 the player has T.
Thus, ~Available(T,I) V ~At(N,I+1) V Has(T,I+1).
In CNF, ~Available(T,I) V ~At(N,I+1) V Has(T,I+1).
For example, ~Available(Gold,1) ^ At(B,2) => Has(Gold,2).
6. If node N has toll T and the player is at N at time I, then the player
no longer has T at time I.
Thus At(N,I) => ~Has(T,I).
In CNF, ~At(N,I) V ~Has(T,I).
For instance ~At(C,3) V ~Has(Gold,3).
7. If treasure T is available at I, and the player is at node N which is
not the home of T at I+1, then T is available at I+1.
Thus Available(T,I) ^ At(N,I+1) => Available(N,I+1).
In CNF, ~Available(T,I) V ~At(N,I+1) V Available(N,I+1).
For example ~Available(Gold,2) V~ At(E,3) V Available(Gold,3).
8. If treasure T is not available at time I, then it is not available at
time I+1. Thus
~Available(T,I) => ~Available(T,I+1).
In CNF, Available(T,I) V ~Available(T,I+1).
For example, Available(Gold,3) V ~Available(Gold,4).
9. If the player has treasure T at time I and is at node N at time I+1, and
N does not require T as a toll, then the player still has T at I+1.
Thus Has(T,I) ^ At(N,I+1) => Has(T,I+1).
In CNF ~Has(T,I) V ~At(N,I+1) V Has(T,I+1).
For example ~Has(Gold,1) ^ At(D,2) V Has(Gold,2).
10. The player is at START at time 0. At(START,0).
11. The treasures are all available at time 0. For each treasure T, Available(T,0).
12. The player is at GOAL at time K. At(GOAL,K).
[In the theory of logic-based planning, axioms such as 1 and 2, which characterize what conditions can hold at a single time, are called "domain constraints". Axioms such as 3 and 4, which characterize what states can follow one another are called "feasibility axioms" or "precondition axioms"; the form of these here is a little unusual since we aren't explicitly representing actions. Axioms such as 5, 6, which characterize what changes between successive states are called "causal axioms". Axioms such as 7, 8, 9, which characterize what remains unchanged between successibe states are called "frame axioms". Axioms 10 and 11 are initial conditions and axiom 12 is a goal condition].
The output from the Davis-Putnam procedure has the following form: First, a list of pairs of atom (a natural number) and truth value (either T or F). Second, a line containing the single value 0. Third, the back matter from the input file, reproduced.
Example: Given the input
1 2 3 -2 3 -3 0 This is a simple example with 3 clauses and 3 atoms.Davis-Putnam will generate the output
1 T 2 F 3 F 0 This is a simple example with 3 clauses and 3 atoms.This corresponds to the clauses
P V Q V R. ~Q V R. ~R.
If the clauses have no solution, then Davis-Putnam outputs a single line containing a 0, followed by the back-matter in the input file.
Note: Your implementation of Davis-Putnam must work on any set of clauses, not just those that are generated by the Maze program.
You may assume that there are no more than 1000 atoms and no more than 10,000 clauses.
The format of the input contains the following elements:
START A B C D E F G GOAL GOLD WAND RUBY 5 START TREASURES TOLLS NEXT A C D A TREASURES TOLLS NEXT START B D B TREASURES GOLD TOLLS WAND NEXT A E C TREASURES TOLLS GOLD NEXT START D GOAL D TREASURES WAND TOLLS NEXT START A C E F E TREASURES TOLLS NEXT B D F G F TREASURES TOLLS RUBY NEXT D E G GOAL G TREASURES RUBY TOLLS WAND NEXT E F GOAL TREASURES TOLLS NEXT C F GOALYou may assume that the input is correctly formatted. You do not have to do any error checking on the input. The output consists of