Programming Assignment 2
Due: Oct. 26
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 at particular
nodes; (2) a list of treasures to get; (3) a maximum number of steps.
The object is to find a path through the maze that collects all these treasures
in at most the given number of steps.
Example: Suppose that the maze is as follows:
If the task is "GOLD, RUBY" in 2 steps, then a solution is
"START, C, F". If the task is "GOLD, WAND, RUBY" in 4 steps,
then a solution is "START,C,D,C,F". If the task is "GOLD,RING"
in two steps, then there is no solution.
You will write three programs.
- An implementation of the Davis-Putnam procedure, which takes
as input a set of clauses and outputs either a satisfying valuation,
or a statement that the clauses cannot be satisfied.
- A front end, which takes as input a maze
problem and outputs a set of clauses that can be input to (1).
- A back end, which takes as input the output of (1) and
translates it into a solution to the original problem.
Compiling the maze puzzle to propositional logic
The MAZE puzzle can be expressed propositionally as follows:
There are two kinds of atoms:
There are 8 kinds of propositions.
- AT_N_I means that the player is on node N at time I. For instance
AT_C_2 means that the player is on node C at time 2. If there is
a maximum of K steps, then there should be one such atom for each node N
and each time I=0 .. K.
- HAS_T_I means that the player has gotten treasure T at time I.
For example, HAS_GOLD_3 means that the player has gotten the gold by
step 3. If there is a maximum of K steps, there should be 1 such atom
HAS_T_I for each treasure T and each I=0 ... K.
- 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), or in CNF
~AT_M_I V ~AT_N_I.
For example, ~AT_C_2 V ~AT_F_2.
- 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 either remains at N or moves to M1 or to M2 ... at time I+1.
Thus AT_N_I => AT_N_(I+1) V AT_M1_(I+1) V ... V AT_Mk_(I+1);
~AT_N_I V AT_N_(I+1) V AT_M1_(I+1) V ... V AT_Mk_(I+1).
For example, ~AT_C_2 V AT_C_3 V AT_START_3 V AT_D_3 V AT_F_3
- Suppose that treasure T is at node N.
If the player is on node N at time I, then
the player has T at time I. AT_N_I => HAS_T_I, or in CNF
~AT_N_I V HAS_T_I.
For instance, ~AT_F_3 V HAS_RUBY_3.
- If the player has treasure T at time I, then he has treasure T
at time I+1, then he has treasure T at time I+1. HAS_T_I => HAS_T_(I+1),
or in CNF, ~HAS_T_I V HAS_T_(I+1). For example, ~HAS_GOLD_3 V HAS_GOLD_4.
- Suppose that treasure T is located at nodes N1, N2 ... Nk.
If the player does not have treasure T at time I-1, and does have treasure
T at time I, then at time I he must be either at N1 or at N2 ... or at Nk.
[~HAS_T_(I-1)] ^ HAS_T_(I+1)] => AT_N1_I V AT_N2_I V ... V AT_Nk_T
HAS_T_(I-1)] V ~HAS_T_(I+1) V AT_N1_I V AT_N2_I V ... V AT_Nk_T
For example, HAS_GOLD_2 V ~HAS_GOLD_3 V AT_B_3 V AT_C_3.
The player is at START at time 0. AT_START_0.
- If T is a treasure that is not at START, then the player does not
have T at time 0.
If the treasures not at START are T1 ... Tk
then we have ~HAS_T1_0 ^ ~HAS_T2_0 ^ ... ^ ~HAS_Tk_0;
In CNF this is the separate clauses:
For instance, in the example problem we have
- At the time limit, the player has all the required treasures. For
instance, in the first problem above, we have
Input / Output.
All three programs take their input from a text file
produce their output to a text file. (If you want, you may use
standard input and output.)
The input to the Davis-Putnam procedure has the following form:
An atom is denoted by a natural number: 1,2,3 ... The literal P
is the same number as atom P; the literal ~P is the negative.
A clause is a line of text containing the integers of the corresponding
literals. After all the clauses have been given, the next line is the
single value 0; anything further in the file is ignored in the
execution of the procedure and reproduced at the end of the output file.
(This is the mechanism we will use to allow the front end to communicate
to the back end.)
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
This is a simple example with 3 clauses and 3 atoms.
Davis-Putnam will generate the output
This is a simple example with 3 clauses and 3 atoms.
This corresponds to the clauses
P V Q V R.
~Q V 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 EXACT SET COVER program.
You may assume that there are no more than 1000 atoms and no more than 10,000
The front end takes as input a specification of a maze and a problem
and generates as
output a set of clauses to be satisfied.
The format of the input contains the following elements:
Thus, the first problem above would be encoded in the following input:
- First line: A list of the nodes, separated by white space.
Each node is a string of up to five characters.
- Second line: A list of the treasures, separated by white space.
Each treasure is a string of up to ten characters.
- Third line; The number of allowed steps and the list of the treasures to be
gotten, separated by white space.
- Remaining lines: The encoding of the maze. Each line consists
of the following parts
- A node N.
- The treasure at node N, or "NONE" if none.
- The nodes connected to N, separated by white space.
START A B C D E F G H
GOLD WAND RUBY RING
2 GOLD RUBY
START NONE A C
A NONE START B D
B GOLD A D E
C GOLD START D F
D WAND A B C E
E GOLD B D G H
F RUBY C G
G NONE E F H
H RING E G
You may assume that the input
is correctly formatted. You do not have to do any error checking on the
The output consists of
- 1. A set of clauses suitable for inputting to Davis-Putnam as described
above. Note that these constraints are already in
clausal form (CNF) and therefore you do not have to implement
a program to translate arbitrary Boolean formulas to CNF.
- 2. A key to allow the back end to translate the numbers used for
propositional atoms in
the clauses into the correct path.
The format of this is up to you.
My suggestion would be that, for each atom AT_N_I, you have a line of
the form "proposition-number N I"
The back end takes as input the output that Davis-Putnam generates when run
on the output of the front end. It generates as output the path
that solves the problem.
If the input indicates that the clauses have no solution, the back end
should output the message "NO SOLUTION".
Another example for Davis-Putnam
The following is the input-output pair just for the Davis-Putnam module ---
not the front and back ends --- corresponding to the example
in the class notes.
A Simpler Maze Example
It would be a mistake to begin your testing using
the above example, with its 39 propositional atoms and hundreds of
propositions. Rather, I would suggest that you start by working on
the following maze, with the problems, "Get the GOLD in 1"; "Get the
GOLD and the WAND in 2" and
"Get the GOLD and the RUBY in 3".
The clauses for the first of these is shown here
Obvious Optimizations for Front End
There are a couple of easy optimizations that can
be made in the front end
which eliminate a substantal number of the atoms. For instance,
or one can eliminate all the treasures that are not to be collected and
their associated atoms
since they obviously don't make any difference to the solution.
Or you can eliminate all the nodes that are not reachable
from START in K steps (or, even better, eliminate all the atoms
AT_N_I where node N is not reachable from START in I steps.)
As we'll discuss in class, that is not an optimization
that Davis-Putnam can find. Whether you want to implement these is up to
you; it is quite likely that they will make the system run noticeably faster.
You should submit by email
(a) the source code; (b) instructions for running it,
if there's anything at all non-obvious about it. Nothing else.
The Davis-Putnam program is worth 60% of the grade; the front end is worth
35%; the back end is worth 5%. In each of these, a program that does
not compile will get a maximum of 10%; a correct program
will get 90%; the remaining 10% is for being well-written and