CS345 PROBLEM SET #2
SOLUTIONS
6 Questions, 100 pts.
Problem 1 (25 pts.)
In this problem we will use two kinds of electronic gates, NAND
gates and INVOR gates.
A NAND gate has inputs X and Y; its output Z is F(alse) if
and only if both X and Y are T(rue).
In this problem the only INVOR gate has inputs X and Y; its
output Z is F(alse) if and only if X is T(rue) and Y is F(alse).
We can represents a circuit of NAND and INVOR gates by the EDB predicate
wireT(Z), meaning that wire Z is a circuit input set to
T(rue), nandGate(X,Y,Z) meaning that there is a NAND gate in
the circuit with input wires X and Y and with output wire Z, and
invorGate(X,Y,Z) meaning that there is an INVOR gate in the
circuit
with input wires X and Y and with output wire Z. The rules describing
a circuit are then the following.
- t(Z) :- wireT(Z).
- t(Z) :- nandGate(X,Y,Z) & NOT t(X).
- t(Z) :- nandGate(X,Y,Z) & NOT t(Y).
- t(Z) :- invorGate(X,Y,Z) & NOT t(X).
- t(Z) :- invorGate(X,Y,Z) & t(Y).
Here is a circuit of NAND and INVOR gates represented by the EDB:
wireT(1), nandGate(6,1,3), nandGate(1,2,4), nandGate(3,5,6),
nandGate(1,4,7), and invorGate(3,1,5).
For this EDB and the logic program above it, find the well-founded
model by repeatedly making positive inferences and looking for
unfounded sets.
Solution
Using the first rule we infer t(1) from t(1) :- wireT(1)
.
Using the last rule we infer t(5) from t(5) :- invorGate(3,1,5)
& t(1)
. Then, we instantiate the rules in all possible ways
and eliminate those with a known false subgoal. For convenience, we
also eliminate true subgoals from the remaining bodies and rules whose
head has already been inferred. The result is
- t(3) :- NOT t(6).
- t(4) :- NOT t(2).
- t(6) :- NOT t(3).
- t(7) :- NOT t(4).
The largest unfounded set is {t(2)} so we infer NOT
t(2). Then we infer t(4) from t(4) :- NOT
t(2)
. After eliminating the last rule because NOT
t(4) is F(alse) the remaining rules are
- t(3) :- NOT t(6).
- t(6) :- NOT t(3).
The largest unfounded set is {t(7)} so we infer NOT
t(7). The remaining rules are the same and there are no
unfounded sets so we are done. The WF model is {t(1),NOT
t(2),t(4),t(5),NOT t(7)}. The value of t(3) and
t(6) is "unknown".
Problem 2 (15pts.)
For the same EDB and program as in Problem 1, find the well-founded
model by the alternating fixedpoint method.
Solution
The table below illustrates finding the well-founded model by the
alternating fixedpoint method.
Round |
0 |
1 |
2 |
3 |
4 |
t(1) |
F |
T |
T |
T |
T |
t(2) |
F |
F |
F |
F |
F |
t(3) |
F |
T |
F |
T |
F |
t(4) |
F |
T |
T |
T |
T |
t(5) |
F |
T |
T |
T |
T |
t(6) |
F |
T |
F |
T |
F |
t(7) |
F |
T |
F |
F |
F |
Note that column 4 is the same as column 2 so we don't need to do any
more iterations. The WF model is {t(1),NOT t(2),t(4),t(5),NOT
t(7)} because t(1),t(4), and t(5) converge to
T(rue), t(2) and t(7) converge to F(alse), and
t(3) and t(6) oscillate.
Problem 3 (20pts.)
For the EDB and program of Problem 1, find all the stable models. Is
there a stable semantics for this EDB and program?
Solution
There are only two stable models: {t(1),NOT t(2),NOT
t(3),t(4),t(5),t(6),NOT t(7)} and {t(1),NOT
t(2),t(3),t(4),t(5),NOT t(6),NOT t(7)}. There is no stable
semantics for the given EDB and program because there is no unique
minimal model.
Problem 4 (15pts.)
Suppose we change the input of terminal 1 to be F(alse); i.e., the
relation wireT is made empty instead of {1}. Show that this
EDB plus the program from Problem 1 is modularly stratified. What are
the "modules"?
Solution
To be supplied later
Problem 5 (10pts.)
Is the program and EDB of Problem 4 locally stratified? Justify your
answer.
Solution
The program and EDB of Problem 4 is not locally stratified because
there is a negative cycle between the ground atoms t(3) and
t(6). The cycle arises from the following two rule
instantiations.
- t(3) :- nandGate(6,1,3) & NOT t(6).
- t(6) :- nandGate(3,5,6) & NOT t(3).
Problem 6 (15pts.)
Compare the well-founded and stable approaches on the following rules
of propositional calculus.
- p :- NOT q.
- q :- NOT r.
- r :- NOT s.
- s :- NOT p.
Solution
The WF model for the above rules is {}, i.e., the value of
p,q,r, and s is "unknown". There is no stable
semantics for the above rules because there are two minimal stable
models {p,NOT q,r,NOT s} and {NOT p,q,NOT r,s}.