This Jupyter notebook acts as supporting material for topics covered in Chapter 6 Logical Agents, Chapter 7 First-Order Logic and Chapter 8 Inference in First-Order Logic of the book Artificial Intelligence: A Modern Approach. We make use of the implementations in the logic.py module. See the intro notebook for instructions.
Let's first import everything from the logic
module.
In [1]:
from utils import *
from logic import *
from notebook import psource
The Expr
class is designed to represent any kind of mathematical expression. The simplest type of Expr
is a symbol, which can be defined with the function Symbol
:
In [2]:
Symbol('x')
Out[2]:
Or we can define multiple symbols at the same time with the function symbols
:
In [3]:
(x, y, P, Q, f) = symbols('x, y, P, Q, f')
We can combine Expr
s with the regular Python infix and prefix operators. Here's how we would form the logical sentence "P and not Q":
In [4]:
P & ~Q
Out[4]:
This works because the Expr
class overloads the &
operator with this definition:
def __and__(self, other): return Expr('&', self, other)
and does similar overloads for the other operators. An Expr
has two fields: op
for the operator, which is always a string, and args
for the arguments, which is a tuple of 0 or more expressions. By "expression," I mean either an instance of Expr
, or a number. Let's take a look at the fields for some Expr
examples:
In [5]:
sentence = P & ~Q
sentence.op
Out[5]:
In [6]:
sentence.args
Out[6]:
In [7]:
P.op
Out[7]:
In [8]:
P.args
Out[8]:
In [9]:
Pxy = P(x, y)
Pxy.op
Out[9]:
In [10]:
Pxy.args
Out[10]:
It is important to note that the Expr
class does not define the logic of Propositional Logic sentences; it just gives you a way to represent expressions. Think of an Expr
as an abstract syntax tree. Each of the args
in an Expr
can be either a symbol, a number, or a nested Expr
. We can nest these trees to any depth. Here is a deply nested Expr
:
In [11]:
3 * f(x, y) + P(y) / 2 + 1
Out[11]:
Here is a table of the operators that can be used to form sentences. Note that we have a problem: we want to use Python operators to make sentences, so that our programs (and our interactive sessions like the one here) will show simple code. But Python does not allow implication arrows as operators, so for now we have to use a more verbose notation that Python does allow: |'==>'|
instead of just ==>
. Alternately, you can always use the more verbose Expr
constructor forms:
Operation | Book | Python Infix Input | Python Output | Python Expr Input |
---|---|---|---|---|
Negation | ¬ P | ~P |
~P |
Expr('~', P) |
And | P ∧ Q | P & Q |
P & Q |
Expr('&', P, Q) |
Or | P ∨ Q | P | Q |
P | Q |
Expr(' |', P, Q) |
Inequality (Xor) | P ≠ Q | P ^ Q |
P ^ Q |
Expr('^', P, Q) |
Implication | P → Q | P |'==>' | Q |
P ==> Q |
Expr('==>', P, Q) |
Reverse Implication | Q ← P | Q |'<==' | P |
Q <== P |
Expr('<==', Q, P) |
Equivalence | P ↔ Q | P |'<=>' | Q |
P <=> Q |
Expr('<=>', P, Q) |
Here's an example of defining a sentence with an implication arrow:
In [12]:
~(P & Q) |'==>'| (~P | ~Q)
Out[12]:
In [13]:
expr('~(P & Q) ==> (~P | ~Q)')
Out[13]:
expr
takes a string as input, and parses it into an Expr
. The string can contain arrow operators: ==>
, <==
, or <=>
, which are handled as if they were regular Python infix operators. And expr
automatically defines any symbols, so you don't need to pre-define them:
In [14]:
expr('sqrt(b ** 2 - 4 * a * c)')
Out[14]:
For now that's all you need to know about expr
. If you are interested, we explain the messy details of how expr
is implemented and how |'==>'|
is handled in the appendix.
PropKB
The class PropKB
can be used to represent a knowledge base of propositional logic sentences.
We see that the class KB
has four methods, apart from __init__
. A point to note here: the ask
method simply calls the ask_generator
method. Thus, this one has already been implemented, and what you'll have to actually implement when you create your own knowledge base class (though you'll probably never need to, considering the ones we've created for you) will be the ask_generator
function and not the ask
function itself.
The class PropKB
now.
__init__(self, sentence=None)
: The constructor __init__
creates a single field clauses
which will be a list of all the sentences of the knowledge base. Note that each one of these sentences will be a 'clause' i.e. a sentence which is made up of only literals and or
s.tell(self, sentence)
: When you want to add a sentence to the KB, you use the tell
method. This method takes a sentence, converts it to its CNF, extracts all the clauses, and adds all these clauses to the clauses
field. So, you need not worry about tell
ing only clauses to the knowledge base. You can tell
the knowledge base a sentence in any form that you wish; converting it to CNF and adding the resulting clauses will be handled by the tell
method.ask_generator(self, query)
: The ask_generator
function is used by the ask
function. It calls the tt_entails
function, which in turn returns True
if the knowledge base entails query and False
otherwise. The ask_generator
itself returns an empty dict {}
if the knowledge base entails query and None
otherwise. This might seem a little bit weird to you. After all, it makes more sense just to return a True
or a False
instead of the {}
or None
But this is done to maintain consistency with the way things are in First-Order Logic, where an ask_generator
function is supposed to return all the substitutions that make the query true. Hence the dict, to return all these substitutions. I will be mostly be using the ask
function which returns a {}
or a False
, but if you don't like this, you can always use the ask_if_true
function which returns a True
or a False
.retract(self, sentence)
: This function removes all the clauses of the sentence given, from the knowledge base. Like the tell
function, you don't have to pass clauses to remove them from the knowledge base; any sentence will do fine. The function will take care of converting that sentence to clauses and then remove those.
In [15]:
wumpus_kb = PropKB()
We define the symbols we use in our clauses.
$P_{x, y}$ is true if there is a pit in [x, y]
.
$B_{x, y}$ is true if the agent senses breeze in [x, y]
.
In [16]:
P11, P12, P21, P22, P31, B11, B21 = expr('P11, P12, P21, P22, P31, B11, B21')
Now we tell sentences based on section 7.4.3
.
There is no pit in [1,1]
.
In [17]:
wumpus_kb.tell(~P11)
A square is breezy if and only if there is a pit in a neighboring square. This has to be stated for each square but for now, we include just the relevant squares.
In [18]:
wumpus_kb.tell(B11 | '<=>' | ((P12 | P21)))
wumpus_kb.tell(B21 | '<=>' | ((P11 | P22 | P31)))
Now we include the breeze percepts for the first two squares leading up to the situation in Figure 7.3(b)
In [19]:
wumpus_kb.tell(~B11)
wumpus_kb.tell(B21)
We can check the clauses stored in a KB
by accessing its clauses
variable
In [20]:
wumpus_kb.clauses
Out[20]:
We see that the equivalence $B_{1, 1} \iff (P_{1, 2} \lor P_{2, 1})$ was automatically converted to two implications which were inturn converted to CNF which is stored in the KB
.
$B_{1, 1} \iff (P_{1, 2} \lor P_{2, 1})$ was split into $B_{1, 1} \implies (P_{1, 2} \lor P_{2, 1})$ and $B_{1, 1} \Longleftarrow (P_{1, 2} \lor P_{2, 1})$.
$B_{1, 1} \implies (P_{1, 2} \lor P_{2, 1})$ was converted to $P_{1, 2} \lor P_{2, 1} \lor \neg B_{1, 1}$.
$B_{1, 1} \Longleftarrow (P_{1, 2} \lor P_{2, 1})$ was converted to $\neg (P_{1, 2} \lor P_{2, 1}) \lor B_{1, 1}$ which becomes $(\neg P_{1, 2} \lor B_{1, 1}) \land (\neg P_{2, 1} \lor B_{1, 1})$ after applying De Morgan's laws and distributing the disjunction.
$B_{2, 1} \iff (P_{1, 1} \lor P_{2, 2} \lor P_{3, 2})$ is converted in similar manner.
A knowledge-based agent is a simple generic agent that maintains and handles a knowledge base.
The knowledge base may initially contain some background knowledge.
The purpose of a KB agent is to provide a level of abstraction over knowledge-base manipulation and is to be used as a base class for agents that work on a knowledge base.
Given a percept, the KB agent adds the percept to its knowledge base, asks the knowledge base for the best action, and tells the knowledge base that it has in fact taken that action.
Our implementation of KB-Agent
is encapsulated in a class KB_AgentProgram
which inherits from the KB
class.
Let's have a look.
In [21]:
psource(KB_AgentProgram)
The helper functions make_percept_sentence
, make_action_query
and make_action_sentence
are all aptly named and as expected,
make_percept_sentence
makes first-order logic sentences about percepts we want our agent to receive,
make_action_query
asks the underlying KB
about the action that should be taken and
make_action_sentence
tells the underlying KB
about the action it has just taken.
In this section we will look at two algorithms to check if a sentence is entailed by the KB
. Our goal is to decide whether $\text{KB} \vDash \alpha$ for some sentence $\alpha$.
It is a model-checking approach which, as the name suggests, enumerates all possible models in which the KB
is true and checks if $\alpha$ is also true in these models. We list the $n$ symbols in the KB
and enumerate the $2^{n}$ models in a depth-first manner and check the truth of KB
and $\alpha$.
In [22]:
psource(tt_check_all)
The algorithm basically computes every line of the truth table $KB\implies \alpha$ and checks if it is true everywhere.
If symbols are defined, the routine recursively constructs every combination of truth values for the symbols and then,
it checks whether model
is consistent with kb
.
The given models correspond to the lines in the truth table,
which have a true
in the KB column,
and for these lines it checks whether the query evaluates to true
result = pl_true(alpha, model)
.
In short, tt_check_all
evaluates this logical expression for each model
pl_true(kb, model) => pl_true(alpha, model)
which is logically equivalent to
pl_true(kb, model) & ~pl_true(alpha, model)
that is, the knowledge base and the negation of the query are logically inconsistent.
tt_entails()
just extracts the symbols from the query and calls tt_check_all()
with the proper parameters.
In [23]:
psource(tt_entails)
Keep in mind that for two symbols P and Q, P => Q is false only when P is True
and Q is False
.
Example usage of tt_entails()
:
In [24]:
tt_entails(P & Q, Q)
Out[24]:
P & Q is True only when both P and Q are True. Hence, (P & Q) => Q is True
In [25]:
tt_entails(P | Q, Q)
Out[25]:
In [26]:
tt_entails(P | Q, P)
Out[26]:
If we know that P | Q is true, we cannot infer the truth values of P and Q. Hence (P | Q) => Q is False and so is (P | Q) => P.
In [27]:
(A, B, C, D, E, F, G) = symbols('A, B, C, D, E, F, G')
tt_entails(A & (B | C) & D & E & ~(F | G), A & D & E & ~F & ~G)
Out[27]:
We can see that for the KB to be true, A, D, E have to be True and F and G have to be False. Nothing can be said about B or C.
Coming back to our problem, note that tt_entails()
takes an Expr
which is a conjunction of clauses as the input instead of the KB
itself.
You can use the ask_if_true()
method of PropKB
which does all the required conversions.
Let's check what wumpus_kb
tells us about $P_{1, 1}$.
In [28]:
wumpus_kb.ask_if_true(~P11), wumpus_kb.ask_if_true(P11)
Out[28]:
Looking at Figure 7.9 we see that in all models in which the knowledge base is True
, $P_{1, 1}$ is False
. It makes sense that ask_if_true()
returns True
for $\alpha = \neg P_{1, 1}$ and False
for $\alpha = P_{1, 1}$. This begs the question, what if $\alpha$ is True
in only a portion of all models. Do we return True
or False
? This doesn't rule out the possibility of $\alpha$ being True
but it is not entailed by the KB
so we return False
in such cases. We can see this is the case for $P_{2, 2}$ and $P_{3, 1}$.
In [29]:
wumpus_kb.ask_if_true(~P22), wumpus_kb.ask_if_true(P22)
Out[29]:
Recall that our goal is to check whether $\text{KB} \vDash \alpha$ i.e. is $\text{KB} \implies \alpha$ true in every model. Suppose we wanted to check if $P \implies Q$ is valid. We check the satisfiability of $\neg (P \implies Q)$, which can be rewritten as $P \land \neg Q$. If $P \land \neg Q$ is unsatisfiable, then $P \implies Q$ must be true in all models. This gives us the result "$\text{KB} \vDash \alpha$ if and only if $\text{KB} \land \neg \alpha$ is unsatisfiable".
This technique corresponds to proof by contradiction, a standard mathematical proof technique. We assume $\alpha$ to be false and show that this leads to a contradiction with known axioms in $\text{KB}$. We obtain a contradiction by making valid inferences using inference rules. In this proof we use a single inference rule, resolution which states $(l_1 \lor \dots \lor l_k) \land (m_1 \lor \dots \lor m_n) \land (l_i \iff \neg m_j) \implies l_1 \lor \dots \lor l_{i - 1} \lor l_{i + 1} \lor \dots \lor l_k \lor m_1 \lor \dots \lor m_{j - 1} \lor m_{j + 1} \lor \dots \lor m_n$. Applying the resolution yields us a clause which we add to the KB. We keep doing this until:
The empty clause is equivalent to False because it arises only from resolving two complementary unit clauses such as $P$ and $\neg P$ which is a contradiction as both $P$ and $\neg P$ can't be True at the same time.
There is one catch however, the algorithm that implements proof by resolution cannot handle complex sentences.
Implications and bi-implications have to be simplified into simpler clauses.
We already know that every sentence of a propositional logic is logically equivalent to a conjunction of clauses.
We will use this fact to our advantage and simplify the input sentence into the conjunctive normal form (CNF) which is a conjunction of disjunctions of literals.
For eg:
$$(A\lor B)\land (\neg B\lor C\lor\neg D)\land (D\lor\neg E)$$
This is equivalent to the POS (Product of sums) form in digital electronics.
Here's an outline of how the conversion is done:
to_cnf
function executes this conversion using helper subroutines.
In [30]:
psource(to_cnf)
to_cnf
calls three subroutines.
eliminate_implications
converts bi-implications and implications to their logical equivalents.
move_not_inwards
removes negations from compound statements and moves them inwards using De Morgan's laws.
distribute_and_over_or
distributes disjunctions over conjunctions.
Run the cell below for implementation details.
In [31]:
psource(eliminate_implications)
psource(move_not_inwards)
psource(distribute_and_over_or)
Let's convert some sentences to see how it works
In [32]:
A, B, C, D = expr('A, B, C, D')
to_cnf(A |'<=>'| B)
Out[32]:
In [33]:
to_cnf(A |'<=>'| (B & C))
Out[33]:
In [34]:
to_cnf(A & (B | (C & D)))
Out[34]:
In [35]:
to_cnf((A |'<=>'| ~B) |'==>'| (C | ~D))
Out[35]:
Coming back to our resolution problem, we can see how the to_cnf
function is utilized here
In [36]:
psource(pl_resolution)
In [37]:
pl_resolution(wumpus_kb, ~P11), pl_resolution(wumpus_kb, P11)
Out[37]:
In [38]:
pl_resolution(wumpus_kb, ~P22), pl_resolution(wumpus_kb, P22)
Out[38]:
Previously, we said we will look at two algorithms to check if a sentence is entailed by the KB
. Here's a third one.
The difference here is that our goal now is to determine if a knowledge base of definite clauses entails a single proposition symbol q - the query.
There is a catch however - the knowledge base can only contain Horn clauses.
Horn clauses can be defined as a disjunction of literals with at most one positive literal.
A Horn clause with exactly one positive literal is called a definite clause.
A Horn clause might look like
$\neg a\lor\neg b\lor\neg c\lor\neg d... \lor z$
This, coincidentally, is also a definite clause.
Using De Morgan's laws, the example above can be simplified to
$a\land b\land c\land d ... \implies z$
This seems like a logical representation of how humans process known data and facts.
Assuming percepts a
, b
, c
, d
... to be true simultaneously, we can infer z
to also be true at that point in time.
There are some interesting aspects of Horn clauses that make algorithmic inference or resolution easier.
AND-OR-Graph-Search
from the chapter on search algorithms.
Implementational details will be explained shortly.pl_fc_entails
implements forward chaining to see if a knowledge base KB
entails a symbol q
.
pl_fc_entails
doesn't use an ordinary KB
instance.
The knowledge base here is an instance of the PropDefiniteKB
class, derived from the PropKB
class,
but modified to store definite clauses.
PropDefiniteKB
that returns a list of clauses in KB that have a given symbol p
in their premise.
In [39]:
psource(PropDefiniteKB.clauses_with_premise)
Let's now have a look at the pl_fc_entails
algorithm.
In [40]:
psource(pl_fc_entails)
The function accepts a knowledge base KB
(an instance of PropDefiniteKB
) and a query q
as inputs.
count
initially stores the number of symbols in the premise of each sentence in the knowledge base.
The conjuncts
helper function separates a given sentence at conjunctions.
inferred
is initialized as a boolean defaultdict.
This will be used later to check if we have inferred all premises of each clause of the agenda.
agenda
initially stores a list of clauses that the knowledge base knows to be true.
The is_prop_symbol
helper function checks if the given symbol is a valid propositional logic symbol.
We now iterate through agenda
, popping a symbol p
on each iteration.
If the query q
is the same as p
, we know that entailment holds.
The agenda is processed, reducing count
by one for each implication with a premise p
.
A conclusion is added to the agenda when count
reaches zero. This means we know all the premises of that particular implication to be true.
clauses_with_premise
is a helpful method of the PropKB
class.
It returns a list of clauses in the knowledge base that have p
in their premise.
Now that we have an idea of how this function works, let's see a few examples of its usage, but we first need to define our knowledge base. We assume we know the following clauses to be true.
In [41]:
clauses = ['(B & F)==>E',
'(A & E & F)==>G',
'(B & C)==>F',
'(A & B)==>D',
'(E & F)==>H',
'(H & I)==>J',
'A',
'B',
'C']
We will now tell
this information to our knowledge base.
In [42]:
definite_clauses_KB = PropDefiniteKB()
for clause in clauses:
definite_clauses_KB.tell(expr(clause))
We can now check if our knowledge base entails the following queries.
In [43]:
pl_fc_entails(definite_clauses_KB, expr('G'))
Out[43]:
In [44]:
pl_fc_entails(definite_clauses_KB, expr('H'))
Out[44]:
In [45]:
pl_fc_entails(definite_clauses_KB, expr('I'))
Out[45]:
In [46]:
pl_fc_entails(definite_clauses_KB, expr('J'))
Out[46]:
The previous segments elucidate the algorithmic procedure for model checking.
In this segment, we look at ways of making them computationally efficient.
The problem we are trying to solve is conventionally called the propositional satisfiability problem, abbreviated as the SAT problem.
In layman terms, if there exists a model that satisfies a given Boolean formula, the formula is called satisfiable.
The SAT problem was the first problem to be proven NP-complete.
The main characteristics of an NP-complete problem are:
This algorithm is very similar to Backtracking-Search.
It recursively enumerates possible models in a depth-first fashion with the following improvements over algorithms like tt_entails
:
In [47]:
psource(dpll)
The algorithm uses the ideas described above to check satisfiability of a sentence in propositional logic.
It recursively calls itself, simplifying the problem at each step. It also uses helper functions find_pure_symbol
and find_unit_clause
to carry out steps 2 and 3 above.
The dpll_satisfiable
helper function converts the input clauses to conjunctive normal form and calls the dpll
function with the correct parameters.
In [48]:
psource(dpll_satisfiable)
Let's see a few examples of usage.
In [49]:
A, B, C, D = expr('A, B, C, D')
In [50]:
dpll_satisfiable(A & B & ~C & D)
Out[50]:
This is a simple case to highlight that the algorithm actually works.
In [51]:
dpll_satisfiable((A & B) | (C & ~A) | (B & ~D))
Out[51]:
If a particular symbol isn't present in the solution, it means that the solution is independent of the value of that symbol. In this case, the solution is independent of A.
In [52]:
dpll_satisfiable(A |'<=>'| B)
Out[52]:
In [53]:
dpll_satisfiable((A |'<=>'| B) |'==>'| (C & ~A))
Out[53]:
In [54]:
dpll_satisfiable((A | (B & C)) |'<=>'| ((A | B) & (A | C)))
Out[54]:
This algorithm is very similar to Hill climbing.
On every iteration, the algorithm picks an unsatisfied clause and flips a symbol in the clause.
This is similar to finding a neighboring state in the hill_climbing
algorithm.
The symbol to be flipped is decided by an evaluation function that counts the number of unsatisfied clauses.
Sometimes, symbols are also flipped randomly to avoid local optima. A subtle balance between greediness and randomness is required. Alternatively, some versions of the algorithm restart with a completely new random assignment if no solution has been found for too long as a way of getting out of local minima of numbers of unsatisfied clauses.
Let's have a look at the algorithm.
In [55]:
psource(WalkSAT)
The function takes three arguments:
clauses
we want to satisfy.
p
of randomly changing a symbol.
max_flips
) the algorithm will run for. If the clauses are still unsatisfied, the algorithm returns None
to denote failure.
In [56]:
A, B, C, D = expr('A, B, C, D')
In [57]:
WalkSAT([A, B, ~C, D], 0.5, 100)
Out[57]:
This is a simple case to show that the algorithm converges.
In [58]:
WalkSAT([A & B, A & C], 0.5, 100)
Out[58]:
In [59]:
WalkSAT([A & B, C & D, C & B], 0.5, 100)
Out[59]:
In [60]:
WalkSAT([A & B, C | D, ~(D | B)], 0.5, 1000)
This one doesn't give any output because WalkSAT did not find any model where these clauses hold. We can solve these clauses to see that they together form a contradiction and hence, it isn't supposed to have a solution.
One point of difference between this algorithm and the dpll_satisfiable
algorithms is that both these algorithms take inputs differently.
For WalkSAT to take complete sentences as input,
we can write a helper function that converts the input sentence into conjunctive normal form and then calls WalkSAT with the list of conjuncts of the CNF form of the sentence.
In [61]:
def WalkSAT_CNF(sentence, p=0.5, max_flips=10000):
return WalkSAT(conjuncts(to_cnf(sentence)), 0, max_flips)
Now we can call WalkSAT_CNF
and DPLL_Satisfiable
with the same arguments.
In [62]:
WalkSAT_CNF((A & B) | (C & ~A) | (B & ~D), 0.5, 1000)
Out[62]:
It works!
Notice that the solution generated by WalkSAT doesn't omit variables that the sentence doesn't depend upon.
If the sentence is independent of a particular variable, the solution contains a random value for that variable because of the stochastic nature of the algorithm.
Let's compare the runtime of WalkSAT and DPLL for a few cases. We will use the %%timeit
magic to do this.
In [63]:
sentence_1 = A |'<=>'| B
sentence_2 = (A & B) | (C & ~A) | (B & ~D)
sentence_3 = (A | (B & C)) |'<=>'| ((A | B) & (A | C))
In [64]:
%%timeit
dpll_satisfiable(sentence_1)
dpll_satisfiable(sentence_2)
dpll_satisfiable(sentence_3)
In [65]:
%%timeit
WalkSAT_CNF(sentence_1)
WalkSAT_CNF(sentence_2)
WalkSAT_CNF(sentence_3)
On an average, for solvable cases, WalkSAT
is quite faster than dpll
because, for a small number of variables,
WalkSAT
can reduce the search space significantly.
Results can be different for sentences with more symbols though.
Feel free to play around with this to understand the trade-offs of these algorithms better.
In this section we show how to make plans by logical inference. The basic idea is very simple. It includes the following three steps:
Lets have a look at the algorithm
In [66]:
psource(SAT_plan)
Let's see few examples of its usage. First we define a transition and then call SAT_plan
.
In [67]:
transition = {'A': {'Left': 'A', 'Right': 'B'},
'B': {'Left': 'A', 'Right': 'C'},
'C': {'Left': 'B', 'Right': 'C'}}
print(SAT_plan('A', transition, 'C', 2))
print(SAT_plan('A', transition, 'B', 3))
print(SAT_plan('C', transition, 'A', 3))
Let us do the same for another transition.
In [68]:
transition = {(0, 0): {'Right': (0, 1), 'Down': (1, 0)},
(0, 1): {'Left': (1, 0), 'Down': (1, 1)},
(1, 0): {'Right': (1, 0), 'Up': (1, 0), 'Left': (1, 0), 'Down': (1, 0)},
(1, 1): {'Left': (1, 0), 'Up': (0, 1)}}
print(SAT_plan((0, 0), transition, (1, 1), 4))
FolKB
The class FolKB
can be used to represent a knowledge base of First-order logic sentences. You would initialize and use it the same way as you would for PropKB
except that the clauses are first-order definite clauses. We will see how to write such clauses to create a database and query them in the following sections.
In this section we create a FolKB
based on the following paragraph.
The law says that it is a crime for an American to sell weapons to hostile nations. The country Nono, an enemy of America, has some missiles, and all of its missiles were sold to it by Colonel West, who is American.
The first step is to extract the facts and convert them into first-order definite clauses. Extracting the facts from data alone is a challenging task. Fortunately, we have a small paragraph and can do extraction and conversion manually. We'll store the clauses in list aptly named clauses
.
In [69]:
clauses = []
“... it is a crime for an American to sell weapons to hostile nations”
The keywords to look for here are 'crime', 'American', 'sell', 'weapon' and 'hostile'. We use predicate symbols to make meaning of them.
Criminal(x)
: x
is a criminalAmerican(x)
: x
is an AmericanSells(x ,y, z)
: x
sells y
to z
Weapon(x)
: x
is a weaponHostile(x)
: x
is a hostile nationLet us now combine them with appropriate variable naming to depict the meaning of the sentence. The criminal x
is also the American x
who sells weapon y
to z
, which is a hostile nation.
$\text{American}(x) \land \text{Weapon}(y) \land \text{Sells}(x, y, z) \land \text{Hostile}(z) \implies \text{Criminal} (x)$
In [70]:
clauses.append(expr("(American(x) & Weapon(y) & Sells(x, y, z) & Hostile(z)) ==> Criminal(x)"))
"The country Nono, an enemy of America"
We now know that Nono is an enemy of America. We represent these nations using the constant symbols Nono
and America
. the enemy relation is show using the predicate symbol Enemy
.
$\text{Enemy}(\text{Nono}, \text{America})$
In [71]:
clauses.append(expr("Enemy(Nono, America)"))
"Nono ... has some missiles"
This states the existence of some missile which is owned by Nono. $\exists x \text{Owns}(\text{Nono}, x) \land \text{Missile}(x)$. We invoke existential instantiation to introduce a new constant M1
which is the missile owned by Nono.
$\text{Owns}(\text{Nono}, \text{M1}), \text{Missile}(\text{M1})$
In [72]:
clauses.append(expr("Owns(Nono, M1)"))
clauses.append(expr("Missile(M1)"))
"All of its missiles were sold to it by Colonel West"
If Nono owns something and it classifies as a missile, then it was sold to Nono by West.
$\text{Missile}(x) \land \text{Owns}(\text{Nono}, x) \implies \text{Sells}(\text{West}, x, \text{Nono})$
In [73]:
clauses.append(expr("(Missile(x) & Owns(Nono, x)) ==> Sells(West, x, Nono)"))
"West, who is American"
West is an American.
$\text{American}(\text{West})$
In [74]:
clauses.append(expr("American(West)"))
We also know, from our understanding of language, that missiles are weapons and that an enemy of America counts as “hostile”.
$\text{Missile}(x) \implies \text{Weapon}(x), \text{Enemy}(x, \text{America}) \implies \text{Hostile}(x)$
In [75]:
clauses.append(expr("Missile(x) ==> Weapon(x)"))
clauses.append(expr("Enemy(x, America) ==> Hostile(x)"))
Now that we have converted the information into first-order definite clauses we can create our first-order logic knowledge base.
In [76]:
crime_kb = FolKB(clauses)
The subst
helper function substitutes variables with given values in first-order logic statements.
This will be useful in later algorithms.
It's implementation is quite simple and self-explanatory.
In [77]:
psource(subst)
Here's an example of how subst
can be used.
In [78]:
subst({x: expr('Nono'), y: expr('M1')}, expr('Owns(x, y)'))
Out[78]:
We sometimes require finding substitutions that make different logical expressions look identical. This process, called unification, is done by the unify
algorithm. It takes as input two sentences and returns a unifier for them if one exists. A unifier is a dictionary which stores the substitutions required to make the two sentences identical. It does so by recursively unifying the components of a sentence, where the unification of a variable symbol var
with a constant symbol Const
is the mapping {var: Const}
. Let's look at a few examples.
In [79]:
unify(expr('x'), 3)
Out[79]:
In [80]:
unify(expr('A(x)'), expr('A(B)'))
Out[80]:
In [81]:
unify(expr('Cat(x) & Dog(Dobby)'), expr('Cat(Bella) & Dog(y)'))
Out[81]:
In cases where there is no possible substitution that unifies the two sentences the function return None
.
In [82]:
print(unify(expr('Cat(x)'), expr('Dog(Dobby)')))
We also need to take care we do not unintentionally use the same variable name. Unify treats them as a single variable which prevents it from taking multiple value.
In [83]:
print(unify(expr('Cat(x) & Dog(Dobby)'), expr('Cat(Bella) & Dog(x)')))
We consider the simple forward-chaining algorithm presented in Figure 9.3. We look at each rule in the knowledge base and see if the premises can be satisfied. This is done by finding a substitution which unifies each of the premise with a clause in the KB
. If we are able to unify the premises, the conclusion (with the corresponding substitution) is added to the KB
. This inferencing process is repeated until either the query can be answered or till no new sentences can be added. We test if the newly added clause unifies with the query in which case the substitution yielded by unify
is an answer to the query. If we run out of sentences to infer, this means the query was a failure.
The function fol_fc_ask
is a generator which yields all substitutions which validate the query.
In [84]:
psource(fol_fc_ask)
Let's find out all the hostile nations. Note that we only told the KB
that Nono was an enemy of America, not that it was hostile.
In [85]:
answer = fol_fc_ask(crime_kb, expr('Hostile(x)'))
print(list(answer))
The generator returned a single substitution which says that Nono is a hostile nation. See how after adding another enemy nation the generator returns two substitutions.
In [86]:
crime_kb.tell(expr('Enemy(JaJa, America)'))
answer = fol_fc_ask(crime_kb, expr('Hostile(x)'))
print(list(answer))
Note: fol_fc_ask
makes changes to the KB
by adding sentences to it.
This algorithm works backward from the goal, chaining through rules to find known facts that support the proof. Suppose goal
is the query we want to find the substitution for. We find rules of the form $\text{lhs} \implies \text{goal}$ in the KB
and try to prove lhs
. There may be multiple clauses in the KB
which give multiple lhs
. It is sufficient to prove only one of these. But to prove a lhs
all the conjuncts in the lhs
of the clause must be proved. This makes it similar to And/Or search.
The OR part of the algorithm comes from our choice to select any clause of the form $\text{lhs} \implies \text{goal}$. Looking at all rules's lhs
whose rhs
unify with the goal
, we yield a substitution which proves all the conjuncts in the lhs
. We use parse_definite_clause
to attain lhs
and rhs
from a clause of the form $\text{lhs} \implies \text{rhs}$. For atomic facts the lhs
is an empty list.
In [87]:
psource(fol_bc_or)
In [88]:
psource(fol_bc_and)
Now the main function fl_bc_ask
calls fol_bc_or
with substitution initialized as empty. The ask
method of FolKB
uses fol_bc_ask
and fetches the first substitution returned by the generator to answer query. Let's query the knowledge base we created from clauses
to find hostile nations.
In [89]:
# Rebuild KB because running fol_fc_ask would add new facts to the KB
crime_kb = FolKB(clauses)
In [90]:
crime_kb.ask(expr('Hostile(x)'))
Out[90]:
You may notice some new variables in the substitution. They are introduced to standardize the variable names to prevent naming problems as discussed in the Unification section
In [91]:
P |'==>'| ~Q
Out[91]:
What is the funny |'==>'|
syntax? The trick is that "|
" is just the regular Python or-operator, and so is exactly equivalent to this:
In [92]:
(P | '==>') | ~Q
Out[92]:
In other words, there are two applications of or-operators. Here's the first one:
In [93]:
P | '==>'
Out[93]:
What is going on here is that the __or__
method of Expr
serves a dual purpose. If the right-hand-side is another Expr
(or a number), then the result is an Expr
, as in (P | Q)
. But if the right-hand-side is a string, then the string is taken to be an operator, and we create a node in the abstract syntax tree corresponding to a partially-filled Expr
, one where we know the left-hand-side is P
and the operator is ==>
, but we don't yet know the right-hand-side.
The PartialExpr
class has an __or__
method that says to create an Expr
node with the right-hand-side filled in. Here we can see the combination of the PartialExpr
with Q
to create a complete Expr
:
In [94]:
partial = PartialExpr('==>', P)
partial | ~Q
Out[94]:
This trick is due to Ferdinand Jamitzky, with a modification by C. G. Vedant, who suggested using a string inside the or-bars.
expr
How does expr
parse a string into an Expr
? It turns out there are two tricks (besides the Jamitzky/Vedant trick):
==>
" with "|'==>'|
" (and likewise for other operators).eval
the resulting string in an environment in which every identifier
is bound to a symbol with that identifier as the op
.In other words,
In [95]:
expr('~(P & Q) ==> (~P | ~Q)')
Out[95]:
is equivalent to doing:
In [96]:
P, Q = symbols('P, Q')
~(P & Q) |'==>'| (~P | ~Q)
Out[96]:
One thing to beware of: this puts ==>
at the same precedence level as "|"
, which is not quite right. For example, we get this:
In [97]:
P & Q |'==>'| P | Q
Out[97]:
which is probably not what we meant; when in doubt, put in extra parens:
In [98]:
(P & Q) |'==>'| (P | Q)
Out[98]:
In [99]:
from notebook import Canvas_fol_bc_ask
canvas_bc_ask = Canvas_fol_bc_ask('canvas_bc_ask', crime_kb, expr('Criminal(x)'))
This notebook by Chirag Vartak and Peter Norvig.