In high school algebra, you saw that while
x3−4x
x
3
4
x
and
x(x−2)(x+2)
x
x
2
x
2
are equivalent, the second form is particularly useful in
letting you quickly know the roots of the equation.
Similarly, in Boolean algebra
there are certain canonical — “normal” — forms
which have nice properties.
A formula in Conjunctive Normal Form, or CNF,
is the conjunction of CNF clauses.
Each clause is a formula of a simple form:
a disjunction of possibly-negated propositions.
c⇒a∧b c
a b
is equivalent to
(a∨¬c)∧(b∨¬c)
a c
b c
.
This latter formula is in CNF, since it is the conjunction of
disjunctions, and each disjunction consists only of propositions
and negated propositions.
The conjunctions and disjunctions need not be binary.
The following formula is also is CNF.
¬a∧(a∨b∨¬c)∧(b∨¬d∨e∨f)
a
a b c
b d e f
Note that its first clause is just one negated proposition.
It is still appropriate to think of this as a disjunction, since
φ≡φ∨φ
φ
φ φ
.
Another format, Disjunctive Normal Form, or DNF
is the dual of conjunctive normal form.
A DNF formula is the disjunction of DNF clauses,
each a conjunction of possibly-negated propositions.
a∧b⇒c
a b
c
is equivalent to
¬a∨¬b∨c
a b
c
which is in DNF: three disjunctions,
each being a clause with only one term.
(It also happens to be in CNF — a single clause with three terms!)
It is also equivalent to the more fleshed out DNF formula
where we insist that each clause include all three variables.
We end up with a formula that includes each possible
clause except
a∧b∧¬ca b c:
That is, the formula
a∧b∧c
∨
a∧¬b∧c
∨
a∧¬b∧¬c
∨
¬a∧b∧c
∨
¬a∧b∧¬c
∨
¬a∧¬b∧c
∨
¬a∧¬b∧¬c
a b c
a b c
a b c
a b c
a b c
a b c
a b c
.
Electrical Engineering courses, coming from more of a circuit perspective,
sometimes call
CNF
product-of-sums,
and call DNF
sum-of-products,
based on
∨∨,
∧∧ being
analogous to
+,*.
Any Boolean function can be represented in CNF and in DNF.
One way to obtain CNF and DNF formulas
is based upon the truth table for the function.
-
A DNF formula results from looking at a truth table,
and focusing on the rows where the function is true:
As if saying “I'm in this row, or in this row, or …”:
For each row where the function is true,
form a conjunction of the propositions.
(E.g., for the row where aa is false,
and bb is true,
form ¬a∧ba b.)
Now, form the disjunction of all those conjunctions.
-
A CNF formula is the pessimistic approach, focusing
on the rows where the function is false:
“I'm not in this row, and not in this row, and …”.
For each row where the function is false,
create a formula for “not in this row”:
(E.g., if in this row aa is false and
bb is true
form ¬(¬a∧b)a b;
then notice that by DeMorgan's law, this is
a∨¬ba b — a disjunct.
Now, form the conjunction of all those disjunctions.
Table 1: Truth table example| aa | bb | cc | Unknown function |
|---|
| false | false | false | false |
| false | false | true | false |
| false | true | false | true |
| false | true | true | true |
| true | false | false | false |
| true | false | true | true |
| true | true | false | false |
| true | true | true | false |
For CNF, the false rows give us the following five clauses:
- a∨b∨ca b c
- a∨b∨¬ca b c
- ¬a∨b∨ca b c
- ¬a∨¬b∨ca b c
- ¬a∨¬b∨¬ca b c
and the full formula is the conjunction of these.
Essentially, each clause rules out one row as being true.
For DNF, the true rows give us the following three clauses:
- ¬a∧b∧¬ca b c
- ¬a∧b∧ca b c
- a∧¬b∧ca b c
and the full formula is the disjunction of these.
Essentially, each clause allows one row to be true.
This shows that, for any arbitrarily complicated WFF, we
can find an equivalent WFF in CNF or DNF. These provide us
with two very regular and relatively uncomplicated forms to use.
The above
example
produced CNF and DNF formulas for a Boolean function, but
they are not the simplest such formulas.
For fun, can you find simpler ones?
-
CNF:
(a∨b)∧(¬a∨b∨c)∧(¬a∨¬b)
a b
a b c
a b
-
DNF:
¬a∧b
∨
a∧¬b∧c
a b
a b c
Karnaugh maps
are a general technique for finding minimal CNF and DNF formulas.
They are most easily used
when only a small number of variables are involved.
We won't worry about minimizing formulas ourselves, though.
Sometimes you'll see the form of CNF and DNF expressed in
a notation with subscripts.
-
DNF is
∨i
ψi
∨iψi,
where each clause
φi
φi
is
∧j
λj
∧jλj,
where each
λλ
is a propositional variable (PropProp),
or a negation of one (¬PropProp).
-
CNF is
∧i
ψi
∧iψi,
where each clause
ψiψi
is
∨j
λj
∨jλj ,
where each
λλ
is again a propositional variable (PropProp),
or a negation of one (¬PropProp).
For example, in the CNF formula
(a∨b)∧(¬a∨b∨c)∧(¬a∨¬b)
a b
a b c
a b
we have
φ2=¬a∨b∨c
φ2
a
b
c
within that clause we have
λ1=¬a
λ1
a
.
One question this notation brings up:
-
What is the disjunction of a single clause?
Well, it's reasonable to say that
ψ≡ψ
ψ
ψ
.
Note that this is also equivalent to
ψ∨falseψ .
-
What is the disjunction of zero clauses?
Well, if we start with
ψ≡ψ∨false
ψ
ψ
and remove the ψψ, that leaves us with false!
Alternately, imagine writing a function which takes a list of booleans,
and returns the ∨∨ of all of them — the natural base case
for this recursive list-processing program turns out to be false.
Indeed, this is the accepted definition of the empty disjunction.
It follows from false being the identity element for ∨∨.
Correspondingly, a conjunction of zero clauses is true.
Actually, that subscript notation above isn't
quite
correct: it forces each clause to be the same length,
which isn't actually required for CNF or DNF.
For fun, you can think about how to patch it up.
(Hint: double-subscripting.)
Note that often one of these forms might be more
concise than the other.
Here are two equivalently verbose ways of encoding true,
in CNF and DNF respectively:
(a∨¬a)∧(b∨¬b)∧…∧(z∨¬z)
a a
b b
…
z z
is equivalent to
a∧b∧c∧…∧y∧z
∨
a∧b∧c∧…∧y∧¬z
∨
a∧b∧c∧…∧¬y∧z
∨…∨
¬a∧¬b∧…∧¬y∧¬z
a
b
c
…
y
z
a
b
c
…
y
z
a
b
c
…
y
z
…
a
b
…
y
z
.
The first version corresponds to enumerating the choices for each location
of a WaterWorld board; it has 26 two-variable clauses.
This may seem like a lot, but compare it to the second version, which
corresponds to enumerating all possible
WaterWorld boards explicitly: it has all possible 26-variable clauses;
there are
226
2
26
≈ 64 billion of them!
