8
a
(A ∨ B)(A ∨ C)
b
(A ∨ B)(A ∨ C)(A ∨ C)(B ∨ C )(B ∨ C )
c
A∨B∨C
A∨B∨C
A∨B ∨C
A∨B∨C
A∨B∨D
A∨B∨D
A∨B ∨D
A∨B∨D
A∨C ∨D
A∨C ∨D
A∨C ∨D
A∨C ∨D
B∨C∨D
B∨C ∨D
B∨C ∨D
B∨C ∨D
9
a
To prove the sa sfiability of the given formula, at least one combina on
of T/F assignment of variables needs to be found, such that the formula
becomes True.
Time complexity of the boolean sa sfiability problem (SAT) is determined
by the the number of variables and the number of connec ves in the
formula. If the number of variables is m and number of connec ves is
then the me complexity should be 2m
n,
⋅ poly(n). Since the number of
variables is not being considered in this scenario, assuming it is fixed, then
the me needed to prove the sa sfiability of the formula should be
poly(n).
b
To prove the unsa sfiability of the given formula, the formula should yield
False for all combina ons of T/F assignment of variables.
Like 8(a), the me complexity should be 2m
⋅ poly(n) where m is the
number of variables and n number of connec ves. Again, since m is not
being considered in this scenario, assuming it is fixed, then the me
needed to prove the sa sfiability of the formula should also be poly(n).
9
a
I can arrange the boolean circuit in a way so that it represents the
formula. Then, for each combina ons of T/F assignment of the variables,
check the output of the circuit. If the circuit outputs True for at least one
combina ons of assignment, the formula is sa sfiable.
b
Like in 8(a) and (b), the number of connec ves contribute to polynomial
me of boolean sa sfiability problem. Assuming the number of variable is
fixed, the me complexity of a formula with n connec veness is Θ(n).
c
Same as 9(b), like in 8(a) and (b), the number of connec ves contribute to
polynomial me of boolean sa sfiability problem. Assuming the number
of variable is fixed, the me complexity of a formula with n
connec veness is Θ(nr ).