Description Usage Arguments Details Value Contributors Author(s) References See Also Examples
This function finds all possibilities for factorizing a configurational expression.
1 2 | factorize(expression, and.split = "", sort.factorizing = FALSE,
sort.factorized = FALSE)
|
expression |
A string representing a configurational expression or a QCA
solution object of class “qca” generated by |
and.split |
The AND-operator (if any). |
sort.factorizing |
Logical, sort results beginning with largest number of factorizing elements. |
sort.factorized |
Logical, sort results beginning with largest number of factorized elements. |
In Boolean algebra, the “*”-operator is distributive over the “+”-
operator such that for any three literals a, b and c, the
following law holds: (a*b) + (a*c) = a*(b + c)
(Hohn 1966, pp.78-80; South 1974, p.12). The 'factorize
' function finds
all possible a for any configurational expression. Factorized versions
of the initial expression(s) can be sorted in decreasing order by the number of
factorizing literals or in decreasing order by the number of factorized literals.
A list with the following components:
initial |
The initial expression. |
factored |
The factorizations of the initial expression. |
Dusa, Adrian | : development, programming, testing |
Thiem, Alrik | : development, documentation, testing |
Alrik Thiem (Personal Website; ResearchGate Website)
Hohn, Franz E. 1966. Applied Boolean Algebra: An Elementary Introduction. 2nd ed. New York: Macmillan.
South, G. F. 1974. Boolean Algebra and Its Uses. New York: Van Nostrand Reinhold.
1 2 3 4 5 6 7 8 9 10 11 12 13 | # factorize a disjunction of two two-way conjunctions;
# if single letters are used, argument "and.split" is not needed
factorize("AB + AC")
# "and.split" is needed in these cases
factorize("one*TWO*four + one*THREE + THREE*four", and.split = "*")
factorize("~ONE*TWO*~FOUR + ~ONE*THREE + THREE*~FOUR", and.split = "*")
factorize("one&TWO&four + one&THREE + THREE&four", and.split = "&")
# factorize solution objects directly
data(d.represent)
KRO.sol <- eQMC(d.represent, outcome = "WNP")
factorize(KRO.sol)
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