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r""" 

Boolean Formulas 

 

Formulas consist of the operators ``&``, ``|``, ``~``, ``^``, ``->``, ``<->``, 

corresponding to ``and``, ``or``, ``not``, ``xor``, ``if...then``, ``if and 

only if``. Operators can be applied to variables that consist of a leading 

letter and trailing underscores and alphanumerics. Parentheses may be used 

to explicitly show order of operation. 

 

EXAMPLES: 

 

Create boolean formulas and combine them with 

:meth:`~sage.logic.boolformula.BooleanFormula.ifthen()` method:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("a&((b|c)^a->c)<->b") 

sage: g = propcalc.formula("boolean<->algebra") 

sage: (f&~g).ifthen(f) 

((a&((b|c)^a->c)<->b)&(~(boolean<->algebra)))->(a&((b|c)^a->c)<->b) 

 

We can create a truth table from a formula:: 

 

sage: f.truthtable() 

a b c value 

False False False True 

False False True True 

False True False False 

False True True False 

True False False True 

True False True False 

True True False True 

True True True True 

sage: f.truthtable(end=3) 

a b c value 

False False False True 

False False True True 

False True False False 

sage: f.truthtable(start=4) 

a b c value 

True False False True 

True False True False 

True True False True 

True True True True 

sage: propcalc.formula("a").truthtable() 

a value 

False False 

True True 

 

Now we can evaluate the formula for a given set of inputs:: 

 

sage: f.evaluate({'a':True, 'b':False, 'c':True}) 

False 

sage: f.evaluate({'a':False, 'b':False, 'c':True}) 

True 

 

And we can convert a boolean formula to conjunctive normal form:: 

 

sage: f.convert_cnf_table() 

sage: f 

(a|~b|c)&(a|~b|~c)&(~a|b|~c) 

sage: f.convert_cnf_recur() 

sage: f 

(a|~b|c)&(a|~b|~c)&(~a|b|~c) 

 

Or determine if an expression is satisfiable, a contradiction, or a tautology:: 

 

sage: f = propcalc.formula("a|b") 

sage: f.is_satisfiable() 

True 

sage: f = f & ~f 

sage: f.is_satisfiable() 

False 

sage: f.is_contradiction() 

True 

sage: f = f | ~f 

sage: f.is_tautology() 

True 

 

The equality operator compares semantic equivalence:: 

 

sage: f = propcalc.formula("(a|b)&c") 

sage: g = propcalc.formula("c&(b|a)") 

sage: f == g 

True 

sage: g = propcalc.formula("a|b&c") 

sage: f == g 

False 

 

It is an error to create a formula with bad syntax:: 

 

sage: propcalc.formula("") 

Traceback (most recent call last): 

... 

SyntaxError: malformed statement 

sage: propcalc.formula("a&b~(c|(d)") 

Traceback (most recent call last): 

... 

SyntaxError: malformed statement 

sage: propcalc.formula("a&&b") 

Traceback (most recent call last): 

... 

SyntaxError: malformed statement 

sage: propcalc.formula("a&b a") 

Traceback (most recent call last): 

... 

SyntaxError: malformed statement 

 

It is also an error to not abide by the naming conventions:: 

 

sage: propcalc.formula("~a&9b") 

Traceback (most recent call last): 

... 

NameError: invalid variable name 9b: identifiers must begin with a letter and contain only alphanumerics and underscores 

 

AUTHORS: 

 

- Chris Gorecki (2006): initial version 

 

- Paul Scurek (2013-08-03): added polish_notation, full_tree, 

updated docstring formatting 

 

- Paul Scurek (2013-08-08): added 

:meth:`~sage.logic.boolformula.BooleanFormula.implies()` 

 

""" 

from __future__ import absolute_import, division 

#***************************************************************************** 

# Copyright (C) 2006 William Stein <wstein.gmail.com> 

# Copyright (C) 2006 Chris Gorecki <chris.k.gorecki@gmail.com> 

# Copyright (C) 2013 Paul Scurek <scurek86@gmail.com> 

# 

# Distributed under the terms of the GNU General Public License (GPL) 

# as published by the Free Software Foundation; either version 2 of 

# the License, or (at your option) any later version. 

# http://www.gnu.org/licenses/ 

#***************************************************************************** 

 

from . import booleval 

from . import logictable 

from . import logicparser 

# import boolopt 

from sage.misc.flatten import flatten 

 

latex_operators = [('&', '\\wedge '), 

('|', '\\vee '), 

('~', '\\neg '), 

('^', '\\oplus '), 

('<->', '\\leftrightarrow '), 

('->', '\\rightarrow ')] 

 

 

class BooleanFormula(object): 

""" 

Boolean formulas. 

 

INPUT: 

 

- ``self`` -- calling object 

 

- ``exp`` -- a string; this contains the boolean expression 

to be manipulated 

 

- ``tree`` -- a list; this contains the parse tree of the expression. 

 

- ``vo`` -- a list; this contains the variables in the expression, in the 

order that they appear; each variable only occurs once in the list 

""" 

__expression = "" 

__tree = [] 

__vars_order = [] 

 

def __init__(self, exp, tree, vo): 

r""" 

Initialize the data fields. 

 

EXAMPLES: 

 

This example illustrates the creation of a statement:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b|~(c|a)") 

sage: s 

a&b|~(c|a) 

""" 

self.__expression = exp.replace(' ', '') 

self.__tree = tree 

self.__vars_order = vo 

 

def __repr__(self): 

r""" 

Return a string representation of this statement. 

 

OUTPUT: 

 

A string representation of calling statement 

 

EXAMPLES:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: propcalc.formula("man->monkey&human") 

man->monkey&human 

""" 

return self.__expression 

 

def _latex_(self): 

r""" 

Return a LaTeX representation of this statement. 

 

OUTPUT: 

 

A string containing the latex code for the statement 

 

EXAMPLES:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("man->monkey&human") 

sage: latex(s) 

man\rightarrow monkey\wedge human 

 

sage: f = propcalc.formula("a & ((~b | c) ^ a -> c) <-> ~b") 

sage: latex(f) 

a\wedge ((\neg b\vee c)\oplus a\rightarrow c)\leftrightarrow \neg b 

""" 

latex_expression = self.__expression 

for old, new in latex_operators: 

latex_expression = latex_expression.replace(old, new) 

return latex_expression 

 

def polish_notation(self): 

r""" 

Convert the calling boolean formula into polish notation. 

 

OUTPUT: 

 

A string representation of the formula in polish notation. 

 

EXAMPLES: 

 

This example illustrates converting a formula to polish notation:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("~~a|(c->b)") 

sage: f.polish_notation() 

'|~~a->cb' 

 

sage: g = propcalc.formula("(a|~b)->c") 

sage: g.polish_notation() 

'->|a~bc' 

 

AUTHORS: 

 

- Paul Scurek (2013-08-03) 

""" 

return ''.join(flatten(logicparser.polish_parse(repr(self)))) 

 

def tree(self): 

r""" 

Return the parse tree of this boolean expression. 

 

OUTPUT: 

 

The parse tree as a nested list 

 

EXAMPLES: 

 

This example illustrates how to find the parse tree of a boolean 

formula:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("man -> monkey & human") 

sage: s.tree() 

['->', 'man', ['&', 'monkey', 'human']] 

 

:: 

 

sage: f = propcalc.formula("a & ((~b | c) ^ a -> c) <-> ~b") 

sage: f.tree() 

['<->', 

['&', 'a', ['->', ['^', ['|', ['~', 'b', None], 'c'], 'a'], 'c']], 

['~', 'b', None]] 

 

.. NOTE:: 

 

This function is used by other functions in the logic module 

that perform semantic operations on a boolean formula. 

""" 

return self.__tree 

 

def full_tree(self): 

r""" 

Return a full syntax parse tree of the calling formula. 

 

OUTPUT: 

 

The full syntax parse tree as a nested list 

 

EXAMPLES: 

 

This example shows how to find the full syntax parse tree 

of a formula:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a->(b&c)") 

sage: s.full_tree() 

['->', 'a', ['&', 'b', 'c']] 

 

sage: t = propcalc.formula("a & ((~b | c) ^ a -> c) <-> ~b") 

sage: t.full_tree() 

['<->', ['&', 'a', ['->', ['^', ['|', ['~', 'b'], 'c'], 'a'], 'c']], ['~', 'b']] 

 

sage: f = propcalc.formula("~~(a&~b)") 

sage: f.full_tree() 

['~', ['~', ['&', 'a', ['~', 'b']]]] 

 

.. NOTE:: 

 

This function is used by other functions in the logic module 

that perform syntactic operations on a boolean formula. 

 

AUTHORS: 

 

- Paul Scurek (2013-08-03) 

""" 

return logicparser.polish_parse(repr(self)) 

 

def __or__(self, other): 

r""" 

Overload the ``|`` operator to 'or' two statements together. 

 

INPUT: 

 

- ``other`` -- a boolean formula; this is the statement 

on the right side of the operator 

 

OUTPUT: 

 

A boolean formula of the form ``self | other``. 

 

EXAMPLES: 

 

This example illustrates combining two formulas with ``|``:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: f = propcalc.formula("c^d") 

sage: s | f 

(a&b)|(c^d) 

""" 

return self.add_statement(other, '|') 

 

def __and__(self, other): 

r""" 

Overload the ``&`` operator to 'and' two statements together. 

 

INPUT: 

 

- ``other`` -- a boolean formula; this is the formula on 

the right side of the operator 

 

OUTPUT: 

 

A boolean formula of the form ``self & other``. 

 

EXAMPLES: 

 

This example shows how to combine two formulas with ``&``:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: f = propcalc.formula("c^d") 

sage: s & f 

(a&b)&(c^d) 

""" 

return self.add_statement(other, '&') 

 

def __xor__(self, other): 

r""" 

Overload the ``^`` operator to 'xor' two statements together. 

 

INPUT: 

 

- ``other`` -- a boolean formula; this is the formula on 

the right side of the operator 

 

OUTPUT: 

 

A boolean formula of the form ``self ^ other``. 

 

EXAMPLES: 

 

This example illustrates how to combine two formulas with ``^``:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: f = propcalc.formula("c^d") 

sage: s ^ f 

(a&b)^(c^d) 

""" 

return self.add_statement(other, '^') 

 

def __pow__(self, other): 

r""" 

Overload the ``^`` operator to 'xor' two statements together. 

 

INPUT: 

 

- ``other`` -- a boolean formula; this is the formula on 

the right side of the operator 

 

OUTPUT: 

 

A boolean formula of the form ``self ^ other``. 

 

EXAMPLES: 

 

This example shows how to combine two formulas with ``^``:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: f = propcalc.formula("c^d") 

sage: s ^ f 

(a&b)^(c^d) 

 

.. TODO:: 

 

This function seems to be identical to ``__xor__``. 

Thus, this function should be replaced with ``__xor__`` everywhere 

that it appears in the logic module. Then it can be deleted 

altogether. 

""" 

return self.add_statement(other, '^') 

 

def __invert__(self): 

r""" 

Overload the ``~`` operator to 'not' a statement. 

 

OUTPUT: 

 

A boolean formula of the form ``~self``. 

 

EXAMPLES: 

 

This example shows how to negate a boolean formula:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: ~s 

~(a&b) 

""" 

exp = '~(' + self.__expression + ')' 

parse_tree, vars_order = logicparser.parse(exp) 

return BooleanFormula(exp, parse_tree, vars_order) 

 

def ifthen(self, other): 

r""" 

Combine two formulas with the ``->`` operator. 

 

INPUT: 

 

- ``other`` -- a boolean formula; this is the formula 

on the right side of the operator 

 

OUTPUT: 

 

A boolean formula of the form ``self -> other``. 

 

EXAMPLES: 

 

This example illustrates how to combine two formulas with '->':: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: f = propcalc.formula("c^d") 

sage: s.ifthen(f) 

(a&b)->(c^d) 

""" 

return self.add_statement(other, '->') 

 

def iff(self, other): 

r""" 

Combine two formulas with the ``<->`` operator. 

 

INPUT: 

 

- ``other`` -- a boolean formula; this is the formula 

on the right side of the operator 

 

OUTPUT: 

 

A boolean formula of the form ``self <-> other``. 

 

EXAMPLES: 

 

This example illustrates how to combine two formulas with '<->':: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: f = propcalc.formula("c^d") 

sage: s.iff(f) 

(a&b)<->(c^d) 

""" 

return self.add_statement(other, '<->') 

 

def __eq__(self, other): 

r""" 

Overload the ``==`` operator to determine logical equivalence. 

 

INPUT: 

 

- ``other`` -- a boolean formula; this is the formula 

on the right side of the comparator 

 

OUTPUT: 

 

A boolean value to be determined as follows: 

 

- ``True`` if ``self`` and ``other`` are logically equivalent 

 

- ``False`` if ``self`` and ``other`` are not logically equivalent 

 

EXAMPLES: 

 

This example shows how to determine logical equivalence:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("(a|b)&c") 

sage: g = propcalc.formula("c&(b|a)") 

sage: f == g 

True 

 

:: 

 

sage: g = propcalc.formula("a|b&c") 

sage: f == g 

False 

""" 

return self.equivalent(other) 

 

def truthtable(self, start=0, end=-1): 

r""" 

Return a truth table for the calling formula. 

 

INPUT: 

 

- ``start`` -- (default: 0) an integer; this is the first 

row of the truth table to be created 

 

- ``end`` -- (default: -1) an integer; this is the laste 

row of the truth table to be created 

 

OUTPUT: 

 

The truth table as a 2-D array 

 

EXAMPLES: 

 

This example illustrates the creation of a truth table:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b|~(c|a)") 

sage: s.truthtable() 

a b c value 

False False False True 

False False True False 

False True False True 

False True True False 

True False False False 

True False True False 

True True False True 

True True True True 

 

We can now create a truthtable of rows 1 to 4, inclusive:: 

 

sage: s.truthtable(1, 5) 

a b c value 

False False True False 

False True False True 

False True True False 

True False False False 

 

.. NOTE:: 

 

Each row of the table corresponds to a binary number, with 

each variable associated to a column of the number, and taking on 

a true value if that column has a value of 1. Please see the 

logictable module for details. The function returns a table that 

start inclusive and end exclusive so ``truthtable(0, 2)`` will 

include row 0, but not row 2. 

 

When sent with no start or end parameters, this is an 

exponential time function requiring `O(2^n)` time, where 

`n` is the number of variables in the expression. 

""" 

max = 2 ** len(self.__vars_order) 

if end < 0: 

end = max 

if end > max: 

end = max 

if start < 0: 

start = 0 

if start > max: 

start = max 

keys, table = [], [] 

vars = {} 

for var in self.__vars_order: 

vars[var] = False 

keys.insert(0, var) 

keys = list(keys) 

for i in range(start, end): 

j = 0 

row = [] 

for key in keys: 

bit = self.get_bit(i, j) 

vars[key] = bit 

j += 1 

row.insert(0, bit) 

row.append(booleval.eval_formula(self.__tree, vars)) 

table.append(row) 

keys.reverse() 

table = logictable.Truthtable(table, keys) 

return table 

 

def evaluate(self, var_values): 

r""" 

Evaluate a formula for the given input values. 

 

INPUT: 

 

- ``var_values`` -- a dictionary; this contains the 

pairs of variables and their boolean values. 

 

OUTPUT: 

 

The result of the evaluation as a boolean. 

 

EXAMPLES: 

 

This example illustrates the evaluation of a boolean formula:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("a&b|c") 

sage: f.evaluate({'a':False, 'b':False, 'c':True}) 

True 

sage: f.evaluate({'a':True, 'b':False, 'c':False}) 

False 

""" 

return booleval.eval_formula(self.__tree, var_values) 

 

def is_satisfiable(self): 

r""" 

Determine if the formula is ``True`` for some assignment of values. 

 

OUTPUT: 

 

A boolean value to be determined as follows: 

 

- ``True`` if there is an assignment of values that makes the 

formula ``True``. 

 

- ``False`` if the formula cannot be made ``True`` by any assignment 

of values. 

 

EXAMPLES: 

 

This example illustrates how to check a formula for satisfiability:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("a|b") 

sage: f.is_satisfiable() 

True 

 

sage: g = f & (~f) 

sage: g.is_satisfiable() 

False 

""" 

table = self.truthtable().get_table_list() 

for row in table[1:]: 

if row[-1] is True: 

return True 

return False 

 

def is_tautology(self): 

r""" 

Determine if the formula is always ``True``. 

 

OUTPUT: 

 

A boolean value to be determined as follows: 

 

- ``True`` if the formula is a tautology. 

 

- ``False`` if the formula is not a tautology. 

 

EXAMPLES: 

 

This example illustrates how to check if a formula is a tautology:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("a|~a") 

sage: f.is_tautology() 

True 

 

sage: f = propcalc.formula("a&~a") 

sage: f.is_tautology() 

False 

 

sage: f = propcalc.formula("a&b") 

sage: f.is_tautology() 

False 

""" 

return not (~self).is_satisfiable() 

 

def is_contradiction(self): 

r""" 

Determine if the formula is always ``False``. 

 

OUTPUT: 

 

A boolean value to be determined as follows: 

 

- ``True`` if the formula is a contradiction. 

 

- ``False`` if the formula is not a contradiction. 

 

EXAMPLES: 

 

This example illustrates how to check if a formula is a contradiction. 

 

:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("a&~a") 

sage: f.is_contradiction() 

True 

 

sage: f = propcalc.formula("a|~a") 

sage: f.is_contradiction() 

False 

 

sage: f = propcalc.formula("a|b") 

sage: f.is_contradiction() 

False 

""" 

return not self.is_satisfiable() 

 

def implies(self, other): 

r""" 

Determine if calling formula implies other formula. 

 

INPUT: 

 

- ``self`` -- calling object 

 

- ``other`` -- instance of :class:`BooleanFormula` 

 

OUTPUT: 

 

A boolean value to be determined as follows: 

 

- ``True`` - if ``self`` implies ``other`` 

 

- ``False`` - if ``self does not imply ``other`` 

 

EXAMPLES: 

 

This example illustrates determining if one formula implies another:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("a<->b") 

sage: g = propcalc.formula("b->a") 

sage: f.implies(g) 

True 

 

:: 

 

sage: h = propcalc.formula("a->(a|~b)") 

sage: i = propcalc.formula("a") 

sage: h.implies(i) 

False 

 

AUTHORS: 

 

- Paul Scurek (2013-08-08) 

""" 

# input validation 

if not isinstance(other, BooleanFormula): 

raise TypeError("implies() takes an instance of the BooleanFormula() class as input") 

 

conditional = self.ifthen(other) 

return (conditional).is_tautology() 

 

def equivalent(self, other): 

r""" 

Determine if two formulas are semantically equivalent. 

 

INPUT: 

 

- ``self`` -- calling object 

 

- ``other`` -- instance of BooleanFormula class. 

 

OUTPUT: 

 

A boolean value to be determined as follows: 

 

True - if the two formulas are logically equivalent 

 

False - if the two formulas are not logically equivalent 

 

EXAMPLES: 

 

This example shows how to check for logical equivalence:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("(a|b)&c") 

sage: g = propcalc.formula("c&(a|b)") 

sage: f.equivalent(g) 

True 

 

sage: g = propcalc.formula("a|b&c") 

sage: f.equivalent(g) 

False 

""" 

return self.iff(other).is_tautology() 

 

def convert_cnf_table(self): 

r""" 

Convert boolean formula to conjunctive normal form. 

 

OUTPUT: 

 

An instance of :class:`BooleanFormula` in conjunctive normal form. 

 

EXAMPLES: 

 

This example illustrates how to convert a formula to cnf:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a ^ b <-> c") 

sage: s.convert_cnf() 

sage: s 

(a|b|~c)&(a|~b|c)&(~a|b|c)&(~a|~b|~c) 

 

We now show that :meth:`convert_cnf` and :meth:`convert_cnf_table` 

are aliases:: 

 

sage: t = propcalc.formula("a ^ b <-> c") 

sage: t.convert_cnf_table(); t 

(a|b|~c)&(a|~b|c)&(~a|b|c)&(~a|~b|~c) 

sage: t == s 

True 

 

.. NOTE:: 

 

This method creates the cnf parse tree by examining the logic 

table of the formula. Creating the table requires `O(2^n)` time 

where `n` is the number of variables in the formula. 

""" 

str = '' 

t = self.truthtable() 

table = t.get_table_list() 

vars = table[0] 

for row in table[1:]: 

if row[-1] is False: 

str += '(' 

for i in range(len(row) - 1): 

if row[i] is True: 

str += '~' 

str += vars[i] 

str += '|' 

str = str[:-1] + ')&' 

self.__expression = str[:-1] 

# in case of tautology 

if len(self.__expression) == 0: 

self.__expression = '(' + self.__vars_order[0] + '|~' + self.__vars_order[0] + ')' 

self.__tree, self.__vars_order = logicparser.parse(self.__expression) 

 

convert_cnf = convert_cnf_table 

 

def convert_cnf_recur(self): 

r""" 

Convert boolean formula to conjunctive normal form. 

 

OUTPUT: 

 

An instance of :class:`BooleanFormula` in conjunctive normal form. 

 

EXAMPLES: 

 

This example hows how to convert a formula to conjunctive normal form:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a^b<->c") 

sage: s.convert_cnf_recur() 

sage: s 

(~a|a|c)&(~b|a|c)&(~a|b|c)&(~b|b|c)&(~c|a|b)&(~c|~a|~b) 

 

.. NOTE:: 

 

This function works by applying a set of rules that are 

guaranteed to convert the formula. Worst case the converted 

expression has an `O(2^n)` increase in size (and time as well), but 

if the formula is already in CNF (or close to) it is only `O(n)`. 

 

This function can require an exponential blow up in space from the 

original expression. This in turn can require large amounts of 

time. Unless a formula is already in (or close to) being in cnf 

:meth:`convert_cnf()` is typically preferred, but results can vary. 

""" 

self.__tree = logicparser.apply_func(self.__tree, self.reduce_op) 

self.__tree = logicparser.apply_func(self.__tree, self.dist_not) 

self.__tree = logicparser.apply_func(self.__tree, self.dist_ors) 

self.convert_expression() 

 

def satformat(self): 

r""" 

Return the satformat representation of a boolean formula. 

 

OUTPUT: 

 

The satformat of the formula as a string. 

 

EXAMPLES: 

 

This example illustrates how to find the satformat of a formula:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: f = propcalc.formula("a&((b|c)^a->c)<->b") 

sage: f.convert_cnf() 

sage: f 

(a|~b|c)&(a|~b|~c)&(~a|b|~c) 

sage: f.satformat() 

'p cnf 3 0\n1 -2 3 0 1 -2 -3 \n0 -1 2 -3' 

 

.. NOTE:: 

 

See www.cs.ubc.ca/~hoos/SATLIB/Benchmarks/SAT/satformat.ps for a 

description of satformat. 

 

If the instance of boolean formula has not been converted to 

CNF form by a call to :meth:`convert_cnf()` or 

:meth:`convert_cnf_recur()`, then :meth:`satformat()` will call 

:meth:`convert_cnf()`. Please see the notes for 

:meth:`convert_cnf()` and :meth:`convert_cnf_recur()` for 

performance issues. 

""" 

self.convert_cnf_table() 

s = '' 

vars_num = {} 

i = 0 

clauses = 0 

for e in self.__vars_order: 

vars_num[e] = str(i + 1) 

i += 1 

i = 0 

w = 1 

while i < len(self.__expression): 

c = self.__expression[i] 

if c == ')': 

clauses += 1 

if c in '()|': 

i += 1 

continue 

if c == '~': 

s += '-' 

elif c == '&': 

s += '0 ' 

else: 

varname = '' 

while i < self.__expression[i] not in '|) ': 

varname += self.__expression[i] 

i += 1 

s += vars_num[varname] + ' ' 

if len(s) >= (w * 15) and s[-1] != '-': 

s += '\n' 

w += 1 

i += 1 

s = 'p cnf ' + str(len(self.__vars_order)) + ' ' + str(clauses) + '\n' + s 

return s[:-1] 

 

# def simplify(self): 

# r""" 

# This function uses the propcalc package to simplify an expression to 

# its minimal form. 

# 

# INPUT: 

# self -- the calling object. 

# 

# OUTPUT: 

# A simplified expression. 

# 

# EXAMPLES:: 

 

# sage: import sage.logic.propcalc as propcalc 

# sage: f = propcalc.formula("a&((b|c)^a->c)<->b") 

# sage: f.truthtable() 

# a b c value 

# False False False True 

# False False True True 

# False True False False 

# False True True False 

# True False False True 

# True False True False 

# True True False True 

# True True True True 

# sage: f.simplify() 

# (~a&~b)|(a&~b&~c)|(a&b) 

# sage: f.truthtable() 

# a b c value 

# False False False True 

# False False True True 

# False True False False 

# False True True False 

# True False False True 

# True False True False 

# True True False True 

# True True True True 

# 

# .. NOTE:: 

# 

# If the instance of boolean formula has not been converted to 

# cnf form by a call to convert_cnf() or convert_cnf_recur() 

# satformat() will call convert_cnf(). Please see the notes for 

# convert_cnf() and convert_cnf_recur() for performance issues. 

# """ 

# exp = '' 

# self.__tree = logicparser.apply_func(self.__tree, self.reduce_op) 

# plf = logicparser.apply_func(self.__tree, self.convert_opt) 

# wff = boolopt.PLFtoWFF()(plf) # convert to positive-normal form 

# wtd = boolopt.WFFtoDNF() 

# dnf = wtd(wff) 

# dnf = wtd.clean(dnf) 

# if(dnf == [] or dnf == [[]]): 

# exp = self.__vars_order[0] + '&~' + self.__vars_order[0] + ' ' 

# opt = boolopt.optimize(dnf) 

# if(exp == '' and (opt == [] or opt == [[]])): 

# exp = self.__vars_order[0] + '|~' + self.__vars_order[0] + ' ' 

# if(exp == ''): 

# for con in opt: 

# s = '(' 

# for prop in con: 

# if(prop[0] == 'notprop'): 

# s += '~' 

# s += prop[1] + '&' 

# exp += s[:-1] + ')|' 

# self.__expression = exp[:-1] 

# self.__tree, self.__vars_order = logicparser.parse(self.__expression) 

# return BooleanFormula(self.__expression, self.__tree, self.__vars_order) 

 

def convert_opt(self, tree): 

r""" 

Convert a parse tree to the tuple form used by :meth:`bool_opt()`. 

 

INPUT: 

 

- ``tree`` -- a list; this is a branch of a 

parse tree and can only contain the '&', '|' 

and '~' operators along with variables 

 

OUTPUT: 

 

A 3-tuple. 

 

EXAMPLES: 

 

This example illustrates the conversion of a formula into its 

corresponding tuple:: 

 

sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser 

sage: s = propcalc.formula("a&(b|~c)") 

sage: tree = ['&', 'a', ['|', 'b', ['~', 'c', None]]] 

sage: logicparser.apply_func(tree, s.convert_opt) 

('and', ('prop', 'a'), ('or', ('prop', 'b'), ('not', ('prop', 'c')))) 

 

.. NOTE:: 

 

This function only works on one branch of the parse tree. To 

apply the function to every branch of a parse tree, pass the 

function as an argument in 

:func:`~sage.logic.logicparser.apply_func()` in 

:mod:`~sage.logic.logicparser`. 

""" 

if not isinstance(tree[1], tuple) and not (tree[1] is None): 

lval = ('prop', tree[1]) 

else: 

lval = tree[1] 

if not isinstance(tree[2], tuple) and not(tree[2] is None): 

rval = ('prop', tree[2]) 

else: 

rval = tree[2] 

if tree[0] == '~': 

return ('not', lval) 

if tree[0] == '&': 

op = 'and' 

if tree[0] == '|': 

op = 'or' 

return (op, lval, rval) 

 

def add_statement(self, other, op): 

r""" 

Combine two formulas with the given operator. 

 

INPUT: 

 

- ``other`` -- instance of :class:`BooleanFormula`; this 

is the formula on the right of the operator 

 

- ``op`` -- a string; this is the operator used to 

combine the two formulas 

 

OUTPUT: 

 

The result as an instance of :class:`BooleanFormula`. 

 

EXAMPLES: 

 

This example shows how to create a new formula from two others:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: f = propcalc.formula("c^d") 

sage: s.add_statement(f, '|') 

(a&b)|(c^d) 

 

sage: s.add_statement(f, '->') 

(a&b)->(c^d) 

""" 

exp = '(' + self.__expression + ')' + op + '(' + other.__expression + ')' 

parse_tree, vars_order = logicparser.parse(exp) 

return BooleanFormula(exp, parse_tree, vars_order) 

 

def get_bit(self, x, c): 

r""" 

Determine if bit ``c`` of the number ``x`` is 1. 

 

INPUT: 

 

- ``x`` -- an integer; this is the number from 

which to take the bit 

 

- ``c`` -- an integer; this is the but number to 

be taken, where 0 is the low order bit 

 

OUTPUT: 

 

A boolean to be determined as follows: 

 

- ``True`` if bit ``c`` of ``x`` is 1. 

 

- ``False`` if bit c of x is not 1. 

 

EXAMPLES: 

 

This example illustrates the use of :meth:`get_bit`:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a&b") 

sage: s.get_bit(2, 1) 

True 

sage: s.get_bit(8, 0) 

False 

 

It is not an error to have a bit out of range:: 

 

sage: s.get_bit(64, 7) 

False 

 

Nor is it an error to use a negative number:: 

 

sage: s.get_bit(-1, 3) 

False 

sage: s.get_bit(64, -1) 

True 

sage: s.get_bit(64, -2) 

False 

 

.. NOTE:: 

 

The 0 bit is the low order bit. Errors should be handled 

gracefully by a return of ``False``, and negative numbers ``x`` 

always return ``False`` while a negative ``c`` will index from the 

high order bit. 

""" 

bits = [] 

while x > 0: 

b = bool(x % 2) 

x = x // 2 

bits.append(b) 

if c > len(bits) - 1: 

return False 

else: 

return bits[c] 

 

def reduce_op(self, tree): 

r""" 

Convert if-and-only-if, if-then, and xor operations to operations 

only involving and/or operations. 

 

INPUT: 

 

- ``tree`` -- a list; this represents a branch 

of a parse tree 

 

OUTPUT: 

 

A new list with no '^', '->', or '<->' as first element of list. 

 

EXAMPLES: 

 

This example illustrates the use of :meth:`reduce_op` with 

:func:`apply_func`:: 

 

sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser 

sage: s = propcalc.formula("a->b^c") 

sage: tree = ['->', 'a', ['^', 'b', 'c']] 

sage: logicparser.apply_func(tree, s.reduce_op) 

['|', ['~', 'a', None], ['&', ['|', 'b', 'c'], ['~', ['&', 'b', 'c'], None]]] 

 

.. NOTE:: 

 

This function only operates on a single branch of a parse tree. 

To apply the function to an entire parse tree, pass the function 

as an argument to :func:`~sage.logic.logicparser.apply_func()` 

in :mod:`~sage.logic.logicparser`. 

""" 

if tree[0] == '<->': 

# parse tree for (~tree[1]|tree[2])&(~tree[2]|tree[1]) 

new_tree = ['&', ['|', ['~', tree[1], None], tree[2]], 

['|', ['~', tree[2], None], tree[1]]] 

elif tree[0] == '^': 

# parse tree for (tree[1]|tree[2])&~(tree[1]&tree[2]) 

new_tree = ['&', ['|', tree[1], tree[2]], 

['~', ['&', tree[1], tree[2]], None]] 

elif tree[0] == '->': 

# parse tree for ~tree[1]|tree[2] 

new_tree = ['|', ['~', tree[1], None], tree[2]] 

else: 

new_tree = tree 

return new_tree 

 

def dist_not(self, tree): 

r""" 

Distribute '~' operators over '&' and '|' operators. 

 

INPUT: 

 

- ``tree`` a list; this represents a branch 

of a parse tree 

 

OUTPUT: 

 

A new list. 

 

EXAMPLES: 

 

This example illustrates the distribution of '~' over '&':: 

 

sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser 

sage: s = propcalc.formula("~(a&b)") 

sage: tree = ['~', ['&', 'a', 'b'], None] 

sage: logicparser.apply_func(tree, s.dist_not) #long time 

['|', ['~', 'a', None], ['~', 'b', None]] 

 

.. NOTE:: 

 

This function only operates on a single branch of a parse tree. 

To apply the function to an entire parse tree, pass the function 

as an argument to :func:`~sage.logic.logicparser.apply_func()` 

in :mod:`~sage.logic.logicparser`. 

""" 

if tree[0] == '~' and isinstance(tree[1], list): 

op = tree[1][0] 

if op != '~': 

if op == '&': 

op = '|' 

else: 

op = '&' 

new_tree = [op, ['~', tree[1][1], None], ['~', tree[1][2], None]] 

return logicparser.apply_func(new_tree, self.dist_not) 

else: 

# cancel double negative 

return tree[1][1] 

else: 

return tree 

 

def dist_ors(self, tree): 

r""" 

Distribute '|' over '&'. 

 

INPUT: 

 

- ``tree`` -- a list; this represents a branch of 

a parse tree 

 

OUTPUT: 

 

A new list. 

 

EXAMPLES: 

 

This example illustrates the distribution of '|' over '&':: 

 

sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser 

sage: s = propcalc.formula("(a&b)|(a&c)") 

sage: tree = ['|', ['&', 'a', 'b'], ['&', 'a', 'c']] 

sage: logicparser.apply_func(tree, s.dist_ors) #long time 

['&', ['&', ['|', 'a', 'a'], ['|', 'b', 'a']], ['&', ['|', 'a', 'c'], ['|', 'b', 'c']]] 

 

.. NOTE:: 

 

This function only operates on a single branch of a parse tree. 

To apply the function to an entire parse tree, pass the function 

as an argument to :func:`~sage.logic.logicparser.apply_func()` 

in :mod:`~sage.logic.logicparser`. 

""" 

if tree[0] == '|' and isinstance(tree[2], list) and tree[2][0] == '&': 

new_tree = ['&', ['|', tree[1], tree[2][1]], 

['|', tree[1], tree[2][2]]] 

return logicparser.apply_func(new_tree, self.dist_ors) 

if tree[0] == '|' and isinstance(tree[1], list) and tree[1][0] == '&': 

new_tree = ['&', ['|', tree[1][1], tree[2]], 

['|', tree[1][2], tree[2]]] 

return logicparser.apply_func(new_tree, self.dist_ors) 

return tree 

 

def to_infix(self, tree): 

r""" 

Convert a parse tree from prefix to infix form. 

 

INPUT: 

 

- ``tree`` -- a list; this represents a branch 

of a parse tree 

 

OUTPUT: 

 

A new list. 

 

EXAMPLES: 

 

This example shows how to convert a parse tree from prefix to 

infix form:: 

 

sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser 

sage: s = propcalc.formula("(a&b)|(a&c)") 

sage: tree = ['|', ['&', 'a', 'b'], ['&', 'a', 'c']] 

sage: logicparser.apply_func(tree, s.to_infix) 

[['a', '&', 'b'], '|', ['a', '&', 'c']] 

 

.. NOTE:: 

 

This function only operates on a single branch of a parse tree. 

To apply the function to an entire parse tree, pass the function 

as an argument to :func:`~sage.logic.logicparser.apply_func()` 

in :mod:`~sage.logic.logicparser`. 

""" 

if tree[0] != '~': 

return [tree[1], tree[0], tree[2]] 

return tree 

 

def convert_expression(self): 

r""" 

Convert the string representation of a formula to conjunctive 

normal form. 

 

EXAMPLES:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("a^b<->c") 

sage: s.convert_expression(); s 

a^b<->c 

""" 

ttree = self.__tree[:] 

ttree = logicparser.apply_func(ttree, self.to_infix) 

self.__expression = '' 

str_tree = str(ttree) 

open_flag = False 

i = 0 

for c in str_tree: 

if i < len(str_tree) - 1: 

op = self.get_next_op(str_tree[i:]) 

if op == '|' and not open_flag: 

self.__expression += '(' 

open_flag = True 

if i < len(str_tree) - 2 and str_tree[i + 1] == '&' and open_flag: 

open_flag = False 

self.__expression += ')' 

if str_tree[i:i + 4] == 'None': 

i += 4 

if i < len(str_tree) and str_tree[i] not in ' \',[]': 

self.__expression += str_tree[i] 

i += 1 

if open_flag is True: 

self.__expression += ')' 

 

def get_next_op(self, str): 

r""" 

Return the next operator in a string. 

 

INPUT: 

 

- ``str`` -- a string; this contains a logical 

expression 

 

OUTPUT: 

 

The next operator as a string. 

 

EXAMPLES: 

 

This example illustrates how to find the next operator in a formula:: 

 

sage: import sage.logic.propcalc as propcalc 

sage: s = propcalc.formula("f&p") 

sage: s.get_next_op("abra|cadabra") 

'|' 

 

.. NOTE:: 

 

The parameter ``str`` is not necessarily the string 

representation of the calling object. 

""" 

i = 0 

while i < len(str) - 1 and str[i] != '&' and str[i] != '|': 

i += 1 

return str[i]