def _handle_irel(self, x, handler):
"""Return either None (if the conditions of self depend only on x) else
a Piecewise expression whose expressions (handled by the handler that
was passed) are paired with the governing x-independent relationals,
e.g. Piecewise((A, a(x) & b(y)), (B, c(x) | c(y)) ->
Piecewise(
(handler(Piecewise((A, a(x) & True), (B, c(x) | True)), b(y) & c(y)),
(handler(Piecewise((A, a(x) & True), (B, c(x) | False)), b(y)),
(handler(Piecewise((A, a(x) & False), (B, c(x) | True)), c(y)),
(handler(Piecewise((A, a(x) & False), (B, c(x) | False)), True))
"""
# identify governing relationals
rel = self.atoms(Relational)
irel = list(
ordered([
r for r in rel
if x not in r.free_symbols and r not in (S.true, S.false)
]))
if irel:
args = {}
exprinorder = []
for truth in product((1, 0), repeat=len(irel)):
reps = dict(zip(irel, truth))
# only store the true conditions since the false are implied
# when they appear lower in the Piecewise args
if 1 not in truth:
cond = None # flag this one so it doesn't get combined
else:
andargs = Tuple(*[i for i in reps if reps[i]])
free = list(andargs.free_symbols)
if len(free) == 1:
from sympy.solvers.inequalities import (
reduce_inequalities, _solve_inequality)
try:
t = reduce_inequalities(andargs, free[0])
# ValueError when there are potentially
# nonvanishing imaginary parts
except (ValueError, NotImplementedError):
# at least isolate free symbol on left
t = And(*[
_solve_inequality(a, free[0], linear=True)
for a in andargs
])
else:
t = And(*andargs)
if t is S.false:
continue # an impossible combination
cond = t
expr = handler(self.xreplace(reps))
if isinstance(expr, self.func) and len(expr.args) == 1:
expr, econd = expr.args[0]
cond = And(econd, True if cond is None else cond)
# the ec pairs are being collected since all possibilities
# are being enumerated, but don't put the last one in since
# its expr might match a previous expression and it
# must appear last in the args
if cond is not None:
args.setdefault(expr, []).append(cond)
# but since we only store the true conditions we must maintain
# the order so that the expression with the most true values
# comes first
exprinorder.append(expr)
# convert collected conditions as args of Or
for k in args:
args[k] = Or(*args[k])
# take them in the order obtained
args = [(e, args[e]) for e in uniq(exprinorder)]
# add in the last arg
args.append((expr, True))
# if any condition reduced to True, it needs to go last
# and there should only be one of them or else the exprs
# should agree
trues = [i for i in range(len(args)) if args[i][1] is S.true]
if not trues:
# make the last one True since all cases were enumerated
e, c = args[-1]
args[-1] = (e, S.true)
else:
assert len(set([e for e, c in [args[i] for i in trues]])) == 1
args.append(args.pop(trues.pop()))
while trues:
args.pop(trues.pop())
return Piecewise(*args)
def _handle_irel(self, x, handler):
"""Return either None (if the conditions of self depend only on x) else
a Piecewise expression whose expressions (handled by the handler that
was passed) are paired with the governing x-independent relationals,
e.g. Piecewise((A, a(x) & b(y)), (B, c(x) | c(y)) ->
Piecewise(
(handler(Piecewise((A, a(x) & True), (B, c(x) | True)), b(y) & c(y)),
(handler(Piecewise((A, a(x) & True), (B, c(x) | False)), b(y)),
(handler(Piecewise((A, a(x) & False), (B, c(x) | True)), c(y)),
(handler(Piecewise((A, a(x) & False), (B, c(x) | False)), True))
"""
# identify governing relationals
rel = self.atoms(Relational)
irel = list(ordered([r for r in rel if x not in r.free_symbols
and r not in (S.true, S.false)]))
if irel:
args = {}
exprinorder = []
for truth in product((1, 0), repeat=len(irel)):
reps = dict(zip(irel, truth))
# only store the true conditions since the false are implied
# when they appear lower in the Piecewise args
if 1 not in truth:
cond = None # flag this one so it doesn't get combined
else:
andargs = Tuple(*[i for i in reps if reps[i]])
free = list(andargs.free_symbols)
if len(free) == 1:
from sympy.solvers.inequalities import (
reduce_inequalities, _solve_inequality)
try:
t = reduce_inequalities(andargs, free[0])
# ValueError when there are potentially
# nonvanishing imaginary parts
except (ValueError, NotImplementedError):
# at least isolate free symbol on left
t = And(*[_solve_inequality(
a, free[0], linear=True)
for a in andargs])
else:
t = And(*andargs)
if t is S.false:
continue # an impossible combination
cond = t
expr = handler(self.xreplace(reps))
if isinstance(expr, self.func) and len(expr.args) == 1:
expr, econd = expr.args[0]
cond = And(econd, True if cond is None else cond)
# the ec pairs are being collected since all possibilities
# are being enumerated, but don't put the last one in since
# its expr might match a previous expression and it
# must appear last in the args
if cond is not None:
args.setdefault(expr, []).append(cond)
# but since we only store the true conditions we must maintain
# the order so that the expression with the most true values
# comes first
exprinorder.append(expr)
# convert collected conditions as args of Or
for k in args:
args[k] = Or(*args[k])
# take them in the order obtained
args = [(e, args[e]) for e in uniq(exprinorder)]
# add in the last arg
args.append((expr, True))
# if any condition reduced to True, it needs to go last
# and there should only be one of them or else the exprs
# should agree
trues = [i for i in range(len(args)) if args[i][1] is S.true]
if not trues:
# make the last one True since all cases were enumerated
e, c = args[-1]
args[-1] = (e, S.true)
else:
assert len(set([e for e, c in [args[i] for i in trues]])) == 1
args.append(args.pop(trues.pop()))
while trues:
args.pop(trues.pop())
return Piecewise(*args)
