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Simplify Complex Arguments to Exp #1332

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Simplify Complex Arguments to Exp #1332

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1716ceb
Fixes #1330
ShikharJ Sep 27, 2017
cc81591
Minor Improvement
ShikharJ Oct 3, 2017
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52 changes: 52 additions & 0 deletions symengine/pow.cpp
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Expand Up @@ -87,6 +87,25 @@ int Pow::compare(const Basic &o) const
return base_cmp;
}

bool exp_mul_helper(const RCP<const Basic> &b)
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How about this function returning -1, 0, 1,2,3 where -1 is when it is not what we want and the others n in the function beloe

{
if (is_a_Complex(*down_cast<const Mul &>(*b).get_coef())) {
const Mul &s = down_cast<const Mul &>(*b);
const map_basic_basic &dict = s.get_dict();
RCP<const Number> coef
= down_cast<const ComplexBase &>(*s.get_coef()).imaginary_part();
RCP<const Basic> arg = mul(coef, integer(2));
if (dict.size() == 1 and is_a<Integer>(*arg)) {
for (const auto &p : dict) {
if (eq(*p.first, *pi) and eq(*p.second, *one)) {
return true;
}
}
}
}
return false;
}

RCP<const Basic> pow(const RCP<const Basic> &a, const RCP<const Basic> &b)
{
if (is_a_Number(*b) and down_cast<const Number &>(*b).is_zero()) {
Expand Down Expand Up @@ -164,6 +183,39 @@ RCP<const Basic> pow(const RCP<const Basic> &a, const RCP<const Basic> &b)
RCP<const Pow> A = rcp_static_cast<const Pow>(a);
return pow(A->get_base(), mul(A->get_exp(), b));
}
if (eq(*a, *E)) {
if (is_a<Mul>(*b) and exp_mul_helper(b)) {
RCP<const Number> coef = down_cast<const ComplexBase &>(
*down_cast<const Mul &>(*b).get_coef())
.imaginary_part();
RCP<const Basic> arg = mul(coef, integer(2));
long n = ((down_cast<const Integer &>(*arg).as_int() % 4) + 4) % 4;
if (!n) {
return one;
} else if (n == 1) {
return I;
} else if (n == 2) {
return minus_one;
} else {
return Complex::from_two_nums(*zero, *minus_one);
}
} else if (is_a<Add>(*b)) {
const umap_basic_num &dict = down_cast<const Add &>(*b).get_dict();
umap_basic_num new_dict;
RCP<const Number> coef = down_cast<const Add &>(*b).get_coef();
RCP<const Basic> s = one;
for (const auto &p : dict) {
if (eq(*p.first, *pi)
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To do this, you don't need to iterate the dictionary. Just use dict.find to find the value pi.

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Is there a way I can remove that particular entry of pi as well? That way, there'd be no need of iteration at all.

and exp_mul_helper(mul(p.first, p.second))) {
s = exp(mul(p.first, p.second));
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Can a call to exp be avoided here?

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@ShikharJ ShikharJ Oct 22, 2017 •
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At the expense of code duplication it can be. I can't seem to think of another way.

} else {
Add::dict_add_term(new_dict, p.second, p.first);
}
}
return mul(s, make_rcp<const Pow>(
a, Add::from_dict(coef, std::move(new_dict))));
}
}
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We need to carefully benchmark these two if branches, before and after this PR. So the first if clause would run for something like E^(a*b). The second one for E^(a+b). Correct?

If so, then we need to benchmark these two things and similar things a lot.

return make_rcp<const Pow>(a, b);
}

Expand Down
38 changes: 38 additions & 0 deletions symengine/tests/basic/test_arit.cpp
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Expand Up @@ -32,6 +32,7 @@ using SymEngine::sub;
using SymEngine::exp;
using SymEngine::E;
using SymEngine::Rational;
using SymEngine::rational;
using SymEngine::Complex;
using SymEngine::Number;
using SymEngine::I;
Expand Down Expand Up @@ -987,6 +988,43 @@ TEST_CASE("Pow: arit", "[arit]")
r1 = pow(mul(sqrt(mul(y, x)), x), i2);
r2 = mul(pow(x, i3), y);
REQUIRE(eq(*r1, *r2));

r1 = exp(mul(I, x));
r2 = pow(E, mul(I, x));
REQUIRE(eq(*r1, *r2));

r1 = exp(mul(I, pi));
r2 = pow(E, mul(I, pi));
REQUIRE(eq(*r1, *r2));

r1 = exp(mul(I, pi));
REQUIRE(eq(*r1, *minus_one));

r1 = exp(mul(mul(I, minus_one), pi));
std::cout << *r1 << std::endl;
REQUIRE(eq(*r1, *minus_one));

r1 = exp(div(mul(I, pi), integer(2)));
REQUIRE(eq(*r1, *I));

r1 = exp(mul(div(mul(I, pi), integer(2)), minus_one));
REQUIRE(eq(*r1, *mul(I, minus_one)));

r1 = div(exp(mul(mul(I, pi), x)), exp(x));
REQUIRE(is_a<Pow>(*r1));
REQUIRE(eq(*down_cast<const Pow &>(*r1).get_base(), *E));
REQUIRE(eq(*down_cast<const Pow &>(*r1).get_exp(),
*add(mul(minus_one, x), mul(x, mul(I, pi)))));

r1 = exp(add(div(mul(I, pi), integer(2)), x));
r2 = mul(I, exp(x));
REQUIRE(eq(*r1, *r2));

r1 = exp(add(div(mul(I, pi), integer(10)), x));
REQUIRE(is_a<Pow>(*r1));
REQUIRE(eq(*down_cast<const Pow &>(*r1).get_base(), *E));
REQUIRE(eq(*down_cast<const Pow &>(*r1).get_exp(),
*add(x, mul(rational(1, 10), mul(I, pi)))));
}

TEST_CASE("Log: arit", "[arit]")
Expand Down