[GiNaC-devel] [PATCH 09/10] gcd_pf_pow: get rid of duplicate code.
Alexei Sheplyakov
varg at theor.jinr.ru
Mon Aug 25 14:57:02 CEST 2008
This function (which helps gcd() handle partially factored expressions)
contained two copies of the same code. Remove one redundant copy.
---
ginac/normal.cpp | 80 ++++++++++++++++++-----------------------------------
1 files changed, 27 insertions(+), 53 deletions(-)
diff --git a/ginac/normal.cpp b/ginac/normal.cpp
index d5319b9..6392e3f 100644
--- a/ginac/normal.cpp
+++ b/ginac/normal.cpp
@@ -1691,65 +1691,39 @@ static ex gcd_pf_pow_pow(const ex& a, const ex& b, ex* ca, ex* cb)
static ex gcd_pf_pow(const ex& a, const ex& b, ex* ca, ex* cb)
{
- if (is_exactly_a<power>(a)) {
- ex p = a.op(0);
- const ex& exp_a = a.op(1);
- if (is_exactly_a<power>(b))
- return gcd_pf_pow_pow(a, b, ca, cb);
- else {
- if (p.is_equal(b)) {
- // a = p^n, b = p, gcd = p
- if (ca)
- *ca = power(p, a.op(1) - 1);
- if (cb)
- *cb = _ex1;
- return p;
- }
+ if (is_exactly_a<power>(a) && is_exactly_a<power>(b))
+ return gcd_pf_pow_pow(a, b, ca, cb);
- ex p_co, bpart_co;
- ex p_gcd = gcd(p, b, &p_co, &bpart_co, false);
+ if (is_exactly_a<power>(b) && (! is_exactly_a<power>(a)))
+ return gcd_pf_pow(b, a, cb, ca);
- if (p_gcd.is_equal(_ex1)) {
- // a(x) = p(x)^n, gcd(p, b) = 1 ==> gcd(a, b) = 1
- if (ca)
- *ca = a;
- if (cb)
- *cb = b;
- return _ex1;
- } else {
- // a(x) = g(x)^n A(x)^n, b(x) = g(x) B(x) ==> gcd(a, b) = g(x) gcd(g(x)^(n-1) A(x)^n, B(x))
- return p_gcd*gcd(power(p_gcd, exp_a-1)*power(p_co, exp_a), bpart_co, ca, cb, false);
- }
- } // is_exactly_a<power>(b)
+ GINAC_ASSERT(is_exactly_a<power>(a));
- } else if (is_exactly_a<power>(b)) {
- ex p = b.op(0);
- if (p.is_equal(a)) {
- // a = p, b = p^n, gcd = p
- if (ca)
- *ca = _ex1;
- if (cb)
- *cb = power(p, b.op(1) - 1);
- return p;
- }
+ ex p = a.op(0);
+ const ex& exp_a = a.op(1);
+ if (p.is_equal(b)) {
+ // a = p^n, b = p, gcd = p
+ if (ca)
+ *ca = power(p, a.op(1) - 1);
+ if (cb)
+ *cb = _ex1;
+ return p;
+ }
- ex p_co, apart_co;
- const ex& exp_b(b.op(1));
- ex p_gcd = gcd(a, p, &apart_co, &p_co, false);
- if (p_gcd.is_equal(_ex1)) {
- // b=p(x)^n, gcd(a, p) = 1 ==> gcd(a, b) == 1
- if (ca)
- *ca = a;
- if (cb)
- *cb = b;
- return _ex1;
- } else {
- // there are common factors:
- // a(x) = g(x) A(x), b(x) = g(x)^n B(x)^n ==> gcd = g(x) gcd(g(x)^(n-1) A(x)^n, B(x))
+ ex p_co, bpart_co;
+ ex p_gcd = gcd(p, b, &p_co, &bpart_co, false);
- return p_gcd*gcd(apart_co, power(p_gcd, exp_b-1)*power(p_co, exp_b), ca, cb, false);
- } // p_gcd.is_equal(_ex1)
+ // a(x) = p(x)^n, gcd(p, b) = 1 ==> gcd(a, b) = 1
+ if (p_gcd.is_equal(_ex1)) {
+ if (ca)
+ *ca = a;
+ if (cb)
+ *cb = b;
+ return _ex1;
}
+ // a(x) = g(x)^n A(x)^n, b(x) = g(x) B(x) ==> gcd(a, b) = g(x) gcd(g(x)^(n-1) A(x)^n, B(x))
+ ex rg = gcd(power(p_gcd, exp_a-1)*power(p_co, exp_a), bpart_co, ca, cb, false);
+ return p_gcd*rg;
}
static ex gcd_pf_mul(const ex& a, const ex& b, ex* ca, ex* cb)
--
1.5.6
Best regards,
Alexei
--
All science is either physics or stamp collecting.
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