Implementar el algoritmo naif de multiplicacion.

[z.facultad/75.29/dale.git] / src / number.h

diff --git a/src/number.h b/src/number.h
index cab115b69526c239c19aa349253f72a4a57c155d..dfd7934dd0d4b02bbeb47e03daee1267bc1eb008 100644 (file)
--- a/src/number.h
+++ b/src/number.h
@@ -324,3 +324,83 @@ number < N, E > karatsuba(const number< N, E > &u, const number< N, E > &v)
 
 }
 
+
+/* Algoritmo "naif" (por no decir "cabeza" o "bruto") de multiplicacion. */
+template < typename N, typename E >
+number < N, E > naif(const number< N, E > &u, const number< N, E > &v)
+{
+       typedef number< N, E > num_type;
+
+       // tomo el chunk size de u (el de v DEBE ser el mismo)
+       typename num_type::size_type chunk_size = u.chunk.size();
+
+       sign_type sign;
+
+       if ( (u.sign == positive && v.sign == positive) ||
+                       (u.sign == negative && v.sign == negative) ) {
+               sign = positive;
+       } else {
+               sign = negative;
+       }
+
+       //printf("naif %d %d\n", u.chunk.size(), v.chunk.size() );
+
+       if (chunk_size == 1)
+       {
+               /* Si llegamos a multiplicar dos de tamaño 1, lo que hacemos
+                * es usar la multiplicacion nativa del tipo N, guardando el
+                * resultado en el tipo E (que sabemos es del doble de tamaño
+                * de N, ni mas ni menos).
+                * Luego, armamos un objeto number usando al resultado como
+                * buffer. Si, es feo.
+                */
+               E tmp;
+               tmp = (E) u.chunk[0] * (E) v.chunk[0];
+               num_type tnum = num_type((N *) &tmp, 2, sign);
+               //std::cout << "T:" << tnum << " " << tmp << "\n";
+               //printf("1: %lu %lu %llu\n", u.chunk[0], v.chunk[0], tmp);
+               return tnum;
+       }
+
+       std::pair< num_type, num_type > u12 = u.split();
+       std::pair< num_type, num_type > v12 = v.split();
+
+       //std::cout << "u:" << u12.first << " - " << u12.second << "\n";
+       //std::cout << "v:" << v12.first << " - " << v12.second << "\n";
+
+       /* m11 = u1*v1
+        * m12 = u1*v2
+        * m21 = u2*v1
+        * m22 = u2*v2
+        */
+       num_type m11 = naif(u12.first, v12.first);
+       num_type m12 = naif(u12.first, v12.second);
+       num_type m21 = naif(u12.second, v12.first);
+       num_type m22 = naif(u12.second, v12.second);
+
+       /*
+       printf("csize: %d\n", chunk_size);
+       std::cout << "11 " << m11 << "\n";
+       std::cout << "12 " << m12 << "\n";
+       std::cout << "21 " << m21 << "\n";
+       std::cout << "22 " << m22 << "\n";
+       */
+
+       /* u*v = (u1*v1) * 2^n + (u1*v2 + u2*v1) * 2^(n/2) + u2*v2
+        * PERO! Como los numeros estan "al reves" nos queda:
+        *     = m22 * 2^n + (m12 + m21) * 2^(n/2) + m11
+        */
+       num_type res;
+       res = m22 << chunk_size;
+       res = res + ((m12 + m21) << (chunk_size / 2));
+       res = res + m11;
+       res.sign = sign;
+       /*
+       std::cout << "r: " << res << "\n";
+       std::cout << "\n";
+       */
+       return res;
+}
+
+
+