From: Alberto Bertogli <albertogli@telpin.com.ar>
Date: Mon, 3 Oct 2005 18:30:30 +0000 (+0000)
Subject: Implementar el algoritmo naif de multiplicacion.
X-Git-Tag: Entrega_1~49
X-Git-Url: https://git.llucax.com/z.facultad/75.29/dale.git/commitdiff_plain/4a91a736401518a4cfe3a7f5e814a2510fb39f40?ds=sidebyside;hp=2a62cec8c48794a36cc4757865bb14baf8b3a006

Implementar el algoritmo naif de multiplicacion.
---

diff --git a/src/number.h b/src/number.h
index cab115b..dfd7934 100644
--- 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;
+}
+
+
+