Bugfix de cast a la C++.

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

diff --git a/src/number.h b/src/number.h
index 28e853e3f244edc2372e7c4222caa26cd066d263..99eec8c22534d6815c8177fda66e77a58d97a4e9 100644 (file)
--- a/src/number.h
+++ b/src/number.h
@@ -10,6 +10,9 @@
 #include <iomanip>
 #include <stdint.h>
 
+enum sign_type { positive, negative };
+
+
 /* sizeof(E) tiene que ser 2*sizeof(N); y son los tipos nativos con los cuales
  * se haran las operaciones mas basicas. */
 
@@ -26,7 +29,6 @@ struct number
        // Tipos
        typedef N native_type;
        typedef E extended_type;
-       enum sign_type { positive, negative };
        typedef typename std::deque< native_type > chunk_type;
        typedef typename chunk_type::size_type size_type;
        typedef typename chunk_type::iterator iterator;
@@ -68,7 +70,8 @@ struct number
 
        // Devuelve referencia a 'átomo' i del chunk (no debería ser necesario
        // si la multiplicación es un método de este objeto).
-       native_type& operator[] (size_type i) {
+       native_type& operator[] (size_type i)
+       {
                return chunk[i];
        }
 
@@ -86,8 +89,8 @@ struct number
        template < typename NN, typename EE >
        friend std::ostream& operator<< (std::ostream& os, const number< NN, EE>& n);
 
-       private:
        // Atributos
+       //private:
        chunk_type chunk;
        sign_type sign;
 
@@ -226,19 +229,17 @@ std::ostream& operator<< (std::ostream& os, const number< N, E >& n)
 template < typename N, typename E >
 number< N, E >& number< N, E >::operator*= (const number< N, E >& n)
 {
-       number < N, E > r_op = n;
-       normalize_length(n);
-       n.normalize_length(*this);
-       *this = divide_n_conquer(*this, n);
+       //number < N, E > r_op = n;
+       //normalize_length(n);
+       //n.normalize_length(*this);
+       *this = naif(*this, n);
        return *this;
 }
 
 template < typename N, typename E >
 number< N, E > operator* (const number< N, E >& n1, const number< N, E >& n2)
 {
-       number< N, E > tmp = n1;
-       tmp *= n2;
-       return tmp;
+       return naif(n1, n2);
 }
 
 template < typename N, typename E >
@@ -289,7 +290,7 @@ std::pair< number< N, E >, number< N, E > > number< N, E >::split() const
 
 // es el algoritmo de división y conquista, que se llama recursivamente
 template < typename N, typename E >
-number < N, E > karatsuba(number< N, E > u, number< N, E > v)
+number < N, E > karatsuba(const number< N, E > &u, const number< N, E > &v)
 {
        typedef number< N, E > num_type;
 
@@ -323,3 +324,83 @@ number < N, E > karatsuba(number< N, E > u, 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 = static_cast< E >(u.chunk[0]) * static_cast< E >(v.chunk[0]);
+               num_type tnum = num_type(reinterpret_cast< 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;
+}
+
+
+