reverse mode automatic differentiation 在C++中。

这是代码:

class Var {
    private:
        double value;
        char character;
        std::vector<std::pair<double, const Var*> > children;

    public:
        Var(const double& _value=0, const char& _character='_') : value(_value), character(_character) {};
        void set_character(const char& character){ this->character = character; }

        // computes the derivative of the current object with respect to 'var'
        double gradient(Var* var) const{
            if(this==var){
                return 1.0;
            }

            double sum=0.0;
            for(auto& pair : children){
                // std::cout << "(" << this->character << " -> " <<  pair.second->character << ", " << this << " -> " << pair.second << ", weight=" << pair.first << ")" << std::endl;
                sum += pair.first*pair.second->gradient(var);
            }
            return sum;
        }

        friend Var operator+(const Var& l, const Var& r){
            Var result(l.value+r.value);
            result.children.push_back(std::make_pair(1.0, &l));
            result.children.push_back(std::make_pair(1.0, &r));
            return result;
        }

        friend Var operator*(const Var& l, const Var& r){
            Var result(l.value*r.value);
            result.children.push_back(std::make_pair(r.value, &l));
            result.children.push_back(std::make_pair(l.value, &r));
            return result;
        }

        friend std::ostream& operator<<(std::ostream& os, const Var& var){
            os << var.value;
            return os;
        }
};

我试着这样运行代码:

int main(int argc, char const *argv[]) {
    Var x(5,'x'), y(6,'y'), z(7,'z');

    Var k = z + x*y;
    k.set_character('k');

    std::cout << "k = " << k << std::endl;
    std::cout << "∂k/∂x = " << k.gradient(&x) << std::endl;
    std::cout << "∂k/∂y = " << k.gradient(&y) << std::endl;
    std::cout << "∂k/∂z = " << k.gradient(&z) << std::endl;

    return 0;
}

应建立的计算图如下:

       x(5)   y(6)              z(7)
         \     /                 /
 ∂w/∂x=y  \   /  ∂w/∂y=x        /
           \ /                 /
          w=x*y               /
             \               /  ∂k/∂z=1
              \             /
      ∂k/∂w=1  \           /
                \_________/
                     |
                   k=w+z

∂k/∂x 例如,我必须乘以边后面的梯度,并对每个边的结果求和。这是递归地通过 double gradient(Var* var) const . 所以我有 ∂k/∂x = ∂k/∂w * ∂w/∂x + ∂k/∂z * ∂z/∂x

问题

如果我有中间计算,比如 x*y 这里,出了点问题。什么时候? std::cout

k = 37
(k -> z, 0x7ffeb3345740 -> 0x7ffeb3345710, weight=1)
(k -> _, 0x7ffeb3345740 -> 0x7ffeb3345770, weight=1)
(_ -> x, 0x7ffeb3345770 -> 0x7ffeb33456b0, weight=0)
(_ -> y, 0x7ffeb3345770 -> 0x7ffeb33456e0, weight=5)
∂k/∂x = 0
(k -> z, 0x7ffeb3345740 -> 0x7ffeb3345710, weight=1)
(k -> _, 0x7ffeb3345740 -> 0x7ffeb3345770, weight=1)
(_ -> x, 0x7ffeb3345770 -> 0x7ffeb33456b0, weight=0)
(_ -> y, 0x7ffeb3345770 -> 0x7ffeb33456e0, weight=5)
∂k/∂y = 5
(k -> z, 0x7ffeb3345740 -> 0x7ffeb3345710, weight=1)
(k -> _, 0x7ffeb3345740 -> 0x7ffeb3345770, weight=1)
(_ -> x, 0x7ffeb3345770 -> 0x7ffeb33456b0, weight=0)
(_ -> y, 0x7ffeb3345770 -> 0x7ffeb33456e0, weight=5)
∂k/∂z = 1

它打印哪个变量连接到哪个变量,然后打印它们的地址,以及连接的权重(应该是梯度)。

问题是 weight=0 x x*y轴 (我指的是 w 我不知道为什么这个是零,而另一个重量与 y .

我注意到的另一件事是,如果你换线 operator*

result.children.push_back(std::make_pair(1.0, &r));
result.children.push_back(std::make_pair(1.0, &l));

那就是 取消的连接。