Question

我试图在java中定义一些基本的数学概念，但不断陷入泛型错误。考虑下面的例子。在这个例子中，我尝试将零映射（所有x的f（x）= 0）定义为常量映射（f（x）= c for all x）和线性映射（f（a * x + b * y）= a * f（x）+ b * f（y））。编译器不允许我这样做，因为语句'零映射是一个向量' 是模糊的（就java而言，但不是数学上的）。

你能为这个问题建议一个干净的解决方案吗？

// A vector in a finite-dimensional vector space over the real numbers.
interface Vector<V extends Vector<?>>
{
    int dimension();
    V plus(V v);
    V times(double c);
}

interface Mapping<U extends Vector<?>, V extends Vector<?>>
// Does not inherit from Vector because the set of all mappings is an
// infinite-dimensional vector space.
{
    V map(U u);
}

// Linear mappings, on the other hand, from one finite-dimensional vector space
// to another, do form a finite-dimensional vector space.
interface LinearMapping<U extends Vector<?>, V extends Vector<?>>
    extends Mapping<U, V>, Vector<LinearMapping<U, V>>
{
}

// All elements of U are mapped to getImage(). If V is finite-dimensional, then
// the set of constant mappings is also a finite-dimensional vector space.
interface ConstMapping<U extends Vector<?>, V extends Vector<?>>
    extends Mapping<U, V>, Vector<ConstMapping<U, V>>
{
    V getImage();
}

// A zero mapping is both constant and linear, but cannot be defined as such.
interface ZeroMapping<U extends Vector<?>, V extends Vector<?>>
    extends LinearMapping<U, V>, ConstMapping<U, V>
// Error: The interface Vector cannot be implemented more than once with
// different arguments: Vector<ConstMapping<U,V>> and Vector<LinearMapping<U,V>>
{
}

Answer 1

问题似乎是ZeroMapping正在扩展Vector<LinearMapping>和Vector<ConstMapping>。

为了解决这个问题，我们将LinearMapping和ConstMapping的通用参数V更改为各个类的下限，以便它们可以扩展通用Vector<V>，从而允许更多灵活性。希望这可以捕获您将LinearMapping设置为Mapping和Vector（仅限其本身）的意图。

编辑：在这种情况下，您可能需要使用3个通用参数：2用于捕获“Mapping”，1用于捕获与其他相同类型的LinearMappings的“Vector”关系。

ZeroMapping遵循相同的格式。

我在下面的回复中提到了@LuiggiMendoza的建议。

interface Vector<V extends Vector<V>>
{
    int dimension();
    V plus(V v);
    V times(double c);
}

interface Mapping<U extends Vector<U>, V extends Vector<V>>
// Does not inherit from Vector because the set of all mappings is an
// infinite-dimensional vector space.
{
    V map(U u);
}

// Linear mappings, on the other hand, from one finite-dimensional vector space
// to another, do form a finite-dimenszional vector space.
interface LinearMapping<U extends Vector<U>, V extends Vector<V>, W extends LinearMapping<U, V, W>>
    extends Mapping<U, V>, Vector<W>
{
}

// All elements of U are mapped to getImage(). If V is finite-dimensional, then
// the set of constant mappings is also a finite-dimensional vector space.
interface ConstMapping<U extends Vector<U>, V extends Vector<V>, W extends ConstMapping<U, V, W>>
    extends Mapping<U, V>, Vector<W>
{
    V getImage();
}

// A zero mapping is both constant and linear, but cannot be defined as such.
interface ZeroMapping<U extends Vector<U>, V extends Vector<V>, W extends ZeroMapping<U, V, W>>
    extends LinearMapping<U, V, W>, ConstMapping<U, V, W>
{
}

如何实现类的示例可以是：

class RealVector implements Vector<RealVector> { // methods here .. // }

class RealLinearMapping implements LinearMapping<RealVector, RealVector, RealLinearMapping>
{
    @Override
    public RealVector map(RealVector u) { ... }

    @Override
    public RealLinearMapping plus(RealLinearMapping v) { ... }
}

用java定义抽象数学

1 个答案: