最优化算法-梯度下降

/*****************************************
 * FileName  : grad.go
 * Author    : fredric
 * Date      : 2017.09.01
 * Note      : 梯度算法
 * History   :
*****************************************/
package grad

import(
    "fmt"
    "math"
)

//无法采用牛顿方法求得极值，主要原因在于无法确定初始值，造成导数偏差很大
func _get_argmin_newton(x1, x2, x3, grad_x1, grad_x2, grad_x3 float64) float64 {

    fmt.Printf("_get_argmin input value %f,%f,%f,%f,%f,%f\n", x1, x2, x3, grad_x1, grad_x2, grad_x3)

    //f(x - a*delta) = (x1 - a * grad_x1 - 4)^4 + (x2 - a * grad_x2 - 3)^2 + 4 * (x3 - a*grad_x3 + 5)^4
    //f‘(x - a*delta) = 4 * grad_x1 * (x1 - a * grad_x1 - 4)^3
    //                + 2 * grad_x2 * (x2 - a * grad_x2 - 3)
    //                + 16* grad_x3 * (x3 - a*grad_x3 + 5)^3
    //f‘‘(x - a*delta)= 12 * grad_x1^2 * (x1 - a * grad_x1 - 4)^2
    //                + 2  * grad_x2^2 * a
    //                + 48 * grad_x3^2 * (x3 - a*grad_x3 + 5)^2
    //采用牛顿法求取f(a)的最小值
    
    //此处的初始值还是比较疑惑，因为初始值取不对，结果差太远
    var a0 float64 = 0.0002
    var a1 float64 = 0.0005
    delta := 0.0005

    for math.Abs(a1 - a0) > delta {

        a0 = a1

        //fmt.Printf("a0: %f\n" , a0)
        //fmt.Printf("grad_x2: %f\n" , grad_x2)
        //fmt.Printf("grad_x2 * a0: %f\n" , grad_x2 * a0)
        //fmt.Printf("grad_x2 * 0.2: %f\n" , grad_x2 * 0.2)

        f_1_v := 4 * grad_x1 * (x1 - a0 * grad_x1 - 4)* (x1 - a0 * grad_x1 - 4)* (x1 - a0 * grad_x1 - 4) + 
        2 * grad_x2 * (x2 - a0 * grad_x2 - 3) + 
        16* grad_x3 * (x3 - a0 * grad_x3 + 5)* (x3 - a0 * grad_x3 + 5) * (x3 - a0 * grad_x3 + 5)


        f_2_v := 12 * grad_x1 * grad_x1 * (x1 - a1 * grad_x1 - 4)* (x1 - a1 * grad_x1 - 4) + 2  * grad_x2* grad_x2 * a1 + 48 * grad_x3* grad_x3 * (x3 - a1 * grad_x3 + 5)* (x3 - a1 * grad_x3 + 5)

        a1 = a0 - f_1_v / f_2_v

        //fmt.Printf("----------abs = %f\n", math.Abs(a1 - a0))

    
        fmt.Printf("step value = %f f_1_v = %f, f_2_v = %f\n", (a0 + a1)/2, f_1_v, f_2_v)
    }

    return (a0 + a1)/2
}

//采用常量方式求极值
func _get_argmin_const(x1, x2, x3, grad_x1, grad_x2, grad_x3 float64) float64{


    /*
    * 不是很搞的清楚，当采用快速下降算法时如何确定固定步长，网上有一个说法实践是正确的
    * 即满足李普希兹条件存在L>0使得|f(x1)-f(x2)|<=L|x1-x2|，步长取1/L
    * 下面这个例子由于存在x3这个高阶，所以如果步长取大的话，完全没有办法计算
    */

    return 0.0004
}

func DoGradAlgorithm(){

    //计算f(x1,x2,x3) = (x1 - 4)^4 + (x2 - 3)^2 + 4*(x3 + 5)^4
    //所谓梯度本质上也是导数，只是针对多维度上，取了各个维度偏导数，组成向量；
    //最速下降法就是在每次迭代时取当前负梯度方向的能获取的函数数最小值

    //初始值x0 = [4, 2, -1]
    x1 := 4.0
    x2 := 2.0
    x3 := -1.0

    //取三次迭代
    for i := 0; i < 4; i++ {

        grad_x1 := 4 * (x1 - 4)*(x1 - 4)*(x1 - 4)
        grad_x2 := 2 * (x2 - 3)
        grad_x3 := 16 * (x3 + 5)* (x3 + 5)* (x3 + 5)

        a := _get_argmin_newton(x1,x2,x3, grad_x1, grad_x2, grad_x3)

        fmt.Printf("grad_x1 = %f, grad_x2 = %f, grad_x3 = %f\n", grad_x1, grad_x2, grad_x3)

        x1 = x1 - a * grad_x1
        x2 = x2 - a * grad_x2
        x3 = x3 - a * grad_x3

        fmt.Printf("x1 = %f, x2 = %f, x3 = %f\n", x1, x2, x3)

    }
}