swi prolog中的优化 [英] Optimisation in swi prolog

问题描述

说我想找到argmax(x，y，z)-1/2(20x ^ 2 + 32xy + 16y ^ 2)+ 2x + 2y.

Say I want to find argmax(x,y,z) -1/2(20x^2+32xy +16y^2)+2x+2y.

受制于: x> = 0，y> = 0，z> = 0和-x-y + z = 0.

subject to: x>=0, y>=0,z>=0 and -x-y+z =0.

我知道被设置为0的偏导数是:

I know the partial derivatives being set to 0 is :

-20x-16y + 2 = 0和-16x-16y + 2 = 0

-20x-16y+2=0 and -16x-16y+2 =0

所以我们可以让x = 0和y = 1/8和z = 1/8.

so we could have x= 0 and y =1/8 and z=1/8.

我该如何在Swi-prolog中做到这一点?我看到有用于线性求解的库单纯形，但这是一个二次问题，而偏导数则不是. (我有点困惑！)

How would I do this in Swi-prolog? I see that there is library simplex for linear solving, but this is a quadratic problem but the partial derivatives are not. (I am a bit confused!)

这就是我所拥有的:

:- use_module(library(simplex)).

my_constraints(S):-
 gen_state(S0),
 constraint([-20*x, -16*y] = 0, S0, S1),
 constraint([-16*x,-16*y] = 0, S1,S2),
 constraint([x] >= 0,S2,S3),
 constraint([y] >= 0,S3,S4),
 constraint([z] >= 0,S4,S5),
 constraint([-x-y+z] = 0,S5,S).

?- my_constraints(S), variable_value(S,x,Val1),variable_value(S,y,Val2).
false.

推荐答案

这里有几个问题.首先，只是为了解决这个问题:library(simplex)仅能处理 linear 约束.因此，是的，至少不能直接—用于解决您的实际问题.

There are several issues here. First, just to get this out of the way: library(simplex) can only handle linear constraints. So yes, it cannot—at least not directly—be used to solve your actual problem.

但是library(simplex)通常无论如何都是有用的，因此我想快速指出以下几点:

But library(simplex) is often useful regardless, and so I want to quickly point out the following:

1. variable_value/3仅适用于已解决表.这意味着您必须先调用maximize/3.

1. variable_value/3 only works on the solved tableau. This means that you must have invoked maximize/3 first.

例如:

?- my_constraints(S), maximize([x,y], S, Max), variable_value(Max, x, X).
S = ...,
Max = ...,
X = 0.

请注意，必须将my_constraint/1的最终目标更改为constraint([-1*x, -1*y,z] = 0, S5, S)，以符合该库所需的语法.

Note that you must change the final goal of my_constraint/1 to constraint([-1*x, -1*y,z] = 0, S5, S) to conform to the syntax required by this library.

话虽这么说，现在让我们进入问题的核心:有一系列使用线性的迭代解决二次优化问题的著名方法.  有关梯度的程序和推理，以使其更接近解决方案.因此，library(simplex)仍可以间接用于解决您的问题.

That being said, let us now get to the core of the issue: There are well-known ways to iteratively solve quadratic optimization problems, using a series of linear programs and reasoning about gradients to get closer to a solution. Thus, library(simplex) can indirectly still be used to solve your problem.

尤其是，请查看 其他程序中可用的最陡峭上升方法. .它包括一个用Prolog编写的小型符号导数计算器.是的，它是符号";-)

In particular, check out the method of steepest ascent available from miscellaneous programs. It includes a small symbolic derivative calculator written in Prolog. Yes, it's "symbolic" ;-)

完成您的任务后，我得到:

Plugging in your task, I get:

?- maximize(- 0.5*(20*x(1)^2 + 32*x(1)*x(2) + 16*x(2)^2) + 2*x(1) + 2*x(2),
   [[-1,0,0],
    [0,-1,0],
    [0,0,-1],
    [-1,-1,1],
    [1,1,-1]],
   [0,0,0,0,0],
   [0,0,0], Max).
Max = [4.298588509886033e-17, 0.125, 0.12500000000000006] ;
false.

我希望您可以使用哪种方法，直到浮点运算难以忍受的麻烦为止.

Which is, up to the unbearable nastiness of floating point arithmetic, something that I hope you can work with.

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