Hill-climbing 算法python 实现

Posted 2022-12-13 MrCharles

Hill-climbing 其实也不是很复杂，在这个博文里面，我假定一个512维度的空间中存在一个点point1 ，我的目标是随机初始化一个点，通过Hill-climbing找到这个目标点point1。

站在当前点，通过探索所有可能的走法去判断下一步哪一个走法是能够朝着目标靠近的。过程其实也不复杂，在512里面的某一个维度，走起来只有两种选择，加一个步长或者减去一个步长。如果下一步找不到比较好的选择，都不能靠近目标点，那么算法结束。

具体代码可以看下面：

__author__ = 'xingbo, it is based on inital version of sidharthgoyal'
import math
from random import *
import random
import numpy
import copy

increment = 0.05
startingPoint = numpy.random.random(512)
point1 = numpy.random.random(512)
print('target:',point1)
# point2 = [6,4]
# point3 = [5,2]
# point4 = [2,1]

def distance(coords1, coords2):

    """ Calculates the euclidean distance between 2 lists of coordinates. """
    # print('coords1',coords1)
    # print('coords2',coords2)
    return numpy.sqrt(numpy.sum((coords1 - coords2)**2))

def sumOfDistances(x, px):
    d1 = distance(x, px)
    return d1

def newDistance(d1, point1):
    d1temp = sumOfDistances(d1, point1 )
    d1 = numpy.append(d1,d1temp) 
    return d1

minDistance = sumOfDistances(startingPoint, point1 )
flag = True

threshold = 0.4
i = 1
lastFitness = 99
while lastFitness > threshold:
    d = []
    old_point = startingPoint
    for index in range(512):
        increment_arr = numpy.zeros(512)
        increment_arr[index] = increment
        newpoint = startingPoint + increment_arr
        d1 = newDistance(newpoint, point1)
        newpoint = startingPoint - increment_arr
        d2 = newDistance(newpoint, point1)
        d.append(d1)
        d.append(d2)
    
    # print(i,' ', startingPoint[:4])
    d = numpy.array(d)
    minimum = min(d[:,512])
    if minimum < minDistance:
        minindex = numpy.argmin(d[:,512])
        startingPoint = d[minindex,:512]
        minDistance = minimum
        print('found ',i,' ', startingPoint[:4],'score',d[minindex,512])
        lastFitness = d[minindex,512]
    else:
        flag = False
        # print('new start poiny ',i)
        startingPoint = startingPoint + numpy.random.random(512) * 0.1
    i+=1

print('target:',point1[:10])
print('result:',startingPoint[:10])

运行结果：

found  3319   [0.54880071 0.14943408 0.16443942 0.67552999] score 0.40665809714898343
found  3320   [0.54880071 0.14943408 0.16443942 0.67552999] score 0.4050587747707967
found  3321   [0.54880071 0.14943408 0.16443942 0.67552999] score 0.4034618667141106
found  3322   [0.54880071 0.14943408 0.16443942 0.67552999] score 0.4018617552269553
found  3323   [0.54880071 0.14943408 0.16443942 0.67552999] score 0.4002772234555598
found  3324   [0.54880071 0.14943408 0.16443942 0.67552999] score 0.3986881730615146
target: [0.54391242 0.16028716 0.15798658 0.6986903  0.74380195 0.25841304
 0.40675962 0.2248158  0.76813796 0.7937874 ]
result: [0.54880071 0.14943408 0.16443942 0.67552999 0.73628524 0.2546786
 0.37020921 0.2397613  0.75941863 0.75618421]

import numpy as np
from scipy.optimize import minimize

xt = np.random.rand(512) * 2
print('x terget:',xt[:10])

def rosen(x):
    """The Rosenbrock function"""
    return np.sqrt(np.sum((x - xt)**2))


# 初始迭代点
x0 = np.random.rand(512) * 2

res = minimize(rosen, x0, method='BFGS', jac=[],
               options='disp': True)

print('res',res.x[:10])