Я читаю код ниже о программе First Search - искусственном интеллекте для робототехники, и я немного застрял с работой этих двух строк ниже:
x2 = x+delta[i][0]
y2 = y+delta[i][1]
Я знаю о дельте для схемы перемещения от узла к соседям, но я не мог понять, как эти две линии работают здесь. Кто-нибудь может мне их объяснить, пожалуйста?
Код ниже:
#grid format
# 0 = navigable space
# 1 = occupied space
grid=[[0,0,1,0,0,0],
[0,0,1,0,0,0],
[0,0,0,0,1,0],
[0,0,1,1,1,0],
[0,0,0,0,1,0]]
init = [0,0]
goal = [len(grid)-1,len(grid[0])-1]
#Below the four potential actions to the single field
delta=[[-1, 0], #up
[ 0,-1], #left
[ 1, 0], #down
[ 0, 1]] #right
delta_name = ['^','<','V','>']
cost = 1
def search():
#open list elements are of the type [g,x,y]
closed = [[0 for row in range(len(grid[0]))] for col in range(len(grid))]
#We initialize the starting location as checked
closed[init[0]][init[1]] = 1
# we assigned the cordinates and g value
x = init[0]
y = init[1]
g = 0
#our open list will contain our initial value
open = [[g,x,y]]
found = False #flag that is set when search complete
resign= False #Flag set if we can't find expand
#print('initial open list:')
#for i in range(len(open)):
#print(' ', open[i])
#print('----')
while found is False and resign is False:
#Check if we still have elements in the open list
if len(open)==0: #If our open list is empty
resign=True
print('Fail')
print('############# Search terminated without success')
else:
#if there is still elements on our list
#remove node from list
open.sort()
open.reverse() #reverse the list
next = open.pop()
#print('list item')
#print('next')
#Then we assign the three values to x,y and g.
x = next[1]
y = next[2]
g = next[0]
#Check if we are done
if x == goal[0] and y == goal[1]:
found = True
print(next) #The three elements above this if
print('############## Search is success')
else:
#expand winning element and add to new open list
for i in range(len(delta)):
x2 = x+delta[i][0]
y2 = y+delta[i][1]
#if x2 and y2 falls into the grid
if x2 >= 0 and x2 < len(grid) and y2 >=0 and y2 <= len(grid[0])-1:
#if x2 and y2 not checked yet and there is not obstacles
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g+cost #we increment the cose
open.append([g2,x2,y2])#we add them to our open list
#print('append list item')
#print([g2,x2,y2])
#Then we check them to never expand again
closed[x2][y2] = 1
search()