如何解决为什么这个R版本的“骑士之旅”不起作用?
我是R的新手,我试图找到从角落开始时骑士在棋盘上移动所需的最少移动次数。
我从此网站使用了Python算法: https://www.geeksforgeeks.org/the-knights-tour-problem-backtracking-1/
并且我试图将其翻译为R。
但是,当我运行该程序时,其输出为:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 0 -1 -1 -1 -1 -1 -1 -1
[2,] -1 -1 -1 -1 -1 -1 -1 -1
[3,] -1 -1 -1 -1 -1 -1 -1 -1
[4,] -1 -1 -1 -1 -1 -1 -1 -1
[5,] -1 -1 -1 -1 -1 -1 -1 -1
[6,] -1 -1 -1 -1 -1 -1 -1 -1
[7,] -1 -1 -1 -1 -1 -1 -1 -1
[8,] -1 -1 -1 -1 -1 -1 -1 -1
[1] "Minimum number of moves: -1"
该如何解决此问题?
这是代码:
chess = rep(-1,times = 64)
board = matrix(data = chess,nrow = 8,ncol = 8,byrow = TRUE)
move_x = c(2,1,-1,-2,2)
move_y = c(1,2,-1)
board[1,1] = 0
pos = 1
valid_move <- function (x,y,board) {
if (x >= 1 && y >= 1 && x <= 8 && y <= 8 && board[x,y] == -1) {
return (T)
}
return (F)
}
solve <- function (board,curr_x,curr_y,move_x,move_y,pos) {
if (pos == 64) {
return (T)
}
for (i in seq(1,8)) {
new_x = curr_x + move_x[i]
new_y = curr_y + move_y[i]
if (valid_move(new_x,new_y,board)) {
board[new_x,new_y] = pos
if (solve(board,new_x,(pos+1))) {
return (TRUE)
}
board[new_x,new_y] = -1
}
}
return (F)
}
main <- function() {
sims = 10
ctr = 0
number_of_moves = c()
solve(board,pos)
for (x in board) {
for (y in board) {
number_of_moves <- c(number_of_moves,board[x,y])
}
}
print(board)
print(paste("Minimum number of moves: ",min(number_of_moves)))
}
main()
解决方法
在R中,当函数对其参数之一进行更改时,它仅对本地副本进行更改,而不对原始变量进行更改。
例如,在此R代码段中,我们可以看到该函数实际上并未修改变量l
。
try_to_modify <- function(l) l[[1]] <- -100
l <- list(1)
try_to_modify(l)
l
#> [[1]]
#> [1] 1
将此与python进行对比,它实际上在 修改了l
。
# (python code)
def try_to_modify(l):
l[0] = -100
l = [1]
try_to_modify(l)
l
#> [-100]
如果您希望函数与调用者进行通讯,则它要么需要修改全局变量(通常不是最佳解决方案),要么需要使用返回值。 (有一些例外,但这通常是它的工作方式。)
因此,您可以返回TRUE
或FALSE
,而不是返回board
或NULL
。
valid_move <- function (x,y,board) {
x >= 1 && y >= 1 && x <= 8 && y <= 8 && board[x,y] == -1
}
solve <- function (board,curr_x,curr_y,move_x,move_y,pos) {
if (pos == 64) {
return (board)
}
for (i in seq(1,8)) {
new_x = curr_x + move_x[i]
new_y = curr_y + move_y[i]
if (valid_move(new_x,new_y,board)) {
board[new_x,new_y] = pos
result <- solve(board,new_x,(pos + 1))
if (!is.null(result)) {
return (result)
}
board[new_x,new_y] = -1
}
}
# Return NULL
# As this is the last result of the function,you don't need to write `return (NULL)`
NULL
}
final_board <- solve(
board = matrix(
c(0,rep_len(-1,63)),nrow = 8,ncol = 8,byrow = TRUE
),curr_x = 1,curr_y = 1,move_x = c(2,1,-1,-2,2),move_y = c(1,2,-1),pos = 1
)
final_board
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 0 59 38 33 30 17 8 63
#> [2,] 37 34 31 60 9 62 29 16
#> [3,] 58 1 36 39 32 27 18 7
#> [4,] 35 48 41 26 61 10 15 28
#> [5,] 42 57 2 49 40 23 6 19
#> [6,] 47 50 45 54 25 20 11 14
#> [7,] 56 43 52 3 22 13 24 5
#> [8,] 51 46 55 44 53 4 21 12
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