求解5x5数字游戏的算法

Question

我一直试图创建一个算法来解决5x5数字游戏(点击规则和评分)。然而，我一直被困在，因为我不知道如何找到最佳位置，以确定数字的去向。到目前为止，我计算的分数，但不能取得进一步的进展。到目前为止，我想出的糟糕代码是：

import numpy as np
import random

# i could use an array to represent board but im more familiar with nested lists
# when i get the vertical score i do have to represent it as a array to do that math
board = [[1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]

# gets the score horizontally
def getHorizontalScore(board):
  score = 0
  # i know it is inefficient but this is what i came up with inorder to figure this out
  for list in board:
    i_zero = list.count(0)
    i_one = list.count(1)
    i_two = list.count(2)
    i_three = list.count(3)
    i_four = list.count(4)
    i_five = list.count(5)
    i_six = list.count(6)
    i_seven = list.count(7)
    i_eight = list.count(8)
    i_nine = list.count(9)
    i_ten = list.count(10)
    if i_zero > 1:
      score += 0 * i_zero
    if i_one > 1:
      score += i_one
    if i_two > 1:
      score += i_two * 2
    if i_three > 1:
      score += i_three * 3
    if i_four > 1:
      score += i_four * 4
    if i_five > 1:
      score += i_five * 5
    if i_six > 1:
      score += i_six * 6
    if i_seven > 1:
      score += i_seven * 7
    if i_eight > 1:
      score += i_eight * 8
    if i_nine > 1:
      score += i_nine * 9
    if i_ten > 1:
      score += i_ten * 10
  return score

# gets the score verticaly (hopefully)
def getVerticalScore(board):
  # represent board as an array
  board = np.asarray(board)
  #score
  score = 0
  # i know it is inefficient but this is what i came up with inorder to figure this out
  # board.T is so useful praise numpy for adding this feature
  for column in board.T:
    i_one = list(column).count(1)
    i_two = list(column).count(2)
    i_three = list(column).count(3)
    i_four = list(column).count(4)
    i_five = list(column).count(5)
    i_six = list(column).count(6)
    i_seven = list(column).count(7)
    i_eight = list(column).count(8)
    i_nine = list(column).count(9)
    i_ten = list(column).count(10)
    if i_one > 1:
      score += i_one
    if i_two > 1:
      score += i_two * 2
    if i_three > 1:
      score += i_three * 3
    if i_four > 1:
      score += i_four * 4
    if i_five > 1:
      score += i_five * 5
    if i_six > 1:
      score += i_six * 6
    if i_seven > 1:
      score += i_seven * 7
    if i_eight > 1:
      score += i_eight * 8
    if i_nine > 1:
      score += i_nine * 9
    if i_ten > 1:
      score += i_ten * 10
    return score


def getScore(board):
  return getVerticalScore(board) + getHorizontalScore(board)
  
def displayBoard(board):
  for list in board:
    print(list)

如果有人能帮我弄清楚如何使算法和压缩我的代码一点点，请帮助我。

Answer 1

您对行和列的分数计算已关闭。

让我们以这个板为例：

board = [[7, 7, 2, 3, 7],     # 14
         [6, 6, 2, 3, 3],     # 18
         [2, 4, 2, 8, 8],     # 16
         [1, 10, 9, 5, 8],    # 0
         [10, 10, 10, 5, 8]]  # 30
#          0  20  6  16  24
# total score = 144

如果我使用您的getScore()函数，我将得到89而不是144

用于获取倍数的代码不考虑数字是否相邻：

for list in board:
#   ...
    i_seven = list.count(7) # this will return 3 although only 2 7's are adjacent

在这种情况下，最好是使用索引遍历您的董事会。看看这个关于访问嵌套列表中的元素的答案。

# Getting each element per row
for row in board:
   for pos in range(5):
      row_num = row[pos]

# Getting each element per column
for col in range(5):
   for pos in range(5):
      col_num = board[pos][col]

你不需要数0，因为它们表示你的棋盘上有一个空字段。所以你的算法知道在一个字段上放一个新的数字是合法的。