Question

我正在尝试使用Python随机生成由ASCII艺术代表的陆块。我有以下代码，主要产生水或大部分土地。我怎么能实现我的最终目标呢？

from random import *

water = "."
land = "#"

waterChance = 40
landChance = 100 - waterChance

width = 40
height = 20

prevTile = water

level = []

for i in range(height):
    level += [[]]
    for j in range(width):

        # change to ternary 
        if prevTile == water:
            waterChance += 10
        elif prevTile == land:
            waterChance -= 10

        if(waterChance == 50):
            waterChance += 5

        tile = randrange(1,101)

        if tile > waterChance:
            print(land, end='')
            prevTile = land
        elif tile <= waterChance:
            print(water, end='')
            prevTile = water
    print()

我正在寻找类似的东西（将X替换为#s并用连字符替换空格。抱歉，我必须使用ASCIIFlow快速执行此操作）

    XXX
    XXXXX
   XXXXXXX
   XXXXXXXX         X
    XXXXXXX      XXXX
        XX       XXXXXX
                 XXXXXXX
                  XXXXXXX
               XXXXXXXXXX
  XXXXXX       XXXXXXX
 XXXXXXX
XXXXXXXX
XXXXXXXXXXX
 XXXXXXXXXXXXXX
  XX XXXXXXXXXXXX
  XXXXXXXXXXXXXXX
   XXXXXXXXXXXXXXX
   XXXXXXXXXXXXXXXXXX
   XXXXXXXXXXXXXXXX
    XXXXXXXXXXXX

Answer 1

这是连接性的loss function。

将总损失作为土地与水的比率+连续性的一些常数。现在，您有一个地图“良好”的指标。地图生成器中的所有硬编码值都是参数。对这些参数执行grid search以获得高“良好”值。

这减少了调整一个值的问题，即损失函数中的常数。这是可取的，因为现在用户可以权衡连通性和土地和水的比率。

修改这是第一个链接的连通性功能。

用布尔2D矩阵计算岛屿的程序

class Graph: def __init__(self, row, col, g): self.ROW = row self.COL = col self.graph = g # A function to check if a given cell # (row, col) can be included in DFS def isSafe(self, i, j, visited): # row number is in range, column number # is in range and value is 1 # and not yet visited return (i >= 0 and i < self.ROW and j >= 0 and j < self.COL and not visited[i][j] and self.graph[i][j]) # A utility function to do DFS for a 2D # boolean matrix. It only considers # the 8 neighbours as adjacent vertices def DFS(self, i, j, visited): # These arrays are used to get row and # column numbers of 8 neighbours # of a given cell rowNbr = [-1, -1, -1, 0, 0, 1, 1, 1]; colNbr = [-1, 0, 1, -1, 1, -1, 0, 1]; # Mark this cell as visited visited[i][j] = True # Recur for all connected neighbours for k in range(8): if self.isSafe(i + rowNbr[k], j + colNbr[k], visited): self.DFS(i + rowNbr[k], j + colNbr[k], visited) # The main function that returns # count of islands in a given boolean # 2D matrix def countIslands(self): # Make a bool array to mark visited cells. # Initially all cells are unvisited visited = [[False for j in range(self.COL)]for i in range(self.ROW)] # Initialize count as 0 and travese # through the all cells of # given matrix count = 0 for i in range(self.ROW): for j in range(self.COL): # If a cell with value 1 is not visited yet, # then new island found if visited[i][j] == False and self.graph[i][j] ==1: # Visit all cells in this island # and increment island count self.DFS(i, j, visited) count += 1 return count graph = [[1, 1, 0, 0, 0], [0, 1, 0, 0, 1], [1, 0, 0, 1, 1], [0, 0, 0, 0, 0], [1, 0, 1, 0, 1]] row = len(graph) col = len(graph[0]) g= Graph(row, col, graph) print "Number of islands is :" print g.countIslands()

如何随机生成ASCII艺术土地？

1 个答案: