使用 Pygame 构建和可视化数独游戏
数独是一个基于逻辑的组合数字放置谜题。目标是用数字填充 9×9 网格，以便构成网格的每一列、每一行和九个 3×3 子网格中的每一个都包含从 1 到 9 的所有数字。
我们将使用 pygame 库在Python中构建数独游戏，并使用回溯算法自动化游戏。
实现的功能：
- 游戏界面玩
- 自动解决
- 自动求解的可视化，即回溯算法可视化
- 选项：重置，清除游戏
先决条件：
- 必须预装 Pygame 库
- 回溯算法知识
实施步骤：
1. Fill the pygame window with Sudoku Board i.e., Construct a 9×9 grid.
2. Fill the board with default numbers.
3. Assign a specific key for each operations and listen it.
4. Integrate the backtracking algorithm into it.
5. Use set of colors to visualize auto solving.
操作说明：
- Press ‘Enter’ To Auto Solve and Visualize.
- To play the game manually,
Place the cursor in any cell you want and enter the number. - At any point, press enter to solve automatically.
以下是实施：
Python3
# import pygame library
import pygame
# initialise the pygame font
pygame.font.init()
# Total window
screen = pygame.display.set_mode((500, 600))
# Title and Icon
pygame.display.set_caption("SUDOKU SOLVER USING BACKTRACKING")
img = pygame.image.load('icon.png')
pygame.display.set_icon(img)
x = 0
y = 0
dif = 500 / 9
val = 0
# Default Sudoku Board.
grid =[
[7, 8, 0, 4, 0, 0, 1, 2, 0],
[6, 0, 0, 0, 7, 5, 0, 0, 9],
[0, 0, 0, 6, 0, 1, 0, 7, 8],
[0, 0, 7, 0, 4, 0, 2, 6, 0],
[0, 0, 1, 0, 5, 0, 9, 3, 0],
[9, 0, 4, 0, 6, 0, 0, 0, 5],
[0, 7, 0, 3, 0, 0, 0, 1, 2],
[1, 2, 0, 0, 0, 7, 4, 0, 0],
[0, 4, 9, 2, 0, 6, 0, 0, 7]
]
# Load test fonts for future use
font1 = pygame.font.SysFont("comicsans", 40)
font2 = pygame.font.SysFont("comicsans", 20)
def get_cord(pos):
global x
x = pos[0]//dif
global y
y = pos[1]//dif
# Highlight the cell selected
def draw_box():
for i in range(2):
pygame.draw.line(screen, (255, 0, 0), (x * dif-3, (y + i)*dif), (x * dif + dif + 3, (y + i)*dif), 7)
pygame.draw.line(screen, (255, 0, 0), ( (x + i)* dif, y * dif ), ((x + i) * dif, y * dif + dif), 7)
# Function to draw required lines for making Sudoku grid
def draw():
# Draw the lines
for i in range (9):
for j in range (9):
if grid[i][j]!= 0:
# Fill blue color in already numbered grid
pygame.draw.rect(screen, (0, 153, 153), (i * dif, j * dif, dif + 1, dif + 1))
# Fill grid with default numbers specified
text1 = font1.render(str(grid[i][j]), 1, (0, 0, 0))
screen.blit(text1, (i * dif + 15, j * dif + 15))
# Draw lines horizontally and verticallyto form grid
for i in range(10):
if i % 3 == 0 :
thick = 7
else:
thick = 1
pygame.draw.line(screen, (0, 0, 0), (0, i * dif), (500, i * dif), thick)
pygame.draw.line(screen, (0, 0, 0), (i * dif, 0), (i * dif, 500), thick)
# Fill value entered in cell
def draw_val(val):
text1 = font1.render(str(val), 1, (0, 0, 0))
screen.blit(text1, (x * dif + 15, y * dif + 15))
# Raise error when wrong value entered
def raise_error1():
text1 = font1.render("WRONG !!!", 1, (0, 0, 0))
screen.blit(text1, (20, 570))
def raise_error2():
text1 = font1.render("Wrong !!! Not a valid Key", 1, (0, 0, 0))
screen.blit(text1, (20, 570))
# Check if the value entered in board is valid
def valid(m, i, j, val):
for it in range(9):
if m[i][it]== val:
return False
if m[it][j]== val:
return False
it = i//3
jt = j//3
for i in range(it * 3, it * 3 + 3):
for j in range (jt * 3, jt * 3 + 3):
if m[i][j]== val:
return False
return True
# Solves the sudoku board using Backtracking Algorithm
def solve(grid, i, j):
while grid[i][j]!= 0:
if i<8:
i+= 1
elif i == 8 and j<8:
i = 0
j+= 1
elif i == 8 and j == 8:
return True
pygame.event.pump()
for it in range(1, 10):
if valid(grid, i, j, it)== True:
grid[i][j]= it
global x, y
x = i
y = j
# white color background\
screen.fill((255, 255, 255))
draw()
draw_box()
pygame.display.update()
pygame.time.delay(20)
if solve(grid, i, j)== 1:
return True
else:
grid[i][j]= 0
# white color background\
screen.fill((255, 255, 255))
draw()
draw_box()
pygame.display.update()
pygame.time.delay(50)
return False
# Display instruction for the game
def instruction():
text1 = font2.render("PRESS D TO RESET TO DEFAULT / R TO EMPTY", 1, (0, 0, 0))
text2 = font2.render("ENTER VALUES AND PRESS ENTER TO VISUALIZE", 1, (0, 0, 0))
screen.blit(text1, (20, 520))
screen.blit(text2, (20, 540))
# Display options when solved
def result():
text1 = font1.render("FINISHED PRESS R or D", 1, (0, 0, 0))
screen.blit(text1, (20, 570))
run = True
flag1 = 0
flag2 = 0
rs = 0
error = 0
# The loop thats keep the window running
while run:
# White color background
screen.fill((255, 255, 255))
# Loop through the events stored in event.get()
for event in pygame.event.get():
# Quit the game window
if event.type == pygame.QUIT:
run = False
# Get the mouse position to insert number
if event.type == pygame.MOUSEBUTTONDOWN:
flag1 = 1
pos = pygame.mouse.get_pos()
get_cord(pos)
# Get the number to be inserted if key pressed
if event.type == pygame.KEYDOWN:
if event.key == pygame.K_LEFT:
x-= 1
flag1 = 1
if event.key == pygame.K_RIGHT:
x+= 1
flag1 = 1
if event.key == pygame.K_UP:
y-= 1
flag1 = 1
if event.key == pygame.K_DOWN:
y+= 1
flag1 = 1
if event.key == pygame.K_1:
val = 1
if event.key == pygame.K_2:
val = 2
if event.key == pygame.K_3:
val = 3
if event.key == pygame.K_4:
val = 4
if event.key == pygame.K_5:
val = 5
if event.key == pygame.K_6:
val = 6
if event.key == pygame.K_7:
val = 7
if event.key == pygame.K_8:
val = 8
if event.key == pygame.K_9:
val = 9
if event.key == pygame.K_RETURN:
flag2 = 1
# If R pressed clear the sudoku board
if event.key == pygame.K_r:
rs = 0
error = 0
flag2 = 0
grid =[
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0]
]
# If D is pressed reset the board to default
if event.key == pygame.K_d:
rs = 0
error = 0
flag2 = 0
grid =[
[7, 8, 0, 4, 0, 0, 1, 2, 0],
[6, 0, 0, 0, 7, 5, 0, 0, 9],
[0, 0, 0, 6, 0, 1, 0, 7, 8],
[0, 0, 7, 0, 4, 0, 2, 6, 0],
[0, 0, 1, 0, 5, 0, 9, 3, 0],
[9, 0, 4, 0, 6, 0, 0, 0, 5],
[0, 7, 0, 3, 0, 0, 0, 1, 2],
[1, 2, 0, 0, 0, 7, 4, 0, 0],
[0, 4, 9, 2, 0, 6, 0, 0, 7]
]
if flag2 == 1:
if solve(grid, 0, 0)== False:
error = 1
else:
rs = 1
flag2 = 0
if val != 0:
draw_val(val)
# print(x)
# print(y)
if valid(grid, int(x), int(y), val)== True:
grid[int(x)][int(y)]= val
flag1 = 0
else:
grid[int(x)][int(y)]= 0
raise_error2()
val = 0
if error == 1:
raise_error1()
if rs == 1:
result()
draw()
if flag1 == 1:
draw_box()
instruction()
# Update window
pygame.display.update()
# Quit pygame window
pygame.quit()输出：