게임 보드 생성을 위해 파이썬에서 파이 게임을 사용하여 여러 선의 모든 교차점 찾기
Aug 21 2020
내 코드에서 줄의 모든 교차점을 찾아야합니다. 이 시점에서 나는 내 게임 조각을 넣고 싶습니다. 게임의 논리는 틱택 토 (tic tac toe)와 비슷합니다. 플레이어가 같은 색의 구슬 3 개를 연속으로 배치하면 순서대로 배열되지 않은 다른 플레이어의 조각을 잡을 수 있습니다.
지금까지 내 코드 :
import pygame
# Define some colors
BLACK = (0, 0, 0)
WHITE = (255, 255, 255)
GREEN = (0, 255, 0)
RED = (255, 0, 0)
# This sets the WIDTH and HEIGHT of each grid location
WIDTH = 20
HEIGHT = 20
# This sets the margin between each cell
MARGIN = 5
# Create a 2 dimensional array. A two dimensional
# array is simply a list of lists.
grid = []
for row in range(19):
# Add an empty array that will hold each cell
# in this row
grid.append([])
for column in range(19):
grid[row].append(0) # Append a cell
# Set row 1, cell 5 to one. (Remember rows and
# column numbers start at zero.)
grid[1][5] = 1
pygame.init()
WINDOW_SIZE = [800, 600]
screen = pygame.display.set_mode(WINDOW_SIZE)
pygame.display.set_caption("Array Backed Grid")
done = False
clock = pygame.time.Clock()
while not done:
# Set the screen background
screen.fill(BLACK)
# Draw the grid
for row in range(19):
for column in range(19):
color = WHITE
if grid[row][column] == 1:
color = GREEN
board_lines = [
( 13,15,462,15 ), ( 13,469,462,469 ), #lin1 and line2,outer rect
( 62,86,409,86 ), ( 62,389,409,389 ), #line3 and l4,mid reect
( 114,186,360,186 ), ( 114,318,360,318 ), #line5,l6,internl rect
( 13,15,13,469 ), ( 462,12,462,469 ), #line9,l10,left and right sides
( 62,86,62,389 ), ( 409,85,409,389 ), #l7,l8left and right sides
( 114,186,114,316), ( 360,187,360,318 ), #l11,lin12left and right sides
( 237,15,237,186 ), ( 237,469,237,320 ), #upper V.line,lowerV
( 13,242,113,242 ), ( 360,242,462,242 ) #rIGHT LEFT hoRIZONTAL LINE
]
for line in board_lines:
line_from = ( line[0], line[1] )
line_to = ( line[2], line[3] )
pygame.draw.line( screen, RED, line_from, line_to, 3)
# Limit to 60 frames per second
clock.tick(60)
pygame.display.flip()
pygame.quit()
답변
1 Rabbid76 Aug 21 2020 at 20:08
다음 함수는 점 ( P0, P1) 및 ( Q0, Q1) 로 주어진 두 선의 교차점을 계산합니다 .
def lineLineIntersect(P0, P1, Q0, Q1):
d = (P1[0]-P0[0]) * (Q1[1]-Q0[1]) + (P1[1]-P0[1]) * (Q0[0]-Q1[0])
if d == 0:
return None
t = ((Q0[0]-P0[0]) * (Q1[1]-Q0[1]) + (Q0[1]-P0[1]) * (Q0[0]-Q1[0])) / d
u = ((Q0[0]-P0[0]) * (P1[1]-P0[1]) + (Q0[1]-P0[1]) * (P0[0]-P1[0])) / d
if 0 <= t <= 1 and 0 <= u <= 1:
return round(P1[0] * t + P0[0] * (1-t)), round(P1[1] * t + P0[1] * (1-t))
return None
알고리즘은 선 교차 계산 문제 에 대한 답변에 자세히 설명되어 있습니다 .
P ... point on the 1. line
R ... normalized direction of the 1. line
Q ... point on the 2. line
S ... normalized direction of the 2. line
alpha ... angle between Q-P and R
beta ... angle between R and S
gamma = 180° - alpha - beta
h = | Q - P | * sin(alpha)
u = h / sin(beta)
t = | Q - P | * sin(gamma) / sin(beta)
t = dot(Q-P, (S.y, -S.x)) / dot(R, (S.y, -S.x)) = determinant(mat2(Q-P, S)) / determinant(mat2(R, S))
u = dot(Q-P, (R.y, -R.x)) / dot(R, (S.y, -S.x)) = determinant(mat2(Q-P, R)) / determinant(mat2(R, S))
X = P + R * t = Q + S * u
선분이 다른 선분과 교차하는지 평가해야합니다. 다음 함수를 사용하여 중첩 된 루프의 행을 반복하고 자신을 제외한 다른 행과의 교차점을 찾습니다 lineLineIntersect.
intersectionPoints = []
for i, line1 in enumerate(board_lines):
for line2 in board_lines[i:]:
isectP = lineLineIntersect(line1[:2], line1[2:], line2[:2], line2[2:])
if isectP:
intersectionPoints.append(isectP)
교차점에 원 그리기 :
for isectP in intersectionPoints:
pygame.draw.circle(screen, GREEN, isectP, 5)
예를 참조하십시오.
import pygame
import math
def lineLineIntersect(P0, P1, Q0, Q1):
d = (P1[0]-P0[0]) * (Q1[1]-Q0[1]) + (P1[1]-P0[1]) * (Q0[0]-Q1[0])
if d == 0:
return None
t = ((Q0[0]-P0[0]) * (Q1[1]-Q0[1]) + (Q0[1]-P0[1]) * (Q0[0]-Q1[0])) / d
u = ((Q0[0]-P0[0]) * (P1[1]-P0[1]) + (Q0[1]-P0[1]) * (P0[0]-P1[0])) / d
if 0 <= t <= 1 and 0 <= u <= 1:
return round(P1[0] * t + P0[0] * (1-t)), round(P1[1] * t + P0[1] * (1-t))
return None
board_lines = [
( 13,15,462,15 ), ( 13,469,462,469 ), #lin1 and line2,outer rect
( 62,86,409,86 ), ( 62,389,409,389 ), #line3 and l4,mid reect
( 114,186,360,186 ), ( 114,318,360,318 ), #line5,l6,internl rect
( 13,15,13,469 ), ( 462,12,462,469 ), #line9,l10,left and right sides
( 62,86,62,389 ), ( 409,85,409,389 ), #l7,l8left and right sides
( 114,186,114,316), ( 360,187,360,318 ), #l11,lin12left and right sides
( 237,15,237,186 ), ( 237,469,237,320 ), #upper V.line,lowerV
( 13,242,113,242 ), ( 360,242,462,242 ) #rIGHT LEFT hoRIZONTAL LINE
]
pygame.init()
intersectionPoints = []
for i, line1 in enumerate(board_lines):
for line2 in board_lines[i:]:
isectP = lineLineIntersect(line1[:2], line1[2:], line2[:2], line2[2:])
if isectP:
intersectionPoints.append(isectP)
# Define some colors
BLACK = (0, 0, 0)
WHITE = (255, 255, 255)
GREEN = (0, 255, 0)
RED = (255, 0, 0)
# This sets the WIDTH and HEIGHT of each grid location
WIDTH = 20
HEIGHT = 20
# This sets the margin between each cell
MARGIN = 5
# Create a 2 dimensional array. A two dimensional
# array is simply a list of lists.
grid = []
for row in range(19):
# Add an empty array that will hold each cell
# in this row
grid.append([])
for column in range(19):
grid[row].append(0) # Append a cell
# Set row 1, cell 5 to one. (Remember rows and
# column numbers start at zero.)
grid[1][5] = 1
WINDOW_SIZE = [800, 600]
screen = pygame.display.set_mode(WINDOW_SIZE)
pygame.display.set_caption("Array Backed Grid")
done = False
clock = pygame.time.Clock()
while not done:
for event in pygame.event.get():
if event.type == pygame.QUIT:
done = True
# Set the screen background
screen.fill(BLACK)
# Draw the grid
for row in range(19):
for column in range(19):
color = WHITE
if grid[row][column] == 1:
color = GREEN
for line in board_lines:
pygame.draw.line(screen, RED, line[:2], line[2:], 3)
for isectP in intersectionPoints:
pygame.draw.circle(screen, GREEN, isectP, 5)
# Limit to 60 frames per second
clock.tick(60)
pygame.display.flip()
pygame.quit()