游戏
您将玩（几乎）Connect-4的标准游戏。不幸的是，这是一个对应游戏，有人从底部开始在第二行上放了黑色胶带，因此您在这些行中看不到对手的任何举动。
在已经满满的栏中进行的任何移动都将视为您的回合，如果游戏进行的时间超过6 * 7回合，则将其视为平局。
挑战规格
您的程序应实现为Python 3函数。第一个参数是棋盘的“视图”，代表已知棋盘的状态，是从下到上的2D行列表，其中1第一个玩家2移动，第二个玩家移动，0空位置或隐藏被你的对手移动。
第二个参数是从索引的转码0，其奇偶性告诉您您是哪个玩家。
最后一个参数是任意状态，初始化为None每个游戏开始时的状态，您可以使用该状态来保留回合之间的状态。
您应该返回要播放的列索引的2元组，并在下一轮返回新状态。
计分
胜利计为+1，平局计为，0损失计为-1。您的目标是在循环锦标赛中获得最高的平均分数。我会尝试根据需要进行尽可能多的比赛，以确定明确的获胜者。
规则
任何竞争者一次最多只能拥有一个竞争机器人，但是如果您进行了改进，可以更新您的参赛作品。请尝试将您的漫游器限制为每转大约1秒的思考时间。
测试中
这是控制器的源代码，以及一些非竞争性示例bot供参考：
import itertools
import random
def get_strides(board, i, j):
yield ((i, k) for k in range(j + 1, 7))
yield ((i, k) for k in range(j - 1, -1, -1))
yield ((k, j) for k in range(i + 1, 6))
yield ((k, j) for k in range(i - 1, -1, -1))
directions = [(1, 1), (-1, -1), (1, -1), (-1, 1)]
def diag(di, dj):
i1 = i
j1 = j
while True:
i1 += di
if i1 < 0 or i1 >= 6:
break
j1 += dj
if j1 < 0 or j1 >= 7:
break
yield (i1, j1)
for d in directions:
yield diag(*d)
DRAWN = 0
LOST = 1
WON = 2
UNDECIDED = 3
def get_outcome(board, i, j):
if all(board[-1]):
return DRAWN
player = board[i][j]
strides = get_strides(board, i, j)
for _ in range(4):
s0 = next(strides)
s1 = next(strides)
n = 1
for s in (s0, s1):
for i1, j1 in s:
if board[i1][j1] == player:
n += 1
if n >= 4:
return WON
else:
break
return UNDECIDED
def apply_move(board, player, move):
for i, row in enumerate(board):
if board[i][move] == 0:
board[i][move] = player
outcome = get_outcome(board, i, move)
return outcome
if all(board[-1]):
return DRAWN
return UNDECIDED
def get_view(board, player):
view = [list(row) for row in board]
for i, row in enumerate(view):
if i % 2:
continue
for j, x in enumerate(row):
if x == 3 - player:
row[j] = 0
return view
def run_game(player1, player2):
players = {1 : player1, 2 : player2}
board = [[0] * 7 for _ in range(6)]
states = {1 : None, 2 : None}
for turn in range(6 * 7):
p = (turn % 2) + 1
player = players[p]
view = get_view(board, p)
move, state = player(view, turn, states[p])
outcome = apply_move(board, p, move)
if outcome == DRAWN:
return DRAWN
elif outcome == WON:
return p
else:
states[p] = state
return DRAWN
def get_score(counts):
return (counts[WON] - counts[LOST]) / float(sum(counts))
def run_tournament(players, rounds=10000):
counts = [[0] * 3 for _ in players]
for r in range(rounds):
for i, player1 in enumerate(players):
for j, player2 in enumerate(players):
if i == j:
continue
outcome = run_game(player1, player2)
if outcome == DRAWN:
for k in i, j:
counts[k][DRAWN] += 1
else:
if outcome == 1:
w, l = i, j
else:
w, l = j, i
counts[w][WON] += 1
counts[l][LOST] += 1
ranks = sorted(range(len(players)), key = lambda i: get_score(counts[i]), reverse=True)
print("Round %d of %d\n" % (r + 1, rounds))
rows = [("Name", "Draws", "Losses", "Wins", "Score")]
for i in ranks:
name = players[i].__name__
score = get_score(counts[i])
rows.append([name + ":"] + [str(n) for n in counts[i]] + ["%6.3f" % score])
lengths = [max(len(s) for s in col) + 1 for col in zip(*rows)]
for i, row in enumerate(rows):
padding = ((n - len(s)) * ' ' for s, n in zip(row, lengths))
print(''.join(s + p for s, p in zip(row, padding)))
if i == 0:
print()
print()
def random_player(view, turn, state):
return random.randrange(0, 7), state
def constant_player(view, turn, state):
return 0, state
def better_random_player(view, turn, state):
while True:
j = random.randrange(0, 7)
if view[-1][j] == 0:
return j, state
def better_constant_player(view, turn, state):
for j in range(7):
if view[-1][j] == 0:
return j, state
players = [random_player, constant_player, better_random_player, better_constant_player]
run_tournament(players)
祝您一切顺利！
临时结果
Name Draws Losses Wins Score
zsani_bot: 40 5377 94583 0.892
better_constant_player: 0 28665 71335 0.427
constant_player: 3 53961 46036 -0.079
normalBot: 38 64903 35059 -0.298
better_random_player: 192 71447 28361 -0.431
random_player: 199 75411 24390 -0.510
您能解释一下为什么查看view [-1] [j] == 0吗？我不完全确定我看到您在哪里填充它们，而我的python知识似乎有点生锈。
—
野蛮人
@ Barbarian772我正在检查该列中是否还有空间。请注意，共有6行，因此最上面一行是完整观察到的。
—
user1502040
您不应该将放置在已满的列中视为通过。许多连接4个游戏的结束时只有一栏未填充，如果一位玩家在该列中输了钱，则这将使游戏平局，否则这将迫使一位玩家获胜。
—
soktinpk
@soktinpk难道不只是增加另一层策略吗？毕竟Connect-4是一款已解决的游戏，因此转弯跳过因素足以满足规则更改的要求，而贡献者不能仅使用标准算法。
—
mypetlion
从底部为零索引，是粘贴的行（0,2,4,6）还是（1,3,5）？某些ASCII艺术将有所帮助。
—
SIGSTACKFAULT