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In this article, the editor introduces in detail "how to achieve Gobang algorithm in python". The content is detailed, the steps are clear, and the details are handled properly. I hope this article "how to achieve Gobang algorithm in python" can help you solve your doubts.

If will win chess if index-fun.

< 0 or pos[index-fun][2] != p[2]: #当前点击的格子是第一行 或 不在第一行,在假设的首个格子中 print('----------第一个格子') for i in range(1,5): # (1-4) if index + (fun * i) = 4: if p[2] == 1: print('最后一下为第一个格子--黑赢') ShowAllChess()#显示所有已经放置了的棋子 ResultShow(1) if p[2] == 2: print('最后一下为第一个格子--白赢') ShowAllChess()#显示所有已经放置了的棋子 ResultShow(2) elif index + fun >

191 or pos [index + fun] [2]! = p [2]: # the grid currently clicked is the last line or not in the last line In the last grid, print ('- last grid') for i in range (1 if 5): # (1-4) if pos [index-(fun * I)] [2] = = p [2]: chessValue = chessValue + 1 if chessValue > = 4: if p [2] = = 1: print ('the last one is the last grid-black wins') ShowAllChess () # shows all the pieces that have been placed ResultShow (1) if p [2] = = 2: print ('the last one is the last grid-win for nothing ') ShowAllChess () # shows all the pieces that have been placed ResultShow (2) else: # not the first and not the last That is, what is currently clicked is the pawn in the middle of the five beads, the print ('- middle grid')''the first cycle to detect whether there are similar pieces in the upper-middle lattice' 'for i in range (1-3): # (1-3) three cycles If index-(fun * I) > = 0 and pos [index-(fun * I)] [2] = = p [2]: chessValue = chessValue + 1 elif index-(fun * I) > = 0 and (pos [index-(fun * I)] [2]! = p [2]\ and pos [index-(fun * I)] [2] ]! = 0): # the logic here is that the top is the opponent chess piece chessValue = 0 # the number of beads is returned to zero 'for i in range (1): # (1-3) three cycles if index + (fun * I)

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