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If you see any posts or comments violating these rules, please report them. I'm having trouble writing functions to check if the winner has won diagonally. Ox is the string passed into the function holding the type of checker we are looking for (black/red) etc. Def isDiagonalWin1(self,ox): for row in range(self.height): for col in range(self.width-3): if self.data[row][col] == self.data[row-1][col - 1] == self.data[row-2][col - 2] == self.data[row-3][col - 3] == ox: return True return False def isDiagonalWin2(self,ox): for row in range(self.height): for col in range(self.width-3): if self.data[row][col] == self.data[row+1][col + 1] == self. Kannada Full Movie 2014 Free Download. data[row+2][col + 2] == self.data[row+3][col + 3] == ox: return True return False • • • • •. Another method, rather than properly bounding your array is to create a table of all the pieces on the board and check those.

Def diagonal_check(game, color, win_count = 4): #search for all the positions that have the color in question #create a listing of their column and row found = [] winner = False for i, row in enumerate(game): for j, item in enumerate(row): if item == color: found.append([i,j]) #now iterate through that list to see if a forward diagonal exists for pair in found: for i in range(1, win_count): if [pair[0]+i, pair[1]+i] in found: winner = True else: winner = False break if winner == True: print found, winner return #no winner? What about reverse diagonal?

For pair in found: for i in range(1, win_count): if [pair[0]+i, pair[1]-i] in found: winner = True else: winner = False break if winner == True: print found, winner return #print the data for debug print found, winner d=[['R',0,'R',0,'R',0,0], [0,0,0,0,'Y',0,'Y'], [0,'Y',0,'R',0,'R',0], [0,0,'R',0,'R',0,0], [0,'Y',0,'Y',0,'R',0], [0,0,'Y',0,0,0,'R']] diagonal_check(d, 'R') • • • •.

I'm still pretty green working with python but I figured making my own game from scratch with what I know would be good practice. I've got this connect four game together and it works in so far as switching between players and 'dropping' their respective pieces.

Now I need a win condition, though honestly, I don't know where to start. I'd prefer some guidance as opposed to straight code for the sake of learning, but of course, any help would be greatly appreciated. Quick note, the current while loop is simply for debugging. I was thinking about setting a variable 'winner' to False and doing the loop 'while winner == False:' and have the win condition set this variable to True. Board = [] #List for holding the board for x in range(6): board.append(['O'] * 7) #builds 7 x 6 board (rows x columns) #function for printing the board def print_board(board): for row in board: print ' '.join(row) print 'Welcome to Connect Four' player_one = raw_input('Player 1. Enter your name: ') player_two = raw_input('Player 2.

Enter your name: ') #Gets players names print '%s vs.%s'% (player_one, player_two) print '--------------' print print_board(board) print 'Player 1 is Red(R) and Player 2 is Black(B)' print 'Let 's play!!' # Game's 'Opening' turn = 0 #Keeps track of turn while turn. You have a 7x6 board, and you need to see if a line of 4 adjacent or diagonal spaces contain the same symbol. This 7x6 board is being stored in a two dimensional array. For a piece placed in the board, you'll have 4 sets of checks you must do to see if it was a winning piece: • Check to the left/right for 3 of your pieces.

Here is a list of the most commonly asked questions about Python ® Security with Remote Start. Most newer systems use a 4 button remote that will allow you to. Connect 4 in Python - or 'How to build a board?' I gotta write a little Connect 4 game in Python. Trouble with a Java connect 4 program - 2 replies.

• Check to the upper left/lower right for 3 of your pieces. • Check to the lower left/upper right for 3 of your pieces. Orbitor Prospector 100 Metal Detector Manual on this page. • Check above and below for 3 of your pieces. If we consider X and Y your current board coordinate, wherever I place a piece, these checks correspond to: • X increases/decreases, Y is constant. • As X increases, Y decreases, and vice versa. • As X increases, Y increases, and vice versa.

• X is constant, Y increases/decreases. You should do a lazy check, checking each direction and going the other way as soon as you find something that isn't your piece. If both directions are complete and your total pieces are not 4, that wasn't a winning vector and go on to the next check. EDIT: Formatting and clarity.