Class5tictactoe #5
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stu-k
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Oct 26, 2017
stu-k
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stu-k
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You're so close! Your logic is #solid for checkForWin() and ticTacToe(), you just have to go for it!
| // Your code here | ||
| // Got help from eddy, I do not completely understand yet | ||
| const checkForWin = () => { | ||
| // if (horizontalWin() || verticalWin() || diagonalWin()) { |
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You can do it! This logic is correct, keep going! Just ignore the turncount thing, and if you look at the tests below, they expect checkForWin() to return either true or false.
| // playerTurn = 'X'; | ||
| // } | ||
| // if (board[row][column] === ' ') { | ||
| // board[row].splice(column, 1, playerTurn); |
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Instead of using unnecessary methods, set that value = to playerTurn.
| it('should alternate between players', () => { | ||
| ticTacToe(0, 0); | ||
| assert.deepEqual(board, [ ['O', ' ', ' '], [' ', 'X', ' '], [' ', ' ', ' '] ]); | ||
| assert.deepEqual(board, [ ['O', '', ' '], [' ', 'X', ' '], [' ', ' ', ' '] ]); |
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Put the space back! The tests need it! (In board[0][1], I think you accidentally took it out).
| // 10 11 12 | ||
| // 20 21 22 | ||
| const horizontalWin = () => { | ||
| if ((board[0][0] === board[0][1] && board[0][1] === board[0][2]) || (board[1][0] === board[1][1] && board[1][1] === board[1][2]) || (board[2][0] === board[2][1] && |
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Yes! Yes yes! You got the short way to reference three cells. Nice work! I mean, SOLID work!
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