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JavaScript Object-Oriented JavaScript: Challenge Adding the Game Logic Build the checkForWin() Method

Birthe Vandermeeren
Birthe Vandermeeren
17,146 Points
Posted by Birthe Vandermeeren
Birthe Vandermeeren
17,146 Points

My solution for the checkForWin() method

I've attempted my own solution for the checkForWin() method. The difference with Ashley's solution: Ashley checks the entire board for a 4 in a row sequence in each of the 4 directions, while I only check for a 4 in a row sequence in the target column in the vertical direction and in the target row in the horizontal column. If there's a 4 in a row sequence in another column/row, that's been detected when that token was dropped, so no need to search the entire board for a 4 in a row sequence. Hope this makes sense.

Here's my code:

checkForWin(target){
         let targetColumn = target.x;
         let targetRow = target.y;
         let owner = target.owner;
         let win = false;

         //vertical
        if ( targetRow <= this.board.rows - 4 ) {
            if ( this.board.spaces[targetColumn][targetRow].owner === owner &&
                this.board.spaces[targetColumn][targetRow + 1].owner === owner &&
                this.board.spaces[targetColumn][targetRow + 2].owner === owner &&
                this.board.spaces[targetColumn][targetRow + 3].owner === owner) {
                    win = true;
            }
        }

         //horizontal
        for ( let x = 0; x <= this.board.columns - 4; x++ ) {
            if ( this.board.spaces[x][targetRow].owner === owner &&
                this.board.spaces[x + 1][targetRow].owner === owner &&
                this.board.spaces[x + 2][targetRow].owner === owner &&
                this.board.spaces[x + 3][targetRow].owner === owner) {
                    win = true;
            }
        }

         //upwardly diagonal
        for ( let x = 0; x <= this.board.columns - 4; x++ ) {
            for ( let y = 3; y <= this.board.rows - 1; y ++ ) {
                if ( this.board.spaces[x][y].owner === owner &&
                    this.board.spaces[x + 1][y - 1].owner === owner &&
                    this.board.spaces[x + 2][y - 2].owner === owner &&
                    this.board.spaces[x + 3][y - 3].owner === owner) {
                        win = true;
                }
            }
        }

         //downwardly diagonal
        for ( let x = 3; x <= this.board.columns - 1; x++ ) {
            for ( let y = 3; y <= this.board.rows - 1; y ++ ) {
                if ( this.board.spaces[x][y].owner === owner &&
                    this.board.spaces[x - 1][y - 1].owner === owner &&
                    this.board.spaces[x - 2][y - 2].owner === owner &&
                    this.board.spaces[x - 3][y - 3].owner === owner) {
                        win = true;
                }
            }
        }
        console.log(win);
        return win;
    }

2 Answers

Steven Parker
Steven Parker
231,271 Points
Steven Parker
Steven Parker
231,271 Points

It's a good idea, and would be an efficiency improvement over the original. Another student was working on a similar concept just last Saturday.

At first glance, it appears you have vertical checks covered, but it looks like horizontal checks only look for wins where the new token is dropped onto the left side. To make the check complete, it should also detect wins where the token is dropped on the right side or inside the row.

If you wanted even more challenge, you could also adapt the diagonal tests to only check the diagonal lines that include the new token.

And good job on finding creative ways to expand your learning! :+1:

Birthe Vandermeeren
Birthe Vandermeeren
17,146 Points
Birthe Vandermeeren
Birthe Vandermeeren
17,146 Points

For the vertical direction I'm just checking if there are 3 more tokens of the same owner right below the spot where the current token is dropped, because that's the only way a 4 in a row sequence can be made in this turn.

For the horizontal direction I'm looping through all spaces which have at least 3 spots behind them in the target row, starting with x=0. Which means if I'm making a 4 in a row sequence by dropping a token in the third column between 2 tokens in front and 1 token behind the token I just dropped, it's already detected in the first iteration, when checking with x=0. So I believe my code is correct.

Doing the diagonal checks in this way seemed to much of a hassle! 😅

Steven Parker
Steven Parker
231,271 Points
Steven Parker
Steven Parker
231,271 Points

My "first glance" was a bit hasty, I see now that the horizontal check is thorough. I was thinking it started with the token position. It could still be optimized a bit:

    // start and end the check no more than 3 away from the new token
    for ( let x = Math.max(0, targetColumn-3);
          x <= Math.min(this.board.columns - 4, targetColumn); x++ ) {
Birthe Vandermeeren
Birthe Vandermeeren
17,146 Points
Birthe Vandermeeren
Birthe Vandermeeren
17,146 Points

You're right, Steven! Thanks for your thoughts on this!