My attempt at a solution for the checkForWin() method

Question

Hi there,

I would really appreciate any feedback on how I approached this - please see my code below. It's not complete but I think maybe repeating the last section for the different directions (with the right adjustments each time) might lead to a functioning solution?

Thanks!

/**
   * Checks if there is a winner on the board after each token drop.
   * @param {array} spaces the 2D array of all spaces @ this.board.spaces
   * @param {Object} activePlayer the active player
   */
  checkForWin(spaces, activePlayer) {

    // flatten 2D array spaces into a 1D array of all spaces:
    const arrayOfAllSpaces = spaces.reduce((accArray, column) => {
      return [...accArray, ...column];
    }, []);

    // filter 1D array of spaces by activePlayer ownership:
    const spacesOwnedByActivePlayer = arrayOfAllSpaces.filter(space => space.owner === activePlayer);

    // use space x and y properties to check for victory streaks in each possible direction:
    // check for vertical down victory:
    const verticalDown = [];
    let incrementor = 1;
    for (space of spacesOwnedByActivePlayer) {
      if (space.y === activeSpace.y + incrementor) {
        verticalDown.push(space);
      }
      incrementor++;
    }
    //check if direction array contains at least 4 in a row:
    if (verticalDown.length >= 4) {
      // call gameover method / winner alert
    }

    // repeat above code (from 'const verticalDown = []' onwards) for the different directions.
  }

Answer 1 · 2019-05-16T21:11:56Z

May 16, 2019 9:11pm

It's not clear what "activeSpace" refers to, but it looks like consecutive pairs of owned spaces will all be added to the "verticalDown" array. So basing the determination on length alone may mean several isolated pairs could be misinterpreted as "win".

Otherwise, the basic idea seems viable.

Answer 2 · 2019-06-05T03:08:14Z

June 5, 2019 3:08am

Hey Chris, sure.

So the function can be said to have three sections

first for loop. This loops through the direction sets in the 'dirs' array. It also sets up the counter at the same time.
the second for loop. this loops through the elements in each row of the 'dirs' vector. Note that the counter i is incrementing at 2. Counter i is to be used for increments in x, and counter j is to be used for increment in y.
the while loop. In conjuction with the if statements, this just propagates the program in the direction specfied in the previous step until one of the following conditions are violated:
- it goes out of bounds
- the owner of the last token scanned differs from the most recent one

You can also see a nested if condition that checks if the win condition has been met (count === 3), if so we have reached our fourth consecutive token, if not, we simply increment the counter by 1.

Hope that helps.

Answer 3 · 2019-05-27T22:07:46Z

May 27, 2019 10:07pm

Here is mine

checkForWin(target) {
        // using the newly placed token as center
        // check if there are four connected
        let dirs = [[1, 0, -1, 0], 
                    [0, 1, 0, -1],
                    [1, 1, -1, -1],
                    [1, -1, -1, 1]];
        // check up down left right and diagonal
        for(let dir of dirs) {
            let i=0;
            let count = 1;
            for(; i<=2; i+=2) {
                let j = i + 1;
                let x = target.x + dir[i];
                let y = target.y + dir[j];
                let test = 0;
                while(x<this.board.columns && x>=0 && y<this.board.rows && y>=0) {
                    if(this.board.spaces[x][y].owner && this.board.spaces[x][y].owner === target.owner) {
                        if(count === 3) {
                            return true;
                        } else {
                            count++;
                        }
                        x += dir[i]; y += dir[j];
                    } else {
                        break;
                    }
                }
            }
        }
        return false;
     }

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George Roberts

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My attempt at a solution for the checkForWin() method

3 Answers

Steven Parker

Steven Parker

George Roberts

George Roberts

Felix Ding

Felix Ding

Felix Ding

Felix Ding

Christopher Gardner

Christopher Gardner

Christopher Gardner

Christopher Gardner

Steven Parker

Steven Parker