Return to Answer

You can also use Ruby's Matrix class to see if player p wins. In the following, square[i][j] is the player in row i, column j.

You can also use Ruby's Matrix and Vector classes to see if player p wins. In the following, square[i][j] is the player in row i, column j.

You can also use Ruby's Matrix class to see if player p wins. In the following, square[i][j] is the player in row i, column j.

[Edit: a couple of days ago an anonymous user proposed an edit that was rejected by two members, both on the grounds that "the edit deviates from the original intent of the post." While I'm sure those members had good intentions, neither appears to have experience with Ruby, so I don't understand how they were able to arrive at that conclusion.

The anonymous user was not just improving my code, he or she was correcting an error I made. It was a good catch and a good fix. I've edited to incorporate the fix (and made a few other small changes).

If you are that anonymous user, please accept my thanks. A small suggestion: it probably would have been better if you had just left a comment, but the main thing is that the code is now fixed. (I hope.)]

You can also use Ruby's Matrix class to see if player p wins. In the following, square[i][j] is the player in row i, column j.

You can also use Ruby's Matrix class to see if player p wins. In the following, square[i][j] is the player in row i, column j.

Code

require 'matrix'
   
def win?(square,p)
  n = square.size
  m = Matrix[*square]
  pvec = Matrix.build(1,n){p}.row(0)
  m.row_vectors.any?    { |r| r == pvec }                ||
    m.column_vectors.any? { |c| c == pvec }              ||
    Matrix.rows([m.each(:diagonal).to_a]).row(0) == pvec ||
    n.times.all? { |i| square[i][n-i-1] == p }
end

Example

square = [[1,3,2],
          [4,2,6],
          [2,8,9]]

win?(square, 2)                    #=> true
win?([[1,3,2],[4,3,6],[2,3,9]], 3) #=> true
win?([[1,3,2],[3,3,3],[2,7,9]], 3) #=> true
win?([[1,3,2],[4,5,6],[7,8,9]], 1) #=> false

Explanation

pvec = Matrix.build(1,n){p}.row(0)

creates a vector for which every element is the value of the player argument p (e.g., pvec # => Vector[2,2,2]).

m.row_vectors.any? { |r| r == pvec }

determines if player p wins in any row,

m.column_vectors.any? { |c| c == pvec }

determines if player p wins in any column,

Matrix.rows([m.each(:diagonal).to_a]).row(0) == pvec

determines if player p wins on the main diagonal, and

(0...n).all? { |i| square[i][n-i-1] == p }

determines if player p wins on the minor diagonal (top right to bottom left).

I was unable to find a way to check the minor diagonal using Matrix class methods.

You can also use Ruby's Matrix class to see if player p wins. In the following, square[i][j] is the player in row i, column j.

Code

require 'matrix'
   
def win?(square,p)
  n = square.size
  m = Matrix[*square]
  pvec = Matrix.build(1,n){p}.row(0)
  m.row_vectors.any?    { |r| r == pvec }      ||
    m.column_vectors.any? { |c| c == pvec }    ||
    Vector[*m.each(:diagonal).to_a] == pvec    ||
    n.times.all? { |i| square[i][n-i-1] == p }
end

Example

square = [[1,3,2],
          [4,2,6],
          [2,8,9]]

win?(square, 2)                    #=> true
win?([[1,3,2],[4,3,6],[2,3,9]], 3) #=> true
win?([[1,3,2],[3,3,3],[2,7,9]], 3) #=> true
win?([[1,3,2],[4,5,6],[7,8,9]], 1) #=> false

Explanation

pvec = Matrix.build(1,n){p}.row(0)

creates a vector for which every element is the value of the player argument p (e.g., pvec # => Vector[2,2,2]).

m.row_vectors.any? { |r| r == pvec }

determines if player p wins in any row,

m.column_vectors.any? { |c| c == pvec }

determines if player p wins in any column,

Vector[*m.each(:diagonal).to_a] == pvec

determines if player p wins on the main diagonal, and

(0...n).all? { |i| square[i][n-i-1] == p }

determines if player p wins on the minor diagonal (top right to bottom left).

I was unable to find a way to check the minor diagonal using Matrix class methods.