We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Implementation
  4. Forming a Magic Square
  5. Discussions

Forming a Magic Square

  • arjit_agarwal_11
     + 0 comments

    Nice explanation. I used this logic and wrote the python code as under. Instead of constructing all magic squares, I found out cost for all magic squares and took minimum.

    def formingMagicSquare(s):
    cost = abs(s[1][1] - 5)
    costArr = [0, 0, 0, 0, 0, 0, 0, 0]
    pickArr = [0, 2, 6, 8]
    for i in range(4):
        n = pickArr[i]
        m = pickArr[i+1] if i % 2 == 0 else pickArr[i-1]
        p = (n+m) / 2
        q = (8-m+n) / 2
        costArr[i*2] += abs(s[n/3][n%3] - 4)
        costArr[i*2] += abs(s[(8-n)/3][(8-n)%3] - 6)
        costArr[i*2] += abs(s[m/3][m%3] - 2)
        costArr[i*2] += abs(s[(8-m)/3][(8-m)%3] - 8)
        costArr[i*2] += abs(s[p/3][p%3] - 9)
        costArr[i*2] += abs(s[(8-p)/3][(8-p)%3] - 1)
        costArr[i*2] += abs(s[q/3][q%3] - 3)
        costArr[i*2] += abs(s[(8-q)/3][(8-q)%3] - 7)
        costArr[i*2+1] += abs(s[n/3][n%3] - 4)
        costArr[i*2+1] += abs(s[(8-n)/3][(8-n)%3] - 6)
        costArr[i*2+1] += abs(s[m/3][m%3] - 8)
        costArr[i*2+1] += abs(s[(8-m)/3][(8-m)%3] - 2)
        costArr[i*2+1] += abs(s[p/3][p%3] - 3)
        costArr[i*2+1] += abs(s[(8-p)/3][(8-p)%3] - 7)
        costArr[i*2+1] += abs(s[q/3][q%3] - 9)
        costArr[i*2+1] += abs(s[(8-q)/3][(8-q)%3] - 1)
    return min(costArr) + cost