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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Forming a Magic Square
  5. Discussions

Forming a Magic Square

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  • wifivop476
     + 0 comments

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  • phamducctrung
     + 0 comments

    Python

    def formingMagicSquare(s):
    all_magic_squares = [
        [[8, 1, 6], [3, 5, 7], [4, 9, 2]],
        [[6, 1, 8], [7, 5, 3], [2, 9, 4]],
        [[4, 9, 2], [3, 5, 7], [8, 1, 6]],
        [[2, 9, 4], [7, 5, 3], [6, 1, 8]],
        [[8, 3, 4], [1, 5, 9], [6, 7, 2]],
        [[4, 3, 8], [9, 5, 1], [2, 7, 6]],
        [[6, 7, 2], [1, 5, 9], [8, 3, 4]],
        [[2, 7, 6], [9, 5, 1], [4, 3, 8]]
    ]
    min_cost = float('inf')
    
    for magic in all_magic_squares:
        current_cost = 0
        for i in range(3):
            for j in range(3):
                current_cost += abs(s[i][j] - magic[i][j])
        if current_cost < min_cost:
            min_cost = current_cost
    return min_cost

  • luisagabrielahp
     + 0 comments

    Mi solución en Python 3:

    def formingMagicSquare(s):
    # Write your code here
    count = [0,0,0,0,0,0,0,0]

    magics = [
        [[8, 1, 6], [3, 5, 7], [4, 9, 2]],
        [[6, 1, 8], [7, 5, 3], [2, 9, 4]],
        [[4, 9, 2], [3, 5, 7], [8, 1, 6]],
        [[2, 9, 4], [7, 5, 3], [6, 1, 8]],
        [[8, 3, 4], [1, 5, 9], [6, 7, 2]],
        [[4, 3, 8], [9, 5, 1], [2, 7, 6]],
        [[6, 7, 2], [1, 5, 9], [8, 3, 4]],
        [[2, 7, 6], [9, 5, 1], [4, 3, 8]]
        ]

    for x in range(0, 8):
        for i in range(0, 3):
            for j in range(0, 3):
                count[x] += abs(s[i][j]-magics[x][i][j])

    return min(count)

  • marco_perozzo
     + 0 comments

    Python 3

    I have created an additional function to calculate all possible combinations for the Magic Square. This challenge should worth more thatn 20 pts.

    from itertools import permutations, chain

# function to list all magic square valid options
def validMagicOptions() -> list:
    seq_valid = []
    for seq in permutations(range(1,10), 9):
        sum3 = sum(seq[0:3])
        # only add records if the sum of all rows, columns and diagonals match
        if     (
            # rows
                sum(seq[3:6])==sum3
            and sum(seq[6:9])==sum3
            # cols
            and sum(seq[0::3])==sum3
            and sum(seq[1::3])==sum3
            and sum(seq[2::3])==sum3
            # diagonals
            and sum(seq[0::4])==sum3
            and sum(seq[2:7:2])==sum3
            ):
            seq_valid.append(list(seq))

    return seq_valid

def formingMagicSquare(s):
    # Write your code here
    
    valid = validMagicOptions()
    sflat = list(chain(*s))
            
    lowest_cost = float('inf')
    for v in valid:
        cost_abs = sum(abs(x-y) for x,y in zip(v, sflat))
        lowest_cost=min(lowest_cost, cost_abs)
        
    return lowest_cost

  • naufalfh19
     + 0 comments

    Python3 Solution:

    def formingMagicSquare(s):
    magicCombination = [
        [2, 7, 6, 9, 5, 1, 4, 3, 8],
        [2, 9, 4, 7, 5, 3, 6, 1, 8],
        [4, 3, 8, 9, 5, 1, 2, 7, 6],
        [4, 9, 2, 3, 5, 7, 8, 1, 6],
        [6, 1, 8, 7, 5, 3, 2, 9, 4],
        [6, 7, 2, 1, 5, 9, 8, 3, 4],
        [8, 1, 6, 3, 5, 7, 4, 9, 2],
        [8, 3, 4, 1, 5, 9, 6, 7, 2]
    ]
    
    res = float("inf")
    
    for magic in magicCombination:
        currRes = 0
        pointer = 0
        for i in range(3):
            for j in range(3):
                val = s[i][j]
                currRes += abs(val - magic[pointer])
                pointer += 1
            
        if currRes < res:
            res = currRes
    
    return res