7
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An unfriendly magic square is a magic square whereby the difference between neighbouring numbers always is subject to a minimum, i.e.

  • Each number in the matrix is unique.
  • Each row, column and the two diagonals add up to the same number (the magic constant).
  • The difference between each pair of neighbouring numbers is greater than or equal to a specified constant value.
  • Difference is defined as the larger number minus the smallest.
  • Neighbour is defined as two numbers that are adjacent horizontally, vertically or diagonally.

In this case, the matrix consists of the numbers 1-25, and the magic constant is 65. The difference must be at least 4*, so 1 and 5 could be adjacent but not 1 and 3.

*No examples for the order 5 magic square exist with a minimum difference of 5 (or greater).

Can you fill in the missing numbers?

\begin{bmatrix}?&?&?&12&?\\?&?&?&?&?\\?&?&?&?&?\\?&?&?&?&?\\?&14&?&?&?\end{bmatrix}

Here is a clue which reveals three more of the numbers:

\begin{bmatrix}?&10&?&12&?\\?&?&?&?&?\\?&?&13&?&?\\?&?&?&?&?\\?&14&?&16&?\end{bmatrix}

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    $\begingroup$ ok @humn I have edited it as such, thank you. If nobody gets it after a while I can add clues. $\endgroup$ – Paul Richards Apr 24 at 14:06
  • $\begingroup$ Sorry, I'm confused by the wording a little. For this puzzle, the minimim difference between adjacent numbers is 4, or is it 5? $\endgroup$ – Galen Apr 24 at 19:39
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    $\begingroup$ Ok thanks for the clarification $\endgroup$ – Galen Apr 24 at 20:23
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    $\begingroup$ Seems close to impossible to solve without computer. Looking forward to see how this can be done. $\endgroup$ – Jens Apr 25 at 14:41
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    $\begingroup$ @Jens Agreed. A computer search found 43 solutions with relaxed constraints (condition for diagonal neighbours ignored). One of those, of course, does meet all conditions. Given this many barely friendly magic squares, finding that one manually is impractical. $\endgroup$ – Daniel Mathias Apr 25 at 17:31
5
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I have found a no-computer solution looking at a few patterns and bringing the possible solution set to 4 trial-and-error solutions, out of which 1 seems to solve it. And I would like to apologise in advance for using the hint. I am extremely sorry because using hints means I lack some understanding somewhere.

We start with the basic $5\times 5$ magic square (called basic hereafter):

\begin{bmatrix} 17 & 24 & 1 & 8 & 15 \\ 23 & 5 & 7 & 14 & 16 \\ 4 & 6 & 13 & 20 & 22 \\ 10 & 12 & 19 & 21 & 3 \\ 11 & 18 & 25 & 2 & 9 \end{bmatrix}

The matrix we want to find and is partially given is (called special hereafter):

\begin{bmatrix}?&10&?&12&?\\?&?&?&?&?\\?&?&13&?&?\\?&?&?&?&?\\?&14&?&16& \end{bmatrix}

Observe that both $10$ and $12$ go from row $4$ of basic of to row $1$ of special and $14$ and $16$ go from row $2$ of basic to row $5$ of special. Let's then follow the assumption that the first row of special is a permutation of the fourth row of basic and similarly the fifth row of special is a permutation of the second row of basic.

We tackle the former case first. Observe that $10$ is in the first column of basic and in the second column of special, whereas $12$ is the second column of basic and in the fourth column of special. Following the pattern, $19$ is in the third column of basic and hence should move to the sixth column, which is the first column modulo $5$ of special. Also, $21$ is in the fourth column of basic and hence should move to the eighth column, which is the third column modulo $5$ of special.

The latter case is similarly solved, which gives us:

\begin{bmatrix}19&10&21&12&3\\?&?&?&?&?\\?&?&13&?&?\\?&?&?&?&?\\23&14&5&16&7\end{bmatrix}

Now look at the primary diagonal of special. It contains $19,13$ and $7$, which are three elements from the third column of basic. Assume that the primary diagonal of special is a permutation of the third column of basic.

Also, for the sake of the arguments that follow, we label some elements of special:

\begin{bmatrix}19&10&21&12&3\\?&c&?&b&?\\?&?&13&?&?\\?&a&?&d&?\\23&14&5&16&7\end{bmatrix}

Then we now have $c=1$ or $25$ and $d=25$ or $1$ respectively. Observe that both solutions satisfy condition three and four of "difference".

We look at possible values of $a$ and $b$ (Note that $a+b=26$). Given condition three and four of difference, the following conditions follow:

$$a\le 10\ \text{or}\ a\ge 18\\ a= 1\ \text{or}\ a\ge 9\\ a\le 9\ \text{or}\ a\ge 17\\ a\le 19$$

This gives that the possible values of $a$ are $\{1,9,18,19\}$. Correspondingly $b$ has possible values $\{25,17,8,10\}$. Since $b\ne 25$ (since $c$ or $d$ has value $25$) and $b\ne 10$ ($10$ has already been used), the possible values of $a$ and $b$ reduce to $\{9,18\}$ and $\{17,8\}$.

Thus we have $4$ cases to check, which are the possible values of the $4$-tuple $(a,b,c,d)$. The first possibility is $(9,17,1,25)$. This gives:

\begin{bmatrix}19&10&21&12&3\\?&1&?&17&?\\?&?&13&?&?\\?&9&?&25&?\\23&14&5&16&7\end{bmatrix}

This is not possible as the remaining element of the second column is $31>25$.

The next possibility we check is $(9,17,25,1)$ which gives:

\begin{bmatrix}19&10&21&12&3\\?&25&?&17&?\\?&?&13&?&?\\?&9&?&1&?\\23&14&5&16&7\end{bmatrix}

This is also not possible because the remaining element of the second column is $7$ which has been used once.

For $(18,8,25,1)$, we have:

\begin{bmatrix}19&10&21&12&3\\?&25&?&8&?\\?&?&13&?&?\\?&18&?&1&?\\23&14&5&16&7\end{bmatrix}

The remaining element of the second column becomes negative, and thus this is discarded. The only remaining solution is $(18,8,1,25)$ which gives:

\begin{bmatrix}19&10&21&12&3\\?&1&?&8&?\\?&?&13&?&?\\?&18&?&25&?\\23&14&5&16&7\end{bmatrix}

Filling in the obvious elements, we have:

\begin{bmatrix}19&10&21&12&3\\?&1&?&8&?\\?&22&13&4&?\\?&18&?&25&?\\23&14&5&16&7\end{bmatrix}

Now the third row of special looks like a permutation of the third row of basic, and we see that the third row first column element of special cannot be $20$ (as $22-20=2<4$) and hence we have:

\begin{bmatrix}19&10&21&12&3\\?&1&?&8&?\\6&22&13&4&20\\?&18&?&25&?\\23&14&5&16&7\end{bmatrix}

Also the third column of special looks like a permutation of the principle diagonal of basic, and we see that the second row third column element of special cannot be $9$ (as $9-8=1<4$) and hence we have:

\begin{bmatrix}19&10&21&12&3\\?&1&17&8&?\\6&22&13&4&20\\?&18&9&25&?\\23&14&5&16&7\end{bmatrix}

The second row of special is a permutation of the first row of basic and the fourth row of special is a permutation of the fifth row of basic. As $15-12=3<4$ and $4-2=2<4$, thus the final special square comes to be:

\begin{bmatrix}19&10&21&12&3\\15&1&17&8&24\\6&22&13&4&20\\2&18&9&25&11\\23&14&5&16&7\end{bmatrix}

One can verify that all the conditions are met by this solution and hence we are done.

This is obviously no algorithm as this does not work in a general situation, and if one asks if we got lucky, we dare say yes...

| improve this answer | |
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    $\begingroup$ Well done for a no-computer solution! $\endgroup$ – Paul Richards Apr 26 at 11:55
  • $\begingroup$ Thank you ! The puzzle was intriguing enough for a hands-only solution, but had you not given the hint, it would have been nigh intractable. $\endgroup$ – Yuzuriha Inori Apr 26 at 11:57
2
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The solution:

 19 10 21 12  3
 15  1 17  8 24
  6 22 13  4 20
  2 18  9 25 11
 23 14  5 16  7

View C code

| improve this answer | |
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With removed, I attempted a brute-force search in Python.

import numpy as np
from itertools import permutations, combinations

best = np.array([[20, 10, 22, 12, 24],
              [15, 6, 2, 8, 4],
              [11, 21, 17, 13, 18],
              [25, 1, 5, 23, 3],
              [7,14,19,9,16]])

assert set(best.flatten()) == set(range(1, 26))

def cost(x):
    min_dist = 4
    total = 0

    if x.shape[0] == x.shape[1]:
        if x[0,3] != 12:
            total += 100
        if x[4, 1] != 14:
            total += 100
        if x[0, 1] != 10:
            total += 100
        if x[2, 2] != 13:
            total += 100
        if x[4, 3] != 16:
            total += 100
        total += 1 if np.abs(65 - np.sum(np.diagonal(x))) else 0
        total += 1 if np.abs(65 - np.sum(np.diag(np.fliplr(x)))) else 0
        for col in x.T:
            total += 1 if np.abs(65 - np.sum(col)) else 0
##    for row in x:
##        total += 1 if np.abs(65 - np.sum(row)) else 0
    for i in range(x.shape[0]):
        for j in range(x.shape[1]):
            if 0 < i: # check above
                total += np.abs(x[i-1,j] - x[i,j]) < min_dist
            if i < x.shape[0] - 1: #check below
                total += np.abs(x[i+1,j] - x[i,j]) < min_dist
            if 0 < j: # check left
                total += np.abs(x[i,j-1] - x[i,j]) < min_dist
            if j < x.shape[1] - 1: #check right
                total += np.abs(x[i,j+1] - x[i,j]) < min_dist
            if 0 < i and 0 < j: # check above left
                total += np.abs(x[i-1,j-1] - x[i,j]) < min_dist
            if i < x.shape[0] - 1 and j < x.shape[1]-1: # check below right
                total += np.abs(x[i+1,j+1] - x[i,j]) < min_dist
            if 0 < i and j < x.shape[1]-1: # check above right
                total += np.abs(x[i-1,j+1] - x[i,j]) < min_dist
            if i < x.shape[0] - 1 and 0 < j: # check above left
                total += np.abs(x[i+1,j-1] - x[i,j]) < min_dist

    return total

all_rows = [perm for perm in permutations(range(1,26), r=5) if sum(perm) == 65]
all_rows = [row for row in all_rows if (10 not in row and 12 not in row) or (10 in row and 12 in row)]
all_rows = [row for row in all_rows if (14 not in row and 16 not in row) or (14 in row and 16 in row)]
all_rows = [row for row in all_rows if (13 not in row) or (13 in row and 10 not in row and 14 not in row)]
all_rows = [row for row in all_rows if (10 not in row) or (row[1] == 10)]
all_rows = [row for row in all_rows if (12 not in row) or (row[3] == 12)]
all_rows = [row for row in all_rows if (13 not in row) or (row[2] == 13)]
all_rows = [row for row in all_rows if (14 not in row) or (row[1] == 14)]
all_rows = [row for row in all_rows if (16 not in row) or (row[3] == 16)]

for j, row1 in enumerate(all_rows):
    print((j+1), len(all_rows))
    if row1[1] == 10 and row1[3] == 12:
        for row2 in all_rows:
            if not set(row1).intersection(set(row2)) and not cost(np.array([row1, row2])):
                for row3 in all_rows:
                    if row3[2] == 13 and np.all([not set(i).intersection(set(row3)) for i in [row1, row2]]) and not cost(np.array([row1, row2, row3])):
                        for row4 in all_rows:
                            if np.all([not set(i).intersection(set(row4)) for i in [row1, row2, row3]]) and not cost(np.array([row1, row2, row3, row4])):
                                for row5 in all_rows:
                                    if row5[1] == 14 and row5[3] == 16 and np.all([not set(i).intersection(set(row5)) for i in [row1, row2, row3, row4]]) and not cost(np.array([row1, row2, row3, row5])):
                                        if cost(best) > cost(np.array([row1, row2, row3, row4, row5])):
                                            best = np.array([row1, row2, row3, row4, row5])
                                            print(cost(best), best, '\n')
                                        if not cost(np.array([row1, row2, row3, row4, row5])):
                                            print(np.array([row1, row2, row3, row4, row5]), '\n')
                                            quit()
                                    else:
                                        continue
                            else:
                                continue
                    else:
                        continue
            else:
                continue
    else:
        continue

And here we finally have the solution:

[[ 19 10 21 12 3] [ 15 1 17 8 24] [ 6 22 13 4 20] [2 18 9 25 11] [ 23 14 5 16 7]]

| improve this answer | |
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  • $\begingroup$ It is a parker square! I'll keep looking. $\endgroup$ – Galen Apr 24 at 17:49
  • $\begingroup$ Yeah, the top row adds to 51 and it needs to be 65. $\endgroup$ – Paul Richards Apr 24 at 18:58

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