6
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Same rules as last time except this time count or calculate the number of possible solutions!

No ordinary magic square

Place 1-9 in the squares below in a manner such that none of the columns, rows or long diagonals have the same sum:

empty 3-by-3 grid

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7
  • $\begingroup$ the other question is not too much old. It was only half an hour old. So you could just add this extra question in that. $\endgroup$
    – manshu
    Jun 9, 2016 at 9:35
  • 1
    $\begingroup$ @manshu If I did that how would I credit the first person to answer? I already marked an answer to my first question as correct so I cant take that back retroactively and credit the correct response to someone who answers the next question can I? $\endgroup$ Jun 9, 2016 at 9:40
  • 7
    $\begingroup$ It seems reasonable to me for this to be a separate question. $\endgroup$
    – Gareth McCaughan
    Jun 9, 2016 at 9:43
  • 1
    $\begingroup$ I agree it should be a separate question. Even though the questions are closely related, the methods one could potentially employ are quite different. $\endgroup$
    – user21939
    Jun 9, 2016 at 9:53
  • 1
    $\begingroup$ Yup. You solve the original one by playing with numbers until you make all the sums distinct. You solve this one by writing a computer program. (Of course you could solve the original one by writing a program too, but that feels like overkill.) $\endgroup$
    – Gareth McCaughan
    Jun 9, 2016 at 9:54

2 Answers 2

8
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According to the simple-minded 7-line Python program I just wrote, there are exactly

24960 solutions.

There is at least a 25% chance that I didn't make any stupid blunders.

Here's the code:

n=0
for p in itertools.permutations([1,2,3,4,5,6,7,8,9]):
  a,b,c,d,e,f,g,h,i = p
  a,b,c,d,e,f,g,h = [a+b+c,d+e+f,g+h+i,a+d+g,b+e+h,c+f+i,a+e+i,c+e+g]
  if a in (b,c,d,e,f,g,h) or b in (c,d,e,f,g,h) or c in (d,e,f,g,h): continue
  if d in (e,f,g,h) or e==f or e==g or e==h or f==g or f==h or g==h: continue
  n += 1

Why two separate lines for the inequality checks? Just because it looks better that way. Why change from x in y to x==... at the end? Out of a vague idea that it might be a bit faster.

It surprised me how quickly this ran -- but 9! isn't really all that large.

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10
  • 1
    $\begingroup$ May you post the code? 7-lines sounds quite interesting. $\endgroup$
    – Inazuma
    Jun 9, 2016 at 9:45
  • $\begingroup$ @gareth I am new here so please advise me what to do. I took advice of another user to edit my first post instead of posting a 2nd question. Is there some way I can delete this puzzle and just leave the first? $\endgroup$ Jun 9, 2016 at 9:45
  • 1
    $\begingroup$ Same result but waaaay more lines ;) $\endgroup$
    – Fabich
    Jun 9, 2016 at 9:46
  • 1
    $\begingroup$ I would suggest that you leave it as two questions. Perhaps put a link in the first question to this one. $\endgroup$
    – Gareth McCaughan
    Jun 9, 2016 at 9:52
  • 1
    $\begingroup$ (But note that I am not an entirely uninterested party -- if you delete this question, I guess I lose the rep I gained from it.) $\endgroup$
    – Gareth McCaughan
    Jun 9, 2016 at 9:52
7
$\begingroup$

My Python code to iterate the solutions and their respective $8$ sums:

def noms():
    sums = set()
    for p in permutations(range(1,10)):
        for e in [p[i:i+3] for i in range(0, 9, 3)]
               + [p[i:9:3] for i in range(3)]
               + [p[0:9:4], p[2:7:2]]:
            s = sum(e)
            if s in sums:
                break
            sums.add(s)
        else:
            yield p, sums
        sums.clear()

Counting them:

>>> sum(1 for solution in noms())
24960

Some more information:

There are $8$ symmetrically isomorphic sets of $3120$ arrangements: $4$ rotations and $2$ lines of symmetry (reflection in one of the diagonals is the same as a rotation plus a reflection).

There are $1538$ different sets of sums distributed as:

Sums  Count  Total
 496    8     3968 
 782   16    12512
 102   24     2448
 106   32     3392
  10   40      400
  36   48     1728
   2   64      128
   4   96      384
------------------
1538         24960

An example of one the $496$ sets of sums that are unique up to symmetry is: $\{9,11,12,13,15,16,19,23\}$
which is yielded by:
2  4  7
5  1  3
8  6  9

2  5  8
4  1  6
7  3  9

7  3  9
4  1  6
2  5  8

7  4  2
3  1  5
9  6  8

8  5  2
6  1  4
9  3  7

8  6  9
5  1  3
2  4  7

9  3  7
6  1  4
8  5  2

9  6  8
3  1  5
7  4  2

The smallest sum of sums is $107$, which necessarily has $1$ or $2$ in the centre) and may be made in $88$ ways, or $11$ up to symmetry ($3\times 1+4\times 2$):
1  7  4
9  2  8
3  6  5  {8, 9, 12, 13, 14, 15, 17, 19}
-----------------------------------------
1  6  5
8  2  9
4  7  3  {6, 11, 12, 13, 14, 15, 17, 19}
-----------------------------------------
2  5  4
8  1  7
3  9  6  {8, 9, 11, 13, 15, 16, 17, 18}
-----------------------------------------
1  7  4
8  2  6
5  9  3  {6, 11, 12, 13, 14, 16, 17, 18}

2  7  4
8  1  5
6  9  3  {6, 11, 12, 13, 14, 16, 17, 18}
-----------------------------------------
1  6  3
9  2  7
4  8  5  {8, 9, 10, 14, 15, 16, 17, 18}

2  5  3
9  1  7
4  8  6  {8, 9, 10, 14, 15, 16, 17, 18}
-----------------------------------------
1  6  3
8  2  9
4  7  5  {8, 9, 10, 13, 15, 16, 17, 19}

2  5  3
7  1  8
4  9  6  {8, 9, 10, 13, 15, 16, 17, 19}
-----------------------------------------
1  6  5
9  2  8
3  7  4  {7, 10, 12, 13, 14, 15, 17, 19}

2  5  6
7  1  9
3  8  4  {7, 10, 12, 13, 14, 15, 17, 19}
-----------------------------------------

The largest sum of sums is $133=8\times30-107$ and is the same set of squares where each of the $9$ digits, $d$, are replaced by $10-d$ and each of the $8$ sums, $s$, are replaced by $3*10-s$.

The lowest possible variance for the sums is $6.234375$, and there are $4\times 8=32$ such squares:
1  7  2
9  4  5
6  3  8  {10, 12, 13, 14, 15, 16, 17  18}

2  5  6
9  8  3
4  1  7  {10, 12, 13, 14, 15, 16, 17, 18}

3  7  4
9  2  5
6  1  8  {10, 12, 13, 14, 15, 16, 17, 18}

2  5  8
7  6  3
4  1  9  {12, 13, 14, 15, 16, 17, 18, 20}

The highest possible variance for the sums is $22.25$, and there are $4\times 8=32$ such squares:
1  2  3
4  6  8
5  7  9  {6, 10, 14, 15, 16, 18, 20, 21}

1  2  3
5  7  9
4  6  8  {6, 10, 14, 15, 16, 18, 20, 21}

1  2  7
3  4  8
5  6  9  {9, 10, 12, 14, 15, 16, 20, 24}

2  1  7
4  3  8
6  5  9  {9, 10, 12, 14, 15, 16, 20, 24}

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