6
$\begingroup$

I'm doing 3x3 magic squares. Here are the squares I'm working on:

|   | 5 |   |
|   |   |   |
| 8 |   |   |

The values must be between 3 and 12, and each line must add to 21.

Here's another one:

|   | 9 |   |
|   |   | 3 |
|   |   |   |

For this one, the boundaries are 3-11. As with the last one, each line total must add to 21.

$\endgroup$
  • 2
    $\begingroup$ Magic squares are usually filled with distinct integers. When he says maximum he means that the range of integers is between those two numbers. 3-12 means that the minimum is 3 and the maximum is 12. $\endgroup$ – LeppyR64 May 11 '15 at 10:02
  • $\begingroup$ I'm voting to close because the duplicated question I found offers the same techniques for an extremely similar puzzle, so I think it will be helpful for future readers to be directed to the other question. $\endgroup$ – Kevin - Reinstate Monica May 18 '15 at 0:28
13
$\begingroup$

In general, any $n\times n$ magic square of range [1, $n^2$] with odd number $n$ can be solved using the following algorithm:

  1. Start at the middle grid in the bottom row. This is your 1.
  2. Move downwards and to the right by one grid. If this move results in a position outside the square, wrap around to the beginning of the row (or column).
  3. If 2 cannot be performed (i.e. the grid is already occupied or you are at the bottom-right corner), move upwards by one grid instead.
  4. Repeat 2 (or 3) until all numbers are filled.

Illustration using $n=3$. (For cells 2 and 3 the position before wrapping is shown in parentheses.)

|   |   |   |
|   |   |   |
|   | 1 |   | // start here

|   |   | 2 |
|   |   |   |
|   | 1 |   | // move to the right and then down.
         (2)  // Because moving down brings us outside the square,
              // we wrap around to the start of the column

|   |   | 2 |
| 3 |   |   | (3)
|   | 1 |   | // move to the right and down.
              // moving right brings us outside, so wrap around to start of row

| 4 |   | 2 |
| 3 |   |   |
|   | 1 |   | // cannot move down and right because that is occupied
              // so we move up instead

| 4 |   | 2 |
| 3 | 5 |   |
|   | 1 |   | // move down and right

| 4 |   | 2 |
| 3 | 5 |   |
|   | 1 | 6 | // move down and right

| 4 |   | 2 |
| 3 | 5 | 7 |
|   | 1 | 6 | // cannot move down and right, so move up

| 4 |   | 2 |
| 3 | 5 | 7 |
| 8 | 1 | 6 | // move down and right, wrap around to start of row

| 4 | 9 | 2 |
| 3 | 5 | 7 |
| 8 | 1 | 6 | // move down and right, wrap around to start of column

Now, if you need to solve your magic square that starts with 3, simply add 2 to all cells of this standard square. Then rotate and/or reflect it until you get one where the numbers match your given ones.

For example, consider your first square (assuming we can omit 12):

|   | 5 |   |
|   |   |   |
| 8 |   |   |

We first add 2 to all elements of our standard square:

| 6  | 11 | 4  |
| 5  | 7  | 9  |
| 10 | 3  | 8  |

Then rotate clockwise by 90 degrees:

| 10 | 5  | 6  |
| 3  | 7  | 11 |
| 8  | 9  | 4  |

Your first square solved.

Further reading: Magic Square from Wolfram MathWorld, which includes methods for solving even squares as well.

$\endgroup$
  • $\begingroup$ Can you expand your answer to include the specific situations given in the question? $\endgroup$ – LeppyR64 May 11 '15 at 10:33
  • $\begingroup$ @JasonLepack Done with the first square. Since it is the same method for the second, and other people have already posted answers, I'll leave that out. $\endgroup$ – user12205 May 11 '15 at 10:38
  • 1
    $\begingroup$ This generates an $n\times n$ square, and fortunately it's unique for the $3\times 3$ case, but it's not unique in general; I wouldn't be surprised if the abstract problem here (given some subset of set cells and an interval, is there a possibly unique $n\times n$ magic square that matches the given values on that set?) is actually computationally hard. $\endgroup$ – Steven Stadnicki May 12 '15 at 3:36
3
$\begingroup$

Here is an alternative technique. Consider your first magic square:

| a | 5 | b |
| c | d | e |
| 8 | f | g |

I filled in the empty cells with the variables $a$ through $g$. We will create a system of equations to solve for these seven variables.

We know that each row must sum to 21. As a result, we can write the following three equations:

  • 1st row: $a + 5 + b = 21$
  • 2nd row: $c + d + e = 21$
  • 3rd row: $8 + f + g = 21$

Likewise, every column and diagonal must sum to 21, giving us five more equations:

  • 1st column: $a + c + 8 = 21$
  • 2nd column: $5 + d + f = 21$
  • 3rd column: $b + e + g = 21$
  • \ diagonal: $a + d + g = 21$
  • / diagonal: $8 + d + b = 21$

So we now have 8 equations with 7 unknowns. We can select any seven of them and apply the usual equation-solving techniques to find the values of $a$, $b$, $c$, $d$, $e$, $f$, and $g$, giving us the solution to the magic square.

$\endgroup$
1
$\begingroup$

The answer for the first puzzle is this:

|10 | 5 | 6 |
| 3 | 7 |11 |
| 8 | 9 | 4 |

| A | B | C |
| D | E | F |
| G | H | I |

| A | 5 | C |
| D | E | F |
| 8 | H | I |

Range 3 - 12.

There is one extra integer.

Neither 11 or 12 can be in the same row, column or diagonal as 8 since the other number would be 1 or 2 which is outside the range. Therefore F is 11 or 12 and the other is the integer that is not included.

Neither 3 nor 4 can be in the same row or column as 5 since the other number would be 12 or 13 and 13 is outside the range and 12 must be in F if it is in the puzzle. Therefore D and I are 3 and 4. Continuing H and A are 10 and 9 and E and C are 6 and 7.

I cannot be 3 since either E would be 6 or 7 and A would be 11 or 12 and 11 and 12 is in F or not in the puzzle. Therfore I = 4. The rest of the puzzle fills itself in from there. The number not included is 12.

$\endgroup$
0
$\begingroup$

For the 1st question:

| 10 | 5 | 6 |

| 3 | 7 | 11 |

| 8 | 9 | 4 |

And Jimmy has given the answer for the 2nd question

For the method. Its very simple. Everyone knows how to solve 3*3 to get 15. But we need 21 i.e is 2 more for every cell. so add 2 in every cell of magic square. that will make 1 as 3 and 9 as 11. And you will then have to just manage the rows.

$\endgroup$
  • 2
    $\begingroup$ The question asks 'How do I solve these 3x3 magic squares?', not 'Solve these 3x3 magic squares'. Your answer needs an explanation. $\endgroup$ – Tryth May 11 '15 at 11:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.