Friday Puzzle -- Magic Square

Hint

Alex Y

The answer requires a little trickiness. Make sure you are not imposing limits on your solution attempts that the statement of the puzzle does not.

2 out of 3

Larry Barrett

It is a magic square.

The number 6 remains where it was.

Unfortunately, all 15 of the other numbers have moved.

.. 3..8..4.15

.13..6.10..1

.14..5..9..2

..0.11..7.12

Re: 2 out of 3

Alex Y

You are much closer than you think.

And I'll leave you with that enigmatic comment for a while.

So close

Alex Y

A rotation of this magic square is a solution.

Re: So close, but still not there

Larry Barrett

I don't know much about magic squares. I knew even less before reading up on them courtesy of Wikipedia - there is an impressive amount of info there and it is an interesting branch of mathematics.

It seems evident that a magic square can be rotated about its center, CW or CCW. So

3 8 4 15

13 6 10 1

14 5 9 2

0 11 7 12

rotated CW 90*, for example, it becomes

0 14 13 3

11 5 6 8

7 9 10 4

12 2 1 15

which is also a magic square, but does not satisfy the problem conditiions.

Also magic squares can be rotated about each major axis and remain a magic square; so

3 8 4 15

13 6 10 1

14 5 9 2

0 11 7 12

rotated about the 0-15 axis becomes

12 2 1 15

7 9 10 4

11 5 6 8

0 14 13 3

which is also a magic square, but again does not satisfy the problem conditions.

I think I have tried the rotations I can think of, but still do not see an answer.

I wonder if there are other ways to rotate magic squares.

Re: So close, but still not there

Alex Y

It seems evident that a magic square can be rotated about its center, CW or CCW.

Re: So close, but still not there

Larry Barrett

Here are the four rotations :

3 8 4 15........ 0 14 13 3

13 6 10 1...... 11 5 6 8

14 5 9 2........ 7 9 10 4

0 11 7 12...... 12 2 1 15

12 7 11 0....... 15 1 2 12

2 9 5 14......... 4 10 9 7

1 10 6 13....... 8 6 5 11

15 4 8 3......... 3 13 14 0

In each case it seems that only one number remains where it was in the original problem.

Now it gets tricky

Alex Y

One of the solutions you gave is indeed a correct solution to the problem as stated

Hint 2

Alex Y

We now have 4 proposed solutions, one of which is indeed a solution.

But none of those magic squares fit the common mental abstraction of the puzzle given.

I don't know if this will help, but try looking at the physical problem given, not your abstraction of it. You don't have to use discs--post-it notes or scraps of paper work just fine.

Re: Hint 2

Larry Barrett

I think trickery has ascended to new heights.

Of the 4 possible rotations, you have already pointed out that the first two are not the answer. This leaves these two:

12 7 11 0....... 15 1 2 12

2 9 5 14......... 4 10 9 7

1 10 6 13....... 8 6 5 11

15 4 8 3......... 3 13 14 0

Of these, I think the last one has the most potential to be resolved by trickery, which begins now.

The 15 1 2 in the top row, and the 3 13 14 in the last row can be moved and reordered in the proper initial positions:

1 2 3 12

4 10 9 7

8 6 5 11

13 14 15 0

Next, the 4 and 10 can be removed from the second row, the 9 can be rotated in place to become a 6; the 8 in the third row can be removed and placed at the end of the second row; the 5 in the third row can be removed and placed at the beginning of the second row; and the 6 in the third row can be rotated in place to become a 9:

1 2 3 12

5 .... 6 7 8

....9.... 11

13 14 15 0

So far there have been 8 moves: 15, 1, 2, 3, 13, 14, 8, and 5. And 4 and 10 have been removed but not replaced.

Lastly, the 12 in the first row can be removed and placed at the end of the third row; the 4 can now be placed at the end of the first row; and the 10 can now be placed after the 9 in the third row.

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 0

This seems to take 11 moves, not 10, but is the best I can do.

Re: Hint 2

Alex Y

Larry said:

I think trickery has ascended to new heights.

Of the 4 possible rotations, you have already pointed out that the first two are not the answer. This leaves these two:

12 7 11 0....... 15 1 2 12

2 9 5 14......... 4 10 9 7

1 10 6 13....... 8 6 5 11

15 4 8 3......... 3 13 14 0

Of these, I think the last one has the most potential to be resolved by trickery,

Correct. But I would wager that you are working with an image of the problem as if it were stated as follows:

The challenge is to rearrange these discs to create a magic square on the same 4x4 grid, where all four rows, all four columns, and both major diagonals add to 30.

The words in red were NOT part of the problem, and imposing them makes a solution impossible.

Re: Hint 2

Larry Barrett

Start with

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 0

Pick up the 3, 4, 5, 8, 10, 12, and 15 (7 moves) to leave this (n used for a blank space, and a blank space added to the left of each row)

n 1 2 n n

n n 6 7 n

n 9 n 11 n

n 13 14 n 0

Rotate the 6 and 9 in place (doesn't count as a move since they remain in place) to leave this

n 1 2 n n

n n 9 7 n

n 6 n 11 n

n 13 14 n 0

Now place the 15 and 12 in spaces in the first row, place the 4 and 10 in spaces in the second row, place the 8 and 5 in spaces in the third row, and place the 3 in a space in the 4th row to form this

15 1 2 12 n

4 10 9 7 n

8 6 5 11 n

3 13 14 n 0

Finally, slide the 0 to the right (the eighth move) to form a perfect square

15 1 2 12

4 10 9 7

8 6 5 11

3 13 14 0

Eight moves, or ten if you count the rotation of the 6 and 9.

  Correct

Alex Y

The 1,2,7,11,13,and 14 stay in place, and all the others are rearranged (moved or spun--I hadn't thought of that possibility) to form the new magic square, which is shifted one position to the left.