Mdthbs

Distribute the digits from 1 to 9 to a 3x3 square, such that you reach as many square numbers as possible.
A valid square number in the 3x3 square is either a single digit square number

or is build with neighbouring number(s) either vertically, horizontally or diagonally.
Example:

9 8 7

6 5 4

1 3 2

In this example the square numbers are 1, 4, 9, 16, 25, 36, 169, 961 - a total of 8 squares.

Bonus:
What is the maximum of squares in a 4x4 square, if using the hexadecimal system with digit 0-9 and A-F?
Note, square numbers here are f.e. 10=4*4 or 2A4=1A*1A.

Best solution I could come up with was,

13 squares:

1 3 7

6 2 5

9 4 8

which includes the squares,

{1, 4, 9, 16, 25, 36, 49, 64, 169, 324, 625, 729, 961}

What I tried,

Mostly trying to get as much squares with 2-digit squares (only missed 81) as possible and swapping around to make 3-digit squares and prioritizing the 169-961 double 3-digit square and other 2-3-digit doubles.

I started with 169 on a column and tried making 625 and 529 on rows, and then only 4 digits are remaining and possible to intuitively add focusing on 2-digits, or even brute-force as there are only 24 possibilities.

I just tried coding this too after seeing OP's comment, and if my program was correct,

This is the maximum and only this arrangement and rotations/reflections gives the answer.

The best I've managed so far is

12 squares

With the following

1 8 3

7 6 4

5 2 9


which has

1, 4, 9, 16, 25, 36, 49, 64, 81, 169, 529, 961

General Strategy

It's not too difficult to include all of the 2-digit squares. After that it's sensible to have 169 in there (as you get 961 for free) and then 529 is also easy to get in as we have 25.

Solution 1 with 2 as center

9 4 8

6 2 5

1 3 7

13 squares { 1,4, 9, 16, 25, 36, 49, 64, 169, 324, 625, 729, 961 }

Strategy: I first write down the all 3 digits squares then I look for the 3 digit squares with a most solid center. means a maximum of 3 digits squares with common center number. I found there are 4 these types of squares {324, 625, 529, 729} where the common center is 2. so I made 2 is my center number and write down all the 3 digits squares. then I tried to cover the other 3 digits squares by rearranging them. then finally I look for all 2 digit squares and be able to get them mostly except 81. Also, one more thing {169 and 961} is also a good combo so don't miss it. if you see the 3 digit square list below that we can use, we can clearly see that 2 makes the most solid center so the most solid solution, after it 3 and 8 can give a better result.

Possible 3 digit squares we can use with solid centers.

1. {324, 529, 625, 729}




2. {169, 361, 961 }




3. {289, 784 }




4. {841}{196}{256}{576}

Solution 2 with 2 as center

5 7 8

3 2 4

1 6 9

same strategy 12 squares { 1,4, 9, 16, 25, 36, 49, 64, 169, 324, 529,
961 }

Solution 3 with 8 as center

2 7 1

5 8 6

3 4 9

Another 12 squares{ 1,4, 9, 16, 25, 49, 64, 81,169, 196, 289, 784 }

Solution 4 as 6 as the center.

1 5 7

8 6 2

3 4 9

12 squares same strategy { 1 ,4, 9, 16, 25, 49, 64, 81,169, 729, 961 }