# Solving Magic Squares via Backtracking

*This post is part of a series.*

Brute force search can often be slow or infeasible when there are lots of possibilities that must be checked.

However, a technique called **backtracking** can often be used to drastically cut down the number of possibilities that must be checked. The idea is that whenever we find ourselves constructing a set of possibilities that we know are hopeless, we skip over them and *backtrack* to the the point right before we started constructing the hopeless possibilities.

## Exercise: Brute Force

Use brute-force search to find all arrangements of the digits $1,$ $2,$ $\ldots,$ $9$ into a $3 \times 3$ *magic square* where all the rows, columns, and diagonals add up to $15$ and no digits are repeated.

You may use $9$ nested for loops if you’d like. It’s ugly but conceptually simple and gets the job done. To illustrate, pseudocode is provided below.

```
digits = [1, 2, ..., 9]
square = [
[None, None, None]
[None, None, None]
[None, None, None]
]
for num1 in digits:
clear out the square and put num1 in
for num2 in digits excluding num1:
clear out the square and put num1, num2 in
for num3 in digits excluding num1, num2:
clear out the square and put num1, num2, num3 in
... and so on
for num9 in digits excluding num1, ..., num8:
clear out the square and put num1, ..., num9 in
if is_valid(square):
print(square)
```

Note that this solution will be rather slow because it will require checking $9! = 362\,880$ arrangements of digits.

## Exercise: Backtracking

Repeat the above exercise, but this time, whenever you reach an arrangement of digits that can no longer become a valid magic square, do not explore that arrangement any further. You can accomplish this by doing the following:

- Write a function
`is_hopeless(square)`

to check whether an incomplete square could ever become a valid square. A square is hopeless if any row, column, or diagonal is filled in and does not sum to $15.$ - Place
`continue`

statements in your $9$ nested for loops so that you skip inner loops whenever you detect a hopeless square.

Below are some examples to illustrate how the `is_hopeless`

function should work.

```
>>> is_hopeless([
[1, 2, None],
[None, 3, None],
[5, 6, 4 ]
])
OUTPUT: True
REASONING
- A diagonal is filled in and it doesn't sum to 15.
- No matter how we fill the rest of the square,
this diagonal still won't sum to 15.
- Therefore, this incomplete arrangement cannot lead
to any valid square.
```

```
>>> is_hopeless([
[1, 2, None],
[None, None, None],
[5, 6, 4 ]
])
OUTPUT: False
REASONING: There are no filled rows, columns, or
diagonals that don't sum to 15.
```

```
>>> is_hopeless([
[None, None, None]
[None, None, None]
[None, None, None]
])
OUTPUT: False
REASONING: There are no filled rows, columns, or
diagonals that don't sum to 15.
```

Below is an illustration of how to skip inner loops using `continue`

statements.

```
for num1 in digits:
clear out the square and put num1 in it
if is_hopeless(square):
continue
for num2 in digits excluding num1:
clear out the square and put num1, num2 in it
if is_hopeless(square):
continue
... and so on
```

To give a concrete demonstration of backtracking, the first several arrangements that get explored are shown below.

```
[[_,_,_],
[_,_,_],
[_,_,_]]
[[1,_,_],
[_,_,_],
[_,_,_]]
[[1,2,_],
[_,_,_],
[_,_,_]]
[[1,2,3],
[_,_,_],
[_,_,_]]
^ hopeless -- do not explore further
[[1,2,4],
[_,_,_],
[_,_,_]]
^ hopeless -- do not explore further
[[1,2,5],
[_,_,_],
[_,_,_]]
^ hopeless -- do not explore further
...
[[1,2,9],
[_,_,_],
[_,_,_]]
^ hopeless -- do not explore further
[[1,3,_],
[_,_,_],
[_,_,_]]
...
```

*This post is part of a series.*