# Constraint Programming with CL Screamer

I’ve been wanting to get into Lisp again and last week it came to my hands a problem that looked like a perfect case for giving CL another try.

The problem is about a very special sudoku with only 4 initial values. The sudoku has, of course special rules:

- The square in the middle is a magic square (all rows,columns, and digonals add up the same).
- Both diagonals of the board can’t have repeated numbers.
- any cell can’t be the same number as any of it’s chess knight-distance peers.

The solution is here.

Anyway, I didn’t want to solve it by hand, so I started thinking how to do it, and remembered a few recent screamer posts I read, so I gave a shoot to it.

Here’s the simple solution for solving the magic square using Constraint Programming.

```
(one-value
(let ((a (an-integer-between 1 9))
(b (an-integer-between 1 9))
(c (an-integer-between 1 9))
(d (an-integer-between 1 9))
(e (an-integer-between 1 9))
(f (an-integer-between 1 9))
(g (an-integer-between 1 9))
(h (an-integer-between 1 9))
(i (an-integer-between 1 9)))
(if (and (= (+ a b c) (+ d e f) (+ g h i)
(+ a d g) (+ b e h) (+ c f i)
(+ a e i) (+ g e c))
(/=v a b c d e f g h i))
(list a b c d e f g h i)
(fail)))) ; => (2 7 6 9 5 1 4 3 8)
```

Very nice to use automatic backtracking for solving any kind of
problem: The `fail`

function, `/=`

, and `one-value`

communicate what’s
the state of the world and decide to try other values.