# Solution: Magic Squares with initial constraints

### The code...

```declare
fun {Nuss N Data}
NN=N*N
L1N={List.number 1 N 1}
in
proc { \$ Square}
fun {Field I J}
Square.((I-1)*N + J)
end
proc {Assert F}
{FD.sum {Map L1N F} '=:' Sum}
end
Sum = {FD.decl}
in
{FD.tuple square NN 1#NN Square}
{FD.distinct Square}
% Diagonals
{Assert fun {\$ I} {Field I I} end}
{Assert fun {\$ I} {Field I N+1-I} end}
% Columns
{For 1 N 1
proc {\$ I} {Assert fun {\$ J} {Field I J} end} end}
% Rows
{For 1 N 1
proc {\$ J} {Assert fun {\$ I} {Field I J} end} end}
% Redundant Constraint
NN*(NN+1) div 2 =: N*Sum
%
{ForAll Data
proc {\$ P}
case P
of p(X Y Val) then Square.((X-1)*N + Y) =: Val
[] l(X Y) then Square.((X-1)*N + Y) <: 9
end
end
}
{FD.distribute split Square}
end
end

Data1=[p(1 4 5) p(2 1 13) p(2 4 8) p(3 1 3) p(4 3 7) l(1 2) l(4 1)]
Data2=[p(2 2 6) p(2 4 4) p(3 1 10) p(4 1 2) p(4 3 3) l(1 2) l(3 4)]
Data3=[p(1 1 7) p(1 3 2) p(2 2 4) p(4 1 1) p(4 2 11)]
Data4=[p(1 2 3) p(2 1 15) p(2 2 6) p(3 3 5) p(4 1 4) l(3 1)]
Data5=[p(2 1 15) p(2 2 1) p(3 4 3) p(4 1 6) p(4 3 7) l(1 4) l(2 3)]
Data6=[p(1 1 7) p(2 4 8) p(3 3 3) p(4 2 5) l(1 2) l(4 1)]

{ExploreAll {Nuss 4 Data1}}

```
The solution for the first square is, for example
```square(16  4  9  5
13  1 12  8
3 15  6 10
2 14  7 11)
```

Markus Löckelt