Prolog get squares of NxN matrix - list

Question

I have a list L = [[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]] Ii. That represents my matrix. The size can change dynamic, so the blocksize can be different, 4x4 = 4 elements, 9x9= 9 elements
I want to obtain the 4 squares that compose the List.(In this case it's a matrix 4 by 4). If I have that matrix:
5 6 7 8
10 11 12 13
1 2 3 4
14 15 16 17
The result should be:
R = [5,6,10,11],[7,8,12,13],[1,2,14,15],[3,4,16,17].
Any suggestions are welcomed. Thanks

Answer 1

The first thing you need is really a lever for turning a list of lists into a matrix. What distinguishes a 2-dimensional matrix from a list of lists? The idea of a coordinate system. So you need a way to relate a coordinate pair with the corresponding value in the matrix.
at(Matrix, X, Y, V) :- nth0(X, Matrix, Row), nth0(Y, Row, V).
This predicate makes it possible to index the matrix at (X,Y) and get the value V. This turns out to be, IMO, a massive demonstration of what makes Prolog powerful, because once you have this one, simple predicate, you gain:
The ability to obtain the value at the point supplied:
?- at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], 1,3, V).
V = 13.
The ability to iterate the entire matrix (only instantiate Matrix and leave the other arguments as variables):
?- at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], X,Y, V).
X = Y, Y = 0,
V = 5 ;
X = 0,
Y = 1,
V = 6 ;
...
X = 3,
Y = 2,
V = 16 ;
X = Y, Y = 3,
V = 17.
The ability to search the matrix for values:
?- at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], X,Y, 14).
X = 3,
Y = 0 ;
false.
So this is a pretty useful lever! In a conventional lanugage, you'd need three different functions to do all these things, but this is different, because in Prolog we just have to define the relationship between things (in this case, a data structure and a coordinate pair) and Prolog can do quite a bit of the heavy lifting.
It's easy to see how we could produce a particular submatrix now, by just defining the sets of X and Y values we'd like to see. For instance, to get the upper-left matrix we would do this:
?- between(0,1,X), between(0,1,Y),
at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], X,Y, V).
X = Y, Y = 0,
V = 5 ;
X = 0,
Y = 1,
V = 6 ;
X = 1,
Y = 0,
V = 10 ;
X = Y, Y = 1,
V = 11.
We can of course use findall/3 to gather up the solutions in one place:
?- findall(V, (between(0,1,X), between(0,1,Y),
at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], X,Y, V)),
Vs).
Vs = [5, 6, 10, 11].
What's left for your problem is basically some arithmetic. Let's see if we have a square matrix:
square_matrix(M, Degree) :-
length(M, Degree),
maplist(length, M, InnerDegrees),
forall(member(I, InnerDegrees), I=Degree).
This is not a perfect predicate, in that it will not generate! But it will tell us whether a matrix is square and if so, what degree it has:
?- square_matrix([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], D).
D = 4.
Once you have that, what you have to do is sort of formulaic:
Make sure the degree is a perfect square
Take the square root of the degree. That's how many rows or columns you have (square root 4 = 2, 2 rows and 2 columns, square root 9 = 3, 3 rows and 3 columns).
Make a relationship between the (row,column) coordinate and a list of (x,y) coordinates for the matrix in that location. For instance in the 4x4 matrix, you have four tiles: (0,0), (0,1), (1,0) and (1,1). The coordinates for (0,0) will be (0,0), (0,1), (1,0), (1,1), but the coordinates for (1,1) will be (2,2),(2,3),(3,2),(3,3). If you do a few of these by hand, you'll see it's going to amount to adding an x and y offset to all the permutations from 0 to row/column count (minus one) for both coordinates.
Now that you have that relationship, you need to do the iteration and assemble your output. I think maplist/N will suffice for this.
Hope this helps!