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| Given a line in one dimension from p1 to p2, we can find the line of length 1/2 (p2-p1) centered around the midpoint of the first line in MLPQ as follows: |
begin%Boxes%
Choice(m,n) :- m=0, n=0.
Choice(m,n) :- m=0, n=1.
Choice(m,n) :- m=0, n=2.
Choice(m,n) :- m=1, n=0.
Choice(m,n) :- m=1, n=2.
Choice(m,n) :- m=2, n=0.
Choice(m,n) :- m=2, n=1.
Choice(m,n) :- m=2, n=2.
D(x,y,z) :- x-y=0, z=0.
D(x,y,z) :- D(x1,y,z1), x-x1=1, z-z1=1.
M(x,y,z) :- x=0, y=0, z=0.
M(x,y,z) :- M(x1,y,z1), D(z,z1,y), x-x1=1.
M(x,y,z) :- M(x,y1,z1), D(z,z1,x), y-y1=1.
/* Creates the first set of base lines length=1,3,9... */
BaseLine(x1,x2):- x1=0, x2=1.
BaseLine(a1,d3):- BaseLine(a1,d1), l=3, M(l,d1,d3).
Box(x,y,s) :- BaseLine(x,b), y-x=0, b-s=0.
Box(x,y,s) :- Box(a,b,l), k=3, M(k,s,l), Choice(m,n),
M(s,m,z1), D(x,a,z1),
M(s,n,z2), D(y,b,z2).
XBox(a1,c1,s) :- Box(a,c,l), k=3, M(k,s,l), D(a1,a,s), D(c1,c,s).
fractal1(id,x,y) :- id=1, Box(a,b,l), x-a>=0, D(x1,l,a), x-x1<=0,
y-b>=0, D(y1,l,b), y-y1<=0.
fractal2(id,x,y) :- id=2, XBox(a,b,l), x-a>=0, D(x1,l,a), x-x1<=0,
y-b>=0, D(y1,l,b), y-y1<=0.
end%Boxes%