--  count triadic couplings in a special set of edges.
--  Idea: three edges form a "triangle" iff they aren't linearly
--  dependent and if they connect exactly three points. This implies
--  an enclosed area > 0. (See the function procedure dep.)

program triads;


	read(F);   -- input F, a set of edges
	--  F is like E, see below, which is the set of 11 - 2 points as
	--  explained on http://www.frank-buss.de/challenge/index.html,
	--  The program computes the Euclidean stuff like "What is the
	--  fraction of the projection of P2 onto the y-axis of the
	--  projection of P6 on the y-axis", but only as auxiliary
	--  expressions, not in a core algorithm, which depends
	--  on the representation of lines as sets of pairs.
	--  (Apologies, I'm a mathematically challenged person...)



	--  Measurement, height and width step 1:
	--
	--     4
	--    3 5
	--   2   6
	--  1     7


	E := {  --  sets of connected points, each representing a line
		{[1,1], [7,1]},
		{[1,1], [6,2]},
		{[1,1], [5,3]},
		{[1,1], [4,4]},
		{[1,1], [3,3]},
		{[1,1], [2,2]},
		{[2,2], [7,1]},
		{[2,2], [2,4]},
		{[2,2], [3,3]},
		{[3,3], [7,1]},
		{[3,3], [4,4]},
		{[4,4], [7,1]},
		{[4,4], [6,2]},
		{[4,4], [5,3]},
		{[5,3], [7,1]},
		{[5,3], [6,2]},
		{[6,2], [7,1]},


		--  points in the middle, computed so they match the
		--  graph on the web page


		{[1,1],   [4, 3/5.0]}, -- ht(P6) / 5 = ht(P2) / 3, see measure
		{[4, 3/5.0],   [7,1]}, -- line (P2, P1), similarly
		{[1,1],   [4, 3/2.0]},
		{[4, 3/2.0],   [7,1]}, -- (P7, P1)

		--  P3 and P4 missing
		};

	assert( #(+/{line: line in E}) = 9 );
	--  there are 9 points in the union of all sets in E

	tri := {l3: l3 in 3 npow E | #(+/l3) = 3 and not dep((+/l3))};
	--   set of all subsets of E of size 3, i.e. three line sets,
	--  such that the ends of lines in l3 are connecting a total of
	--  3 points, not all on a straight line

	print(#tri);

	--    done, output is number of elements of "three liners"
	--  enclosing an area


	--  ad lib: print all triangles found:

	for dbg in tri loop
		print(dbg);
	end loop;



	procedure dep(triset);

		-- three points in a line?
		--  (a  c)(x) = (e)
		--  (b  d)(y) = (f)

		[a,b] from triset;
		[c,d] from triset;
		[e,f] from triset;

		return a*(f - d) + b*(c - e) = 0;
	end dep;

end triads;