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							62 lines
						
					
					
						
							1.9 KiB
						
					
					
				
								/* MFASP, Minimum Feedback Arc Set Problem */
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								/* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */
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								/* The Minimum Feedback Arc Set Problem for a given directed graph
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								   G = (V, E), where V is a set of vertices and E is a set of arcs, is
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								   to find a minimal subset of arcs, which being removed from the graph
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								   make it acyclic.
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								   Reference:
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								   Garey, M.R., and Johnson, D.S. (1979), Computers and Intractability:
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								   A guide to the theory of NP-completeness [Graph Theory, Covering and
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								   Partitioning, Minimum Feedback Arc Set, GT9]. */
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								param n, integer, >= 0;
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								/* number of vertices */
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								set V, default 1..n;
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								/* set of vertices */
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								set E, within V cross V,
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								default setof{i in V, j in V: i <> j and Uniform(0,1) <= 0.15} (i,j);
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								/* set of arcs */
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								printf "Graph has %d vertices and %d arcs\n", card(V), card(E);
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								var x{(i,j) in E}, binary;
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								/* x[i,j] = 1 means that (i->j) is a feedback arc */
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								/* It is known that a digraph G = (V, E) is acyclic if and only if its
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								   vertices can be assigned numbers from 1 to |V| in such a way that
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								   k[i] + 1 <= k[j] for every arc (i->j) in E, where k[i] is a number
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								   assigned to vertex i. We may use this condition to require that the
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								   digraph G = (V, E \ E'), where E' is a subset of feedback arcs, is
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								   acyclic. */
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								var k{i in V}, >= 1, <= card(V);
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								/* k[i] is a number assigned to vertex i */
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								s.t. r{(i,j) in E}: k[j] - k[i] >= 1 - card(V) * x[i,j];
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								/* note that x[i,j] = 1 leads to a redundant constraint */
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								minimize obj: sum{(i,j) in E} x[i,j];
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								/* the objective is to minimize the cardinality of a subset of feedback
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								   arcs */
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								solve;
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								printf "Minimum feedback arc set:\n";
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								printf{(i,j) in E: x[i,j]} "%d %d\n", i, j;
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								data;
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								/* The optimal solution is 3 */
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								param n := 15;
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								set E := 1 2, 2 3, 3 4, 3 8, 4 9, 5 1, 6 5, 7 5, 8 6, 8 7, 8 9, 9 10,
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								         10 11, 10 14, 11 15, 12 7, 12 8, 12 13, 13 8, 13 12, 13 14,
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								         14 9, 15 14;
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								end;
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