[1] | 1 | /* MFASP, Minimum Feedback Arc Set Problem */ |
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| 2 | |
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| 3 | /* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */ |
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| 4 | |
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| 5 | /* The Minimum Feedback Arc Set Problem for a given directed graph |
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| 6 | G = (V, E), where V is a set of vertices and E is a set of arcs, is |
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| 7 | to find a minimal subset of arcs, which being removed from the graph |
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| 8 | make it acyclic. |
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| 9 | |
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| 10 | Reference: |
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| 11 | Garey, M.R., and Johnson, D.S. (1979), Computers and Intractability: |
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| 12 | A guide to the theory of NP-completeness [Graph Theory, Covering and |
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| 13 | Partitioning, Minimum Feedback Arc Set, GT9]. */ |
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| 14 | |
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| 15 | param n, integer, >= 0; |
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| 16 | /* number of vertices */ |
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| 17 | |
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| 18 | set V, default 1..n; |
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| 19 | /* set of vertices */ |
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| 20 | |
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| 21 | set E, within V cross V, |
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| 22 | default setof{i in V, j in V: i <> j and Uniform(0,1) <= 0.15} (i,j); |
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| 23 | /* set of arcs */ |
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| 24 | |
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| 25 | printf "Graph has %d vertices and %d arcs\n", card(V), card(E); |
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| 26 | |
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| 27 | var x{(i,j) in E}, binary; |
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| 28 | /* x[i,j] = 1 means that (i->j) is a feedback arc */ |
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| 29 | |
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| 30 | /* It is known that a digraph G = (V, E) is acyclic if and only if its |
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| 31 | vertices can be assigned numbers from 1 to |V| in such a way that |
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| 32 | k[i] + 1 <= k[j] for every arc (i->j) in E, where k[i] is a number |
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| 33 | assigned to vertex i. We may use this condition to require that the |
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| 34 | digraph G = (V, E \ E'), where E' is a subset of feedback arcs, is |
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| 35 | acyclic. */ |
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| 36 | |
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| 37 | var k{i in V}, >= 1, <= card(V); |
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| 38 | /* k[i] is a number assigned to vertex i */ |
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| 39 | |
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| 40 | s.t. r{(i,j) in E}: k[j] - k[i] >= 1 - card(V) * x[i,j]; |
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| 41 | /* note that x[i,j] = 1 leads to a redundant constraint */ |
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| 42 | |
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| 43 | minimize obj: sum{(i,j) in E} x[i,j]; |
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| 44 | /* the objective is to minimize the cardinality of a subset of feedback |
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| 45 | arcs */ |
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| 46 | |
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| 47 | solve; |
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| 48 | |
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| 49 | printf "Minimum feedback arc set:\n"; |
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| 50 | printf{(i,j) in E: x[i,j]} "%d %d\n", i, j; |
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| 51 | |
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| 52 | data; |
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| 53 | |
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| 54 | /* The optimal solution is 3 */ |
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| 55 | |
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| 56 | param n := 15; |
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| 57 | |
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| 58 | 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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| 59 | 10 11, 10 14, 11 15, 12 7, 12 8, 12 13, 13 8, 13 12, 13 14, |
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| 60 | 14 9, 15 14; |
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| 61 | |
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| 62 | end; |
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