COIN-OR::LEMON - Graph Library

source: lemon-project-template-glpk/deps/glpk/examples/mfasp.mod

subpack-glpk
Last change on this file was 9:33de93886c88, checked in by Alpar Juttner <alpar@…>, 13 years ago

Import GLPK 4.47

File size: 1.9 KB
RevLine 
[9]1/* MFASP, Minimum Feedback Arc Set Problem */
2
3/* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */
4
5/* The Minimum Feedback Arc Set Problem for a given directed graph
6   G = (V, E), where V is a set of vertices and E is a set of arcs, is
7   to find a minimal subset of arcs, which being removed from the graph
8   make it acyclic.
9
10   Reference:
11   Garey, M.R., and Johnson, D.S. (1979), Computers and Intractability:
12   A guide to the theory of NP-completeness [Graph Theory, Covering and
13   Partitioning, Minimum Feedback Arc Set, GT9]. */
14
15param n, integer, >= 0;
16/* number of vertices */
17
18set V, default 1..n;
19/* set of vertices */
20
21set E, within V cross V,
22default setof{i in V, j in V: i <> j and Uniform(0,1) <= 0.15} (i,j);
23/* set of arcs */
24
25printf "Graph has %d vertices and %d arcs\n", card(V), card(E);
26
27var x{(i,j) in E}, binary;
28/* x[i,j] = 1 means that (i->j) is a feedback arc */
29
30/* It is known that a digraph G = (V, E) is acyclic if and only if its
31   vertices can be assigned numbers from 1 to |V| in such a way that
32   k[i] + 1 <= k[j] for every arc (i->j) in E, where k[i] is a number
33   assigned to vertex i. We may use this condition to require that the
34   digraph G = (V, E \ E'), where E' is a subset of feedback arcs, is
35   acyclic. */
36
37var k{i in V}, >= 1, <= card(V);
38/* k[i] is a number assigned to vertex i */
39
40s.t. r{(i,j) in E}: k[j] - k[i] >= 1 - card(V) * x[i,j];
41/* note that x[i,j] = 1 leads to a redundant constraint */
42
43minimize obj: sum{(i,j) in E} x[i,j];
44/* the objective is to minimize the cardinality of a subset of feedback
45   arcs */
46
47solve;
48
49printf "Minimum feedback arc set:\n";
50printf{(i,j) in E: x[i,j]} "%d %d\n", i, j;
51
52data;
53
54/* The optimal solution is 3 */
55
56param n := 15;
57
58set 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,
59         10 11, 10 14, 11 15, 12 7, 12 8, 12 13, 13 8, 13 12, 13 14,
60         14 9, 15 14;
61
62end;
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