examples/maxcut.mod
author Alpar Juttner <alpar@cs.elte.hu>
Mon, 06 Dec 2010 13:09:21 +0100
changeset 1 c445c931472f
permissions -rw-r--r--
Import glpk-4.45

- Generated files and doc/notes are removed
     1 /* MAXCUT, Maximum Cut Problem */
     2 
     3 /* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */
     4 
     5 /* The Maximum Cut Problem in a network G = (V, E), where V is a set
     6    of nodes, E is a set of edges, is to find the partition of V into
     7    disjoint sets V1 and V2, which maximizes the sum of edge weights
     8    w(e), where edge e has one endpoint in V1 and other endpoint in V2.
     9 
    10    Reference:
    11    Garey, M.R., and Johnson, D.S. (1979), Computers and Intractability:
    12    A guide to the theory of NP-completeness [Network design, Cuts and
    13    Connectivity, Maximum Cut, ND16]. */
    14 
    15 set E, dimen 2;
    16 /* set of edges */
    17 
    18 param w{(i,j) in E}, >= 0, default 1;
    19 /* w[i,j] is weight of edge (i,j) */
    20 
    21 set V := (setof{(i,j) in E} i) union (setof{(i,j) in E} j);
    22 /* set of nodes */
    23 
    24 var x{i in V}, binary;
    25 /* x[i] = 0 means that node i is in set V1
    26    x[i] = 1 means that node i is in set V2 */
    27 
    28 /* We need to include in the objective function only that edges (i,j)
    29    from E, for which x[i] != x[j]. This can be modeled through binary
    30    variables s[i,j] as follows:
    31 
    32       s[i,j] = x[i] xor x[j] = (x[i] + x[j]) mod 2,                  (1)
    33 
    34    where s[i,j] = 1 iff x[i] != x[j], that leads to the following
    35    objective function:
    36 
    37       z = sum{(i,j) in E} w[i,j] * s[i,j].                           (2)
    38 
    39    To describe "exclusive or" (1) we could think that s[i,j] is a minor
    40    bit of the sum x[i] + x[j]. Then introducing binary variables t[i,j],
    41    which represent a major bit of the sum x[i] + x[j], we can write:
    42 
    43       x[i] + x[j] = s[i,j] + 2 * t[i,j].                             (3)
    44 
    45    An easy check shows that conditions (1) and (3) are equivalent.
    46 
    47    Note that condition (3) can be simplified by eliminating variables
    48    s[i,j]. Indeed, from (3) it follows that:
    49 
    50       s[i,j] = x[i] + x[j] - 2 * t[i,j].                             (4)
    51 
    52    Since the expression in the right-hand side of (4) is integral, this
    53    condition can be rewritten in the equivalent form:
    54 
    55       0 <= x[i] + x[j] - 2 * t[i,j] <= 1.                            (5)
    56 
    57    (One might note that (5) means t[i,j] = x[i] and x[j].)
    58 
    59    Substituting s[i,j] from (4) to (2) leads to the following objective
    60    function:
    61 
    62       z = sum{(i,j) in E} w[i,j] * (x[i] + x[j] - 2 * t[i,j]),       (6)
    63 
    64    which does not include variables s[i,j]. */
    65 
    66 var t{(i,j) in E}, binary;
    67 /* t[i,j] = x[i] and x[j] = (x[i] + x[j]) div 2 */
    68 
    69 s.t. xor{(i,j) in E}: 0 <= x[i] + x[j] - 2 * t[i,j] <= 1;
    70 /* see (4) */
    71 
    72 maximize z: sum{(i,j) in E} w[i,j] * (x[i] + x[j] - 2 * t[i,j]);
    73 /* see (6) */
    74 
    75 data;
    76 
    77 /* In this example the network has 15 nodes and 22 edges. */
    78 
    79 /* Optimal solution is 20 */
    80 
    81 set E :=
    82    1 2, 1 5, 2 3, 2 6, 3 4, 3 8, 4 9, 5 6, 5 7, 6 8, 7 8, 7 12, 8 9,
    83    8 12, 9 10, 9 14, 10 11, 10 14, 11 15, 12 13, 13 14, 14 15;
    84 
    85 end;