/* MAXCUT, Maximum Cut Problem */

 /* Written in GNU MathProg by Andrew Makhorin <mao@gnu.org> */

 /* The Maximum Cut Problem in a network G = (V, E), where V is a set

    of nodes, E is a set of edges, is to find the partition of V into

    disjoint sets V1 and V2, which maximizes the sum of edge weights

    w(e), where edge e has one endpoint in V1 and other endpoint in V2.

    Reference:

    Garey, M.R., and Johnson, D.S. (1979), Computers and Intractability:

    A guide to the theory of NP-completeness [Network design, Cuts and

    Connectivity, Maximum Cut, ND16]. */

 set E, dimen 2;

 /* set of edges */

 param w{(i,j) in E}, >= 0, default 1;

 /* w[i,j] is weight of edge (i,j) */

 set V := (setof{(i,j) in E} i) union (setof{(i,j) in E} j);

 /* set of nodes */

 var x{i in V}, binary;

 /* x[i] = 0 means that node i is in set V1

    x[i] = 1 means that node i is in set V2 */

 /* We need to include in the objective function only that edges (i,j)

    from E, for which x[i] != x[j]. This can be modeled through binary

    variables s[i,j] as follows:

       s[i,j] = x[i] xor x[j] = (x[i] + x[j]) mod 2,                  (1)

    where s[i,j] = 1 iff x[i] != x[j], that leads to the following

    objective function:

       z = sum{(i,j) in E} w[i,j] * s[i,j].                           (2)

    To describe "exclusive or" (1) we could think that s[i,j] is a minor

    bit of the sum x[i] + x[j]. Then introducing binary variables t[i,j],

    which represent a major bit of the sum x[i] + x[j], we can write:

       x[i] + x[j] = s[i,j] + 2 * t[i,j].                             (3)

    An easy check shows that conditions (1) and (3) are equivalent.

    Note that condition (3) can be simplified by eliminating variables

    s[i,j]. Indeed, from (3) it follows that:

       s[i,j] = x[i] + x[j] - 2 * t[i,j].                             (4)

    Since the expression in the right-hand side of (4) is integral, this

    condition can be rewritten in the equivalent form:

       0 <= x[i] + x[j] - 2 * t[i,j] <= 1.                            (5)

    (One might note that (5) means t[i,j] = x[i] and x[j].)

    Substituting s[i,j] from (4) to (2) leads to the following objective

    function:

       z = sum{(i,j) in E} w[i,j] * (x[i] + x[j] - 2 * t[i,j]),       (6)

    which does not include variables s[i,j]. */

 var t{(i,j) in E}, binary;

 /* t[i,j] = x[i] and x[j] = (x[i] + x[j]) div 2 */

 s.t. xor{(i,j) in E}: 0 <= x[i] + x[j] - 2 * t[i,j] <= 1;

 /* see (4) */

 maximize z: sum{(i,j) in E} w[i,j] * (x[i] + x[j] - 2 * t[i,j]);

 /* see (6) */

 data;

 /* In this example the network has 15 nodes and 22 edges. */

 /* Optimal solution is 20 */

 set E :=

    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,

    8 12, 9 10, 9 14, 10 11, 10 14, 11 15, 12 13, 13 14, 14 15;

 end;