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/* 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;
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