// special adaptation of the one-max problem
// must find maximal number of 1-terminals
// optimal solution = 243, number of solutions 3^243
PROBLEM OneMaxTernary

NONTERMINALS
  S<<out int n>>.
  U<<out int n>>.
  V<<out int n>>.
  W<<out int n>>.
  X<<out int n>>.
  T<<out int n>>.

TERMINALS
  A. B. C.

RULES
  S<<out int n>> =                                LOCAL << int n1, n2, n3; >>
    U<<out n1>> U<<out n2>> U<<out n3>>           SEM << n = n1 + n2 + n3; >>
  .
  U<<out int n>> =                                LOCAL << int n1, n2, n3; >>
    V<<out n1>> V<<out n2>> V<<out n3>>           SEM << n = n1 + n2 + n3; >>
  .
  V<<out int n>> =                                LOCAL << int n1, n2, n3; >>
    W<<out n1>> W<<out n2>> W<<out n3>>           SEM << n = n1 + n2 + n3; >>
  .
  W<<out int n>> =                                LOCAL << int n1, n2, n3; >>
    X<<out n1>> X<<out n2>> X<<out n3>>           SEM << n = n1 + n2 + n3; >>
  .

  X<<out int n>> =                                LOCAL << int n1, n2, n3; >>
    T<<out n1>> T<<out n2>> T<<out n3>>           SEM << n = n1 + n2 + n3; >>
  .

  T<<out int n>> =     
    A                                             SEM << n = 0; >>
    | B                                           SEM << n = 1; >>
    | C                                           SEM << n = 0; >>
  .

MAXIMIZE
  << 
    int n;
    S(out n);
    return (double) n;
  >>
END OneMaxTernary.