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我想知道如果上限（最小化问题）在“x”秒后没有变得更好，CP Optimizer 12.10 中是否存在停止搜索的方法。当试图解决一个困难的 NP-Hard 问题实例并且搜索停留在一个现有的解决方案中时，这将特别有用。

我知道存在一些参数来控制搜索cp.param.TimeLimit（我正在使用它）或者cp.param.FailLimit但这不是我的问题所需要的。

任何帮助将不胜感激。

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1 回答 1

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如果您依赖 OPL API，您可以做的是使用主块（流控制），并且在此主块中，您给出 x 秒的时间限制，然后您将获得边界和当前目标，如果您认为可以保存它们如果您对解决方案没问题，您应该走得更远或停下来。

让我分享一个来自 OPL 产品中 lifegame 示例的完整示例。并再次记住约翰康威。

using CP;





int n=20;
int Half=n div 2;
range FirstHalf = 1..Half;
range LastHalf = n-Half+1..n; 
range States = 0..1;
range Bord = 0..(n+1);
range Interior = 1..n;

range obj = 0..(n*n);

tuple neighbors {
   int row;
   int col;
}

{neighbors} Neighbor = 
  {<(-1),(-1)>,<(-1),0>,<(-1),1>,<0,(-1)>,<0,1>,<1,(-1)>,<1,0>,<1,1>};

dvar int Life[Bord][Bord] in States;
dvar int Obj in obj;

dvar int x[0..(n+2)*(n+2)-1];

maximize Obj;

subject to {

forall(i,j in Bord) Life[i][j]==x[i*(n+2)+j];

  //ct1:
    Obj == sum( i , j in Bord ) 
      Life[i][j];

  forall( i , j in Interior ) {
    ct21: 
      2*Life[i][j] - sum( nb in Neighbor ) 
        Life[i+nb.row][j+nb.col] <= 0;
    ct22:
      3*Life[i][j] + sum( nb in Neighbor ) 
        Life[i+nb.row][j+nb.col] <= 6;
    forall( ordered n1 , n2 , n3 in Neighbor ) {
      ct23: 
        -Life[i][j]+Life[i+n1.row][j+n1.col]
                   +Life[i+n2.row][j+n2.col]
                   +Life[i+n3.row][j+n3.col]
        -sum( nb in Neighbor : nb!=n1 && nb!=n2 && nb!=n3 ) 
          Life[i+nb.row][j+nb.col] <= 2;
    }
  }
  forall( j in Bord ) {
    ct31:
      Life[0][j] == 0;
    ct32:  
      Life[j][0] == 0;
    ct33:  
      Life[j][n+1] == 0;
    ct34:  
      Life[n+1][j] == 0;
  }
  forall( i in Bord : i<n ) {
    ct41:
      Life[i][1]+Life[i+1][1]+Life[i+2][1] <= 2;
    ct42:
      Life[1][i]+Life[1][i+1]+Life[1][i+2] <= 2;
    ct43:
      Life[i][n]+Life[i+1][n]+Life[i+2][n] <= 2;
    ct44:
      Life[n][i]+Life[n][i+1]+Life[n][i+2] <= 2;
  }
  ct5:
    sum( i in FirstHalf , j in Bord ) 
      Life[i][j] >= 
    sum( i in LastHalf , j in Bord ) 
      Life[i][j];
  ct6:
    sum( i in Bord , j in FirstHalf ) 
      Life[i][j] >= 
    sum( i in Bord , j in LastHalf ) 
      Life[i][j];   
}


main
{

thisOplModel.generate();
cp.param.timelimit=10;
var obj=0;
var bound=0;
var oldobj=0;
var oldbound=0;

while (1==1)
{
  cp.solve();
  obj=cp.getObjValue();
  bound=cp.getObjBound();
  writeln("bound and obj =",bound," ",obj);
  if (Opl.abs((oldobj-oldbound-obj+bound))<=0) break;
  oldobj=obj;
  oldbound=bound;
}  

}

这使

bound and obj =398 181
bound and obj =398 180
bound and obj =398 180
// Script execution finished with status 1.
于 2020-06-05T06:24:49.570 回答