将布尔型FlatZinc转换为CNF

Question

为了解决布尔方程组问题，我正在使用以下输入对Constraint-Programming Solver MiniZinc进行实验：

%  Solve system of Brent's equations modulo 2

%  Matrix dimensions
int: aRows = 3;
int: aCols = 3;
int: bCols = 3;
int: noOfProducts = 23;

%  Dependent parameters
int: bRows = aCols;
int: cRows = aRows;
int: cCols = bCols;
set of int: products = 1..noOfProducts;

%  Corefficients are stored in arrays
array[1..aRows, 1..aCols, products] of var bool: A;
array[1..bRows, 1..bCols, products] of var bool: B;
array[1..cRows, 1..cCols, products] of var bool: C;

constraint
    forall(rowA in 1..aRows, colA in 1..aCols) (
        forall(rowB in 1..bRows, colB in 1..bCols) (
            forall (rowC in 1..cRows, colC in 1..cCols) (
                xorall (k in products) (
                    A[rowA, colA, k] /\ B[rowB, colB, k] /\ C[rowC, colC, k]
                ) == ((rowA == rowC) /\ (colB == colC) /\ (colA == rowB))
            )
        )
    );

solve satisfy;

%  Output solution as table of variable value assignments
output 
["\nSolution for <" ++ show(aRows) ++ ", " ++ show(aCols) ++ 
                 ", " ++ show(bCols) ++ "> " ++ show(noOfProducts) ++ " products:"] ++
["\nF" ++ show(100*rowA+10*colA+k) ++ " = " ++ 
show(bool2int(A[rowA, colA, k])) |
rowA in 1..aRows, colA in 1..aCols, k in products] ++

["\nG" ++ show(100*rowB+10*colB+k) ++ " = " ++ 
show(bool2int(B[rowB, colB, k])) |
rowB in 1..bRows, colB in 1..bCols, k in products] ++

["\nD" ++ show(100*rowC+10*colC+k) ++ " = " ++ 
show(bool2int(C[rowC, colC, k])) |
rowC in 1..cRows, colC in 1..cCols, k in products];

MiniZinc确实为小参数(rows=cols=2, products=7)找到了一种解决方案，但并没有随着稍微增加的参数而结束。我想将生成的FlatZinc模型输入到卫星解算器中，比如隐孢子虫、凌岭或卡环。我希望这些工具的性能可能优于现有的MiniZinc后端。

我的问题：

是否有任何工具可以将纯布尔型FlatZinc模型转换为CNF (DIMACS)

当一些xorall()后端似乎不支持它时，我能做什么来替换MiniZinc谓词呢？

Answer 1

我不知道有什么工具可以将FlatZinc文件转换为和CNF (DIMACS)文件。( MiniZinc发行版有一个程序将平面锌转换为XCSP格式。也许有一个XCSP到CNF的工具？)

然而，有一些基于SAT的/受启发的解决方案可能更好，例如minicsp、fzn2smt。问题是，正如您提到的，它们不支持非常新的xorall()函数。

另外，使用有标签的搜索可能是个好主意，比如这样的搜索(注意bool_search)

  solve :: bool_search(
       [A[i,j,k] | i in 1..aRows, j in 1..aCols, k in products],
       first_fail,
       indomain_min,
       complete)
     satisfy;

另外，我建议您进行测试，将其转换为基于0..1的模型，在该模型中，可以测试这些求解器以及其他模型。

下面是我转换的模型，我刚刚将var bool更改为var 0.1，并将xorall()替换为sum()和bool2int()，我希望转换正确。更新:我已经更改为Axel建议的版本。

 %  Solve system of Brent's equations modulo 2

 %  Matrix dimensions
 int: aRows = 3;
 int: aCols = 3;
 int: bCols = 3;
 int: noOfProducts = 23;

 %  Dependent parameters
 int: bRows = aCols;
 int: cRows = aRows;
 int: cCols = bCols;
 set of int: products = 1..noOfProducts;

 %  Corefficients are stored in arrays
 array[1..aRows, 1..aCols, products] of var 0..1: A; % hakank: change to 0..1
 array[1..bRows, 1..bCols, products] of var 0..1: B;
 array[1..cRows, 1..cCols, products] of var 0..1: C;

constraint
     forall(rowA in 1..aRows, colA in 1..aCols) (
         forall(rowB in 1..bRows, colB in 1..bCols) (
             forall (rowC in 1..cRows, colC in 1..cCols) (
                 % hakank: changed
                 sum (k in products) (
                     bool2int(A[rowA, colA, k]=1/\ B[rowB, colB, k]=1 /\ C[rowC, colC, k]=1)
                ) == 
                     %% bool2int(rowA == rowC)+ bool2int(colB == colC) + bool2int(colA == rowB)
                     bool2int((rowA == rowC)/\(colB == colC)/\(colA == rowB))
             )
         )
     );


     solve :: int_search(
         [A[i,j,k] | i in 1..aRows, j in 1..aCols, k in products] ++ 
         [B[i,j,k] | i in 1..aRows, j in 1..aCols, k in products] ++ 
         [C[i,j,k] | i in 1..aRows, j in 1..aCols, k in products] 
         ,
         first_fail,
         indomain_min,
         complete)
     satisfy;

    %  Output solution as table of variable value assignments
    output 
    ["\nSolution for <" ++ show(aRows) ++ ", " ++ show(aCols) ++ 
             ", " ++ show(bCols) ++ "> " ++ show(noOfProducts) ++ " products:"] ++
    ["\nF" ++ show(100*rowA+10*colA+k) ++ " = " ++ 
        show(A[rowA, colA, k]) |
        rowA in 1..aRows, colA in 1..aCols, k in products] ++

    ["\nG" ++ show(100*rowB+10*colB+k) ++ " = " ++ 
       show(B[rowB, colB, k]) |
       rowB in 1..bRows, colB in 1..bCols, k in products] ++

    ["\nD" ++ show(100*rowC+10*colC+k) ++ " = " ++ 
       show(C[rowC, colC, k]) |
       rowC in 1..cRows, colC in 1..cCols, k in products];

下面是模型：2.mzn。

更新:这些时间是早期的，错误的，模型。该模型中的问题实例用3s的最小解(包括平坦)、5s的Opturion CPX求解器和6s的fzn2smt求解。该模型还可以通过贴标等进行进一步的调整。

与所提到的解决者的链接：

• 光学CPX：http://www.opturion.com/cpx.html
• MiniCSP：http://www.inra.fr/mia/T/katsirelos/minicsp.html
• fzn2smt：http://ima.udg.edu/Recerca/ESLIP/fzn2smt/index.html

还可以查看我的MiniZinc页面，以获得更长的FlatZinc解决程序列表：http://www.hakank.org/minizinc/。