Wolfram 语言™

用积和式求解组合问题

积和式 (permanent) 与行列式 (determinant) 类似，除了所有项都为正以外.

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Permanent[\!\(\* TagBox[ RowBox[{"(", "", GridBox[{ { SubscriptBox["a", RowBox[{"1", ",", "1"}]], SubscriptBox["a", RowBox[{"1", ",", "2"}]]}, { SubscriptBox["a", RowBox[{"2", ",", "1"}]], SubscriptBox["a", RowBox[{"2", ",", "2"}]]} }, GridBoxAlignment->{ "Columns" -> {{Left}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}, "Items" -> {}, "ItemsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}, "Items" -> {}, "ItemsIndexed" -> {}}], "", ")"}], Function[BoxForm`e$, MatrixForm[BoxForm`e$]]]\)]

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Permanent[\!\(\* TagBox[ RowBox[{"(", "", GridBox[{ { SubscriptBox["a", RowBox[{"1", ",", "1"}]], SubscriptBox["a", RowBox[{"1", ",", "2"}]], SubscriptBox["a", RowBox[{"1", ",", "3"}]]}, { SubscriptBox["a", RowBox[{"2", ",", "1"}]], SubscriptBox["a", RowBox[{"2", ",", "2"}]], SubscriptBox["a", RowBox[{"2", ",", "3"}]]}, { SubscriptBox["a", RowBox[{"3", ",", "1"}]], SubscriptBox["a", RowBox[{"3", ",", "2"}]], SubscriptBox["a", RowBox[{"3", ",", "3"}]]} }, GridBoxAlignment->{ "Columns" -> {{Left}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> {}, "Items" -> {}, "ItemsIndexed" -> {}}, GridBoxSpacings->{"Columns" -> { Offset[0.27999999999999997`], { Offset[0.7]}, Offset[0.27999999999999997`]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.4]}, Offset[0.2]}, "RowsIndexed" -> {}, "Items" -> {}, "ItemsIndexed" -> {}}], "", ")"}], Function[BoxForm`e$, MatrixForm[BoxForm`e$]]]\)]

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因此，将 Permanent 应用于所有项都等于 1 的矩阵虽然有趣但并不是计算阶乘函数的有效方法.

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Table[Permanent[ConstantArray[1, {n, n}]], {n, 10}]

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积和式可用于求解以下更有趣的组合问题：给定 n 个集合，每个都包含一个 子集，有多少种方法可以从每个子集中选出一个不同元素？首先，创建矩阵 m，其中当子集 i 含有 j 时，(i, j) 位置含有一个 1，其他时候则为 0.

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sets = {{3, 5, 6, 7}, {3, 7}, {1, 2, 4, 5, 7}, {3}, {1, 3, 6}, {1, 5, 7}, {1, 2, 3, 6}}

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m = Table[If[MemberQ[sets[[i]], j], 1, 0] , {i, 7}, {j, 7}]; m // MatrixForm

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m 的积和式的便是该问题的解.

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Permanent[m]

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通过明确构建所有元组确认答案.

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Select[Tuples[sets], DuplicateFreeQ]

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