在 Mathematica 中以编程方式创建多元函数 [英] Programmatically creating multivariate functions in Mathematica

问题描述

这是对之前问题的讨论的分离.

This is a split from discussion on earlier question.

假设我需要定义一个函数 f 来检查给定的图形标签是否正确着色.换句话说，我们为每个节点分配了一个整数，并且没有两个相邻的节点得到相同的答案.例如，对于 {"Path",3}，f[{1,2,3}] 返回 True，f[{1,1,2}] 返回 False.我将如何为任意图形创建这样的函数?

Suppose I need to define a function f which checks if given labeling of a graph is a proper coloring. In other words, we have an integer assigned to every node and no two adjacent nodes get the same answer. For instance, for {"Path",3}, f[{1,2,3}] returns True and f[{1,1,2}] returns False. How would I go about creating such a function for arbitrary graph?

以下内容基本上满足了我的需要，但会生成零件警告.

The following does essentially what I need, but generates Part warnings.

g[edges_] := Function @@ {{x}, And @@ (x[[First[#]]] != x[[Last[#]]] & /@ edges)}
f = g[GraphData[{"Path", 3}, "EdgeIndices"]];
f[{1, 2, 1}]==False

这是我经常遇到的一个玩具实例问题——我需要以编程方式创建一个多元函数 f，并最终得到 1) 部分警告 2) 推迟对 g 的评估直到对 f 的评估

This is a toy instance problem I regularly come across -- I need to programmatically create a multivariate function f, and end up with either 1) part warning 2) deferring evaluation of g until evaluation of f

推荐答案

这里有一些东西.当没有其他工作时，Hold 和规则通常可以完成工作.我不确定它会产生正确的结果 w.r.t.您的图形着色问题，但希望为您提供一个起点.我最终使用 Slot 而不是命名变量，因为存在一些范围问题(也出现在我之前的建议中，x$ vs. x)当我使用了一个我没有花时间尝试解决的命名变量时.

Here's something. When nothing else is working, Hold and rules can usually get the job done. I'm not sure it produces the correct results w.r.t. your graph-coloring question but hopefully gives you a starting place. I ended up using Slot instead of named variable because there were some scoping issues (also present in my previous suggestion, x$ vs. x) when I used a named variable that I didn't spend the time trying to work around.

g[edges_] := 
 With[{ors = (Hold @@ edges) /. {a_, b_} :> #[[a]] == #[[b]]},
  Function[!ors] /. Hold -> Or
  ]

In[90]:= f = g[GraphData[{"Path", 3}, "EdgeIndices"]]
Out[90]= !(#1[[1]] == #1[[2]] || #1[[2]] == #1[[3]]) &

In[91]:= f[{1, 2, 3}]
Out[91]= True

In[92]:= f[{1, 1, 2}]
Out[92]= False

我觉得它缺乏典型的 Mathematica 优雅，但它确实有效.如果我受到更美的启发，我会更新.

I feel like it lacks typical Mathematica elegance, but it works. I'll update if I'm inspired with something more beautiful.

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