Detection and styling of multiple functions in Mathematica's Plot

Question

This question started me thinking about how Mathematica detects multiple functions being plotted. I find that I really do not understand the process.

Consider:

Answer 1

Well, it knows that there three arguments just so:

In[13]:= Function[x, Length[Unevaluated[x]], HoldAll][{1, 
  Sequence[2, 3], 4}]

Out[13]= 3

If x is allowed to evaluate, then

In[14]:= Function[x, Length[x], HoldAll][{1, Sequence[2, 3], 4}]

Out[14]= 4

EDIT: One sees it better with:

In[15]:= Hold[{1, Sequence[2, 3], 4}]

Out[15]= Hold[{1, Sequence[2, 3], 4}]

in other words, flattening of Sequence requires evaluator.

EDIT 2: I clearly missed the real question posed and will try to answer it now.

Once Plot determines the number of argument it builds {{ style1, Line ..}, {style2, Line..}, ... }. In the case of {1,Sequence[2,3],4} we get the following structure:

In[23]:= Cases[
  Plot[{1, Sequence[2, 3], 4}, {x, 0, 1}, 
   PlotRange -> {0, 5}], {_Hue, __Line}, 
  Infinity] /. {x_Line :> Line, _Hue -> Hue}

Out[23]= {{Hue, Line}, {Hue, Line, Line}, {Hue, Line}}

When plotting {1,{2,3},4} we get a different structure:

In[24]:= Cases[
  Plot[{1, List[2, 3], 4}, {x, 0, 1}, 
   PlotRange -> {0, 5}], {_Hue, __Line}, 
  Infinity] /. {x_Line :> Line, _Hue -> Hue}

Out[24]= {{Hue, Line}, {Hue, Line}, {Hue, Line}, {Hue, Line}}

because lists would be flattened, just not using the evaluator. So as you see the tagging in the same color occurs because Sequence[2,3] is treated as a black-box function which returns a list of two elements:

In[25]:= g[x_?NumberQ] := {2, 3}

In[26]:= Cases[
  Plot[{1, g[x], 4}, {x, 0, 1}, PlotRange -> {0, 5}], {_Hue, __Line}, 
  Infinity] /. {x_Line :> Line, _Hue -> Hue}

Out[26]= {{Hue, Line}, {Hue, Line, Line}, {Hue, Line}}

I was trying to build a top-level implementation which would build such a structure, but one has to fight the evaluator. For example:

In[28]:= Thread /@ Function[x,
   Thread[{Hold @@ {Range[Length[Unevaluated[x]]]}, Hold[x]}, Hold]
   , HoldAll][{1, Sequence[2, 3], 4}]

Out[28]= Hold[Thread[{{1, 2, 3}, {1, Sequence[2, 3], 4}}]]

Now we have to evaluate the Thread without evaluating its arguments, which would give {{1, 1}, {2, Sequence[2,3]}, {3, 4}}, where the first element of the list is a tag, and the subsequent once are functions to be sampled.

Hope this helps.