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I have a complicated function $f$ which turns an input (let's say a single number, to simplify) into either 1, -1, -2 or -3, each value corresponding to different categories.

I can plot the categories as a function of the input, and that's a typical results:

Mathematica graphics

What I am interested in is $f^{-1}(\{1\})$. For example, for the plot above it would be something like $[-11.1,-10.9]\cup [-9.7,7.6]\cup[-7.4,-0.5]\cup\dots$, but much more accurately (typically, with five digits).

One possibility would be to increase PlotPoints and/or MaxRecursions, but this will also refine the parts in which I am not intersted, i.e. the transitions between -1 and -2, etc.


Here is a simplified example; f is a very simple function given in the end, but assume it is not known explicitly.

plot = Plot[f[x], {x, -5, 5}, Exclusions -> None, PlotPoints -> 10, AspectRatio -> 1/5];

gives

Mathematica graphics

The points can be extracted from the plot and the transitions obtained with BobHanlon's code:

transitions = Cases[plot, Line[pts_] :> pts, Infinity][[1]] //.
                  {s___, {x1_, y1_}, {x2_, y1_}, e___} :> {s, {x1, y1}, e}

(* {{-5., -2.}, {-3.13482, 1.}, {0.0167145, -1.}, {3.14997, -2.}} *)

Since I'm interested in the $x$ values of image $1$, I simply select:

pos = Position[transitions, _?(Last[#] == 1 &), {1}];
Take[transitions[[All, 1]], Flatten@{pos, pos + 1}]

(* {-3.13482, 0.0167145} *)

Which gives me an estimation of $f^{-1}(\{1\})$ of $[-3.13482, 0.0167145]$. In the present case, f is build up so that the exact transitions are known: the solution is -Pi,0. My goal is to get more accurate values, in an efficient manner.

One possibility is to increase PlotPoints and MaxRecusions: for example, same code with PlotPoints -> 2000, MaxRecursion -> 5:

 plot = Plot[f[x], {x, -5, 5}, Exclusions -> None, AspectRatio -> 1/5, 
   ImageSize -> 700, PlotPoints -> 2000, MaxRecursion -> 5];
transitions = 
 Cases[plot, Line[pts_] :> pts, 
    Infinity][[1]] //. {s___, {x1_, y1_}, {x2_, y1_}, 
    e___} :> {s, {x1, y1}, e}
pos = Position[transitions, _?(Last[#] == 1 &), {1}];
Take[transitions[[All, 1]], Flatten@{pos, pos + 1}]

(* {-3.14151, 0.0000912312} *)

It's much better... but I'm looking for a more efficient way to do this, in particular without refining the transitions which are not of interest. In other words, focus on making very fine transitions only when the functions reaches or leaves the value 1.

I am also interested in approaches which are not plot-based; I just want to get the intervals of image 1 as accurate as possible.


Function f used in the illustration code:

f[x_] = -UnitStep[x - Pi] + 3 UnitStep[x + Pi] - 2 UnitStep[x] - 2;
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  • $\begingroup$ For Mathematica v10.2 or later, convert your data into steps and use ListStepPlot $\endgroup$
    – Bob Hanlon
    Nov 11, 2016 at 17:51
  • $\begingroup$ @BobHanlon What data? It's a "continuum". I used UnitStep as a trivial example but in reality is a much more complicated function. $\endgroup$
    – anderstood
    Nov 11, 2016 at 18:09
  • $\begingroup$ Knowing how f is defined might help. Is it always a linear combination of unit-step functions? $\endgroup$
    – Michael E2
    Nov 11, 2016 at 23:10
  • $\begingroup$ @MichaelE2 No, actually it's really complicated: it takes a value $x$, makes a List from a list of interpolating functions evaluated at $x$, then applies another function to the list and check some criteria such as "does the function exceeds this value" or "is the derivative positive at x=3". $\endgroup$
    – anderstood
    Nov 11, 2016 at 23:13

2 Answers 2

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Plot has an suboption "ControlValue" which specify the max angle between successive points (in Radians)

f[x_] := -UnitStep[x - Pi] + 3 UnitStep[x + Pi] - 2 UnitStep[x] - 2;
    gr = Plot[f[x], {x, -5, 5},
      Method -> {"Refinement" -> {"ControlValue" -> 2 Pi/100}}, 
      Mesh -> All, MaxRecursion -> 15, PlotPoints -> 10, 
      MeshStyle -> PointSize[0.015]]

enter image description here

intervals = Cases[Normal[gr], Point[___], {1, Infinity}] //
      First /@ # & //
     Sort //
    Split[#, Abs[#1[[2]] - #2[[2]]] < 0.1 &] & //
   #[[All, {1, -1}, 1]] &;

intervals // Grid

enter image description here

error around -Pi :

-3.1415587451696543` - -3.141626562009932`

0.0000678168

There are a lot of related informations here and here

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  • $\begingroup$ Does not this refine everywhere and not just around y=1? Which is precisely what I'd like to avoid. $\endgroup$
    – anderstood
    Nov 14, 2016 at 20:56
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f[x_] = -UnitStep[x - 2] + 3 UnitStep[2 x + 4] - 2 UnitStep[x] - 2;

transitions = 
 Cases[Plot[f[x], {x, -5, 5}, Exclusions -> None], Line[pts_] :> pts, 
    Infinity][[1]] //.
  {s___, {x1_, y1_}, {x2_, y1_}, e___} :>
   {s, {x1, y1}, e}

(*  {{-5., -2.}, {-1.99783, 1.}, {0.000769648, -1.}, {2.00061, -2.}}  *)

ListStepPlot[transitions]

enter image description here

EDIT:

Transitions occur at discontinuities in function's derivative. Select transition pairs for which function has desired value between those transitions.

intervals[func_, val_: 1] :=
 Module[{x, transitions},
  transitions = 
   SortBy[x /. Solve[#, x][[1]] & /@ 
     Cases[func'[x], {Indeterminate, rel_} :> rel, Infinity], N[#] &];
  Select[
   Partition[transitions, 2, 1],
   func[Mean[#]] == val &]]

f[x_] = -UnitStep[x - 2] + 3 UnitStep[2 x + 4] - 2 UnitStep[x] - 2;

intervals[f]

(*  {{-2, 0}}  *)

f2[x_] = -UnitStep[x - Pi] + 3 UnitStep[x + Pi] - 2 UnitStep[x] - 2;

intervals[f2]

(*  {{-π, 0}}  *)
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  • $\begingroup$ I don't think that answers my question: it does not seem to refine around the transitions to 1. As I understand, what you added (essentially //. {s___, {x1_, y1_}, {x2_, y1_}, e___} :> {s, {x1, y1}, e}) extracts transitions (which I can do too, in a less elegant manner) but does not refine. Or am I missing something? $\endgroup$
    – anderstood
    Nov 11, 2016 at 18:54
  • $\begingroup$ From your example, the reverse image of $\{1\}$ is $[-1.99783,0.000769648]$. I would like to refine it, in the sense that it should be closer to $[-2,0]$ (without using the properties of the trivial function f). $\endgroup$
    – anderstood
    Nov 11, 2016 at 19:11
  • $\begingroup$ Unfortunately I don't know them, "-2" and "0" are trivial transitions here, but actually they are numerical values which I am looking for with as many digits as possible (5 should be enough). $\endgroup$
    – anderstood
    Nov 11, 2016 at 21:00
  • $\begingroup$ I just clarified (I hope!) the question. $\endgroup$
    – anderstood
    Nov 11, 2016 at 21:49

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