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I'm facing a problem with RandomReal when using ListPlot to plot some constrained points. I found that at every run of MA kernel the graph's points changes. In details, suppose for instance, functions:

k1[s_, d_] := 3 s + 5 d

k2[s_, d_] := 5 s - 7 d

I'm using the following to plot conditional k1 and k2 in (s,d) plan 

ClearAll[ps]

ps = Transpose[{RandomReal[{0.1, 2.}, 1000], RandomReal[{-1, 3}, 1000]}];

styleps = 
  Style[{##}, PointSize[.01], 
     Piecewise[{{Blue, 0 < k1[#, #2] <= 1 && 0 < k2[#, #2] <= 2}}, 
      White]] & @@@ ps;

ListPlot[styleps, DataRange -> {{-1, 1}, {-1, 1}}, Frame -> True, 
 GridLines -> {Table[i, {i, 0, 2, 0.1}], Table[i, {i, -1, 3, 0.2}]}, 
 ImageSize -> 500, Axes -> False, GridLinesStyle -> Lighter[Gray]]

Now for the first run this gives:

enter image description here

While if I make ClearAll[ps] and rerun or quit the kernel and started again, this gives totally different points, such as:

enter image description here

This is expected from RandomReal because it generates random points each time, but this is totally confusing in this case , because how one can determine all real points which satisfy the required condition if in each run only some points are appear ?

So any suggestions to improve this code to can plot whole points of of s versus d for which k1 and k2 conditions are satisfied for any run?

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  • 3
    $\begingroup$ Add a SeedRandom call with an appropriate seed to get identical output from RandomReal every time you run it. $\endgroup$
    – MarcoB
    Commented Jun 22, 2016 at 13:35
  • $\begingroup$ The problem is that even I fixed RandomReal for each run, it won't generate whole required points for this fix but it will be upon my choice such as setting length of genertaed points, etc. Can I avoid RandomReal from the beginning and generate points in more efficient way ? $\endgroup$
    – S.S.
    Commented Jun 22, 2016 at 13:52
  • $\begingroup$ @MarcoB have you an idea how to use SeedRandom in my example ? $\endgroup$
    – S.S.
    Commented Jun 22, 2016 at 13:58

2 Answers 2

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Here is an approach that generates many more points than you had, selects the ones for which your conditions are met, and plots them:

k1[s_, d_] := 3 s + 5 d
k2[s_, d_] := 5 s - 7 d
ps = RandomVariate[UniformDistribution[{{0.1, 2}, {-1, 3}}], 1000000];

valid = Select[ps, 0 < k1[Sequence @@ #] <= 1 && 0 < k2[Sequence @@ #] <= 2 &];

ListPlot[
 valid,
 Frame -> True,
 GridLines -> {Table[i, {i, 0, 2, 0.1}], Table[i, {i, -1, 3, 0.2}]},
 Axes -> False, GridLinesStyle -> Lighter[Gray]
]

Mathematica graphics

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  • $\begingroup$ That's interesting @MarcoB .. thanx $\endgroup$
    – S.S.
    Commented Jun 22, 2016 at 23:19
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Your approach is a dead end, you can't determine all points by picking them and checking conditions. Because there are infinitely many of them.

ImplicitRegion[
  0 < k1[x, y] <= 1 && 0 < k2[x, y] <= 2, 
  {x, y}
] // RegionPlot

enter image description here

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  • $\begingroup$ Hi @Kuba, it's pleasant to know my point is not reachable .. $\endgroup$
    – S.S.
    Commented Jun 22, 2016 at 13:56
  • $\begingroup$ @ Kuba I still have a question, how can we imply set length as you mentioned condition[x_] := .1 < x < .5; Module[{i = 0}, First@Last@ Reap@While[i < 5, If[condition[#], i++; Sow[#]] &@RandomReal[]]] to my example ? Also do you think SeedRandom mentioned above can do the same act with RandomReal ? sorry I wish if I'm an expert in MA ..` $\endgroup$
    – S.S.
    Commented Jun 22, 2016 at 14:07
  • $\begingroup$ @S.S. I don't understand, the cardinality of set of points fulfilling your conditions is the cardinality of the continuum. The problem isn't about MMA skills I think. p.s. SeedRandom and RandomReal are related but they do different things, press F1 and read more. $\endgroup$
    – Kuba
    Commented Jun 22, 2016 at 14:24
  • $\begingroup$ @ Kuba . I mean I try to use that Module in plotting for my example, if you have an idea.. $\endgroup$
    – S.S.
    Commented Jun 22, 2016 at 16:01
  • $\begingroup$ @S.S. replace i with number of points you want to enerate, RandomReal[1, {2}] or something. I really don't know what is the goal, you can't get that way all the points. $\endgroup$
    – Kuba
    Commented Jun 22, 2016 at 17:01

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