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Plotting maxima within a simplex


Plotting multivalued function using ParametricPlotColouring of multiple data sets in DiscretePlotHow to draw a tropical surface?Storing Variables in “Loops” and Point PlottingPlotting a manifold within an existing surfaceIncreasing the thickness of minimum of two curves while keeping their coloring the samePlotting a 2D plot along the maxima of a 3D plotPlanetary motion of a curve within a regionPlotting the plastic zone — Tresca criterionPlotting Complex Numbers as “Arrows” on the Complex Plane






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


I am trying to draw a particular maximum plot within a simplex. I have functions based on two variables:



Write p,q for the two variables where p,q geq 0 and p + q leq 1. This defines the simplex I am interested in. (Think of, say, p on the x-axis and q on the y-axis.)



The functions are then of the following form:



f(p,q) = 22.5p + 10(1-p)



g(p,q) = 40(1-p-q) + 15(p+q)



h(p,q) = 10(1-q) + 22.5q.



(I am giving examples of functions---there are several I want to look at.)



I am interested in which function is "highest" for any given (p,q). So, for any given (p,q) in the simplex, I want to look at the max f, g, h. (The functions are such that almost always the maximum will be unique.) I then want it plot the point (p,q) as, say, red if f is the maximum function, blue if g is the maximum function, and green if h is the maximum function.



Does anyone have any idea how this can be done within mathematica?










share|improve this question











$endgroup$


















    2












    $begingroup$


    I am trying to draw a particular maximum plot within a simplex. I have functions based on two variables:



    Write p,q for the two variables where p,q geq 0 and p + q leq 1. This defines the simplex I am interested in. (Think of, say, p on the x-axis and q on the y-axis.)



    The functions are then of the following form:



    f(p,q) = 22.5p + 10(1-p)



    g(p,q) = 40(1-p-q) + 15(p+q)



    h(p,q) = 10(1-q) + 22.5q.



    (I am giving examples of functions---there are several I want to look at.)



    I am interested in which function is "highest" for any given (p,q). So, for any given (p,q) in the simplex, I want to look at the max f, g, h. (The functions are such that almost always the maximum will be unique.) I then want it plot the point (p,q) as, say, red if f is the maximum function, blue if g is the maximum function, and green if h is the maximum function.



    Does anyone have any idea how this can be done within mathematica?










    share|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      I am trying to draw a particular maximum plot within a simplex. I have functions based on two variables:



      Write p,q for the two variables where p,q geq 0 and p + q leq 1. This defines the simplex I am interested in. (Think of, say, p on the x-axis and q on the y-axis.)



      The functions are then of the following form:



      f(p,q) = 22.5p + 10(1-p)



      g(p,q) = 40(1-p-q) + 15(p+q)



      h(p,q) = 10(1-q) + 22.5q.



      (I am giving examples of functions---there are several I want to look at.)



      I am interested in which function is "highest" for any given (p,q). So, for any given (p,q) in the simplex, I want to look at the max f, g, h. (The functions are such that almost always the maximum will be unique.) I then want it plot the point (p,q) as, say, red if f is the maximum function, blue if g is the maximum function, and green if h is the maximum function.



      Does anyone have any idea how this can be done within mathematica?










      share|improve this question











      $endgroup$




      I am trying to draw a particular maximum plot within a simplex. I have functions based on two variables:



      Write p,q for the two variables where p,q geq 0 and p + q leq 1. This defines the simplex I am interested in. (Think of, say, p on the x-axis and q on the y-axis.)



      The functions are then of the following form:



      f(p,q) = 22.5p + 10(1-p)



      g(p,q) = 40(1-p-q) + 15(p+q)



      h(p,q) = 10(1-q) + 22.5q.



      (I am giving examples of functions---there are several I want to look at.)



      I am interested in which function is "highest" for any given (p,q). So, for any given (p,q) in the simplex, I want to look at the max f, g, h. (The functions are such that almost always the maximum will be unique.) I then want it plot the point (p,q) as, say, red if f is the maximum function, blue if g is the maximum function, and green if h is the maximum function.



      Does anyone have any idea how this can be done within mathematica?







      plotting color simplex






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 8 hours ago









      user64494

      4,1472 gold badges14 silver badges23 bronze badges




      4,1472 gold badges14 silver badges23 bronze badges










      asked 9 hours ago









      AmandaAmanda

      756 bronze badges




      756 bronze badges




















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          ClearAll[f, g, h]

          f[p_, q_] := 22.5 p + 10 (1 - p)
          g[p_, q_] := 40 (1 - p - q) + 15 (p + q)
          h[p_, q_] := 10 (1 - q) + 22.5 q

          max[p_, q_] := Max[f[p, q], g[p, q], h[p, q]]


          Plot3D[max[p, q], p, q ∈ SSSTriangle[1, 1, 1],
          Filling -> Bottom, Exclusions -> None]


          enter image description here



          rf, rg, rh = Quiet @
          Reduce[#[p, q] >= max[p, q] , p, q ∈ SSSTriangle[1, 1, 1]] & /@ f, g, h;

          RegionPlot[rf, rg, rh, p, 0, 1, q, 0, 1, PlotStyle -> Red, Blue, Green]


          enter image description here






          share|improve this answer











          $endgroup$




















            3












            $begingroup$

            Piggybacking on @kglr's answer, but using a right triangle rather than equilateral, so we can see p and q on the axes and to highlight the symmetry between f and h.



            How about:



            Plot3D[max[p, q], p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
            ColorFunction ->
            (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2], Green, #3 == h[#1, #2], Blue] &),
            ColorFunctionScaling -> False, PlotPoints -> 100]


            Mathematica graphics



            A dirty trick to get a 2D plot is to move the ViewPoint to 0, 0, [Infinity]:



            Plot3D[max[p, q], p, q [Element] 
            Triangle[0, 0, 1, 0, 0, 1],
            ColorFunction -> (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2],
            Green, #3 == h[#1, #2], Blue] &), ColorFunctionScaling -> False,
            PlotPoints -> 100, ViewPoint -> 0, 0, [Infinity],
            Axes -> True, True, False, Mesh -> None]


            Mathematica graphics



            Here's another possible 2D solution:



            DensityPlot[
            Which[
            max[p, q] == f[p, q], 1,
            max[p, q] == g[p, q], 2,
            max[p, q] == h[p, q], 3
            ],
            p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
            PlotPoints -> 100,
            ColorFunction -> (Which[#1 == 1, Red, #1 == 2, Green, #1 == 3, Blue] &),
            ColorFunctionScaling -> False
            ]


            (similar output)



            Of course hard coding the number of functions is ugly; it'd be nice to have this accept an arbitrary number of functions to compare!






            share|improve this answer











            $endgroup$












            • $begingroup$
              These pictures are beautiful. But I was rather hoping for a 2-Dimensional version of this---with say p on the x-axis and q on the y-axis. Is there any way to project the 3D figure onto 2D? Or draw it originally as 2D?
              $endgroup$
              – Amanda
              6 hours ago










            • $begingroup$
              @kglr any ideas on making your solution 2D?
              $endgroup$
              – Chris K
              5 hours ago










            • $begingroup$
              @ChrisK, found something that works-- could be cleaner.
              $endgroup$
              – kglr
              3 hours ago










            • $begingroup$
              @kglr Yeah, too bad RegionPlot leaves a big hole in the middle by default.
              $endgroup$
              – Chris K
              3 hours ago













            Your Answer








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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            ClearAll[f, g, h]

            f[p_, q_] := 22.5 p + 10 (1 - p)
            g[p_, q_] := 40 (1 - p - q) + 15 (p + q)
            h[p_, q_] := 10 (1 - q) + 22.5 q

            max[p_, q_] := Max[f[p, q], g[p, q], h[p, q]]


            Plot3D[max[p, q], p, q ∈ SSSTriangle[1, 1, 1],
            Filling -> Bottom, Exclusions -> None]


            enter image description here



            rf, rg, rh = Quiet @
            Reduce[#[p, q] >= max[p, q] , p, q ∈ SSSTriangle[1, 1, 1]] & /@ f, g, h;

            RegionPlot[rf, rg, rh, p, 0, 1, q, 0, 1, PlotStyle -> Red, Blue, Green]


            enter image description here






            share|improve this answer











            $endgroup$

















              3












              $begingroup$

              ClearAll[f, g, h]

              f[p_, q_] := 22.5 p + 10 (1 - p)
              g[p_, q_] := 40 (1 - p - q) + 15 (p + q)
              h[p_, q_] := 10 (1 - q) + 22.5 q

              max[p_, q_] := Max[f[p, q], g[p, q], h[p, q]]


              Plot3D[max[p, q], p, q ∈ SSSTriangle[1, 1, 1],
              Filling -> Bottom, Exclusions -> None]


              enter image description here



              rf, rg, rh = Quiet @
              Reduce[#[p, q] >= max[p, q] , p, q ∈ SSSTriangle[1, 1, 1]] & /@ f, g, h;

              RegionPlot[rf, rg, rh, p, 0, 1, q, 0, 1, PlotStyle -> Red, Blue, Green]


              enter image description here






              share|improve this answer











              $endgroup$















                3












                3








                3





                $begingroup$

                ClearAll[f, g, h]

                f[p_, q_] := 22.5 p + 10 (1 - p)
                g[p_, q_] := 40 (1 - p - q) + 15 (p + q)
                h[p_, q_] := 10 (1 - q) + 22.5 q

                max[p_, q_] := Max[f[p, q], g[p, q], h[p, q]]


                Plot3D[max[p, q], p, q ∈ SSSTriangle[1, 1, 1],
                Filling -> Bottom, Exclusions -> None]


                enter image description here



                rf, rg, rh = Quiet @
                Reduce[#[p, q] >= max[p, q] , p, q ∈ SSSTriangle[1, 1, 1]] & /@ f, g, h;

                RegionPlot[rf, rg, rh, p, 0, 1, q, 0, 1, PlotStyle -> Red, Blue, Green]


                enter image description here






                share|improve this answer











                $endgroup$



                ClearAll[f, g, h]

                f[p_, q_] := 22.5 p + 10 (1 - p)
                g[p_, q_] := 40 (1 - p - q) + 15 (p + q)
                h[p_, q_] := 10 (1 - q) + 22.5 q

                max[p_, q_] := Max[f[p, q], g[p, q], h[p, q]]


                Plot3D[max[p, q], p, q ∈ SSSTriangle[1, 1, 1],
                Filling -> Bottom, Exclusions -> None]


                enter image description here



                rf, rg, rh = Quiet @
                Reduce[#[p, q] >= max[p, q] , p, q ∈ SSSTriangle[1, 1, 1]] & /@ f, g, h;

                RegionPlot[rf, rg, rh, p, 0, 1, q, 0, 1, PlotStyle -> Red, Blue, Green]


                enter image description here







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 hours ago

























                answered 7 hours ago









                kglrkglr

                206k10 gold badges235 silver badges467 bronze badges




                206k10 gold badges235 silver badges467 bronze badges























                    3












                    $begingroup$

                    Piggybacking on @kglr's answer, but using a right triangle rather than equilateral, so we can see p and q on the axes and to highlight the symmetry between f and h.



                    How about:



                    Plot3D[max[p, q], p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
                    ColorFunction ->
                    (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2], Green, #3 == h[#1, #2], Blue] &),
                    ColorFunctionScaling -> False, PlotPoints -> 100]


                    Mathematica graphics



                    A dirty trick to get a 2D plot is to move the ViewPoint to 0, 0, [Infinity]:



                    Plot3D[max[p, q], p, q [Element] 
                    Triangle[0, 0, 1, 0, 0, 1],
                    ColorFunction -> (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2],
                    Green, #3 == h[#1, #2], Blue] &), ColorFunctionScaling -> False,
                    PlotPoints -> 100, ViewPoint -> 0, 0, [Infinity],
                    Axes -> True, True, False, Mesh -> None]


                    Mathematica graphics



                    Here's another possible 2D solution:



                    DensityPlot[
                    Which[
                    max[p, q] == f[p, q], 1,
                    max[p, q] == g[p, q], 2,
                    max[p, q] == h[p, q], 3
                    ],
                    p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
                    PlotPoints -> 100,
                    ColorFunction -> (Which[#1 == 1, Red, #1 == 2, Green, #1 == 3, Blue] &),
                    ColorFunctionScaling -> False
                    ]


                    (similar output)



                    Of course hard coding the number of functions is ugly; it'd be nice to have this accept an arbitrary number of functions to compare!






                    share|improve this answer











                    $endgroup$












                    • $begingroup$
                      These pictures are beautiful. But I was rather hoping for a 2-Dimensional version of this---with say p on the x-axis and q on the y-axis. Is there any way to project the 3D figure onto 2D? Or draw it originally as 2D?
                      $endgroup$
                      – Amanda
                      6 hours ago










                    • $begingroup$
                      @kglr any ideas on making your solution 2D?
                      $endgroup$
                      – Chris K
                      5 hours ago










                    • $begingroup$
                      @ChrisK, found something that works-- could be cleaner.
                      $endgroup$
                      – kglr
                      3 hours ago










                    • $begingroup$
                      @kglr Yeah, too bad RegionPlot leaves a big hole in the middle by default.
                      $endgroup$
                      – Chris K
                      3 hours ago















                    3












                    $begingroup$

                    Piggybacking on @kglr's answer, but using a right triangle rather than equilateral, so we can see p and q on the axes and to highlight the symmetry between f and h.



                    How about:



                    Plot3D[max[p, q], p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
                    ColorFunction ->
                    (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2], Green, #3 == h[#1, #2], Blue] &),
                    ColorFunctionScaling -> False, PlotPoints -> 100]


                    Mathematica graphics



                    A dirty trick to get a 2D plot is to move the ViewPoint to 0, 0, [Infinity]:



                    Plot3D[max[p, q], p, q [Element] 
                    Triangle[0, 0, 1, 0, 0, 1],
                    ColorFunction -> (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2],
                    Green, #3 == h[#1, #2], Blue] &), ColorFunctionScaling -> False,
                    PlotPoints -> 100, ViewPoint -> 0, 0, [Infinity],
                    Axes -> True, True, False, Mesh -> None]


                    Mathematica graphics



                    Here's another possible 2D solution:



                    DensityPlot[
                    Which[
                    max[p, q] == f[p, q], 1,
                    max[p, q] == g[p, q], 2,
                    max[p, q] == h[p, q], 3
                    ],
                    p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
                    PlotPoints -> 100,
                    ColorFunction -> (Which[#1 == 1, Red, #1 == 2, Green, #1 == 3, Blue] &),
                    ColorFunctionScaling -> False
                    ]


                    (similar output)



                    Of course hard coding the number of functions is ugly; it'd be nice to have this accept an arbitrary number of functions to compare!






                    share|improve this answer











                    $endgroup$












                    • $begingroup$
                      These pictures are beautiful. But I was rather hoping for a 2-Dimensional version of this---with say p on the x-axis and q on the y-axis. Is there any way to project the 3D figure onto 2D? Or draw it originally as 2D?
                      $endgroup$
                      – Amanda
                      6 hours ago










                    • $begingroup$
                      @kglr any ideas on making your solution 2D?
                      $endgroup$
                      – Chris K
                      5 hours ago










                    • $begingroup$
                      @ChrisK, found something that works-- could be cleaner.
                      $endgroup$
                      – kglr
                      3 hours ago










                    • $begingroup$
                      @kglr Yeah, too bad RegionPlot leaves a big hole in the middle by default.
                      $endgroup$
                      – Chris K
                      3 hours ago













                    3












                    3








                    3





                    $begingroup$

                    Piggybacking on @kglr's answer, but using a right triangle rather than equilateral, so we can see p and q on the axes and to highlight the symmetry between f and h.



                    How about:



                    Plot3D[max[p, q], p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
                    ColorFunction ->
                    (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2], Green, #3 == h[#1, #2], Blue] &),
                    ColorFunctionScaling -> False, PlotPoints -> 100]


                    Mathematica graphics



                    A dirty trick to get a 2D plot is to move the ViewPoint to 0, 0, [Infinity]:



                    Plot3D[max[p, q], p, q [Element] 
                    Triangle[0, 0, 1, 0, 0, 1],
                    ColorFunction -> (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2],
                    Green, #3 == h[#1, #2], Blue] &), ColorFunctionScaling -> False,
                    PlotPoints -> 100, ViewPoint -> 0, 0, [Infinity],
                    Axes -> True, True, False, Mesh -> None]


                    Mathematica graphics



                    Here's another possible 2D solution:



                    DensityPlot[
                    Which[
                    max[p, q] == f[p, q], 1,
                    max[p, q] == g[p, q], 2,
                    max[p, q] == h[p, q], 3
                    ],
                    p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
                    PlotPoints -> 100,
                    ColorFunction -> (Which[#1 == 1, Red, #1 == 2, Green, #1 == 3, Blue] &),
                    ColorFunctionScaling -> False
                    ]


                    (similar output)



                    Of course hard coding the number of functions is ugly; it'd be nice to have this accept an arbitrary number of functions to compare!






                    share|improve this answer











                    $endgroup$



                    Piggybacking on @kglr's answer, but using a right triangle rather than equilateral, so we can see p and q on the axes and to highlight the symmetry between f and h.



                    How about:



                    Plot3D[max[p, q], p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
                    ColorFunction ->
                    (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2], Green, #3 == h[#1, #2], Blue] &),
                    ColorFunctionScaling -> False, PlotPoints -> 100]


                    Mathematica graphics



                    A dirty trick to get a 2D plot is to move the ViewPoint to 0, 0, [Infinity]:



                    Plot3D[max[p, q], p, q [Element] 
                    Triangle[0, 0, 1, 0, 0, 1],
                    ColorFunction -> (Which[#3 == f[#1, #2], Red, #3 == g[#1, #2],
                    Green, #3 == h[#1, #2], Blue] &), ColorFunctionScaling -> False,
                    PlotPoints -> 100, ViewPoint -> 0, 0, [Infinity],
                    Axes -> True, True, False, Mesh -> None]


                    Mathematica graphics



                    Here's another possible 2D solution:



                    DensityPlot[
                    Which[
                    max[p, q] == f[p, q], 1,
                    max[p, q] == g[p, q], 2,
                    max[p, q] == h[p, q], 3
                    ],
                    p, q [Element] Triangle[0, 0, 1, 0, 0, 1],
                    PlotPoints -> 100,
                    ColorFunction -> (Which[#1 == 1, Red, #1 == 2, Green, #1 == 3, Blue] &),
                    ColorFunctionScaling -> False
                    ]


                    (similar output)



                    Of course hard coding the number of functions is ugly; it'd be nice to have this accept an arbitrary number of functions to compare!







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 2 hours ago

























                    answered 6 hours ago









                    Chris KChris K

                    9,0002 gold badges23 silver badges49 bronze badges




                    9,0002 gold badges23 silver badges49 bronze badges











                    • $begingroup$
                      These pictures are beautiful. But I was rather hoping for a 2-Dimensional version of this---with say p on the x-axis and q on the y-axis. Is there any way to project the 3D figure onto 2D? Or draw it originally as 2D?
                      $endgroup$
                      – Amanda
                      6 hours ago










                    • $begingroup$
                      @kglr any ideas on making your solution 2D?
                      $endgroup$
                      – Chris K
                      5 hours ago










                    • $begingroup$
                      @ChrisK, found something that works-- could be cleaner.
                      $endgroup$
                      – kglr
                      3 hours ago










                    • $begingroup$
                      @kglr Yeah, too bad RegionPlot leaves a big hole in the middle by default.
                      $endgroup$
                      – Chris K
                      3 hours ago
















                    • $begingroup$
                      These pictures are beautiful. But I was rather hoping for a 2-Dimensional version of this---with say p on the x-axis and q on the y-axis. Is there any way to project the 3D figure onto 2D? Or draw it originally as 2D?
                      $endgroup$
                      – Amanda
                      6 hours ago










                    • $begingroup$
                      @kglr any ideas on making your solution 2D?
                      $endgroup$
                      – Chris K
                      5 hours ago










                    • $begingroup$
                      @ChrisK, found something that works-- could be cleaner.
                      $endgroup$
                      – kglr
                      3 hours ago










                    • $begingroup$
                      @kglr Yeah, too bad RegionPlot leaves a big hole in the middle by default.
                      $endgroup$
                      – Chris K
                      3 hours ago















                    $begingroup$
                    These pictures are beautiful. But I was rather hoping for a 2-Dimensional version of this---with say p on the x-axis and q on the y-axis. Is there any way to project the 3D figure onto 2D? Or draw it originally as 2D?
                    $endgroup$
                    – Amanda
                    6 hours ago




                    $begingroup$
                    These pictures are beautiful. But I was rather hoping for a 2-Dimensional version of this---with say p on the x-axis and q on the y-axis. Is there any way to project the 3D figure onto 2D? Or draw it originally as 2D?
                    $endgroup$
                    – Amanda
                    6 hours ago












                    $begingroup$
                    @kglr any ideas on making your solution 2D?
                    $endgroup$
                    – Chris K
                    5 hours ago




                    $begingroup$
                    @kglr any ideas on making your solution 2D?
                    $endgroup$
                    – Chris K
                    5 hours ago












                    $begingroup$
                    @ChrisK, found something that works-- could be cleaner.
                    $endgroup$
                    – kglr
                    3 hours ago




                    $begingroup$
                    @ChrisK, found something that works-- could be cleaner.
                    $endgroup$
                    – kglr
                    3 hours ago












                    $begingroup$
                    @kglr Yeah, too bad RegionPlot leaves a big hole in the middle by default.
                    $endgroup$
                    – Chris K
                    3 hours ago




                    $begingroup$
                    @kglr Yeah, too bad RegionPlot leaves a big hole in the middle by default.
                    $endgroup$
                    – Chris K
                    3 hours ago

















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