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Construct, in some manner, a four-dimensional "RegionPlot"

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Construct, in some manner, a four-dimensional “RegionPlot”


Labeling distinct objects produced by Show[RegionPlot3D's]How to create this four-dimensional cube animation?Four-way logarithmic plotRegionPlot not plotting some regionsList of Inequalities in RegionPlot with different colorsHow can I create a four dimensional plot (3D space + color) of the data provided?Creating a graphic with four rectangles and four pointsHow to visualize four-dimensional data?Plot four dimensional data consisting of discrete and continuous variables?RegionPlot misses a corner when plotting a two-dimensional regionHow to use ColorSlider for some objects?






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








4












$begingroup$


Let me abuse some Mathematica notation and formulate the following "command":



Show[RegionPlot4D[(Q1 - Q4)^2 < 16 Q3^2 && 
Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0, Q1, 0,
6/61, Q2, 0, 2/9, Q3, 0, 1/32, Q4, 0, 1/6],
RegionPlot4D[
Q4 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 4 (Q2 + Q3) + 3 Q4 < 1 &&
4 Q2 + 9 Q4 < Q1, Q1, 0, 6/61, Q2, 0, 2/9, Q3, 0, 1/32, Q4,
0, 1/6],
RegionPlot4D[
Q4 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 4 (Q2 + Q3) + 3 Q4 < 1 &&
2 (Q2 + Q3) + 3 Q4 < Q1, Q1, 0, 6/61, Q2, 0, 2/9, Q3, 0,
1/32, Q4, 0, 1/6]]


(Of course, there is a RegionPlot3D command, but no RegionPlot4D one.)



Can this be processed/interpreted in some manner? (use of coloring,...)



Also, these three "RegionPlot"s could be considered individually (challenging enough).



These pertain to certain quantum-information-theoretic problems concerned with probabilities of (bound) entanglement.



The problem as put is very much a direct 4D analogue of the 3D problem



Labeling distinct objects produced by Show[RegionPlot3D's]



that kglr answered. So, perhaps I should just try fixing (in various ways) one of the four coordinates and approaching the problem in the very same manner as there. (In fact, the constraints are set up in the same order both times, with the first one each times being the "PPT" one. Incidentally, the "PPT" body should be convex, but not the other two.)










share|improve this question











$endgroup$


















    4












    $begingroup$


    Let me abuse some Mathematica notation and formulate the following "command":



    Show[RegionPlot4D[(Q1 - Q4)^2 < 16 Q3^2 && 
    Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
    4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0, Q1, 0,
    6/61, Q2, 0, 2/9, Q3, 0, 1/32, Q4, 0, 1/6],
    RegionPlot4D[
    Q4 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 4 (Q2 + Q3) + 3 Q4 < 1 &&
    4 Q2 + 9 Q4 < Q1, Q1, 0, 6/61, Q2, 0, 2/9, Q3, 0, 1/32, Q4,
    0, 1/6],
    RegionPlot4D[
    Q4 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 4 (Q2 + Q3) + 3 Q4 < 1 &&
    2 (Q2 + Q3) + 3 Q4 < Q1, Q1, 0, 6/61, Q2, 0, 2/9, Q3, 0,
    1/32, Q4, 0, 1/6]]


    (Of course, there is a RegionPlot3D command, but no RegionPlot4D one.)



    Can this be processed/interpreted in some manner? (use of coloring,...)



    Also, these three "RegionPlot"s could be considered individually (challenging enough).



    These pertain to certain quantum-information-theoretic problems concerned with probabilities of (bound) entanglement.



    The problem as put is very much a direct 4D analogue of the 3D problem



    Labeling distinct objects produced by Show[RegionPlot3D's]



    that kglr answered. So, perhaps I should just try fixing (in various ways) one of the four coordinates and approaching the problem in the very same manner as there. (In fact, the constraints are set up in the same order both times, with the first one each times being the "PPT" one. Incidentally, the "PPT" body should be convex, but not the other two.)










    share|improve this question











    $endgroup$














      4












      4








      4





      $begingroup$


      Let me abuse some Mathematica notation and formulate the following "command":



      Show[RegionPlot4D[(Q1 - Q4)^2 < 16 Q3^2 && 
      Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
      4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0, Q1, 0,
      6/61, Q2, 0, 2/9, Q3, 0, 1/32, Q4, 0, 1/6],
      RegionPlot4D[
      Q4 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 4 (Q2 + Q3) + 3 Q4 < 1 &&
      4 Q2 + 9 Q4 < Q1, Q1, 0, 6/61, Q2, 0, 2/9, Q3, 0, 1/32, Q4,
      0, 1/6],
      RegionPlot4D[
      Q4 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 4 (Q2 + Q3) + 3 Q4 < 1 &&
      2 (Q2 + Q3) + 3 Q4 < Q1, Q1, 0, 6/61, Q2, 0, 2/9, Q3, 0,
      1/32, Q4, 0, 1/6]]


      (Of course, there is a RegionPlot3D command, but no RegionPlot4D one.)



      Can this be processed/interpreted in some manner? (use of coloring,...)



      Also, these three "RegionPlot"s could be considered individually (challenging enough).



      These pertain to certain quantum-information-theoretic problems concerned with probabilities of (bound) entanglement.



      The problem as put is very much a direct 4D analogue of the 3D problem



      Labeling distinct objects produced by Show[RegionPlot3D's]



      that kglr answered. So, perhaps I should just try fixing (in various ways) one of the four coordinates and approaching the problem in the very same manner as there. (In fact, the constraints are set up in the same order both times, with the first one each times being the "PPT" one. Incidentally, the "PPT" body should be convex, but not the other two.)










      share|improve this question











      $endgroup$




      Let me abuse some Mathematica notation and formulate the following "command":



      Show[RegionPlot4D[(Q1 - Q4)^2 < 16 Q3^2 && 
      Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
      4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0, Q1, 0,
      6/61, Q2, 0, 2/9, Q3, 0, 1/32, Q4, 0, 1/6],
      RegionPlot4D[
      Q4 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 4 (Q2 + Q3) + 3 Q4 < 1 &&
      4 Q2 + 9 Q4 < Q1, Q1, 0, 6/61, Q2, 0, 2/9, Q3, 0, 1/32, Q4,
      0, 1/6],
      RegionPlot4D[
      Q4 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 4 (Q2 + Q3) + 3 Q4 < 1 &&
      2 (Q2 + Q3) + 3 Q4 < Q1, Q1, 0, 6/61, Q2, 0, 2/9, Q3, 0,
      1/32, Q4, 0, 1/6]]


      (Of course, there is a RegionPlot3D command, but no RegionPlot4D one.)



      Can this be processed/interpreted in some manner? (use of coloring,...)



      Also, these three "RegionPlot"s could be considered individually (challenging enough).



      These pertain to certain quantum-information-theoretic problems concerned with probabilities of (bound) entanglement.



      The problem as put is very much a direct 4D analogue of the 3D problem



      Labeling distinct objects produced by Show[RegionPlot3D's]



      that kglr answered. So, perhaps I should just try fixing (in various ways) one of the four coordinates and approaching the problem in the very same manner as there. (In fact, the constraints are set up in the same order both times, with the first one each times being the "PPT" one. Incidentally, the "PPT" body should be convex, but not the other two.)







      plotting graphics color dimension-reduction






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 5 hours ago







      Paul B. Slater

















      asked 8 hours ago









      Paul B. SlaterPaul B. Slater

      7634 silver badges14 bronze badges




      7634 silver badges14 bronze badges




















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          You can define a 4D region with



          R = ImplicitRegion[(Q1 - Q4)^2 < 16 Q3^2 && 
          Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
          4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0,
          Q1, Q2, Q3, Q4]


          and then check for region membership of any point. For example, make a list of lots of points in 4D and pick out those that lie inside of R:



          P = Select[Tuples[Range[0, 1/4, 1/128], 4], Element[#, R] &];
          Length[P]
          (* 84579 *)


          These can be plotted in many ways, for example by projecting out the fourth dimension and using only the first, second, third dimension as coordinate axes:



          ListPointPlot3D[P[[All, 1, 2, 3]]]


          enter image description here



          For a convex set, you can construct the convex hull in 3D for such a projection, for better visibility than the point cloud:



          ConvexHullMesh[P[[All, 1, 2, 3]], Boxed -> True, Axes -> True]


          enter image description here






          share|improve this answer









          $endgroup$












          • $begingroup$
            Could one use ListPointPlot3D with multiple point sets, using different colors?
            $endgroup$
            – Paul B. Slater
            4 hours ago










          • $begingroup$
            @PaulB.Slater Yes you can.
            $endgroup$
            – Roman
            4 hours ago


















          0












          $begingroup$

          Using Graphics3D with VertexColors based on the fourth column is much faster than using ListPointPlot3D.



          With a smaller version of Roman's P (to stay within my cloud credit limits):



          P = Select[Tuples[Range[0, 1/4, 1/64], 4], Element[#, R] &];

          Graphics3D[PointSize[Small], Point[P[[All, ;; 3]],
          VertexColors -> (Hue /@ P[[All, 4]])]] // RepeatedTiming


          enter image description here



          versus two alternative ways to use ListPointPlot3D:



          ListPointPlot3D[Style[#[[;;3]], Hue @ #[[4]]]& /@ P,
          BaseStyle -> PointSize[Small]] // RepeatedTiming


          enter image description here



          ListPointPlot3D[List /@ P[[All, ;; 3]], 
          PlotStyle -> (Hue /@ P[[All, 4]]),
          BaseStyle -> PointSize[Small]] // RepeatedTiming


          enter image description here






          share|improve this answer









          $endgroup$















            Your Answer








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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            You can define a 4D region with



            R = ImplicitRegion[(Q1 - Q4)^2 < 16 Q3^2 && 
            Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
            4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0,
            Q1, Q2, Q3, Q4]


            and then check for region membership of any point. For example, make a list of lots of points in 4D and pick out those that lie inside of R:



            P = Select[Tuples[Range[0, 1/4, 1/128], 4], Element[#, R] &];
            Length[P]
            (* 84579 *)


            These can be plotted in many ways, for example by projecting out the fourth dimension and using only the first, second, third dimension as coordinate axes:



            ListPointPlot3D[P[[All, 1, 2, 3]]]


            enter image description here



            For a convex set, you can construct the convex hull in 3D for such a projection, for better visibility than the point cloud:



            ConvexHullMesh[P[[All, 1, 2, 3]], Boxed -> True, Axes -> True]


            enter image description here






            share|improve this answer









            $endgroup$












            • $begingroup$
              Could one use ListPointPlot3D with multiple point sets, using different colors?
              $endgroup$
              – Paul B. Slater
              4 hours ago










            • $begingroup$
              @PaulB.Slater Yes you can.
              $endgroup$
              – Roman
              4 hours ago















            4












            $begingroup$

            You can define a 4D region with



            R = ImplicitRegion[(Q1 - Q4)^2 < 16 Q3^2 && 
            Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
            4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0,
            Q1, Q2, Q3, Q4]


            and then check for region membership of any point. For example, make a list of lots of points in 4D and pick out those that lie inside of R:



            P = Select[Tuples[Range[0, 1/4, 1/128], 4], Element[#, R] &];
            Length[P]
            (* 84579 *)


            These can be plotted in many ways, for example by projecting out the fourth dimension and using only the first, second, third dimension as coordinate axes:



            ListPointPlot3D[P[[All, 1, 2, 3]]]


            enter image description here



            For a convex set, you can construct the convex hull in 3D for such a projection, for better visibility than the point cloud:



            ConvexHullMesh[P[[All, 1, 2, 3]], Boxed -> True, Axes -> True]


            enter image description here






            share|improve this answer









            $endgroup$












            • $begingroup$
              Could one use ListPointPlot3D with multiple point sets, using different colors?
              $endgroup$
              – Paul B. Slater
              4 hours ago










            • $begingroup$
              @PaulB.Slater Yes you can.
              $endgroup$
              – Roman
              4 hours ago













            4












            4








            4





            $begingroup$

            You can define a 4D region with



            R = ImplicitRegion[(Q1 - Q4)^2 < 16 Q3^2 && 
            Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
            4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0,
            Q1, Q2, Q3, Q4]


            and then check for region membership of any point. For example, make a list of lots of points in 4D and pick out those that lie inside of R:



            P = Select[Tuples[Range[0, 1/4, 1/128], 4], Element[#, R] &];
            Length[P]
            (* 84579 *)


            These can be plotted in many ways, for example by projecting out the fourth dimension and using only the first, second, third dimension as coordinate axes:



            ListPointPlot3D[P[[All, 1, 2, 3]]]


            enter image description here



            For a convex set, you can construct the convex hull in 3D for such a projection, for better visibility than the point cloud:



            ConvexHullMesh[P[[All, 1, 2, 3]], Boxed -> True, Axes -> True]


            enter image description here






            share|improve this answer









            $endgroup$



            You can define a 4D region with



            R = ImplicitRegion[(Q1 - Q4)^2 < 16 Q3^2 && 
            Q1^2 + 4 Q1 Q2 + 16 Q2 (Q2 + Q3) + 12 Q2 Q4 + Q4^2 <
            4 Q2 + 2 Q1 Q4 && Q1 > 0 && Q2 > 0 && Q3 > 0 && Q4 > 0,
            Q1, Q2, Q3, Q4]


            and then check for region membership of any point. For example, make a list of lots of points in 4D and pick out those that lie inside of R:



            P = Select[Tuples[Range[0, 1/4, 1/128], 4], Element[#, R] &];
            Length[P]
            (* 84579 *)


            These can be plotted in many ways, for example by projecting out the fourth dimension and using only the first, second, third dimension as coordinate axes:



            ListPointPlot3D[P[[All, 1, 2, 3]]]


            enter image description here



            For a convex set, you can construct the convex hull in 3D for such a projection, for better visibility than the point cloud:



            ConvexHullMesh[P[[All, 1, 2, 3]], Boxed -> True, Axes -> True]


            enter image description here







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 8 hours ago









            RomanRoman

            13.8k1 gold badge19 silver badges51 bronze badges




            13.8k1 gold badge19 silver badges51 bronze badges











            • $begingroup$
              Could one use ListPointPlot3D with multiple point sets, using different colors?
              $endgroup$
              – Paul B. Slater
              4 hours ago










            • $begingroup$
              @PaulB.Slater Yes you can.
              $endgroup$
              – Roman
              4 hours ago
















            • $begingroup$
              Could one use ListPointPlot3D with multiple point sets, using different colors?
              $endgroup$
              – Paul B. Slater
              4 hours ago










            • $begingroup$
              @PaulB.Slater Yes you can.
              $endgroup$
              – Roman
              4 hours ago















            $begingroup$
            Could one use ListPointPlot3D with multiple point sets, using different colors?
            $endgroup$
            – Paul B. Slater
            4 hours ago




            $begingroup$
            Could one use ListPointPlot3D with multiple point sets, using different colors?
            $endgroup$
            – Paul B. Slater
            4 hours ago












            $begingroup$
            @PaulB.Slater Yes you can.
            $endgroup$
            – Roman
            4 hours ago




            $begingroup$
            @PaulB.Slater Yes you can.
            $endgroup$
            – Roman
            4 hours ago













            0












            $begingroup$

            Using Graphics3D with VertexColors based on the fourth column is much faster than using ListPointPlot3D.



            With a smaller version of Roman's P (to stay within my cloud credit limits):



            P = Select[Tuples[Range[0, 1/4, 1/64], 4], Element[#, R] &];

            Graphics3D[PointSize[Small], Point[P[[All, ;; 3]],
            VertexColors -> (Hue /@ P[[All, 4]])]] // RepeatedTiming


            enter image description here



            versus two alternative ways to use ListPointPlot3D:



            ListPointPlot3D[Style[#[[;;3]], Hue @ #[[4]]]& /@ P,
            BaseStyle -> PointSize[Small]] // RepeatedTiming


            enter image description here



            ListPointPlot3D[List /@ P[[All, ;; 3]], 
            PlotStyle -> (Hue /@ P[[All, 4]]),
            BaseStyle -> PointSize[Small]] // RepeatedTiming


            enter image description here






            share|improve this answer









            $endgroup$

















              0












              $begingroup$

              Using Graphics3D with VertexColors based on the fourth column is much faster than using ListPointPlot3D.



              With a smaller version of Roman's P (to stay within my cloud credit limits):



              P = Select[Tuples[Range[0, 1/4, 1/64], 4], Element[#, R] &];

              Graphics3D[PointSize[Small], Point[P[[All, ;; 3]],
              VertexColors -> (Hue /@ P[[All, 4]])]] // RepeatedTiming


              enter image description here



              versus two alternative ways to use ListPointPlot3D:



              ListPointPlot3D[Style[#[[;;3]], Hue @ #[[4]]]& /@ P,
              BaseStyle -> PointSize[Small]] // RepeatedTiming


              enter image description here



              ListPointPlot3D[List /@ P[[All, ;; 3]], 
              PlotStyle -> (Hue /@ P[[All, 4]]),
              BaseStyle -> PointSize[Small]] // RepeatedTiming


              enter image description here






              share|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                Using Graphics3D with VertexColors based on the fourth column is much faster than using ListPointPlot3D.



                With a smaller version of Roman's P (to stay within my cloud credit limits):



                P = Select[Tuples[Range[0, 1/4, 1/64], 4], Element[#, R] &];

                Graphics3D[PointSize[Small], Point[P[[All, ;; 3]],
                VertexColors -> (Hue /@ P[[All, 4]])]] // RepeatedTiming


                enter image description here



                versus two alternative ways to use ListPointPlot3D:



                ListPointPlot3D[Style[#[[;;3]], Hue @ #[[4]]]& /@ P,
                BaseStyle -> PointSize[Small]] // RepeatedTiming


                enter image description here



                ListPointPlot3D[List /@ P[[All, ;; 3]], 
                PlotStyle -> (Hue /@ P[[All, 4]]),
                BaseStyle -> PointSize[Small]] // RepeatedTiming


                enter image description here






                share|improve this answer









                $endgroup$



                Using Graphics3D with VertexColors based on the fourth column is much faster than using ListPointPlot3D.



                With a smaller version of Roman's P (to stay within my cloud credit limits):



                P = Select[Tuples[Range[0, 1/4, 1/64], 4], Element[#, R] &];

                Graphics3D[PointSize[Small], Point[P[[All, ;; 3]],
                VertexColors -> (Hue /@ P[[All, 4]])]] // RepeatedTiming


                enter image description here



                versus two alternative ways to use ListPointPlot3D:



                ListPointPlot3D[Style[#[[;;3]], Hue @ #[[4]]]& /@ P,
                BaseStyle -> PointSize[Small]] // RepeatedTiming


                enter image description here



                ListPointPlot3D[List /@ P[[All, ;; 3]], 
                PlotStyle -> (Hue /@ P[[All, 4]]),
                BaseStyle -> PointSize[Small]] // RepeatedTiming


                enter image description here







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 32 mins ago









                kglrkglr

                205k10 gold badges233 silver badges463 bronze badges




                205k10 gold badges233 silver badges463 bronze badges



























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                    Кастелфранко ди Сопра Становништво Референце Спољашње везе Мени за навигацију43°37′18″ СГШ; 11°33′32″ ИГД / 43.62156° СГШ; 11.55885° ИГД / 43.62156; 11.5588543°37′18″ СГШ; 11°33′32″ ИГД / 43.62156° СГШ; 11.55885° ИГД / 43.62156; 11.558853179688„The GeoNames geographical database”„Istituto Nazionale di Statistica”проширитиууWorldCat156923403n850174324558639-1cb14643287r(подаци)