Plotting maxima within a simplexPlotting 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
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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;
$begingroup$
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?
plotting color simplex
$endgroup$
add a comment |
$begingroup$
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?
plotting color simplex
$endgroup$
add a comment |
$begingroup$
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?
plotting color simplex
$endgroup$
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
plotting color simplex
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
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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]

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]

$endgroup$
add a comment |
$begingroup$
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]

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]

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!
$endgroup$
$begingroup$
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 badRegionPlotleaves a big hole in the middle by default.
$endgroup$
– Chris K
3 hours ago
add a comment |
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$begingroup$
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]

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]

$endgroup$
add a comment |
$begingroup$
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]

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]

$endgroup$
add a comment |
$begingroup$
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]

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]

$endgroup$
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]

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]

edited 2 hours ago
answered 7 hours ago
kglrkglr
206k10 gold badges235 silver badges467 bronze badges
206k10 gold badges235 silver badges467 bronze badges
add a comment |
add a comment |
$begingroup$
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]

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]

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!
$endgroup$
$begingroup$
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 badRegionPlotleaves a big hole in the middle by default.
$endgroup$
– Chris K
3 hours ago
add a comment |
$begingroup$
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]

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]

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!
$endgroup$
$begingroup$
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 badRegionPlotleaves a big hole in the middle by default.
$endgroup$
– Chris K
3 hours ago
add a comment |
$begingroup$
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]

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]

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!
$endgroup$
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]

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]

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!
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 badRegionPlotleaves a big hole in the middle by default.
$endgroup$
– Chris K
3 hours ago
add a comment |
$begingroup$
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 badRegionPlotleaves 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
add a comment |
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
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Post as a guest
Required, but never shown
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown