How can I plot a list returned by the mathematica solution to in bounded integer equations - wolfram-mathematica

Question

So I have a set of bounded diophantine equations that specify lines on the plane. I want to make mathematica plot the intersection of two of these equations so I can see what they look like.
So far I have something like:
Solve[0 < x - y < 3 && -1 < 2 x - y < 2, {x, y}, Integers]
which returns some structure like:
{{x -> -2, y -> -4}, {x -> -1, y -> -3}, {x -> -1, y -> -2}, {x -> 0,
y -> -1}}
but how can I now make mathematica plot this so I can see the resulting shape. Preferably I would like the plot to consider every 'point' to be a 1x1 square.
Also, I wonder if there is a better way to do such things. Thanks.

Answer 1

Define the data you wish to plot by transforming the list Solve[] returns. This can done as
data = {x, y} /. Solve[0 < x - y < 3 && -1 < 2 x - y < 2, {x, y}, Integers]
More generally, you can make Solve return the solution in a list format (rather than as a set of rules) using the following trick:
data = Solve[0 < x - y < 3 && -1 < 2 x - y < 2, {x, y}, Integers] /. Rule[a_,b_]->b
For plotting, among many alternatives, you can use ListPlot as
ListPlot[data, PlotMarkers -> {Style["\[FilledSquare]", FontSize -> 16]}]
to get the following output
You can further refine it using many styling and other options of ListPlot. For example, you can join the points
ListPlot[data, PlotMarkers -> {Style["\[FilledSquare]", FontSize -> 16]},
Joined -> True]
to get
EDIT: To play with the marker placement and size there are several alternatives. Using ListPlot you can get what you need in either of the two ways:
(* Alternative 1: use fontsize to change the marker size *)
lp1 := ListPlot[{#} & /# #1,
PlotMarkers -> {Style["\[FilledSquare]", FontSize -> Scaled[#2]]},
AspectRatio -> 1, AxesOrigin -> {0, 0},
PlotRange -> {{-5, 1}, {-5, 1}},
PlotStyle -> Hue /# RandomReal[1, {Length##1}],
Epilog -> {GrayLevel[.3], PointSize[.02], Point##1, Thick,
Line##1}, Frame -> True, FrameTicks -> All] &;
(* usage example *)
lp1 ## {data, .30}
(* Alternative 2: use the second parameter of PlotMarkers to control scaled size *)
lp2 := ListPlot[{#} & /# #1,
PlotMarkers -> {Graphics#{Rectangle[]}, #2}, AspectRatio -> 1,
AxesOrigin -> {0, 0}, PlotRange -> {{-5, 1}, {-5, 1}},
PlotStyle -> Hue /# RandomReal[1, {Length##1}],
Epilog -> {GrayLevel[.3], PointSize[.02], Point##1, Thick,
Line##1}, Frame -> True, FrameTicks -> All] &
(* usage example *)
lp2 ## {data, 1/5.75}
In both cases, you need to use Epilog, otherwise the lines joining points are occluded by the markers. Both alternatives produce the following output:
Alternatively, you can use Graphics, RegionPlot, ContourPlot, BubbleChart with appropriate transformations of data to get results similar to the one in ListPlot output above.
Using Graphics primitives:
(* data transformation to define the regions *)
trdataG[data_, size_] := data /. {a_, b_} :>
{{a - size/2, b - size/2}, {a + size/2, b + size/2}};
(* plotting function *)
gr := Graphics[
{
{Hue[RandomReal[]], Rectangle[##]} & ### trdataG ## {#1, #2},
GrayLevel[.3], PointSize[.02], Thick, Point##1, Line##1},
PlotRange -> {{-5, 1}, {-5, 1}
},
PlotRangePadding -> 0, Axes -> True, AxesOrigin -> {0, 0},
Frame -> True, FrameTicks -> All] &
(* usage example *)
gr ## {data, .99}
Using BubbleChart:
(* Transformation of data to a form that BubbleChart expects *)
dataBC[data_] := data /. {a_, b_} :> {a, b, 1};
(* custom markers *)
myMarker[size_][{{xmin_, xmax_}, {ymin_, ymax_}}, ___] :=
{EdgeForm[], Rectangle[{(1/2) (xmin + xmax) - size/2, (1/2) (ymin + ymax) -
size/2}, {(1/2) (xmin + xmax) + size/2, (1/2) (ymin + ymax) + size/2}]};
(* charting function *)
bc := BubbleChart[dataBC[#1], ChartElementFunction -> myMarker[#2],
ChartStyle -> Hue /# RandomReal[1, {Length##1}], Axes -> True,
AxesOrigin -> {0, 0}, PlotRange -> {{-5, 1}, {-5, 1}},
PlotRangePadding -> 0, AspectRatio -> 1, FrameTicks -> All,
Epilog -> {GrayLevel[.3], PointSize[.02], Point##1, Thick, Line##1}] &
(* usage example *)
bc ## {data, .99}
Using RegionPlot:
(* Transformation of data to a form that RegionPlot expects *)
trdataRP[data_, size_] := data /. {a_, b_} :>
a - size/2 <= x <= a + size/2 && b - size/2 <= y <= b + size/2
(* charting function *)
rp := RegionPlot[Evaluate#trdataRP[#1, #2], {x, -5, 1}, {y, -5, 1},
AspectRatio -> 1, Axes -> True, AxesOrigin -> {0, 0},
PlotRange -> {{-5, 1}, {-5, 1}},
PlotStyle -> Hue /# RandomReal[1, {Length##1}], FrameTicks -> All,
PlotPoints -> 100, BoundaryStyle -> None,
Epilog -> {GrayLevel[.3], PointSize[.02], Point##1, Thick, Line##1}] &
(* usage example *)
rp ## {data, .99}
Using ContourPlot:
(* Transformation of data to a form that ContourPlot expects *)
trdataRP[data_, size_] := data /. {a_, b_} :>
a - size/2 <= x <= a + size/2 && b - size/2 <= y <= b + size/2;
trdataCP[data_, size_] := Which ## Flatten#
Thread[{trdataRP[data, size], Range#Length#data}];
(* charting function *)
cp := ContourPlot[trdataCP[#1, #2], {x, -5, 1}, {y, -5, 1},
AspectRatio -> 1, Axes -> True, AxesOrigin -> {0, 0},
PlotRange -> {{-5, 1}, {-5, 1}}, FrameTicks -> All,
ExclusionsStyle -> None, PlotPoints -> 100,
ColorFunction -> Hue,
Epilog -> {GrayLevel[.3], PointSize[.02], Point##1, Thick, Line##1}] &
(* usage example *)
cp ## {data, .99}

Answer 2

may be
sol = Solve[0 < x - y < 3 && -1 < 2 x - y < 2, {x, y}, Integers];
pts = Cases[sol, {_ -> n_, _ -> m_} :> {n, m}];
ListPlot[pts, Mesh -> All, Joined -> True, AxesOrigin -> {0, 0},
PlotMarkers -> {Automatic, 10}]
Can also extract the points to plot using
{#[[1, 2]], #[[2, 2]]} & /# sol