bugs - RegionIntersection fails for some values with polygons

Bug introduced in 10.1 and fixed in 10.3.0

I want to find the intersection (region) of 2 polygons. I use RegionInstersection but it only works for some sizes, for example, when c=.5 it gives the correct result, but with c=.4 it only finds a line as intersection.

Area[r3[.5]] works fine...

Area[r3[.4]] does not work...

What is wrong?

b = 15.0;
d = 4.0;
lbp = 40.0;
d1E = 2.;
l1E = 7.0;

d1D = d1E;
l1D = 5.0;

r1 = Polygon[{{0, 0}, {lbp/2 - l1D, 0}, {lbp/2, d1D}, {lbp/2, 
     d}, {-(lbp/2), d}, {-(lbp/2), d1E}, {-(lbp/2) + l1E, 0}}];
r2[c_] := Rectangle[{-1.1 lbp/2, -1.5}, {1.1 lbp/2, c}];
r3[c_] := RegionIntersection[r1, r2[c]];

Answer

This bug has been fixed as of Mathematica 10.3.

b = 15.0; d = 4.0; lbp = 40.0; d1E = 2.; l1E = 7.0; d1D = d1E; l1D = 5.0;

r1 = Polygon[{{0, 0}, {lbp/2 - l1D, 0}, {lbp/2, d1D}, {lbp/2, d}, 
              {-(lbp/2), d}, {-(lbp/2), d1E}, {-(lbp/2) + l1E, 0}}];
r2[c_] := Rectangle[{-1.1 lbp/2, -1.5}, {1.1 lbp/2, c}];
r3[c_] := RegionIntersection[r1, r2[c]];
{Area[r3[.5]], Area[r3[.4]]}

(* {14.75, 11.68} *)

Comments

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

Blog

Search This Blog