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jnthn
  • Member for 12 years, 3 months
  • Last seen more than 6 years ago
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Converting polygon to graph in V7
replaced \[DirectedEdge] with ->
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Converting polygon to graph in V7
In[]:= Head[GraphPlot[e] Out[]:= Graphics, so it is not a graph. I cant use Degrees[] on it.
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Converting polygon to graph in V7
same errors as above
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Does Graphics`Mesh`SimplePolygonQ[] work for you?
@Silvia: My version wasn't important until that function was added. I'll check out your links. Thanks again! :)
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Does Graphics`Mesh`SimplePolygonQ[] work for you?
@J.M: I assume you are saying it works for you with ver. 8. Thanks.
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Does Graphics`Mesh`SimplePolygonQ[] work for you?
@J.M.: p = Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}] GraphicsMeshSimplePolygonQ[p] Out[]:= GraphicsMeshSimplePolygonQ[Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}]]
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How can I draw a polygon from a set of angles?
Very very nice! After looking at the "house" shape (polygon #10 in the first EDIT section) more closely I realized it was being drawn correctly. I can now see that I will have to add an area check to weed out results that have area not equal to n times a unit triangle. Again, thanks so much Silvia.
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How can I draw a polygon from a set of angles?
Maybe I didn't make this clear...the triangles need to be of equal size. I will add an image to my original post showing you the problem I see..
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How can I draw a polygon from a set of angles?
π/180 got rid of the Floor[] errors. ty. If you use the order-3 sets as input, you will see that the squares with the 'X' in the center have hypotenuses of length 1, rather than length $\sqrt 2$. And there is a similar problem with the "house" shape. Is that error coming from the If[] statement, do you think?