On Wed, Jul 3, 2013 at 8:11 AM, Ondřej Čertík <[email protected]> wrote: > On Tue, Jul 2, 2013 at 3:42 PM, Amit Saha <[email protected]> wrote: >> On Wed, Jul 3, 2013 at 2:29 AM, Ondřej Čertík <[email protected]> >> wrote: >>> On Tue, Jul 2, 2013 at 8:33 AM, Amit Saha <[email protected]> wrote: >>>> Hi Ondřej, >>>> >>>> On Tue, Jul 2, 2013 at 1:30 AM, Ondřej Čertík <[email protected]> >>>> wrote: >>>>> Hi Amit, >>>>> >>>>> On Sat, Jun 29, 2013 at 4:32 AM, Amit Saha <[email protected]> wrote: >>>>>> Hello, >>>>>> >>>>>> Currently the is_perpendicular() method is defined, like so: >>>>>> >>>>>> a1, b1, c1 = l1.coefficients >>>>>> a2, >>>>>> b2, c2 = l2.coefficients >>>>>> return >>>>>> bool(simplify(a1*a2 + b1*b2) == 0) >>>>>> >>>>>> Now it is possible (as I have just found) that the above sum may >>>>>> equate to something tlike 1e-15, but not 0 (same with the product of >>>>>> the slope coming to -0.9994). >>>>>> >>>>>> Does it make sense to have the provision for adding an epsilon value >>>>>> such that anything below that is deemed to be 0? >>>>> >>>>> Can you post your code that gives 1e-15? The idea is that when only >>>>> symbolic >>>>> values are used (i.e. no floating point numbers), you can compare >>>>> directly to 0. >>>> >>>> Okay, this is the code: >>>> >>>>>>> from sympy import Circle, Point, Segment >>>>>>> c = Point(0,0) >>>>>>> r=1 >>>>>>> circle = Circle(c,r) >>>>>>> tp = circle.random_point() >>>>>>> tangent = circle.tangent_lines(tp) >>>>>>> tangent = tangent[0] >>>> >>>>>>> rad = Segment(circle.center, tp) >>>> >>>>>>> (tangent.slope * rad.slope).evalf() >>>> -1.00000000000000 >>>>>>> tangent.is_perpendicular(rad) >>>> True >>> >>> See my notebook where I explain it: >>> >>> http://nbviewer.ipython.org/5910821 >>> >>>> >>>> So, in this case as you can see, both "tests" tell me that they are >>>> perpendicular, as they should be. >>>> >>>> However, for another point (when I originally wrote my query), the >>>> product (tangent.slope * rad.slope).evalf() resulted >>>> in -0.999999 (I am sorry what I meant by 1e-15 was a made up figure to >>>> illustrate the infinitesimal difference between -1 and the product and >>>> *not* the product). >>> >>> The evalf() method returns a floating point, which of course cannot be >>> directly compared to 0. And you would have to use eps, as you >>> suggested. However, the point is that you should not call evalf() on >>> it. >> >> Thanks for the notebook. I am clear about what I was doing there. >> >>> >>>> >>>> Also, the is_perpendicular() method returned False (the value was >>>> close to 0 but not quite). >>> >>> Can you post a code which does that? >> >> Sure, here it is: >> >> ''' >> script for multiple tangent/radius >> ''' >> from sympy import Point, Segment, Circle, simplify >> c = Point(0,0) >> r=1 >> circle = Circle(c,r) >> >> for i in range(20): >> tp = circle.random_point() >> tangent = circle.tangent_lines(tp) >> tangent = tangent[0] >> rad = Segment(circle.center, tp) >> >> a1, b1, c1 = tangent.coefficients >> a2, b2, c2 = rad.coefficients >> print(tangent.is_perpendicular(rad), simplify(a1*a2 + b1*b2)) >> >> >> >> I am basically generating 20 random points on the circle and checking >> for the validity of the radius and the tangent line being >> perpendicular to each other. You may not see one False for one run, >> but you may in another. >> >> For example on one of my runs, I got a False value with the following >> value of a1*a2 + b1*b2: >> >> False >> -3*2**(109/331)*3**(441/662)*5**(311/331)*7**(267/331)*sqrt(7822160432578995537131522933)/14000000000000000 >> + >> 128582730964399*sqrt(23466481297736986611394568799)/40000000000000000000000000000 >> + 23466481297736986611394568799/40000000000000000000000000000 >> >> (Sorry about that). >> >> That actually evaluates to >> >>>>> val.evalf() >> 1.40310612431989e-17 >> >> which is perhaps what I emailed in my original query, but forgot about >> it later :) > > I see. That indeed is a bug if it gives False. Can you try to do > val.evalf(100), this prints 100 digits. I expect the number to be > ~1e-100. If you print 1000 digits I expect the number to be ~1e-1000. > In other words, "val" is zero, but obviously once you convert to > floating point, you only get approximate zero. > > Finally, this clearly needs some simplification. Can you apply > "simplify" on val? > > Also, can you create some definite test case (without random numbers), > so that we can add it to our test suite and fix it? I can help you if > you get stuck.
Okay, here is something funny. I ran the above code using the SymPy from the GitHub clone (using the 2to3 script for Python 3 install) and didn't get a False for 1000 random iterations. Then what I did was I noted a point for which I was getting a failure (False) and fed it into the system where I was using the library from the latest clone, and the perpendicular test passes. So has something been fixed since 0.7.2? It seems to me so.. Is there any other test I can do to help? Thanks, Amit. -- http://echorand.me -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/sympy. For more options, visit https://groups.google.com/groups/opt_out.
