diff options
Diffstat (limited to 'src/2geom/sbasis-roots.cpp')
| -rw-r--r-- | src/2geom/sbasis-roots.cpp | 14 |
1 files changed, 7 insertions, 7 deletions
diff --git a/src/2geom/sbasis-roots.cpp b/src/2geom/sbasis-roots.cpp index e3e5e4441..244b7efe4 100644 --- a/src/2geom/sbasis-roots.cpp +++ b/src/2geom/sbasis-roots.cpp @@ -46,11 +46,11 @@ * the gsl poly roots finder is faster than bernstein too, but we don't use it for 3 reasons: - a) it requires convertion to poly, which is numerically unstable + a) it requires conversion to poly, which is numerically unstable b) it requires gsl (which is currently not a dependency, and would bring in a whole slew of unrelated stuff) - c) it finds all roots, even complex ones. We don't want to accidently treat a nearly real root as a real root + c) it finds all roots, even complex ones. We don't want to accidentally treat a nearly real root as a real root From memory gsl poly roots was about 10 times faster than bernstein in the case where all the roots are in [0,1] for polys of order 5. I spent some time working out whether eigenvalue root finding @@ -85,7 +85,7 @@ namespace Geom{ /** Find the smallest interval that bounds a \param a sbasis function - \returns inteval + \returns interval */ @@ -113,7 +113,7 @@ OptInterval bounds_exact(SBasis const &a) { /** Find a small interval that bounds a \param a sbasis function - \returns inteval + \returns interval */ // I have no idea how this works, some clever bounding argument by jfb. @@ -231,7 +231,7 @@ static void multi_roots_internal(SBasis const &f, } return; } -////usefull? +////useful? // if (f.size()==1){ // int idxa=upper_level(levels,fa); // int idxb=upper_level(levels,fb); @@ -390,8 +390,8 @@ static std::vector<Interval> fuseContiguous(std::vector<Interval> const &sets, d -compute f at both ends of the given segment [a,b]. -compute bounds m<df(t)<M for df on the segment. Suppose f(a) is between two 'levels' c and C. Then - f wont enter c before a + (f(a)-c.max())/m - f wont enter C before a + (C.min()-f(a))/M + f won't enter c before a + (f(a)-c.max())/m + f won't enter C before a + (C.min()-f(a))/M From this we conclude nothing happens before a'=a+min((f(a)-c.max())/m,(C.min()-f(a))/M). We do the same for b: compute some b' such that nothing happens in (b',b]. -if [a',b'] is not empty, repeat the process with [a',(a'+b')/2] and [(a'+b')/2,b']. |