Computing bezier curves of degree n with recursive functions












1














I have been looking at homemade CNC and have been wondering how curves are drawn, so I looked into it and found this cool article. I then decided to try coming up with a bezier curve algorithm in C. It seems to work ok, and while I haven't tried plotting points with this exact implementation, it seems to match up with results from a previous implementation that I did plot points with.



#include <stdint.h>

static long double bbezier(long double t, long double p0, long double p1){
return ((p1 - p0) * t) + p0;
}

long double bezier(long double t, uint64_t *points, uint64_t n){
long double p0 = points[0], p1 = points[1];
if(n == 1) return bbezier(t, p0, p1);

long double q0 = bezier(t, points, n - 1),
q1 = bezier(t, points + 1, n - 1);
return bbezier(t, q0, q1);
}


I then quickly wrote this test program in C++.



#include <iostream>

extern "C" long double bezier(long double, uint64_t *, uint64_t);

uint64_t pointsx = {
0, 40, 100, 200
};
uint64_t pointsy = {
0, 150, 60, 100
};

int main(){
for(uint64_t i = 0; i <= 10000; ++i){
long double ii = i;
long double j = ii/10000;

long double x = bezier(j, pointsx, 3);
long double y = bezier(j, pointsy, 3);

std::cout << "X: " << x << ", Y: " << y << 'n';
}
return 0;
}


I wrote an implementation running in javascript from a lightly modified w3 schools canvas tutorial here to understand how bezier curves work but it only supports 3rd degree curves. It does plot the points though, and that's what I based the above implementation on.



It doesn't make any checks to ensure t is between 0 and 1 and n != 0 but I'm not too worried. The only thing I'm worried about is segfaults in cases where n is so high that you get a stack overflow but that will be a pretty crazy curve. Anyway, how does it look?










share|improve this question



























    1














    I have been looking at homemade CNC and have been wondering how curves are drawn, so I looked into it and found this cool article. I then decided to try coming up with a bezier curve algorithm in C. It seems to work ok, and while I haven't tried plotting points with this exact implementation, it seems to match up with results from a previous implementation that I did plot points with.



    #include <stdint.h>

    static long double bbezier(long double t, long double p0, long double p1){
    return ((p1 - p0) * t) + p0;
    }

    long double bezier(long double t, uint64_t *points, uint64_t n){
    long double p0 = points[0], p1 = points[1];
    if(n == 1) return bbezier(t, p0, p1);

    long double q0 = bezier(t, points, n - 1),
    q1 = bezier(t, points + 1, n - 1);
    return bbezier(t, q0, q1);
    }


    I then quickly wrote this test program in C++.



    #include <iostream>

    extern "C" long double bezier(long double, uint64_t *, uint64_t);

    uint64_t pointsx = {
    0, 40, 100, 200
    };
    uint64_t pointsy = {
    0, 150, 60, 100
    };

    int main(){
    for(uint64_t i = 0; i <= 10000; ++i){
    long double ii = i;
    long double j = ii/10000;

    long double x = bezier(j, pointsx, 3);
    long double y = bezier(j, pointsy, 3);

    std::cout << "X: " << x << ", Y: " << y << 'n';
    }
    return 0;
    }


    I wrote an implementation running in javascript from a lightly modified w3 schools canvas tutorial here to understand how bezier curves work but it only supports 3rd degree curves. It does plot the points though, and that's what I based the above implementation on.



    It doesn't make any checks to ensure t is between 0 and 1 and n != 0 but I'm not too worried. The only thing I'm worried about is segfaults in cases where n is so high that you get a stack overflow but that will be a pretty crazy curve. Anyway, how does it look?










    share|improve this question

























      1












      1








      1







      I have been looking at homemade CNC and have been wondering how curves are drawn, so I looked into it and found this cool article. I then decided to try coming up with a bezier curve algorithm in C. It seems to work ok, and while I haven't tried plotting points with this exact implementation, it seems to match up with results from a previous implementation that I did plot points with.



      #include <stdint.h>

      static long double bbezier(long double t, long double p0, long double p1){
      return ((p1 - p0) * t) + p0;
      }

      long double bezier(long double t, uint64_t *points, uint64_t n){
      long double p0 = points[0], p1 = points[1];
      if(n == 1) return bbezier(t, p0, p1);

      long double q0 = bezier(t, points, n - 1),
      q1 = bezier(t, points + 1, n - 1);
      return bbezier(t, q0, q1);
      }


      I then quickly wrote this test program in C++.



      #include <iostream>

      extern "C" long double bezier(long double, uint64_t *, uint64_t);

      uint64_t pointsx = {
      0, 40, 100, 200
      };
      uint64_t pointsy = {
      0, 150, 60, 100
      };

      int main(){
      for(uint64_t i = 0; i <= 10000; ++i){
      long double ii = i;
      long double j = ii/10000;

      long double x = bezier(j, pointsx, 3);
      long double y = bezier(j, pointsy, 3);

      std::cout << "X: " << x << ", Y: " << y << 'n';
      }
      return 0;
      }


      I wrote an implementation running in javascript from a lightly modified w3 schools canvas tutorial here to understand how bezier curves work but it only supports 3rd degree curves. It does plot the points though, and that's what I based the above implementation on.



      It doesn't make any checks to ensure t is between 0 and 1 and n != 0 but I'm not too worried. The only thing I'm worried about is segfaults in cases where n is so high that you get a stack overflow but that will be a pretty crazy curve. Anyway, how does it look?










      share|improve this question













      I have been looking at homemade CNC and have been wondering how curves are drawn, so I looked into it and found this cool article. I then decided to try coming up with a bezier curve algorithm in C. It seems to work ok, and while I haven't tried plotting points with this exact implementation, it seems to match up with results from a previous implementation that I did plot points with.



      #include <stdint.h>

      static long double bbezier(long double t, long double p0, long double p1){
      return ((p1 - p0) * t) + p0;
      }

      long double bezier(long double t, uint64_t *points, uint64_t n){
      long double p0 = points[0], p1 = points[1];
      if(n == 1) return bbezier(t, p0, p1);

      long double q0 = bezier(t, points, n - 1),
      q1 = bezier(t, points + 1, n - 1);
      return bbezier(t, q0, q1);
      }


      I then quickly wrote this test program in C++.



      #include <iostream>

      extern "C" long double bezier(long double, uint64_t *, uint64_t);

      uint64_t pointsx = {
      0, 40, 100, 200
      };
      uint64_t pointsy = {
      0, 150, 60, 100
      };

      int main(){
      for(uint64_t i = 0; i <= 10000; ++i){
      long double ii = i;
      long double j = ii/10000;

      long double x = bezier(j, pointsx, 3);
      long double y = bezier(j, pointsy, 3);

      std::cout << "X: " << x << ", Y: " << y << 'n';
      }
      return 0;
      }


      I wrote an implementation running in javascript from a lightly modified w3 schools canvas tutorial here to understand how bezier curves work but it only supports 3rd degree curves. It does plot the points though, and that's what I based the above implementation on.



      It doesn't make any checks to ensure t is between 0 and 1 and n != 0 but I'm not too worried. The only thing I'm worried about is segfaults in cases where n is so high that you get a stack overflow but that will be a pretty crazy curve. Anyway, how does it look?







      c recursion






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      share|improve this question











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      asked yesterday









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          2 Answers
          2






          active

          oldest

          votes


















          2















          • I don't like bbezier. It collided too much with bezier, and is not very informative. It performs a linear interpolation, so why not call it interpolate?



          • The p0 and p1 just add noise. Consider



                if (n == 1) {
            return interpolate(t, points[0], points[1]);
            }


            I would seriously consider getting rid of q0 and q1:



                return interpolate(t,
            bezier(t, points, n - 1),
            bezier(t, points + 1, n - 1));


            Don't take it as a recommendation.



          • The recursion leads to the exponential time complexity. Way before you start having memory problems you'd face a performance problem. Consider computing the Bernstein form instead. It gives you linear time, and no memory problems.







          share|improve this answer





















          • "It gives you linear time, and no memory problems." Evaluating them in Bernstein form through de Casteljau's algorithm is inefficient: $n(n+1)/2$ additions and $n(n+1)$ multiplications to calculate a point on a curve of degree $n$. Rather, I'd prefer Wang-Ball form. For reference: "Efficient algorithms for Bézier curves."
            – esote
            23 hours ago












          • @esote Did I ever mention de Casteljau? A not-so-naive implementation is linear. Two multiplications and two divisions per term. It will have some numerical issues with large n, sure.
            – vnp
            23 hours ago












          • Fair enough, I should've read your answer more carefully.
            – esote
            23 hours ago





















          2














          Recursion



          [Edit]

          In bezier(), the 2 recursive calls to bezier() is inefficient as it exponential grows with O(2n) and only O(n2) operations are needed. I suspect better efficiency (linear) can be had with a pre-computed weighing of the d terms below.



          The concern about seg faulting due to excessive recursion would be mitigated with the above improvement.



          I also change function bbezier() to code.



          long double bezier_alt1(long double t, const uint64_t *points, size_t n) {
          assert(n);
          long double omt = 1.0 - t;
          long double d[n]; // Save in between calculations.

          for (size_t i = 0; i < n; i++) {
          d[i] = omt * points[i] + t * points[i + 1];
          }
          while (n > 1) {
          n--;
          for (size_t i = 0; i < n; i++) {
          d[i] = omt * d[i] + t * d[i + 1];
          }
          }
          return d[0];
          }


          [Edit2]



          A linear solution O(n) is possible with O(1) additional memory.



          long double bezier_alt2(long double t, const uint64_t *points, size_t n) {
          assert(n);
          long double omt = 1.0 - t;
          long double power_t = 1.0;
          long double power_omt = powl(omt,n);
          long double omt_div = omt != 0.0 ? 1.0/omt : 0.0;

          long double sum = 0.0;
          unsigned long term_n = 1;
          unsigned long term_d = 1;
          for (size_t i = 0; i < n; i++) {
          long double y = power_omt*power_t*points[i]*term_n/term_d;
          sum += y;
          power_t *= t;
          power_omt *= omt_div;
          term_n *= (n-i);
          term_d *= (i+1);
          }
          power_omt = 1.0;
          long double y = power_omt*power_t*points[n]*term_n/term_d;
          sum += y;
          return sum;
          }


          Additional linear simplifications possible - perhaps another day.





          Minor stuff



          Use const



          A const in the referenced date allows for some optimizations, wider application and better conveys code's intent.



          // long double bezier(long double t, uint64_t *points, uint64_t n){
          long double bezier(long double t, const uint64_t *points, uint64_t n){


          Excessive wide type



          With uint64_t n, there is no reasonable expectation that such an iteration will finish for large n.



          Fortunately, n indicates the size of the array. For array indexing and sizing, using size_t. It is the right size - not too narrow, nor too wide a type.



          // long double bezier(long double t, uint64_t *points, uint64_t n){
          long double bezier(long double t, uint64_t *points, size_t n){


          For this application, certainly unsigned would alway suffice.



          static



          Good use of static in static long double bbezier() to keep that function local.



          Missing "bezier.h"



          I'd expect a bezier() declarations in a .h file and implementation in the .c file instead of extern "C" long double bezier(long double, uint64_t *, uint64_t); in main.c



          // extern "C" long double bezier(long double, uint64_t *, uint64_t);
          #include "bezier.h".


          n range check



          Perhaps in a debug build, test n.



          long double bezier(long double t, const uint64_t *points, uint64_t n){
          assert( n > 0); // Low bound
          assert( n < 1000); // Maybe an sane upper limit too





          share|improve this answer























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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

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            votes






            active

            oldest

            votes









            2















            • I don't like bbezier. It collided too much with bezier, and is not very informative. It performs a linear interpolation, so why not call it interpolate?



            • The p0 and p1 just add noise. Consider



                  if (n == 1) {
              return interpolate(t, points[0], points[1]);
              }


              I would seriously consider getting rid of q0 and q1:



                  return interpolate(t,
              bezier(t, points, n - 1),
              bezier(t, points + 1, n - 1));


              Don't take it as a recommendation.



            • The recursion leads to the exponential time complexity. Way before you start having memory problems you'd face a performance problem. Consider computing the Bernstein form instead. It gives you linear time, and no memory problems.







            share|improve this answer





















            • "It gives you linear time, and no memory problems." Evaluating them in Bernstein form through de Casteljau's algorithm is inefficient: $n(n+1)/2$ additions and $n(n+1)$ multiplications to calculate a point on a curve of degree $n$. Rather, I'd prefer Wang-Ball form. For reference: "Efficient algorithms for Bézier curves."
              – esote
              23 hours ago












            • @esote Did I ever mention de Casteljau? A not-so-naive implementation is linear. Two multiplications and two divisions per term. It will have some numerical issues with large n, sure.
              – vnp
              23 hours ago












            • Fair enough, I should've read your answer more carefully.
              – esote
              23 hours ago


















            2















            • I don't like bbezier. It collided too much with bezier, and is not very informative. It performs a linear interpolation, so why not call it interpolate?



            • The p0 and p1 just add noise. Consider



                  if (n == 1) {
              return interpolate(t, points[0], points[1]);
              }


              I would seriously consider getting rid of q0 and q1:



                  return interpolate(t,
              bezier(t, points, n - 1),
              bezier(t, points + 1, n - 1));


              Don't take it as a recommendation.



            • The recursion leads to the exponential time complexity. Way before you start having memory problems you'd face a performance problem. Consider computing the Bernstein form instead. It gives you linear time, and no memory problems.







            share|improve this answer





















            • "It gives you linear time, and no memory problems." Evaluating them in Bernstein form through de Casteljau's algorithm is inefficient: $n(n+1)/2$ additions and $n(n+1)$ multiplications to calculate a point on a curve of degree $n$. Rather, I'd prefer Wang-Ball form. For reference: "Efficient algorithms for Bézier curves."
              – esote
              23 hours ago












            • @esote Did I ever mention de Casteljau? A not-so-naive implementation is linear. Two multiplications and two divisions per term. It will have some numerical issues with large n, sure.
              – vnp
              23 hours ago












            • Fair enough, I should've read your answer more carefully.
              – esote
              23 hours ago
















            2












            2








            2







            • I don't like bbezier. It collided too much with bezier, and is not very informative. It performs a linear interpolation, so why not call it interpolate?



            • The p0 and p1 just add noise. Consider



                  if (n == 1) {
              return interpolate(t, points[0], points[1]);
              }


              I would seriously consider getting rid of q0 and q1:



                  return interpolate(t,
              bezier(t, points, n - 1),
              bezier(t, points + 1, n - 1));


              Don't take it as a recommendation.



            • The recursion leads to the exponential time complexity. Way before you start having memory problems you'd face a performance problem. Consider computing the Bernstein form instead. It gives you linear time, and no memory problems.







            share|improve this answer













            • I don't like bbezier. It collided too much with bezier, and is not very informative. It performs a linear interpolation, so why not call it interpolate?



            • The p0 and p1 just add noise. Consider



                  if (n == 1) {
              return interpolate(t, points[0], points[1]);
              }


              I would seriously consider getting rid of q0 and q1:



                  return interpolate(t,
              bezier(t, points, n - 1),
              bezier(t, points + 1, n - 1));


              Don't take it as a recommendation.



            • The recursion leads to the exponential time complexity. Way before you start having memory problems you'd face a performance problem. Consider computing the Bernstein form instead. It gives you linear time, and no memory problems.








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 23 hours ago









            vnpvnp

            38.7k13097




            38.7k13097












            • "It gives you linear time, and no memory problems." Evaluating them in Bernstein form through de Casteljau's algorithm is inefficient: $n(n+1)/2$ additions and $n(n+1)$ multiplications to calculate a point on a curve of degree $n$. Rather, I'd prefer Wang-Ball form. For reference: "Efficient algorithms for Bézier curves."
              – esote
              23 hours ago












            • @esote Did I ever mention de Casteljau? A not-so-naive implementation is linear. Two multiplications and two divisions per term. It will have some numerical issues with large n, sure.
              – vnp
              23 hours ago












            • Fair enough, I should've read your answer more carefully.
              – esote
              23 hours ago




















            • "It gives you linear time, and no memory problems." Evaluating them in Bernstein form through de Casteljau's algorithm is inefficient: $n(n+1)/2$ additions and $n(n+1)$ multiplications to calculate a point on a curve of degree $n$. Rather, I'd prefer Wang-Ball form. For reference: "Efficient algorithms for Bézier curves."
              – esote
              23 hours ago












            • @esote Did I ever mention de Casteljau? A not-so-naive implementation is linear. Two multiplications and two divisions per term. It will have some numerical issues with large n, sure.
              – vnp
              23 hours ago












            • Fair enough, I should've read your answer more carefully.
              – esote
              23 hours ago


















            "It gives you linear time, and no memory problems." Evaluating them in Bernstein form through de Casteljau's algorithm is inefficient: $n(n+1)/2$ additions and $n(n+1)$ multiplications to calculate a point on a curve of degree $n$. Rather, I'd prefer Wang-Ball form. For reference: "Efficient algorithms for Bézier curves."
            – esote
            23 hours ago






            "It gives you linear time, and no memory problems." Evaluating them in Bernstein form through de Casteljau's algorithm is inefficient: $n(n+1)/2$ additions and $n(n+1)$ multiplications to calculate a point on a curve of degree $n$. Rather, I'd prefer Wang-Ball form. For reference: "Efficient algorithms for Bézier curves."
            – esote
            23 hours ago














            @esote Did I ever mention de Casteljau? A not-so-naive implementation is linear. Two multiplications and two divisions per term. It will have some numerical issues with large n, sure.
            – vnp
            23 hours ago






            @esote Did I ever mention de Casteljau? A not-so-naive implementation is linear. Two multiplications and two divisions per term. It will have some numerical issues with large n, sure.
            – vnp
            23 hours ago














            Fair enough, I should've read your answer more carefully.
            – esote
            23 hours ago






            Fair enough, I should've read your answer more carefully.
            – esote
            23 hours ago















            2














            Recursion



            [Edit]

            In bezier(), the 2 recursive calls to bezier() is inefficient as it exponential grows with O(2n) and only O(n2) operations are needed. I suspect better efficiency (linear) can be had with a pre-computed weighing of the d terms below.



            The concern about seg faulting due to excessive recursion would be mitigated with the above improvement.



            I also change function bbezier() to code.



            long double bezier_alt1(long double t, const uint64_t *points, size_t n) {
            assert(n);
            long double omt = 1.0 - t;
            long double d[n]; // Save in between calculations.

            for (size_t i = 0; i < n; i++) {
            d[i] = omt * points[i] + t * points[i + 1];
            }
            while (n > 1) {
            n--;
            for (size_t i = 0; i < n; i++) {
            d[i] = omt * d[i] + t * d[i + 1];
            }
            }
            return d[0];
            }


            [Edit2]



            A linear solution O(n) is possible with O(1) additional memory.



            long double bezier_alt2(long double t, const uint64_t *points, size_t n) {
            assert(n);
            long double omt = 1.0 - t;
            long double power_t = 1.0;
            long double power_omt = powl(omt,n);
            long double omt_div = omt != 0.0 ? 1.0/omt : 0.0;

            long double sum = 0.0;
            unsigned long term_n = 1;
            unsigned long term_d = 1;
            for (size_t i = 0; i < n; i++) {
            long double y = power_omt*power_t*points[i]*term_n/term_d;
            sum += y;
            power_t *= t;
            power_omt *= omt_div;
            term_n *= (n-i);
            term_d *= (i+1);
            }
            power_omt = 1.0;
            long double y = power_omt*power_t*points[n]*term_n/term_d;
            sum += y;
            return sum;
            }


            Additional linear simplifications possible - perhaps another day.





            Minor stuff



            Use const



            A const in the referenced date allows for some optimizations, wider application and better conveys code's intent.



            // long double bezier(long double t, uint64_t *points, uint64_t n){
            long double bezier(long double t, const uint64_t *points, uint64_t n){


            Excessive wide type



            With uint64_t n, there is no reasonable expectation that such an iteration will finish for large n.



            Fortunately, n indicates the size of the array. For array indexing and sizing, using size_t. It is the right size - not too narrow, nor too wide a type.



            // long double bezier(long double t, uint64_t *points, uint64_t n){
            long double bezier(long double t, uint64_t *points, size_t n){


            For this application, certainly unsigned would alway suffice.



            static



            Good use of static in static long double bbezier() to keep that function local.



            Missing "bezier.h"



            I'd expect a bezier() declarations in a .h file and implementation in the .c file instead of extern "C" long double bezier(long double, uint64_t *, uint64_t); in main.c



            // extern "C" long double bezier(long double, uint64_t *, uint64_t);
            #include "bezier.h".


            n range check



            Perhaps in a debug build, test n.



            long double bezier(long double t, const uint64_t *points, uint64_t n){
            assert( n > 0); // Low bound
            assert( n < 1000); // Maybe an sane upper limit too





            share|improve this answer




























              2














              Recursion



              [Edit]

              In bezier(), the 2 recursive calls to bezier() is inefficient as it exponential grows with O(2n) and only O(n2) operations are needed. I suspect better efficiency (linear) can be had with a pre-computed weighing of the d terms below.



              The concern about seg faulting due to excessive recursion would be mitigated with the above improvement.



              I also change function bbezier() to code.



              long double bezier_alt1(long double t, const uint64_t *points, size_t n) {
              assert(n);
              long double omt = 1.0 - t;
              long double d[n]; // Save in between calculations.

              for (size_t i = 0; i < n; i++) {
              d[i] = omt * points[i] + t * points[i + 1];
              }
              while (n > 1) {
              n--;
              for (size_t i = 0; i < n; i++) {
              d[i] = omt * d[i] + t * d[i + 1];
              }
              }
              return d[0];
              }


              [Edit2]



              A linear solution O(n) is possible with O(1) additional memory.



              long double bezier_alt2(long double t, const uint64_t *points, size_t n) {
              assert(n);
              long double omt = 1.0 - t;
              long double power_t = 1.0;
              long double power_omt = powl(omt,n);
              long double omt_div = omt != 0.0 ? 1.0/omt : 0.0;

              long double sum = 0.0;
              unsigned long term_n = 1;
              unsigned long term_d = 1;
              for (size_t i = 0; i < n; i++) {
              long double y = power_omt*power_t*points[i]*term_n/term_d;
              sum += y;
              power_t *= t;
              power_omt *= omt_div;
              term_n *= (n-i);
              term_d *= (i+1);
              }
              power_omt = 1.0;
              long double y = power_omt*power_t*points[n]*term_n/term_d;
              sum += y;
              return sum;
              }


              Additional linear simplifications possible - perhaps another day.





              Minor stuff



              Use const



              A const in the referenced date allows for some optimizations, wider application and better conveys code's intent.



              // long double bezier(long double t, uint64_t *points, uint64_t n){
              long double bezier(long double t, const uint64_t *points, uint64_t n){


              Excessive wide type



              With uint64_t n, there is no reasonable expectation that such an iteration will finish for large n.



              Fortunately, n indicates the size of the array. For array indexing and sizing, using size_t. It is the right size - not too narrow, nor too wide a type.



              // long double bezier(long double t, uint64_t *points, uint64_t n){
              long double bezier(long double t, uint64_t *points, size_t n){


              For this application, certainly unsigned would alway suffice.



              static



              Good use of static in static long double bbezier() to keep that function local.



              Missing "bezier.h"



              I'd expect a bezier() declarations in a .h file and implementation in the .c file instead of extern "C" long double bezier(long double, uint64_t *, uint64_t); in main.c



              // extern "C" long double bezier(long double, uint64_t *, uint64_t);
              #include "bezier.h".


              n range check



              Perhaps in a debug build, test n.



              long double bezier(long double t, const uint64_t *points, uint64_t n){
              assert( n > 0); // Low bound
              assert( n < 1000); // Maybe an sane upper limit too





              share|improve this answer


























                2












                2








                2






                Recursion



                [Edit]

                In bezier(), the 2 recursive calls to bezier() is inefficient as it exponential grows with O(2n) and only O(n2) operations are needed. I suspect better efficiency (linear) can be had with a pre-computed weighing of the d terms below.



                The concern about seg faulting due to excessive recursion would be mitigated with the above improvement.



                I also change function bbezier() to code.



                long double bezier_alt1(long double t, const uint64_t *points, size_t n) {
                assert(n);
                long double omt = 1.0 - t;
                long double d[n]; // Save in between calculations.

                for (size_t i = 0; i < n; i++) {
                d[i] = omt * points[i] + t * points[i + 1];
                }
                while (n > 1) {
                n--;
                for (size_t i = 0; i < n; i++) {
                d[i] = omt * d[i] + t * d[i + 1];
                }
                }
                return d[0];
                }


                [Edit2]



                A linear solution O(n) is possible with O(1) additional memory.



                long double bezier_alt2(long double t, const uint64_t *points, size_t n) {
                assert(n);
                long double omt = 1.0 - t;
                long double power_t = 1.0;
                long double power_omt = powl(omt,n);
                long double omt_div = omt != 0.0 ? 1.0/omt : 0.0;

                long double sum = 0.0;
                unsigned long term_n = 1;
                unsigned long term_d = 1;
                for (size_t i = 0; i < n; i++) {
                long double y = power_omt*power_t*points[i]*term_n/term_d;
                sum += y;
                power_t *= t;
                power_omt *= omt_div;
                term_n *= (n-i);
                term_d *= (i+1);
                }
                power_omt = 1.0;
                long double y = power_omt*power_t*points[n]*term_n/term_d;
                sum += y;
                return sum;
                }


                Additional linear simplifications possible - perhaps another day.





                Minor stuff



                Use const



                A const in the referenced date allows for some optimizations, wider application and better conveys code's intent.



                // long double bezier(long double t, uint64_t *points, uint64_t n){
                long double bezier(long double t, const uint64_t *points, uint64_t n){


                Excessive wide type



                With uint64_t n, there is no reasonable expectation that such an iteration will finish for large n.



                Fortunately, n indicates the size of the array. For array indexing and sizing, using size_t. It is the right size - not too narrow, nor too wide a type.



                // long double bezier(long double t, uint64_t *points, uint64_t n){
                long double bezier(long double t, uint64_t *points, size_t n){


                For this application, certainly unsigned would alway suffice.



                static



                Good use of static in static long double bbezier() to keep that function local.



                Missing "bezier.h"



                I'd expect a bezier() declarations in a .h file and implementation in the .c file instead of extern "C" long double bezier(long double, uint64_t *, uint64_t); in main.c



                // extern "C" long double bezier(long double, uint64_t *, uint64_t);
                #include "bezier.h".


                n range check



                Perhaps in a debug build, test n.



                long double bezier(long double t, const uint64_t *points, uint64_t n){
                assert( n > 0); // Low bound
                assert( n < 1000); // Maybe an sane upper limit too





                share|improve this answer














                Recursion



                [Edit]

                In bezier(), the 2 recursive calls to bezier() is inefficient as it exponential grows with O(2n) and only O(n2) operations are needed. I suspect better efficiency (linear) can be had with a pre-computed weighing of the d terms below.



                The concern about seg faulting due to excessive recursion would be mitigated with the above improvement.



                I also change function bbezier() to code.



                long double bezier_alt1(long double t, const uint64_t *points, size_t n) {
                assert(n);
                long double omt = 1.0 - t;
                long double d[n]; // Save in between calculations.

                for (size_t i = 0; i < n; i++) {
                d[i] = omt * points[i] + t * points[i + 1];
                }
                while (n > 1) {
                n--;
                for (size_t i = 0; i < n; i++) {
                d[i] = omt * d[i] + t * d[i + 1];
                }
                }
                return d[0];
                }


                [Edit2]



                A linear solution O(n) is possible with O(1) additional memory.



                long double bezier_alt2(long double t, const uint64_t *points, size_t n) {
                assert(n);
                long double omt = 1.0 - t;
                long double power_t = 1.0;
                long double power_omt = powl(omt,n);
                long double omt_div = omt != 0.0 ? 1.0/omt : 0.0;

                long double sum = 0.0;
                unsigned long term_n = 1;
                unsigned long term_d = 1;
                for (size_t i = 0; i < n; i++) {
                long double y = power_omt*power_t*points[i]*term_n/term_d;
                sum += y;
                power_t *= t;
                power_omt *= omt_div;
                term_n *= (n-i);
                term_d *= (i+1);
                }
                power_omt = 1.0;
                long double y = power_omt*power_t*points[n]*term_n/term_d;
                sum += y;
                return sum;
                }


                Additional linear simplifications possible - perhaps another day.





                Minor stuff



                Use const



                A const in the referenced date allows for some optimizations, wider application and better conveys code's intent.



                // long double bezier(long double t, uint64_t *points, uint64_t n){
                long double bezier(long double t, const uint64_t *points, uint64_t n){


                Excessive wide type



                With uint64_t n, there is no reasonable expectation that such an iteration will finish for large n.



                Fortunately, n indicates the size of the array. For array indexing and sizing, using size_t. It is the right size - not too narrow, nor too wide a type.



                // long double bezier(long double t, uint64_t *points, uint64_t n){
                long double bezier(long double t, uint64_t *points, size_t n){


                For this application, certainly unsigned would alway suffice.



                static



                Good use of static in static long double bbezier() to keep that function local.



                Missing "bezier.h"



                I'd expect a bezier() declarations in a .h file and implementation in the .c file instead of extern "C" long double bezier(long double, uint64_t *, uint64_t); in main.c



                // extern "C" long double bezier(long double, uint64_t *, uint64_t);
                #include "bezier.h".


                n range check



                Perhaps in a debug build, test n.



                long double bezier(long double t, const uint64_t *points, uint64_t n){
                assert( n > 0); // Low bound
                assert( n < 1000); // Maybe an sane upper limit too






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 20 hours ago

























                answered 23 hours ago









                chuxchux

                12.8k21344




                12.8k21344






























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