Simple basic spiral code implementation 2 with tolerance 0.1

Simple basic spiral code implementation 2 with tolerance 0.1

This is the sequel of the previous code implementation of spiral, that time I made a very simple archimedean spiral with the same step over. But sometime, we want more than a step over, a distance between each point of a spiral (tolerance).

Here is my code implementation in c.

 /**********************************************************************  
  *  
  *  
  *    @par    Copyright(C)  
  *    2018     Tokyo University of Agriculture, Saville Lab.  
  *    All rights reserved.  
  *  
  *    @par    History:  
  *    - 2018/07/15    Saville Ramadhona  
  *        - first code.  
  *  
  ***********************************************************************/  
 /*  
  * Program to make spiral curve and put the point on spiral  
  * based on tollerance (point distance).  
  * Not based on parameter of spiral curve.  
  * eg. When tollerance is o.1, distance between adjacent points on spiral   
  * must not exceed the tollerance.   
  *  
  */  
   
   
 #include<stdio.h>  
 #include<math.h>  
   
 #define M_PI 3.14159265358979323846  
   
 /**  
  *    function to evaluate/compute point on spiral  
  *  
  *    @param[in]    param    spiral parameter  
  *    @param[in]    theta    angle in computed point (radians)  
  *    @param[out]    pt        array coordinate of computed point on spiral (x,y)   
  *    @return                radius of center point of spiral to computed point  
  */  
 double evalSpiralpt(double param, double theta, double pt[2])  
 {  
     double r = param*theta;  
     pt[0] = r * cos(theta);  
     pt[1] = r * sin(theta);  
     return r;  
 }  
   
 /**  
  *    function to check distance square between adjacent points  
  *    we do not use sqrt to make computation faster  
  *  
  *    @param[in]    point        newly computed point on spiral   
  *    @param[in]    prevpoint    previously computed point on spiral   
  *    @return                    distance square obetween adjacent points  
  */  
 double checkDistpt2(double point[2], double prevpoint[2])  
 {  
     return (point[0]-prevpoint[0])*(point[0]-prevpoint[0])   
             + (point[1]-prevpoint[1])*(point[1]-prevpoint[1]);  
 }  
   
 //spiral definition r = a + bθ  
 void main()  
 {  
     //remove file if exist  
   int status;  
   status = remove("C:\\Users\\BA17655\\Documents\\spiral.csv");  
     
   //prepare csv file to see the computation result  
   FILE *fp;  
   fp = fopen("C:\\Users\\BA17655\\Documents\\spiral.csv", "a");  
   
   fprintf(fp, "rad, theta, r, x, y, d\n");  
   
     double pi_2 = 2* M_PI;  
     int numTurns = 5;                                      //automatically determine by detecting outer edge, later!  
     double stepover = 0.1;                                 //tool path stepover (USER INPUT)  
     double distanceBetweenTurns = stepover * (1 / (pi_2)); //b  
     double theta = 0.0;                                    //starting angle  
     double offset = 0.0;                                   //a (offset of starting point)  
   
     /*  
     double pointsPerTurn = 20; //to be modified  
     for(int i=0; i<(pointsPerTurn*numTurns+1); i++)  
     {  
         double r = offset + (distanceBetweenTurns*theta);  
         double point[2];  
         point[0] = r * cos(theta);  
         point[1] = r * sin(theta);  
   
         printf("%f %f %f %f \n", theta*(180.0/M_PI), r, point[0], point[1]);  
         theta += pi_2 / pointsPerTurn;  
     }  
     */  
   
     double dist_thres2 = 0.01;          //distance threshold(control point tollerance - USER INPUT)  
     int cnt = 0;  
     double point[2];                    //x, y coordinate control point on spiral  
     double prevpoint[2];  
     double prevtheta;  
     double degree;  
     double sizelimit = numTurns * 360.0; //to sizelimit the size of spiral  
     do  
     {  
         //evaluate point in spiral   
         double r = evalSpiralpt(distanceBetweenTurns, theta, point);  
   
         if( cnt < 1 )  
         {  
             fprintf(fp, "%f, %f, %f, %f, %f\n", theta, theta*(180.0/M_PI), r, point[0], point[1]);  
             theta += pi_2 * dist_thres2;  
         }  
         if( cnt > 0 )  
         {  
             //check distance of control point  
             double dist2 = checkDistpt2(point, prevpoint);  
   
             //check if point distance exceed tolerance  
             if( dist2 > dist_thres2 )   
             { //if exeec tollerance, compute new control point  
                 do   
                 {  
                     //make the new point to be located in the middle of previous and current point  
                     theta = (theta + prevtheta) * 0.5;  
                     r = evalSpiralpt(distanceBetweenTurns, theta, point);  
                     dist2 = checkDistpt2(point, prevpoint);  
   
                     //when above is not enough, do it one more time!  
                     if( dist2 > dist_thres2 )  
                     {  
                         theta = (theta + prevtheta) * 0.5;  
                         r = evalSpiralpt(distanceBetweenTurns, theta, point);  
                         dist2 = checkDistpt2(point, prevpoint);  
                     }  
                 }while( dist2 > dist_thres2 ); //when point distance lies under the tollerance, exit loop  
             }  
   
             fprintf(fp, "%f, %f, %f, %f, %f, %f\n", theta, theta*(180.0/M_PI), r, point[0], point[1], dist2);  
             prevtheta = theta;  
             theta += pi_2 * dist_thres2;  
         }          
           
         cnt++;  
                   
         prevpoint[0] = point[0];  
         prevpoint[1] = point[1];  
         degree = theta*(180.0/M_PI);  
   
     }while( degree <= sizelimit );  
   
 }  

After running the code, it will print out the angle (radian), angle (degree), r of each point(interpreted as 1/curvature), x, y coordinate of each point, and the distance of each point.
The plotted spiral from the code in excel is as shown below.

Pretty simple right!

spiral 2

Simple basic archimedean spiral code implementation

Recently, I have an interest to draw spiral with the same step over (distance between one spiral cycle).
Actually, it is pretty simple to draw basic archimedean spiral. The mathematical formula for archimedean spiral is
r = a + bθ,
with r is radius at each spiral point, a is offset of starting point of spiral and b is the coefficient of angle θ. Then for the point at spiral, it is the same like parametric circle
x coordinate = r * cos(theta) and
y coordinate = r * sin(theta).
Here is my code implementation of simple basic archimedean spiral.

 #include<stdio.h>  
 #include<math.h>  
   
 #define M_PI 3.14159265358979323846  
   
 //spiral definition r = a + bθ  
 void main()  
 {  
     int numTurns = 5;                                      //number of spiral cycle 
     double stepover = 0.1;                                 //distance between one spiral cycle   
     double distanceBetweenTurns = stepover * (1/(2*M_PI)); //b 2*PI is 360deg  
     double theta = 0.0;                                    //starting angle, all in radian  
     double offset = 0.0;                                   //a (offset of starting point)  
       
     double pointsPerTurn = 20; //number of point in each spiral cycle  
     printf("rad, theta, r, x, y\n");  
     for(int i=0; i<(pointsPerTurn*numTurns+1); i++)  
     {  
         double r = offset + (distanceBetweenTurns*theta);  
         double point[2];  
         point[0] = r * cos(theta);  
         point[1] = r * sin(theta);  
   
         //print out the point to be plotted   
         printf("%f, %f, %f, %f, %f\n",theta, theta*(180.0/M_PI), r, point[0], point[1]);  
         theta += 2*M_PI / pointsPerTurn;  
     }      
 }  

After running the code, it will print out the angle (radian), angle (degree), r of each point(interpreted as 1/curvature), x and y coordinate of each point.
The plotted spiral from the code in excel is as shown bellow.
Pretty simple right!

*NB bonus:This kind of spiral are used for making spiral toolpath. Usually, projected onto target surface.
But, of course we have to modify the point at the spiral, because distance between each point gets larger and larger as the spiral grow. For this I am going to save the code for the next post.

Browser Comparison: Firefox quantum vs Chrome

when working with a quite memory usage program, sometimes i also need to browse, that was when the idea of browser comparison came up
i usually have both of firefox and chrome just in case one of them have an issue or something, so let's compare them
yes, i know that microsoft also have its internet exploler or the newest microsoft edge, but as we know that both of them are out of the league, i didn't included them
for opera, i know opera is quite good as well but let's stick with firefox and chrome for righ now

quick snap result
chrome's speed is slightly faster (to be honest, the speed difference is not that significant to be written here)
but, firefox uses MUCH less memory (this is what i need)

both are the newest version up to this date (2018/06/18)
firefox quantum 60.0.2 (64 bit)
chrome 67.0.3396.87 (64 bit)

during the test i already open several applications which take 3.5gb
(clover 9 tabs, sticky notes windows native, power point, excel, outlook, cygwin)

then i opened firefox and chrome with 20 tabs of the same address

after opening firefox, the memory usage became 3.9gb, and close the firefox
but after opening chrome, the memory usage became 5.4gb!
i also opened both of firefox and chrome using the exact same 20 tabs, the memory usage became 6.0gb

for me, who consider less memory usage, firefox is the clear winner (for now)
let's see what the big g is going to do with the next chrome


----------------
test environment
i3-6100 3.7ghz
win 7 64 bit sp 1
8gb memory


ブラウザー比較：ファイヤーフォックスクァンタム　vs　クローム

結論
クロームの速度はファイヤーフォックスよりも少し上回るが(速度はここに記載するまでの差はない)、
ファイヤーフォックスはクロームよりも非常に少ないメモリを使用


2018/06/18時点で両ブラウザーとも最新のバージョン
ファイヤーフォックスクァンタム 60.0.2 (64 bit)
クローム 67.0.3396.87 (64 bit)

テストをしていた時には既に3.5gbのメモリを使用

ファイヤーフォックスを20タブ開いたら3.9gbのメモリ使用になり、ファイヤーフォックスを閉じて、クロームを同じ20タブを開いたらなんと5.4gbのメモリ使用になった！

現時点ではファイヤーフォックスクァンタムの勝ち
グーグルはファイヤーフォックスクァンタムに対し、今後何をするだろうかをお楽しみに

1. playing with string manipulation, 2. function to sort dynamically alocated string array in visual c++

 #include "stdafx.h"  
 #include <string.h>     //header for string manipulation  
 #include <cstdlib>     //header for malloc  
   
 #define COL 300  
 bool sortStr(char(*str)[COL], int n)  
 {     //function to sort string array   
      if( str == NULL )     return false;  
      char tmp[COL];  
      for(int i=0; i<n-1; ++i)  
      {  
           for(int j=i+1; j<n; ++j)  
           {  
                if( strcmp(str[i], str[j]) > 0 )  
                {                      
                     strcpy_s(tmp, str[i]);  
                     strcpy_s(str[i], str[j]);  
                     strcpy_s(str[j], tmp);  
                }  
           }  
      }  
      return true;  
 }  
 int _tmain(int argc, _TCHAR* argv[])  
 {  
      int ncp = 4;                     //for string array  
      //dynamic allocation of string array
      char(*str0)[COL] = (char(*)[COL])malloc(sizeof(char)*ncp*COL);  
   
      int id0 = 9, id1 = 12;       
      strcpy_s(str0[0], "s");          //input your string  
      strcpy_s(str0[1], "e");  
      char tmp[COL];  
      sprintf_s(tmp, "%d", id1);      //input int value (id1) into string tmp  
      strcat_s(str0[0], tmp);         //concatenate multiple strings into str[0]  
      sprintf_s(tmp, "%d", id0);  
      strcat_s(str0[1], tmp);  
      strcpy_s(str0[2], "e1");  
      strcpy_s(str0[3], "s2");       
      for(int i=0; i<ncp; ++i)  
           printf("str0[%d]:%s\n", i, str0[i]);  
      printf("--\n");  
             
      char(*str1)[COL] = (char(*)[COL])malloc(sizeof(char)*ncp*COL);  
      strcpy_s(str1[0], "e9");  
      strcpy_s(str1[1], "s12");  
      strcpy_s(str1[2], "e1");  
      strcpy_s(str1[3], "s3");  
   
      for(int i=0; i<ncp; ++i)  
           printf("str1[%d]:%s\n", i, str1[i]);  
   
      printf("\n=======compare=======\n");  
      printf("====before sorting===\n");  
      for(int i=0; i<ncp; ++i)  
      {  
           int res = strcmp(str0[i], str1[i]);     //compare 2 strings value based on ASCII  
           printf("str[%d]: %d\n", i, res);  
      }  
   
      printf("\n=======sort=======\n");  
        
      if( sortStr(str0, ncp) == true )  
      {  
           for(int i=0; i<ncp; ++i)  
                printf("str0[%d]:%s\n", i, str0[i]);  
      }  
             
      printf("--\n");  
        
      if( sortStr(str1, ncp) == true )  
      {  
           for(int i=0; i<ncp; ++i)  
                printf("str0[%d]:%s\n", i, str1[i]);  
      }       
   
      printf("\n=======compare=======\n");  
      printf("====after sorting====\n");  
      for(int i=0; i<ncp; ++i)  
      {  
           int res = strcmp(str0[i], str1[i]);  
           printf("str[%d]: %d\n", i, res);  
      }  
 } 

result

str0[0]:s12
str0[1]:e9
str0[2]:e1
str0[3]:s2
--
str1[0]:e9
str1[1]:s12
str1[2]:e1
str1[3]:s3

=======compare=======
====before sorting===
str[0]: 1
str[1]: -1
str[2]: 0
str[3]: -1

=======sort=======
str0[0]:e1
str0[1]:e9
str0[2]:s12
str0[3]:s2
--
str0[0]:e1
str0[1]:e9
str0[2]:s12
str0[3]:s3

=======compare=======
====after sorting====
str[0]: 0
str[1]: 0
str[2]: 0
str[3]: -1

centripetal parametrization for curve fitting or interpolation

very useful to compute parameter for curve fitting or interpolation
the detail explanation is written here


and here is the code for three dimensional parametric curve

 void GetCentripetal_Param(  
       int ncp,                                                  //number of control point  
       double point[][3],                                        //control points of curve  
       double t[]                                                //array of t parameter (output) 
       )  
 {  
      double *dcp = (double*)malloc(sizeof(double) * (ncp-1));   //CP0-CP1, CP2-CP1, ..., CP[ncp-1][ncp]  
      double *sum = (double*)malloc(sizeof(double) * ncp);  
             
      int i, j;  
      double dcptotal = 0.0;  
      for(i=0; i<ncp-1; i++)  
           dcp[i] = 0.0;  
        
      sum[0] = 0.0;  
      for(i=0; i<ncp-1; i++)  
      {  
        for(j=0; j<3; j++)  
          dcp[i] += (point[i+1][j]-point[i][j]) * (point[i+1][j]-point[i][j]);        
       
        dcp[i] = sqrt(sqrt(dcp[i]));  
        dcptotal += dcp[i];  
        sum[i+1] = dcptotal;  
      }  
      sum[0] = 0.0;  
      double inv_dcptotal = 1 / dcptotal;  
      for(i=0; i<ncp; i++)  
         t[i] = (sum[i] * inv_dcptotal);  
        
      if(dcp) free(dcp);  
      if(sum) free(sum);  
 }     

simple example of using operator overloading in c++

 #include <stdio.h>  
 #include <stdlib.h>  
   
 //3d point  
 struct Pt3d  
 {  
      double x,y,z;  
   
      //define operator "+" for Pt3d  
      Pt3d operator+(Pt3d &p)  
      {  
           Pt3d temp = {  
                this->x + p.x,   
                this->y + p.y,   
                this->z + p.z,   
           };  
           return temp;  
      }  
 };  
   
   
 int _tmain(int argc, _TCHAR* argv[])  
 {  
      Pt3d a, b, c;  
      a.x = 1.0;  
      a.y = 1.1;  
      a.z = 1.2;  
   
      b.x = 2.0;  
      b.y = 2.1;  
      b.z = 2.2;  
        
      //by only using this, you can get addition of point a and b  
      c = a + b;  
        
      //let's check it  
      //result should be 3.00, 3.20, 3.40  
      printf("coordinate of c in a, y, z is as follow:\n");  
      printf("%.2f, %.2f, %.2f \n", c.x, c.y, c.z);  
      return 0;  
 } 

result:

 
3.00, 3.20, 3.40 

compute derivative coefficient of n-th degree polynomial equation

 #include <stdio.h>  
 #define ELE 4  
   
 int main()  
 {  
  int i;  
  int g = 3; //degree  
  double c[ELE][3] = {{1, 5, 9},   
                      {2, 6, 10},  
                      {3, 7, 11},  
                      {4, 8, 12}};   
    
  //ELE coefficient of parametric equation  
  //eg ELE == 0 4x^3 + 3x^2 + 2x + 1 2d element of array is x, y, z  
    
  for(int i=0; i<ELE; i++)  
  {  
   for(int j=0; j<3; j++)  
     printf("%.1f ", c[i][j]);  
   printf("\n");  
  }  
  printf("\n--\n");  
   
  //derivative  
  int n = 2; //n-th derivative   
  for(int drv=1; drv<=n; drv++)  
   for(i=g; i>=0; i--)  
     for(int axs=0; axs<3; axs++) //if we want the derivative not be 3 dimensional parametric equation delete this loop  
      c[i][axs]=c[i][axs]*(i-(drv-1));  
  //derivative  
   
  for(int i=0; i<ELE; i++)  
  {  
   for(int j=0; j<3; j++)  
     printf("%.1f ", c[i][j]);  
   printf("\n");  
  }  
   

2d dynamic array with known number of column element

how to make dynamic array with known number of column element
for example for a 3d coordinate.

suppose i need to make 2d dynamic array for a 3d coordinate, because the element of 3d coordinate is x, y, z so the number column is 3.
let's say i need to have
double point[some_number][3];
and that number is suppose unknown until we read the file, lets say ptscount.

one simple way to allocate the dynamic 2d array is as follow. 


double (*point)[3] = (double (*)[3])malloc(sizeof(double) * ptscount*3);

brief explanation
- double (*point)[3]: double variable of  "point" that has 3 array element, this will make point[ptscount][3]

- (double (*)[3]): casting to double of pointer that has 3 array element

- malloc(sizeof(double) * ptscount*3): memory allocation for (the size of double variable) times(×) (ptscount (number or row of the array) with 3 column)

after using it you have to free the memory
free (point);

use malloc for struct

 #include <stdio.h>  
 #include <stdlib.h>  
   
 struct Healdirect  
 {  
    double m_tors;  
    double curvedir[3];  
 };  
   
 void showme(struct Healdirect *healdeci)  
 {     
    struct Healdirect test;  
   
    test.m_tors = healdeci->m_tors;  
    test.curvedir[0] = healdeci->curvedir[0];  
    test.curvedir[1] = healdeci->curvedir[1];  
    test.curvedir[2] = healdeci->curvedir[2];  
   
    printf("torsion: %.2f \n", test.m_tors);  
    printf("curvature direction: ");  
    for(int i=0; i<3; i++)  
       printf("%.2f ", test.curvedir[i]);  
    printf("\n");  
 }  
   
 void main()  
 {  
    //example of struct using pointer and malloc  
    struct Healdirect *hd = malloc(sizeof(hd));  
      
    //sample data   
    hd->m_tors = 0.24;  
   hd->curvedir[0] = 0.99;  
   hd->curvedir[1] = 0.88;  
   hd->curvedir[2] = 0.77;  
   
   //show sample data after the memory is filled  
   showme(hd);  
   
   //free the memory  
   if(hd) free(hd);  
 }  
   

dynimcally allocated array 2

 #include <stdio.h>  
 #include <stdlib.h>  
   
 void main()  
 {  
    double *ary;   //will be your dynamic 1d array  
    int num;       //your array element  
   
    //input your number of array element  
    printf("num: ");  
    scanf("%d", &num);  
   
    //syntax: number of block address times byte needed (sizeof) for double variable  
    ary = malloc(num * sizeof(double));     
   
    //fill the array (can be changed to any number) and display them  
    for(int i=0; i<num; i++)  
    {  
       ary[i] = i * i * 2.8;  
       printf("%.2f\n", ary[i]);  
    }  
   
    //must free the memory allocation to not make memory leak  
    if(ary) free(ary);  
 }  
   

dynamically allocated 2d array using an array of pointers

 #include <stdio.h>   
 #include <stdlib.h>   
     
  void main()   
  {   
    int row = 5, col = 3;   
    int i, j;   
    double *ar[row]; //will be your 2d array   
   
    for(i=0; i<row; i++)  //make it into 2d array
         ar[i] = malloc(sizeof(int)*col);  //get the column number of array
   
    for(i=0; i<row; i++)  
    {  
         for(j=0; j<col; j++)  
         {  
              ar[i][j] = (i + j) * 1.0;  //filling the array element
         }  
    }  
     
    for(i=0; i<row; i++)  
    {  
         for(j=0; j<col; j++)  
         {  
          printf("[%d][%d]=%.2f ", i, j, ar[i][j]);   //print the array after filling it
         }  
         printf("\n");  
    }  
   
    //must free the memorry allocation after using it  
    for(i=0; i<row; i++)  
         free(ar[i]);      
  } 

result:

[0][0]=0.00 [0][1]=1.00 [0][2]=2.00
[1][0]=1.00 [1][1]=2.00 [1][2]=3.00
[2][0]=2.00 [2][1]=3.00 [2][2]=4.00
[3][0]=3.00 [3][1]=4.00 [3][2]=5.00
[4][0]=4.00 [4][1]=5.00 [4][2]=6.00

dynamically allocated array

 #include <stdio.h>  
 #include <stdlib.h>  
   
 void main()  
 {  
     double *ar; //will be your array  
     double elem[] = {0.1, 0.2 ,0.3 ,0.4 ,0.5};   
     int num = 5, i;  
   
     printf("array number: %d\n", num);  
     for(i=0; i<num; i++)  
         printf("[%d]:%.2f\n", i, elem[i]);  
   
     printf("lets double it and put in your new array using malloc\n");  
     printf("--\n");  
   
     ar = malloc(sizeof(double)*num); //allocate the memory for your array  
   
     for(i=0; i<num; i++){  
       *(ar+i) = elem[i] * 2;  
     }  
   
     for(i=0; i<num; i++)   
         printf("array[%d]=%.2f\n", i, ar[i]);  
   
 free(ar); //must free the memorry allocation after using it  
 } 

result：

-------------- 

 array number: 5
[0]:0.10
[1]:0.20
[2]:0.30
[3]:0.40
[4]:0.50
lets double it and put in your new array using malloc
--
array[0]=0.20
array[1]=0.40
array[2]=0.60
array[3]=0.80
array[4]=1.00 

binomial coefficient c code

 #include＜stdio.h＞  
   
 //binomial coefficient function  
 double Ni(  
         int deg,    //degree   
         int b       //for binomial  
         )  
 {  
     if( deg < 0 )  
         return 0.0;  
     else if ( deg == 0 )  
         return 1.0;  
     else{  
         double temp = 1.0;  
         for(int i=1; i<=b;i++ ){  
             temp = temp* (double)(deg-i+1)/i;  
             printf("%f\n", temp);  
         }  
         return temp;  
     }  
 }  
   
   
 void main ()  
 {  
     int a=10, b=3;  
     double bin;  
     bin = Ni(a, b);  
     printf("\n=============\n");  
     printf("%f\n", bin);  
 }  

comparison of secant and bisection method

   
 #include<stdio.h>  
 #include<math.h>  
 #include<time.h>  
   
 //example of function f(t), you can make any function you like  
 float f(float t)  
 {  
   return(((48.2291-30.6208*t-43) * (48.2291-30.6208*t-43)) + 35*t - 25.0);   
 }  
   
 void main()  
 {  
     
   float a = 0.1, b = 0.3, c = 0, e = 7.5e-3;  
   //a, b interval  
   //e error  
   
   int count = 1, n = 50; //n=maximum iteration  
     
   //  
   //secant  
   //  
   clock_t begin = clock();  
   do  
   {      
     c = (a*f(b) - b*f(a)) / (f(b) - f(a));  
     //c = b - (f(b)*(b-a) / (f(b) - f(a)));  
     a=b;  
     b=c;  
     printf("i No-%d  t=%f  d=%f\n",count,c, f(c));  
     count++;  
     if(count==n) break;  
   } while(fabs(f(c))&gt;e);  
   
   clock_t end = clock();  
   double time_spent = (double)(end - begin) / CLOCKS_PER_SEC;  
   printf("time:%f", time_spent);  
   
   printf("\nThe required solution is %f\n",c);  
   printf("d = %f\n", f(c) + 25);  
   
   //  
   //bisection  
   //  
   double fa, fb, fc;  
   a = 0.1; b = 0.3; c = 0; e = 7.5e-3;  
   count = 1;  
   clock_t begin2 = clock();  
   do{  
     c = (a+b)/2;  
     fa = f(a);  
     fb = f(b);  
     fc = f(c);  
   
     if(fa*fc &lt; 0){  
       b = c;  
     }  
     else{  
       a = c;  
     }  
     printf("i No-%d  t=%f  d=%f\n", count, c, fc);  
     count++;  
     if(count==n) break;  
   } while(fabs(fc)&gt;e);  
   
   clock_t end2 = clock();  
   double time_spent2 = (double)(end2 - begin2) / CLOCKS_PER_SEC;  
   printf("time:%f", time_spent2);  
   
   printf("\nThe required solution is %f\n",c);  
   printf("d = %f\n", fc + 25);  
   
    
 }  

dynamic 2d array using double data type

 #include <stdio.h>  
 #include <iostream>  
   
   
 int main() {  
  // dimensions  
  int N = 74;  
  int M = 3;  
   
  int i, j;  
    
  // dynamic allocation  
  double** ary = new double*[N];   //2d array  
  for(i = 0; i < N; ++i)         //↑  
    ary[i] = new double[M];      //1d array  
    
  for(i=0; i<N; i++){  
       for(j=0; j<M; j++){  
          ary[i][j] = i+j;  
          printf("point[%d][%d] = %.2f \n", i, j, ary[i][j]);  
       }  
    }  
    
  // free  
  for(i = 0; i < N; ++i)  
   delete [] ary[i];  
  delete [] ary;  
    
  return 0;  
 }  

//reffer to [http: codeformatter.blogspot.jp]

passing and returning array in c language

#include                                         //standard library
#include

//first derivative function
static double* GetDerivative(double t, double A[], double B[], double C[])       
                                                              //the function must be pointer double*
                                                              //for returning array

{
    //P・(t) = 3At^2 + 2Bt + C
    static double Pdot[3] = {0};                      //must be static
   
    Pdot[0] = 3*A[0]*t*t + 2*B[0]*t + C[0];
    Pdot[1] = 3*A[1]*t*t + 2*B[1]*t + C[1];
    Pdot[2] = 3*A[2]*t*t + 2*B[2]*t + C[2];
   
    return Pdot;                                       //return the array
}
//first derivative function

int main ()
{
    //P(t) = At3 + Bt2 + Ct + D
    ////P・(t) = 3At^2 + 2Bt + C
    double *derivative = 0;                        //must be pointer double *
                                                          //because the function is a pointer as well

    //double t[4] = {0.0, 0.25, 0.5, 0.75};
    double t[1] = {1};
    double A[] = {1, 2, 3};
    double B[] = {4, 6, 6};
    double C[] = {7, 8, 9};
    double D[] = {10, 11, 12};

    derivative = GetDerivative(t[0], A, B, C);    //put the return function to derivative
                                                              //plus make it as an array
                                                              //do not add [] even if it is array,
                                                              // unless you want to point the specific
                                                              // address of the array like t[0]
   
    printf("%ft2 + %ft + %f \n", derivative[0], derivative[1], derivative[2]);

    return 0;
}

//output 18.000000t2 + 26.000000t + 30.000000

make group (aggregate) data set in R

Suppose you have a data set called sort2013

And you want to calculate the group/aggregate of each Head_age

> sort2013

  Head_age P2_age P3_age P4_age P5_age P6_age P7_age P8_age

1       84     75      0      0      0      0      0      0

2       84     75      0      0      0      0      0      0

3       84     75      0      0      0      0      0      0

4       84     75      0      0      0      0      0      0

5       84     75      0      0      0      0      0      0

6       85     76      0      0      0      0      0      0

Etc...

Do the aggregate or grouping the data frame with aggregate()

> group2013 = aggregate(. ~ Head_age, sort2013, mean)

#aggregate or group of .(all column) ~(by) Head_age, in sort2013 data frame, and calculate the mean of all column. Then input into group2013 data frame

The result would be

> group2013

  Head_age   P2_age   P3_age   P4_age P5_age P6_age P7_age P8_age

1       16  0.00000 0.000000 0.000000      0      0      0      0

2       17  5.00000 0.000000 0.000000      0      0      0      0

3       18 13.44444 0.000000 0.000000      0      0      0      0

4       19 22.38095 7.000000 0.000000      0      0      0      0

5       20 17.45455 0.000000 0.000000      0      0      0      0

6       21 42.60000 6.666667 8.466667      0      0      0      0

Etc...

The result would be automatically sorted as you can see above

sort data in each row in R data set

Suppose we have this kind of data A

And want to sort from the maximum until the minimum value each row

> A

    [1]      [2]          [3]          [4]

[1] 3       7             9              5

[2] 7       9            11             3

[3] 5       3            7               8

We want to sort A by row and resulting like this

    [1]      [2]          [3]          [4]

[1] 3       5             7             9

[2] 3       7             9            11

Etc....

We can sort by row by doing this

>Asorted = t(apply(A,1,sort))  

#sort the data A each row and input to Asorted data

apply(A,1,sort) means apply for A data set, in 1 (row), and sort them
*1: row, 2: column and (1:2): both of row and column

Yet it will be like this. It will be sorted from the minimum value to the maximum value in each row  (reverse)

    [1]      [2]          [3]          [4]

[1] 3       5             7             9

[2] 3       7             9            11

Etc....

What we need to do next is just reverse it by this command

>Asorted = Asorted[ ,ncol(Asorted):1]    

#reverse sequence in column index, and input it (rewrite) to Asorted data

There you go, the result would be like this

>Asorted

    [1]      [2]          [3]          [4]

[1] 9       7             5            3

[2] 11     9             7            3

Etc....

never trust excel!

Conclusion: Never trust excel, even for a bit complicated statistics!

As a "wanna-be-researcher", of course, I am quite close with data and number. Well, sometimes I use Excel to do some simple things on regards to the data, like drawing a graph, and so on.
But you know what?!
Today my professor and I did several things using excel, and we got real surprised. Imagine... We only did the simple linear trendline, it was only a super simple thing, I mean a real simple thing. The result was y = 29018x - 31865. That was okay, but then when we tested, it went completely wrong!
After that we decided to manually calculate the slope and the intercept. And yes, it was wrong, it should be -318658. We found out that the last digit was missing!
What the heck!

No more excel for this kind of thing!



結論から言うとちょっと複雑統計やればはエクセル信用できません！

”研究者に目指す”わたくしが、当然、データー、数値等に接することが多いです。シンプルなことならたまにはエクセルでグラフ描いたり、シンプルな計算したりするのです。
今日先生と相談していた時に、いろんなことエクセルでやってみたのですが、変なことがあって、びっくりしましたよ。想像して欲しいんです…散布図描いて、線形近似曲線追加したら y = 29018x - 31865という結果が出ました。出た時には「あぁ、出たね」と思っただけ、チェックしたら「あれ？！全然NGじゃんかよ！」っと
そしたら手動でスロープと切片計算したら、全然違いましたよ、マイクロソフトさん。正しかったのは　- 318658で、最後の桁がなかったんじゃないですか。
アホか！

こんなことはやっぱりエクセルでNG！

change R language

As you know that R is a super powerful software

The story was started since I installed R software like a year ago and never use it

Then I started to use it for some period and I realized something, "the language is in Japanese!"　Well, I am okay with Japanese, but what if I got result and need to put it in the paper?!

After searching, I got some inspiration in my dream <<== a big liar

皆さんご存知だと思いますが、Rっというものはスーパーパワーフールソフトでございま~す。

実は一年前にインストールしましたが、全然使わなくてですね…

それで必要なので使い始めましたが、”えぇ？！これ日本語じゃないですか！”まぁ、日本語はまあまあなんとなく読めますし、問題ないかと思っていたら…Rの結果を論文に載せようとしたらまずいですね。

それである時に、夢でインスピレーションが来ました <<== 嘘に決まってるでしょう。

とにかくこういうやり方です。

This is the Japanese screen shot

Well, what you need to do is close the R, go to your desktop, right click on it, choose property and put language=en in the property link and click OK

en is for English, it means Italian and so on.

Rを閉じて、デスクトップにRアイコンを見つけて、右クリック、プロパティ、リンク先のところに「language=en」と入れておいてください。それで、OKをクリックしてください。

en=英語、it=イタリア語等

as you can see there⇩⇩⇩

Open your R

voila! 

Rを立ち上げてくください

じゃじゃじゃ~ン

Your R is in English man!

 

 でございます。

Okay then, I need to back to my reality, work!

あぁ！仕事に戻らないといけないですね。