#12 (03/18/2024)

Structures

#include <stdio.h>

int main()
{
	int a[4][2]={{89, 76},{65, 91}, {56, 78}, {49, 56}};

        return 0;
}

#include <stdio.h>

struct student{
        char *ID;
        int Midterm;
        int Final;
        char Grade;
		};
int main()
{
  struct student smith={"1000123456", 89, 98, 'A'},
                 doe={"1000123457", 45, 53, 'F'},
                 jones={"1000123458", 67, 80, 'C'};

  printf("%s\n", smith.ID);
  printf("%c\n", jones.Grade);
  printf("%d\n", doe.Midterm);

  doe.Midterm=79;
  doe.Grade='D';

  printf("%c\n", doe.Grade);

  return 0;
}
As seen in the example above, a member in a structure can be accessed by using the dot (.). 
#include <stdio.h>

struct student{
        char *Name;
        int Midterm;
        int Final;
        char Grade;
		};
int main()
{
  struct student fall[15];
  int i;
  FILE *data;
  data=fopen("junk.dat", "r");

  for (i=0;i<15;i++)
   fscanf(data, "%s %d %d %c", 
        fall[i].Name, &fall[i].Midterm, &fall[i].Final, &fall[i].Grade);
  fclose(data);

  printf("%d\n", fall[0].Midterm);
  printf("%s\n", fall[0].Name);

  return 0;
}

Doe 45 47 F
Jones 89 65 B
Smith 76 79 B
........

#include <stdio.h>

struct student{
        char *Name;
        int Midterm;
        int Final;
        char Grade;
		};
int main()
{
  struct student fall[3]= { {"Jones", 45, 65, 'D'}, {"Doe", 98, 79, 'B'}, {"Smith", 67, 76, 'C'}};
/*
..................
*/
  printf("%s %d %d %c\n", fall[0].Name, fall[0].Midterm, fall[0].Final, fall[0].Grade);
  return 0;
}
It is also possible to use a pointer to a structure as 
#include <stdio.h>

struct student{
        char *Name;
        int Midterm;
        int Final;
        char Grade;
		};
int main()
{
  struct student Smith={"David Smith", 12, 45, 'F'}, *ptr;
  ptr = &Smith;
/*
..................
*/

  printf("%s %d %d %c\n", ptr->Name, ptr->Midterm, ptr->Final, ptr->Grade);
  return 0;
}
Note that a member in a structure can be referred by using ->.
Finally, by using typedef, you can define a structure and declare a variable as that structure type just like integer or float. 
#include <stdio.h>
typedef struct {
        char *Name;
        int Midterm;
        int Final;
        char Grade;
      } student;

int main()
{
  student Jones={"Jones", 12, 45, 'F'}, *ptr;
  ptr = &Jones;
/*
..................
*/

 printf("%s\n", ptr->Name);
  return 0;
}
The concept of structure is extended to the concept of class which plays an essential role in C++ (and Java).
The following code defines the complex type (a + b i) and computes addition of two complex numbers: 
#include <stdio.h>

typedef struct
 {float Real; float Im;} Complex;

Complex ComplexAdd(Complex z1, Complex z2)
{
	Complex z;
	z.Real = z1.Real + z2.Real;
	z.Im = z1.Im + z2.Im;
	return z;
}

int main()
{
	Complex z1, z2, z;

	printf("Enter real and imaginary parts of z1 separated by space = ");
	scanf("%f %f", &z1.Real, &z1.Im);

        printf("Enter real and imaginary parts of z2 separated by space = ");
	scanf("%f %f", &z2.Real, &z2.Im);

	z = ComplexAdd(z1, z2);

	printf("%f + %f I \n", z.Real, z.Im);
	return 0;
}

Closure on C syntax

  1. What happened to the arguments of main() ?
    See Lecture note #10.
  2. What does "return 0;" really mean ?
    The OS requires a return value when you run a program. Depending on the return value, the OS knows whether the program you run exit successfully or not.

Roots of f(x) = 0

  1. Algebraic equations (Example: x5 + 3 x2 − 2 x = 0)
  2. Bisection method (slow, steady)
  3. Newton-Raphson method (fast, risk)
  1. Algebraic equations
    According to the Galois theorem, it is not possible to solve general algebraic equations of the fifth order and beyond in closed form.
    1. Quadratic equations
      a x2 + b x + c = 0,     x =
      −b ±
      √
      b2−4ac
      2a
      		 .
      		(1)
      Example:
      x2 + 200000 x − 3 = 0.
      		(2)
      The exact solutions are
      x1 = −200000,     x2 = 0.000015,
      		(3)
      #include <stdio.h>
#include <math.h>
int main()
{
float a, b, c, disc,x1, x2;
a=1.0;b=200000;c=-3;
disc=b*b-4*a*c;
x1=(-b-sqrt(disc))/(2*a);
x2=(-b+sqrt(disc))/(2*a);
printf("x1 = %f, x2 = %f\n",x1,x2); 
return 0;
}
      What went wrong ??
      From eq.(1), one obtains
      D
      =
      200002 − 4×1.0 ×(−3)
      		(4)
      =
      40000000012,
      		(5)
      √D
      =
      200000.00003
      		(6)
      On the other hand,
      b = 200000.0
      		(7)
      Subtraction of two numbers with similar magnitude may result in cancellation error.
      Solution 1 Change float to double
      Solution 2 Noting that
      a x2 + b x +c
      =
      a(x−x1) (x−x2)
      		(8)
      =
      a (x2 − (x1+x2) x + x1 x2)
      		(9)
      =
      a x2 − a (x1+x2) x+ a x1 x2,
      		(10)
      so

      x1 x2 = c
      a
      		 ,
      		(11)
      hence
      x2 = c
      x1 a
      		 = −3
      −200000
      		 = 0.000015.
      		(12)
      Solution 3 Rationalization:

      x2
      =
      −b + √D
      2a
      		(13)
      =
      (−b−√D)(−b+√D)
      2a (−b − √D)
      		(14)
      =
      b2 − D
      2a (−b − √D)
      		 ,
      		(15)
      so
      x2 = −2000002 − 40000000012
      2 (−200000 − 200000)
      		 = 0.000015.
      		(16)
    2. Cubic equations
      x3 + a x2 + b x+ c = 0.
      		(17)
      (Try MathCheat.)
    3. Quartic equations
      x4 + a x3 + b x2+ c x + d = 0.
      		(18)
      (Try MathCheat.)
  2. Bisection method
    The mean value theorem:
    If f(x) is continuous over x1 ≤ x ≤ x2 and f′(x) exists over x1 < x < x2 and f(x1) f(x2) < 0, there is at least one point x between x1 and x2 such that f(x)=0.
    bisection1.gif
    1. Choose x1 and x2 such that f(x1) f(x2) < 0.
    2. Set x3 ← (x1+x2)/2.
    3. If f(x1) f(x3) < 0, then, set x2 ← x3.
    4. Else set x1 ← x3.
    5. Until |x1 − x2| < ϵ or f(x3) = 0, repeat b-d.
    bisection2.gif 
    /* Compute the square root of 2 */
#include <stdio.h>
#include <math.h>
#define EPS 1.0e-10
#define N 100

double f(double x)
{
return pow(x,2)-2;
}

/* start of main */
int main()
{
double x1, x2, x3;
int count;

printf("Enter xleft and xright separated by space =");
scanf("%lf %lf", &x1, &x2);

/* sanity check */
if (f(x1)*f(x2)>0) {printf("Invalid x1 and x2 !\n"); return 0;}

/* bisection start */
for (count=0;count< N; count++)
 {
 x3= (x1+x2)/2.0;
 if (f(x1)*f(x3)<0 ) x2=x3; else x1=x3;
 if ( f(x3)==0.0 || fabs(x1-x2)< EPS ) break;
}

printf("iteration = %d\n", count);
printf("x= %lf\n", x1);
return 0;
}
    Notes:
    1. fabs is a function to return the absolute value of the argument in math.h (list of functions)
    2. Try plotting a graph by gnuplot first.
    Example: By solving ex=2, obtain approximation of ln 2.
  3. Newton-Raphson Method
    newton1.gif
    The equation of a straight line that passes (a,b) with a slope of m is
    y − b = m (x − a),
    newton2.gif
    so the tangent line that passes (x1, f(x1)) with the slope of f ′(x1) is

    y − f(x1) = f ′(x1) (x−x1).
    newton3.gif

    0 − f(x1) = f ′(x1) (x−x1),

    x = x1 − f(x1)
    f ′(x1)
    		 ,

    x2 = x1 − f(x1)
    f ′(x1)
    		 .

    x3 = x2 − f(x2)
    f ′(x2)
    		 ,
    General forms:

    xn+1 = xn − f(xn)
    f ′(xn)
    		 .
    Example: √2
    By solving f(x) ≡ x2 − 2 = 0 one can obtain a numerical value of √2.

    f(x)
    =
    x2 −2,
    f ′(x)
    =
    2 x.
    		(19)
    Start with x1=2.0 (initial guess), then,
    x1
    =
    2
    		(20)
    x2
    =
    2− f(2)
    f ′(2)
    		 = 2 − 2
    4
    		 = 1.5
    		(21)
    x3
    =
    1.5 − f(1.5)
    f ′(1.5)
    		 = 1.5 − 0.25
    3
    		 = 1.41667
    		(22)
    x4
    =
    1.4167− f(1.4167)
    f ′(1.4167)
    		 = … = 1.4142.
    		(23)
    The convergence is attained after only 4 iterations.
    Algorithm:
    1. Pick an initial guess, x1.
    2. Make sure f ′(x1) ≠ 0.
    3. Repeat
      1. x2 ← x1 − f(x1)/f ′(x1)
      2. x1 ← x2
    4. Until |x1−x2| ≤ ϵ.
    The Newton-Raphson method fails when f ′(xn) is zero. 
    

    #include <stdio.h>
#include <math.h>
#define EPS 1.0e-10

double f(double x)
{
	return x*x-2;
}

double fp(double x)
{
	return 2*x;
}

double newton(double x)
{
	return x - f(x)/fp(x);
}

int main()
{
	double x1, x2;
	int i;

	printf("Enter initial guess  =");
	scanf("%lf", &x1);

	if (fp(x1)==0.0) {
		printf("No convergence.\n");
		return 0;
	}

	for (i=0;i<100;i++)
	{
		x2=newton(x1);
		if (fabs(x1-x2)< EPS) break;
		x1=x2;
	}

	printf("iteration = %d\n", i);
	printf("x= %lf\n", x1);
	return 0;
}

    Some interesting examples

    The following examples are useful in emergency when you do not have an access to a computer.
    1. Square root of a ( = √a)
      Let f(x) = x2 − a, the Newton-Raphson method yields

      xn+1
      =
      xn − xn2 − a
      2 xn
      		(24)
      =
      1
      2
      		 ⎛
      ⎝       		xn + a
      xn
      		⎞
      ⎠       		.
      		(25)
      Example (√3 = 1.732 …)
      x1=1,     x2 = 1
      2
      		 (1 + 3
      1
      		 ) = 2,    x3 = 1
      2
      		 (2 + 3
      2
      		 ) = 1.75,    x4 = 1
      2
      		 (1.75 + 3
      1.75
      		 ) = 1.732.
    2. Inverse of a ( = 1/a)
      Let f(x) = 1/x − a, the Newton-Raphson method yields
      xn+1
      =
      xn −
      1
      xn
      		 −a
      −1
      xn2
      		(26)
      =
      xn (2 − a xn).
      		(27)
      In case you forgot how to do divisions, you can use this formula.
      Example (1/6 = 0.16667…)

      x1 = 0.2,     x2 = 0.2 (2 − 6×0.2)=0.16,    x3 = 0.16 (2 − 6×0.16) = 0.1664,    x4 = 0.1664 (2 − 6×0.1664) = 0.166666.
    3. 1/√a
      Let f(x) = 1/x2 − a,

      xn+1
      =
      xn − 1/xn2 − a
      − 2/xn3
      		(28)
      =
      xn (3 − a xn2)/2.
      		(29)

Summary



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On 17 Mar 2024, 16:18.