Python program - determining odd or even number

#determine the odd or even number by python program
#Odd or Even
num = input("Enter a Number: \n")
if num%2 == 0:
print "Even"
else:
print "Odd"

Python program for sum of numbers, area of a triangle

#Sum of numbers from 1 to 100
sum = 0
for n in range(1,101):
sum = sum + n
print "Sum = ", sum

Result:
Sum = 5050

#Area of a Triangle by Heron's formula
import math
a, b, theta = input('Enter a, b & angle in degree: \n')
theta = math.pi*theta/180 #Convert into radian
c = math.sqrt(a*a + b*b - 2*a*b*math.cos(theta))
s =(a + b + c)*0.5
area = math.sqrt(s*(s - a)*(s - b)*(s - c))
print 'Area = ', area

Python programming for beginner

Writing a Python program is very easy and close to the human language and logic. Python
comes with minimal syntaxes.
Some mathematical operations:-
Let, variable- a=8,b=2
Addition (+): a + b = 10
Subtraction (-): a – b = 6
Multiplication (*): a*b = 16
Division(/): a/b = 4
Modulus (%): a%b = 0 [Remainder, as a is divided by b]
Power or Exponent (**): a**b = 64
Rounding off (//): Example, 7.5//2 = 3.0, where 7.5/2 = 3.75
Variables in python:-
Name of a variable (or any identifier such as class, function, module or other object) can be
anything combined with alphabets, letters and some symbols in the keyboard (except @, $, %
etc.), beginning with a letter (the upper case or lower case letters from A to Z). The names of
the variables are, of course, case sensitive. The variable names, ‘Aa’, ‘aa’, ‘aA' are all different.
But the names cannot be the words given in the following table, used for system commands
and functions. Those are Python keywords (in lower case letters) reserved for the use by the
system.
Some reserved words:
and, assert, break, class, continue, def, del, elif, else,
except, exec, finally, for, from, global, if, import, in,
is, lambda, not, or, pass, print, raise, return, try,
while, with, yield

Numbers:-
Real or Floating [Examples: 19.25, 0.0, -3.23, 1.9e-20, 3.75e2 ]
 Integer [Examples: 6, -23]
 Complex [Examples: 70j, 1.9j, 7+9j, 9.3e-21j]
 Long [Examples: Any number ends with "l" or "L". 123L]

Input/Output(I/O) Statements:
input -
x = 56
y = -190.8
z = 54
x,y,z = 56,-190.8,54
xx = "name" [Alphabetic characters, put inside quotes]
a = input("Give some value")
output-
print x, y, xx, Aa
print “Some statement”
print 'x = ', x

Reading/ Writing in a File
To Open a file"
f1 = open("filename", "w")                             #"w" for writing
f2 = open("filename", "r")                              #"r" for reading
To write in a file
print >> f1, x, y                                            #Writing data: x, y in file indexed ‘f1’
To Close the file
f1.close()

ratio of consecutive terms of fibonocci series - c program

/*ratio of consecutive terms of fibonocci series*/

#include<stdio.h>

#include<math.h>

void

main()

{

                int i;

                float r0,r1,f[100];

                i=0;

                f[0]=1;

                f[1]=1;

                r1=1.00;

label:

                r0=r1;

                i++;

                f[i+1]=f[i]+f[(i-1)];

                r1=f[i+1]/f[i];

                if(fabs(r1-r0)>0.001)

                                goto label;

                printf("the value of the ratio is %f\n",r1);

}

RESULT:

the value of the ratio is 1.618182

Press any key to continue

C program for least square fitting of a straight line

/*least square fitting of a straight line */

#include<stdio.h>

#include<math.h>

void

main()

{

int n,i;

float a,b,x[20],y[20],sx=0.0,sxx=0.0,sy=0.0,sxy=0.0;

printf("enter the total number of data:");

scanf("%d",&n);

printf("enter the value of x:");

for(i=0;i<n;i++)

{

                scanf("%f",&x[i]);

}

printf("enter the values of y:");

for(i=0;i<n;i++)

{

                scanf("%f",&y[i]);

}

for(i=0;i<n;i++)

{

                sx=sx+x[i];

                sxx=sxx+(x[i]*x[i]);

                sy=sy+y[i];

                sxy=sxy+(x[i]*y[i]);

}

a=(n*sxy-sx*sy)/(n*sxx-sx*sx);

b=(sxx*sy-sx*sxy)/(n*sxx-sx*sx);

printf("the line of least fit is:\n y=%fx+%f\n",a,b);

}

Result:

enter the total number of data:5

enter the value of x:1 2 3 4 5

enter the values of y:2.03 4.26 6.59 8.12 10.28

the line of least fit is:

 y=2.036000x+0.148000

Press any key to continue

C program for least square fitting of aexp(bx)

/*least square fitting of aexp(bx) */

#include<stdio.h>

#include<math.h>

void

main()

{

int n,i;

float a,b,x[20],y[20],x1[20],y1[20],sx=0.0,sxx=0.0,sy=0.0,sxy=0.0,c,m,e=2.71828;

printf("enter the total number of data:");

scanf("%d",&n);

printf("enter the value of x:");

for(i=0;i<n;i++)

{

                scanf("%f",&x[i]);

}

printf("enter the values of y:");

for(i=0;i<n;i++)

{

                scanf("%f",&y[i]);

}

for(i=0;i<n;i++)

{ x1[i]=x[i];

y1[i]=log(y[i]);

                sx=sx+x1[i];

                sxx=sxx+(x1[i]*x1[i]);

                sy=sy+y1[i];

                sxy=sxy+(x1[i]*y1[i]);

}

m=(n*sxy-sx*sy)/(n*sxx-sx*sx);

c=(sxx*sy-sx*sxy)/(n*sxx-sx*sx);

a=pow(e,c);

b=m;

printf("the line of least fit is:\n y=%fe^(%fx)\n",a,b);

}

RESULT:

enter the total number of data:5

enter the value of x:1 2 3 4 5

enter the values of y:70 259 920 3340 12000

the line of least fit is:

 y=19.563120e^(1.284523x)

Press any key to continue

C program - Newton Raphson method

/*newton raphson method*/

#include<stdio.h>

#include<math.h>

void

main()

{

                int count=0;

                float x0,xn,fx,dfx;

                printf("enter the guess value:");

                scanf("%f",&x0);

                /*iteration process*/

label: fx=log(x0)-x0+2;

                   dfx=pow(x0,-1)-1;

                   xn=x0-(fx/dfx);

                   if((fabs(xn-x0))<0.001)

                                   printf("\n the final root is %.4f \n",xn);

                   else

                   {x0=xn;

                   count=count+1;

                   if (count>100)

                   {

                                   printf("the solution does not converge");

                                   printf("enter a different guess value:");

                   }

                   else

                                   goto label;

                   }

}

Result1:

enter the guess value:2

 the final root is 3.1462

Press any key to continue

Result2:

enter the guess value:10

 the final root is 3.1462

Press any key to continue

/*least square fitting of a(x^b)*/

#include<stdio.h>

#include<math.h>

void

main()

{

int n,i;

float a,b,x[20],y[20],x1[20],y1[20],sx=0.0,sxx=0.0,sy=0.0,sxy=0.0,c,m,e=2.71828;

printf("enter the total number of data:");

scanf("%d",&n);

printf("enter the value of x:");

for(i=0;i<n;i++)

{

                scanf("%f",&x[i]);

}

printf("enter the values of y:");

for(i=0;i<n;i++)

{

                scanf("%f",&y[i]);

}

for(i=0;i<n;i++)

{

x1[i]=log(x[i]);

y1[i]=log(y[i]);

                sx=sx+x1[i];

                sxx=sxx+(x1[i]*x1[i]);

                sy=sy+y1[i];

                sxy=sxy+(x1[i]*y1[i]);

}

m=(n*sxy-sx*sy)/(n*sxx-sx*sx);

c=(sxx*sy-sx*sxy)/(n*sxx-sx*sx);

a=pow(e,c);

b=m;

printf("the line of least fit is:\n y=(%f)(x,(%f))\n",a,b);

}

Result:

enter the total number of data:5

enter the value of x:1 2 3 4 5

enter the values of y:20 30 38 43 48

the line of least fit is:

 y=(20.295511)(x,(0.546196))

Press any key to continue

C programming for bisection method

/*bisection method*/

#include<stdio.h>

#include<math.h>

#define f(x) ((x*sin(x))+((x-2)*cos(x)))

void

main()

{

                int count;

                float a,b,x,y,y1,y2;

    printf("enter the valu of the intervals:");

                scanf("%f %f",&a,&b);

                y1=f(a);

                y2=f(b);

                count=1;

                if(y1*y2>0)

                                printf("\n invalid interval\n");

                                else

                                {label: x=(a+b)/2;

                                y=f(x);

                                if(y1*y<0)

                                {               b=x;

                                                y2=y;

                                }

                                else

                                {               a=x;

                                                y1=y;

                                }

                                printf("%d\t",count);

                                printf("%f\n",x);

                                count=count+1;

                                if(fabs(b-a)>0.001)

                                                goto label;

                                }

}

RESULT:

enter the valu of the intervals:2 5

1       3.500000

2       2.750000

3       3.125000

4       2.937500

5       2.843750

6       2.890625

7       2.867188

8       2.855469

9       2.849609

10      2.852539

11      2.851074

12      2.851807

Press any key to continue

C program for Newton Raphson method

/*newton raphson method*/

#include<stdio.h>

#include<math.h>

void

main()

{

                int count=0;

                float x0,xn,fx,dfx;

                printf("enter the guess value:");

                scanf("%f",&x0);

                /*iteration process*/

label: fx=(((1-x0*x0)*tan(x0))-x0);

                   dfx=(((1-x0*x0)*(pow(cos(x0),-2)))-(1+(2*x0*tan(x0))));

                   xn=x0-(fx/dfx);

                   if((fabs(xn-x0))<0.001)

                                   printf("\n the final root is %.4f \n",xn);

                   else

                   {x0=xn;

                   count=count+1;

                   if (count>100)

                   {

                                   printf("the solution does not converge");

                                   printf("enter a different guess value:");

                   }

                   else

                                   goto label;

                   }

}

RESULT:

enter the guess value:2 3

 the final root is 2.7437

Press any key to continue

Programming for integration by simpson's 1/3 rule

/*integration by simpson's 1/3 rule*/

#include<stdio.h>

#include<math.h>

float f(float x)

{

float d;

d=(x/(1+pow(x,5)));

return d;

}

void

main()

{

int i,n;

float a,b,h,s=0.0,c;

printf("enter the value of lower and upper limit:");

scanf("%f %f",&a,&b);

printf("enter the number of intervals:");

scanf("%d",&n);

label:

h=(b-a)/n;

c=s;

s=f(a)+f(b)+4*f(a+h);

for(i=3;i<=(n-1);i=i+2)

{

                s=s+(4*f(a+i*h))+(2*f(a+(i-1)*h));

}

s=(h/3.0)*s;

if(fabs (s-c)/s>0.001)

{

                n=n+10;

                goto label;

}

printf("the value of the interval is %.3f and is converged %d intervals\n",s,n);

}

RESULT:

enter the value of lower and upper limit:0 3

enter the number of intervals:10

the value of the interval is 0.648 and is converged 20 intervals

Press any key to continue

Programming for Lagrange's interpolation method

/*lagrange's interpolation method*/

#include<stdio.h>

#include<math.h>

void

main()

{int i,j,n;

float x[20],y[20],x0,sum=0.0,num,deno;

printf("enter the total number of inputs:");

scanf("%d",&n);

printf("enter the value of x:");

for(i=0;i<n;i++)

{

                scanf("%f",& x[i]);

}

printf("enter the corresponding value of y:");

for(i=0;i<n;i++)

{

                scanf("%f",&y[i]);

}

printf("enter the value of x at which y is calculated:");

scanf("%f",& x0);

for(i=0;i<n;i++)

{

                num=1.0;

                deno=1.0;

                for(j=0;j<n;j++)

                {

                                if(j!=i)

                                {

                                                num=num*(x0-x[j]);

                                                deno=deno*(x[i]-x[j]);

                                }

                }

                sum=sum+(num/deno)*y[i];

}

printf("the required value of y is=%f\n",sum);

}

RESULT:

enter the total number of inputs:7

enter the value of x:2 3 4 5 6 7 8

enter the corresponding value of y:2 7 14 23 34 47 62

enter the value of x at which y is calculated:5.5

the required value of y is=28.250000

Press any key to continue

Programming for Trapizoidal rule

/*trapizoidal rule*/

#include<stdio.h>

#include<math.h>

float fun(float x)

{

                float d;

                d=sqrt(1-(x*x));

                return(d);

}

void

main()

{

                float a,b,sum=0.0,integral=1.0,c=0.0,h;

                int i,n;

                printf("enter the lower and upper limits:");

                scanf("%f%f",&a,&b);

                printf("enter the total number of intervals:");

                scanf("%d",&n);

label:

                c=integral;

                h=(b-a)/n;

                sum=fun(a)+fun(b);

                for(i=1;i<n;i++)

                {

                                sum=sum+(2*fun(a+(i*h)));

                }

                integral=(h*0.5)*sum;

                if(fabs((integral-c)/integral)>0.001)

                {

                                n=n+10;

                                goto label;

                }

                printf("the value of the integral is %f and is converged for %d intervals",integral,n);

}

Result:

enter the lower and upper limits:0 1

enter the total number of intervals:10

the value of the integral is 0.784237 and is converged for 40 intervals

Press any key to continue

/*trapizoidal rule*/

#include<stdio.h>

#include<math.h>

float fun(float x)

{

                float d;

                d=(x*cos(x));

                return(d);

}

void

main()

{

                float a,b,sum=0.0,integral=1.0,c=0.0,h;

                int i,n;

                printf("enter the lower and upper limits in degree:");

                scanf("%f%f",&a,&b);

                printf("enter the total number of intervals:");

                scanf("%d",&n);

                a=a*(3.1415/180);

                b=b*(3.1415/180);

label:

                c=integral;

                h=(b-a)/n;

                sum=fun(a)+fun(b);

                for(i=1;i<n;i++)

                {

                                sum=sum+(2*fun(a+(i*h)));

                }

                integral=(h*0.5)*sum;

                if(fabs((integral-c)/integral)>0.001)

                {

                                n=n+10;

                                goto label;

                }

                printf("the value of the integral is %f and is converged for %d intervals",integral,n);

}

Result:

enter the lower and upper limits in degree:90 180

enter the total number of intervals:10

the value of the integral is -2.570212 and is converged for 20 intervals

Press any key to continue

Program - Gauss seidel method | convergence test

/*gauss seidel method*/

#include<stdio.h>

#include<math.h>

void

main()

{

float a[20][20],b[20],x[20],error,sum=0.0,e,s=0.0;

int n,i,j,c=0,f=0;

/*taking inputs*/

printf("\n enter the number of variables:");

 scanf("%d",&n);

 printf("\n enter the coefficient matrix rowwise\n");

 for(j=0;j<n;j++)

 {

                 for(i=0;i<n;i++)

                 {

                                 printf("\n\t a[%d][%d]=",j,i);

                                 scanf("%f",&a[j][i]);

                 }

 }

 printf("\n enter the constants:\n");

for(i=0;i<n;i++)

{

                printf("\n\t b[%d]=",i);

                scanf("%f",&b[i]);

}

/*convergence test*/

for(j=0;j<n;j++)

{

                sum=0;

                for(i=0;i<n;i++)

                {

                                if(i!=j)

                                                sum=sum+fabs(a[i][j]);

                }

if(fabs(a[j][j]>sum))

{

                f++;

}

}

if(f==0)

{

                printf("\n the solution does not converge");

                goto end;

}

else

{

                printf("\n\t %d equations satisfy the conditions",f);

}

/*iteration process*/

printf("\n enter the guess value of the solution \n");

for(i=0;i<n;i++)

{

                printf("\n\t x[%d]=",i);

                scanf("%f",&x[i]);

}

f=0;

printf("\n enter the required amount of accuracy");

scanf("%f",&error);

/*iteration loop*/

label:

{

                e=error;

                for(j=0;j<n;j++)

                {

                                s=0;

                                for(i=0;i<n;i++)

                                {

                                                if(i!=j)

                                                {

                                                                s=s+a[j][i]*x[i];

                                                }

                                }

                                sum=(b[j]-s)/a[j][j];

                                if(fabs(sum-x[j]<error))

                                {

                                                x[j]=sum;

                                }

                                else

                                {

                                                e=fabs(sum-x[j]);

                                                x[j]=sum;

                                }

                }

                c++;

                printf("\n number of iteration is %d \n",c);

                for(i=0;i<n;i++)

                {

                                printf("\n\t x[%d]=%f\n",i,x[i]);

                }

}

if(e>error)

goto label;

end:;

}

RESULT1:

 enter the number of variables:3

 enter the coefficient matrix rowwise

         a[0][0]=12

         a[0][1]=3

         a[0][2]=-5

         a[1][0]=1

         a[1][1]=5

         a[1][2]=3

         a[2][0]=3

         a[2][1]=7

         a[2][2]=13

 enter the constants:

         b[0]=1

         b[1]=28

         b[2]=76

         2 equations satisfy the conditions

 enter the guess value of the solution

         x[0]=1

         x[1]=1

         x[2]=1

 enter the required amount of accuracy0.001

 number of iteration is 1

         x[0]=0.250000

         x[1]=4.950000

         x[2]=3.123077

 number of iteration is 2

         x[0]=0.147116

         x[1]=3.696731

         x[2]=3.821657

 number of iteration is 3

         x[0]=0.751508

         x[1]=3.156704

         x[2]=3.972965

 number of iteration is 4

         x[0]=0.949559

         x[1]=3.026309

         x[2]=3.997474

 number of iteration is 5

         x[0]=0.992370

         x[1]=3.003042

         x[2]=4.000123

 number of iteration is 6

         x[0]=0.999291

         x[1]=3.000068

         x[2]=4.000127

 number of iteration is 7

         x[0]=1.000036

         x[1]=2.999917

         x[2]=4.000037

Press any key to continue

RESULT2:

 enter the number of variables:4

 enter the coefficient matrix rowwise

         a[0][0]=10

         a[0][1]=-2

         a[0][2]=-1

         a[0][3]=-1

         a[1][0]=-2

         a[1][1]=10

         a[1][2]=-1

         a[1][3]=-1

         a[2][0]=-1

         a[2][1]=-1

         a[2][2]=10

         a[2][3]=-2

         a[3][0]=-1

         a[3][1]=-1

         a[3][2]=-2

         a[3][3]=10

 enter the constants:

         b[0]=3

         b[1]=15

         b[2]=27

         b[3]=-9

         4 equations satisfy the conditions

 enter the guess value of the solution

         x[0]=1

         x[1]=1

         x[2]=1

         x[3]=1

 enter the required amount of accuracy0.001

 number of iteration is 1

         x[0]=0.700000

         x[1]=1.840000

         x[2]=3.154000

         x[3]=-0.015200

 number of iteration is 2

         x[0]=0.981880

         x[1]=2.010256

         x[2]=2.996174

         x[3]=-0.001552

 number of iteration is 3

         x[0]=1.001513

         x[1]=1.999765

         x[2]=2.999818

         x[3]=0.000091

 number of iteration is 4

         x[0]=0.999944

         x[1]=1.999980

         x[2]=3.000011

         x[3]=-0.000006

Press any key to continue