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Showing posts with label New Technologies. Show all posts
Showing posts with label New Technologies. Show all posts

Wednesday, 21 December 2011

Two Earth- size planets orbiting a star outside the solar system


Scientists have found two Earth- sized planets orbiting a star outside the solar system, an encouraging sign for prospects of finding life elsewhere.

The discovery shows that such planets exist and that they can be detected by the Kepler spacecraft, said Francois Fressin of the Harvard-Smithsonian Center for Astrophysics in Cambridge, Massachusetts.

They are the smallest planets found so far that orbit a star resembling our sun. Scientists are seeking Earth-sized planets as potential homes for extraterrestrial life, said Fressin, who reports the new findings in a paper published online yesterday by the journal Nature.

One planet’s diameter is only 3 per cent larger than Earth’s, while the other’s diameter is about nine-tenths that of Earth. They appear to be rocky, like our planet.

But they are too hot to contain life as we know it, with calculated temperatures of about 1,400 degrees and 800 degrees Fahrenheit (760 Celsius and 425 Celsius), he said.

Any life found on another planet may not be intelligent; it could be bacteria or mold or some completely unknown form.

Since it was launched in 2009, NASA’s planet-hunting Kepler telescope has found evidence of dozens of possible Earth-sized planets.

But Fressin’s report is the first to provide confirmation, said Alan Boss of the Carnegie Institution for Science in Washington. He’s a member of the Kepler science team but not an author of the paper.

The researchers ruled out a possible alternative explanation for the signals that initially indicated the planets were orbiting the star Kepler-20. The star is 950 light-years from Earth in the direction of the constellation Lyra.

The planets, called Kepler-20e and Kepler-20f, are part of a five-planet system around the star, and their location challenges current understanding of how planets form, scientists said.

In our own solar system, the small rocky planets are closest to the sun, while gaseous giants are on the periphery.

But the five-planet system has no such dividing line; big and small planets alternate as one moves away from the star.

That’s ‘crazy,’ and unexplained by current understanding of how planets form around stars, said study co-author and Harvard scientist David Charbonneau.

Earlier this month, scientists said they’d found a planet around another distant star with a life-friendly surface temperature of about 72 degrees Fahrenheit (22 Celsius). But it was too big to suggest life on its surface. (DC)

Thursday, 15 December 2011

Top 20 Important C Language Programs, with explanation

 

Program:
#include <stdio.h>
int main(){
int a, b, c;
printf("Enter a,b,c: \n");
scanf("%d %d %d", &a, &b, &c);
if (a > b && a > c) {
printf("a is Greater than b and c");
}
else if (b > a && b > c) {
printf("b is Greater than a and c");
}
else if (c > a && c > b) {
printf("c is Greater than a and b");
}
else {
printf("all are equal or any two values are equal");
}
return 0;
}
Output:
Enter a,b,c: 3 5 8
c is Greater than a and b
Explanation with examples:
Consider three numbers a=5,b=4,c=8
if(a>b && a>c) then a is greater than b and c
now check this condition for the three numbers 5,4,8 i.e.
if(5>4 && 5>8) /* 5>4 is true but 5>8 fails */
so the control shifts to else if condition
else if(b>a && b>c) then b is greater than a and c
now checking this condition for 5,4,8 i.e.
else if(4>5 && 4>8) /* both the conditions fail */
now the control shifts to the next else if condition
else if(c>a && c>b) then c is greater than a and b
now checking this condition for 5,4,8 i.e.
else if(8>5 && 8>4) /* both conditions are satisfied */
Thus c is greater than a and b.


Program:
#include <stdio.h>
int main() {
int a[10];
int i;
int greatest;
printf("Enter ten values:");
//Store 10 numbers in an array
for (i = 0; i < 10; i++) {
scanf("%d", &a[i]);
}
//Assume that a[0] is greatest
greatest = a[0];
for (i = 0; i < 10; i++) {
if (a[i] > greatest) {
greatest = a[i];
}
}
printf("\nGreatest of ten numbers is %d", greatest);
return 0;
}
Output:
Enter ten values: 2 53 65 3 88 8 14 5 77 64 Greatest of ten numbers is 88
Explanation with example:
Entered values are 2, 53, 65, 3, 88, 8, 14, 5, 77, 64
They are stored in an array of size 10. let a[] be an array holding these values.
/* how the greatest among ten numbers is found */
Let us consider a variable 'greatest'. At the beginning of the loop, variable 'greatest' is assinged with the value of
first element in the array greatest=a[0]. Here variable 'greatest' is assigned 2 as a[0]=2.
Below loop is executed until end of the array 'a[]';.
for(i=0; i<10; i++)
{
if(a[i]>greatest)
{
greatest= a[i];
}
}
For each value of 'i', value of a[i] is compared with value of variable 'greatest'. If any value greater than the value
of 'greatest' is encountered, it would be replaced by a[i]. After completion of 'for' loop, the value of variable
'greatest' holds the greatest number in the array. In this case 88 is the greatest of all the numbers.


A prime number is a natural number that has only one and itself as factors. Examples: 2, 3, 13 are prime
numbers.
Program:
#include <stdio.h>
main() {
int n, i, c = 0;
printf("Enter any number n: \n");
scanf("%d", &n);
/*logic*/
for (i = 1; i <= n; i++) {
if (n % i == 0) {
c++;
}
}
if (c == 2) {
printf("n is a Prime number");
}
else {
printf("n is not a Prime number");
}
return 0;
}
Output:
Enter any number n: 7
n is Prime
Explanation with examples:
consider a number n=5
for(i=0;i<=n;i++) /* for loop is executed until the n value equals i */
i.e. for(i=0;i<=5;i++) /* here the for loop is executed until i is equal to n */
1st iteration: i=1;i<=5;i++
here i is incremented i.e. i value for next iteration is 2
now if(n%i==0) then c is incremented
i.e.if(5%1==0)then c is incremented, here 5%1=0 thus c is incremented.
now c=1;
2nd iteration: i=2;i<=5;i++
here i is incremented i.e. i value for next iteration is 3
now if(n%i==0) then c is incremented
i.e.if(5%2==0) then c is incremented, but 5%2!=0 and so c is not incremented, c remains 1
c=1;
3rd iteration: i=3;i<=5;i++
here i is incremented i.e. i value for next iteration is 4
now if(n%i==0) then c is incremented
i.e.if(5%3==0) then c ic incremented, but 5%3!=0 and so c is not incremented, c remains 1
c=1;
4th iteration: i=4;i<=5;i++
here i is incremented i.e. i value for next iteration is 5
now if(n%i==0) then c is incremented
i.e. if(5%4==0) then c is incremented, but 5%4!=0 and so c is not incremented, c remains 1
c=1;
5th iteration: i=5;i<=5;i++
here i is incremented i.e. i value for next iteration is 6
now if(n%i==0) then c is incremented
i.e. if(5%5==0) then c is incremented, 5%5=0 and so c is incremented.
i.e. c=2
6th iteration: i=6;i<=5;i++
here i value is 6 and 6<=5 is false thus the condition fails and control leaves the for loop.
now if(c==2) then n is a prime number
we have c=2 from the 5th iteration and thus n=5 is a Prime number.


If a number, which when read in both forward and backward way is same, then such a number is called a
palindrome number.
Program:
#include <stdio.h>
int main() {
int n, n1, rev = 0, rem;
printf("Enter any number: \n");
scanf("%d", &n);
n1 = n;
/* logic */
while (n > 0){
rem = n % 10;
rev = rev * 10 + rem;
n = n / 10;
}
if (n1 == rev){
printf("Given number is a palindromic number");
}
else{
printf("Given number is not a palindromic number");
}
return 0;
}
Output:
Enter any number: 121
Given number is a palindrome
Explanation with an example:
Consider a number n=121, reverse=0, remainder;
number=121
now the while loop is executed /* the condition (n>0) is satisfied */
/* calculate remainder */
remainder of 121 divided by 10=(121%10)=1;
now reverse=(reverse*10)+remainder
=(0*10)+1 /* we have initialized reverse=0 */
=1
number=number/10
=121/10
=12
now the number is 12, greater than 0. The above process is repeated for number=12.
remainder=12%10=2;
reverse=(1*10)+2=12;
number=12/10=1;
now the number is 1, greater than 0. The above process is repeated for number=1.
remainder=1%10=1;
reverse=(12*10)+1=121;
number=1/10 /* the condition n>0 is not satisfied,control leaves the while loop */
Program stops here. The given number=121 equals the reverse of the number. Thus the given number is a
palindrome number.


Palindrome is a string, which when read in both forward and backward way is same.
Example: radar, madam, pop, lol, rubber, etc.,
Program:
#include <stdio.h>
#include <string.h>
int main() {
char string1[20];
int i, length;
int flag = 0;
printf("Enter a string: \n");
scanf("%s", string1);
length = strlen(string1);
for(i=0;i < length ;i++){
if(string1[i] != string1[length-i-1]){
flag = 1;
break;
}
}
if (flag) {
printf("%s is not a palindrome\n", string1);
}
else {
printf("%s is a palindrome\n", string1);
}
return 0;
}
Output:
Enter a string: radar
"radar" is a palindrome
Explanation with example:
To check if a string is a palindrome or not, a string needs to be compared with the reverse of itself.
Consider a palindrome string: "radar",
---------------------------
index: 0 1 2 3 4
value: r a d a r
---------------------------
To compare it with the reverse of itself, the following logic is used:
0th character in the char array, string1 is same as 4th character in the same string.
1st character is same as 3rd character.
2nd character is same as 2nd character.
. . . .
ith character is same as 'length-i-1'th character.
If any one of the above condition fails, flag is set to true(1), which implies that the string is not a palindrome.
By default, the value of flag is false(0). Hence, if all the conditions are satisfied, the string is a palindrome.

Top 20 Important C Language Programs with explanation



Recursion: A function is called 'recursive' if a statement within the body of a function calls the same function. It
is also called 'circular definition'. Recursion is thus a process of defining something in terms of itself.
Program: To calculate the factorial value using recursion.
#include <stdio.h>
int fact(int n);
int main() {
int x, i;
printf("Enter a value for x: \n");
scanf("%d", &x);
i = fact(x);
printf("\nFactorial of %d is %d", x, i);
return 0;
} int fact(int n) {
/* n=0 indicates a terminating condition */
if (n <= 0) {
return (1);
} else {
/* function calling itself */
return (n * fact(n - 1));
/*n*fact(n-1) is a recursive expression */
}
}
Output:
Enter a value for x:
4
Factorial of 4 is 24
Explanation:
fact(n) = n * fact(n-1)
If n=4
fact(4) = 4 * fact(3) there is a call to fact(3)
fact(3) = 3 * fact(2)
fact(2) = 2 * fact(1)
fact(1) = 1 * fact(0)
fact(0) = 1
fact(1) = 1 * 1 = 1
fact(2) = 2 * 1 = 2
fact(3) = 3 * 2 = 6
Thus fact(4) = 4 * 6 = 24
Terminating condition(n <= 0 here;) is a must for a recursive program. Otherwise the program enters into an
infinite loop.


Program:
#include <stdio.h>
int main() {
int a;
printf("Enter a: \n");
scanf("%d", &a);
/* logic */
if (a % 2 == 0) {
printf("The given number is EVEN\n");
}
else {
printf("The given number is ODD\n");
}
return 0;
}
Output:
Enter a: 2
The given number is EVEN
Explanation with examples:
Example 1: If entered number is an even number
Let value of 'a' entered is 4
if(a%2==0) then a is an even number, else odd.
i.e. if(4%2==0) then 4 is an even number, else odd.
To check whether 4 is even or odd, we need to calculate (4%2).
/* % (modulus) implies remainder value. */
/* Therefore if the remainder obtained when 4 is divided by 2 is 0, then 4 is even. */
4%2==0 is true
Thus 4 is an even number.
Example 2: If entered number is an odd number.
Let value of 'a' entered is 7
if(a%2==0) then a is an even number, else odd.
i.e. if(7%2==0) then 4 is an even number, else odd.
To check whether 7 is even or odd, we need to calculate (7%2).
7%2==0 is false /* 7%2==1 condition fails and else part is executed */
Thus 7 is an odd number.

Swapping interchanges the values of two given variables.
Logic:
step1: temp=x;
step2: x=y;
step3: y=temp;
Example:
if x=5 and y=8, consider a temporary variable temp.
step1: temp=x=5;
step2: x=y=8;
step3: y=temp=5;
Thus the values of the variables x and y are interchanged.
Program:
#include <stdio.h>
int main() {
int a, b, temp;
printf("Enter the value of a and b: \n");
scanf("%d %d", &a, &b);
printf("Before swapping a=%d, b=%d \n", a, b);
/*Swapping logic */
temp = a;
a = b;
b = temp;
printf("After swapping a=%d, b=%d", a, b);
return 0;
}
Output:
Enter the values of a and b: 2 3
Before swapping a=2, b=3
After swapping a=3, b=2
Swapping interchanges the values of two given variables.
Logic:
step1: x=x+y;
step2: y=x-y;
step3: x=x-y;
Example:
if x=7 and y=4
step1: x=7+4=11;
step2: y=11-4=7;
step3: x=11-7=4;
Thus the values of the variables x and y are interchanged.
Program:
#include <stdio.h>
int main() {
int a, b;
printf("Enter values of a and b: \n");
scanf("%d %d", &a, &b);
printf("Before swapping a=%d, b=%d\n", a,b);
/*Swapping logic */
a = a + b;
b = a - b;
a = a - b;
printf("After swapping a=%d b=%d\n", a, b);
return 0;
}
Output:
Enter values of a and b: 2 3
Before swapping a=2, b=3
The values after swapping are a=3 b=2


Program:
#include <stdio.h>
int main() {
int i = 65;
int k = 120;
printf("\n value of i=%d k=%d before swapping", i, k);
i = i ^ k;
k = i ^ k;
i = i ^ k;
printf("\n value of i=%d k=%d after swapping", i, k);
return 0;
}
Explanation:
i = 65; binary equivalent of 65 is 0100 0001
k = 120; binary equivalent of 120 is 0111 1000
i = i^k;
i...0100 0001
k...0111 1000
---------
val of i = 0011 1001
---------
k = i^k
i...0011 1001
k...0111 1000
---------
val of k = 0100 0001 binary equivalent of this is 65
---------(that is the initial value of i)
i = i^k
i...0011 1001
k...0100 0001
---------
val of i = 0111 1000 binary equivalent of this is 120
--------- (that is the initial value of k)

**FRESHER JOBS**