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rshinde70
October 24th, 2010, 03:59 PM
please check out the attachment,
I've written an program for finding two variables but it doesn't works :(

Also I've inserted an picture of the terminal showing the output for this program.

Friends.. I'm waiting for your replies ....

krishnandu.sarkar
October 24th, 2010, 04:05 PM
I think you should start a and b with 1, because anything multiplying with 0 returns 0. So the condition is getting true each time.

BTW remember initializing those variables with 1 will never make your condition true, so you won't get any output.

shankhs
October 24th, 2010, 04:06 PM
As I understand you want to find a,b for:

(a+b)^2 = a^2 + b^2

This is only possible if a=0 ( any value of b) or b=0 ( any value of a )

Am I missing anything?

darshan123
October 24th, 2010, 04:09 PM
Actually, a*a + 2*a*b + b*b = (a+b) * (a+b)

in ur case, the condition is satified only when b == 0 . hence the output.

change the formula and things will be fine.

Darshan

arubislander
October 24th, 2010, 04:19 PM
As I understand you want to find a,b for:

(a+b)^2 = a^2 + b^2

This is only possible if a=0 ( any value of b) or b=0 ( any value of a )

Am I missing anything?

Yep you are: 2*a*b

rshinde70
October 24th, 2010, 04:21 PM
@krishnandu.sarkar (http://ubuntuforums.org/member.php?u=1063048) you are true,
but Friends, somebody told me that, there are really two numbers, and those are non-zero.
so i thought, to make a program to search it.

look at the output carefully, you will get noticed that, only one variable is getting incremented while both should get incremented..
like..

for a=0, b should increment from 0 to 999,
again, for a=1, b from 0-999, and so on upto a becomes 999.

please help.

](*,) :confused: ](*,)

shankhs
October 24th, 2010, 04:23 PM
(a+b)^2 = a^2+b^2
=>a^2+b^2+2ab=a^2+b^2
=>2ab=0

So either a=0 or b=0
:)

rshinde70
October 24th, 2010, 04:31 PM
@ shankhs,

(a+b)^2 = a^2+b^2

is not possible,
actually, the correct formula is :



(a+b)^2 = a^2+b^2 + 2*a*b

krishnandu.sarkar
October 24th, 2010, 04:44 PM
@krishnandu.sarkar (http://ubuntuforums.org/member.php?u=1063048) you are true,
but Friends, somebody told me that, there are really two numbers, and those are non-zero.
so i thought, to make a program to search it.

look at the output carefully, you will get noticed that, only one variable is getting incremented while both should get incremented..
like..

for a=0, b should increment from 0 to 999,
again, for a=1, b from 0-999, and so on upto a becomes 999.

please help.

](*,) :confused: ](*,)


Well...how would we know what formula or what do you want to implement. I just corrected the logic.

krishnandu.sarkar
October 24th, 2010, 05:03 PM
Do you want this thing??


#include<stdio.h>
main()
{


// A PROGRAM TO FIND a AND b IF (a*a)+(2*a*b)+(b*b) = (a*a)+(b*b)

int i;
int a=0, b=0;
int x, y;

for(i=0;i<1000;i++)
{

x=(a*a)+(2*a*b)+(b*b);

y=(a*a)+(b*b);

if(x==y)
{
printf("\na=%d\tb=%d",a,b);
a++;
b++;
}

}

if(a==999 && b==999)
printf("\nElements not found!");


}

arubislander
October 24th, 2010, 05:13 PM
Pencil, paper and highschool math reveal:

(a+b)*(a+b)=a*a + b*b
<=> a*a + 2*a*b + b*b = a*a + b*b
<=> 2*a*b = 0
so: a=0 or b=0!

rshinde70
October 25th, 2010, 01:18 PM
](*,)](*,)](*,)](*,)](*,)](*,)](*,)](*,)


It doesn't works !

I want this to be work as like password breaker software.
Its loops should work like that.

for example :

If i want to look for a and b then =====>

it should work like :

a=0, b=0
a=0, b=1
a=0, b=2
.
.
.
.
a=0, b=999;

and then,

a=1, b=0
a=1, b=1
a=1, b=2
.
.
.
a=1, b=999
.
.
.
.
.
a=999, b=999


please help somebody..

look carefully in " for "

Vaphell
October 25th, 2010, 02:37 PM
but it does that, x==y comparison is done exactly 1000000 times
you don't see it because you printf(a, b) only when x==y and that happens exactly 1999 times: (0,0-999) and (0-999,0) minus 1 ((0,0) is in both sets)

besides no program in the world will help you when you can't see the obviousness of your problem. It's a basic math.

run this

#include<stdio.h>
main()
{
// A PROGRAM TO FIND a AND b IF (a*a)+(2*a*b)+(b*b) = (a*a)+(b*b)

int a, b;
int x, y;
int counter = 0;
int counter_if_any_zero = 0;
int total_tests = 0;

for(a=0;a<1000;a++)
{
for(b=0;b<1000;b++)
{
total_tests++;
x=((a*a)+(2*a*b)+(b*b));
y=(a*a)+(b*b);

if(x==y)
{
printf("a=%d\tb=%d\n",a,b);
counter++;
if(!(a && b))
counter_if_any_zero++;
}
}
}
printf( "number of tests: %d\n", total_tests );
printf( "found: %d\n", counter );
printf( "found with a=0 or b=0: %d\n", counter_if_any_zero );
if(!counter)
printf("Elements not found!");
}

VernonA
October 26th, 2010, 02:47 PM
@OP. Your program is working fine, and is printing out all the correct solutions. It is checking a million combinations of a and b, and is correct when it says that the only ones which are valid are those in which a is zero or b is zero. Your screen capture indeed shows the end of the list, where a is non-zero and b is 0.

Multiple posters have noted that the formula for your x variable is just (a+b)^2 but since C doesn't provide an exponentiation operator you have to expand it using schoolbook algebra to the formula you are actually using. By adding some whitespace to your formulae, you will see you are computing:

x=((a*a)+(2*a*b)+(b*b));
y= (a*a)+ (b*b) ;
and it should be clear that the only way these two will be equal is if the 2*a*b term is zero.

Finally, if you still doubt your own code, add an else clause to the 'if', eg:

if (x==y)
printf("\na=%d\tb=%d",a,b);
else
printf("\na=%d\tb=%d is NOT a solution",a,b);
This will at least show the loops are working as you intended, although the output is going to be pretty long-winded!

rshinde70
October 27th, 2010, 01:26 PM
@Vaphell (http://ubuntuforums.org/member.php?u=347382)
you are right.
I just wanted to confirm that, those for loops are working correctly or not.
But yeah, that structure is correct. Thank you Very Very Much Vaphell.

=;=;=;=;=;=;=;