Solved

Posted on 2003-11-21

How to find if a number is a perfect square?

I am NOT looking for the solution which says check if sqrt is integer or not

I am NOT looking for the solution which says check if sqrt is integer or not

9 Comments

Comment Utility

Could you give us more information? How about a sample of what you are looking for? When is this due? ;)

Comment Utility

for those who need clarification

I don't want this solution

int n;

cin >> n;

if (sqrt (N) == Whole Number)

cout << "A perfect square"

else

cout << "Not a perfect square"

I was hoping more in terms of bit wise operators (>>,<<,&,|) etc

I actually read the solution in terms of bit wise operators many weeks ago. I could not recall the solution.

This is not due as in for assignment or hw. Hence I alotted only few points. Take your time and hopefully provide a solution i am looking for.

K

I don't want this solution

int n;

cin >> n;

if (sqrt (N) == Whole Number)

cout << "A perfect square"

else

cout << "Not a perfect square"

I was hoping more in terms of bit wise operators (>>,<<,&,|) etc

I actually read the solution in terms of bit wise operators many weeks ago. I could not recall the solution.

This is not due as in for assignment or hw. Hence I alotted only few points. Take your time and hopefully provide a solution i am looking for.

K

Comment Utility

Perhaps then you'd be interested in researching assembly square root solutions and convert them manually? Here is one for z80 which has a similar feel to x86:

http://www.ticalc.org/pub/83plus/asm/source/routines/_fastsqrt_.zip

http://www.ticalc.org/pub/

Comment Utility

I came across this website which might be closer to what you are looking for.

http://www.fifi.org/cgi-bin/info2www?(gmp)Perfect+Square+Algorithm

I came across this website which might be closer to what you are looking for.

http://www.fifi.org/cgi-bi

Comment Utility

Please disregard my question if it is irrelevant, but are you sure you saw the biwise solution for N^2?

It is possible that you remember seeing the bitwise solution for checking whether the number is power of 2 - it would be indeed simple in bitwise terms:

M is 2^N if and only if (M & (M-1)) equals zero (assuming N>0).

For N^2 I cannot remember anything simple (if I do, I'll let you know, of course).

Regards

It is possible that you remember seeing the bitwise solution for checking whether the number is power of 2 - it would be indeed simple in bitwise terms:

M is 2^N if and only if (M & (M-1)) equals zero (assuming N>0).

For N^2 I cannot remember anything simple (if I do, I'll let you know, of course).

Regards

Comment Utility

there is a very easy to do it using mod... unfortunatly i cant quite remember it but if i do ill tell u

l8rz... suma

l8rz... suma

Comment Utility

No comment has been added to this question in more than 21 days, so it is now classified as abandoned.

I will leave the following recommendation for this question in the Cleanup topic area:

PAQ - no points refunded (of 50)

Any objections should be posted here in the next 4 days. After that time, the question will be closed.

Venabili

EE Cleanup Volunteer

I will leave the following recommendation for this question in the Cleanup topic area:

PAQ - no points refunded (of 50)

Any objections should be posted here in the next 4 days. After that time, the question will be closed.

Venabili

EE Cleanup Volunteer

Comment Utility

Question finalized per recommendation.

RomMod

Community Support Moderator

RomMod

Community Support Moderator

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