Solved

find if a number is perfect square

Posted on 2003-11-21
9
2,156 Views
Last Modified: 2013-11-13
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
0
Comment
Question by:bsarvanikumar
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
9 Comments
 
LVL 8

Expert Comment

by:Rog
ID: 9800610
Could you give us more information?  How about a sample of what you are looking for?  When is this due? ;)
0
 

Author Comment

by:bsarvanikumar
ID: 9800691
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
0
 
LVL 8

Expert Comment

by:_corey_
ID: 9801897
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
0
Technology Partners: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 1

Expert Comment

by:Skytzo
ID: 9802069

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

0
 

Expert Comment

by:mem_fam
ID: 9805828
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
0
 
LVL 1

Expert Comment

by:suma_ds
ID: 9830135
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
0
 
LVL 20

Expert Comment

by:Venabili
ID: 10962993
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
0
 

Accepted Solution

by:
RomMod earned 0 total points
ID: 11001885
Question finalized per recommendation.

RomMod
Community Support Moderator
0

Featured Post

Technology Partners: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Whether you’re a college noob or a soon-to-be pro, these tips are sure to help you in your journey to becoming a programming ninja and stand out from the crowd.
Although it can be difficult to imagine, someday your child will have a career of his or her own. He or she will likely start a family, buy a home and start having their own children. So, while being a kid is still extremely important, it’s also …
An introduction to basic programming syntax in Java by creating a simple program. Viewers can follow the tutorial as they create their first class in Java. Definitions and explanations about each element are given to help prepare viewers for future …

733 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question