Mathimatical Problem

[Tryfon]

Hi guys
am using an 80x86 assembler mips and i have a little problem.
There is a number(int) "n" and i want to find the number(int) wich when you square it you will get either the number n or the int before

Best regards
Tryfon

 

[Salem]

> an 80x86 assembler mips
This makes no sense.
80x86 is a processor series made by

http://www.intel.com/homepage/index.htm

MIPS is a processor series made by

http://www.mips.com/

Their instruction sets are not compatible.


--
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.

 

[strettonman]

I'm not sure it makes sense mathematically either. To paraphrase:

There is a number(int), 7 say, and i want to find the number(int) wich when you square it you will get either the number 7 or the int before it (i.e. 6 ).

Well neither 6 or 7 have integer square roots.

 

[Tryfon]

No strettoman ... let me explain more clearly this time

For example there is an int 6 and i need to find to find an int which when you square it you ll get an int less or equal than 6 i.e the integer am looking for in the case of 6 is 2(2x2=4 which 4 holds the argument of <=6)

 

[lionelhill]

This is a standard text-book question, because all you're looking for is a rounded-down integer square root. Try any basic algorithm text, or web-searching for square root methods.

 

[strettonman]

If you are using an Intel 80x86 processor, at least from the 486 onwards, you should be able to find some floating point instructions which will let you store a decimal square root to memory, and then read it back as a (rounded down) integer square root.

 

[fancom]

well, i am triing to learn asm. But If I didn't missunderstand you the basic code is ;

input "enter number";n%  'get an integer

r = 0 'set the root for zero

do                'loop
r = r + 0.1       'increase root by 0.1
loop until r * r >= n%  'check if the r * r >= n%
                        'if it is, then exit loop

if r * r  > n% then r = r - 0.1  'if the square of 
                                 'square roots 
                                 'bigger than n%
                                 'move 1 step back  

nr% = 0

do
if r < 1 then exit do else nr% = nr% + 1
r = r - 1
loop



print "The square root of ";n% is;r
print "And the nearest square is";nr%;"x";nr%;nr%xnr%

where n%, nr% are integer and r is a single (float)




lets n% = 8

so the root will be something bigger than 2.5 and we want only 2

first step ;

nr% = 1 r = 1.5


second :

nr% = 2 r = 0.5

third :

nr% = 2 r is smaller than 1 so exit


I hope the code works. It is a handmade square root program. And I would be pleased if you write the explained asm code...

 

[fancom]

sorry for the wrong answer, so here you are ;

The nearest integer that is square rott of any integer in basic

Input "Enter An Integer";n%

r% = 0

do
r%=r%+1
loop until r% * r% > n%

r% = r% - 1

the integer r% is what you want...