find Sqrt witout math.h

[shehata]

i've created this code but gives me wrong values ! So what is wrong ??

#include <iostream.h>
#include <conio.h>

main()
{ 
double n,end,start=0,mid;

cin>>n;
end=n/2;

while(start<end)
{
mid=(start+end)/2;


if ((n-(mid*mid)<=0.001) && (n-(mid*mid)>=0))
{cout<<mid;break; }

 if(n>mid*mid)
{start=mid+0.001 ;
end=end;}

if(n<mid*mid)
{end=mid-0.001;
 start=start;}


 }

 getch();
 }

[lanadr96]

Why are you dividing n by 2 at the beginning?

[MiiNiPaa]

Your compiler is extremely outdated. update it ASAP.

Let me show what happens in your code:
Imagine that you have entered 88.
in the begunning:

start = 0;
end = 44;
mid = 22;

First check: if(n-(mid*mid)<=0.001) → if(88 - (22*22) <= 0.001) → true

[shehata]

Because the Sqrt of any number(n) can't be greater than (n/2) .. so i decreased the Possibilities

[shehata]

@MiiNiPaa ok i edit it and the new if condition beacome

if ((n-(mid*mid)<=0.001) && (n-(mid*mid)>=0))

but still gives worng values or not gives anything!

[MiiNiPaa]

worng values

How are they wrong? show an example?

not gives anything

Precision problems. For example 402 = 1600 and 40.001 = 1600,080001. Difference of 0.08: it might skip over checked range.
Also: difference of -0.000000001 will not be taken by your program.

[shehata]

ok no wrong values ..
but there is no results!! i tried to find Sqrt of number from (2 to 20)
guess what only 16 gives Sqrt =4 ?!

[MiiNiPaa]

1) if you enter 2, you will check values in range [0..1] when real value is ≈1.7

Values of mid and calcualted difference, which are checked for input = 25:

1 2 3 4 5 6 7 8 9 10 11 12 13

6.25 3.1244999999999998 4.6872499999999997 5.4686249999999994 5.0779374999999991 4.8825937499999998 4.9802656249999995 5.0291015624999993 5.0046835937499994 4.9924746093749999 4.9985791015625001 5.0016313476562502 5.0001052246093751

-14.0625 15.237499750000001 3.0296874375000016 4.9058593906249932 0.78544925390623987 1.1602782724609391 0.19695430444336637 -0.29186252593993345 -0.046857873550408158 0.075197274745942622 0.014206965422630446 -0.016316137857678825 -0.0010522571659699054

As you can see no values suits your condition.

[shehata]

@MiiNiPaa So what could i do for the range to make it possible to find Sqrt of any numbers with accuracy 0.001 ??

[shehata]

Help!

[MiiNiPaa]

Use another algorithm. You can use Tailor series for that for example.

[shehata]

So u say that there is nothing to do to make this code right ?

[MiiNiPaa]

Well, you can adjust step size and comparsion precision depending on input value. But you need to mathematically analyse your algorithm and deduct number/constants relations

[shehata]

@MiiNiPaa Thank u 4 Helping

[2AoiNwbp]

I also think that Taylor's suits better for finding a square root. You can google Taylor`s formula which divides a function by its derivative to get a good approximation. And that`s all you`ll get, an approximation, not an exact value. But let`s take a look at your algorithm.

With n/2 what you`re doing is to reduce the search space in halves, so you are using binary search, which is good to get a good approximation to a square root.

On the other hand, you also seem to be using something like a brute force algorithm, increassing start by epsilon (0.001) in line 20, which doesn`t sound good for a square root, in terms of computing costs.

Why are you testing (n-(mid*mid)>=0) in line 16? just for clarification, it might be abs()...? if you cannot use math.h, then, define your own abs. It will clarify your code.

If you can`t find anwers for numbers between 0 and 1, then you might want to take a look at the search space you are searching for.

regards,
Alejandro