For ye Maths and Programming wizards

Pretence

Heres a puzzle that a friend and I were trying to work out last night, we were very drunk

We all know the term Square Root and call functions in most languages to implement it on a certain number. But can anyone actually figure out who it works. Can anyone actually define in simple steps what a number has to go through to derive the square root of that number.

Is there a simple algorithm to represent this function????????????

Karla

The easiest way is to loop through a load of numbers, squaring it, and seeing if it's the same as your given number. Of course, that's only gonna work for ints.

You can also use the Dividing and Averaging method:
Use the above method to see which number is closest to the actual square root of the number i.e.
To get the square root of 40 try 6X6=36, 7X7=49, 6X6 is closest.
Now you know that the square root of 40 starts with a 6 followed by whatevers after the decimal point.
Now divide 40 by 6 to the nearest 10th. 40/6 = 6.6
Get the average of 6 and 6.6 then square that.

Repeat.

Pretence

Won't you just get an infinite loop using that method..........no??

shank

Glad I'm not one of your drinkin buddies

Blitzkrieger

I think there's a formula but it's very long-winded. I think most programs look up a table like you get in log books (only bigger)

Hecate

and they say drink kills braincells

I remember being in the pub with some guy who was quite sloshed, for want of a better word, and spent the whole night roaring out the cube roots of real and integer numbers.

You could say the craic was 90.6879920111

riam

see below...............
didnt type this had it saved from before

_____________
| 98 25 71

Example of square root of 982,571


Begin by finding the largest digit whose square is less than the first group of numbers, 98 in this case, which is greater than 9^2 = 81. So use 9 to be the leading digit of the square root. Now subtract 9^2 = 81 from the leading pair 98, getting a remainder of 17. Bring down the next pair of digits, 25, so we have 1725 get one digit for each pair of digits in the input, but after the first digit, each digit involves a trial divisor & remainder process. Take twice the "root quotient" you have found so far (twice 9 is 18) and leave room for one more digit to the right: 18x into 1725 Find a digit x as large as possible so that 18x times x is less than or equal to 1725. Now 189 times 9 is 1701, so the next digit we want will be 9 also. Subtract 1701 from 1725: 1725 - 1701 = 24
and bring down the last two digits to put with the 24:
2471
The last trial divisor is twice the "root" extracted so far, twice 99 is 198, with one additional digit:
198x into 2471 goes x times
this can only go one time:
2471 - 1981 = 490
991^2 = 982081
982571 - (991^2) = 490
sqrt(982571) = 991.2471941952723630922423...

Hobbes

I recall years ago someone explaining to me that the square root can be done in assembly using binary and a cool formula. Or was it logs? I forget. I just remember it was kind of cool and long winded and something my brain didn't want to remember.

It didn't use the method above, it had something to do with shifting bits.

Vincent

All this numbers are giving me a headache

Pretence

I agree, not the well defined, SIMPLE , algorithm I was hoping for.

Good though......

Trojan

Originally posted by Hecate:
You could say the craic was 90.6879920111

On a Pentium perhaps

Al.