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 2008-03-06, 22:18 #1 petrw1 1976 Toyota Corona years forever!     "Wayne" Nov 2006 Saskatchewan, Canada 3×29×53 Posts "ODD" Square root Algorithm Odd to me anyway ... some of you may be familiar with it or why it works. http://pullmoll.stop1984.com/sqrt.html
2008-03-08, 20:45   #2
m_f_h

Feb 2007

24·33 Posts

Quote:
 Originally Posted by petrw1 Odd to me anyway ... some of you may be familiar with it or why it works. http://pullmoll.stop1984.com/sqrt.html
Did you read at the end of the page:
Quote:
 More information After asking on the usenet group sci.math.num-analysis I was pointed in the right direction and was told, that the algorithm was known and used in mechanical calculators ("Friede desk calculator") in the early 60s at Boeing already. Later Hewlett Packard sold calculators such as the HP9100, which used a similiar or identical algorithm utilizing BCD (binary coded decimals). I found this reference to the inner workings of the HP Model 9100A and the mathematical background for the algorithm. The background is that: n Σ 2*i - 1 = n2 i=1 so it says that the sum of all 2*i-1 for i starting with 1 up to n is equal to the square of n. I would be interested to learn how to write the corresponding formula for the cubic numbers, any hints?

 2008-03-08, 22:49 #3 petrw1 1976 Toyota Corona years forever!     "Wayne" Nov 2006 Saskatchewan, Canada 10010000000112 Posts Yes I did in fact read that but didn't take enough time to absorb how that extends to the algorithm he describes. For example to calculate sqrt(1024) only requires adding 1+3+5 and later 61+63, rather than all odds from 1 to 63. (1+3+5) * 100 = 900 = sum of odds to 60. Cool... As I continue to investigate it starts to make sense .... so let me at it for a while and I'll be fine. Thanks
 2008-03-15, 07:35 #4 nibble4bits     Nov 2005 2×7×13 Posts Do I understand right? That this is based on this formula for quadratics: (n+1)^2=n^2+2n+1; let m=n+1 m^2-n^2-1=2n Thus consecutive squares are always 2n+1 appart. That formule you mentioned is equivilent to this code: Code: Let x = 0 For i = 1 to n x = x + i + i - 1 Next i REM Returns in x, the square of n using recursive addition instead of multiplication. The optimizes trivially without multiplication to: Code: Let x=0 For i = 1 to n x = x + i + i Next i x = x - n You can further remove the For...Next loop using an identity but you get this: Let x = 2 * n * (n + 1) / 2 - n Or: Code: Let x = n * (n + 1) - n Or: Code: Let x = n^2 + n - n As you can see, this leads you back to your original equation of x=n^2. EDIT: You should note that for m=n+1: m^3 - n^3 - 1 = (n+1)^3 - n^3 - 1 = n^3 + 3n^2 + 3n + 1 - n^3 - 1 = 3(n^2 + n) = 3n(n+1) Last fiddled with by nibble4bits on 2008-03-15 at 07:44 Reason: Added note

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