Ex Parte HO et al - Page 2




            Appeal No. 2004-0115                                                          Page 2              
            Application No. 09/363,637                                                                        


            reciprocal square root.  More specifically, the Newton-Raphson algorithm for                      
            approximating the reciprocal square root of N is expressed as follows:                            
                   Xi+1 = (3 - N*Xi *Xi)*Xi/2,                                                                
            where Xi is an approximation of the reciprocal square root of N at the ith iteration, \ is        
            greater than or equal to 1, and Xi+1 is a more accurate approximation.  (Id. at 3.)               


                   According to the appellants' method for approximating the reciprocal square root           
            of a number (N), a reciprocal square root of N is estimated as Xi.  The estimate and N            
            are multiplied to produce a first intermediate result (IR1).  A second intermediate               
            result (IR2) is determined according to the equation: IR2 = (1-Xi*IR1)/2.  The second             
            intermediate result and the estimate are multiplied to produce a third intermediate result.       
            The third intermediate result and the estimate are added to produce an approximation of           
            the reciprocal square root of the number.  (Id. at 4.)                                            


                   According to the appellants, the order in which their method performs                      
            multiplication "makes it likely that all results will be in normalized form."  (Id.)  Although    
            the Newton-Raphson algorithm requires one to determine the product of: N*Xi*Xi, they              
            explain that their method first multiplies N by Xi to produce an intermediate result, which       
            will likely be in normalized form.  Their method then multiplies the intermediate result by       
            Xi to produce: (N*Xi)*Xi.  If their method first multiplied Xi and Xi to produce the              








Page:  Previous  1  2  3  4  5  6  7  8  9  Next 

Last modified: November 3, 2007