Another Method for Extracting Cube Roots. Brian J. Shelburne Dept of Math and Computer Science Wittenberg University

Transcription

1 Another Method for Extracting Cube Roots Brian J. Shelburne Dept of Math and Computer Science Wittenberg University Probably the easiest way (?) to compute the cube root of a is to use the iteration formula xn xn+ =. However, while checing out an article in the December 958 issue of the xn Communications of the ACM I ran across a paper that demonstrated another method for extracting cube roots. The method was useful (?) since it was done using fixed point arithmetic, it required no division and could be modified so only minimal multiplication was needed (both expensive operations for computers in the 950 s). It s similar to the method of extracting a square root by hand. The Theory The theory behind this technique is simple. Let be the number whose cube root we want. Let the number we want, be represented by the sum i= d 0 n i i and 9 and n is the number of digits above the decimal point. Letting where di is a digit between 0 a i = d we can 0 n i i write = ( a +... n +...). We will show how to compute each term a or more exactly each digit di in an iterative fashion.. Observe the following For positive integer, ( a a... a ) + + pproximates and so ( a +... ) approximates. Using the binomial theorem we can expand ( a +... ) in terms of ( a +... ) and a, specifically ( a +... ) = ( a +... ) + ( a +... ) a + ( a +... ) a For example Continuing we have ( a ) a a a a a = + a + +. ( a ) = ( a ) + ( a ) a + ( a ) a etc. The idea is to wor by stages so that given ( a +... ) at one stage we choose a so that ( a +... ) is a better approximation to.

2 . Exactly how do we use ( a +... ) to obtain ( a +... )? Begin by finding the largest value of a such that the error = a is non-negative, or equivalently finding the largest value of a such that a is less than or equal to. Knowing the value of a find the largest value of a such that the error = a a + aa is non-negative or equivalently such that a a + aa is less than or equal to. Note that = ( a ) In general given ( a +... ) find the largest value of a such that the error = ( a +... ) a + ( a +... ) a is non-negative. Note that = ( a +... ).. A Notational Refinement! Before going further, let s introduce some notation to mae our formulas above less cumbersome and, as we will see, to simplify our calculations. Let and let u = ( a +... ) v = u = ( a +... ) So given u and v = u find the largest value of a such that the error = v a + u a is non-negative! Again note that = u + An Example Unfortunately it s not clear how to apply the above theory in a practical and efficient way to actually compute a cube root; that s what the following example will show. For example compute 790. We begin by partitioning the digits of 790 into groups of three (i.e ) starting at the decimal point. At each stage we will bring down for consideration the next group of three digits. Essentially we are scaling our calculations by 000, the cube of 0. Observe that (a + ) 0 so (a + ) is the largest digit approximation less than or equal to the cube root of.

3 We begin by only woring with the left-most group of digits, essentially treating 790 lie This allows us to wor with single digit values between 0 and 9; that is each a we find will be a digit d between 0 and 9. Stage. Find the largest digit d such that d 79. This is 4 (it helps to now the cubes of the first nine integers:, 8, 7, 64, 5, 6, 4, 5, 79). Subtract 64 from 79 and bring down the next three digits = Stage. We need to find the largest a such that the error a a + aa is non-negative. Because bringing down the next three digits scales by 000, the current cube root approximation 4 needs to be scaled by 0, i.e. a = 4 0. We need to compute a = (4 0) = 0 and a = (4 0) = a will be a digit between 0 and 9 In finding the largest digit d such that 4800 d + 0 d + d is less than 50, seems lie a good estimate for d but it s too high. So d equals = 0088 which when subtracted from 50 results in = Note that 4 is the largest integer less than or equal to 790. Shift in three more digits and repeat with = Stage. Because of the way we scale our values ( a ) = u = ( ) = 40. We need to compute (a ) = (40) = 60 and ( a ) = v = (40) = 59,00 and find the largest digit d such that the error 59, 00 d + 60 d + d is nonnegative. We estimate d = 9 which wors since 59, = 4,865, 589 so subtracting =

4 With d = 9 our approximation to 790 is 4.9. Shift in three more digits and repeat with = 47,4,000 The TANSTAAFL Effect The obvious draw bacs to this method are ) the computational complexity of woring with large integer values, ) the need to compute the square of a large integer, i.e. ( a +... ) and, ) the need to find the largest digit d such that the error at the th stage ( a +... ) a + ( a +... ) a is non-negative, usually by trial and error. Reducing Computational Complexity - I We can simplify the squaring of ( a +... ) at the th step by using previously computed values. Recall that u is the value of ( a +... ) and v = u is the value of ( a +... ) at the th step. It s not difficult to derive a pair of interlocing formulas to compute u and v. and u 0 if = 0 = ( u + d ) 0 if > 0 v 0 if = 0 = ( v + u d + d ) 00 if > 0 Given u and v the computational cost of computing v = ( a +... ) is reduced to an addition followed by a multiplication by 0 to obtain u plus two additions and three simple multiplications (by, by the digit d and by 00) to obtain v (assuming a simple table looup to find the square of the digit d ). Thus we can simplify our calculations if at each step we also calculate and use u and v as seen below where the 4 th digit for obtained. There Ain t No Such Thing As A Free Lunch from The Moon is A Harsh Mistress by Robert Heinlein The article in CACM mentioned that this method had been successfully and efficiently implemented on an IBM 650 a decimal computer where multiplications by powers of ten were easily implemented by left shifts. Even on a binary machine multiplication by ten can be efficiently implemented by a left shift by bits added to a left shift by bit essentially x*8 + x*. The IBM 650 also had a table loop-up operation. 4

5 d u v v ²d +u d ²+d ³ You can easily chec that v = ( v + u d + d ) 00 = ( ) 00 = Reducing Computational Complexity II The difference between adjacent terms of the form be written as v m u m m + + for m =,, etc. can ( u m ) v m + u m + m v ( m ) + ( m ) + ( ) = v + u (m ) + m where m = m ( m ) denotes the difference between adjacent cubes and (m ) is the difference between adjacent squares. The computational cost of estimating the largest d such that v d + u d + d is non-negative can be simplified by subtracting v + u + from and continuing to subtract expressions of the form v + u (m ) + m until a negative value is obtained (then restoring the previous value). The difference between adjacent cubes (i.e., 7, 9, 7, 6, 9, 7, 69, 7, 7) is easily found by table loo up. For example, the calculation for = 4 above (where v4 = 55,,00 and u =,870 ) can be redone using the following table. 4 d (-)+d = = = = = = Computationally for each row you need only one multiplication (by an odd integer), two additions and one table loo up for the first difference of the cube. 5

6 Computing a Cube Root by Hand Finding the cube root by hand can be done in a way similar to the method used for extracting square roots. For example, suppose we want to find. We start with a division lie structure Begin by finding the largest integer cube less than or equal to (8) and putting its cube root above where the quotient goes, subtracting the cube from the dividend and bringing down the next three zeros Now find the largest digit d such that 00 d + 0 d + d = 00 d + 60 d + d is less than or equal to 5,000. Since the first term (00) dominates this expression, we estimate 8 for d. To simply the calculation of 00 d + 0 d + d = 00 d + 60 d + d we re going to factor out the d and express the remaining factor as a sum of three terms: 00, 60 d and d (We express 00 below as 400 for reasons we ll explain at the next step.). We then multiply this sum by 8 and subtract from our dividend and bring down the next three digits = d = d = Alternately tae the current quotient times 0, square it and multiply by. Then again tae the quotient times 0 multiply by and by d (i.e. 8). Finally square d and add all three terms together. 6

7 At the next step complexity becomes apparent in the calculation of the expression v d + ud + d or more precisely v + u d + d where u = ( a +... ) and v = u. So for the next round we use the recursive formula for v to obtain v = ( ) 00 = Since v = 500 dominates the expression above we estimate 4 as the next digit = d = d = Observe that once the value of d - and the rest is just so much routine calculation. v An Afterword - So What Good is This? is obtained, it is fairly easy to estimate a value for the digit Given the ubiquity of cheap powerful calculators that can easily do division, why use this technique? There is no good reason but it is interesting that with this technique you could calculate a cube root by hand 4 or program a very unsophisticated computer to do so using only minimal multiplication and no division. The technique to obtain a cube root is an extension of the standard technique to obtain a square root. In this case given = ( a +... n +..) observe that ( a +... ) can be expanded as ( a +... ) = ( a +... ) + [ ( a +... ) a ]. Applying the technique discussed above to square roots (except we group digits by twos), at the th step we find the largest digit d such that the error at the th stage = u d + d where u = ( a +... ) is non-negative. This is usually expressed as tae the current quotient, double and multiply by 0 and using this trial divisor and divide into the dividend to estimate the next digit which is then added to the trial divisor i.e. u + d. Now multiply the 4 Isaac Asimov s short story The Feeling of Power tells the tale where all calculations are done my hand-held calculators and people have forgotten how to do calculations by hand until one day someone rediscovers hand calculations. The story has a tragic ending as the military immediately sees certain "applications" for hand calculation! 7

8 digit by the augmented trial divisor which gives you dividend (the result should be positive). Repeat. u d d + ) and subtract from the Example: Find Partition the digits into groups of two and begin by finding the largest digit whose square less than or equal to the leading group. Put the digit on top, subtract the square from the dividend and bring down the next two digits Tae the quotient (4) double and multiply by 0 (80) and use this as trial divisor. Put the digit result on top, add the digit to the trial divisor, multiply the trial divisor by the digit and subtract Repeat: Double 47 and multiply by 0 to obtain 940 as a trial divisor. 940 goes into 900 approximately 9 times. Thus 4.79 approximates The basic theory given above can also be applied to roots higher than the cube, e.g. fourth and fifth etc. For the 4 th root case observe ( a +... ) = ( a +... ) + [4 w a + 6 v a + 4 u a ] Where u and v are defined as above and w = u! The same method can be applied to find the 4 th root (group digits by 4) but the computational complexity problems are even more severe! 8

9 References Sugai, Iwao; Extraction of Roots by Repeated Subtractions for Digital Computers ; Comm. of A.C.M.; Vol No ; December 958; pp 6 8. Asimov, Isaac; "The Feeling of Power"; The Mathematical Magpie, Clifton Fadiman, ed.; Simon and Schuster; New Yor; 96. 9