Since it was much easier for the calculator to detect an actual overdraft than to detect that an overdraft would occur, this algorithm was more efficient all around. It then kept this last term (which is one beyond the last term that could be subtracted and thus contained the correct count) and cleared the 5. This was very handy because, instead of doing the shift/add on each iteration to build the new s from the old s and d, the calculator could produce the new s' directly by simply subtracting until the negative number appeared (an overdraft) and then adding that last term back into the remainder. In the scaled by 5 case, if the 5 in the term is cleared, the digit left in the term is the count minus 1. Notice an interesting property of the terms being subtracted: Further Refinement: Computing the Next S Directly The bottom row of keys on the Friden keyboard indicated the decimal position in the number and started the square root process. Placing the decimal in the result yields 15.8113 for 250 or 158.10. Placing the decimal yields 5 or 50 respectively.Īnd so on. It does require some care with the decimal points however, since the roots of have the same digits as do 20, but the digits in the square roots of 25 and 250 are different. The above example uses integers but the algorithm works just as well for fractional numbers and roots. (so d is 5 which is appended to s yielding 25) from the leftmost digits: (s started at 0) Then it would start subtracting 100s+5 100s+15. To take the square root of 625, the calculator would multiply by 5 to yield: 3125. Thus, the calculator could compute d by successively subtracting 100s+5, 100s+15 until a negative number would result. Note that this is also the sum of a (scaled by 5) odd digits series: Appending the next digit would cause a change in the remainder which could be expressed (scaled by 5) as: So, given an s (root so far), we need to find the next digit d. If we square s and subtract it from the original number n, we are left with some remainder r. Thus, you want to proceed from left to right as you did in school.īegin by assuming we have already found the square root of some of the leading digits of the number n. but that would take a very long time for large numbers. The above is all you need to start brute-forcing square roots by subtracting 1, 3. This was more efficient for the machine because it turned multiplications by 20 into multiplications by 100 which can be done by shifting the carriage.
#Friden calculator division series
To make the following algorithm work well on a calculating machine, Friden used the same series multiplied by 5:ĥ×(4 2) = 5×(1 + 3 + 5 + 7) = 5 + 15 + 25 + 35ĥ×(n 2) = 5×(1 + 3 +. Each squared number is the previous square with one more layer which is always two more than the one before it. Here's a picture of how this series works:Īnd so on. Since a calculating machine would be hard pressed to deal with an algorithm with involved "finding some d such that.", it used the odd integer series in which the square of a number n can be computed by the sum of the odd integers from 1 to 2n-1 i.e.: The Friden's square root extraction was based on essentially the same "long division" algorithm but unlike poor human calculators, the Friden could extract the root of a 10-digit number in an amazing 9 seconds. At each iteration you were probably told to "find some next digit d for which ("the root so far" × 20+d) multiplied by d could be subtracted from the remainder. The reader will probably (vaguely) remember being taught to extract square roots in school by a method that looked a lot like "long division". How the Friden Extracted Square Roots It's Basically What You Learned in School
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