Percy Ludgate(en.wikipedia.org)
en.wikipedia.org
Percy Ludgate
https://en.wikipedia.org/wiki/Percy_Ludgate
18 comments
The algorithm is described in more details in https://academic.oup.com/comjnl/article/14/3/317/420539?logi... (around page 323, page 318, ..)
I haven't thought about it at all, but isn't it somewhat similar to Newton Raphson, in how it approximates a first reciprocal (here with a table), and then converges on better estimates in a few iterations?
Arguably the method for converging looks like it could be different at first glance. You could add it per WP:BOLD and see if anyone objects, the descriptions above are good enough to cite, such that there's no original work in describing the algorithm
I haven't thought about it at all, but isn't it somewhat similar to Newton Raphson, in how it approximates a first reciprocal (here with a table), and then converges on better estimates in a few iterations?
Arguably the method for converging looks like it could be different at first glance. You could add it per WP:BOLD and see if anyone objects, the descriptions above are good enough to cite, such that there's no original work in describing the algorithm
That representation's even nicer – the logarithmic complexity is immediately evident:
1/(1 + x) = 1 - x + x^2 - x^3 ...
= (1 - x) (1 + x^2) (1 + x^4) (1 + x^8) ...
double recip(double y) {
long double x = ((long double) y) - 1.0;
long double u = 1.0 - x;
long double v = x;
for (int j=1; j<=5; ++j) {
v = v * v; // v = x^(2^j)
u = u * (1.0 + v);
}
return (double)u; // approx. 1.0/y
}Not only similar, it's identical to Newton Raphson in ideal arithmetic!
The nth order product approximation for the reciprocal of d = 1 - x is y_n = product_{i=1}{n} (1 + x^{2^(i-1)}) = sum_{i=0}^{2^n - 1} x^i = (1 - x^(2 n)) / (1 - x).
The Newton-Raphson iteration for the reciprocal of d is z_{n+1} = z_n * (2 - (1-x) * z_n).
Induction with the base case z_0 = 1 = y_0 shows the sequences are equal by inserting y_n for z_n in the Newton Raphson iteration: z_{n+1} = (1 - x^(2 n)) * (2 - (1 - x^(2 n))) / (1 - x) = ( 1 - x^(4 n) ) / ( 1 - x) = y_{n+1}.
So I guess you could call it an explicit product formula for the Newton Raphson reciprocal.
The nth order product approximation for the reciprocal of d = 1 - x is y_n = product_{i=1}{n} (1 + x^{2^(i-1)}) = sum_{i=0}^{2^n - 1} x^i = (1 - x^(2 n)) / (1 - x).
The Newton-Raphson iteration for the reciprocal of d is z_{n+1} = z_n * (2 - (1-x) * z_n).
Induction with the base case z_0 = 1 = y_0 shows the sequences are equal by inserting y_n for z_n in the Newton Raphson iteration: z_{n+1} = (1 - x^(2 n)) * (2 - (1 - x^(2 n))) / (1 - x) = ( 1 - x^(4 n) ) / ( 1 - x) = y_{n+1}.
So I guess you could call it an explicit product formula for the Newton Raphson reciprocal.
This 100% checks out! I didn't recognize this at all.
- "...in ideal arithmetic!"
Comparing the two, Newton-Raphson seems rather better conditioned. I guess it's the (1.0 + x^(2^i)) part that holds back the explicit-product version (adding a large float with a small one).
- "...in ideal arithmetic!"
Comparing the two, Newton-Raphson seems rather better conditioned. I guess it's the (1.0 + x^(2^i)) part that holds back the explicit-product version (adding a large float with a small one).
I don't know where you got the idea that there is a first guess that comes from a table. The table lookup just brings x into a suitable range, given A.
Shades of "Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out?"— a Taylor series can only do so much!
Shades of "Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out?"— a Taylor series can only do so much!
This is basically how division on slide rules works.
"If we want to calculate a/b we can also calculate it as a * (1/b); that is the division of two numbers is the equivalent of the first number multiplied by its reciprocal."
http://www.sliderules.info/a-to-z/mul-div.htm
"If we want to calculate a/b we can also calculate it as a * (1/b); that is the division of two numbers is the equivalent of the first number multiplied by its reciprocal."
http://www.sliderules.info/a-to-z/mul-div.htm
Multiplication by the reciprocal isn't the interesting part. Computing the reciprocal using a convergent series is. (Which is not a process that is useful or even feasible on a slide rule, owing to the addition steps required, and the fact that slide rules can trivially perform division as you point out.)
Ah, I see what you mean. Still, though, how is this that different from a slide rule + log table? e is the limit of a convergent series too.
No log table lookups are required for this algorithm.
Working a day job and then spending your nights working on an engineering project (like Ludgate did) will always be the most inspiring part for me.
Yeah. The "day job" has always been nothing more than a means to
better ends. Everything worthwhile, exciting, and satisfying I've
achieved in life was in the gaps between doing something I could use
my time better for. Like Twain said - "don't let schooling get in the
way of your education", similarly I think ones "work" happens despite
employment, rarely through it. In this way "capitalism" is a horrible
waste of human capital. Imagine what Ludgate might have achieved if
not counting penny corns. I hope for a world for children who'll get a
UBI, and space in life to develop their true selves.
Maybe the habits of discipline that come from formal schooling and work are necessary for a man like Ludgate, or maybe not. We are products of our constraints to some extent
[deleted]
However, it also means we do not know how many hobbyists has tinkered with analytical engines between Babbage and WWII, and whether any of these attempts were earlier or more advanced.
In a slightly different timeline, WWI Britain adopts his machines for ballistics, etc. and starts the computer revolution 20 years early.
Babbage was never able to complete any of his machines due to conflicts with his chief engineer and insufficient funding. It wasn't until 1941 that the first general purpose computer, the Z3, was built, more than a century after Babbage proposed the pioneering Analytical Engine in 1837.
Am I overlooking it, or is this method missing from Wikipedia's "Division Algorithm" entry?
https://en.wikipedia.org/wiki/Division_algorithm?useskin=vec...
How it works is easy enough to figure from that description. You rescale your divisor to fit in the range (0.5, 1.0], then expand it as the series 1/(1 - x) = 1 + x + x^2..., which converges, and does so linearly at >1 bit per term (since x < 1/2). What's useful about this, is you can calculate the finite sums with a circuit that's logarithmic depth, or maybe better. You could do the repeated squarings to get: x, x^2, x^4, x^8...; then you have a bunch of parallel hardware multipliers filling out all the intermediate terms, which are products of zero- or one- of each of these basic terms; then you just find the sum with a logarithmic-sized tree of adders.
Somehow, I can't find the name of this algorithm (?)