Digits Of E
June 19, 2012
We gave an algorithm for computing the digits of π in a previous exercise. Today, we look at two algorithms for computing the digits of e.
We begin with an algorithm due to Stanley Rabinowitz and Stan Wagon:
Algorithm e-spigot: compute the first n decimal digits of e:
1. Initialize: Let the first digit be 2 and initialize an array A of length n + 1 to (1, 1, 1, . . . , 1).
2. Repeat n − 1 times:
Multiply by 10: Multiply each entry of A by 10.
Take the fractional part: Starting from the right, reduce the ith entry of A modulo i + 1, carrying the quotient one place left.
Output the next digit: The final quotient is the next digit of e.
We give an example of the calculation for the first ten digits of e on the next page, where we compute the first ten digits of e as 2 7 1 8 2 8 1 8 2 6. As you can see, this algorithm suffers from the fact that the last digit may be wrong (it should be 8, not 6); in fact, as the paper suggests, there are circumstances where several of the last digits may be wrong. The algorithm also suffers from the fact that it is bounded, meaning that the number of digits must be specified in advance, and it needs space proportional to n2.
A different algorithm comes from Jeremy Gibbons, and is both unbounded and requires only constant space; Gibbons gave the sequence for π, but Tom Moertel adapts it to e. When I was unable to work out the algorithm from Moertel’s description, I asked Remco Niemeijer, a regular contributor to Programming Praxis, for help, and he responded with this gorgeous hunk of Haskell code, which computes both π and e:
stream :: Integral a => (a, a) -> (a, a, a) -> [(a, a, a)] -> [a]
stream (lo, hi) z ~(x:xs) = if lbound == approx z hi
then lbound : stream (lo, hi) (mul (10, -10*lbound, 1) z) (x:xs)
else stream (lo, hi) (mul z x) xs
where lbound = approx z lo
approx (a,b,c) n = div (a*n + b) c
mul (a,b,c) (d,e,f) = (a*d, a*e + b*f, c*f)
streamDigits :: (Num a, Integral a, Enum a) => (a -> (a, a, a)) -> (a, a) -> [a]
streamDigits f range = stream range (1,0,1) [(n, a*d, d) | (n,d,a) <- map f [1..]]
stream_pi, stream_e :: [Integer]
stream_pi = streamDigits (\k -> (k, 2*k + 1, 2)) (3, 4)
stream_e = streamDigits (\k -> (1, k , 1)) (1, 2)
main :: IO ()
main = do print $ take 30 stream_pi
print $ take 30 stream_e
Niemeijer explained that he had adapted the algorithm in Gibbons’ paper as follows:
–
unitis a constant used in only one place. Remove the definition and just use the value directly.–
compis basic 2×2 matrix multiplication. Rename tomulto make purpose clearer.–
extras defined in the paper didn’t compile for me: it should befromInteger (q * x)instead offromInteger q * x. But there’s more to be gained: the only placeextris used is preceded byfloor, so essentially we havefloor (a / b), which is equal todiv a b. Since it’s used to approximate the real value I named itapprox.– The code used 2×2 matrices expressed as a 4-tuple. However, the bottom-left element is always 0. Remove this element everywhere, leaving 3-tuples and making
mulandapproxsimpler.– For both π and e, the
nextandsafefunctions passed tostreamare identical save for the bounds used, so pass in the bounds and integrate the functions instream.– The y used in
streamis the lower bound. Rename accordingly.– The
conspassed to stream is alwayscomp(or in my casemul). Remove the argument and usemuldirectly.–
prodis also the same for π and e, so remove the argument and inline the function.– The definition of π from the paper calls
streamwith some starting arguments. Since e will be doing the same, make a functionstreamDigitsto abstract this out.– Since
streamnow takes far fewer parameters, there’s no longer a reason to name them all.– According to the hulver site, each term in the π function is 2 + k/(2k + 1)x. For e, each term is 1 + (1/k)x (or rather, it defines e − 1 starting from k = 2, but if you start from k = 1 you add the term 1 + 1/1x, which gives you 1 + 1/1*(e-1), which of course is e). So both are of the form a + (n/d)x. The matrix used in
lftsis (n, a*d, 0, d). I didn’t like having to do the multiplication myself, so I madestreamDigitstake a function that produces n, d and a for a given term to stay closer to the mathemetical definition. This function is used to produce the actual matrix.– Add
stream_eandstream_pifunctions.
Your task is to write the two spigot functions for e described above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
[…] today’s Programming Praxis exercise, our goal is to implement two algorithms to calculate the digits of e […]
Here’s my Haskell solution for the first algorithm (since my solution for the second one has already been posted in the exercise). A version with comments can be found at http://bonsaicode.wordpress.com/2012/06/19/programming-praxis-digits-of-e/ .
import Data.List spigot_e :: Int -> [Int] spigot_e n = 2 : take (n - 1) (f $ replicate (n + 1) 1) where f = (\(d,xs) -> d : f xs) . mapAccumR (\a (i,x) -> divMod (10*x+a) i) 0 . zip [2..][…] First we reuse the unbounded spigot algorithm for calculating e from the last exercise; […]
Does anybody share a Java or C# code for this exercises?
Basically a direct translation of the haskell code into Python 2.7. I create the input stream and initialize the state vector (z) in ‘stream()’ and eliminated ‘streamDigits()’.
from itertools import count, imap def stream(lo, hi, f): def approx((a,b,c), n): return (a*n + b)//c def mul((a,b,c),(d,e,f)): return a*d, a*e + b*f, c*f xs = ((n, a*d, d) for n,d,a in imap(f, count(1))) z = 1, 0, 1 while True: lbound = approx(z, lo) if lbound == approx(z, hi): yield lbound z = mul((10, -10*lbound, 1), z) else: z = mul(z, next(xs)) def pi_digits(): return stream(3, 4, lambda k: (k, 2*k + 1, 2)) def e_digits(): return stream(1, 2, lambda k: (1, k, 1)) # test from itertools import islice print ''.join(str(d) for d in islice(pi_digits, 10)) # returns "3141592653" print ''.join(imap(str, islice(e_digits, 14))) # returns "27182818284590"Here is FORTH code for the first algorithm (by Stan & Stanley) Though space is proportional to n, not sure why you mention n**2.
100 constant #digits : int-array create cells allot does> swap cells + ; #digits 1+ int-array e-digits[] : init-e ( -- ) [ #digits 1+ ] literal 0 DO 1 i e-digits[] ! LOOP cr ." 2." ; : .e ( -- ) init-e [ #digits 1- ] literal 0 DO 0 \ carry 0 #digits DO i e-digits[] dup @ 10 * rot + i 2 + /mod -rot swap ! -1 +LOOP 0 .r LOOP ; .e 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427 okHi Mike, when I tried your code it gave me error can’t iterate on a function, fixed the problem by changing the last two lines to
print ”.join(str(d) for d in islice(pi_digits(), 10))
print ”.join(imap(str, islice(e_digits(), 14)))
in other words replacing the function pi_digits with pi_digits() which invokes the function and returns an array, similarly replacin e_digits with e_digits()