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 espigot: 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 n^{2}.
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:

unit
is a constant used in only one place. Remove the definition and just use the value directly.–
comp
is basic 2×2 matrix multiplication. Rename tomul
to make purpose clearer.–
extr
as 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 placeextr
is 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 4tuple. However, the bottomleft element is always 0. Remove this element everywhere, leaving 3tuples and making
mul
andapprox
simpler.– For both π and e, the
next
andsafe
functions passed tostream
are identical save for the bounds used, so pass in the bounds and integrate the functions instream
.– The y used in
stream
is the lower bound. Rename accordingly.– The
cons
passed to stream is alwayscomp
(or in my casemul
). Remove the argument and usemul
directly.–
prod
is also the same for π and e, so remove the argument and inline the function.– The definition of π from the paper calls
stream
with some starting arguments. Since e will be doing the same, make a functionstreamDigits
to abstract this out.– Since
stream
now 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*(e1), which of course is e). So both are of the form a + (n/d)x. The matrix used in
lfts
is (n, a*d, 0, d). I didn’t like having to do the multiplication myself, so I madestreamDigits
take 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_e
andstream_pi
functions.
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.
[…] 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/programmingpraxisdigitsofe/ .
[…] 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()’.
Here is FORTH code for the first algorithm (by Stan & Stanley) Though space is proportional to n, not sure why you mention n**2.