Digits Of E

June 19, 2012

We begin with the algorithm by Rabinowitz and Wagon:

(define-generator (e-spigot n)
  (define (times10 x) (* 10 x))
  (yield 2)
  (let loop1 ((ts (make-list n 10)) (k n))
    (if (< 1 k)
      (let loop2 ((ts ts) (rs (list)) (i (+ n 1)) (carry 0))
        (if (= i 1)
            (begin (yield carry) (loop1 (map times10 (reverse rs)) (- k 1)))
            (let* ((x (+ (car ts) carry)) (q (quotient x i)) (r (remainder x i)))
              (loop2 (cdr ts) (cons r rs) (- i 1) q))))))
  (let loop3 () (yield #f) (loop3)))

The outer loop processes the rows of the tableau one by one, producing another digit each time, and the inner loop processes each row. The third loop returns an indication that processing is complete each time it is called.

The Gibbons algorithm is nearly as pretty in Scheme as in Haskell; like Niemeijer, I made a few changes in translation:

(define-generator (make-spigot f lo hi)
  (define (split v)
    (values (vector-ref v 0) (vector-ref v 1) (vector-ref v 2)))
  (define (approx abc n)
    (let-values (((a b c) (split abc)))
      (quotient (+ (* a n) b) c)))
  (define (mul abc def)
    (let-values (((a b c) (split abc)) ((d e f) (split def)))
      (vector (* a d) (+ (* a e) (* b f)) (* c f))))
  (define (g k) (let-values (((n d a) (f k))) (vector n (* a d) d)))
  (let loop ((z (vector 1 0 1)) (k 1))
    (let ((lbound (approx z lo)))
      (cond ((= lbound (approx z hi))
              (yield lbound)
              (loop (mul (vector 10 (* -10 lbound) 1) z) k))
      (else (loop (mul z (g k)) (+ k 1)))))))

(define pi-spigot (make-spigot (lambda (k) (values k (+ k k 1) 2)) 3 4))
(define e-spigot (make-spigot (lambda (k) (values 1 k 1)) 1 2))

We use a three-slot vector for the tuple, and values to split tuples into pieces; Scheme’s lack of pattern-matching hurts us here.

We used generators from a previous exercise. You can run the program at http://programmingpraxis.codepad.org/my56jfCf.


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7 Responses to “Digits Of E”

  1. […] today’s Programming Praxis exercise, our goal is to implement two algorithms to calculate the digits of e […]

  2. 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..]
  3. […] First we reuse the unbounded spigot algorithm for calculating e from the last exercise; […]

  4. Miracle said

    Does anybody share a Java or C# code for this exercises?

  5. Mike said

    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)
                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"
  6. David said

    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[]  !
       cr ." 2." ;
    : .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 ;
    2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427 ok
  7. Lucio said

    Hi 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()

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