Digits Of E
June 19, 2012
| |
modulus |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
| |
|
| 2 |
initialization |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
| |
|
| |
times ten |
10 |
10 |
10 |
10 |
10 |
10 |
10 |
10 |
10 |
10 |
| |
carry |
4 |
3 |
2 |
1 |
1 |
1 |
1 |
1 |
0 |
|
| |
sum |
14 |
13 |
12 |
11 |
11 |
11 |
11 |
11 |
10 |
10 |
| 7 |
quotient |
7 |
4 |
3 |
2 |
1 |
1 |
1 |
1 |
1 |
0 |
| |
remainder |
0 |
1 |
0 |
1 |
5 |
4 |
3 |
2 |
0 |
10 |
| |
|
| |
times ten |
0 |
10 |
0 |
10 |
50 |
40 |
30 |
20 |
0 |
100 |
| |
carry |
3 |
0 |
3 |
9 |
6 |
4 |
2 |
0 |
9 |
|
| |
sum |
3 |
10 |
3 |
19 |
56 |
44 |
32 |
20 |
9 |
100 |
| 1 |
quotient |
1 |
3 |
0 |
3 |
9 |
6 |
4 |
2 |
0 |
9 |
| |
remainder |
1 |
1 |
3 |
4 |
2 |
2 |
0 |
2 |
9 |
1 |
| |
|
| |
times ten |
10 |
10 |
30 |
40 |
20 |
20 |
0 |
20 |
90 |
10 |
| |
carry |
6 |
9 |
8 |
3 |
2 |
0 |
3 |
9 |
0 |
|
| |
sum |
16 |
19 |
38 |
43 |
22 |
20 |
3 |
29 |
90 |
10 |
| 8 |
quotient |
8 |
6 |
9 |
8 |
3 |
2 |
0 |
3 |
9 |
0 |
| |
remainder |
0 |
1 |
2 |
3 |
4 |
6 |
3 |
2 |
0 |
10 |
| |
|
| |
times ten |
0 |
10 |
20 |
30 |
40 |
60 |
30 |
20 |
0 |
100 |
| |
carry |
5 |
6 |
7 |
8 |
9 |
4 |
2 |
0 |
9 |
|
| |
sum |
5 |
16 |
27 |
38 |
49 |
64 |
32 |
20 |
9 |
100 |
| 2 |
quotient |
2 |
5 |
6 |
7 |
8 |
9 |
4 |
2 |
0 |
9 |
| |
remainder |
1 |
1 |
3 |
3 |
1 |
1 |
0 |
2 |
9 |
1 |
| |
|
| |
times ten |
10 |
10 |
30 |
30 |
10 |
10 |
0 |
20 |
90 |
10 |
| |
carry |
6 |
9 |
6 |
1 |
1 |
0 |
3 |
9 |
0 |
|
| |
sum |
16 |
19 |
36 |
31 |
11 |
10 |
3 |
29 |
90 |
10 |
| 8 |
quotient |
8 |
6 |
9 |
6 |
1 |
1 |
0 |
3 |
9 |
0 |
| |
remainder |
0 |
1 |
0 |
1 |
5 |
3 |
3 |
2 |
0 |
10 |
| |
|
| |
times ten |
0 |
10 |
0 |
10 |
50 |
30 |
30 |
20 |
0 |
100 |
| |
carry |
3 |
0 |
3 |
9 |
4 |
4 |
2 |
0 |
9 |
|
| |
sum |
3 |
10 |
3 |
19 |
54 |
34 |
32 |
20 |
9 |
100 |
| 1 |
quotient |
1 |
3 |
0 |
3 |
9 |
4 |
4 |
2 |
0 |
9 |
| |
remainder |
1 |
1 |
3 |
4 |
0 |
6 |
0 |
2 |
9 |
1 |
| |
|
| |
times ten |
10 |
10 |
30 |
40 |
0 |
60 |
0 |
20 |
90 |
10 |
| |
carry |
6 |
9 |
8 |
1 |
8 |
0 |
3 |
9 |
0 |
|
| |
sum |
16 |
19 |
38 |
41 |
8 |
60 |
3 |
29 |
90 |
10 |
| 8 |
quotient |
8 |
6 |
9 |
8 |
1 |
8 |
0 |
3 |
9 |
0 |
| |
remainder |
0 |
1 |
2 |
1 |
2 |
4 |
3 |
2 |
0 |
10 |
| |
|
| |
times ten |
0 |
10 |
20 |
10 |
20 |
40 |
30 |
20 |
0 |
100 |
| |
carry |
5 |
5 |
2 |
4 |
6 |
4 |
2 |
0 |
9 |
|
| |
sum |
5 |
15 |
22 |
14 |
26 |
44 |
32 |
20 |
9 |
100 |
| 2 |
quotient |
2 |
5 |
5 |
2 |
4 |
6 |
4 |
2 |
0 |
9 |
| |
remainder |
1 |
0 |
2 |
4 |
2 |
2 |
0 |
2 |
9 |
1 |
| |
|
| |
times ten |
10 |
0 |
20 |
40 |
20 |
20 |
0 |
20 |
90 |
10 |
| |
carry |
2 |
7 |
8 |
3 |
2 |
0 |
3 |
9 |
0 |
|
| |
sum |
12 |
7 |
28 |
43 |
22 |
20 |
3 |
29 |
90 |
10 |
| 6 |
quotient |
6 |
2 |
7 |
8 |
3 |
2 |
0 |
3 |
9 |
0 |
| |
remainder |
0 |
1 |
0 |
3 |
4 |
6 |
3 |
2 |
0 |
10 |
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Posted by programmingpraxis
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()