-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- The three rules of Ruby Quiz: 1. Please do not post any solutions or spoiler discussion for this quiz until 48 hours have elapsed from the time this message was sent. 2. Support Ruby Quiz by submitting ideas and responses as often as you can. 3. Enjoy! Suggestion: A [QUIZ] in the subject of emails about the problem helps everyone on Ruby Talk follow the discussion. Please reply to the original quiz message, if you can. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - RSS Feed: http://rubyquiz.strd6.com/quizzes.rss Suggestions?: http://rubyquiz.strd6.com/suggestions -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- ## Digits of e (#226) Wayumbe Rubyists, The mathematical constant e is the unique real number such that the value of the derivative (slope of the tangent line) of the function f(x) = e^x at the point x = 0 is exactly 1. The function e^x so defined is called the exponential function, and its inverse is the natural logarithm, or logarithm to base e.[1] e is one of the most important numbers in mathematics, alongside the additive and multiplicative identities 0 and 1, the constant ð, and the imaginary unit i. These are the five constants appearing in one formulation of [Euler's identity][2]. This week¢s quiz is to write a Ruby program that can compute the first 100,000 digits of e. Have fun! [1]: http://en.wikipedia.org/wiki/E_(mathematical_constant) [2]: http://en.wikipedia.org/wiki/Euler's_identity

on 2010-01-08 06:27

on 2010-01-10 17:51

Daniel Moore wrote: > This week¢s quiz is to write a Ruby program that can compute the first > 100,000 digits of e. > > Have fun! > > [1]: http://en.wikipedia.org/wiki/E_(mathematical_constant) > [2]: http://en.wikipedia.org/wiki/Euler's_identity I used the fast converging continued fraction series from the first Wikipedia article: # compute e*10**digits as a rounded integer def calc_e(digits) limit = 10**((digits+3)/2) p = [2, 3] q = [2, 1] n = 1 begin if p.length.even? a = 4*(4*n-1) else a = 4*n+1 n += 1 end p << a*p[-1] + p[-2] q << a*q[-1] + q[-2] end while p.last <= limit (p.last*10**(digits+3)/q.last+500)/1000 end if __FILE__ == $0 p calc_e(if ARGV.length.zero? then 100_000 else Integer(ARGV[0]) end) end

on 2010-01-10 18:33

Hello, 2010/1/8 Daniel Moore <yahivin@gmail.com>: > This week’s quiz is to write a Ruby program that can compute the first > 100,000 digits of e. > > Have fun! Well I had :o) I got inspired from Quiz Digits of Pi (#202) and tried to estimate how long the standard library call could take (I didn't want to waste too much CPU on this): require 'bigdecimal' require 'bigdecimal/math' include Math include BigMath d = [Time.now] 1000.step(20000,1000) do |i| E(i) d << Time.now puts i.to_s + " needing #{d[-1] - d[-2]} s" end which gives 1000 needing 0.147869 s 2000 needing 0.956326 s 3000 needing 2.993516 s 4000 needing 6.738271 s 5000 needing 12.78967 s 6000 needing 21.584826 s 7000 needing 34.154156 s 8000 needing 49.223259 s 9000 needing 69.57902 s 10000 needing 94.507271 s 11000 needing 123.327777 s 12000 needing 158.839553 s 13000 needing 198.976605 s 14000 needing 246.62736 s 15000 needing 300.471167 s 16000 needing 364.013845 s ^C almost scaling as the cube of the digit number you ask for. So that it should take around 10**5 s to compute E(100_000). I then tried my own implementation using the definition exp(x) = \Sum{n=0}{\infty} x^n/n! applied in x=1 and checking from E(number) that it was correct. It's converging pretty fast (in term of iteration number, only 34_000 to get 100_000 digits) but it's quite exactly 10 times slower than the standard method (and I didn't want to wait 10**6 s to get all the 100_000 digits :o). The needed number of iterations is guessed using Stirling formula and I discovered that each step does not take the same time (it gets slower as time goes) whereas it does not depend on the initial number of digits asked for (that is the 1000 first iterations take approximately the same time when you ask for 10_000 digits or for 100_000 digits) which I couldn't really understand. I first thought that the slowing-down was due to the fact that for more digits you need bigger integers to play with but the latter observations seems to invalidate this argument.. If anybody have an explanation, I will be happy to hear it. require 'rational' require 'enumerator' require 'bigdecimal' require 'bigdecimal/math' include Math include BigMath d1 = Time.now precision = (ARGV[0] || 1000).to_i iterations = precision /(log10(precision)) + precision/(log10(precision))*log10(log10(precision)) puts 'Approximate number of iterations needed: ' + iterations.to_i.to_s accuracy = 10**precision.to_i fact = 1 final = accuracy other = false d = [d1] 1.upto(iterations) do |i| if i%100 == 0 d << Time.now puts i.to_s + " needing #{d[-1] - d[-2]} s" end fact *= i final_old = final final += Rational(accuracy,fact) end d2 = Time.now puts "Time elapsed in computation: " + (d2 - d1).to_s + ' s' puts "Computation terminated: computing E(#{precision}) now." puts "Delta * 10**(#{precision}): " + (final.round - E(precision)*accuracy).to_s d3 = Time.now puts "Time elapsed calling E(#{precision}): " + (d3 - d2).to_s + ' s' Last but not least, the same method could be called as one line in irb using inject and getting the result in the form of a Rational: fact = 1 ; (1..34_000).inject(1) {|sum,i| fact *= i ; sum + Rational(1,fact)} but you will rather want to try with a smaller number of steps (say 500 to get 1000 digits, remember: 34_000 steps will take around 10**6 s...) Cheers,

on 2010-01-11 14:58

Hello, based on the spigot algorithm of Rabonitz and Wagon [1]: #!/usr/bin/ruby def spigot n e = [] arr = [2] + Array.new(n+1,1) (n-1).times do |i| (n+1).downto 0 do |j| arr[j] *= 10 end q = 0 (n+1).downto 0 do |j| arr[j] += q q = arr[j]/(j+1) arr[j] %= j+1 end e << q end e[0]/= 10.0 puts e.join('') end if $0 == __FILE__ spigot ARGV[0].to_i end [1] http://www.mathpropress.com/stan/bibliography/spigot.pdf : citing the easier case of e Am 08.01.2010 06:26, schrieb Daniel Moore:

on 2010-01-11 15:00

Sorry, Rabonitz meant Rabinowitz, ugly typo. Am 11.01.2010 14:57, schrieb Thorsten Hater:

on 2010-01-13 22:59

After recovering some vaguely remembered math, and trying not to look to closely at any code on the net, here is my solution. The reference data for testing, from Project Gutenberg, has been truncated for this posting. The program will report "Incorrect" for more than a couple of hundred digits. Bill Rutiser wrutiser AT gmail DOT com # Produce digits of e # wrutiser AT gmail DOT com # # The Loop Invariant requires the following conceptual numbers to sum to _e_. # (1) The number represented by the decimal digits previously generated. # # (2) An intermediate fraction having the value of the sum of several # terms taken from the infinite series expansion that have yet to be # included in (1) # # (3) The tail of the infinite series expansion. # # The generated digits are contained in the variable _z_ but are not used # in the generation of the subsequent bits. # # The current term of the infinite series is represented by the variable _i_. # The term itself, given by the expression "1/factorial( i )", is never actually # computed. # # The sum of the intermediate terms is represented by the variables _nn_, # _dd_, and _mm_. While the sum itself is never actually computed, it is # equivalent to the expression "nn / (mm * dd)". # # _mm_ is a power of 10, increasing as each digit is extracted. # _dd_ is maintained equal to factorial( i ). # # _nn_ is the numerator of the conceptional rational number. # # Each loop interation: # removes the first remaining term from the series # # adds the removed term to the intermediate fraction # # computes the leading two digits of the fraction # # if these digits are the same as those computed for the # previous term, one digit is extracted. # def digits_int( n_terms ) nn = 0 dd = 1 mm = 1 q1 = -1 z = "2.\n" digits = 0 2.upto(n_terms) do |i| dd *= i nn *= i nn += mm q2, r2 = (nn * 100).divmod( dd ) if( q1 == q2 ) q, r = (nn * 10).divmod( dd ) z << q.to_s digits += 1 z << " " if 0 == digits % 10 z << "\n" if 0 == digits % 50 mm *= 10 nn = r q1 = -1 else q1 = q2 end end puts( "\ndigits: #{digits} n_terms: #{n_terms}" \ " terms/digit: #{Float(n_terms) /digits}" ) puts( "dd.size: #{dd.size} nn.size: #{nn.size}" ) puts z if compare?( $gutenberg_digits, z ) puts "Correct!!" else puts "Not correct." end end def compare?( ssx, ssy ) sx = ssx.delete( " \n" ) sy = ssy.delete( " \n" ) sx.start_with?( sy ) end # Digits from http://www.gutenberg.org/files/127/127.txt <http://www.gutenberg.org/files/127/127.txt> # Computed by Robert Nemiroff and Jerry Bonnell. # See the file itself for details, license, copyrights, etc. $gutenberg_digits = <<HERE 2. 7182818284590452353602874713526624977572470936999595749669676277240766303535 47594571382178525166427427466391932003059921817413596629043572900334295260595630 73813232862794349076323382988075319525101901157383418793070215408914993488416750 #### thousands of digits removed from email posting #### HERE $gutenberg_digits = $gutenberg_digits.delete( " \n" ) puts "gutenberg_digits.length: #{$gutenberg_digits.length}"

on 2010-01-14 18:46

Hello, 2010/1/8 Daniel Moore <yahivin@gmail.com> > > Suggestions?: http://rubyquiz.strd6.com/suggestions > defined is called the exponential function, and its inverse is the > Have fun! > > [1]: http://en.wikipedia.org/wiki/E_(mathematical_const... > [2]: http://en.wikipedia.org/wiki/Euler's_identity<http... > > -- > -Daniel > http://rubyquiz.strd6.com > > I got some fun playing with computing e. I didn't get any formal solution (it definitely takes too long ...) So, e.rb contains 8 basic methods to compute e, using basic Float of ruby(so don't exepect to get a ot of precision). Anyway, we can see the continuous fraction is good, and it's the only one who worked easily with BigDecimal(I got only 113 digits, shame on me ...). Here is the output of e.rb ------------------------------------------------------- 2.718281828459045 lim(n->âˆž) (1+1/n)**n with n = 100000000 2.7182817983473577 3.011168736577474e-08 lim(n->0) (1+n)**(1/n) with n = 1.0e-08 2.7182817983473577 3.011168736577474e-08 Î£(n=0,âˆž) 1/n! with n = 17 2.7182818284590455 -4.440892098500626e-16 lim(n->âˆž) n/(âˆš(n,n!)) with n : 170 2.663087878748024 0.055193949711020984 [[2;1,2,1,1,4,1,1,6,1,1,8,1,1,...,2n,1,1,...]] with 23 numbers 2.718281828459045 0.0 [[1,0,1,1,2,1,1,4,1,1,6,1,1,8,1,1,...]] with 25 numbers 2.718281828459045 0.0 [[1,0.5,12,5,28,9,44,13,60,17,...,4(4n-1),4n+1,...]] with 10 numbers 2.718281828459045 0.0 Global maximum of f(x) = âˆš(x,x) with p = 16 2.7182818284590446 4.440892098500626e-16 ------------------------------------------------------- And I got also a strange thing. I saw somewhere it's possible to compute e with lim(n->inf) ((2n+1)/(2n-1))**n, but changing the limit to 0 gives Ï€ (hum, the complex part of the result, divided by n), awesome :D (or this is surprising me at least :) ) include Math p E # 2.718281828459045 puts formula = -> n { ((2*n+1)/(2*n-1).to_f)**n } puts "lim(n->inf) ((2n+1)/(2n-1))**n with n = #{n = 100_000}" puts "e" p e = formula[n] # 2.718281828493031 p e - E # 3.398570314061544e-11 puts "lim(n->0) ((2n+1)/(2n-1))**n with n = #{n = 1.0/100_000_000}" puts "Ï€" p pi = formula[n] # (1.0+3.141592653589794e-08i) p pi = pi.imaginary / n # 3.1415926535897936 p pi - PI # 4.440892098500626e-16 Cheers, B.D.

on 2010-01-18 05:20

> ## Digits of e (#226) > > This weekï¿½s quiz is to write a Ruby program that can compute the first > 100,000 digits of e. $ cat e.rb digits = 100000 fudge = 10 unity = 10**(digits + fudge) e = unity n = unity i = 0 while (n>0) i += 1 n /= i e += n end e /= 10**fudge p e $ time ruby e.rb > /dev/null real 0m4.023s user 0m3.940s sys 0m0.087s This uses the taylor series definition of e which actually converges quite fast. -----Jay

on 2010-03-09 00:11

I decided to work on an old quiz. ---------------------- > This week's quiz is to write a Ruby program that can compute the first > 100,000 digits of e. Here is what I came up with: --------------------------------------------------------- PLACES = 100_000 euler = 0 big_one = 10**(PLACES+5) nxt_one = big_one n = 1 while (nxt_one != 0) n += 1 nxt_one /= n euler += nxt_one end puts "2."+euler.to_s[0..PLACES-1] --------------------------------------------------------- It ended up looking a lot like Jay's solution.

on 2010-04-22 23:24

This quiz harkens back to another mathematically themed computation competition Digits of Pi (#202)[1]. Many of the same techniques that can be applied to the computation of pi can be applied to the computation of e, and there are some surprising results. Jack Rouse used the fast converging continued fraction series from Wikipedia[2] to create a short program that quickly calculates e to 100_000 digits. Jack's solution is quite quick, yielding the output within several seconds. Jean-Julien Fleck started with a benchmark of the the standard library call and estimated that it would take 10**5 seconds to compute the first 100_000 digits. Jean-Julien provides a handy one liner that can be used in irb: fact = 1 ; (1..34_000).inject(1) {|sum,i| fact *= i ; sum + Rational(1,fact)} This computation makes use of the Taylor series definition. The results use Rational, so can be quite slow, it is advised to start with a smaller number of iterations, possibly 500 or so. Thorsten Hater created a program based on the spigot algorithm from Rabonitz and Wagon[3]. From the paper: > This algorithm is a "spigot" algorithm: it pumps out digits one > at a time and does not use the digits after they are computed [...]; > the entire algorithm uses only ordinary integer arithmetic on > relatively small integers. The paper is actually about computing the digits of pi with the same method, but in doing so covers the case of e, which is simpler. An interesting approach, and worth looking at if you are into mathematics. Benoit Daloze explored eight different methods of computing e with a focus on breadth rather than depth. See them all in the attached solutions supplement. Benoit also came across this interesting bit of information: lim(n->inf) ((2n+1)/(2n-1))**n => e lim(n->0) ((2n+1)/(2n-1))**n => Ï€ Jay Anderson and and David Springer used the Taylor series definition of e, which is quite fast at converging and delivering 100_000 digits. These solutions are much faster that other Taylor series implementations because they use only integer arithmatic and a very simple loop. Here is Jay Anderson's solution: digits = 100000 fudge = 10 unity = 10**(digits + fudge) e = unity n = unity i = 0 while (n>0) i += 1 n /= i e += n end e /= 10**fudge p e Setting unity to be 10^digits shifts everything into the integers so that floating point math won't be required. The extra fudge factor ensures that there aren't rounding errors near the end. Each step of the loop requires only an increment, an integer division, and an addition. Notice how the division in the loop accumulates the factorial, because the result of 1/2 is stored when it is divided by 3 it is set equal to 1/3!. The loop terminates when the series term becomes 0, i.e. too small to add anything that will matter within the chosen number of digits. Thanks everyone for your solutions to the quiz! [Digits of e (#226) - Solutions][4] P.S. Brian Candler linked to an xkcd comic involving e and Ï€[5], not all contributions need to be code. [1] http://rubyquiz.strd6.com/quizzes/202-digits-of-pi [2] http://en.wikipedia.org/wiki/E_(mathematical_constant) [3] http://www.mathpropress.com/stan/bibliography/spigot.pdf [4] http://rubyquiz.strd6.com/quizzes/226.tar.gz [5] http://xkcd.com/217/