Calculate the sum of all odd digits of pi in range [duplicate]

Question

Introduction

In the not so distant future with the AI revolution, we will need a way to solve their captchas to prove we're not humans.

This challenge was inspired by CommitStrip.

Challenge

One such captcha is the challenge of calculating the sum of all odd digits of Pi from digit 1 to digit n, including the first digit, being 3.

Output

• The calculated sum of all odd digits in the range 1 to n.

Rules

• All the digits of pi must be calculated by your program at runtime, how you implement that is up to you.
• Hard-coded constants of pi are not allowed.
• This is code-golf, so shortest bytes win.

Example

If n = 6, we must find the sum of the odd digits in [3, 1, 4, 1, 5, 9] meaning an output of 19

Other

Since this questions keeps getting marked as a duplicate of this, allow me to explain how this is different.

This question asks for the sum of a range of numbers, not finding the nth decimal.

Answer 1

Python3, 346 bytes

Based on a recursive implementation of the Chudnovsky algorithm, one of the fastest algorithms to estimate pi. For each iteration, roughly 14 digits are estimated (take a look here for further details).

def f(n):import decimal as d;d.getcontext().prec=n+10;q=lambda n,k=6,m=1,l=13591409,x=1,i=1:(d.Decimal(((k**3-16*k)*m//i**3)*(l+545140134))/(x*-262537412640768000)+q(n,k+12,(k**3-16*k)*m//i**3,l+545140134,x*-262537412640768000,i+1)if i<n else 13591409);return sum([ord(i)%2 and int(i)for i in str(426880*d.Decimal(10005).sqrt()/q(n//14+1))[:n+1]])

Ungolfed version

def f(n):
    import decimal as d
    d.getcontext().prec = n + 10
    q=lambda n, k=6, m=1, l=13591409, x=1, i=1:\
        (d.Decimal(((k ** 3 - 16 * k) * m // i**3) * (l + 545140134)) / (x * -262537412640768000) +
         q(n, k+12, (k ** 3 - 16 * k) * m // i**3, l + 545140134, x * -262537412640768000, i+1)if i < n else 13591409)
    return sum([ord(i) % 2 and int(i)for i in
                str(426880 * d.Decimal(10005).sqrt() / q(n // 14+1))[:n + 1]])

Function q computes the value of S(see the reference), than it is used to divide the term 426880 * d.Decimal(10005).sqrt() and get the value of the approximation of pi. The precision is controlled by setting the value of d.getcontext().prec.

A a little bit longer (372 bytes), more golfed but all-in-one version:

f=lambda n,k=6,m=1,l=13591409,x=1,i=0:not i and(exec('global d;import decimal as d;d.getcontext().prec=%d+10'%n)or sum([ord(i)%2 and int(i)for i in str(426880*d.Decimal(10005).sqrt()/f(n//14+1,k,m,l,x,1))[:n+1]]))or i<n and d.Decimal(((k**3-16*k)*m//i**3)*(l+545140134))/(x*-262537412640768000)+f(n,k+12,(k**3-16*k)*m//i**3,l+545140134,x*-262537412640768000,i+1)or 13591409

Thanks to O.O.Balance to push me to go deeper into this challenge: it was very fun and interesting.

Old answer, based on math.pi (74 bytes)

import math;f=lambda n:sum([ord(i)%2and int(i)for i in"%.*G"%(n,math.pi)])

Try it online!

N.B.: math.pi is "The mathematical constant π = 3.141592…, to available precision.", but the digits for the challenge are computed on the fly through the rounding and truncating operations. So, technically this is not a violation of the rules: "all the digits of pi must be calculated by your program at runtime."

Answer 2

Mathematica, 53 bytes

Total@DeleteCases[(RealDigits@N[Pi,#])[[1]],_?EvenQ]&

Mathematica noob, please excuse me.

Answer 3

Ruby, 78 bytes

f=->n{"#{eval'x=2*10**n+p*x/=2+p+p-=1;'*x=p=4*n}"[0,n].bytes.sum{|c|c%2*c%16}}

Computes only to necessary precision.

Try it online!

Answer 4

Husk, 7 bytes

As you can see here¹ the following program, does indeed compute all the digits itself:

Σf%2↑İπ

Try it online!

Explanation

Σf%2↑İπ  -- implicit input N, eg: 8
     İπ  -- compute all digits of pi: [3,1,4,1,5,9,2,6,5,3,...
    ↑    -- take N: [3,1,4,1,5,9,2,6]
 f%2     -- filter out even ones: [3,1,1,5,9]
Σ        -- sum: 19

^{1: This is a link to the Haskell code which is generated by Husk, in case you're lost (it's quite long) just search for func_intseq 'π'..}

Answer 5

Java 10, 275 bytes

import java.math.*;
n->{int j,t,i=0,s=3;for(;i<n-1;++i){var a=BigInteger.TEN.pow(10010).multiply(new BigInteger("2"));var p=a;for(j=1;a.signum()>0;p=p.add(a))a=a.multiply(new BigInteger(j+"")).divide(new BigInteger((2*j+++1)+""));s+=((t=(p+"").charAt(i+1)-48)%2)*t;}return s;}

Uses the approach from this answer by Kevin Cruijssen to calculate the digits. Try it online here.

Answer 6

Python 3, 119 bytes

y=sum([int(i)for i in str(sum([4/j if j%4%3 else-4/j for j in range(10**8)if j%2]))[0:n+1].replace('.','')if int(i)%2])

My first code-golf submit, so please be gentle ._.

This computes pi on the fly using the Gregory-Leibniz series. X may prove insufficient for big n, but be aware of the RAMifications of choosing a large x.

Explanation:

Computation of Pi:

i.e. sum( [4/i if i%4==1 else -4/i for i in range(x) if i%2==1] ) with x as large as sensible, also i%4==1 is one byte larger than i%4%3, so that's nice...

Casting a lot and eventual sum:

I'm sure there are better possibilities than all this, but I cast the float to string, take the first n*1 chars (including the dec point), replace the point, then cast each char in my string to int for the modulo check at the very end of the code, and take the sum of all i that are odd to assign them to y.

Room for improvement:

I'm sure there's lots, but I am the least happy with the series to compute pi, and also with the multiple int() casts ad the need to replace the dec point... HELP

Answer 7

SmallTalk – 299 bytes

Relies on the identity tan⁻¹(x) = x − x³/3 + x⁵/5 − x⁷/7 ..., and that π = 16⋅tan⁻¹(1/5) − 4⋅tan⁻¹(1/239), so no “cheats”.

|l a b c d e f g h p t s|l:=stdin nextLine asInteger. a:=1/5. b:=1/239. c:=a. d:=b. e:=a. f:=b. g:=3. h:=-1. s:=0. l timesRepeat:[c:=c*a*a. d:=d*b*b. e:=h*c/g+e. f:=h*d/g+f. g:=g+2. h:=0-h]. p:=4*e-f*4. l timesRepeat:[t:=p floor. p:=(p-t)*10. t odd ifTrue:[s:=s+t]]. Transcript show:s printString;cr

Save as pi.st. You provide the number of digits from the console when you run it, or through the <<< operator.

$ gst -q pi.st <<< 6
19