Find the nth decimal of pi

Question

There are already 30 challenges dedicated to pi but not a single one asks you to find the nth decimal, so...

Challenge

For any integer in the range of 0 <= n <= 10000 display the nth decimal of pi.

Rules

• Decimals are every number after 3.
• Your program may be a function, or a full program
• You must output the result in base 10
• You may get n from any suitable input method (stdin, input(), function parameters, ...), but not hardcoded
• You may use 1-based indexing if that's native to your language of choice
• You don't have to deal with invalid input (n == -1, n == 'a' or n == 1.5)
• Builtins are allowed, if they support up to at least 10k decimals
• Runtime doesn't matter, since this is about the shortest code and not the fastest code
• This is code-golf, shortest code in bytes wins

Test cases

f(0)     == 1
f(1)     == 4 // for 1-indexed languages f(1) == 1
f(2)     == 1 // for 1-indexed languages f(2) == 4
f(3)     == 5
f(10)    == 8
f(100)   == 8
f(599)   == 2
f(760)   == 4
f(1000)  == 3
f(10000) == 5

For reference, here are the first 100k digits of pi.

Answer 1

05AB1E, 3 bytes

žs¤

Explained

žs   # push pi to N digits
  ¤  # get last digit

Try it online

Uses 1-based indexing.
Supports up to 100k digits.

Answer 2

Python 2, 66 bytes

n=input()+9
x=p=5L**7
while~-p:x=p/2*x/p+10**n;p-=2
print`x/5`[-9]

Input is taken from stdin.

---

Sample Usage

$ echo 10 | python pi-nth.py
8

$ echo 100 | python pi-nth.py
8

$ echo 1000 | python pi-nth.py
3

$ echo 10000 | python pi-nth.py
5

Answer 3

Bash + coreutils, 60 49 bytes

echo "scale=10100;4*a(1)"|bc -l|tr -d '\\\n'|cut -c$(($1+2))

bc -l<<<"scale=$1+9;4*a(1)-3"|tr -dc 0-9|cut -c$1

Improved by Dennis. Thanks!

The index is one-based.

Answer 4

Python 2, 73 71 73 bytes

thanks to @aditsu for increasing my score by 2 bytes

Finally an algorithm that can complete under 2 seconds.

n=10**10010
a=p=2*n
i=1
while a:a=a*i/(2*i+1);p+=a;i+=1
lambda n:`p`[n+1]

Ideone it!

Uses the formula pi = 4*arctan(1) while computing arctan(1) using its taylor series.

Answer 5

MATL, 11 10 bytes

1 byte saved thanks to @Luis

YPiEY$GH+)

This solution utilizes 1-based indexing

Try it Online

All test cases

Explanation

YP  % Pre-defined literal for pi
iE  % Grab the input and multiply by 2 (to ensure we have enough digits to work with)
Y$  % Compute the first (iE) digits of pi and return as a string
G   % Grab the input again
H+  % Add 2 (to account for '3.') in the string
)   % And get the digit at that location
    % Implicitly display the result

Answer 6

Mathematica 30 bytes

RealDigits[Pi,10,1,-#][[1,1]]&

---

f=%

f@0
f@1
f@2
f@3
f@10
f@100
f@599
f@760
f@1000
f@10000

1
4
1
5
8
8
2
4
3
5

Answer 7

CJam, 32

7e4,-2%{2+_2/@*\/2e10005+}*sq~)=

Try it online (it's a bit slow)

Answer 8

Sage, 32 25 bytes

lambda d:`n(pi,9^5)`[d+2]

My first answer in a language of this kind.

n rounds pi to 17775 digits.

Answer 9

Mathematica, 23 21 bytes

⌊10^# Pi⌋~Mod~10&

SageMath, 24 bytes

lambda n:int(10^n*pi)%10

Answer 10

J, 19 15 bytes

10([|<.@o.@^)>:

Takes an integer n and outputs the n^th digit of pi. Uses zero-based indexing. To get the n^th digit, compute pi times 10ⁿ⁺¹, take the floor of that value, and then take it modulo 10.

Usage

The input is an extended integer.

   f =: 10([|<.@o.@^)>:
   (,.f"0) x: 0 1 2 3 10 100 599 760 1000
   0 1
   1 4
   2 1
   3 5
  10 8
 100 8
 599 2
 760 4
1000 3
   timex 'r =: f 10000x'
1100.73
   r
5

On my machine, it takes about 18 minutes to compute the 10000^th digit.

Explanation

10([|<.@o.@^)>:  Input: n
             >:  Increment n
10               The constant n
           ^     Compute 10^(n+1)
        o.@      Multiply by pi
     <.@         Floor it
   [             Get 10
    |            Take the floor modulo 10 and return

Answer 11

Clojure, 312 bytes

(fn[n](let[b bigdec d #(.divide(b %)%2(+ n 4)BigDecimal/ROUND_HALF_UP)m #(.multiply(b %)%2)a #(.add(b %)%2)s #(.subtract % %2)](-(int(nth(str(reduce(fn[z k](a z(m(d 1(.pow(b 16)k))(s(s(s(d 4(a 1(m 8 k)))(d 2(a 4(m 8 k))))(d 1(a 5(m 8 k))))(d 1(a 6(m 8 k)))))))(bigdec 0)(map bigdec(range(inc n)))))(+ n 2)))48)))48)))

So, as you can probably tell, I have no idea what I'm doing. This ended up being more comical than anything. I Google'd "pi to n digits", and ended up on the Wikipedia page for the Bailey–Borwein–Plouffe formula. Knowing just barely enough Calculus(?) to read the formula, I managed to translate it into Clojure.

The translation itself wasn't that difficult. The difficulty came from handling precision up to n-digits, since the formula requires (Math/pow 16 precision); which gets huge really fast. I needed to use BigDecimal everywhere for this to work, which really bloated things up.

Ungolfed:

(defn nth-pi-digit [n]
  ; Create some aliases to make it more compact
  (let [b bigdec
        d #(.divide (b %) %2 (+ n 4) BigDecimal/ROUND_HALF_UP)
        m #(.multiply (b %) %2)
        a #(.add (b %) %2)
        s #(.subtract % %2)]
    (- ; Convert the character representation to a number...
      (int ; by casting it using `int` and subtracting 48
         (nth ; Grab the nth character, which is the answer
           (str ; Convert the BigDecimal to a string
             (reduce ; Sum using a reduction
               (fn [sum k]
                 (a sum ; The rest is just the formula
                       (m
                         (d 1 (.pow (b 16) k))
                         (s
                           (s
                             (s
                               (d 4 (a 1 (m 8 k)))
                               (d 2 (a 4 (m 8 k))))
                             (d 1 (a 5 (m 8 k))))
                           (d 1 (a 6 (m 8 k)))))))
               (bigdec 0)
               (map bigdec (range (inc n))))) ; Create an list of BigDecimals to act as k
           (+ n 2)))
      48)))

Needless to say, I'm sure there's an easier way to go about this if you know any math.

(for [t [0 1 2 3 10 100 599 760 1000 10000]]
  [t (nth-pi-digit t)])

([0 1] [1 4] [2 1] [3 5] [10 8] [100 8] [599 2] [760 4] [1000 3] [10000 5])

Answer 12

Clojure, 253 bytes

(defmacro q[& a] `(with-precision ~@a))(defn h[n](nth(str(reduce +(map #(let[p(+(* n 2)1)a(q p(/ 1M(.pow 16M %)))b(q p(/ 4M(+(* 8 %)1)))c(q p(/ 2M(+(* 8 %)4)))d(q p(/ 1M(+(* 8 %)5)))e(q p(/ 1M(+(* 8 %)6)))](* a(-(-(- b c)d)e)))(range(+ n 9)))))(+ n 2)))

Calculate number pi using this formula. Have to redefine macro with-precision as it's used too frequently.

You can see the output here: https://ideone.com/AzumC3 1000 and 10000 takes exceeds time limit used on ideone, shrugs

Answer 13

Python 3, 338 bytes

This implementation is based on 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).

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'%(n+7))or str(426880*d.Decimal(10005).sqrt()/f(n//14+1,k,m,l,x,1))[n+2])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

Try it online!

Answer 14

Smalltalk – 270 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). SmallTalk uses unlimited precision integer arithmetic so it will work for large inputs, if you're willing to wait!

|l a b c d e f g h p t|l:=stdin nextLine asInteger+1. a:=1/5. b:=1/239. c:=a. d:=b. e:=a. f:=b. g:=3. h:=-1. 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]. Transcript show:t printString;cr

Save as pi.st and run as in the following test cases. Indexing is one based.

$ gst -q pi.st <<< 1
1
$ gst -q pi.st <<< 2
4
$ gst -q pi.st <<< 3
1
$ gst -q pi.st <<< 4
5
$ gst -q pi.st <<< 11
8
$ gst -q pi.st <<< 101
8
$ gst -q pi.st <<< 600
2
$ gst -q pi.st <<< 761
4
$ gst -q pi.st <<< 1001
3
$ gst -q pi.st <<< 10001 -- wait a long time!
5

Answer 15

JavaScript (Node.js) (Chrome 67+), 75 73 67 63 bytes

n=>`${eval(`for(a=c=100n**++n*20n,d=1n;a*=d;)c+=a/=d+++d`)}`[n]

Try it online!

Using \$\pi/2=\sum_{k=0}^{\infty}k!/(2k+1)!!\$ (same logic used by Leaky Nun's Python answer, but thanks to the syntax of JS that makes this shorter). Input is passed to the function as a BigInt. 2 bytes can be removed if 1-based indexing is used:

n=>`${eval(`for(a=c=100n**n*20n,d=1n;a*=d;)c+=a/=d+++d`)}`[n]

JavaScript (Node.js) (Chrome 67+), 90 89 bytes

n=>`${eval(`for(a=100n**++n*2n,b=a-a/3n,c=0n,d=1n;w=a+b;a/=-4n,b/=-9n,d+=2n)c+=w/d`)}`[n]

Try it online!

Using \$\pi/4=\arctan(1/2)+\arctan(1/3)\$. Input is passed to the function as a BigInt. 2 bytes can be removed if 1-based indexing is used:

n=>`${eval(`for(a=100n**n*2n,b=a-a/3n,c=0n,d=1n;w=a+b;a/=-4n,b/=-9n,d+=2n)c+=w/d`)}`[n]

Answer 16

Java 7, 262 260 bytes

import java.math.*;int c(int n){BigInteger p,a=p=BigInteger.TEN.pow(10010).multiply(new BigInteger("2"));for(int i=1;a.compareTo(BigInteger.ZERO)>0;p=p.add(a))a=a.multiply(new BigInteger(i+"")).divide(new BigInteger((2*i+++1)+""));return(p+"").charAt(n+1)-48;}

Used @LeakyNun's Python 2 algorithm.

Ungolfed & test code:

Try it here.

import java.math.*;
class M{
  static int c(int n){
    BigInteger p, a = p = BigInteger.TEN.pow(10010).multiply(new BigInteger("2"));
    for(int i = 1; a.compareTo(BigInteger.ZERO) > 0; p = p.add(a)){
      a = a.multiply(new BigInteger(i+"")).divide(new BigInteger((2 * i++ + 1)+""));
    }
    return (p+"").charAt(n+1) - 48;
  }

  public static void main(String[] a){
    System.out.print(c(0)+", ");
    System.out.print(c(1)+", ");
    System.out.print(c(2)+", ");
    System.out.print(c(3)+", ");
    System.out.print(c(10)+", ");
    System.out.print(c(100)+", ");
    System.out.print(c(599)+", ");
    System.out.print(c(760)+", ");
    System.out.print(c(1000)+", ");
    System.out.print(c(10000));
  }
}

Output:

1, 4, 1, 5, 8, 8, 2, 4, 3, 5

Answer 17

R, 59 bytes

m=scan()+2;substr(gsub("\\.","",numbers::dropletPi(m)),m,m)

Try it online!

Took an idea from a similar challenge...

Answer 18

Maple, 24 bytes

 trunc(10^(n+1)*Pi)mod 10

Test cases:

> f:=n->trunc(10^(n+1)*Pi)mod 10;
> f(0);
  1
> f(1);
  4
> f(2);
  1
> f(3);
  5
> f(10);
  8
> f(100);
  8
> f(599);
  2
> f(760);
  4
> f(1000);
  3
> f(10000);
  5