trjhtr

I have written code in Kotlin with the objective of computing Pi in few enough lines so that it looks good on a t-shirt.

Can be cut and paste into http://try.kotlin.org under "My Programs" and run from browser - I just tested it with version 1.2.41 and it's working.

import kotlin.math.*
import java.math.BigInteger 

fun main(args: Array<String>) 
 val r = (4*(4*arccot(5) - arccot(239))).toString()
 println("314 digits of Pi $r[0].$r.substring(1).dropLast(5)")


fun arccot(x:BigInteger):BigInteger 
 var precision = 10.toBigInteger().pow(319) / x
 var total = precision
 var divisor = 1.toBigInteger()

 while(precision.abs() >= divisor) 
 precision = -precision / x.pow(2)
 divisor += 2
 total += precision / divisor
 
 return total


fun arccot(x:Int) = arccot(x.toBigInteger())
operator fun Int.times(x: BigInteger) = this.toBigInteger() * x
operator fun BigInteger.plus(x: Int) = this + x.toBigInteger()

Currently it's longer than I'd like. I would like to shorten without making it less understandable. My vision is to have code that is readable enough it wouldn't be out of place in a production code base.

To give an idea, here's the significantly shorter Python version (which has been printed on a t-shirt and in my opinion looks good and is short enough but also quite readable). Can be run in browser here: https://repl.it/@sek/314-Digits - (there's also a link from there to the t-shirt if you are curious how that looks - the length in question isn't only the number of lines but also the width of the longest line as that determines the font size that can be used)

def pi():
 r = 4*(4*arccot(5) - arccot(239))
 return str(r)[0]+'.'+str(r)[1:-5]

def arccot(x):
 total = power = 10**319 // x
 divisor = 1
 while abs(power) >= divisor:
 power = -power // x**2
 divisor += 2
 total += power // divisor
 return total

print("314 digits of Pi " + pi())

I noticed that you are calculating to 314 decimal places instead of 314 digits, so drop 6 instead of 5.

• 
  

  You really don't need the extra functions. You can remove

  
  
  

  fun arccot(x:Int) = arccot(x.toBigInteger())
  

  
  
  

  if you convert Int to BigInteger inside your arctan method.

  
  


• 
  

  You can remove

  
  
  

  operator fun Int.times(x: BigInteger) = this.toBigInteger() * x
  

  
  
  

  if you shl(2) instead of * 4.

  
  


• 
  

  You can remove

  
  
  

  operator fun BigInteger.plus(x: Int) = this + x.toBigInteger()
  

  
  
  

  if you change

  
  
  

  divisor += 2
  

  
  
  

  to

  
  
  

  divisor += BigInteger("2")
  

  
  


• 
  

  You can shorten

  
  
  

  println("314 digits of Pi $r[0].$r.substring(1).dropLast(5)")
  

  
  
  

  to

  
  
  

  println("314 digits of Pi $r[0].$r.substring(1,314)")
  

  
  


• 
  

  Now, you can change the imports to

  
  
  

  import java.math.*
  

  
  



• You don't have to declare the type for val r.

The final code I came up with by doing that is:

import java.math.*
fun main(args: Array<String>) 
 val r = (acot(5).shl(2)-acot(239)).shl(2).toString()
 println("314 digits of Pi $r[0].$r.substring(1,314)")

fun acot(x:Int):BigInteger 
 var precision = BigInteger.TEN.pow(319)/x.toBigInteger()
 var total = precision
 var divisor = BigInteger.ONE;
 while(precision.abs() >= divisor) 
 precision = -precision/(x.toBigInteger().pow(2))
 divisor += BigInteger("2")
 total += precision / divisor
 
 return total

I also came up with shorter code, with decreased width, by calculating a different way:

import java.math.*
fun main(args:Array<String>) 
 val b4 = BigDecimal(4)
 val r = ((atan(5)*b4-atan(239))*b4).toString()
 println("314 digits of Pi $r.substring(0,315)")

fun atan(xInv:Int):BigDecimal 
 var x = BigDecimal(1).divide(BigDecimal(xInv),330,3)
 var (numer, total) = arrayOf(x, x)
 for (i in 3..450 step 2) 
 numer = -numer * x * x
 total += numer / BigDecimal(i)
 
 return total