# "Bytes of Pi" by David Fraser

This is a talk on how to calculate Pi. It was originally done for bright high school maths students - so it is also an example of an approach to interesting such students in a programming career. It includes a brief tour of the algorithm development behind calculating Pi to many decimal places, and an example of making a functioning web page that does this. And a mug. All with an iPython notebook to illustrate things along the way.

October 08, 2020

## Transcript

3. ### 3 Outline | hexagon.com • Hand-Calculating Pi • Polygons, Pythagoras,

Archimedes • Infinite Series • Decimal, Binary and Bases • Computer Calculations • Spigot Algorithms • Coffee Mugs 2020 10 = 1 1 1 1 1 1 0 0 1 0 0 2 = 210 +29+28+27+26+25 +22 = 7 E 4 16 = 7x163 + 14x162 + 24x160

Pi by Hand

8. ### 8 Like Archimedes, but squares Working out increasingly accurate values

Exponent Sides Lower Bound Upper Bound 2 4 2.828427 4.000000 3 8 3.061467 3.313708 4 16 3.121445 3.182598 5 32 3.136548 3.151725 11 2048 3.141591 3.141595

10. ### 10 Srinivasa Ramanujan Infinite Series 1 3.141592730013305523328881463385 2 3.141592653589794004176383168669 3

3.141592653589793115997963468544 4 3.141592653589793115997963468544 5 3.141592653589793115997963468544
11. ### 11 Decimal, Binary and Computers Number Systems and Bases 2020

10 = 1 1 1 1 1 1 0 0 1 0 0 2 = 210 +29+28+27+26+25 +22 = 7 E 4 16 = 7x163 + 14x162 + 24x160
12. ### 12 Fixed bits, negative numbers, etc Storing Numbers in Computers

2020 10 in 16 bits: = 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 2 -2020 10 , using a sign bit = 1 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 2 -2020 10 , using two’s complement: invert bits, +1 = 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 2 calculating 2020 10 +(-2020 10 ): 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 2 + 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 2 = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2
13. ### 13 Floating point, etc Storing Numbers in Computers sign bit

exponent mantissa 0 10000000 10010010000111111011011 2 Value = (-1)sign x 2exponent-127 x (1 + 0.mantissa) How does pi look in a floating point number? sign bit exponent mantissa = 0 10000000 10010010000111111011011 2 which translates to = (-1)0 x 21 x (1 + 0.5707963705062866) = 3.1415927410125732
14. ### 14 Srinivasa Ramanujan Infinite Series 1 3.1415927300133056603139961890252155185995816071102 2 3.1415926535897938779989058263060130942166450293230 3

3.1415926535897932384626490657027588981566774804625 4 3.1415926535897932384626433832795552731599742104206 5 3.1415926535897932384626433832795028841976638181331 6 3.1415926535897932384626433832795028841971693993799
15. ### 15 Record # of digits of Pi calculated Era of

Computers Year Record-holder Time Digits of Pi 1949 G W Reitwiesner et al 70 hours 2,037 1958 Francois Genuys 1.7 hours 10,000 1961 Daniel Shanks & John Wrensch 8.7 hours 100,000 1973 Hean Guilloud and M. Dichampt 28 hours 1,001,250 1983 Yasumasa Kanada, Sayaka Yoshino and Yoshiaki Tamura 16,777,206 1987 Yasumasa Kanada, Yoshiaki Tamura, Yoshinobu Kubo et al 134,214,700 1989 Gregory V. Chudnovsky & David V. Chudnovsky 480,000,000 1991 Gregory V. Chudnovsky & David V. Chudnovsky 2,260,000,000 1997 Yasumasa Kanada and Daisuke Takahashi 51,539,600,000 1999 Yasumasa Kanada and Daisuke Takahashi 206,158,430,000 2002 Yasumasa Kanada & 9 person team (64 computing nodes) 600 hours 1,241,100,000,000 2009 Fabrice Bellard (with home computer) 131 days 2,699,999,990,000 2010 Shigeru Kondo (with home computer) 90 days 5,000,000,000,000 2011 Shigeru Kondo (with home computer) 371 days 10,000,000,000,050 2019 Emma Haruka Iwao (using cloud computer) 121 days 31,415,926,535,897 2020 Timothy Mullican (using home computers) 303 days 50,000,000,000,000

17. ### 17 Multi-Radix, Time and Irrationality Varying-Base Number Systems 2020-10-08 16:30:00

2020*365 + 10*30 + 8 + 16/24 + 30/(24*60) + 00/(24*60*60) = 2020 365/30 1030 8 1624 3060 060

19. ### 19 Some actual code Spigot Algorithms import sys write =

sys.stdout.write n = 1000 l = 10*n//3 a = [2 for j in range(0, l)] nines, predigit, started = 0, 0, False for j in range(0, n): q = 0 for i in range(l-1, -1, -1): x = 10*a[i] + q*(i+1) a[i] = x % (2*i+1) q = x // (2*i+1) q, a = divmod(q, 10) if q == 9: nines += 1 elif q == 10: write(str(predigit+1) + '0'*nines) nines, predigit = 0, 0 else: if started or predigit or nines: write(str(predigit) + '9'*nines) if not started: sys.stdout.write('.') started = True nines, predigit = 0, q write(str(predigit) + '\n') 3.1415926535897932384626433832795028841971693993751058209749445 923078164062862089986280348253421170679821480865132823066470938 446095505822317253594081284811174502841027019385211055596446229 489549303819644288109756659334461284756482337867831652712019091 456485669234603486104543266482133936072602491412737245870066063 155881748815209209628292540917153643678925903600113305305488204 665213841469519415116094330572703657595919530921861173819326117 931051185480744623799627495673518857527248912279381830119491298 336733624406566430860213949463952247371907021798609437027705392 171762931767523846748184676694051320005681271452635608277857713 427577896091736371787214684409012249534301465495853710507922796 892589235420199561121290219608640344181598136297747713099605187 072113499999983729780499510597317328160963185950244594553469083 026425223082533446850352619311881710100031378387528865875332083 814206171776691473035982534904287554687311595628638823537875937 51957781857780532171226806613001927876611195909216420198