Bonus Puzzle #1 Random Doubles [0,5]

1. Bonus Puzzle #1Random Doubles [0,5] Instructor: Scott Kristjanson CMPT 125/125 SFU Burnaby, Fall 2013

2. Bonus Puzzle #1 Write an expression to calculate double random numbers from [0, 5.000000000] inclusive Note: Generating the range: [0, 5.0) is easy: generator.nextFloat()*5

3. Honourable Mentions • Some good tries by: • Amritpaul Gill • Gavin Haynes • Adam Tuck • A working solution by Raymond Zeng • But slow convergence, not guaranteed to ever complete

4. And the Winner is…. • Serena Read: • Simple elegant well commented solution • Returns entire range with [0,5] inclusive • Fast convergence, but still may never terminate public static float ranNum() { Random ran = new Random(); int a; float f, r; do{ a = ran.nextInt(6); f = ran.nextFloat(); r = (float)a - f; } while(r<0); return r; }

5. Scott’s solution • Can we improve on Serena’s and Raymond’s solution? • Want guaranteed convergence • Want to eliminate loops • Want to cover the whole range uniformly (this last bullet is much trickier than you think!)

6. BUT DO WE WANT TO SOLVE THIS? • What is the practical difference between [0,5) and [0,5] ? • Remember when you compare floats, we useTolerances • That means 4.999999 equals 5.000000 within a tolerance • How many real numbers are in between 4.999999 and 5.000000? ∞ • How many java floating point numbers are in this gap? ONE : 4.9999995 • That is because floats only have 32 bits to represent the number • There is limited precision, and thus limited resolution. • Do we WANT to put all this effort into something whose probability of occurring approaches zero in the limit? It took over 2 million tries for Serena’s algorithm to return 5.000000

7. Representing Real numbers in Java [2] • Java uses the IEEE 754 standard:This uses 1-plus form of the binary normalized fraction (rounded). • The fraction part is called the mantissa. • 1-plus normalized scientific notation base two is then: • N = ± (1.b1b2b3b4 ...)2 x 2+E • For a great overview of this standard, see [2] at: • http://people.uncw.edu/tompkinsj/133/Numbers/Reals.htm float double

8. Ok, so you really want to solve this… • Only missing a single finite range with [0,5) • If we know the size of this gap, • We can add it in with 50/50 probability to close the gap uniformly generator.nextFloat()*5 + ((generator.nextInt() >= 0) ? epsilon : 0); • What’s epsilon? • It is based on the resolution (# bits in real number) plus the exponent (smaller magnitude numbers have finer resolution). • epsilon = (.5)23 = 0.00000000000000000000001B

9. But this assumes nextFloat works as expected… • Yes, nextFloat works, but is does not cover the entire [0,1) space uniformly. • nextFloat produces a random number based on 224 possible floats within this range (see [3] for details). • However there are really 230 possible floats in the range [0,1) using the IEEE 754 standard (see [2] for details). • That’s only about 1.6% of all possible floats that get returned. • If you really want to get them all, you will need to add in a 6 bit random number to the end of what nextFloat returns. • Instead of using a constant Epsilon as in previous slide, it needs to become a 6-bit random variable.

10. References: • J. Lewis, P. DePasquale, and J. Chase., Java Foundations: Introduction to Program Design & Data Structures. Addison-Wesley, Boston, Massachusetts, 3rd edition, 2014, ISBN 978-0-13-337046-1 • Tompkins, J.A. , Java Primitive Data Types - Reals - IEEE754http://people.uncw.edu/tompkinsj/133/Numbers/Reals.htm • Oracle Reference Pages on the Random Class • http://docs.oracle.com/javase/7/docs/api/java/util/Random.html#nextFloat%28%29