/** This program tests floating-point timing for a large number of digits. It also calculates the Big-O notation (which should be O(n^something)) where n is the number of digits being calculated. */ runs = million lastTime = undef lastIterations = undef lastPrecision = undef n1 = 1. d1 = 7. n2 = 2 d2 = 17. for n = 1 to 3 { lastIterations = 0 lastTime = 0 s lastPrec = 0 multifor [p, m] = [1 to 7, [1,2]] { prec = m 10^p iterations = ceil[10^(7-p) / m] setPrecision[prec] a = n1/d1 b = n2/d2 start = now[] for r = 1 to iterations { c = a / b } end = now[] setPrecision[20] time = end-start print["precision=\$prec "] if m == 2 print[" "] print["iterations=\$iterations "] print["Time:" (time -> "ms")] // Calculate Big-O exponent O(n^x) if lastTime != undef and lastTime != 0 s and time != 0 s and lastIterations != 0 and iterations != 0 and lastPrec != 0 { iterTime = time/iterations lastIterTime = lastTime / lastIterations x = log[iterTime/lastIterTime] / log[prec/lastPrec] // t = c prec^x (solve for c) // Constant factor c solved via // https://frinklang.org/fsp/solve.fsp?eq=t+%3D+c+prec%5Ex&solveFor=c c = prec^(-x) iterTime println[" O(n^" + formatFix[x,1,3] + ")\t" + formatSig[c,"s",6] + " n^" + formatFix[x,1,3]] } else println[] lastTime = time lastPrec = prec lastIterations = iterations } }