arrieta/benchmark-poly.py

## benchmark-poly.py
"""Simple re-arrangement can yield significant performance improvements in
naive, low-order polynomial evaluation. The interested reader may wish to read
about "Horner's method."

(C) 2018 Nabla Zero Labs
MIT License
"""
import time

def p0(a, x):
    """First strategy: naively calculate the polynomial.
    Three additions, three multiplications, one pow(2), one pow(3)."""
    for xi in x:
        y = a[0] + a[1] * xi + a[2] * xi**2 + a[3] * xi**3
    return y

def p1(a, x):
    """Second strategy: save multiplications at the cost of memory.
    Three additions, five multiplications, two extra variables"""
    for xi in x:
        xi2 = xi * xi
        xi3 = xi2 * xi
        y = a[0] + a[1] * xi + a[2] * xi2 + a[3] * xi3
    return y

def p2(a, x):
    """Third strategy: save multiplication with no cost.
    Three additions, three multiplications.
    """
    for xi in x:
        y = a[0] + xi * (a[1] + xi * ( a[2] + a[3] * xi))
    return y

def cpu(fun, *args):
    a, x = args
    tbeg = time.perf_counter()
    y = fun(a, x)
    tend = time.perf_counter()
    elap = tend - tbeg
    thro = len(x) / elap
    print(f"{fun.__name__}: {elap:.6f} sec ({thro:.0f} eval/sec)",
          f"last value: {y:23.16e}", sep=" ")

if __name__ == "__main__":
    import random

    a = [random.uniform(-10, 10) for _ in range(4)]
    x = [random.uniform(-1, 1) for _ in range(1000000)]

    for _ in range(10):
        cpu(p0, a, x)
        cpu(p1, a, x)
        cpu(p2, a, x)
	"""Simple re-arrangement can yield significant performance improvements in
	naive, low-order polynomial evaluation. The interested reader may wish to read
	about "Horner's method."

	(C) 2018 Nabla Zero Labs
	MIT License
	"""
	import time

	def p0(a, x):
	"""First strategy: naively calculate the polynomial.
	Three additions, three multiplications, one pow(2), one pow(3)."""
	for xi in x:
	y = a[0] + a[1] * xi + a[2] * xi*2 + a[3] xi**3
	return y

	def p1(a, x):
	"""Second strategy: save multiplications at the cost of memory.
	Three additions, five multiplications, two extra variables"""
	for xi in x:
	xi2 = xi * xi
	xi3 = xi2 * xi
	y = a[0] + a[1] * xi + a[2] * xi2 + a[3] * xi3
	return y

	def p2(a, x):
	"""Third strategy: save multiplication with no cost.
	Three additions, three multiplications.
	"""
	for xi in x:
	y = a[0] + xi * (a[1] + xi * ( a[2] + a[3] * xi))
	return y

	def cpu(fun, *args):
	a, x = args
	tbeg = time.perf_counter()
	y = fun(a, x)
	tend = time.perf_counter()
	elap = tend - tbeg
	thro = len(x) / elap
	print(f"{fun.__name__}: {elap:.6f} sec ({thro:.0f} eval/sec)",
	f"last value: {y:23.16e}", sep=" ")

	if __name__ == "__main__":
	import random

	a = [random.uniform(-10, 10) for _ in range(4)]
	x = [random.uniform(-1, 1) for _ in range(1000000)]

	for _ in range(10):
	cpu(p0, a, x)
	cpu(p1, a, x)
	cpu(p2, a, x)