On Tue, 26 May 2015 05:21 am, ravas wrote:
> I read an interesting comment:
> """
> The coolest thing I've ever discovered about Pythagorean's Theorem is an
> alternate way to calculate it. If you write a program that uses the
> distance form c = sqrt(a^2 + b^2) you will suffer from the lose of half of
> your available precision because the square root operation is last. A more
> accurate calculation is c = a * sqrt(1 + b^2 / a^2). If a is less than b,
> you should swap them and of course handle the special case of a = 0. """
Let's compare three methods.
import math
import random
def naive(a, b):
return math.sqrt(a**2 + b**2)
def alternate(a, b):
a, b = min(a, b), max(a, b)
if a == 0: return b
if b == 0: return a
return a * math.sqrt(1 + b**2 / a**2)
counter = 0
print("Type Ctrl-C to exit")
while True:
counter += 1
a = random.uniform(0, 1000)
b = random.uniform(0, 1000)
d1 = naive(a, b)
d2 = alternate(a, b)
d3 = math.hypot(a, b)
if not (d1 == d2 == d3):
print("mismatch after %d trials" % counter)
print("naive:", d1)
print("alternate:", d2)
print("hypot:", d3)
break
When I run that, I get:
mismatch after 3 trials
naive: 1075.6464259886257
alternate: 1075.6464259886257
hypot: 1075.6464259886254
A second run gives:
mismatch after 3 trials
naive: 767.3916150255787
alternate: 767.3916150255789
hypot: 767.3916150255787
which shows that:
(1) It's not hard to find mismatches;
(2) It's not obvious which of the three methods is more accurate.
--
Steven
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