Infinite loop when trying to find cube root of a non-perfect cube through bisection search

问题

This code gives pretty accurate result when deltanum = 0.0000000000001 but gets into an infinite loop when deltanum = 0.00000000000001(adding another zero into deltanum).

It occurs only for non-perfect cubes, it works fine for perfect cubes like 1000. Why?

I am new to programming, following OSSU.

num = 100
high = num
low = 0
icount = 0
cuberoot = (high + low)/2      #cuberoot of num
deltanum = 0.00000000000001
while abs(cuberoot**3 - num)>=deltanum:
    icount+=1
    print(icount)
    if cuberoot**3 > num:
        high = cuberoot
    elif cuberoot**3 < num:
        low = cuberoot
    else:
        break
    cuberoot = (high + low)/2
print("Cube root: " + str(cuberoot))
print("Number of iterations: " + str(icount))

回答1:

You are using floats. float math is flawed precision wise - your delta might be to small to work correctly and your solution flipps between values without ever reaching your while conditions limit. See Is floating point math broken? for more reasoning about why float is sometimes "broken".

You can limit it to a certain amount of repeats as well:

num = 100
high = num
low = 0
icount = 0
maxcount = 100000
cuberoot = (high + low)/2      #cuberoot of num
deltanum = 0.00000000000001
while abs(cuberoot**3 - num)>=deltanum:
    icount+=1
    print(icount)
    if cuberoot**3 > num:
        high = cuberoot
    elif cuberoot**3 < num:
        low = cuberoot
    else:
        break
    cuberoot = (high + low)/2

    if icount > maxcount:
        print("Unstable solution reached after ",maxcount, "tries")
        break
print("Cube root: " + str(cuberoot) + " yields " + str(cuberoot**3))
print("Number of iterations: " + str(icount))

Output:

Unstable solution reached after  100000 tries
Cube root: 4.641588833612779 yields 100.00000000000003
Number of iterations: 100001

回答2:

Most people would name that epsilon, and would use delta for cuberoot**3 - num.

Toward the end, you're hoping that this expression,

    cuberoot = (high + low)/2

will buy you approximately one more bit of precision in your answer on each iteration. (Near the beginning it approximately cuts the number of error bits in half each time.)

You're complaining that an IEEE-754 double has limited precision when computing cubes, and also differences. Fifty-three bits gives you almost sixteen decimal digits, and your epsilon is 1e-14. But an input num that is a few digits long will eat away at your margin, as you found.

For higher precision computations, you might prefer to use Decimal. Or, look into the gmp library.

You believe a certain loop invariant holds, that the quanities cuberoot and cuberoot ** 3 will change on each iteration. It's simple to verify. Just assign them to temp variables, and verify they change. Terminate the loop early if they don't. More generally, to detect oscillation between a handful of limiting values, put previous values in a set and terminate early upon seeing a repeat value.

来源：https://stackoverflow.com/questions/55315519/infinite-loop-when-trying-to-find-cube-root-of-a-non-perfect-cube-through-bisect

标签

python

python-3.x

bisection