Explain the logic and meaning of this Python code.

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Python

x = 25
epsilon = 0.01
step = epsilon**2
numGuesses = 0
ans = 0.0

while (abs(ans**2 - x)) >= epsilon and ans <= x:
    ans += step
    numGuesses += 1

print('numGuesses = ' + str(numGuesses))

if abs(ans**2-x) >= epsilon:
    print('Failed on square root of ' + str(x))

else:
    print(str(ans) + ' is close to the square root of ' + str(x))

What I have tried:

How does this program determine the square root?
Ans: I understood it. It reduces x by a small value i.e. 0.0001 i.e. step until x is less than epsilon. ans is incremented by ans += step. So when it reaches to the value greater than or equal to x, loop terminates.

I am not sure about this ans I wrote, tell me is this right or wrong

Posted 22-Jun-16 1:35am

Member 12597839

Updated 22-Jun-16 10:38am

Patrice T

v5

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Richard MacCutchan 22-Jun-16 8:47am

That's a math question, not a Python one.

Sergey Alexandrovich Kryukov 22-Jun-16 10:02am

It is, essentially, not a Python question, but this is a question on a numeric algorithm, which I think is one of the topics of this forum. Please see Solution 1.
—SA

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Sergey Alexandrovich Kryukov · Answer 1 · 2016-06-22T03:44:00

What's wrong? The whole idea. You use epsilon = 0.01 and write about the use of 0.0001. Fine, then try to find a square root not from 25, but from 0.000025 and see what happens. Of course, you can reduce epsilon, to make it a small fraction input value, but you cannot get any acceptable accuracy anyway. Two reasons for that are: 1) you only increase your "answer" ans, never decrease, 2) you use fixed step, that is, in your iterations, the accuracy of each iteration is not adopted as ans**2 gets closer to the input value.

And you even did not define any function for calculation of the root.

This is what works: Methods of computing square roots.

—SA

Patrice T · Answer 2 · 2016-06-22T10:32:00

Quote:
Ans: I understood it. It reduces x by a small value i.e. 0.0001 i.e. step until x is less than epsilon. ans is incremented by ans += step. So when it reaches to the value greater than or equal to x, loop terminates.

Wrong.
This program search for the square root of x by trying every possible ans in sequence starting at 0 and incrementing with a small value 0.0001 .
On every loop, the square of actual ans is compared to x until the difference is less than epsilon and ans is less that x.

The method is very naive and have many problems.
- it is slow, very slow, waaay slow.
- for x greater than 10000, you start to find values where ans never comes within the epsilon tolerance.

Explain the logic and meaning of this Python code.

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