Tell us what’s happening: i cant pass tests where code is printing the result and tests with 0.001 and 0.25. I don’t really understand the essence of the task because binary search and the bisection method are DIFFERENT things. There should be an interval, but only one number is given. Your code so…

Easy to misunderstand what is meant by tolerance here. I’ve tried to explain it here: [image] Implement the Bisection Method - Step 8-9 Python I also misunderstood this one at first. The mistake is assuming what “tolerance” is. Look at this line: Look at t…

ok, but there’s another problem with floating dot: python can’t properly count result on printing stage. i was trying to solve the problem, but nothing changed

I don’t understand what you mean

at the end of function python giving me this: “The square root of 225 is approximately 0.0“. so on output i cant give right number ( if square_root_bisection(225, 1e-5, 100)), because it just refuses to count as needed. also because of this problem i can’t solve tests №8-9

You aren’t really doing the bisection algorithm. Your add variable n is not part of the bisection algorithm. You should directly use the tol to shop iterating instead of creating this n.

[image] def square_root_bisection(tofind, toler=0.01, times=100): if tofind < 0: raise ValueError("Square root of negative number is not defined in real numbers") if tofind == 0 or tofind == 1: print(f"The square root of {tofind} is {tofind}") return tofind**0.5 …

You’re still using the variable n. You need to get rid of it. Also, the termination condition really should be related to the tolerance, not anything else.

def square_root_bisection(tofind, toler=0.01, times=100): if tofind < 0: raise ValueError("Square root of negative number is not defined in real numbers") if tofind == 0 or tofind == 1: print(f"The square root of {tofind} is {tofind}") return tofind**1 interv = …

[image] MrHampter: interv[1]-interv[0] <= toler*2: This is not how the instructions say to use the tolerance For example, if the tolerance is 0.01, the bisection method will keep halving the interval until the difference between the upper and lower bounds is less than or equal to 0.01.

oh, fixed on interv[1]-interv[0] <= toler:, but i still don’t know how to cut it right

Implement the Bisection Method - Implement the Bisection Method

Curriculum Help

pkdvalis January 4, 2026, 3:48pm 2

Easy to misunderstand what is meant by tolerance here.

I’ve tried to explain it here:

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Implement the Bisection Method - Implement the Bisection Method

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