Implement the Bisection Method - Implement the Bisection Method

Python
1

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 far

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

    #recounting tolerance
    n=0
    i = toler
    while i != 1:
        i = i * 10
        n +=1

    interv = [0, tofind] #[low, high]    
    itertimes = times #separating iterations
    
    while itertimes > 0:
        
        mid = (interv[0]+interv[1])/2
        
        if mid ** 2 > tofind:
            interv[1] = mid
            interv[1] = round(interv[1], n)

        elif mid ** 2 < tofind:
            interv[0] = mid
            interv[0] = round(interv[0], n)

        elif mid**2 == tofind or (interv[1]*10**n-interv[0]*10**n)/10**n <= toler*1:
            interv[0] = round(interv[0], n)
            interv[1] = round(interv[1], n)
            print(interv)
            print(f"The square root of {tofind} is approximately {(interv[1]*10**n-interv[0]*10**n)/10**n}")
            return mid
        
        
        #print(f"{(interv[1]*10**n-interv[0]*10**n)/10**n} / {toler*2}\n",interv, "\n")
        itertimes -= 1
    print(f"Failed to converge within {times} iterations")
    return None

print(square_root_bisection(225, 1e-5, 100))

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Challenge Information:

Implement the Bisection Method - Implement the Bisection Method

2

Easy to misunderstand what is meant by tolerance here.

I’ve tried to explain it here:

3

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

4

I don’t understand what you mean

5

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

6

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.

7

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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

    #recounting tolerance
    n=0
    i = toler
    while i != 1:
        i = i * 10
        n +=1

    interv = [0, tofind] #[low, high]    
    itertimes = times #separating iterations
    
    while itertimes > 0:
        
        mid = (interv[0]+interv[1])/2
        
        if mid ** 2 > tofind:
            interv[1] = mid
            #interv[1] = round(interv[1], n)

        elif mid ** 2 < tofind:
            interv[0] = mid
            #interv[0] = round(interv[0], n)

        elif mid**2 == tofind or interv[1]-interv[0] <= toler*2:
            #interv[0] = round(interv[0], n)
            #interv[1] = round(interv[1], n)
            print(interv)
            print(f"The square root of {tofind} is approximately {interv[1]-interv[0]}")
            return mid
        
        
        print(f"{(interv[1]-interv[0]} / {toler*2}\n",interv, "\n")
        itertimes -= 1
    print(f"Failed to converge within {times} iterations")
    return None

print(square_root_bisection(225, 1e-7, 100))

still the same

8

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.

9 
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 = [0, tofind] #[low, high]    
    itertimes = times #separating iterations
    
    while itertimes > 0:
        
        mid = (interv[0]+interv[1])/2
        
        if mid**2 == tofind or interv[1]-interv[0] <= toler*2:
            print(interv)
            print(f"The square root of {tofind} is approximately {interv[1]-interv[0]}")
            return mid
        
        elif mid ** 2 < tofind:
            interv[0] = mid

        else:
            interv[1] = mid
        itertimes -= 1
        
    print(f"Failed to converge within {times} iterations")
    return None

print(square_root_bisection(81, 0.001, 100))

still not. i don’t understand how i should cut the result

10

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.

11

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

12

I’m not sure which tests are failing with the phrase ‘cut it right’

13

How should I return a value between two numbers when truncating, for example 14.999999, gives 15.0? (square_root_bisection(81, 1e-3, 50) should print The square root of 81 is approximately X, where X is a number between 8.999 and 9.001)

14

Did you implement this yet?

Maybe it’s related to your question?

Try to translate this instruction into a while loop

15

Were do the instructions talk about truncation?

16

i did it. if i implement it in while loop, then None will be returnet, like Failed to converge within {times} iterations

17

nowhere… but everything tells what i should to round the final result

18

Where do the instructions say to round though?

19

I would look at which tests are failing and which tests are passing.

20

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like “0.4999999 and 0.5000001”, “8.999 and 9.001”