curriculum/challenges/english/blocks/learn-the-bisection-method-by-finding-the-square-root-of-a-number/65ef1cb031de30c2cde64c12.md
If the difference is within the specified tolerance, set the value of root to mid and break out of the loop.
You should remove the existing pass statement.
({
test: () =>
{
assert.isFalse(runPython(`_Node(_code).find_function("square_root_bisection").find_ifs()[1].find_bodies()[2].find_for_loops()[0].find_ifs()[0].find_bodies()[0].has_pass()`))
}
})
You should assign the value of mid to root and break out of the loop.
({
test: () =>
{
assert(runPython(`_Node(_code).find_function("square_root_bisection").find_ifs()[1].find_bodies()[2].find_for_loops()[0].find_ifs()[0].find_bodies()[0].is_equivalent("root = mid\\nbreak")`))
}
})
def square_root_bisection(square_target, tolerance=1e-7, max_iterations=100):
if square_target < 0:
raise ValueError('Square root of negative number is not defined in real numbers')
if square_target == 1:
root = 1
print(f'The square root of {square_target} is 1')
elif square_target == 0:
root = 0
print(f'The square root of {square_target} is 0')
else:
low = 0
high = max(1, square_target)
root = None
for _ in range(max_iterations):
mid = (low + high) / 2
square_mid = mid**2
--fcc-editable-region--
if abs(square_mid - square_target) < tolerance:
pass
--fcc-editable-region--