TheAlgorithms/Python · error · ValueError

You cannot supply more or less than 2 values

Error message

You cannot supply more or less than 2 values

What it means

Raised by shear_stress(stress, tangential_force, area) in physics/shear_stress.py when the number of arguments equal to 0 among the three is not exactly one. Like gravitational_law, it is a solver: you supply two known values and 0 as the placeholder for the unknown (stress = F/A family), and it returns a (name, value) tuple.

Source

Thrown at physics/shear_stress.py:32

    tangential_force: float,
    area: float,
) -> tuple[str, float]:
    """
    This function can calculate any one of the three -
    1. Shear Stress
    2. Tangential Force
    3. Cross-sectional Area
    This is calculated from the other two provided values
    Examples -
    >>> shear_stress(stress=25, tangential_force=100, area=0)
    ('area', 4.0)
    >>> shear_stress(stress=0, tangential_force=1600, area=200)
    ('stress', 8.0)
    >>> shear_stress(stress=1000, tangential_force=0, area=1200)
    ('tangential_force', 1200000)
    """
    if (stress, tangential_force, area).count(0) != 1:
        raise ValueError("You cannot supply more or less than 2 values")
    elif stress < 0:
        raise ValueError("Stress cannot be negative")
    elif tangential_force < 0:
        raise ValueError("Tangential Force cannot be negative")
    elif area < 0:
        raise ValueError("Area cannot be negative")
    elif stress == 0:
        return (
            "stress",
            tangential_force / area,
        )
    elif tangential_force == 0:
        return (
            "tangential_force",
            stress * area,
        )
    else:
        return (

View on GitHub (pinned to f5988cc097)

Solutions

1. Pass exactly one argument as 0 for the quantity you want computed, e.g. shear_stress(stress=0, tangential_force=1600, area=200) -> ('stress', 8.0)
2. Unpack the returned tuple: name, value = shear_stress(...)
3. If validating known triples, write your own stress == tangential_force/area check instead of calling the solver

Example fix

# before
shear_stress(stress=25, tangential_force=100, area=50)
# ValueError: You cannot supply more or less than 2 values

# after
name, area = shear_stress(stress=25, tangential_force=100, area=0)
# -> ('area', 4.0)

n = sum(v == 0 for v in (stress, tangential_force, area))
if n != 1:
    raise ValueError(f"exactly one argument must be 0 (the unknown), got {n}")
name, value = shear_stress(stress, tangential_force, area)

def has_single_zero(*vals) -> bool:
    return sum(v == 0 for v in vals) == 1

try:
    result = shear_stress(s, f, a)
except ValueError as e:
    if "more or less than 2" in str(e):
        result = None  # re-ask user for exactly two knowns + one 0
    else:
        raise

When it happens

Trigger: shear_stress(stress=25, tangential_force=100, area=50) — no zero, nothing to solve; shear_stress(stress=0, tangential_force=0, area=200) — two zeros; shear_stress(stress=0, tangential_force=0, area=0).

Common situations: Expecting a plain calculator (pass all three knowns) instead of a solver; UIs mapping multiple empty fields to 0; porting code that used None as the unknown sentinel.

Related errors

• One and only one argument must be 0
• Expected a_coeffs to have {self.order + 1} elements for {sel
• n must not be negative
• Candidates list should not be empty
• Depth cannot be less than 0

AI-assisted analysis of TheAlgorithms/Python@f5988cc097 (2026-08-14). Data as JSON: /api/errors/bc35191d11fdbc60. Report an issue: GitHub.