Minimizing scalar functions (minimize_scalar) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of minimize_scalar on a quadratic function

What is the output of the following code snippet using minimize_scalar from scipy.optimize?

SciPy

from scipy.optimize import minimize_scalar

result = minimize_scalar(lambda x: (x - 3)**2 + 4)
print(round(result.x, 2))

A 0.0

B 3.0

C 4.0

D 1.5

💡 Hint

Think about where the function (x - 3)^2 + 4 reaches its lowest value.

The function (x - 3)^2 + 4 has its minimum at x = 3 because the squared term is zero there, making the function value 4, which is the lowest possible.

❓ data_output

intermediate

Number of iterations in minimize_scalar

How many iterations does minimize_scalar take to minimize the function f(x) = (x - 1)**4 using the default method?

SciPy

from scipy.optimize import minimize_scalar

result = minimize_scalar(lambda x: (x - 1)**4)
print(result.nit)

A 20

B 22

C 15

D 18

💡 Hint

Check the nit attribute in the result object.

The default method 'brent' typically takes around 20 iterations for this smooth quartic function to converge.

🔧 Debug

advanced

Identify the error in minimize_scalar usage

What error does the following code raise when run?

SciPy

from scipy.optimize import minimize_scalar

def f(x, y):
    return (x - 2)**2 + (y - 3)**2

result = minimize_scalar(f)
print(result.x)

A TypeError: f() missing 1 required positional argument: 'y'

B No error, prints the minimum x value

C ValueError: too many input arguments

D TypeError: f() takes 2 positional arguments but 1 was given

💡 Hint

minimize_scalar expects a function of a single variable.

minimize_scalar only works with functions of one variable. Here, f expects two arguments, but minimize_scalar passes only one, causing a TypeError about argument count.

🧠 Conceptual

advanced

Choosing method in minimize_scalar

Which method in minimize_scalar is best suited for minimizing a smooth unimodal function without bounds?

A 'bounded'

B 'golden'

C 'brent'

D 'newton'

💡 Hint

The default method is often a good choice for smooth functions.

The 'brent' method is the default and is efficient for smooth unimodal functions without bounds. 'bounded' requires bounds, 'golden' is slower, and 'newton' requires derivative info.

🚀 Application

expert

Find minimum of a noisy function with bounds

You want to minimize the function f(x) = (x - 2)**2 + noise where noise is random normal noise with mean 0 and std 0.1. You want to restrict x between 0 and 4. Which code snippet correctly uses minimize_scalar to find the minimum?

SciPy

import numpy as np
from scipy.optimize import minimize_scalar

def f(x):
    noise = np.random.normal(0, 0.1)
    return (x - 2)**2 + noise

A minimize_scalar(f, bounds=(0, 4), method='bounded')

B minimize_scalar(f, method='brent')

C minimize_scalar(f, bounds=(0, 4))

D minimize_scalar(f, method='golden', bounds=(0, 4))

💡 Hint

To use bounds, you must specify the 'bounded' method explicitly.

The 'bounded' method requires bounds and is the only method that accepts the bounds argument. Other methods ignore bounds or raise errors if bounds are given without 'bounded'.