Preconditioners in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the preconditioner module from scipy.

SciPy

from scipy.sparse.linalg import [1]

Drag options to blanks, or click blank then click option'

Aaslinearoperator

Bcg

CLinearOperator

Dspilu

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing 'cg' which is a solver, not a preconditioner.

Choosing 'LinearOperator' which is a class, not a preconditioner function.

The spilu function is used to create an incomplete LU preconditioner.

2fill in blank

medium

Complete the code to create an incomplete LU preconditioner for matrix A.

SciPy

M_x = spilu([1])

Drag options to blanks, or click blank then click option'

AA

Bx

Cb

DM

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the right-hand side vector b instead of the matrix.

Passing the preconditioner variable M before it is defined.

The spilu function requires the matrix A to compute the preconditioner.

3fill in blank

hard

Fix the error in creating a LinearOperator for the preconditioner.

SciPy

M = LinearOperator(A.shape, matvec=[1].solve)

Drag options to blanks, or click blank then click option'

Ab

BA

CM_x

Dspilu

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the matrix A instead of the preconditioner object.

Passing the function spilu itself instead of its result.

The matvec parameter needs a function that applies the preconditioner, which is M_x.solve.

4fill in blank

hard

Fill both blanks to solve the linear system using the preconditioner.

SciPy

x, info = cg([1], [2], M=M)

Drag options to blanks, or click blank then click option'

AA

Bb

Cx0

DM_x

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the matrix and vector arguments.

Passing the preconditioner as a positional argument.

The conjugate gradient solver cg takes the matrix A and the right-hand side vector b as first arguments.

5fill in blank

hard

Fill all three blanks to create and apply a preconditioner in a solver.

SciPy

M_x = spilu([1])
M = LinearOperator([2], matvec=M_x.solve)
x, info = cg([3], b, M=M)

Drag options to blanks, or click blank then click option'

AA

BA.shape

Db

Attempts:

3 left

💡 Hint

Common Mistakes

Using the vector b instead of the matrix A for preconditioner.

Passing the wrong shape to LinearOperator.

Passing b instead of A to the solver.

First, create the preconditioner from matrix A. Then create a LinearOperator with shape A.shape. Finally, solve the system with matrix A and vector b.

Practice

1.

What is the main purpose of a preconditioner in scipy when solving linear systems?

easy

A. To speed up the convergence of iterative solvers

B. To increase the size of the matrix

C. To change the solution of the system

D. To make the matrix non-square

Solution

1. Step 1: Understand the role of preconditioners

   Preconditioners are used to improve the efficiency of iterative methods by transforming the system into an easier one to solve.

2. Step 2: Identify the effect on convergence

   They help iterative solvers like Conjugate Gradient converge faster by approximating the inverse of the matrix.

3. Final Answer:

   To speed up the convergence of iterative solvers -> Option A

4. Quick Check:

   Preconditioner purpose = speed up convergence [OK]

Hint: Preconditioners help iterative solvers run faster [OK]

Common Mistakes:

• Thinking preconditioners change the solution
• Believing preconditioners increase matrix size
• Confusing preconditioners with matrix transformations that alter shape

2.

Which of the following is the correct way to create a simple Jacobi preconditioner using scipy.sparse.linalg.LinearOperator?

import numpy as np
from scipy.sparse.linalg import LinearOperator

A = np.array([[4, 1], [1, 3]])
M = LinearOperator(shape=A.shape, matvec=lambda x: ...)

easy

A. matvec=lambda x: x / np.diag(A)

B. matvec=lambda x: np.dot(A, x)

C. matvec=lambda x: x * np.diag(A)

D. matvec=lambda x: np.linalg.solve(A, x)

Solution

1. Step 1: Recall Jacobi preconditioner definition

   Jacobi preconditioner uses the inverse of the diagonal elements of matrix A.

2. Step 2: Implement matvec for Jacobi

   Applying the preconditioner means dividing each element of x by the corresponding diagonal element of A.

3. Final Answer:

   matvec=lambda x: x / np.diag(A) -> Option A

4. Quick Check:

   Jacobi preconditioner = divide by diagonal [OK]

Hint: Jacobi preconditioner divides vector by matrix diagonal [OK]

Common Mistakes:

• Using matrix multiplication instead of division
• Trying to solve full system instead of diagonal scaling
• Multiplying by diagonal instead of dividing

3.

Given the following code, what will be the output of print(M.matvec(b))?

import numpy as np
from scipy.sparse.linalg import LinearOperator

A = np.array([[2, 0], [0, 5]])
b = np.array([4, 10])
M = LinearOperator(shape=A.shape, matvec=lambda x: x / np.diag(A))
print(M.matvec(b))

medium

A. [0.5, 2.0]

B. [8.0, 50.0]

C. [2.0, 2.0]

D. [4.0, 10.0]

Solution

1. Step 1: Calculate diagonal of A

   Diagonal elements are [2, 5].

2. Step 2: Apply matvec function

   Divide each element of b by corresponding diagonal: [4/2, 10/5] = [2.0, 2.0].

3. Final Answer:

   [2.0, 2.0] -> Option C

4. Quick Check:

   Vector divided by diagonal = [2.0, 2.0] [OK]

Hint: Divide vector elements by diagonal elements to get output [OK]

Common Mistakes:

• Multiplying instead of dividing
• Confusing vector and matrix multiplication
• Using wrong diagonal values

4.

Identify the error in the following code that attempts to create a Jacobi preconditioner:

import numpy as np
from scipy.sparse.linalg import LinearOperator

A = np.array([[3, 1], [1, 4]])
M = LinearOperator(shape=A.shape, matvec=lambda x: np.diag(A) * x)
print(M.matvec(np.array([1, 2])))

medium

A. Using np.diag(A) incorrectly as a matrix

B. Multiplying by diagonal instead of dividing

C. Shape of LinearOperator is wrong

D. Input vector has wrong size

Solution

1. Step 1: Understand Jacobi preconditioner operation

   Jacobi preconditioner divides vector elements by diagonal elements of A.

2. Step 2: Check given matvec function

   Code multiplies vector by diagonal instead of dividing, which is incorrect.

3. Final Answer:

   Multiplying by diagonal instead of dividing -> Option B

4. Quick Check:

   Jacobi requires division, not multiplication [OK]

Hint: Jacobi preconditioner divides vector by diagonal, not multiply [OK]

Common Mistakes:

• Confusing multiplication with division
• Ignoring element-wise operations
• Not verifying mathematical definition

5.

You want to speed up solving a large sparse system Ax = b using Conjugate Gradient in scipy. Which approach best uses a preconditioner?

from scipy.sparse.linalg import cg, LinearOperator
import numpy as np

# A is large sparse matrix
# b is known vector

# Option 1: Use identity preconditioner
M1 = LinearOperator(A.shape, matvec=lambda x: x)

# Option 2: Use Jacobi preconditioner
diag = A.diagonal()
M2 = LinearOperator(A.shape, matvec=lambda x: x / diag)

# Option 3: Use incomplete Cholesky (not shown)

x, info = cg(A, b, M=M2)

Why is Option 2 preferred over Option 1?

hard

A. Because identity preconditioner is not a LinearOperator

B. Because identity preconditioner changes the solution

C. Because Jacobi preconditioner makes matrix larger

D. Because Jacobi preconditioner approximates inverse and speeds convergence

Solution

1. Step 1: Understand identity preconditioner effect

   Identity preconditioner does nothing; it returns the vector unchanged, so no speedup.

2. Step 2: Understand Jacobi preconditioner effect

   Jacobi approximates the inverse of the diagonal, improving convergence speed of iterative solver.

3. Final Answer:

   Because Jacobi preconditioner approximates inverse and speeds convergence -> Option D

4. Quick Check:

   Jacobi preconditioner = faster convergence [OK]

Hint: Jacobi preconditioner speeds up solver by approximating inverse [OK]

Common Mistakes:

• Thinking identity preconditioner changes solution
• Believing Jacobi increases matrix size
• Confusing LinearOperator requirements