2. Problem 2: In this problem you will test our "in class" version of Gaussian elimination against the version in numpy. Use both codes to solve some randomly chosen linear systems of size N = 10, 20, 40, 80, 160, 320, 640, 1280, 2560, and 5120. Depending on your computer, the last two may or may not work. If they don't, just say so in your report. If you computer can do N = 10240 then try this as well. Note that in each case N = 10 * 2k for k = 1,..., 10. That is, cach time we are doubling the size of the matrix. - Plot the results as two curves. Our code in red and numPy in bluc. (Or whatever colors you prefer). Can you guess the function which governs the runtime? Based on this guess, how long would it take to 40960 using numPy? N 819202 do N -
2. Problem 2: In this problem you will test our "in class" version of Gaussian elimination against the version in numpy. Use both codes to solve some randomly chosen linear systems of size N = 10, 20, 40, 80, 160, 320, 640, 1280, 2560, and 5120. Depending on your computer, the last two may or may not work. If they don't, just say so in your report. If you computer can do N = 10240 then try this as well. Note that in each case N = 10 * 2k for k = 1,..., 10. That is, cach time we are doubling the size of the matrix. - Plot the results as two curves. Our code in red and numPy in bluc. (Or whatever colors you prefer). Can you guess the function which governs the runtime? Based on this guess, how long would it take to 40960 using numPy? N 819202 do N -
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Transcribed Image Text:**Problem 2:** In this problem, you will test our “in class” version of Gaussian elimination against the version in NumPy. Use both codes to solve some randomly chosen linear systems of size \( N = 10, 20, 40, 80, 160, 320, 640, 1280, 2560, \) and \( 5120 \). Depending on your computer, the last two may or may not work. If they don’t, just say so in your report. If your computer can do \( N = 10240 \) then try this as well. Note that in each case \( N = 10 \times 2^k \) for \( k = 1, \ldots, 10 \). That is, each time we are doubling the size of the matrix.
Plot the results as two curves. Our code in red and NumPy in blue. (Or whatever colors you prefer). Can you guess the function which governs the runtime? Based on this guess, how long would it take to do \( N = 40960 \) using NumPy? \( N = 81920 \)?
![```python
import time
import math
import matplotlib.pyplot as plt
import numpy as np
# N = 3
A = np.zeros((3,4))
A[0,0] = 0.4
A[0,1] = 0.2
A[0,2] = -0.3
A[0,3] = 2.3
A[1,0] = 1.2
A[1,1] = 1.2
A[1,2] = -0.8
A[1,3] = 4.9
A[2,0] = -0.3
A[2,1] = -1.4
A[2,2] = -3.4
A[2,3] = 1.3
B = A.copy()
for n in range(0,2):
leadingCoef = A[n, n]
for m in range(0, 4):
A[n, m] = A[n,m]/leadingCoef
for m in range(n+1,3):
thisCoef = A[m,n]
for k in range(0,4):
A[m, k] = A[m,k] - thisCoef*A[n,k]
lastCoef = A[2,2]
for k in range(0,4):
A[2,k] = A[2,k]/lastCoef
for n in range(0, 2):
for m in range(0,2-n):
leadingCoef = A[m,2-n]
for k in range(0,4):
A[m,k] = A[m,k] - leadingCoef*A[2-n,k]
sol = np.zeros((3,1))
for n in range(0,3):
```
**Explanation:**
This code represents a numerical method for solving a system of linear equations using Gaussian elimination with back substitution. The steps are as follows:
1. **Imports:**
- `time` and `math`: Imported but not used in this snippet.
- `matplotlib.pyplot` as `plt`: Also imported but not used here.
- `numpy` as `np`: Used for numerical operations.
2. **Matrix Setup:**
- A \(3 \times 4\) zero matrix `A` is initialized.
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F34c2a7c6-20d8-45e7-8c90-b1c224068cda%2Fef511b8b-fbf9-421d-ae79-a3e97e8d5aa7%2F34z47hn_processed.png&w=3840&q=75)
Transcribed Image Text:```python
import time
import math
import matplotlib.pyplot as plt
import numpy as np
# N = 3
A = np.zeros((3,4))
A[0,0] = 0.4
A[0,1] = 0.2
A[0,2] = -0.3
A[0,3] = 2.3
A[1,0] = 1.2
A[1,1] = 1.2
A[1,2] = -0.8
A[1,3] = 4.9
A[2,0] = -0.3
A[2,1] = -1.4
A[2,2] = -3.4
A[2,3] = 1.3
B = A.copy()
for n in range(0,2):
leadingCoef = A[n, n]
for m in range(0, 4):
A[n, m] = A[n,m]/leadingCoef
for m in range(n+1,3):
thisCoef = A[m,n]
for k in range(0,4):
A[m, k] = A[m,k] - thisCoef*A[n,k]
lastCoef = A[2,2]
for k in range(0,4):
A[2,k] = A[2,k]/lastCoef
for n in range(0, 2):
for m in range(0,2-n):
leadingCoef = A[m,2-n]
for k in range(0,4):
A[m,k] = A[m,k] - leadingCoef*A[2-n,k]
sol = np.zeros((3,1))
for n in range(0,3):
```
**Explanation:**
This code represents a numerical method for solving a system of linear equations using Gaussian elimination with back substitution. The steps are as follows:
1. **Imports:**
- `time` and `math`: Imported but not used in this snippet.
- `matplotlib.pyplot` as `plt`: Also imported but not used here.
- `numpy` as `np`: Used for numerical operations.
2. **Matrix Setup:**
- A \(3 \times 4\) zero matrix `A` is initialized.
-
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