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
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Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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Chapter1: Introduction
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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 cach case N 10 * 2k for k = 1,..., 10. That is, each time we
are doubling the size of the matrix.
abone
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
do N 40960 using numPy? N=81920?
Transcribed Image Text: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 cach case N 10 * 2k for k = 1,..., 10. That is, each time we are doubling the size of the matrix. abone 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 do N 40960 using numPy? N=81920?
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
= 1.3
A[2,3]
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):
this Coef = A[m, n]
for k in range (0,4):
A[m, k] A[m, 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]
=
sol = np.zeros((3,1))
for n in range(0,3):
this Coef*A [n, k]
-
leadingCoef*A [ 2-n, k]
Transcribed Image Text: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 = 1.3 A[2,3] 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): this Coef = A[m, n] for k in range (0,4): A[m, k] A[m, 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] = sol = np.zeros((3,1)) for n in range(0,3): this Coef*A [n, k] - leadingCoef*A [ 2-n, k]
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