Jacobi Iteration in numerical methods

In the following code I implemented Jacobi iteration. My code right now runs through $25$ iterations. If I wanted to change it to just run through one iteration how would I do that? Would I need to change my jacobi function?

from pprint import pprint

from numpy import array, zeros, diag, diagflat, dot



def jacobi(A,b,N=25,x=None):

    """Solves the equation Ax=b via the Jacobi iterative method."""

    # Create an initial guess if needed                                                                                                                                                            

    if x is None:

        x = zeros(len(A[0]))



    # Create a vector of the diagonal elements of A                                                                                                                                                

    # and subtract them from A                                                                                                                                                                     

    D = diag(A)

    R = A - diagflat(D)



    # Iterate for N times                                                                                                                                                                          

    for i in range(N):

        x = (b - dot(R,x)) / D

    return x



A = array([[2.0,1.0],[5.0,7.0]])

b = array([11.0,13.0])

guess = array([1.0,1.0])



sol = jacobi(A,b,N=25,x=guess)



print ("A:")

pprint(A)



print ("b:")

pprint(b)



print ("x:")

pprint(sol)

asked Dec 4 '18 at 16:10

mt12345

958

I'm confused. You wrote the code and you do not know where does it control the number of iteration?
– Siong Thye Goh
Dec 4 '18 at 16:14

just didn't know if I had to change both iteration counts
– mt12345
Dec 4 '18 at 16:15

add a comment |

from pprint import pprint

from numpy import array, zeros, diag, diagflat, dot



def jacobi(A,b,N=25,x=None):

    """Solves the equation Ax=b via the Jacobi iterative method."""

    # Create an initial guess if needed                                                                                                                                                            

    if x is None:

        x = zeros(len(A[0]))



    # Create a vector of the diagonal elements of A                                                                                                                                                

    # and subtract them from A                                                                                                                                                                     

    D = diag(A)

    R = A - diagflat(D)



    # Iterate for N times                                                                                                                                                                          

    for i in range(N):

        x = (b - dot(R,x)) / D

    return x



A = array([[2.0,1.0],[5.0,7.0]])

b = array([11.0,13.0])

guess = array([1.0,1.0])



sol = jacobi(A,b,N=25,x=guess)



print ("A:")

pprint(A)



print ("b:")

pprint(b)



print ("x:")

pprint(sol)

asked Dec 4 '18 at 16:10

mt12345

958

I'm confused. You wrote the code and you do not know where does it control the number of iteration?
– Siong Thye Goh
Dec 4 '18 at 16:14

just didn't know if I had to change both iteration counts
– mt12345
Dec 4 '18 at 16:15

add a comment |

from pprint import pprint

from numpy import array, zeros, diag, diagflat, dot



def jacobi(A,b,N=25,x=None):

    """Solves the equation Ax=b via the Jacobi iterative method."""

    # Create an initial guess if needed                                                                                                                                                            

    if x is None:

        x = zeros(len(A[0]))



    # Create a vector of the diagonal elements of A                                                                                                                                                

    # and subtract them from A                                                                                                                                                                     

    D = diag(A)

    R = A - diagflat(D)



    # Iterate for N times                                                                                                                                                                          

    for i in range(N):

        x = (b - dot(R,x)) / D

    return x



A = array([[2.0,1.0],[5.0,7.0]])

b = array([11.0,13.0])

guess = array([1.0,1.0])



sol = jacobi(A,b,N=25,x=guess)



print ("A:")

pprint(A)



print ("b:")

pprint(b)



print ("x:")

pprint(sol)

asked Dec 4 '18 at 16:10

mt12345

958

from pprint import pprint

from numpy import array, zeros, diag, diagflat, dot



def jacobi(A,b,N=25,x=None):

    """Solves the equation Ax=b via the Jacobi iterative method."""

    # Create an initial guess if needed                                                                                                                                                            

    if x is None:

        x = zeros(len(A[0]))



    # Create a vector of the diagonal elements of A                                                                                                                                                

    # and subtract them from A                                                                                                                                                                     

    D = diag(A)

    R = A - diagflat(D)



    # Iterate for N times                                                                                                                                                                          

    for i in range(N):

        x = (b - dot(R,x)) / D

    return x



A = array([[2.0,1.0],[5.0,7.0]])

b = array([11.0,13.0])

guess = array([1.0,1.0])



sol = jacobi(A,b,N=25,x=guess)



print ("A:")

pprint(A)



print ("b:")

pprint(b)



print ("x:")

pprint(sol)

linear-algebra numerical-methods

asked Dec 4 '18 at 16:10

mt12345

958

asked Dec 4 '18 at 16:10

mt12345

958

asked Dec 4 '18 at 16:10

mt12345

958

asked Dec 4 '18 at 16:10

mt12345

958

asked Dec 4 '18 at 16:10

mt12345

958

I'm confused. You wrote the code and you do not know where does it control the number of iteration?
– Siong Thye Goh
Dec 4 '18 at 16:14

just didn't know if I had to change both iteration counts
– mt12345
Dec 4 '18 at 16:15

add a comment |

I'm confused. You wrote the code and you do not know where does it control the number of iteration?
– Siong Thye Goh
Dec 4 '18 at 16:14

just didn't know if I had to change both iteration counts
– mt12345
Dec 4 '18 at 16:15

I'm confused. You wrote the code and you do not know where does it control the number of iteration?
– Siong Thye Goh
Dec 4 '18 at 16:14

just didn't know if I had to change both iteration counts
– mt12345
Dec 4 '18 at 16:15

add a comment |

1 Answer
1

active

oldest

votes

Nothing, just call the function jacobi with a different argument

sol = jacobi(A, b, N = 1, x = guess)

This should give you

A:

array([[2., 1.],

       [5., 7.]])

b:

array([11., 13.])

x:

array([5.        , 1.14285714])

answered Dec 4 '18 at 16:13

caverac

13.9k21130

so I wouldn't need to change the original jacobi function?
– mt12345
Dec 4 '18 at 16:14

1

@mt12345 No, you don't need to change anything
– caverac
Dec 4 '18 at 16:15

thanks for your help!
– mt12345
Dec 4 '18 at 16:15

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Nothing, just call the function jacobi with a different argument

sol = jacobi(A, b, N = 1, x = guess)

This should give you

A:

array([[2., 1.],

       [5., 7.]])

b:

array([11., 13.])

x:

array([5.        , 1.14285714])

answered Dec 4 '18 at 16:13

caverac

13.9k21130

so I wouldn't need to change the original jacobi function?
– mt12345
Dec 4 '18 at 16:14

1

@mt12345 No, you don't need to change anything
– caverac
Dec 4 '18 at 16:15

thanks for your help!
– mt12345
Dec 4 '18 at 16:15

add a comment |

Nothing, just call the function jacobi with a different argument

sol = jacobi(A, b, N = 1, x = guess)

This should give you

A:

array([[2., 1.],

       [5., 7.]])

b:

array([11., 13.])

x:

array([5.        , 1.14285714])

answered Dec 4 '18 at 16:13

caverac

13.9k21130

so I wouldn't need to change the original jacobi function?
– mt12345
Dec 4 '18 at 16:14

1

@mt12345 No, you don't need to change anything
– caverac
Dec 4 '18 at 16:15

thanks for your help!
– mt12345
Dec 4 '18 at 16:15

add a comment |

Nothing, just call the function jacobi with a different argument

sol = jacobi(A, b, N = 1, x = guess)

This should give you

A:

array([[2., 1.],

       [5., 7.]])

b:

array([11., 13.])

x:

array([5.        , 1.14285714])

answered Dec 4 '18 at 16:13

caverac

13.9k21130

Nothing, just call the function jacobi with a different argument

sol = jacobi(A, b, N = 1, x = guess)

This should give you

A:

array([[2., 1.],

       [5., 7.]])

b:

array([11., 13.])

x:

array([5.        , 1.14285714])

answered Dec 4 '18 at 16:13

caverac

13.9k21130

answered Dec 4 '18 at 16:13

caverac

13.9k21130

answered Dec 4 '18 at 16:13

caverac

13.9k21130

answered Dec 4 '18 at 16:13

caverac

13.9k21130

so I wouldn't need to change the original jacobi function?
– mt12345
Dec 4 '18 at 16:14

1

@mt12345 No, you don't need to change anything
– caverac
Dec 4 '18 at 16:15

thanks for your help!
– mt12345
Dec 4 '18 at 16:15

add a comment |

so I wouldn't need to change the original jacobi function?
– mt12345
Dec 4 '18 at 16:14

1

@mt12345 No, you don't need to change anything
– caverac
Dec 4 '18 at 16:15

thanks for your help!
– mt12345
Dec 4 '18 at 16:15

so I wouldn't need to change the original jacobi function?
– mt12345
Dec 4 '18 at 16:14

@mt12345 No, you don't need to change anything
– caverac
Dec 4 '18 at 16:15

thanks for your help!
– mt12345
Dec 4 '18 at 16:15

add a comment |

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