What is the computational complexity of $Ax = b$ and $AX = B$?
Suppose $A, A_i$ are $mtimes m$-matrices, $x_i, b_i$ are $m times 1$, $X$ and $B$ are $m times n$-matrices.
What is the total computational burden of solving $A_ix_i = b_i$ for $i = 1, 2, dots, n$ versus $AX = B$ (where we might assume $A_1 =A_2 = dots A_n = A$).
calculus linear-algebra computational-complexity
edited Nov 26 '18 at 7:59
Björn Friedrich
2,58961831
asked Nov 26 '18 at 5:43
shani
1208
add a comment |
Suppose $A, A_i$ are $mtimes m$-matrices, $x_i, b_i$ are $m times 1$, $X$ and $B$ are $m times n$-matrices.
What is the total computational burden of solving $A_ix_i = b_i$ for $i = 1, 2, dots, n$ versus $AX = B$ (where we might assume $A_1 =A_2 = dots A_n = A$).
calculus linear-algebra computational-complexity
edited Nov 26 '18 at 7:59
Björn Friedrich
2,58961831
asked Nov 26 '18 at 5:43
shani
1208
add a comment |
Suppose $A, A_i$ are $mtimes m$-matrices, $x_i, b_i$ are $m times 1$, $X$ and $B$ are $m times n$-matrices.
What is the total computational burden of solving $A_ix_i = b_i$ for $i = 1, 2, dots, n$ versus $AX = B$ (where we might assume $A_1 =A_2 = dots A_n = A$).
calculus linear-algebra computational-complexity
edited Nov 26 '18 at 7:59
Björn Friedrich
2,58961831
asked Nov 26 '18 at 5:43
shani
1208
Suppose $A, A_i$ are $mtimes m$-matrices, $x_i, b_i$ are $m times 1$, $X$ and $B$ are $m times n$-matrices.
What is the total computational burden of solving $A_ix_i = b_i$ for $i = 1, 2, dots, n$ versus $AX = B$ (where we might assume $A_1 =A_2 = dots A_n = A$).
calculus linear-algebra computational-complexity
calculus linear-algebra computational-complexity
edited Nov 26 '18 at 7:59
Björn Friedrich
2,58961831
asked Nov 26 '18 at 5:43
shani
1208
edited Nov 26 '18 at 7:59
Björn Friedrich
2,58961831
asked Nov 26 '18 at 5:43
shani
1208
edited Nov 26 '18 at 7:59
Björn Friedrich
2,58961831
edited Nov 26 '18 at 7:59
Björn Friedrich
2,58961831
edited Nov 26 '18 at 7:59
Björn Friedrich
2,58961831
2,58961831
asked Nov 26 '18 at 5:43
shani
1208
asked Nov 26 '18 at 5:43
shani
1208
asked Nov 26 '18 at 5:43
shani
1208
1208
add a comment |
add a comment |
1 Answer
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This is the general question of linear programming, and there are a number of works on the question. In short, linear programming is in P.
This is an older paper surveying some results: Megiddo, Nimrod. On the complexity of linear programming. IBM Thomas J. Watson Research Division, 1986. https://theory.stanford.edu/~megiddo/pdf/wcongres_4.0.a.pdf
answered Nov 26 '18 at 8:36
David Manheim
175111
(I assumed you were not asking how much longer it takes to solve N different problems as it does to solve 1. Obviously it takes approximately N times as long to solve N different problems as it does to solve 1.)
– David Manheim
Nov 26 '18 at 8:40
I am asking how much computation time can be saved by doing $AX=B$ vs solving it as n number of $Ax=b$. Is there any computational advantage?
– shani
Nov 26 '18 at 9:08
The second way is MUCH more complex. In essence, the question you are asking is about the computational complexity of combining the answers to many $Ax=b$. I think, intuitively, that it will be roughly as high as solving $AX=B$, meaning all the computation you do to solve the individual equations is wasted - but I don't know how to show this.
– David Manheim
Nov 27 '18 at 6:58
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is the general question of linear programming, and there are a number of works on the question. In short, linear programming is in P.
This is an older paper surveying some results: Megiddo, Nimrod. On the complexity of linear programming. IBM Thomas J. Watson Research Division, 1986. https://theory.stanford.edu/~megiddo/pdf/wcongres_4.0.a.pdf
answered Nov 26 '18 at 8:36
David Manheim
175111
(I assumed you were not asking how much longer it takes to solve N different problems as it does to solve 1. Obviously it takes approximately N times as long to solve N different problems as it does to solve 1.)
– David Manheim
Nov 26 '18 at 8:40
I am asking how much computation time can be saved by doing $AX=B$ vs solving it as n number of $Ax=b$. Is there any computational advantage?
– shani
Nov 26 '18 at 9:08
The second way is MUCH more complex. In essence, the question you are asking is about the computational complexity of combining the answers to many $Ax=b$. I think, intuitively, that it will be roughly as high as solving $AX=B$, meaning all the computation you do to solve the individual equations is wasted - but I don't know how to show this.
– David Manheim
Nov 27 '18 at 6:58
add a comment |
This is the general question of linear programming, and there are a number of works on the question. In short, linear programming is in P.
This is an older paper surveying some results: Megiddo, Nimrod. On the complexity of linear programming. IBM Thomas J. Watson Research Division, 1986. https://theory.stanford.edu/~megiddo/pdf/wcongres_4.0.a.pdf
answered Nov 26 '18 at 8:36
David Manheim
175111
(I assumed you were not asking how much longer it takes to solve N different problems as it does to solve 1. Obviously it takes approximately N times as long to solve N different problems as it does to solve 1.)
– David Manheim
Nov 26 '18 at 8:40
I am asking how much computation time can be saved by doing $AX=B$ vs solving it as n number of $Ax=b$. Is there any computational advantage?
– shani
Nov 26 '18 at 9:08
The second way is MUCH more complex. In essence, the question you are asking is about the computational complexity of combining the answers to many $Ax=b$. I think, intuitively, that it will be roughly as high as solving $AX=B$, meaning all the computation you do to solve the individual equations is wasted - but I don't know how to show this.
– David Manheim
Nov 27 '18 at 6:58
add a comment |
This is the general question of linear programming, and there are a number of works on the question. In short, linear programming is in P.
This is an older paper surveying some results: Megiddo, Nimrod. On the complexity of linear programming. IBM Thomas J. Watson Research Division, 1986. https://theory.stanford.edu/~megiddo/pdf/wcongres_4.0.a.pdf
answered Nov 26 '18 at 8:36
David Manheim
175111
This is the general question of linear programming, and there are a number of works on the question. In short, linear programming is in P.
This is an older paper surveying some results: Megiddo, Nimrod. On the complexity of linear programming. IBM Thomas J. Watson Research Division, 1986. https://theory.stanford.edu/~megiddo/pdf/wcongres_4.0.a.pdf
answered Nov 26 '18 at 8:36
David Manheim
175111
answered Nov 26 '18 at 8:36
David Manheim
175111
answered Nov 26 '18 at 8:36
David Manheim
175111
answered Nov 26 '18 at 8:36
David Manheim
175111
175111
(I assumed you were not asking how much longer it takes to solve N different problems as it does to solve 1. Obviously it takes approximately N times as long to solve N different problems as it does to solve 1.)
– David Manheim
Nov 26 '18 at 8:40
I am asking how much computation time can be saved by doing $AX=B$ vs solving it as n number of $Ax=b$. Is there any computational advantage?
– shani
Nov 26 '18 at 9:08
The second way is MUCH more complex. In essence, the question you are asking is about the computational complexity of combining the answers to many $Ax=b$. I think, intuitively, that it will be roughly as high as solving $AX=B$, meaning all the computation you do to solve the individual equations is wasted - but I don't know how to show this.
– David Manheim
Nov 27 '18 at 6:58
add a comment |
(I assumed you were not asking how much longer it takes to solve N different problems as it does to solve 1. Obviously it takes approximately N times as long to solve N different problems as it does to solve 1.)
– David Manheim
Nov 26 '18 at 8:40
I am asking how much computation time can be saved by doing $AX=B$ vs solving it as n number of $Ax=b$. Is there any computational advantage?
– shani
Nov 26 '18 at 9:08
The second way is MUCH more complex. In essence, the question you are asking is about the computational complexity of combining the answers to many $Ax=b$. I think, intuitively, that it will be roughly as high as solving $AX=B$, meaning all the computation you do to solve the individual equations is wasted - but I don't know how to show this.
– David Manheim
Nov 27 '18 at 6:58
(I assumed you were not asking how much longer it takes to solve N different problems as it does to solve 1. Obviously it takes approximately N times as long to solve N different problems as it does to solve 1.)
– David Manheim
Nov 26 '18 at 8:40
(I assumed you were not asking how much longer it takes to solve N different problems as it does to solve 1. Obviously it takes approximately N times as long to solve N different problems as it does to solve 1.)
– David Manheim
Nov 26 '18 at 8:40
I am asking how much computation time can be saved by doing $AX=B$ vs solving it as n number of $Ax=b$. Is there any computational advantage?
– shani
Nov 26 '18 at 9:08
I am asking how much computation time can be saved by doing $AX=B$ vs solving it as n number of $Ax=b$. Is there any computational advantage?
– shani
Nov 26 '18 at 9:08
The second way is MUCH more complex. In essence, the question you are asking is about the computational complexity of combining the answers to many $Ax=b$. I think, intuitively, that it will be roughly as high as solving $AX=B$, meaning all the computation you do to solve the individual equations is wasted - but I don't know how to show this.
– David Manheim
Nov 27 '18 at 6:58
The second way is MUCH more complex. In essence, the question you are asking is about the computational complexity of combining the answers to many $Ax=b$. I think, intuitively, that it will be roughly as high as solving $AX=B$, meaning all the computation you do to solve the individual equations is wasted - but I don't know how to show this.
– David Manheim
Nov 27 '18 at 6:58
add a comment |
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