Strassen algorithm for matrix multiplication complexity analysis
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I see everywhere that the recursive equation for the complexity of Strassen alg is:
$$T(n) = 7T(tfrac{n}{2})+O(n^2).$$ This is not so clear to me.
The parameter $n$ is supposed to be the size of the input, but it seems that here it is one dimension of a matrix while the input size is actually $n^2$.
Also, each matrix of the input is divided to 4 sub matrices so it seems the recursive equation should be $$T(n) = 7T(tfrac{n}{4}) + O(n).$$
algorithms complexity-theory divide-and-conquer matrix
edited Dec 17 '18 at 18:15
OmG
1,456514
asked Dec 16 '18 at 17:22
dafnahaktanadafnahaktana
1443
$endgroup$
add a comment |
$begingroup$
I see everywhere that the recursive equation for the complexity of Strassen alg is:
$$T(n) = 7T(tfrac{n}{2})+O(n^2).$$ This is not so clear to me.
The parameter $n$ is supposed to be the size of the input, but it seems that here it is one dimension of a matrix while the input size is actually $n^2$.
Also, each matrix of the input is divided to 4 sub matrices so it seems the recursive equation should be $$T(n) = 7T(tfrac{n}{4}) + O(n).$$
algorithms complexity-theory divide-and-conquer matrix
edited Dec 17 '18 at 18:15
OmG
1,456514
asked Dec 16 '18 at 17:22
dafnahaktanadafnahaktana
1443
$endgroup$
add a comment |
$begingroup$
I see everywhere that the recursive equation for the complexity of Strassen alg is:
$$T(n) = 7T(tfrac{n}{2})+O(n^2).$$ This is not so clear to me.
The parameter $n$ is supposed to be the size of the input, but it seems that here it is one dimension of a matrix while the input size is actually $n^2$.
Also, each matrix of the input is divided to 4 sub matrices so it seems the recursive equation should be $$T(n) = 7T(tfrac{n}{4}) + O(n).$$
algorithms complexity-theory divide-and-conquer matrix
edited Dec 17 '18 at 18:15
OmG
1,456514
asked Dec 16 '18 at 17:22
dafnahaktanadafnahaktana
1443
$endgroup$
I see everywhere that the recursive equation for the complexity of Strassen alg is:
$$T(n) = 7T(tfrac{n}{2})+O(n^2).$$ This is not so clear to me.
The parameter $n$ is supposed to be the size of the input, but it seems that here it is one dimension of a matrix while the input size is actually $n^2$.
Also, each matrix of the input is divided to 4 sub matrices so it seems the recursive equation should be $$T(n) = 7T(tfrac{n}{4}) + O(n).$$
algorithms complexity-theory divide-and-conquer matrix
algorithms complexity-theory divide-and-conquer matrix
edited Dec 17 '18 at 18:15
OmG
1,456514
asked Dec 16 '18 at 17:22
dafnahaktanadafnahaktana
1443
edited Dec 17 '18 at 18:15
OmG
1,456514
asked Dec 16 '18 at 17:22
dafnahaktanadafnahaktana
1443
edited Dec 17 '18 at 18:15
OmG
1,456514
edited Dec 17 '18 at 18:15
OmG
1,456514
edited Dec 17 '18 at 18:15
OmG
1,456514
1,456514
asked Dec 16 '18 at 17:22
dafnahaktanadafnahaktana
1443
asked Dec 16 '18 at 17:22
dafnahaktanadafnahaktana
1443
asked Dec 16 '18 at 17:22
dafnahaktanadafnahaktana
1443
1443
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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It's true that the parameter $n$ usually denotes the size of the input, but this is not always the case. For square matrix multiplication, $n$ denotes the number of rows (or columns). For graphs, $n$ often denotes the number of vertices, and $m$ the number of edges. For algorithms on Boolean functions, $n$ denotes the number of inputs, though the truth table itself has size $2^n$. There are many other examples.
answered Dec 16 '18 at 19:04
Yuval FilmusYuval Filmus
194k14183347
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add a comment |
$begingroup$
It's back to the size of the matrix. Suppose the original matrix is $ntimes n$. Hence we will consider $T(n)$ as a computation of two matrix with size of $ntimes n$. When we divide the original matrix to 4 part, size of each part is $frac{n}{2}times frac{n}{2}$. Hence, the computation cost of multiplication of two matrices with this size is $T(frac{n}{2})$.
answered Dec 16 '18 at 17:55
OmGOmG
1,456514
$endgroup$
add a comment |
$begingroup$
Time complexity is often based on the input size, but it is not an absolute requirement. In this case, for the multiplication of n x n matrices, it seems most natural to count the number of operations based on n, not on the problem size n x n.
answered Dec 17 '18 at 21:04
gnasher729gnasher729
11.1k1217
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's true that the parameter $n$ usually denotes the size of the input, but this is not always the case. For square matrix multiplication, $n$ denotes the number of rows (or columns). For graphs, $n$ often denotes the number of vertices, and $m$ the number of edges. For algorithms on Boolean functions, $n$ denotes the number of inputs, though the truth table itself has size $2^n$. There are many other examples.
answered Dec 16 '18 at 19:04
Yuval FilmusYuval Filmus
194k14183347
$endgroup$
add a comment |
$begingroup$
It's true that the parameter $n$ usually denotes the size of the input, but this is not always the case. For square matrix multiplication, $n$ denotes the number of rows (or columns). For graphs, $n$ often denotes the number of vertices, and $m$ the number of edges. For algorithms on Boolean functions, $n$ denotes the number of inputs, though the truth table itself has size $2^n$. There are many other examples.
answered Dec 16 '18 at 19:04
Yuval FilmusYuval Filmus
194k14183347
$endgroup$
add a comment |
$begingroup$
It's true that the parameter $n$ usually denotes the size of the input, but this is not always the case. For square matrix multiplication, $n$ denotes the number of rows (or columns). For graphs, $n$ often denotes the number of vertices, and $m$ the number of edges. For algorithms on Boolean functions, $n$ denotes the number of inputs, though the truth table itself has size $2^n$. There are many other examples.
answered Dec 16 '18 at 19:04
Yuval FilmusYuval Filmus
194k14183347
$endgroup$
It's true that the parameter $n$ usually denotes the size of the input, but this is not always the case. For square matrix multiplication, $n$ denotes the number of rows (or columns). For graphs, $n$ often denotes the number of vertices, and $m$ the number of edges. For algorithms on Boolean functions, $n$ denotes the number of inputs, though the truth table itself has size $2^n$. There are many other examples.
answered Dec 16 '18 at 19:04
Yuval FilmusYuval Filmus
194k14183347
answered Dec 16 '18 at 19:04
Yuval FilmusYuval Filmus
194k14183347
answered Dec 16 '18 at 19:04
Yuval FilmusYuval Filmus
194k14183347
answered Dec 16 '18 at 19:04
Yuval FilmusYuval Filmus
194k14183347
194k14183347
add a comment |
add a comment |
$begingroup$
It's back to the size of the matrix. Suppose the original matrix is $ntimes n$. Hence we will consider $T(n)$ as a computation of two matrix with size of $ntimes n$. When we divide the original matrix to 4 part, size of each part is $frac{n}{2}times frac{n}{2}$. Hence, the computation cost of multiplication of two matrices with this size is $T(frac{n}{2})$.
answered Dec 16 '18 at 17:55
OmGOmG
1,456514
$endgroup$
add a comment |
$begingroup$
It's back to the size of the matrix. Suppose the original matrix is $ntimes n$. Hence we will consider $T(n)$ as a computation of two matrix with size of $ntimes n$. When we divide the original matrix to 4 part, size of each part is $frac{n}{2}times frac{n}{2}$. Hence, the computation cost of multiplication of two matrices with this size is $T(frac{n}{2})$.
answered Dec 16 '18 at 17:55
OmGOmG
1,456514
$endgroup$
add a comment |
$begingroup$
It's back to the size of the matrix. Suppose the original matrix is $ntimes n$. Hence we will consider $T(n)$ as a computation of two matrix with size of $ntimes n$. When we divide the original matrix to 4 part, size of each part is $frac{n}{2}times frac{n}{2}$. Hence, the computation cost of multiplication of two matrices with this size is $T(frac{n}{2})$.
answered Dec 16 '18 at 17:55
OmGOmG
1,456514
$endgroup$
It's back to the size of the matrix. Suppose the original matrix is $ntimes n$. Hence we will consider $T(n)$ as a computation of two matrix with size of $ntimes n$. When we divide the original matrix to 4 part, size of each part is $frac{n}{2}times frac{n}{2}$. Hence, the computation cost of multiplication of two matrices with this size is $T(frac{n}{2})$.
answered Dec 16 '18 at 17:55
OmGOmG
1,456514
answered Dec 16 '18 at 17:55
OmGOmG
1,456514
answered Dec 16 '18 at 17:55
OmGOmG
1,456514
answered Dec 16 '18 at 17:55
OmGOmG
1,456514
1,456514
add a comment |
add a comment |
$begingroup$
Time complexity is often based on the input size, but it is not an absolute requirement. In this case, for the multiplication of n x n matrices, it seems most natural to count the number of operations based on n, not on the problem size n x n.
answered Dec 17 '18 at 21:04
gnasher729gnasher729
11.1k1217
$endgroup$
add a comment |
$begingroup$
Time complexity is often based on the input size, but it is not an absolute requirement. In this case, for the multiplication of n x n matrices, it seems most natural to count the number of operations based on n, not on the problem size n x n.
answered Dec 17 '18 at 21:04
gnasher729gnasher729
11.1k1217
$endgroup$
add a comment |
$begingroup$
Time complexity is often based on the input size, but it is not an absolute requirement. In this case, for the multiplication of n x n matrices, it seems most natural to count the number of operations based on n, not on the problem size n x n.
answered Dec 17 '18 at 21:04
gnasher729gnasher729
11.1k1217
$endgroup$
Time complexity is often based on the input size, but it is not an absolute requirement. In this case, for the multiplication of n x n matrices, it seems most natural to count the number of operations based on n, not on the problem size n x n.
answered Dec 17 '18 at 21:04
gnasher729gnasher729
11.1k1217
answered Dec 17 '18 at 21:04
gnasher729gnasher729
11.1k1217
answered Dec 17 '18 at 21:04
gnasher729gnasher729
11.1k1217
answered Dec 17 '18 at 21:04
gnasher729gnasher729
11.1k1217
11.1k1217
add a comment |
add a comment |
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