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 $$T(n) = 7T(tfrac{n}{4}) + O(n).$$
complexity-theory divide-and-conquer matrix
edited 1 hour ago
Yuval Filmus
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up vote
3
down vote
favorite
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 $$T(n) = 7T(tfrac{n}{4}) + O(n).$$
complexity-theory divide-and-conquer matrix
edited 1 hour ago
Yuval Filmus
188k12177339
asked 2 hours ago
user97899
161
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user97899 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 $$T(n) = 7T(tfrac{n}{4}) + O(n).$$
complexity-theory divide-and-conquer matrix
edited 1 hour ago
Yuval Filmus
188k12177339
asked 2 hours ago
user97899
161
New contributor
user97899 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 $$T(n) = 7T(tfrac{n}{4}) + O(n).$$
complexity-theory divide-and-conquer matrix
complexity-theory divide-and-conquer matrix
edited 1 hour ago
Yuval Filmus
188k12177339
asked 2 hours ago
user97899
161
New contributor
user97899 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
Yuval Filmus
188k12177339
asked 2 hours ago
user97899
161
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user97899 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
Yuval Filmus
188k12177339
edited 1 hour ago
Yuval Filmus
188k12177339
edited 1 hour ago
Yuval Filmus
188k12177339
188k12177339
asked 2 hours ago
user97899
161
New contributor
user97899 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 hours ago
user97899
161
asked 2 hours ago
user97899
161
161
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user97899 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
user97899 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
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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 2 hours ago
OmG
1,363412
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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 1 hour ago
Yuval Filmus
188k12177339
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2 Answers
2
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oldest
votes
2 Answers
2
active
oldest
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active
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active
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up vote
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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 2 hours ago
OmG
1,363412
add a comment |
up vote
3
down vote
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 2 hours ago
OmG
1,363412
add a comment |
up vote
3
down vote
up vote
3
down vote
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 2 hours ago
OmG
1,363412
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 2 hours ago
OmG
1,363412
answered 2 hours ago
OmG
1,363412
answered 2 hours ago
OmG
1,363412
answered 2 hours ago
OmG
1,363412
1,363412
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add a comment |
up vote
0
down vote
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 1 hour ago
Yuval Filmus
188k12177339
add a comment |
up vote
0
down vote
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 1 hour ago
Yuval Filmus
188k12177339
add a comment |
up vote
0
down vote
up vote
0
down vote
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 1 hour ago
Yuval Filmus
188k12177339
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 1 hour ago
Yuval Filmus
188k12177339
answered 1 hour ago
Yuval Filmus
188k12177339
answered 1 hour ago
Yuval Filmus
188k12177339
answered 1 hour ago
Yuval Filmus
188k12177339
188k12177339
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add a comment |
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