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:
$$T(n) = 7T(tfracn2)+O(n^2).$$ 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 $$T(n) = 7T(tfracn4) + O(n).$$
algorithms complexity-theory divide-and-conquer matrix
edited Dec 17 at 18:15
OmG
1,400412
asked Dec 16 at 17:22
dafnahaktana
1442
add a comment |
I see everywhere that the recursive equation for the complexity of Strassen alg is:
$$T(n) = 7T(tfracn2)+O(n^2).$$ 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 $$T(n) = 7T(tfracn4) + O(n).$$
algorithms complexity-theory divide-and-conquer matrix
edited Dec 17 at 18:15
OmG
1,400412
asked Dec 16 at 17:22
dafnahaktana
1442
add a comment |
I see everywhere that the recursive equation for the complexity of Strassen alg is:
$$T(n) = 7T(tfracn2)+O(n^2).$$ 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 $$T(n) = 7T(tfracn4) + O(n).$$
algorithms complexity-theory divide-and-conquer matrix
edited Dec 17 at 18:15
OmG
1,400412
asked Dec 16 at 17:22
dafnahaktana
1442
I see everywhere that the recursive equation for the complexity of Strassen alg is:
$$T(n) = 7T(tfracn2)+O(n^2).$$ 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 $$T(n) = 7T(tfracn4) + O(n).$$
algorithms complexity-theory divide-and-conquer matrix
algorithms complexity-theory divide-and-conquer matrix
edited Dec 17 at 18:15
OmG
1,400412
asked Dec 16 at 17:22
dafnahaktana
1442
edited Dec 17 at 18:15
OmG
1,400412
asked Dec 16 at 17:22
dafnahaktana
1442
edited Dec 17 at 18:15
OmG
1,400412
edited Dec 17 at 18:15
OmG
1,400412
edited Dec 17 at 18:15
OmG
1,400412
1,400412
asked Dec 16 at 17:22
dafnahaktana
1442
asked Dec 16 at 17:22
dafnahaktana
1442
asked Dec 16 at 17:22
dafnahaktana
1442
1442
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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 at 19:04
Yuval Filmus
189k12177340
add a comment |
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 $fracn2times fracn2$. Hence, the computation cost of multiplication of two matrices with this size is $T(fracn2)$.
answered Dec 16 at 17:55
OmG
1,400412
add a comment |
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 at 21:04
gnasher729
10.2k1115
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
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 at 19:04
Yuval Filmus
189k12177340
add a comment |
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 at 19:04
Yuval Filmus
189k12177340
add a comment |
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 at 19:04
Yuval Filmus
189k12177340
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 at 19:04
Yuval Filmus
189k12177340
answered Dec 16 at 19:04
Yuval Filmus
189k12177340
answered Dec 16 at 19:04
Yuval Filmus
189k12177340
answered Dec 16 at 19:04
Yuval Filmus
189k12177340
189k12177340
add a comment |
add a comment |
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 $fracn2times fracn2$. Hence, the computation cost of multiplication of two matrices with this size is $T(fracn2)$.
answered Dec 16 at 17:55
OmG
1,400412
add a comment |
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 $fracn2times fracn2$. Hence, the computation cost of multiplication of two matrices with this size is $T(fracn2)$.
answered Dec 16 at 17:55
OmG
1,400412
add a comment |
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 $fracn2times fracn2$. Hence, the computation cost of multiplication of two matrices with this size is $T(fracn2)$.
answered Dec 16 at 17:55
OmG
1,400412
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 $fracn2times fracn2$. Hence, the computation cost of multiplication of two matrices with this size is $T(fracn2)$.
answered Dec 16 at 17:55
OmG
1,400412
answered Dec 16 at 17:55
OmG
1,400412
answered Dec 16 at 17:55
OmG
1,400412
answered Dec 16 at 17:55
OmG
1,400412
1,400412
add a comment |
add a comment |
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 at 21:04
gnasher729
10.2k1115
add a comment |
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 at 21:04
gnasher729
10.2k1115
add a comment |
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 at 21:04
gnasher729
10.2k1115
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 at 21:04
gnasher729
10.2k1115
answered Dec 17 at 21:04
gnasher729
10.2k1115
answered Dec 17 at 21:04
gnasher729
10.2k1115
answered Dec 17 at 21:04
gnasher729
10.2k1115
10.2k1115
add a comment |
add a comment |
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