Mixing
Fast parallel multiplication method
$begingroup$
In page 6 of A very fast multiplication algorithm , how do the PAR-allel multiplication work ?
Why is 12021011 equivalent to 299 ?
Parallel Multiplication method
linear-algebra algebra-precalculus discrete-mathematics optimization binary
asked Jan 28 at 7:00
kevinkevin
42
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add a comment |
$begingroup$
In page 6 of A very fast multiplication algorithm , how do the PAR-allel multiplication work ?
Why is 12021011 equivalent to 299 ?
Parallel Multiplication method
linear-algebra algebra-precalculus discrete-mathematics optimization binary
asked Jan 28 at 7:00
kevinkevin
42
$endgroup$
add a comment |
$begingroup$
In page 6 of A very fast multiplication algorithm , how do the PAR-allel multiplication work ?
Why is 12021011 equivalent to 299 ?
Parallel Multiplication method
linear-algebra algebra-precalculus discrete-mathematics optimization binary
asked Jan 28 at 7:00
kevinkevin
42
$endgroup$
In page 6 of A very fast multiplication algorithm , how do the PAR-allel multiplication work ?
Why is 12021011 equivalent to 299 ?
Parallel Multiplication method
linear-algebra algebra-precalculus discrete-mathematics optimization binary
linear-algebra algebra-precalculus discrete-mathematics optimization binary
asked Jan 28 at 7:00
kevinkevin
42
asked Jan 28 at 7:00
kevinkevin
42
asked Jan 28 at 7:00
kevinkevin
42
asked Jan 28 at 7:00
kevinkevin
42
asked Jan 28 at 7:00
kevinkevin
42
42
add a comment |
add a comment |
1 Answer
1
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$begingroup$
12021011 is $100101011_2 = 299_{10}$ before the carry.
The rough idea of algorithm:
(1) divide the factors in chunks,
(2) do in parallel the product of chunks,
[now, we must shift and add the (many) chunks]
(3) do in parallel the shifts,
(4) take the (say) S shifted chunks in subsets of 2 and do in parallel the S/2 sums,
(5) repeat the procedure with the S/2 sums
...
answered Jan 28 at 7:42
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
$begingroup$
Why is 12021011 equivalent to 100101011 in binary ?
$endgroup$
– kevin
Jan 28 at 7:47
$begingroup$
Ok, I got it now. But why is this parallel multiplication method faster than carry save array multiplier ? Note that those '2' also needs additions by itself ....
$endgroup$
– kevin
Jan 28 at 7:57
$begingroup$
@kevin it is hardware implementation. You should look at Wallace tree, carry save adder, carry skip adder to understand better.
$endgroup$
– kelalaka
Jan 28 at 10:17
$begingroup$
@kelalaka Wallace tree is no longer used now due to its irregular and famously long routing.
$endgroup$
– kevin
Jan 28 at 10:54
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
12021011 is $100101011_2 = 299_{10}$ before the carry.
The rough idea of algorithm:
(1) divide the factors in chunks,
(2) do in parallel the product of chunks,
[now, we must shift and add the (many) chunks]
(3) do in parallel the shifts,
(4) take the (say) S shifted chunks in subsets of 2 and do in parallel the S/2 sums,
(5) repeat the procedure with the S/2 sums
...
answered Jan 28 at 7:42
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
$begingroup$
Why is 12021011 equivalent to 100101011 in binary ?
$endgroup$
– kevin
Jan 28 at 7:47
$begingroup$
Ok, I got it now. But why is this parallel multiplication method faster than carry save array multiplier ? Note that those '2' also needs additions by itself ....
$endgroup$
– kevin
Jan 28 at 7:57
$begingroup$
@kevin it is hardware implementation. You should look at Wallace tree, carry save adder, carry skip adder to understand better.
$endgroup$
– kelalaka
Jan 28 at 10:17
$begingroup$
@kelalaka Wallace tree is no longer used now due to its irregular and famously long routing.
$endgroup$
– kevin
Jan 28 at 10:54
add a comment |
$begingroup$
12021011 is $100101011_2 = 299_{10}$ before the carry.
The rough idea of algorithm:
(1) divide the factors in chunks,
(2) do in parallel the product of chunks,
[now, we must shift and add the (many) chunks]
(3) do in parallel the shifts,
(4) take the (say) S shifted chunks in subsets of 2 and do in parallel the S/2 sums,
(5) repeat the procedure with the S/2 sums
...
answered Jan 28 at 7:42
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
$begingroup$
Why is 12021011 equivalent to 100101011 in binary ?
$endgroup$
– kevin
Jan 28 at 7:47
$begingroup$
Ok, I got it now. But why is this parallel multiplication method faster than carry save array multiplier ? Note that those '2' also needs additions by itself ....
$endgroup$
– kevin
Jan 28 at 7:57
$begingroup$
@kevin it is hardware implementation. You should look at Wallace tree, carry save adder, carry skip adder to understand better.
$endgroup$
– kelalaka
Jan 28 at 10:17
$begingroup$
@kelalaka Wallace tree is no longer used now due to its irregular and famously long routing.
$endgroup$
– kevin
Jan 28 at 10:54
add a comment |
$begingroup$
12021011 is $100101011_2 = 299_{10}$ before the carry.
The rough idea of algorithm:
(1) divide the factors in chunks,
(2) do in parallel the product of chunks,
[now, we must shift and add the (many) chunks]
(3) do in parallel the shifts,
(4) take the (say) S shifted chunks in subsets of 2 and do in parallel the S/2 sums,
(5) repeat the procedure with the S/2 sums
...
answered Jan 28 at 7:42
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
12021011 is $100101011_2 = 299_{10}$ before the carry.
The rough idea of algorithm:
(1) divide the factors in chunks,
(2) do in parallel the product of chunks,
[now, we must shift and add the (many) chunks]
(3) do in parallel the shifts,
(4) take the (say) S shifted chunks in subsets of 2 and do in parallel the S/2 sums,
(5) repeat the procedure with the S/2 sums
...
answered Jan 28 at 7:42
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
edited Jan 28 at 12:17
answered Jan 28 at 7:42
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
answered Jan 28 at 7:42
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
answered Jan 28 at 7:42
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
34.8k42971
$begingroup$
Why is 12021011 equivalent to 100101011 in binary ?
$endgroup$
– kevin
Jan 28 at 7:47
$begingroup$
Ok, I got it now. But why is this parallel multiplication method faster than carry save array multiplier ? Note that those '2' also needs additions by itself ....
$endgroup$
– kevin
Jan 28 at 7:57
$begingroup$
@kevin it is hardware implementation. You should look at Wallace tree, carry save adder, carry skip adder to understand better.
$endgroup$
– kelalaka
Jan 28 at 10:17
$begingroup$
@kelalaka Wallace tree is no longer used now due to its irregular and famously long routing.
$endgroup$
– kevin
Jan 28 at 10:54
add a comment |
$begingroup$
Why is 12021011 equivalent to 100101011 in binary ?
$endgroup$
– kevin
Jan 28 at 7:47
$begingroup$
Ok, I got it now. But why is this parallel multiplication method faster than carry save array multiplier ? Note that those '2' also needs additions by itself ....
$endgroup$
– kevin
Jan 28 at 7:57
$begingroup$
@kevin it is hardware implementation. You should look at Wallace tree, carry save adder, carry skip adder to understand better.
$endgroup$
– kelalaka
Jan 28 at 10:17
$begingroup$
@kelalaka Wallace tree is no longer used now due to its irregular and famously long routing.
$endgroup$
– kevin
Jan 28 at 10:54
$begingroup$
Why is 12021011 equivalent to 100101011 in binary ?
$endgroup$
– kevin
Jan 28 at 7:47
$begingroup$
Why is 12021011 equivalent to 100101011 in binary ?
$endgroup$
– kevin
Jan 28 at 7:47
$begingroup$
Ok, I got it now. But why is this parallel multiplication method faster than carry save array multiplier ? Note that those '2' also needs additions by itself ....
$endgroup$
– kevin
Jan 28 at 7:57
$begingroup$
Ok, I got it now. But why is this parallel multiplication method faster than carry save array multiplier ? Note that those '2' also needs additions by itself ....
$endgroup$
– kevin
Jan 28 at 7:57
$begingroup$
@kevin it is hardware implementation. You should look at Wallace tree, carry save adder, carry skip adder to understand better.
$endgroup$
– kelalaka
Jan 28 at 10:17
$begingroup$
@kevin it is hardware implementation. You should look at Wallace tree, carry save adder, carry skip adder to understand better.
$endgroup$
– kelalaka
Jan 28 at 10:17
$begingroup$
@kelalaka Wallace tree is no longer used now due to its irregular and famously long routing.
$endgroup$
– kevin
Jan 28 at 10:54
$begingroup$
@kelalaka Wallace tree is no longer used now due to its irregular and famously long routing.
$endgroup$
– kevin
Jan 28 at 10:54
add a comment |
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