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Tim Wescott
Guest

Fri Sep 02, 2016 1:53 am

There's a method that I know, but can't remember the name. And now I
want to tell someone to Google for it.

It basically starts with the notion of multiplying by shift-and-add, but
uses the fact that if you shift and then either add or subtract, you can

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

--
Tim Wescott
Control systems, embedded software and circuit design
I'm looking for work! See my website if you're interested
http://www.wescottdesign.com

Guest

Fri Sep 02, 2016 1:53 am

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev Tim Wescott:
Quote:
There's a method that I know, but can't remember the name. And now I
want to tell someone to Google for it.

It basically starts with the notion of multiplying by shift-and-add, but
uses the fact that if you shift and then either add or subtract, you can

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

Guest

Fri Sep 02, 2016 1:53 am

Den fredag den 2. september 2016 kl. 00.40.28 UTC+2 skrev rickman:
Quote:
On 9/1/2016 4:24 PM, Tim Wescott wrote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev Tim Wescott:
There's a method that I know, but can't remember the name. And now I
want to tell someone to Google for it.

It basically starts with the notion of multiplying by shift-and-add,
but uses the fact that if you shift and then either add or subtract,

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the utility. It
would be useful for multiplying by constants, but otherwise how would
this be used to advantage? It would save add/subtract operations, but I
can't think of another situation where this would be useful.

If the algorithm is doing an add and shift, the add does not increase
the time or the hardware. If building the full multiplier, an adder is
included for each stage, it is either used or not used. When done in
software, the same applies. It is easier to do the add than to skip
over it.

you only need half the stages so it is half the size* and the critical path through your adders are only half as long

* you need a few multiplexors to choose between x1 and x2, subtract is invert and carry in

-Lasse

Kevin Neilson
Guest

Fri Sep 02, 2016 1:53 am

Quote:

It basically starts with the notion of multiplying by shift-and-add, but
uses the fact that if you shift and then either add or subtract, you can

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Canonical Signed Digit (CSD) representation. The CSD form of any number has at least 1/3 of the bits cleared.

244 = 011110100 = +000-0100

(Replace 01111 with +000-)

Kevin Neilson
Guest

Fri Sep 02, 2016 1:53 am

Quote:

Canonical Signed Digit (CSD) representation. The CSD form of any number has at least 1/3 of the bits cleared.

244 = 011110100 = +000-0100

(Replace 01111 with +000-)

I meant:
244 = 011110100 = +000-0+00

Tim Wescott
Guest

Fri Sep 02, 2016 2:24 am

On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen wrote:

Quote:
Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev Tim Wescott:
There's a method that I know, but can't remember the name. And now I
want to tell someone to Google for it.

It basically starts with the notion of multiplying by shift-and-add,
but uses the fact that if you shift and then either add or subtract,

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

--
Tim Wescott
Control systems, embedded software and circuit design
I'm looking for work! See my website if you're interested
http://www.wescottdesign.com

rickman
Guest

Fri Sep 02, 2016 4:40 am

On 9/1/2016 4:24 PM, Tim Wescott wrote:
Quote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev Tim Wescott:
There's a method that I know, but can't remember the name. And now I
want to tell someone to Google for it.

It basically starts with the notion of multiplying by shift-and-add,
but uses the fact that if you shift and then either add or subtract,

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the utility. It
would be useful for multiplying by constants, but otherwise how would
this be used to advantage? It would save add/subtract operations, but I
can't think of another situation where this would be useful.

If the algorithm is doing an add and shift, the add does not increase
the time or the hardware. If building the full multiplier, an adder is
included for each stage, it is either used or not used. When done in
software, the same applies. It is easier to do the add than to skip
over it.

--

Rick C

Tim Wescott
Guest

Fri Sep 02, 2016 5:09 am

On Thu, 01 Sep 2016 18:40:25 -0400, rickman wrote:

Quote:
On 9/1/2016 4:24 PM, Tim Wescott wrote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev Tim
Wescott:
There's a method that I know, but can't remember the name. And now I
want to tell someone to Google for it.

It basically starts with the notion of multiplying by shift-and-add,
but uses the fact that if you shift and then either add or subtract,

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the utility. It
would be useful for multiplying by constants, but otherwise how would
this be used to advantage? It would save add/subtract operations, but I
can't think of another situation where this would be useful.

If the algorithm is doing an add and shift, the add does not increase
the time or the hardware. If building the full multiplier, an adder is
included for each stage, it is either used or not used. When done in
software, the same applies. It is easier to do the add than to skip
over it.

I asked here and on comp.arch.embedded. It's for a guy who's doing
assembly-language programming on a PIC12xxx -- for that guy, and for a
small range of constants (4 through 7), it can save time over a full-
blown multiplication algorithm.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

I'm looking for work -- see my website!

rickman
Guest

Fri Sep 02, 2016 7:30 am

On 9/1/2016 7:09 PM, Tim Wescott wrote:
Quote:
On Thu, 01 Sep 2016 18:40:25 -0400, rickman wrote:

On 9/1/2016 4:24 PM, Tim Wescott wrote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev Tim
Wescott:
There's a method that I know, but can't remember the name. And now I
want to tell someone to Google for it.

It basically starts with the notion of multiplying by shift-and-add,
but uses the fact that if you shift and then either add or subtract,

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the utility. It
would be useful for multiplying by constants, but otherwise how would
this be used to advantage? It would save add/subtract operations, but I
can't think of another situation where this would be useful.

If the algorithm is doing an add and shift, the add does not increase
the time or the hardware. If building the full multiplier, an adder is
included for each stage, it is either used or not used. When done in
software, the same applies. It is easier to do the add than to skip
over it.

I asked here and on comp.arch.embedded. It's for a guy who's doing
assembly-language programming on a PIC12xxx -- for that guy, and for a
small range of constants (4 through 7), it can save time over a full-
blown multiplication algorithm.

Oh, sure, I use that all the time for constant multiplication. No magic
involved. In some cases (such at multiplying by 4) it is simpler to
just use a simple shift and add (which becomes trivial for a single
bit). Even for multipliers with two bits set, it is easier to use a
simple add the shifted values. In fact, the only case of multiplying by
numbers in the range of 4 to 7 where Booth's algorithm simplifies the
work is 7. Since there are only four cases, I expect this could easily
be implemented as four special cases.

--

Rick C

rickman
Guest

Fri Sep 02, 2016 12:39 pm

On 9/1/2016 7:22 PM, lasselangwadtchristensen_at_gmail.com wrote:
Quote:
Den fredag den 2. september 2016 kl. 00.40.28 UTC+2 skrev rickman:
On 9/1/2016 4:24 PM, Tim Wescott wrote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen
wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev Tim
Wescott:
There's a method that I know, but can't remember the name.
And now I want to tell someone to Google for it.

It basically starts with the notion of multiplying by
shift-and-add, but uses the fact that if you shift and then
operations.

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the utility.
It would be useful for multiplying by constants, but otherwise how
operations, but I can't think of another situation where this would
be useful.

If the algorithm is doing an add and shift, the add does not
increase the time or the hardware. If building the full
multiplier, an adder is included for each stage, it is either used
or not used. When done in software, the same applies. It is
easier to do the add than to skip over it.

you only need half the stages so it is half the size* and the

I think you need to look at the algorithm again. The degenerate case is
a multiplier with alternating ones and zeros. An add or subtract is
needed at each 1->0 or 0->1 transition. Since every bit is a transition
you still need an adder/subtractor for every bit.

Of course you could add logic to detect these cases and do fewer adder
ops, but then that is not Booth's algorithm anymore and is much more
complex. Booth's algorithm looks at pairs of bits in the multiplier,
this would require looking at more bits.

If you are thinking in terms of constant multiplication then again, this
is a modified method that combines Booth's with straight adds.

Quote:
* you need a few multiplexors to choose between x1 and x2, subtract
is invert and carry in

Multiplexers are not low cost in any sense in many technologies, but it
doesn't matter. Booth's algorithm doesn't use multiplexers.

--

Rick C

Tauno Voipio
Guest

Fri Sep 02, 2016 4:59 pm

On 1.9.16 22:53, Tim Wescott wrote:
Quote:
There's a method that I know, but can't remember the name. And now I
want to tell someone to Google for it.

It basically starts with the notion of multiplying by shift-and-add, but
uses the fact that if you shift and then either add or subtract, you can

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

It seems that you're after Booth's algorithm:
<https://en.wikipedia.org/wiki/Booth%27s_multiplication_algorithm>.

--

-TV

Guest

Fri Sep 02, 2016 10:01 pm

Den fredag den 2. september 2016 kl. 08.39.20 UTC+2 skrev rickman:
Quote:
On 9/1/2016 7:22 PM, lasselangwadtchristensen_at_gmail.com wrote:
Den fredag den 2. september 2016 kl. 00.40.28 UTC+2 skrev rickman:
On 9/1/2016 4:24 PM, Tim Wescott wrote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen
wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev Tim
Wescott:
There's a method that I know, but can't remember the name.
And now I want to tell someone to Google for it.

It basically starts with the notion of multiplying by
shift-and-add, but uses the fact that if you shift and then
operations.

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the utility.
It would be useful for multiplying by constants, but otherwise how
operations, but I can't think of another situation where this would
be useful.

If the algorithm is doing an add and shift, the add does not
increase the time or the hardware. If building the full
multiplier, an adder is included for each stage, it is either used
or not used. When done in software, the same applies. It is
easier to do the add than to skip over it.

you only need half the stages so it is half the size* and the

I think you need to look at the algorithm again. The degenerate case is
a multiplier with alternating ones and zeros. An add or subtract is
needed at each 1->0 or 0->1 transition. Since every bit is a transition
you still need an adder/subtractor for every bit.

Of course you could add logic to detect these cases and do fewer adder
ops, but then that is not Booth's algorithm anymore and is much more
complex. Booth's algorithm looks at pairs of bits in the multiplier,
this would require looking at more bits.

you are right, I was thinking of "modified Booth" it looks at 3 bits at
a time,

http://www.ellab.physics.upatras.gr/~bakalis/Eudoxus/mbm8.gif

Quote:
If you are thinking in terms of constant multiplication then again, this
is a modified method that combines Booth's with straight adds.

* you need a few multiplexors to choose between x1 and x2, subtract
is invert and carry in

Multiplexers are not low cost in any sense in many technologies, but it
doesn't matter. Booth's algorithm doesn't use multiplexers.

may not be low cost, but compared to a full adder?

and since the inputs come from the multiplicand and the multiplier not from other intermediate results it shouldn't be in the critical path

-Lasse

Guest

Sat Sep 03, 2016 2:08 am

Den lørdag den 3. september 2016 kl. 01.37.56 UTC+2 skrev rickman:
Quote:
On 9/2/2016 4:01 PM, lasselangwadtchristensen_at_gmail.com wrote:
Den fredag den 2. september 2016 kl. 08.39.20 UTC+2 skrev rickman:
On 9/1/2016 7:22 PM, lasselangwadtchristensen_at_gmail.com wrote:
Den fredag den 2. september 2016 kl. 00.40.28 UTC+2 skrev
rickman:
On 9/1/2016 4:24 PM, Tim Wescott wrote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen
wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev
Tim Wescott:
There's a method that I know, but can't remember the
name. And now I want to tell someone to Google for it.

It basically starts with the notion of multiplying by
shift-and-add, but uses the fact that if you shift and
operations.

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the
utility. It would be useful for multiplying by constants, but
otherwise how would this be used to advantage? It would save
add/subtract operations, but I can't think of another situation
where this would be useful.

If the algorithm is doing an add and shift, the add does not
increase the time or the hardware. If building the full
multiplier, an adder is included for each stage, it is either
used or not used. When done in software, the same applies. It
is easier to do the add than to skip over it.

you only need half the stages so it is half the size* and the

I think you need to look at the algorithm again. The degenerate
case is a multiplier with alternating ones and zeros. An add or
subtract is needed at each 1->0 or 0->1 transition. Since every
bit is a transition you still need an adder/subtractor for every
bit.

Of course you could add logic to detect these cases and do fewer
adder ops, but then that is not Booth's algorithm anymore and is
much more complex. Booth's algorithm looks at pairs of bits in the
multiplier, this would require looking at more bits.

you are right, I was thinking of "modified Booth" it looks at 3 bits
at a time,

http://www.ellab.physics.upatras.gr/~bakalis/Eudoxus/mbm8.gif

If you are thinking in terms of constant multiplication then again,
this is a modified method that combines Booth's with straight

* you need a few multiplexors to choose between x1 and x2,
subtract is invert and carry in

Multiplexers are not low cost in any sense in many technologies,
but it doesn't matter. Booth's algorithm doesn't use
multiplexers.

may not be low cost, but compared to a full adder?

In an FPGA the unit of size is the Logic Cell (LC), a LUT and a FF.
Because there is extra carry logic in the LC one bit of an adder is the
same logic as one bit of a multiplexer. The only disadvantage of an
designed you end up with one ripple cascade through one adder and then

sure most fpgas have fast carry chains or build in multipliers so hand
coding "fancy" multiplier structures might not come out ahead

Quote:

and since the inputs come from the multiplicand and the multiplier
not from other intermediate results it shouldn't be in the critical
path

???

I don't see how you use multiplexers instead of adders. If the
multiplier changes from one calculation to the next you need adders in
every position. If the multiplier is fixed the combinations of sums is
fixed and no multiplexers are needed.

with a regular multiplier you have to go through N layers of adders
with a modified Booth you only have to go through N/2 adders

-Lasse

rickman
Guest

Sat Sep 03, 2016 5:37 am

On 9/2/2016 4:01 PM, lasselangwadtchristensen_at_gmail.com wrote:
Quote:
Den fredag den 2. september 2016 kl. 08.39.20 UTC+2 skrev rickman:
On 9/1/2016 7:22 PM, lasselangwadtchristensen_at_gmail.com wrote:
Den fredag den 2. september 2016 kl. 00.40.28 UTC+2 skrev
rickman:
On 9/1/2016 4:24 PM, Tim Wescott wrote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen
wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev
Tim Wescott:
There's a method that I know, but can't remember the
name. And now I want to tell someone to Google for it.

It basically starts with the notion of multiplying by
shift-and-add, but uses the fact that if you shift and
operations.

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the
utility. It would be useful for multiplying by constants, but
otherwise how would this be used to advantage? It would save
add/subtract operations, but I can't think of another situation
where this would be useful.

If the algorithm is doing an add and shift, the add does not
increase the time or the hardware. If building the full
multiplier, an adder is included for each stage, it is either
used or not used. When done in software, the same applies. It
is easier to do the add than to skip over it.

you only need half the stages so it is half the size* and the

I think you need to look at the algorithm again. The degenerate
case is a multiplier with alternating ones and zeros. An add or
subtract is needed at each 1->0 or 0->1 transition. Since every
bit is a transition you still need an adder/subtractor for every
bit.

Of course you could add logic to detect these cases and do fewer
adder ops, but then that is not Booth's algorithm anymore and is
much more complex. Booth's algorithm looks at pairs of bits in the
multiplier, this would require looking at more bits.

you are right, I was thinking of "modified Booth" it looks at 3 bits
at a time,

http://www.ellab.physics.upatras.gr/~bakalis/Eudoxus/mbm8.gif

If you are thinking in terms of constant multiplication then again,
this is a modified method that combines Booth's with straight

* you need a few multiplexors to choose between x1 and x2,
subtract is invert and carry in

Multiplexers are not low cost in any sense in many technologies,
but it doesn't matter. Booth's algorithm doesn't use
multiplexers.

may not be low cost, but compared to a full adder?

In an FPGA the unit of size is the Logic Cell (LC), a LUT and a FF.
Because there is extra carry logic in the LC one bit of an adder is the
same logic as one bit of a multiplexer. The only disadvantage of an
designed you end up with one ripple cascade through one adder and then

Quote:
and since the inputs come from the multiplicand and the multiplier
not from other intermediate results it shouldn't be in the critical
path

???

I don't see how you use multiplexers instead of adders. If the
multiplier changes from one calculation to the next you need adders in
every position. If the multiplier is fixed the combinations of sums is
fixed and no multiplexers are needed.

--

Rick C

rickman
Guest

Sat Sep 03, 2016 7:30 am

On 9/2/2016 8:08 PM, lasselangwadtchristensen_at_gmail.com wrote:
Quote:
Den lørdag den 3. september 2016 kl. 01.37.56 UTC+2 skrev rickman:
On 9/2/2016 4:01 PM, lasselangwadtchristensen_at_gmail.com wrote:
Den fredag den 2. september 2016 kl. 08.39.20 UTC+2 skrev rickman:
On 9/1/2016 7:22 PM, lasselangwadtchristensen_at_gmail.com wrote:
Den fredag den 2. september 2016 kl. 00.40.28 UTC+2 skrev
rickman:
On 9/1/2016 4:24 PM, Tim Wescott wrote:
On Thu, 01 Sep 2016 13:19:09 -0700, lasselangwadtchristensen
wrote:

Den torsdag den 1. september 2016 kl. 21.53.33 UTC+2 skrev
Tim Wescott:
There's a method that I know, but can't remember the
name. And now I want to tell someone to Google for it.

It basically starts with the notion of multiplying by
shift-and-add, but uses the fact that if you shift and
operations.

I.e., 255 = 256 - 1, 244 = 256 - 16 + 4, etc.

Booth?

-Lasse

That's it. Thanks.

That is very familiar from college, but I don't recall the
utility. It would be useful for multiplying by constants, but
otherwise how would this be used to advantage? It would save
add/subtract operations, but I can't think of another situation
where this would be useful.

If the algorithm is doing an add and shift, the add does not
increase the time or the hardware. If building the full
multiplier, an adder is included for each stage, it is either
used or not used. When done in software, the same applies. It
is easier to do the add than to skip over it.

you only need half the stages so it is half the size* and the

I think you need to look at the algorithm again. The degenerate
case is a multiplier with alternating ones and zeros. An add or
subtract is needed at each 1->0 or 0->1 transition. Since every
bit is a transition you still need an adder/subtractor for every
bit.

Of course you could add logic to detect these cases and do fewer
adder ops, but then that is not Booth's algorithm anymore and is
much more complex. Booth's algorithm looks at pairs of bits in the
multiplier, this would require looking at more bits.

you are right, I was thinking of "modified Booth" it looks at 3 bits
at a time,

http://www.ellab.physics.upatras.gr/~bakalis/Eudoxus/mbm8.gif

If you are thinking in terms of constant multiplication then again,
this is a modified method that combines Booth's with straight

* you need a few multiplexors to choose between x1 and x2,
subtract is invert and carry in

Multiplexers are not low cost in any sense in many technologies,
but it doesn't matter. Booth's algorithm doesn't use
multiplexers.

may not be low cost, but compared to a full adder?

In an FPGA the unit of size is the Logic Cell (LC), a LUT and a FF.
Because there is extra carry logic in the LC one bit of an adder is the
same logic as one bit of a multiplexer. The only disadvantage of an
designed you end up with one ripple cascade through one adder and then

sure most fpgas have fast carry chains or build in multipliers so hand
coding "fancy" multiplier structures might not come out ahead

and since the inputs come from the multiplicand and the multiplier
not from other intermediate results it shouldn't be in the critical
path

???

I don't see how you use multiplexers instead of adders. If the
multiplier changes from one calculation to the next you need adders in
every position. If the multiplier is fixed the combinations of sums is
fixed and no multiplexers are needed.

with a regular multiplier you have to go through N layers of adders
with a modified Booth you only have to go through N/2 adders

and N/2 multiplexers which have the same delay... PLUS the selectable
inverter which may or may not be combined with the adder. What's the
point? A simple adder has N stages of delay in an FPGA, same as the
much more complicated modified Booth's adder. In an ASIC there may be
some advantage. In software, I expect the much more complicated control
will make the modified Booth's algorithm the slowest of the three.

People so often forget that multiplexers are not trivial logic.

--

Rick C

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