a * b

[Vols]

If we cannot use 'multiply' because of the memory limitation, how to
implement "a*b" ?
I know 3*7 can be written as 3<<3 - 3, but what is the general
solution? thanks.

Vol.

[Victor Bazarov]

Re: a * b

Vols wrote:

If we cannot use 'multiply' because of the memory limitation, how to
implement "a*b" ?

Huh? Memory limitation? What do you mean, exactly?

I know 3*7 can be written as 3<<3 - 3,

Try it, you'll be surprised at the result.

but what is the general
solution? thanks.

3*7

or you could try

7*3

..

V
--
Please remove capital 'A's when replying by e-mail
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[Vols]

Re: a * b

OK. let's put this way:
How to implement ' a*b ' ( a multiply with b ) without using ' * ',
for example
3* 7 = 21, we can do this way 3<<3 - 3 = 21,
but what is the general solution for ' a * b' ?
Thanks.
Vol


Victor Bazarov wrote:

Vols wrote:

If we cannot use 'multiply' because of the memory limitation, how to
implement "a*b" ?

>
Huh? Memory limitation? What do you mean, exactly?
>

I know 3*7 can be written as 3<<3 - 3,

>
Try it, you'll be surprised at the result.
>

but what is the general
solution? thanks.

>
3*7
>
or you could try
>
7*3
>
.
>
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Siam]

Re: a * b

for(int i=0; i<b-1; i++)
{
a += a;
}

or

for(int i=0; i<b-1; i++)
{
b += b;
}

That would work when multiply by a positive number... Easy to adapt for
negatives too...

Vols wrote:

OK. let's put this way:
How to implement ' a*b ' ( a multiply with b ) without using ' * ',
for example
3* 7 = 21, we can do this way 3<<3 - 3 = 21,
but what is the general solution for ' a * b' ?
Thanks.
Vol
>
>
Victor Bazarov wrote:

Vols wrote:

If we cannot use 'multiply' because of the memory limitation, how to
implement "a*b" ?

Huh? Memory limitation? What do you mean, exactly?

I know 3*7 can be written as 3<<3 - 3,

Try it, you'll be surprised at the result.

but what is the general
solution? thanks.

3*7

or you could try

7*3

.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Siam]

Re: a * b

Oops, the second solution should of course be:

for(int i=0; i<a-1; i++)
{
b += b;
}

[Victor Bazarov]

Re: a * b

Vols wrote:

OK. let's put this way:

Please don't top-post.

How to implement ' a*b ' ( a multiply with b ) without using ' * ',
for example
3* 7 = 21, we can do this way 3<<3 - 3 = 21,

No, *we* cannot.

(3<<3 - 3) will *not* yield 21, at least in C++. Try it and you'll see.

but what is the general solution for ' a * b' ?

If multiplication is not available on your hardware, then you create a loop
in which you add 'a' to 'a' (b-1) times. That's what they were supposed to
teach you in grade school.

But that only works for _numbers_, not for matrices, for example.
Operations
mean different things for different types, so there can be no general
solution.

Thanks.
Vol
>
>
Victor Bazarov wrote:

>Vols wrote:

>>If we cannot use 'multiply' because of the memory limitation, how to
>>implement "a*b" ?

>>
>Huh? Memory limitation? What do you mean, exactly?
>>

>>I know 3*7 can be written as 3<<3 - 3,

>>
>Try it, you'll be surprised at the result.
>>

>>but what is the general
>>solution? thanks.

>>
> 3*7
>>
>or you could try
>>
> 7*3
>>
>.
>>
>V
>--
>Please remove capital 'A's when replying by e-mail
>I do not respond to top-posted replies, please don't ask

--
Please remove capital 'A's when replying by e-mail
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[Howard]

Re: a * b

Please don't top-post! [re-arranged]

>>
>Victor Bazarov wrote:

Vols wrote:
If we cannot use 'multiply' because of the memory limitation, how to
implement "a*b" ?
>
Huh? Memory limitation? What do you mean, exactly?
>
I know 3*7 can be written as 3<<3 - 3,
>
Try it, you'll be surprised at the result.
>
but what is the general
solution? thanks.
>
3*7
>
or you could try
>
7*3
>

Vols wrote:

>OK. let's put this way:
>How to implement ' a*b ' ( a multiply with b ) without using ' * ',
>for example
>3* 7 = 21, we can do this way 3<<3 - 3 = 21,

Really? Better test it. (And then check your operator precedence table.)

>but what is the general solution for ' a * b' ?

There is no "general solution" which involves shifting or any other trick
like that. There are an infinite number of ways to generate a given value,
once you know the value you want.

Multiplication is simply repeated addition. The solutions below show one
way (well, two ways) to use addition to get your answer.

"Siam" <siamraf@gmail. comwrote in message
news:1154448203 .629948.306460@ s13g2000cwa.goo glegroups.com.. .

for(int i=0; i<b-1; i++)
{
a += a;
}
>
or
>
for(int i=0; i<b-1; i++)

I take it you meant "i < a-1"? (And it might be more easily understood if
you started at 1.)

{
b += b;
}
>
That would work when multiply by a positive number... Easy to adapt for
negatives too...
>

What about for 0?

-Howard

[Vols]

Re: a * b

How to implement ' a*b ' ( a multiply with b ) without using ' * ',
for example
3* 7 = 21, we can do this way 3<<3 - 3 = 21,

>
No, *we* cannot.

Yes, we are, please try :unsigned a = (3<<3)-3;

[Vols]

Re: a * b

Vols wrote:

OK. let's put this way:

>
>
Please don't top-post.

What is top-post?

How to implement ' a*b ' ( a multiply with b ) without using ' * ',
for example
3* 7 = 21, we can do this way 3<<3 - 3 = 21,

>
No, *we* cannot.

Yes, YOU cannot, others can with (3<<3)-3

(3<<3 - 3) will *not* yield 21, at least in C++. Try it and you'll see.
>

but what is the general solution for ' a * b' ?

>
If multiplication is not available on your hardware, then you create a loop
in which you add 'a' to 'a' (b-1) times. That's what they were supposed to
teach you in grade school.

Congruatulation s! You graduated from primary school.
There could be some other solutions.

But that only works for _numbers_, not for matrices, for example.
Operations
mean different things for different types, so there can be no general
solution.
>

Thanks.
Vol


Victor Bazarov wrote:

Vols wrote:
>If we cannot use 'multiply' because of the memory limitation, how to
>implement "a*b" ?
>
Huh? Memory limitation? What do you mean, exactly?
>
>I know 3*7 can be written as 3<<3 - 3,
>
Try it, you'll be surprised at the result.
>
>but what is the general
>solution? thanks.
>
3*7
>
or you could try
>
7*3
>
.
>
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

>
--
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[Jakob Bieling]

Re: a * b

Vols <volunteers@gma il.comwrote:

>>How to implement ' a*b ' ( a multiply with b ) without using ' * ',
>>for example
>>3* 7 = 21, we can do this way 3<<3 - 3 = 21,

>No, *we* cannot.

Yes, we are, please try :unsigned a = (3<<3)-3;

But that was not the code in question. The original code does *not*
work, while yours does. '3<<3 - 3' is different from '(3<<3)-3'.

regards
--
jb

(reply address in rot13, unscramble first)

[Victor Bazarov]

Re: a * b

Vols wrote:

>Vols wrote:

>>OK. let's put this way:

>>
>>
>Please don't top-post.

>
What is top-post?

What you did before and what you're not doing any more. You're learning.
That's good!

>>How to implement ' a*b ' ( a multiply with b ) without using ' * ',
>>for example
>>3* 7 = 21, we can do this way 3<<3 - 3 = 21,

>>
>No, *we* cannot.

Yes, YOU cannot, others can with (3<<3)-3

Parentheses make a difference, don't they?

>(3<<3 - 3) will *not* yield 21, at least in C++. Try it and you'll
>see.
>>

>>but what is the general solution for ' a * b' ?

>>
>If multiplication is not available on your hardware, then you create
>a loop in which you add 'a' to 'a' (b-1) times. That's what they
>were supposed to teach you in grade school.

Congruatulation s! You graduated from primary school.
There could be some other solutions.

There could. But why bother?

[...excessive quoting removed...]

V
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[Ivan Vecerina]

Re: a * b

"Vols" <volunteers@gma il.comwrote in message
news:1154447730 .841957.250560@ m79g2000cwm.goo glegroups.com.. .
: OK. let's put this way:
: How to implement ' a*b ' ( a multiply with b ) without using ' * ',
: for example
: 3* 7 = 21, we can do this way 3<<3 - 3 = 21,
: but what is the general solution for ' a * b' ?
You'll find a complete and definitive answer
in Knuth's TAOCP.

But a reasonably easy first improvement over the
addition of a repeated b times looks somewhat like:
sum = 0;
while( b>0 ) {
if( b&1 ) sum += a;
b>>=1; a<<=1;
}

But instead of reinventing the wheel, I would
recommend looking for an existing "big int"
library, there are many around...

hth -Ivan
--
http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form
Brainbench MVP for C++ <http://www.brainbench.com

[Thomas J. Gritzan]

Re: a * b

Siam schrieb:

for(int i=0; i<b-1; i++)
{
a += a;
}
>
or
>
for(int i=0; i<b-1; i++)

....; i<a-1; ...

{
b += b;
}
>
That would work when multiply by a positive number... Easy to adapt for
negatives too...

Did you try it?
I think it would calculate a * 2^b and the other b * 2^a.

To multiply a and b, you would need another variable:

// works for positive b.
int sum = 0;
for(int i=0; i<b; i++)
sum += a;

--
Thomas

[Howard]

Re: a * b


"Howard" <alicebt@hotmai l.comwrote in message
news:e7Lzg.5266 89$Fs1.328019@b gtnsc05-news.ops.worldn et.att.net...

>
Multiplication is simply repeated addition. The solutions below show one
way (well, two ways) to use addition to get your answer.
>

Actually they're not, as Thomas pointed out. :-(

-Howard

[Thomas J. Gritzan]

Re: a * b

Thomas J. Gritzan schrieb:

Siam schrieb:

>for(int i=0; i<b-1; i++)
>{
> a += a;
>}
>>
>or
>>
>for(int i=0; i<b-1; i++)

>
...; i<a-1; ...
>

>{
> b += b;
>}
>>
>That would work when multiply by a positive number... Easy to adapt for
>negatives too...

>
Did you try it?
I think it would calculate a * 2^b and the other b * 2^a.

Just for clarification:
2^b in this case is the b-th exponent of 2, and not 2 xor b.

--
Thomas