multiplication circuit by using only sumators

[spini]

Hi,
how can I make for example a circuit which multiply a number by 10, only by using (built-in) sumators and simple logical gates? (I'm using cedar logic soft)

Thanks.

[MK14]

Hi,
how can I make for example a circuit which multiply a number by 10, only by using (built-in) sumators and simple logical gates? (I'm using cedar logic soft)

Thanks.

It might be a homework question, so I will give a hint, rather than a full answer.

If X is some number.

What is (X shifted to the left 3 bits) plus (using adders) (X shifted left just one bit) ?

Example:
17

17 = 10001 in binary

So 10001000 <shifted left 3 bits, so had 000 appended to it> + 100010 <shifted 1 bit>

10001000
00100010 + in binary

10101010 = binary
= 170

[spini]

Thanks for response,

how did U know that it must be shifted in this example (multiplication by 10) to the left 3 bits + shifted 1 bit?
I get it how to make circuit by Ur explanation but, I don't understand how did U invent it 

[MK14]

Thanks for response,

how did U know that it must be shifted in this example (multiplication by 10) to the left 3 bits + shifted 1 bit?
I get it how to make circuit by Ur explanation but, I don't understand how did U invent it 

There was some FPGA discussions on here, a while ago. It used that method, as a very high speed way of multiplying by 10.
If I am misremembering that, then it would have been something like programming in C, where that method was used, to speed up the program, as multiplication, tends to be a little bit slow. Especially on very old processors.

[JohnnyMalaria]

Think how you do it in decimal:

e.g., 23 x 7 = [(2 x 10) x 7] + [3 x 7]

Multiplying by 10 in decimal is equivalent to multiplying by 2 in binary - which is just a shift.

And in binary, multiplication is easily done with adders. So, in the binary case, the above can be achieved with just adders and shifters.

Search for "binary multiplier gate implementation"

[Nominal Animal]

Do you know long multiplication?  Binary multiplication works exactly like that.

If you want to multiply by some constant C, convert C to binary representation.  Ignore all zeros in the binary representation.  The ones represent the shifts (of the other multiplicand) you need to sum together.

For example, consider C = 6 = 110₂ = 4 + 2.   To multiply any number X by C=6 we do
    6X = (4 + 2)X = 4X + 2X = (X << 2) + (X << 1)

Just remember that 1<<0=2⁰=1, 1<<1=2¹=2, 1<<2=2²=4, 1<<3=2³=8, 1<<4=2⁴=16, 1<<5=2⁵=32, and so on.

[ledtester]

And don't forget the trick of implementing division with multiplication:

https://ridiculousfish.com/blog/posts/labor-of-division-episode-i.html

So now you can implement division by a constant with just adders.

[rstofer]

2³ = 8 so when we shift right 3 places we are multiplying by 8 so shift X left 3 places
2¹ = 2 so when we shift left   1 place   we are multiplying by 2 sp shift X left 1 place

Finally, 8 times X plus 2 times X => 10 times X

[rstofer]

If we can't use shifts but only adders, we need a couple of temporary values, call them A + B
A = X + X  -> A := 2X
B = A + A  -> B := 4X   
ans = (B + B) + A  -> ans := 4X + 4X + 2X := 10X

[gnuarm]

If we can't use shifts but only adders, we need a couple of temporary values, call them A + B
A = X + X  -> A := 2X
B = A + A  -> B := 4X   
ans = (B + B) + A  -> ans := 4X + 4X + 2X := 10X

It would be a very odd situation where you could not use a shift.  But then I think in terms of hardware where shifting is just a matter of how you connect things.