Given the numbers $x = 123$ and $y = 100$ how to apply the Karatsuba algorithm to multiply these numbers ?
The formula is
xy=10^n(ac)+10^n/2(ad+bc)+bd
As I understand $n = 3$ (number of digits) and I tried writing the numbers as
x = 10*12+3 , y = 10*10 +0 thus a = 12 , b = 3 , c = 10 , d = 0
or
x = 100*1+23 , y = 100*1 +0 thus a = 1 , b = 23 , c = 1 , d = 0
I looked at some explanations of the algorithm and tried it successfully for other numbers , but I don't know how to solve it in this particular case.
Is this example part of a more general case of the algorithm (like 3-digit numbers)? I found a question that may be related (Karatsuba multiplication with integers of size 3) and from the answer I gather that it's impossible , so is it that Karatsuba can't multiply $2$ numbers of $3$-digits or is there a way to do this ?
What you have writtem is almost correct. The Karatsuba Algorithm expands the multiplication of $123$ and $100$ to $$\begin{aligned} (12\times 10+3)(10\times10+0)&= 10^2\times(12\times10) + 10^1\times (12\times0 + 3 \times 10) + 3\times 10\\ &= 12000 + 300 + 0\\ &= 12300 \end{aligned}$$
The Wikipedia page describes the algorithm and how it is used to multiply integers of any length. The only caveat is that the two numbers being multiplied are of the same length.