Comments on: Ethiopian Multiplication

By: Ξ

Thu, 06 Sep 2012 12:32:09 +0000

Steve, you’re absolutely right! That would be the right name.

shuka — I know that there are examples of this from Egypt, but I don’t actually know enough about the documentation for this being Ethiopian.

And jack, my own view is that I’m undecided on this. I don’t think remembering doubles is inherently any harder than remembering what 4×2 is, say, although certainly I could do 14×12 in my head and that part would be harder with doubles. (Although I’d probably do it on my fingers, doing 100+60+8).

By: jack

jack — Thu, 06 Sep 2012 07:38:19 +0000

It’s an interesting method this one, but it doesn’t work too fast.
You have to remember all the “doubles” along the way…

Compared with 14 x 12 = 14 x 10 + 14 x 2, for example…

By: shuka

shuka — Thu, 06 Sep 2012 06:35:04 +0000

Very interesting, I am an Ethiopian but I don’t know this method.

By: Steve

Steve — Thu, 06 Sep 2012 02:19:30 +0000

“I think it would be fantastic to write a book called 25 ways to multiply”

Shouldn’t that be 5×5 ways to multiply?

By: Alexander Bogomolny

Alexander Bogomolny — Sun, 24 Apr 2011 19:42:55 +0000

Here’s a Java applet implementation:

http://www.cut-the-knot.org/Curriculum/Algebra/PeasantMultiplication.shtml

By: Rick

Rick — Sun, 14 Jun 2009 15:57:07 +0000

Here’s another viewpoint without the word “binary”: 12*14 = 12*(8 + 4 + 2) = 96 + 48 + 24 = 168

By: Ξ

Sat, 13 Jun 2009 12:20:49 +0000

I really like all the different ways that people look at this problem and see why it works. I usually justify this method by nothing that if you double one number and halve the other exactly, then the product is the same. So 14×12 is the the same as (14/2)x(12×2). The problem comes when you ignore a remainder.

14×12 is the same as 7×24, and that would be the same as 3.5×48. But we’re only left with 3×48. That means that the new product will be “off” by 0.5×48, which is 24 — not coincidentally, the number, above the 48. So whatever is across from 1 at the end is most of the product, and you have to add on the missing pieces (like that 24), which are always across from odd numbers since that’s when you run into trouble from dividing.

(And really, this is the same as the binary explanations without using the word binary.)

By: Rick Regan

Rick Regan — Fri, 12 Jun 2009 19:38:52 +0000

Just another take on the binary angle…

14 is 1110 in binary and 12 is 1100. Multiply the two in binary: 1100 x 1110 (think of 1110 as being on the bottom — I can’t draw it that way since the formatting won’t work). The partial products, in order, are 0000 (0), 11000 (24), 110000 (48), 1100000 (96). The nonzero products are copies of the top number, 12, shifted left — doubled — an appropriate number of times. Adding the partial products gives 10101000 (168).

By: Barry Leiba

Barry Leiba — Thu, 11 Jun 2009 05:01:04 +0000

You’re actually doing it in binary, generating (2^(N-1))*B on the right at the Nth stage. Then by taking the ones where the left side is odd, you’re taking the ones where the corresponding position in the binary representation of A is 1: reading from bottom up, you have 1 1 1 0, the binary representation of 14.

So there’s no surprise at all that this works… it’s basically the way a digital computer does it.

By: Chris

Chris — Thu, 11 Jun 2009 03:22:17 +0000

This method can be approached recursively. Each step, except the final one, is identical to the previous.

(define (mult a b)
(if (= a 1) b
(+ (if (odd a) b 0) (mult (/ a 2) (* b 2)))))

The integer division and multiplication can be replaced by single bitwise right and left shifts. (I hope this pseudo-code displays ok.)