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Hewitt,

come on, wise Mrs White may quite well be pushing the flowers by now, no?

If so, let her rest in peace. :) . And, if she had known the Vedic / Butterfly method

and would have been somewhat adventurous she might quite well have taught you
this method instead!

Actually, for Doniol you may use whatever method you like: The issue here is
not one of spherical geometry or the like. It is one of numerical multiplication.

Vedic / Butterfly does exactly the same thing as your longhand method
and Napiers bones, etc  - just in a different way, and, I say, in a more time saving
and less error prone way.

These advantages come from two sources:

1. For the sake of this discussion, you are given 2 numbers 4 digits each, with all
   digits different from 0 and need the product, a frequent situation in in CelNav.
   Then the product will have at least seven, often 8 digits. But, again in CelNav also
   rather frequently, you need only 4 digits of those hard earned 7 or 8  digits
   of the product and you will round the result. Yes, you will make a small error,
   often quite acceptable and, to some extend, correctable. More about that detail later.

   The rest will go to the bucket. What a loss of time and effort!

   Wouldn't you rather like a method that yielded only these 4 leading digits of the product?

   And that is what Vedic / Butterfly can do for you.

2. The scheme of the work for VB does not require memorizing carries or scribbling them

    somewhere above, below or in between the digits to be multiplied. They are recorded

    in a list easy to read, they will not be forgotten and can therefore be checked reliably later.

    Further, the partial products will never exceed 2 digits which come from the simple

    1 digit by 1 digit multiplication table Mrs White made you remember in your sleep.

For many people, whom I greatly envy, these advantages mean nothing. Greg Rudzinski,
Peter Monta, my wife, and perhaps you, race through any multiplication in seconds and make very

rarely errors. Remember, when we discussed making a cos() table by hand? Peter used

the Taylor series which requires a number of multiplications for each entry and proposed
that as a speedy way to get the cos(). Incredible, fantastic, unbelievable, but he showed

us how he would actually do that. And so these people may look down on VB. Let them.

However, for those slackers who have the same problems I do, VB is the way to go.

Before I continue, I'd like you to choose an actual problem and the way, in all detail, you

used to generate the product. I will respond with my VB solution. I will  also show you
how to deal with the rounding or cut-off errors.

So we will have an actual problem which makes the our discussion a little more realistic, right?

H

  

On Wed, Jun 17, 2015 at 10:01 AM, Hewitt Schlereth <NoReply_Schlereth@fer3.com> wrote:

Hanno, I think it's a case of "so easy when you know how." To abandon what my beloved Mrs White toiled to make clear to me in grade school ... Well, it would feel like a betrayal.

I remember looking up Napier's Bones on Wikipedia. They are a grid with a column of the numbers 1 thru 9 on the left; a row 1 thru 9 across the top. Each box in the grid has the number you get by multiplying across from left to right down the columns.

The the number in each box is printed into its 10s and 1s piece. 10s piece in the NW corner, 1s in the SE. They are separated by a diagonal line running NE-SW.

Then you make a copy of this grid (table) with the row across the top consisting of the digits of the number you want to multiply.

Here's the example given on the Internet site

http://mathworld.wolfram.com/NapiersBones.html

"To multiply two numbers, arrange the bones as described above. The above illustration shows this process for . The computation proceeds from right to left, starting with the rightmost bone in the row determined by themultiplier. In this case, the last digit in the 7s row of the 6-bone is 2, so write down 2. Now add the two adjacent numbers in the same row to the left (i.e., the ones in the parallelogram) to obtain , which is the next digit, so we now have 72. The next sum is , so write down the 2 two obtain 272 and carry the 1. Proceeding to the next digit, it is  (because of the carry), so write down the 4 to obtain  and carry the 1. The leftmost digit is then  (from the carry), giving the final answer ."

If I understand this correctly, for Doniol you put one number across the top. Then you go down the left column beginning with the rightmost digit of the number you are multiplying by, and multiply each digit in the top row, stacking and shifting the results to the left, just as my dear old teacher showed me.

I just don't see where the savings in time or accuracy come from. I have to 1) draw up a matrix (table, grid), 2) multiply row and column digit by digit, 3) stack the results, just as in my old method, 4) finally, add the columns of the stacks, just as in my old method. 

For me, the blunders come in the last step. So, for me Napier doesn't offer any advantage. Beyond step 1, everything is the same. So, for me the table is, A) extra work, B) one more place to blunder.

Besides, I couldn't dump Mrs White for "Bones" Napier. It would break her heart. :-)

Hewitt

On Jun 15, 2015, at 11:41 PM, Hanno Ix <NoReply_HannoIx@fer3.com> wrote:

Hewitt,

sorry, I do not understand - a table for / of what, please?

And, well, you would not need "cheating" as you professed, Vedic can be done w/o a calculator easily. If I, the slowest multiplicator on the list, can you surely can multiply with Vedic easily, too. Ever tried it?

H

On Mon, Jun 15, 2015 at 12:26 PM, Hewitt Schlereth <NoReply_Schlereth@fer3.com> wrote:

Hanno, don't these methods require you to make a table before you do the multiplying - ie, a preliminary chance to make blunders even ante getting to to the problem?

I'm with Gary and Greg on this one. Though I'm very slow at long multiplication, I do get there. (Then of course there's my backup - baggies of batteries and two-bit calculators.)

Hewitt