Non-conventional mathematics

tekesta · #1

Lately I've been brushing up on my algebra. It's been 13 years since I last studied it seriously. In fact, I've never considered myself to be proficient at math.

Then in the last few days I've (re)discovered abacus, discovered Trachtenberg speed math and Vedic maths - which I have just begun to practice - and "Ethiopian binary math." This last one really grabbed my attention. Below is a problem from an article published by the University of Houston.

A goat herder is in talks with a farmer over the sale of his 34 goats. They both agree on a price of $7.00, but none of them know how to multiply. So they go to a shaman to find out the total market price for all 34 goats. To do this, the shaman digs in the ground two rows of six holes each. Then, he takes his bag of pebbles and begins to add them to each hole as shown below.

On the left column, a hole with an even number of pebbles is considered "evil", while a hole with an odd number of pebbles is considered "good". Only the pebbles in the holes directly adjacent to the good ones are added to the final count. In this case the good holes are the one with 17 pebbles and the one with 1 pebble. Next to the former is a hole with 14 pebbles and next to the latter is a hole with 224 pebbles. These are added together, resulting in 238 pebbles.

[For source material, click here: http://www.uh.edu/engines/epi504.htm]

Now I am convinced of the existence of alternative mathematical methods

At least math does not appear so difficult any more.

ez4u · #2

I am too feckless to read the article myself. How would this have worked if we had started with 35 goats?

shapenaji · #3

ez4u wrote:

I am too feckless to read the article myself. How would this have worked if we had started with 35 goats?

The first pile would have been "good" and would have constituted a remainder. Basically that's just what the piles are doing, divide by 2, if there's a remainder, add that to the final total.

It appears to be an accurate, if not particularly powerful, method.

Addendum:

It works based on the fact that you can determine any positive integer from a sum of powers of 2.

Additional Addendum:

The real question is, how does the holy man know how many holes to dig?

DrStraw · #4

I have also heard that as referred to as the Russian peasant method. As shapenaj says, it is merely based on the remainders after binary division. Basically it is binary counting in disguise.

SmoothOper · #5

Looks vaguely like the way an Arithmetic Logic Unit implemented with shift registers would do it, dividing by two is easy in binary.

http://en.wikipedia.org/wiki/Binary_multiplier

jug · #6

shapenaji wrote:

ez4u wrote:

I am too feckless to read the article myself. How would this have worked if we had started with 35 goats?

The first pile would have been "good" and would have constituted a remainder. Basically that's just what the piles are doing, divide by 2, if there's a remainder, add that to the final total.
[...]
Additional Addendum:
The real question is, how does the holy man know how many holes to dig?

Well, if they can't multiply then the probably also cannot divide. Also I don't think the shaman "knows" how many holes to dig, otherwise he needs to know how to apply a logarithm (base 2).

But he can do it like this ("divide by counting"):

1. Start with one hole, put in 34 pebbles (for the goats).
2. Dig the next hole and split the former pebbles into two new "buckets" A/B by counting one pebble into A, one into B till there's one or zero pebbles left in the former hole. A is the next hole, and put as much stones as you have in A also into B (to have the original amount there).
3. Repeat step 1/2 till there's only one pebble in the "new" hole.

lemmata · #7

The shaman seems to be obfuscating the matter to artificially inflate his value. (Like an analyst at Lehman Brothers? Somethings never change ;-)

) He needs more than 238 pebbles to calculate 34 times 7. If this were a computer algorithm, then we might also say that he wastes memory by digging so many holes. All children have a simpler way of doing multiplication visually. A friend of mine from Japan with wild hair claims that this way is forgotten after we become adults because we start to view mathematics as the sequences of symbolic manipulations used to solve a problem rather than the real meaning of those symbolic manipulations. Are any of you "childish" enough to realize how a child would solve this problem?

Mef · #8

shapenaji wrote:

It appears to be an accurate, if not particularly powerful, method.

If memory serves, it is valuable to computer science as it allows you to perform multiplication using very simple binary operations (register shifts, single bit comparisons, etc).

tekesta · #9

In Vedic maths they have something known as the "ten-point circle", which functions much like the number line in Western maths. One goes around the circle, starting with 0 at the top of the circle. Once 10 is reached, the circle begins all over again so that 11 sits over 1, 12 sits over 2, etc. For the numbers 1-10, numbers that together make a sum of 10 are connected by a chord line through the circle. From top to bottom, 1 + 9, 2 + 8, 3 + 7, and 4 + 6 are the connected pairs. 10 at the top represents 10 + 0 and 5 at the bottom represents 5 + 5. This I did not learn in school.

Another thing I did not learn in school was the following method of multiplying any two digit numbers up to 99. Let's say we want to multiply 95 by 66. First we multiply the left side digits together and write this sub-product into the product as the left hand figure of the answer. Second, we make a space in the middle for the missing digit, multiply the right side digits together, and write this sub-product in as the right hand figure of the answer. Then, to find the missing middle digit, we multiply outer digit by outer digit, inner digit by inner digit, and add these sub-products to make the middle digit. Then everything is added as follows.

Code:

95 x 66

9x6=54 [LH digits]
5x6=30 [RH digits]
9x6=54 [outer digits]
5x6=30 [inner digits]

  5400 [LH digit]
  0840 [inner digits + outer digits]
+ 0030 [RH digit]
-------
  6270

Now we check to ensure the answer is correct.

Code:

    95
  x 66
  ----
   570
+ 5700
------
  6270

This method of multiplication can be visually depicted with lines on a grid rotated 45º to the right. Here is a YouTube video showing how:
http://www.youtube.com/watch?v=_AJvshZmYPs

phillip1882 · **#10**

so with a multiplication like 47 * 15

Code:

cool.

DrStraw · **#11**

tekesta wrote:

Code:

95 x 66

9x6=54 [LH digits]
5x6=30 [RH digits]
9x6=54 [outer digits]
5x6=30 [inner digits]

  5400 [LH digit]
  0840 [inner digits + outer digits]
+ 0030 [RH digit]
-------
  6270

This is nothing more than the FOIL of high school algebra classes and it is the way I have multiplied two digit numbers together for over 50 years.

phillip1882 · **#12**

see if you can figure out my method.

Code:

0   35
-  105
0  315
0  945
+ 2835
    ------
78*35=2730

-   21
+   63
-  189
-  567
+ 1701
----------
 47*21=987

try 63 * 31.

shapenaji · **#13**

phillip1882 wrote:

see if you can figure out my method.

Code:

0   35
-  105
0  315
0  945
+ 2835
    ------
78*35=2730

-   21
+   63
-  189
-  567
+ 1701
----------
 47*21=987

try 63 * 31.

Ternary division this time, if it divides exactly: -, remainder 2: 0, remainder 1: +

phillip1882 · **#14**

not quite ternary but very very close.
note that 78 is evenly divisable by 3, but got a 0.

Non-conventional mathematics

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