### Ancient Computers, a paper of mine posted 7/15/2010 on IEEE Global History Network. It largely supercedes other material on this site. It determines:

1. that the Romans used The Salamis Tablet structure as their line abacus and used it for both duodecimal and decimal calculations;
2. that the Babylonians designed The Salamis Tablet abacus structure to accomodate their cuneform sexagesimal numbers;
3. how the pebbles were manipulated to do all four arithmetic operations in all three bases (10, 12, 60); and
4. that the methods of use are very rapid, the total number of pebbles needed very small, and the line abacus very portable.
=====

### An Ancient Base-60 Calculator?

by

#### Stephen K. Stephenson

B.S., M.Eng.(Elect.), M.Ed.
High School Math Teacher
sks23cu AT gmail.com

© 2005-2008 by Stephen K. Stephenson. Some Rights Reserved

#### Contents

Overview

Structure of Base 60 Counting Boards

Operations on Base 60 Counting Boards

Representing Digits as Sums of Five, Ten, Three, or Six Complements

Yale Tablet YBC 7289 Calculation (Sqrt(2)/2)

#### Overview

How did the Babylonians actually do their base-60 numeric calculations? This paper builds on a previous one, Ancient Scientific Calculators (Stephenson), that demonstrated methods to use copies of The Salamis Tablet counting board for base 10 calculations with numbers formed in what we call scientific notation. This paper describes and demonstrates extending those methods to do base-60 calculations as well. As a tutorial example, the length of a diagonal of a square whose sides are ½ units long is calculated, duplicating the results found on Yale tablet YBC 7289.

Further extensions into other multiple base counting boards are also possible, but not explored here. These could have aided calculations in various metrological systems.

#### Structure of Base 60 Counting Boards

Figure 1 shows two possibilities for structuring a base-60 counting board following the format of The Salamis Tablet. Both structures assume a pebble on a space between lines is worth ½ a pebble on the line above it. In support of that assumption, the Sumerians had words for numbers that were 5 times multiples of 10 (Ifrah, p.94) and Greek and Roman written numerals have symbols for them (Ifrah, pp.182-200). The Salamis Tablet itself has inscriptions of Greek Attic (or Acrophonic) Numerals that include these 5 times multiples of 10 numbers. The Roman Hand Abacus and its more modern copy, the Japanese Soroban, have at each digit position four one-count beads and one five-count bead. A five-count bead, or pebble, can also be thought of as a ½-count of the next higher digit, and it is that concept that we carry over to the base-60 counting board. The practical impact of using ½-count pebbles is to significantly reduce the number of pebbles needed both to represent a number on the counting board, thereby freeing up space on the board, and those needed to carry around to do calculations.

From the structure of their cuneiform numerals, the Babylonians obviously chose the second base-60 counting board structure. Numbers are entered on the board in the form ±a×60±b where (0,0;0,0,0,1) ≤ a < (3,0;0,0,0,0) and (0,0,0;) ≤ b < (3,0,0;). The a value is entered in the large table with the most significant digit at the top near the half circle. The b value is entered in the small table with the least significant digit at the bottom near the half circle. Any requirement for documenting leading or trailing zero digits is avoided, and the probability of zeros occurring between non-zero digits is low (1/60 for any one digit, 1/3600 for two consecutive digits, etc.). Perhaps numbers with b > 0 were interpreted as large units, like we use kilometer for 103 meters.

#### Operations on Base 60 Counting Boards

To add or subtract numbers their exponent parts must be equal, so one or both may have to be multiplied by appropriate powers of 60. This is accomplished by moving the pebble representation of the a part of a number up or down the table by full sexagesimal digits and adding the number of digits moved to the exponent part, b; an extremely similar process to adding numbers in our scientific notation.

Positive numbers have their pebbles added to the right of the median (vertical line), negative numbers have their pebbles added to the left of the median. Subtraction is accomplished by adding opposites: x - y = x + (-y). The Chinese documented the use of "false" number rods (negative numbers) about 200 B.C.E. (Ifrah, p.287, & Burton, p.236); and the Babylonians actually represented negative numbers before 1300 B.C.E. (Burton, p.66).

Multiplication and division are accomplished by adding or subtracting partial products until the multiplier or dividend is exhausted, forming the product or quotient in the process. See Ancient Scientific Calculators (Stephenson) for an example of multiplication on a base-10 counting board. The partial products are formed mostly by doubling or halving powers of 60 times the multiplicand or divisor. Since doubling or halving each pebble is pretty simple, no multiplication table is really needed; but sometimes other multiples of the four pebble locations are useful (even they are simple; see Figure 2). That the ancients used doubling and halving is well documented in the case of the Egyptians (Burton, pp.35-40), so the technique was probably known to other ancients, including the Babylonians.

The MacTutor History of Mathematics site has an An overview of Babylonian Mathematics page (retrieved 7/7/2005) that states:

Perhaps the most amazing aspect of the Babylonian's calculating skills was their construction of tables to aid calculation. Two tablets ... give squares of the numbers up to 59 and cubes of the numbers up to 32. ... The Babylonians used the ... formula ab = [(a + b)2 - (a - b)2]/4 which shows that a table of squares is all that is necessary to multiply numbers, simply taking the difference of the two squares that were looked up in the table then taking a quarter of the answer.

To use this formula, and a table of squares of numbers from (;1) to (;59), to multiply two five digit sexagesimal numbers we would proceed as follows:

(;a1,a2,a3,a4,a5) (;b1,b2,b3,b4,b5)

= (;a1) (;b1,b2,b3,b4,b5) + (;0,a2) (;b1,b2,b3,b4,b5) + (;0,0,a3) (;b1,b2,b3,b4,b5) + (;0,0,0,a4) (;b1,b2,b3,b4,b5) + (;0,0,0,0,a5) (;b1,b2,b3,b4,b5)

= (;a1) (;b1) + (;a1) (;0,b2) + (;a1) (;0,0,b3) + (;a1) (;0,0,0,b4) + (;a1) (;0,0,0,0,b5)
+ (;0,a2) (;b1) + (;0,a2) (;0,b2) + (;0,a2) (;0,0,b3) + (;0,a2) (;0,0,0,b4) + (;0,a2) (;0,0,0,0,b5)
+ (;0,0,a3) (;b1) + (;0,0,a3) (;0,b2) + (;0,0,a3) (;0,0,b3) + (;0,0,a3) (;0,0,0,b4) + (;0,0,a3) (;0,0,0,0,b5)
+ (;0,0,0,a4) (;b1)+(;0,0,0,a4) (;0,b2)+(;0,0,0,a4) (;0,0,b3)+(;0,0,0,a4) (;0,0,0,b4)+(;0,0,0,a4) (;0,0,0,0,b5)
+ (;0,0,0,0,a5) (;b1)+(;0,0,0,0,a5) (;0,b2)+(;0,0,0,0,a5) (;0,0,b3)+(;0,0,0,0,a5) (;0,0,0,b4)+(;0,0,0,0,a5) (;0,0,0,0,b5)

= (1;) [(;a1) (;b1)] + (;1) [(;a1) (;b2) + (;a2) (;b1)] + (;0,1) [(;a1) (;b3) + (;a2) (;b2) + (;a3) (;b1)] + (;0,0,1) [(;a1) (;b4) + (;a2) (;b3) + (;a3) (;b2) + (;a4) (;b1)] + (;0,0,0,1) [(;a1) (;b5) + (;a2) (;b4) + (;a3) (;b3) + (;a4) (;b2) + (;a5) (;b1)] + (;0,0,0,0,1) [(;a2) (;b5) + (;a3) (;b4) + (;a4) (;b3) + (;a5) (;b2)] + (;0,0,0,0,0,1) [(;a3) (;b5) + (;a4) (;b4) + (;a5) (;b3)] + (;0,0,0,0,0,0,1) [(;a4) (;b5) + (;a5) (;b4)] + (;0,0,0,0,0,0,0,1) [(;a5) (;b5)]

If we only wanted the five most significant figures in the result, we could throw away the terms in red. Then we have nineteen products to calculate using the formula aibj = [(ai + bj)2 - (ai - bj)2]/4. In (ai + bj)2 the sum will be of the form (1;cij) half the time, so that (ai + bj)2 = [(1;) + (;cij)]2 = (1;) + 2(;cij) + (;cij)2. On average, the calculation of each product would then require: two additions, two table lookup, half a doubling, two subtractions, and two halvings. Combining the nineteen products will require another 18 additions, being careful to add into the proper place value. In all, there are 56 additions, 38 table lookups, 8 doublings, 38 subtractions, and 38 halvings; a total of 178 operations!

How would you keep track of all this in cuneiform? How many errors would you make? How would you find them? Is this a simpler process than using counting boards to multiply, like the example in Ancient Scientific Calculators (Stephenson)?

#### Representing Digits as Sums of Five, Ten, Three, or Six Complements

Whenever pebble counts are reduced on lines and spaces between lines, digits should be represented as a sum of their five, ten, three, or six complements, as in Figure 3. Ancient Scientific Calculators (Stephenson) includes an analysis of the striking impact this has on reducing pebble count and increasing pebble efficiency. The board space freed by using these representations allow adding a second number onto a board without immediately combining it with the number already there. The new number can then be double checked before combining with the old. This reduces errors considerably.

Subtractive numeral notation was used by the Sumerians before 2500 B.C.E. (Ifrah, p.89), Babylonians (Burton, p.22), Etruscans (Ifrah, p.190, 38 = XXXIIX), and Romans (e.g., 4 = IV), with many other cultures using "back-counting" in their number names (Menninger, p.74).

The official methods to use the Japanese Soroban make extensive use of five and ten complements in order to "mechanize" operations "to minimize … mental labor … and let the result form … mechanically and naturally on the board" (Kojima, p.42-44).

The same thing should happen when using the operations described for a Salamis Tablet counting board.

#### Yale Tablet YBC 7289 Calculation (Sqrt(2)/2)

To demonstrate the use of base-60 counting boards, let's calculate the length of a diagonal of a square whose sides are ½ units long (Yale tablet YBC 7289). If the length of the diagonal is d, then we know, as did the Babylonians, that d2 = (;30)2+(;30)2 = 2(;30)(;30) = (;60)(;30) = (1;)(;30) = (;30). If we now think of d as the side of another square, that square must have an area of A := (;30). Squares are rectangles whose width and length are equal. If we look at rectangles that have area A and find one whose width and length are equal, then we have found d. We start with a guess for d, g0 = (1;). The two dimensions of the rectangle must then be g0 and A/g0. Our next and subsequent guesses for d will be the average of the previous two dimensions, gi= (gi-1+A/gi-1)/2. We will continue this iterative process until we achieve the desired number of digits, or run out of counting board space; see Figures 4 through 45. This is a lengthy example designed to demonstrate serious and practical use of counting boards; it goes much more rapidly with actual counting boards and tokens/pebbles. If you'd like to work along, you can paste together two of these for each counting board and use pennies for the tokens.

#### Bibliography

Burton, D. M.
1999     The History of Mathematics: An introduction. Fourth Edition. New York: McGraw-Hill.
Ifrah, G. [Bellos, D., Harding, E.F., Wood, S. & Monk, I., French Translators].
2000     The Universal History of Numbers: From prehistory to the invention of the computer. New York: John Wiley & Sons, Inc.
Kojima, T.
1954     The Japanese Abacus: Its Use and Theory. Tokyo, Japan: Charles E. Tuttle Co.
Menninger, K.
1969     Number Words and Number Symbols: A Cultural History of Numbers. Cambridge, Massachusetts: M.I.T. Press.
Stephenson, S. K.
2005     "Ancient Scientific Calculators"

© 2006-2008 Stephen Kent Stephenson. Some Rights Reserved.