A History of Mathematics From Mesopotamia to Modernity

ChineseMathematics 89

Here is the text of chapter 8, problem 1, together with the ‘Array Rule’ which solves it.

Now given 3 bundles of top grade paddy, 2 bundles of medium-grade paddy, [and] 1 bundle of low grade paddy. Yield:

39 douof grain. 2 bundles of top grade paddy, 3 bundles of medium-grade paddy, [and] 1 bundle of low grade paddy,

yield 34dou. 1 bundle of top grade paddy, 2 bundles of medium-grade paddy, [and] 3 bundles of low grade paddy, yield

26 dou. Tell: how much paddy does one bundle of each grade yield?

Answer: Top grade paddy yields 9^14 dou[per bundle]; medium grade paddy 4^14 dou; [and] low grade paddy 2^34 dou.

The Array [Fangcheng] Rule

[Let Problem 1 serve as an example,] lay down in the right column 3 bundles of top grade paddy, 2 bundles of medium

grade paddy, [and] 1 bundle of low grade paddy. Yield: 39douof grain. Similarly for the middle and left column.

Use [the number of bundles of] top grade paddy in the right column to multiply the middle column then merge. Again

multiply the next [and] follow by pivoting. Then use the remainder of the medium grade paddy in the middle column

to multiply the left column and pivot. The remainder of the low grade paddy in the left column is the divisor, the entry

below is the dividend. The quotient is the yield of the low grade paddy...

The above quotation is enough (a) to compute the basic solution, the yield of low-grade paddy
(the others can be found by substitution), (b) to show how the method is described in theNine
Chapters. For us, the description of what to do is unclear unless you have had the terms, for example,
‘merge’ and ‘pivot’, explained to you. The process begins as follows. First, imagine counting rods
laid out (perhaps on a board) to represent the numbers used; in matrix terms:




123

232

311

26 34 39







The textbook method today would be to get enough zeros in the matrix (‘triangular form’) by
subtracting multiples of rows or columns from each other. This is nearly what the method described
does. Here is Liu’s explanation for the cryptic ‘merge’:
The meaning of this rule is: subtract the column with smallest [top entry] repeatedly from the columns with larger
[top entries], then the top entry must vanish. With the top entry gone, the column has one item absent.

How does this work? First note that we have (from the method above) multiplied the middle column

by 3=number of bundles of top-grade. We now subtract the right column (the smaller) repeatedly

from the middle (the larger). The stages are:









163

292

331

26 102 39







→









103

252

311

26 24 39









The differences from today’s procedure are fairly trivial. Most probably:

1. we would look for a column, for example, the left one, which could be subtracted from others
   without first having to ‘multiply them up’;


2. we would say we were subtracting twice column 3 rather than saying that we were subtracting
   it repeatedly.