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The Laws of Cryptography with Java Code.

This is a lesson from prehistoric times. It brings back nostalgic memories. Before calculators, one used printed tables to carry out calculations. The example in the main section was to calculate 23.427 * 23.427 * 3.1416. To do this, one first needed the logarithms (base 10) of the two numbers. In red bold below are the actual table entries (using a book of tables dating from 1957) -- everything else you had to do mentally or on paper:

Number   Logarithm

2342     36959    using interpolation entry:
23427    369716      7th entry under 18 is 12.6
2343     36977    take 369590 + 126 to get 369716

This means that log(2.3427) = 0.369716 approximately. Then log(23.427) = log(2.3427 * 10) = log(2.3427) + log(10) = 0.369716 + 1 = 1.369716

Similarly, look up 3.1416:

Number   Logarithm

3141     49707    using interpolation entry:
31416    497154      6th entry under 14 is 8.4
3142     49721    take 497070 + 84 to get 497154

This means that log(3.1416) = 0.497154 approximately.

Form the sum: 1.369716 + 1.369716 + 0.497154 = 3.236586 (this must be done by hand, with pencil and paper).

Now finally, one has to look up the ``anti-log'' in the same table:

Number   Logarithm

1724     23654    using interpolation entry:
17242    23659      2nd entry under 25 is 5.0
1724     23679    take 17240 + 2 to get 17242

This means that log(1.7242) = 0.23659 approximately, so 3.23659 = 3 + 0.23659 = log(1000) + log(1.7242) = log(1000 * 1.7242) = log(1724.2), or (finally), the answer is 1724.2 approximately. So the area of a circle of radius 23.427 is approximately 1724.2.

All this pain just to multiply 3 numbers together, to get 4 or 5 digits of accuracy in the answer. The next two lessons from our primitive ancestors: how to use tables of the logarithms of trig functions (to save one lookup), and how to use a slide rule. (Just kidding.)