The reader should recognise the table below. The first row consists of the key, and therefore the pseudo-random string (PRS) starts with the first iteration on row 2. (The author invites opinion regarding which iteration is the first that can be said to fully meet the requirements of pseudo-randomness):
The reader should also recognise the encoded string below:
11,09,2,9,3,2,3,2,7,02,5,5,5,0,5,0,5,0,11,6,14,7,02,3,4,6,05,10,13,05,17
The encoded string (Code) is combined with the PRS by addition as below to produce the Product, the least signification digit of which becomes the Output. (When deciphering it is important to remember that the PRS is subtracted from the ciphertext, not the other way around; also, where the digit in the ciphertext is smaller than the digit in the PRS, 10 must be added to the ciphertext digit before the subtraction takes place.)
| Code | PRS | Product | Output |
| 1 | 1 | 2 | 2 |
| 1 | 6 | 7 | 7 |
| 0 | 4 | 4 | 4 |
| 9 | 8 | 17 | 7 |
| 2 | 4 | 6 | 6 |
| 9 | 4 | 13 | 3 |
| 3 | 2 | 5 | 5 |
| 2 | 8 | 10 | 0 |
| 3 | 1 | 4 | 4 |
| 2 | 4 | 6 | 6 |
| 7 | 1 | 8 | 8 |
| 0 | 4 | 4 | 4 |
| 2 | 0 | 2 | 2 |
| 5 | 2 | 7 | 7 |
| 5 | 7 | 12 | 2 |
| 5 | 5 | 10 | 0 |
| 0 | 0 | 0 | 0 |
| 5 | 5 | 10 | 0 |
| 0 | 2 | 2 | 2 |
| 5 | 3 | 8 | 8 |
| 0 | 3 | 3 | 3 |
| 1 | 5 | 6 | 6 |
| 1 | 5 | 6 | 6 |
| 6 | 7 | 13 | 3 |
| 1 | 3 | 4 | 4 |
| 4 | 8 | 12 | 2 |
| 7 | 5 | 12 | 2 |
| 0 | 1 | 1 | 1 |
| 2 | 3 | 5 | 5 |
| 3 | 4 | 7 | 7 |
| 4 | 2 | 6 | 6 |
| 6 | 5 | 11 | 1 |
| 0 | 1 | 1 | 1 |
| 5 | 5 | 10 | 0 |
| 1 | 9 | 10 | 0 |
| 0 | 9 | 9 | 9 |
| 1 | 9 | 10 | 0 |
| 3 | 1 | 4 | 4 |
| 0 | 8 | 8 | 8 |
| 5 | 5 | 10 | 0 |
| 1 | 7 | 8 | 8 |
| 7 | 8 | 15 | 5 |
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