r. Here, we find that the inverse of 7 is 28. Proof: 7 X 28 = 196 = 1 (modulo 39). Consider the second example, which concerns the search of 5-1 mod 16272. obtain Table Y2. In steps 1 and 2, we fill out the table as shown. We first divide 16272 by 5, to find that the integer part of the result is [16272/5] = 3254. In step 3, we multiply line 2 by 3254, and in step 4 we subtract the result of line 3 from that of line 1. This completes the first round. Consider next the ratio 5/2, whose integer part is [5/2] = 2. In step 5, we multiply the results of line 4 by 2, and in step 6, we subtract the result of line 5 from that of line 2. This completes the search, since we have obtained 1 in the first column (line 6). However, the number in the right column, b = 26035 is negative. In this case, the inverse is the complement km + b (k > 1), namely 5-1 = 16272 - 26035 = 6509. Proof: 5 X 6509 = 32 545 = (2 X 16 272) + 1 = 1 (modulo 16 272). Polynomial byte representation and modular multiplication I shall now implement the EEA with polynomials modulo m(x). For instance, we want the inverse of p(x) = x5 + x3 + 1 modulo m(x) = x8+x4+x3+ Table Y3 shows the different calculation steps: Table Y3 The steps in calculating p-1(x). x8tx4tx3+x+1 x8tx4tx3+x+1 x5+x3+1 =x8+x4+1 Since we obtained 1 in the last line of left column, the result in the right column, x 8 + x + + 1, is the inverse p-1(x). We can readily check that p-1(x) = = (x5 + x 3 - 1 ) X (x8+x4+ mod m(x) = 1 mod m(x). Index A-law, 592 poly (-) substitution, 526 Action table (Turing machine), 97, 107 substitution, 526 Adaptive coding, see Coding ALU (algorithmic and logical unit), 283, 288, 289 Adder (plain), 299 American Addition (quantum operation), 326 Automobile Association (AA), 159 Additive noise, see Noise Civil War, 132 Adleman L., 424, 536 wheel, 170 ADSL, 225 Ammeter, 360 Advanced encryption standard, see AES Ampersand AES (advanced encryption standard), 523, 541-3, Amplified coherent light, see Coherent 555, 563, 597, 672 Analog-to-digital voice conversion, see S