This page demonstrates how to generate ECC blocks by performing polynomial divison on a message polynomial. Just enter the coefficients of the message polynomial and the desired number of ECC blocks, then click Perform Division.
If the steps are unclear, please read the error correction generation section of the tutorial, which includes a detailed explanation of these steps.
Polynomial Division Steps
The first step to the division is to prepare the message polynomial for the division. The full message polynomial is:
70x15 + 247x14 + 118x13 + 86x12 + 194x11 + 6x10 + 151x9 + 50x8 + 16x7 + 236x6 + 17x5 + 236x4 + 17x3 + 236x2 + 17x1 + 236
To make sure that the exponent of the lead term doesn't become too small during the division, multiply the message polynomial by xn where n is the number of error correction codewords that are needed. In this case n is 18, for 18 error correction codewords, so multiply the message polynomial by x18, which gives us:
70x33 + 247x32 + 118x31 + 86x30 + 194x29 + 6x28 + 151x27 + 50x26 + 16x25 + 236x24 + 17x23 + 236x22 + 17x21 + 236x20 + 17x19 + 236x18
The lead term of the generator polynomial should also have the same exponent, so multiply by x15 to get
ɑ0x33 + ɑ215x32 + ɑ234x31 + ɑ158x30 + ɑ94x29 + ɑ184x28 + ɑ97x27 + ɑ118x26 + ɑ170x25 + ɑ79x24 + ɑ187x23 + ɑ152x22 + ɑ148x21 + ɑ252x20 + ɑ179x19 + ɑ5x18 + ɑ98x17 + ɑ96x16 + ɑ153x15
Now it is possible to perform the repeated division steps. The number of steps in the division must equal the number of terms in the message polynomial. In this case, the division will take 16 steps to complete. This will result in a remainder that has 18 terms. These terms will be the 18 error correction codewords that are required.
Step 1a: Multiply the Generator Polynomial by the Lead Term of the Message Polynomial
The first step is to multiply the generator polynomial by the lead term of the message polynomial. The lead term in this case is 70x33. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 70x33 to alpha notation. According to the log antilog table, for the integer value 70, the alpha exponent is 48. Therefore 70 = ɑ48. Multiply the generator polynomial by ɑ48:
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ48x33 + ɑ(263 % 255)x32 + ɑ(282 % 255)x31 + ɑ206x30 + ɑ142x29 + ɑ232x28 + ɑ145x27 + ɑ166x26 + ɑ218x25 + ɑ127x24 + ɑ235x23 + ɑ200x22 + ɑ196x21 + ɑ(300 % 255)x20 + ɑ227x19 + ɑ53x18 + ɑ146x17 + ɑ144x16 + ɑ201x15The result is:
ɑ48x33 + ɑ8x32 + ɑ27x31 + ɑ206x30 + ɑ142x29 + ɑ232x28 + ɑ145x27 + ɑ166x26 + ɑ218x25 + ɑ127x24 + ɑ235x23 + ɑ200x22 + ɑ196x21 + ɑ45x20 + ɑ227x19 + ɑ53x18 + ɑ146x17 + ɑ144x16 + ɑ201x15
Now, convert this to integer notation:
70x33 + 29x32 + 12x31 + 83x30 + 42x29 + 247x28 + 77x27 + 63x26 + 43x25 + 204x24 + 235x23 + 28x22 + 200x21 + 193x20 + 144x19 + 40x18 + 154x17 + 168x16 + 56x15Step 1b: XOR the result with the message polynomial
Since this is the first division step, XOR the result from 1a with the message polynomial.
(70 ⊕ 70)x33 + (247 ⊕ 29)x32 + (118 ⊕ 12)x31 + (86 ⊕ 83)x30 + (194 ⊕ 42)x29 + (6 ⊕ 247)x28 + (151 ⊕ 77)x27 + (50 ⊕ 63)x26 + (16 ⊕ 43)x25 + (236 ⊕ 204)x24 + (17 ⊕ 235)x23 + (236 ⊕ 28)x22 + (17 ⊕ 200)x21 + (236 ⊕ 193)x20 + (17 ⊕ 144)x19 + (236 ⊕ 40)x18 + (0 ⊕ 154)x17 + (0 ⊕ 168)x16 + (0 ⊕ 56)x15
The result is:
0x33 + 234x32 + 122x31 + 5x30 + 232x29 + 241x28 + 218x27 + 13x26 + 59x25 + 32x24 + 250x23 + 240x22 + 217x21 + 45x20 + 129x19 + 196x18 + 154x17 + 168x16 + 56x15
Discard the lead 0 term to get:
234x32 + 122x31 + 5x30 + 232x29 + 241x28 + 218x27 + 13x26 + 59x25 + 32x24 + 250x23 + 240x22 + 217x21 + 45x20 + 129x19 + 196x18 + 154x17 + 168x16 + 56x15
Step 2a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 234x32. Convert 234x32 to alpha notation. According to the log antilog table, for the integer value 234, the alpha exponent is 22. Therefore 234 = ɑ22. Multiply the generator polynomial by ɑ22:
(ɑ22 * ɑ0)x32 + (ɑ22 * ɑ215)x31 + (ɑ22 * ɑ234)x30 + (ɑ22 * ɑ158)x29 + (ɑ22 * ɑ94)x28 + (ɑ22 * ɑ184)x27 + (ɑ22 * ɑ97)x26 + (ɑ22 * ɑ118)x25 + (ɑ22 * ɑ170)x24 + (ɑ22 * ɑ79)x23 + (ɑ22 * ɑ187)x22 + (ɑ22 * ɑ152)x21 + (ɑ22 * ɑ148)x20 + (ɑ22 * ɑ252)x19 + (ɑ22 * ɑ179)x18 + (ɑ22 * ɑ5)x17 + (ɑ22 * ɑ98)x16 + (ɑ22 * ɑ96)x15 + (ɑ22 * ɑ153)x14
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ22x32 + ɑ237x31 + ɑ(256 % 255)x30 + ɑ180x29 + ɑ116x28 + ɑ206x27 + ɑ119x26 + ɑ140x25 + ɑ192x24 + ɑ101x23 + ɑ209x22 + ɑ174x21 + ɑ170x20 + ɑ(274 % 255)x19 + ɑ201x18 + ɑ27x17 + ɑ120x16 + ɑ118x15 + ɑ175x14The result is:
ɑ22x32 + ɑ237x31 + ɑ1x30 + ɑ180x29 + ɑ116x28 + ɑ206x27 + ɑ119x26 + ɑ140x25 + ɑ192x24 + ɑ101x23 + ɑ209x22 + ɑ174x21 + ɑ170x20 + ɑ19x19 + ɑ201x18 + ɑ27x17 + ɑ120x16 + ɑ118x15 + ɑ175x14
Now, convert this to integer notation:
234x32 + 139x31 + 2x30 + 150x29 + 248x28 + 83x27 + 147x26 + 132x25 + 130x24 + 34x23 + 162x22 + 241x21 + 215x20 + 90x19 + 56x18 + 12x17 + 59x16 + 199x15 + 255x14Step 2b: XOR the result with the result from step 1b
Use the result from step 1b to perform the next XOR.
(234 ⊕ 234)x32 + (122 ⊕ 139)x31 + (5 ⊕ 2)x30 + (232 ⊕ 150)x29 + (241 ⊕ 248)x28 + (218 ⊕ 83)x27 + (13 ⊕ 147)x26 + (59 ⊕ 132)x25 + (32 ⊕ 130)x24 + (250 ⊕ 34)x23 + (240 ⊕ 162)x22 + (217 ⊕ 241)x21 + (45 ⊕ 215)x20 + (129 ⊕ 90)x19 + (196 ⊕ 56)x18 + (154 ⊕ 12)x17 + (168 ⊕ 59)x16 + (56 ⊕ 199)x15 + (0 ⊕ 255)x14
The result is:
0x32 + 241x31 + 7x30 + 126x29 + 9x28 + 137x27 + 158x26 + 191x25 + 162x24 + 216x23 + 82x22 + 40x21 + 250x20 + 219x19 + 252x18 + 150x17 + 147x16 + 255x15 + 255x14
Discard the lead 0 term to get:
241x31 + 7x30 + 126x29 + 9x28 + 137x27 + 158x26 + 191x25 + 162x24 + 216x23 + 82x22 + 40x21 + 250x20 + 219x19 + 252x18 + 150x17 + 147x16 + 255x15 + 255x14
Step 3a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 241x31. Convert 241x31 to alpha notation. According to the log antilog table, for the integer value 241, the alpha exponent is 174. Therefore 241 = ɑ174. Multiply the generator polynomial by ɑ174:
(ɑ174 * ɑ0)x31 + (ɑ174 * ɑ215)x30 + (ɑ174 * ɑ234)x29 + (ɑ174 * ɑ158)x28 + (ɑ174 * ɑ94)x27 + (ɑ174 * ɑ184)x26 + (ɑ174 * ɑ97)x25 + (ɑ174 * ɑ118)x24 + (ɑ174 * ɑ170)x23 + (ɑ174 * ɑ79)x22 + (ɑ174 * ɑ187)x21 + (ɑ174 * ɑ152)x20 + (ɑ174 * ɑ148)x19 + (ɑ174 * ɑ252)x18 + (ɑ174 * ɑ179)x17 + (ɑ174 * ɑ5)x16 + (ɑ174 * ɑ98)x15 + (ɑ174 * ɑ96)x14 + (ɑ174 * ɑ153)x13
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ174x31 + ɑ(389 % 255)x30 + ɑ(408 % 255)x29 + ɑ(332 % 255)x28 + ɑ(268 % 255)x27 + ɑ(358 % 255)x26 + ɑ(271 % 255)x25 + ɑ(292 % 255)x24 + ɑ(344 % 255)x23 + ɑ253x22 + ɑ(361 % 255)x21 + ɑ(326 % 255)x20 + ɑ(322 % 255)x19 + ɑ(426 % 255)x18 + ɑ(353 % 255)x17 + ɑ179x16 + ɑ(272 % 255)x15 + ɑ(270 % 255)x14 + ɑ(327 % 255)x13The result is:
ɑ174x31 + ɑ134x30 + ɑ153x29 + ɑ77x28 + ɑ13x27 + ɑ103x26 + ɑ16x25 + ɑ37x24 + ɑ89x23 + ɑ253x22 + ɑ106x21 + ɑ71x20 + ɑ67x19 + ɑ171x18 + ɑ98x17 + ɑ179x16 + ɑ17x15 + ɑ15x14 + ɑ72x13
Now, convert this to integer notation:
241x31 + 218x30 + 146x29 + 60x28 + 135x27 + 136x26 + 76x25 + 74x24 + 225x23 + 71x22 + 52x21 + 188x20 + 194x19 + 179x18 + 67x17 + 75x16 + 152x15 + 38x14 + 101x13Step 3b: XOR the result with the result from step 2b
Use the result from step 2b to perform the next XOR.
(241 ⊕ 241)x31 + (7 ⊕ 218)x30 + (126 ⊕ 146)x29 + (9 ⊕ 60)x28 + (137 ⊕ 135)x27 + (158 ⊕ 136)x26 + (191 ⊕ 76)x25 + (162 ⊕ 74)x24 + (216 ⊕ 225)x23 + (82 ⊕ 71)x22 + (40 ⊕ 52)x21 + (250 ⊕ 188)x20 + (219 ⊕ 194)x19 + (252 ⊕ 179)x18 + (150 ⊕ 67)x17 + (147 ⊕ 75)x16 + (255 ⊕ 152)x15 + (255 ⊕ 38)x14 + (0 ⊕ 101)x13
The result is:
0x31 + 221x30 + 236x29 + 53x28 + 14x27 + 22x26 + 243x25 + 232x24 + 57x23 + 21x22 + 28x21 + 70x20 + 25x19 + 79x18 + 213x17 + 216x16 + 103x15 + 217x14 + 101x13
Discard the lead 0 term to get:
221x30 + 236x29 + 53x28 + 14x27 + 22x26 + 243x25 + 232x24 + 57x23 + 21x22 + 28x21 + 70x20 + 25x19 + 79x18 + 213x17 + 216x16 + 103x15 + 217x14 + 101x13
Step 4a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 221x30. Convert 221x30 to alpha notation. According to the log antilog table, for the integer value 221, the alpha exponent is 204. Therefore 221 = ɑ204. Multiply the generator polynomial by ɑ204:
(ɑ204 * ɑ0)x30 + (ɑ204 * ɑ215)x29 + (ɑ204 * ɑ234)x28 + (ɑ204 * ɑ158)x27 + (ɑ204 * ɑ94)x26 + (ɑ204 * ɑ184)x25 + (ɑ204 * ɑ97)x24 + (ɑ204 * ɑ118)x23 + (ɑ204 * ɑ170)x22 + (ɑ204 * ɑ79)x21 + (ɑ204 * ɑ187)x20 + (ɑ204 * ɑ152)x19 + (ɑ204 * ɑ148)x18 + (ɑ204 * ɑ252)x17 + (ɑ204 * ɑ179)x16 + (ɑ204 * ɑ5)x15 + (ɑ204 * ɑ98)x14 + (ɑ204 * ɑ96)x13 + (ɑ204 * ɑ153)x12
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ204x30 + ɑ(419 % 255)x29 + ɑ(438 % 255)x28 + ɑ(362 % 255)x27 + ɑ(298 % 255)x26 + ɑ(388 % 255)x25 + ɑ(301 % 255)x24 + ɑ(322 % 255)x23 + ɑ(374 % 255)x22 + ɑ(283 % 255)x21 + ɑ(391 % 255)x20 + ɑ(356 % 255)x19 + ɑ(352 % 255)x18 + ɑ(456 % 255)x17 + ɑ(383 % 255)x16 + ɑ209x15 + ɑ(302 % 255)x14 + ɑ(300 % 255)x13 + ɑ(357 % 255)x12The result is:
ɑ204x30 + ɑ164x29 + ɑ183x28 + ɑ107x27 + ɑ43x26 + ɑ133x25 + ɑ46x24 + ɑ67x23 + ɑ119x22 + ɑ28x21 + ɑ136x20 + ɑ101x19 + ɑ97x18 + ɑ201x17 + ɑ128x16 + ɑ209x15 + ɑ47x14 + ɑ45x13 + ɑ102x12
Now, convert this to integer notation:
221x30 + 198x29 + 196x28 + 104x27 + 119x26 + 109x25 + 159x24 + 194x23 + 147x22 + 24x21 + 79x20 + 34x19 + 175x18 + 56x17 + 133x16 + 162x15 + 35x14 + 193x13 + 68x12Step 4b: XOR the result with the result from step 3b
Use the result from step 3b to perform the next XOR.
(221 ⊕ 221)x30 + (236 ⊕ 198)x29 + (53 ⊕ 196)x28 + (14 ⊕ 104)x27 + (22 ⊕ 119)x26 + (243 ⊕ 109)x25 + (232 ⊕ 159)x24 + (57 ⊕ 194)x23 + (21 ⊕ 147)x22 + (28 ⊕ 24)x21 + (70 ⊕ 79)x20 + (25 ⊕ 34)x19 + (79 ⊕ 175)x18 + (213 ⊕ 56)x17 + (216 ⊕ 133)x16 + (103 ⊕ 162)x15 + (217 ⊕ 35)x14 + (101 ⊕ 193)x13 + (0 ⊕ 68)x12
The result is:
0x30 + 42x29 + 241x28 + 102x27 + 97x26 + 158x25 + 119x24 + 251x23 + 134x22 + 4x21 + 9x20 + 59x19 + 224x18 + 237x17 + 93x16 + 197x15 + 250x14 + 164x13 + 68x12
Discard the lead 0 term to get:
42x29 + 241x28 + 102x27 + 97x26 + 158x25 + 119x24 + 251x23 + 134x22 + 4x21 + 9x20 + 59x19 + 224x18 + 237x17 + 93x16 + 197x15 + 250x14 + 164x13 + 68x12
Step 5a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 42x29. Convert 42x29 to alpha notation. According to the log antilog table, for the integer value 42, the alpha exponent is 142. Therefore 42 = ɑ142. Multiply the generator polynomial by ɑ142:
(ɑ142 * ɑ0)x29 + (ɑ142 * ɑ215)x28 + (ɑ142 * ɑ234)x27 + (ɑ142 * ɑ158)x26 + (ɑ142 * ɑ94)x25 + (ɑ142 * ɑ184)x24 + (ɑ142 * ɑ97)x23 + (ɑ142 * ɑ118)x22 + (ɑ142 * ɑ170)x21 + (ɑ142 * ɑ79)x20 + (ɑ142 * ɑ187)x19 + (ɑ142 * ɑ152)x18 + (ɑ142 * ɑ148)x17 + (ɑ142 * ɑ252)x16 + (ɑ142 * ɑ179)x15 + (ɑ142 * ɑ5)x14 + (ɑ142 * ɑ98)x13 + (ɑ142 * ɑ96)x12 + (ɑ142 * ɑ153)x11
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ142x29 + ɑ(357 % 255)x28 + ɑ(376 % 255)x27 + ɑ(300 % 255)x26 + ɑ236x25 + ɑ(326 % 255)x24 + ɑ239x23 + ɑ(260 % 255)x22 + ɑ(312 % 255)x21 + ɑ221x20 + ɑ(329 % 255)x19 + ɑ(294 % 255)x18 + ɑ(290 % 255)x17 + ɑ(394 % 255)x16 + ɑ(321 % 255)x15 + ɑ147x14 + ɑ240x13 + ɑ238x12 + ɑ(295 % 255)x11The result is:
ɑ142x29 + ɑ102x28 + ɑ121x27 + ɑ45x26 + ɑ236x25 + ɑ71x24 + ɑ239x23 + ɑ5x22 + ɑ57x21 + ɑ221x20 + ɑ74x19 + ɑ39x18 + ɑ35x17 + ɑ139x16 + ɑ66x15 + ɑ147x14 + ɑ240x13 + ɑ238x12 + ɑ40x11
Now, convert this to integer notation:
42x29 + 68x28 + 118x27 + 193x26 + 203x25 + 188x24 + 22x23 + 32x22 + 186x21 + 69x20 + 137x19 + 53x18 + 156x17 + 66x16 + 97x15 + 41x14 + 44x13 + 11x12 + 106x11Step 5b: XOR the result with the result from step 4b
Use the result from step 4b to perform the next XOR.
(42 ⊕ 42)x29 + (241 ⊕ 68)x28 + (102 ⊕ 118)x27 + (97 ⊕ 193)x26 + (158 ⊕ 203)x25 + (119 ⊕ 188)x24 + (251 ⊕ 22)x23 + (134 ⊕ 32)x22 + (4 ⊕ 186)x21 + (9 ⊕ 69)x20 + (59 ⊕ 137)x19 + (224 ⊕ 53)x18 + (237 ⊕ 156)x17 + (93 ⊕ 66)x16 + (197 ⊕ 97)x15 + (250 ⊕ 41)x14 + (164 ⊕ 44)x13 + (68 ⊕ 11)x12 + (0 ⊕ 106)x11
The result is:
0x29 + 181x28 + 16x27 + 160x26 + 85x25 + 203x24 + 237x23 + 166x22 + 190x21 + 76x20 + 178x19 + 213x18 + 113x17 + 31x16 + 164x15 + 211x14 + 136x13 + 79x12 + 106x11
Discard the lead 0 term to get:
181x28 + 16x27 + 160x26 + 85x25 + 203x24 + 237x23 + 166x22 + 190x21 + 76x20 + 178x19 + 213x18 + 113x17 + 31x16 + 164x15 + 211x14 + 136x13 + 79x12 + 106x11
Step 6a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 181x28. Convert 181x28 to alpha notation. According to the log antilog table, for the integer value 181, the alpha exponent is 42. Therefore 181 = ɑ42. Multiply the generator polynomial by ɑ42:
(ɑ42 * ɑ0)x28 + (ɑ42 * ɑ215)x27 + (ɑ42 * ɑ234)x26 + (ɑ42 * ɑ158)x25 + (ɑ42 * ɑ94)x24 + (ɑ42 * ɑ184)x23 + (ɑ42 * ɑ97)x22 + (ɑ42 * ɑ118)x21 + (ɑ42 * ɑ170)x20 + (ɑ42 * ɑ79)x19 + (ɑ42 * ɑ187)x18 + (ɑ42 * ɑ152)x17 + (ɑ42 * ɑ148)x16 + (ɑ42 * ɑ252)x15 + (ɑ42 * ɑ179)x14 + (ɑ42 * ɑ5)x13 + (ɑ42 * ɑ98)x12 + (ɑ42 * ɑ96)x11 + (ɑ42 * ɑ153)x10
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ42x28 + ɑ(257 % 255)x27 + ɑ(276 % 255)x26 + ɑ200x25 + ɑ136x24 + ɑ226x23 + ɑ139x22 + ɑ160x21 + ɑ212x20 + ɑ121x19 + ɑ229x18 + ɑ194x17 + ɑ190x16 + ɑ(294 % 255)x15 + ɑ221x14 + ɑ47x13 + ɑ140x12 + ɑ138x11 + ɑ195x10The result is:
ɑ42x28 + ɑ2x27 + ɑ21x26 + ɑ200x25 + ɑ136x24 + ɑ226x23 + ɑ139x22 + ɑ160x21 + ɑ212x20 + ɑ121x19 + ɑ229x18 + ɑ194x17 + ɑ190x16 + ɑ39x15 + ɑ221x14 + ɑ47x13 + ɑ140x12 + ɑ138x11 + ɑ195x10
Now, convert this to integer notation:
181x28 + 4x27 + 117x26 + 28x25 + 79x24 + 72x23 + 66x22 + 230x21 + 121x20 + 118x19 + 122x18 + 50x17 + 174x16 + 53x15 + 69x14 + 35x13 + 132x12 + 33x11 + 100x10Step 6b: XOR the result with the result from step 5b
Use the result from step 5b to perform the next XOR.
(181 ⊕ 181)x28 + (16 ⊕ 4)x27 + (160 ⊕ 117)x26 + (85 ⊕ 28)x25 + (203 ⊕ 79)x24 + (237 ⊕ 72)x23 + (166 ⊕ 66)x22 + (190 ⊕ 230)x21 + (76 ⊕ 121)x20 + (178 ⊕ 118)x19 + (213 ⊕ 122)x18 + (113 ⊕ 50)x17 + (31 ⊕ 174)x16 + (164 ⊕ 53)x15 + (211 ⊕ 69)x14 + (136 ⊕ 35)x13 + (79 ⊕ 132)x12 + (106 ⊕ 33)x11 + (0 ⊕ 100)x10
The result is:
0x28 + 20x27 + 213x26 + 73x25 + 132x24 + 165x23 + 228x22 + 88x21 + 53x20 + 196x19 + 175x18 + 67x17 + 177x16 + 145x15 + 150x14 + 171x13 + 203x12 + 75x11 + 100x10
Discard the lead 0 term to get:
20x27 + 213x26 + 73x25 + 132x24 + 165x23 + 228x22 + 88x21 + 53x20 + 196x19 + 175x18 + 67x17 + 177x16 + 145x15 + 150x14 + 171x13 + 203x12 + 75x11 + 100x10
Step 7a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 20x27. Convert 20x27 to alpha notation. According to the log antilog table, for the integer value 20, the alpha exponent is 52. Therefore 20 = ɑ52. Multiply the generator polynomial by ɑ52:
(ɑ52 * ɑ0)x27 + (ɑ52 * ɑ215)x26 + (ɑ52 * ɑ234)x25 + (ɑ52 * ɑ158)x24 + (ɑ52 * ɑ94)x23 + (ɑ52 * ɑ184)x22 + (ɑ52 * ɑ97)x21 + (ɑ52 * ɑ118)x20 + (ɑ52 * ɑ170)x19 + (ɑ52 * ɑ79)x18 + (ɑ52 * ɑ187)x17 + (ɑ52 * ɑ152)x16 + (ɑ52 * ɑ148)x15 + (ɑ52 * ɑ252)x14 + (ɑ52 * ɑ179)x13 + (ɑ52 * ɑ5)x12 + (ɑ52 * ɑ98)x11 + (ɑ52 * ɑ96)x10 + (ɑ52 * ɑ153)x9
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ52x27 + ɑ(267 % 255)x26 + ɑ(286 % 255)x25 + ɑ210x24 + ɑ146x23 + ɑ236x22 + ɑ149x21 + ɑ170x20 + ɑ222x19 + ɑ131x18 + ɑ239x17 + ɑ204x16 + ɑ200x15 + ɑ(304 % 255)x14 + ɑ231x13 + ɑ57x12 + ɑ150x11 + ɑ148x10 + ɑ205x9The result is:
ɑ52x27 + ɑ12x26 + ɑ31x25 + ɑ210x24 + ɑ146x23 + ɑ236x22 + ɑ149x21 + ɑ170x20 + ɑ222x19 + ɑ131x18 + ɑ239x17 + ɑ204x16 + ɑ200x15 + ɑ49x14 + ɑ231x13 + ɑ57x12 + ɑ150x11 + ɑ148x10 + ɑ205x9
Now, convert this to integer notation:
20x27 + 205x26 + 192x25 + 89x24 + 154x23 + 203x22 + 164x21 + 215x20 + 138x19 + 92x18 + 22x17 + 221x16 + 28x15 + 140x14 + 245x13 + 186x12 + 85x11 + 82x10 + 167x9Step 7b: XOR the result with the result from step 6b
Use the result from step 6b to perform the next XOR.
(20 ⊕ 20)x27 + (213 ⊕ 205)x26 + (73 ⊕ 192)x25 + (132 ⊕ 89)x24 + (165 ⊕ 154)x23 + (228 ⊕ 203)x22 + (88 ⊕ 164)x21 + (53 ⊕ 215)x20 + (196 ⊕ 138)x19 + (175 ⊕ 92)x18 + (67 ⊕ 22)x17 + (177 ⊕ 221)x16 + (145 ⊕ 28)x15 + (150 ⊕ 140)x14 + (171 ⊕ 245)x13 + (203 ⊕ 186)x12 + (75 ⊕ 85)x11 + (100 ⊕ 82)x10 + (0 ⊕ 167)x9
The result is:
0x27 + 24x26 + 137x25 + 221x24 + 63x23 + 47x22 + 252x21 + 226x20 + 78x19 + 243x18 + 85x17 + 108x16 + 141x15 + 26x14 + 94x13 + 113x12 + 30x11 + 54x10 + 167x9
Discard the lead 0 term to get:
24x26 + 137x25 + 221x24 + 63x23 + 47x22 + 252x21 + 226x20 + 78x19 + 243x18 + 85x17 + 108x16 + 141x15 + 26x14 + 94x13 + 113x12 + 30x11 + 54x10 + 167x9
Step 8a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 24x26. Convert 24x26 to alpha notation. According to the log antilog table, for the integer value 24, the alpha exponent is 28. Therefore 24 = ɑ28. Multiply the generator polynomial by ɑ28:
(ɑ28 * ɑ0)x26 + (ɑ28 * ɑ215)x25 + (ɑ28 * ɑ234)x24 + (ɑ28 * ɑ158)x23 + (ɑ28 * ɑ94)x22 + (ɑ28 * ɑ184)x21 + (ɑ28 * ɑ97)x20 + (ɑ28 * ɑ118)x19 + (ɑ28 * ɑ170)x18 + (ɑ28 * ɑ79)x17 + (ɑ28 * ɑ187)x16 + (ɑ28 * ɑ152)x15 + (ɑ28 * ɑ148)x14 + (ɑ28 * ɑ252)x13 + (ɑ28 * ɑ179)x12 + (ɑ28 * ɑ5)x11 + (ɑ28 * ɑ98)x10 + (ɑ28 * ɑ96)x9 + (ɑ28 * ɑ153)x8
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ28x26 + ɑ243x25 + ɑ(262 % 255)x24 + ɑ186x23 + ɑ122x22 + ɑ212x21 + ɑ125x20 + ɑ146x19 + ɑ198x18 + ɑ107x17 + ɑ215x16 + ɑ180x15 + ɑ176x14 + ɑ(280 % 255)x13 + ɑ207x12 + ɑ33x11 + ɑ126x10 + ɑ124x9 + ɑ181x8The result is:
ɑ28x26 + ɑ243x25 + ɑ7x24 + ɑ186x23 + ɑ122x22 + ɑ212x21 + ɑ125x20 + ɑ146x19 + ɑ198x18 + ɑ107x17 + ɑ215x16 + ɑ180x15 + ɑ176x14 + ɑ25x13 + ɑ207x12 + ɑ33x11 + ɑ126x10 + ɑ124x9 + ɑ181x8
Now, convert this to integer notation:
24x26 + 125x25 + 128x24 + 110x23 + 236x22 + 121x21 + 51x20 + 154x19 + 7x18 + 104x17 + 239x16 + 150x15 + 227x14 + 3x13 + 166x12 + 39x11 + 102x10 + 151x9 + 49x8Step 8b: XOR the result with the result from step 7b
Use the result from step 7b to perform the next XOR.
(24 ⊕ 24)x26 + (137 ⊕ 125)x25 + (221 ⊕ 128)x24 + (63 ⊕ 110)x23 + (47 ⊕ 236)x22 + (252 ⊕ 121)x21 + (226 ⊕ 51)x20 + (78 ⊕ 154)x19 + (243 ⊕ 7)x18 + (85 ⊕ 104)x17 + (108 ⊕ 239)x16 + (141 ⊕ 150)x15 + (26 ⊕ 227)x14 + (94 ⊕ 3)x13 + (113 ⊕ 166)x12 + (30 ⊕ 39)x11 + (54 ⊕ 102)x10 + (167 ⊕ 151)x9 + (0 ⊕ 49)x8
The result is:
0x26 + 244x25 + 93x24 + 81x23 + 195x22 + 133x21 + 209x20 + 212x19 + 244x18 + 61x17 + 131x16 + 27x15 + 249x14 + 93x13 + 215x12 + 57x11 + 80x10 + 48x9 + 49x8
Discard the lead 0 term to get:
244x25 + 93x24 + 81x23 + 195x22 + 133x21 + 209x20 + 212x19 + 244x18 + 61x17 + 131x16 + 27x15 + 249x14 + 93x13 + 215x12 + 57x11 + 80x10 + 48x9 + 49x8
Step 9a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 244x25. Convert 244x25 to alpha notation. According to the log antilog table, for the integer value 244, the alpha exponent is 230. Therefore 244 = ɑ230. Multiply the generator polynomial by ɑ230:
(ɑ230 * ɑ0)x25 + (ɑ230 * ɑ215)x24 + (ɑ230 * ɑ234)x23 + (ɑ230 * ɑ158)x22 + (ɑ230 * ɑ94)x21 + (ɑ230 * ɑ184)x20 + (ɑ230 * ɑ97)x19 + (ɑ230 * ɑ118)x18 + (ɑ230 * ɑ170)x17 + (ɑ230 * ɑ79)x16 + (ɑ230 * ɑ187)x15 + (ɑ230 * ɑ152)x14 + (ɑ230 * ɑ148)x13 + (ɑ230 * ɑ252)x12 + (ɑ230 * ɑ179)x11 + (ɑ230 * ɑ5)x10 + (ɑ230 * ɑ98)x9 + (ɑ230 * ɑ96)x8 + (ɑ230 * ɑ153)x7
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ230x25 + ɑ(445 % 255)x24 + ɑ(464 % 255)x23 + ɑ(388 % 255)x22 + ɑ(324 % 255)x21 + ɑ(414 % 255)x20 + ɑ(327 % 255)x19 + ɑ(348 % 255)x18 + ɑ(400 % 255)x17 + ɑ(309 % 255)x16 + ɑ(417 % 255)x15 + ɑ(382 % 255)x14 + ɑ(378 % 255)x13 + ɑ(482 % 255)x12 + ɑ(409 % 255)x11 + ɑ235x10 + ɑ(328 % 255)x9 + ɑ(326 % 255)x8 + ɑ(383 % 255)x7The result is:
ɑ230x25 + ɑ190x24 + ɑ209x23 + ɑ133x22 + ɑ69x21 + ɑ159x20 + ɑ72x19 + ɑ93x18 + ɑ145x17 + ɑ54x16 + ɑ162x15 + ɑ127x14 + ɑ123x13 + ɑ227x12 + ɑ154x11 + ɑ235x10 + ɑ73x9 + ɑ71x8 + ɑ128x7
Now, convert this to integer notation:
244x25 + 174x24 + 162x23 + 109x22 + 47x21 + 115x20 + 101x19 + 182x18 + 77x17 + 80x16 + 191x15 + 204x14 + 197x13 + 144x12 + 57x11 + 235x10 + 202x9 + 188x8 + 133x7Step 9b: XOR the result with the result from step 8b
Use the result from step 8b to perform the next XOR.
(244 ⊕ 244)x25 + (93 ⊕ 174)x24 + (81 ⊕ 162)x23 + (195 ⊕ 109)x22 + (133 ⊕ 47)x21 + (209 ⊕ 115)x20 + (212 ⊕ 101)x19 + (244 ⊕ 182)x18 + (61 ⊕ 77)x17 + (131 ⊕ 80)x16 + (27 ⊕ 191)x15 + (249 ⊕ 204)x14 + (93 ⊕ 197)x13 + (215 ⊕ 144)x12 + (57 ⊕ 57)x11 + (80 ⊕ 235)x10 + (48 ⊕ 202)x9 + (49 ⊕ 188)x8 + (0 ⊕ 133)x7
The result is:
0x25 + 243x24 + 243x23 + 174x22 + 170x21 + 162x20 + 177x19 + 66x18 + 112x17 + 211x16 + 164x15 + 53x14 + 152x13 + 71x12 + 0x11 + 187x10 + 250x9 + 141x8 + 133x7
Discard the lead 0 term to get:
243x24 + 243x23 + 174x22 + 170x21 + 162x20 + 177x19 + 66x18 + 112x17 + 211x16 + 164x15 + 53x14 + 152x13 + 71x12 + 0x11 + 187x10 + 250x9 + 141x8 + 133x7
Step 10a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 243x24. Convert 243x24 to alpha notation. According to the log antilog table, for the integer value 243, the alpha exponent is 233. Therefore 243 = ɑ233. Multiply the generator polynomial by ɑ233:
(ɑ233 * ɑ0)x24 + (ɑ233 * ɑ215)x23 + (ɑ233 * ɑ234)x22 + (ɑ233 * ɑ158)x21 + (ɑ233 * ɑ94)x20 + (ɑ233 * ɑ184)x19 + (ɑ233 * ɑ97)x18 + (ɑ233 * ɑ118)x17 + (ɑ233 * ɑ170)x16 + (ɑ233 * ɑ79)x15 + (ɑ233 * ɑ187)x14 + (ɑ233 * ɑ152)x13 + (ɑ233 * ɑ148)x12 + (ɑ233 * ɑ252)x11 + (ɑ233 * ɑ179)x10 + (ɑ233 * ɑ5)x9 + (ɑ233 * ɑ98)x8 + (ɑ233 * ɑ96)x7 + (ɑ233 * ɑ153)x6
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ233x24 + ɑ(448 % 255)x23 + ɑ(467 % 255)x22 + ɑ(391 % 255)x21 + ɑ(327 % 255)x20 + ɑ(417 % 255)x19 + ɑ(330 % 255)x18 + ɑ(351 % 255)x17 + ɑ(403 % 255)x16 + ɑ(312 % 255)x15 + ɑ(420 % 255)x14 + ɑ(385 % 255)x13 + ɑ(381 % 255)x12 + ɑ(485 % 255)x11 + ɑ(412 % 255)x10 + ɑ238x9 + ɑ(331 % 255)x8 + ɑ(329 % 255)x7 + ɑ(386 % 255)x6The result is:
ɑ233x24 + ɑ193x23 + ɑ212x22 + ɑ136x21 + ɑ72x20 + ɑ162x19 + ɑ75x18 + ɑ96x17 + ɑ148x16 + ɑ57x15 + ɑ165x14 + ɑ130x13 + ɑ126x12 + ɑ230x11 + ɑ157x10 + ɑ238x9 + ɑ76x8 + ɑ74x7 + ɑ131x6
Now, convert this to integer notation:
243x24 + 25x23 + 121x22 + 79x21 + 101x20 + 191x19 + 15x18 + 217x17 + 82x16 + 186x15 + 145x14 + 46x13 + 102x12 + 244x11 + 213x10 + 11x9 + 30x8 + 137x7 + 92x6Step 10b: XOR the result with the result from step 9b
Use the result from step 9b to perform the next XOR.
(243 ⊕ 243)x24 + (243 ⊕ 25)x23 + (174 ⊕ 121)x22 + (170 ⊕ 79)x21 + (162 ⊕ 101)x20 + (177 ⊕ 191)x19 + (66 ⊕ 15)x18 + (112 ⊕ 217)x17 + (211 ⊕ 82)x16 + (164 ⊕ 186)x15 + (53 ⊕ 145)x14 + (152 ⊕ 46)x13 + (71 ⊕ 102)x12 + (0 ⊕ 244)x11 + (187 ⊕ 213)x10 + (250 ⊕ 11)x9 + (141 ⊕ 30)x8 + (133 ⊕ 137)x7 + (0 ⊕ 92)x6
The result is:
0x24 + 234x23 + 215x22 + 229x21 + 199x20 + 14x19 + 77x18 + 169x17 + 129x16 + 30x15 + 164x14 + 182x13 + 33x12 + 244x11 + 110x10 + 241x9 + 147x8 + 12x7 + 92x6
Discard the lead 0 term to get:
234x23 + 215x22 + 229x21 + 199x20 + 14x19 + 77x18 + 169x17 + 129x16 + 30x15 + 164x14 + 182x13 + 33x12 + 244x11 + 110x10 + 241x9 + 147x8 + 12x7 + 92x6
Step 11a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 234x23. Convert 234x23 to alpha notation. According to the log antilog table, for the integer value 234, the alpha exponent is 22. Therefore 234 = ɑ22. Multiply the generator polynomial by ɑ22:
(ɑ22 * ɑ0)x23 + (ɑ22 * ɑ215)x22 + (ɑ22 * ɑ234)x21 + (ɑ22 * ɑ158)x20 + (ɑ22 * ɑ94)x19 + (ɑ22 * ɑ184)x18 + (ɑ22 * ɑ97)x17 + (ɑ22 * ɑ118)x16 + (ɑ22 * ɑ170)x15 + (ɑ22 * ɑ79)x14 + (ɑ22 * ɑ187)x13 + (ɑ22 * ɑ152)x12 + (ɑ22 * ɑ148)x11 + (ɑ22 * ɑ252)x10 + (ɑ22 * ɑ179)x9 + (ɑ22 * ɑ5)x8 + (ɑ22 * ɑ98)x7 + (ɑ22 * ɑ96)x6 + (ɑ22 * ɑ153)x5
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ22x23 + ɑ237x22 + ɑ(256 % 255)x21 + ɑ180x20 + ɑ116x19 + ɑ206x18 + ɑ119x17 + ɑ140x16 + ɑ192x15 + ɑ101x14 + ɑ209x13 + ɑ174x12 + ɑ170x11 + ɑ(274 % 255)x10 + ɑ201x9 + ɑ27x8 + ɑ120x7 + ɑ118x6 + ɑ175x5The result is:
ɑ22x23 + ɑ237x22 + ɑ1x21 + ɑ180x20 + ɑ116x19 + ɑ206x18 + ɑ119x17 + ɑ140x16 + ɑ192x15 + ɑ101x14 + ɑ209x13 + ɑ174x12 + ɑ170x11 + ɑ19x10 + ɑ201x9 + ɑ27x8 + ɑ120x7 + ɑ118x6 + ɑ175x5
Now, convert this to integer notation:
234x23 + 139x22 + 2x21 + 150x20 + 248x19 + 83x18 + 147x17 + 132x16 + 130x15 + 34x14 + 162x13 + 241x12 + 215x11 + 90x10 + 56x9 + 12x8 + 59x7 + 199x6 + 255x5Step 11b: XOR the result with the result from step 10b
Use the result from step 10b to perform the next XOR.
(234 ⊕ 234)x23 + (215 ⊕ 139)x22 + (229 ⊕ 2)x21 + (199 ⊕ 150)x20 + (14 ⊕ 248)x19 + (77 ⊕ 83)x18 + (169 ⊕ 147)x17 + (129 ⊕ 132)x16 + (30 ⊕ 130)x15 + (164 ⊕ 34)x14 + (182 ⊕ 162)x13 + (33 ⊕ 241)x12 + (244 ⊕ 215)x11 + (110 ⊕ 90)x10 + (241 ⊕ 56)x9 + (147 ⊕ 12)x8 + (12 ⊕ 59)x7 + (92 ⊕ 199)x6 + (0 ⊕ 255)x5
The result is:
0x23 + 92x22 + 231x21 + 81x20 + 246x19 + 30x18 + 58x17 + 5x16 + 156x15 + 134x14 + 20x13 + 208x12 + 35x11 + 52x10 + 201x9 + 159x8 + 55x7 + 155x6 + 255x5
Discard the lead 0 term to get:
92x22 + 231x21 + 81x20 + 246x19 + 30x18 + 58x17 + 5x16 + 156x15 + 134x14 + 20x13 + 208x12 + 35x11 + 52x10 + 201x9 + 159x8 + 55x7 + 155x6 + 255x5
Step 12a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 92x22. Convert 92x22 to alpha notation. According to the log antilog table, for the integer value 92, the alpha exponent is 131. Therefore 92 = ɑ131. Multiply the generator polynomial by ɑ131:
(ɑ131 * ɑ0)x22 + (ɑ131 * ɑ215)x21 + (ɑ131 * ɑ234)x20 + (ɑ131 * ɑ158)x19 + (ɑ131 * ɑ94)x18 + (ɑ131 * ɑ184)x17 + (ɑ131 * ɑ97)x16 + (ɑ131 * ɑ118)x15 + (ɑ131 * ɑ170)x14 + (ɑ131 * ɑ79)x13 + (ɑ131 * ɑ187)x12 + (ɑ131 * ɑ152)x11 + (ɑ131 * ɑ148)x10 + (ɑ131 * ɑ252)x9 + (ɑ131 * ɑ179)x8 + (ɑ131 * ɑ5)x7 + (ɑ131 * ɑ98)x6 + (ɑ131 * ɑ96)x5 + (ɑ131 * ɑ153)x4
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ131x22 + ɑ(346 % 255)x21 + ɑ(365 % 255)x20 + ɑ(289 % 255)x19 + ɑ225x18 + ɑ(315 % 255)x17 + ɑ228x16 + ɑ249x15 + ɑ(301 % 255)x14 + ɑ210x13 + ɑ(318 % 255)x12 + ɑ(283 % 255)x11 + ɑ(279 % 255)x10 + ɑ(383 % 255)x9 + ɑ(310 % 255)x8 + ɑ136x7 + ɑ229x6 + ɑ227x5 + ɑ(284 % 255)x4The result is:
ɑ131x22 + ɑ91x21 + ɑ110x20 + ɑ34x19 + ɑ225x18 + ɑ60x17 + ɑ228x16 + ɑ249x15 + ɑ46x14 + ɑ210x13 + ɑ63x12 + ɑ28x11 + ɑ24x10 + ɑ128x9 + ɑ55x8 + ɑ136x7 + ɑ229x6 + ɑ227x5 + ɑ29x4
Now, convert this to integer notation:
92x22 + 163x21 + 103x20 + 78x19 + 36x18 + 185x17 + 61x16 + 54x15 + 159x14 + 89x13 + 161x12 + 24x11 + 143x10 + 133x9 + 160x8 + 79x7 + 122x6 + 144x5 + 48x4Step 12b: XOR the result with the result from step 11b
Use the result from step 11b to perform the next XOR.
(92 ⊕ 92)x22 + (231 ⊕ 163)x21 + (81 ⊕ 103)x20 + (246 ⊕ 78)x19 + (30 ⊕ 36)x18 + (58 ⊕ 185)x17 + (5 ⊕ 61)x16 + (156 ⊕ 54)x15 + (134 ⊕ 159)x14 + (20 ⊕ 89)x13 + (208 ⊕ 161)x12 + (35 ⊕ 24)x11 + (52 ⊕ 143)x10 + (201 ⊕ 133)x9 + (159 ⊕ 160)x8 + (55 ⊕ 79)x7 + (155 ⊕ 122)x6 + (255 ⊕ 144)x5 + (0 ⊕ 48)x4
The result is:
0x22 + 68x21 + 54x20 + 184x19 + 58x18 + 131x17 + 56x16 + 170x15 + 25x14 + 77x13 + 113x12 + 59x11 + 187x10 + 76x9 + 63x8 + 120x7 + 225x6 + 111x5 + 48x4
Discard the lead 0 term to get:
68x21 + 54x20 + 184x19 + 58x18 + 131x17 + 56x16 + 170x15 + 25x14 + 77x13 + 113x12 + 59x11 + 187x10 + 76x9 + 63x8 + 120x7 + 225x6 + 111x5 + 48x4
Step 13a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 68x21. Convert 68x21 to alpha notation. According to the log antilog table, for the integer value 68, the alpha exponent is 102. Therefore 68 = ɑ102. Multiply the generator polynomial by ɑ102:
(ɑ102 * ɑ0)x21 + (ɑ102 * ɑ215)x20 + (ɑ102 * ɑ234)x19 + (ɑ102 * ɑ158)x18 + (ɑ102 * ɑ94)x17 + (ɑ102 * ɑ184)x16 + (ɑ102 * ɑ97)x15 + (ɑ102 * ɑ118)x14 + (ɑ102 * ɑ170)x13 + (ɑ102 * ɑ79)x12 + (ɑ102 * ɑ187)x11 + (ɑ102 * ɑ152)x10 + (ɑ102 * ɑ148)x9 + (ɑ102 * ɑ252)x8 + (ɑ102 * ɑ179)x7 + (ɑ102 * ɑ5)x6 + (ɑ102 * ɑ98)x5 + (ɑ102 * ɑ96)x4 + (ɑ102 * ɑ153)x3
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ102x21 + ɑ(317 % 255)x20 + ɑ(336 % 255)x19 + ɑ(260 % 255)x18 + ɑ196x17 + ɑ(286 % 255)x16 + ɑ199x15 + ɑ220x14 + ɑ(272 % 255)x13 + ɑ181x12 + ɑ(289 % 255)x11 + ɑ254x10 + ɑ250x9 + ɑ(354 % 255)x8 + ɑ(281 % 255)x7 + ɑ107x6 + ɑ200x5 + ɑ198x4 + ɑ255x3The result is:
ɑ102x21 + ɑ62x20 + ɑ81x19 + ɑ5x18 + ɑ196x17 + ɑ31x16 + ɑ199x15 + ɑ220x14 + ɑ17x13 + ɑ181x12 + ɑ34x11 + ɑ254x10 + ɑ250x9 + ɑ99x8 + ɑ26x7 + ɑ107x6 + ɑ200x5 + ɑ198x4 + ɑ0x3
Now, convert this to integer notation:
68x21 + 222x20 + 231x19 + 32x18 + 200x17 + 192x16 + 14x15 + 172x14 + 152x13 + 49x12 + 78x11 + 142x10 + 108x9 + 134x8 + 6x7 + 104x6 + 28x5 + 7x4 + 1x3Step 13b: XOR the result with the result from step 12b
Use the result from step 12b to perform the next XOR.
(68 ⊕ 68)x21 + (54 ⊕ 222)x20 + (184 ⊕ 231)x19 + (58 ⊕ 32)x18 + (131 ⊕ 200)x17 + (56 ⊕ 192)x16 + (170 ⊕ 14)x15 + (25 ⊕ 172)x14 + (77 ⊕ 152)x13 + (113 ⊕ 49)x12 + (59 ⊕ 78)x11 + (187 ⊕ 142)x10 + (76 ⊕ 108)x9 + (63 ⊕ 134)x8 + (120 ⊕ 6)x7 + (225 ⊕ 104)x6 + (111 ⊕ 28)x5 + (48 ⊕ 7)x4 + (0 ⊕ 1)x3
The result is:
0x21 + 232x20 + 95x19 + 26x18 + 75x17 + 248x16 + 164x15 + 181x14 + 213x13 + 64x12 + 117x11 + 53x10 + 32x9 + 185x8 + 126x7 + 137x6 + 115x5 + 55x4 + 1x3
Discard the lead 0 term to get:
232x20 + 95x19 + 26x18 + 75x17 + 248x16 + 164x15 + 181x14 + 213x13 + 64x12 + 117x11 + 53x10 + 32x9 + 185x8 + 126x7 + 137x6 + 115x5 + 55x4 + 1x3
Step 14a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 232x20. Convert 232x20 to alpha notation. According to the log antilog table, for the integer value 232, the alpha exponent is 11. Therefore 232 = ɑ11. Multiply the generator polynomial by ɑ11:
(ɑ11 * ɑ0)x20 + (ɑ11 * ɑ215)x19 + (ɑ11 * ɑ234)x18 + (ɑ11 * ɑ158)x17 + (ɑ11 * ɑ94)x16 + (ɑ11 * ɑ184)x15 + (ɑ11 * ɑ97)x14 + (ɑ11 * ɑ118)x13 + (ɑ11 * ɑ170)x12 + (ɑ11 * ɑ79)x11 + (ɑ11 * ɑ187)x10 + (ɑ11 * ɑ152)x9 + (ɑ11 * ɑ148)x8 + (ɑ11 * ɑ252)x7 + (ɑ11 * ɑ179)x6 + (ɑ11 * ɑ5)x5 + (ɑ11 * ɑ98)x4 + (ɑ11 * ɑ96)x3 + (ɑ11 * ɑ153)x2
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ11x20 + ɑ226x19 + ɑ245x18 + ɑ169x17 + ɑ105x16 + ɑ195x15 + ɑ108x14 + ɑ129x13 + ɑ181x12 + ɑ90x11 + ɑ198x10 + ɑ163x9 + ɑ159x8 + ɑ(263 % 255)x7 + ɑ190x6 + ɑ16x5 + ɑ109x4 + ɑ107x3 + ɑ164x2The result is:
ɑ11x20 + ɑ226x19 + ɑ245x18 + ɑ169x17 + ɑ105x16 + ɑ195x15 + ɑ108x14 + ɑ129x13 + ɑ181x12 + ɑ90x11 + ɑ198x10 + ɑ163x9 + ɑ159x8 + ɑ8x7 + ɑ190x6 + ɑ16x5 + ɑ109x4 + ɑ107x3 + ɑ164x2
Now, convert this to integer notation:
232x20 + 72x19 + 233x18 + 229x17 + 26x16 + 100x15 + 208x14 + 23x13 + 49x12 + 223x11 + 7x10 + 99x9 + 115x8 + 29x7 + 174x6 + 76x5 + 189x4 + 104x3 + 198x2Step 14b: XOR the result with the result from step 13b
Use the result from step 13b to perform the next XOR.
(232 ⊕ 232)x20 + (95 ⊕ 72)x19 + (26 ⊕ 233)x18 + (75 ⊕ 229)x17 + (248 ⊕ 26)x16 + (164 ⊕ 100)x15 + (181 ⊕ 208)x14 + (213 ⊕ 23)x13 + (64 ⊕ 49)x12 + (117 ⊕ 223)x11 + (53 ⊕ 7)x10 + (32 ⊕ 99)x9 + (185 ⊕ 115)x8 + (126 ⊕ 29)x7 + (137 ⊕ 174)x6 + (115 ⊕ 76)x5 + (55 ⊕ 189)x4 + (1 ⊕ 104)x3 + (0 ⊕ 198)x2
The result is:
0x20 + 23x19 + 243x18 + 174x17 + 226x16 + 192x15 + 101x14 + 194x13 + 113x12 + 170x11 + 50x10 + 67x9 + 202x8 + 99x7 + 39x6 + 63x5 + 138x4 + 105x3 + 198x2
Discard the lead 0 term to get:
23x19 + 243x18 + 174x17 + 226x16 + 192x15 + 101x14 + 194x13 + 113x12 + 170x11 + 50x10 + 67x9 + 202x8 + 99x7 + 39x6 + 63x5 + 138x4 + 105x3 + 198x2
Step 15a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 23x19. Convert 23x19 to alpha notation. According to the log antilog table, for the integer value 23, the alpha exponent is 129. Therefore 23 = ɑ129. Multiply the generator polynomial by ɑ129:
(ɑ129 * ɑ0)x19 + (ɑ129 * ɑ215)x18 + (ɑ129 * ɑ234)x17 + (ɑ129 * ɑ158)x16 + (ɑ129 * ɑ94)x15 + (ɑ129 * ɑ184)x14 + (ɑ129 * ɑ97)x13 + (ɑ129 * ɑ118)x12 + (ɑ129 * ɑ170)x11 + (ɑ129 * ɑ79)x10 + (ɑ129 * ɑ187)x9 + (ɑ129 * ɑ152)x8 + (ɑ129 * ɑ148)x7 + (ɑ129 * ɑ252)x6 + (ɑ129 * ɑ179)x5 + (ɑ129 * ɑ5)x4 + (ɑ129 * ɑ98)x3 + (ɑ129 * ɑ96)x2 + (ɑ129 * ɑ153)x1
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ129x19 + ɑ(344 % 255)x18 + ɑ(363 % 255)x17 + ɑ(287 % 255)x16 + ɑ223x15 + ɑ(313 % 255)x14 + ɑ226x13 + ɑ247x12 + ɑ(299 % 255)x11 + ɑ208x10 + ɑ(316 % 255)x9 + ɑ(281 % 255)x8 + ɑ(277 % 255)x7 + ɑ(381 % 255)x6 + ɑ(308 % 255)x5 + ɑ134x4 + ɑ227x3 + ɑ225x2 + ɑ(282 % 255)x1The result is:
ɑ129x19 + ɑ89x18 + ɑ108x17 + ɑ32x16 + ɑ223x15 + ɑ58x14 + ɑ226x13 + ɑ247x12 + ɑ44x11 + ɑ208x10 + ɑ61x9 + ɑ26x8 + ɑ22x7 + ɑ126x6 + ɑ53x5 + ɑ134x4 + ɑ227x3 + ɑ225x2 + ɑ27x1
Now, convert this to integer notation:
23x19 + 225x18 + 208x17 + 157x16 + 9x15 + 105x14 + 72x13 + 131x12 + 238x11 + 81x10 + 111x9 + 6x8 + 234x7 + 102x6 + 40x5 + 218x4 + 144x3 + 36x2 + 12x1Step 15b: XOR the result with the result from step 14b
Use the result from step 14b to perform the next XOR.
(23 ⊕ 23)x19 + (243 ⊕ 225)x18 + (174 ⊕ 208)x17 + (226 ⊕ 157)x16 + (192 ⊕ 9)x15 + (101 ⊕ 105)x14 + (194 ⊕ 72)x13 + (113 ⊕ 131)x12 + (170 ⊕ 238)x11 + (50 ⊕ 81)x10 + (67 ⊕ 111)x9 + (202 ⊕ 6)x8 + (99 ⊕ 234)x7 + (39 ⊕ 102)x6 + (63 ⊕ 40)x5 + (138 ⊕ 218)x4 + (105 ⊕ 144)x3 + (198 ⊕ 36)x2 + (0 ⊕ 12)x1
The result is:
0x19 + 18x18 + 126x17 + 127x16 + 201x15 + 12x14 + 138x13 + 242x12 + 68x11 + 99x10 + 44x9 + 204x8 + 137x7 + 65x6 + 23x5 + 80x4 + 249x3 + 226x2 + 12x1
Discard the lead 0 term to get:
18x18 + 126x17 + 127x16 + 201x15 + 12x14 + 138x13 + 242x12 + 68x11 + 99x10 + 44x9 + 204x8 + 137x7 + 65x6 + 23x5 + 80x4 + 249x3 + 226x2 + 12x1
Step 16a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step
Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 18x18. Convert 18x18 to alpha notation. According to the log antilog table, for the integer value 18, the alpha exponent is 224. Therefore 18 = ɑ224. Multiply the generator polynomial by ɑ224:
(ɑ224 * ɑ0)x18 + (ɑ224 * ɑ215)x17 + (ɑ224 * ɑ234)x16 + (ɑ224 * ɑ158)x15 + (ɑ224 * ɑ94)x14 + (ɑ224 * ɑ184)x13 + (ɑ224 * ɑ97)x12 + (ɑ224 * ɑ118)x11 + (ɑ224 * ɑ170)x10 + (ɑ224 * ɑ79)x9 + (ɑ224 * ɑ187)x8 + (ɑ224 * ɑ152)x7 + (ɑ224 * ɑ148)x6 + (ɑ224 * ɑ252)x5 + (ɑ224 * ɑ179)x4 + (ɑ224 * ɑ5)x3 + (ɑ224 * ɑ98)x2 + (ɑ224 * ɑ96)x1 + (ɑ224 * ɑ153)x0
The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:
ɑ224x18 + ɑ(439 % 255)x17 + ɑ(458 % 255)x16 + ɑ(382 % 255)x15 + ɑ(318 % 255)x14 + ɑ(408 % 255)x13 + ɑ(321 % 255)x12 + ɑ(342 % 255)x11 + ɑ(394 % 255)x10 + ɑ(303 % 255)x9 + ɑ(411 % 255)x8 + ɑ(376 % 255)x7 + ɑ(372 % 255)x6 + ɑ(476 % 255)x5 + ɑ(403 % 255)x4 + ɑ229x3 + ɑ(322 % 255)x2 + ɑ(320 % 255)x1 + ɑ(377 % 255)x0The result is:
ɑ224x18 + ɑ184x17 + ɑ203x16 + ɑ127x15 + ɑ63x14 + ɑ153x13 + ɑ66x12 + ɑ87x11 + ɑ139x10 + ɑ48x9 + ɑ156x8 + ɑ121x7 + ɑ117x6 + ɑ221x5 + ɑ148x4 + ɑ229x3 + ɑ67x2 + ɑ65x1 + ɑ122x0
Now, convert this to integer notation:
18x18 + 149x17 + 224x16 + 204x15 + 161x14 + 146x13 + 97x12 + 127x11 + 66x10 + 70x9 + 228x8 + 118x7 + 237x6 + 69x5 + 82x4 + 122x3 + 194x2 + 190x1 + 236Step 16b: XOR the result with the result from step 15b
Use the result from step 15b to perform the next XOR.
(18 ⊕ 18)x18 + (126 ⊕ 149)x17 + (127 ⊕ 224)x16 + (201 ⊕ 204)x15 + (12 ⊕ 161)x14 + (138 ⊕ 146)x13 + (242 ⊕ 97)x12 + (68 ⊕ 127)x11 + (99 ⊕ 66)x10 + (44 ⊕ 70)x9 + (204 ⊕ 228)x8 + (137 ⊕ 118)x7 + (65 ⊕ 237)x6 + (23 ⊕ 69)x5 + (80 ⊕ 82)x4 + (249 ⊕ 122)x3 + (226 ⊕ 194)x2 + (12 ⊕ 190)x1 + (0 ⊕ 236)x0
The result is:
0x18 + 235x17 + 159x16 + 5x15 + 173x14 + 24x13 + 147x12 + 59x11 + 33x10 + 106x9 + 40x8 + 255x7 + 172x6 + 82x5 + 2x4 + 131x3 + 32x2 + 178x1 + 236
Discard the lead 0 term to get:
235x17 + 159x16 + 5x15 + 173x14 + 24x13 + 147x12 + 59x11 + 33x10 + 106x9 + 40x8 + 255x7 + 172x6 + 82x5 + 2x4 + 131x3 + 32x2 + 178x1 + 236
Use the terms of the remainder as the error correction codewords
The division has been performed 16 times, which is the number of terms in the message polynomial. This means that the division is complete and the terms of the above polynomial are the error correction codewords to use for the original message polynomial:
235 159 5 173 24 147 59 33 106 40 255 172 82 2 131 32 178 236