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:
16x8 + 32x7 + 12x6 + 86x5 + 97x4 + 128x3 + 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 17, for 17 error correction codewords, so multiply the message polynomial by x17, which gives us:
16x25 + 32x24 + 12x23 + 86x22 + 97x21 + 128x20 + 236x19 + 17x18 + 236x17
The lead term of the generator polynomial should also have the same exponent, so multiply by x8 to get
ɑ0x25 + ɑ43x24 + ɑ139x23 + ɑ206x22 + ɑ78x21 + ɑ43x20 + ɑ239x19 + ɑ123x18 + ɑ206x17 + ɑ214x16 + ɑ147x15 + ɑ24x14 + ɑ99x13 + ɑ150x12 + ɑ39x11 + ɑ243x10 + ɑ163x9 + ɑ136x8
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 9 steps to complete. This will result in a remainder that has 17 terms. These terms will be the 17 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 16x25. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 16x25 to alpha notation. According to the log antilog table, for the integer value 16, the alpha exponent is 4. Therefore 16 = ɑ4. Multiply the generator polynomial by ɑ4:
The exponents of the alphas are added together. The result is:
ɑ4x25 + ɑ47x24 + ɑ143x23 + ɑ210x22 + ɑ82x21 + ɑ47x20 + ɑ243x19 + ɑ127x18 + ɑ210x17 + ɑ218x16 + ɑ151x15 + ɑ28x14 + ɑ103x13 + ɑ154x12 + ɑ43x11 + ɑ247x10 + ɑ167x9 + ɑ140x8
Now, convert this to integer notation:
16x25 + 35x24 + 84x23 + 89x22 + 211x21 + 35x20 + 125x19 + 204x18 + 89x17 + 43x16 + 170x15 + 24x14 + 136x13 + 57x12 + 119x11 + 131x10 + 126x9 + 132x8Step 1b: XOR the result with the message polynomial
Since this is the first division step, XOR the result from 1a with the message polynomial.
(16 ⊕ 16)x25 + (32 ⊕ 35)x24 + (12 ⊕ 84)x23 + (86 ⊕ 89)x22 + (97 ⊕ 211)x21 + (128 ⊕ 35)x20 + (236 ⊕ 125)x19 + (17 ⊕ 204)x18 + (236 ⊕ 89)x17 + (0 ⊕ 43)x16 + (0 ⊕ 170)x15 + (0 ⊕ 24)x14 + (0 ⊕ 136)x13 + (0 ⊕ 57)x12 + (0 ⊕ 119)x11 + (0 ⊕ 131)x10 + (0 ⊕ 126)x9 + (0 ⊕ 132)x8
The result is:
0x25 + 3x24 + 88x23 + 15x22 + 178x21 + 163x20 + 145x19 + 221x18 + 181x17 + 43x16 + 170x15 + 24x14 + 136x13 + 57x12 + 119x11 + 131x10 + 126x9 + 132x8
Discard the lead 0 term to get:
3x24 + 88x23 + 15x22 + 178x21 + 163x20 + 145x19 + 221x18 + 181x17 + 43x16 + 170x15 + 24x14 + 136x13 + 57x12 + 119x11 + 131x10 + 126x9 + 132x8
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 3x24. Convert 3x24 to alpha notation. According to the log antilog table, for the integer value 3, the alpha exponent is 25. Therefore 3 = ɑ25. Multiply the generator polynomial by ɑ25:
(ɑ25 * ɑ0)x24 + (ɑ25 * ɑ43)x23 + (ɑ25 * ɑ139)x22 + (ɑ25 * ɑ206)x21 + (ɑ25 * ɑ78)x20 + (ɑ25 * ɑ43)x19 + (ɑ25 * ɑ239)x18 + (ɑ25 * ɑ123)x17 + (ɑ25 * ɑ206)x16 + (ɑ25 * ɑ214)x15 + (ɑ25 * ɑ147)x14 + (ɑ25 * ɑ24)x13 + (ɑ25 * ɑ99)x12 + (ɑ25 * ɑ150)x11 + (ɑ25 * ɑ39)x10 + (ɑ25 * ɑ243)x9 + (ɑ25 * ɑ163)x8 + (ɑ25 * ɑ136)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:
ɑ25x24 + ɑ68x23 + ɑ164x22 + ɑ231x21 + ɑ103x20 + ɑ68x19 + ɑ(264 % 255)x18 + ɑ148x17 + ɑ231x16 + ɑ239x15 + ɑ172x14 + ɑ49x13 + ɑ124x12 + ɑ175x11 + ɑ64x10 + ɑ(268 % 255)x9 + ɑ188x8 + ɑ161x7The result is:
ɑ25x24 + ɑ68x23 + ɑ164x22 + ɑ231x21 + ɑ103x20 + ɑ68x19 + ɑ9x18 + ɑ148x17 + ɑ231x16 + ɑ239x15 + ɑ172x14 + ɑ49x13 + ɑ124x12 + ɑ175x11 + ɑ64x10 + ɑ13x9 + ɑ188x8 + ɑ161x7
Now, convert this to integer notation:
3x24 + 153x23 + 198x22 + 245x21 + 136x20 + 153x19 + 58x18 + 82x17 + 245x16 + 22x15 + 123x14 + 140x13 + 151x12 + 255x11 + 95x10 + 135x9 + 165x8 + 209x7Step 2b: XOR the result with the result from step 1b
Use the result from step 1b to perform the next XOR.
(3 ⊕ 3)x24 + (88 ⊕ 153)x23 + (15 ⊕ 198)x22 + (178 ⊕ 245)x21 + (163 ⊕ 136)x20 + (145 ⊕ 153)x19 + (221 ⊕ 58)x18 + (181 ⊕ 82)x17 + (43 ⊕ 245)x16 + (170 ⊕ 22)x15 + (24 ⊕ 123)x14 + (136 ⊕ 140)x13 + (57 ⊕ 151)x12 + (119 ⊕ 255)x11 + (131 ⊕ 95)x10 + (126 ⊕ 135)x9 + (132 ⊕ 165)x8 + (0 ⊕ 209)x7
The result is:
0x24 + 193x23 + 201x22 + 71x21 + 43x20 + 8x19 + 231x18 + 231x17 + 222x16 + 188x15 + 99x14 + 4x13 + 174x12 + 136x11 + 220x10 + 249x9 + 33x8 + 209x7
Discard the lead 0 term to get:
193x23 + 201x22 + 71x21 + 43x20 + 8x19 + 231x18 + 231x17 + 222x16 + 188x15 + 99x14 + 4x13 + 174x12 + 136x11 + 220x10 + 249x9 + 33x8 + 209x7
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 193x23. Convert 193x23 to alpha notation. According to the log antilog table, for the integer value 193, the alpha exponent is 45. Therefore 193 = ɑ45. Multiply the generator polynomial by ɑ45:
(ɑ45 * ɑ0)x23 + (ɑ45 * ɑ43)x22 + (ɑ45 * ɑ139)x21 + (ɑ45 * ɑ206)x20 + (ɑ45 * ɑ78)x19 + (ɑ45 * ɑ43)x18 + (ɑ45 * ɑ239)x17 + (ɑ45 * ɑ123)x16 + (ɑ45 * ɑ206)x15 + (ɑ45 * ɑ214)x14 + (ɑ45 * ɑ147)x13 + (ɑ45 * ɑ24)x12 + (ɑ45 * ɑ99)x11 + (ɑ45 * ɑ150)x10 + (ɑ45 * ɑ39)x9 + (ɑ45 * ɑ243)x8 + (ɑ45 * ɑ163)x7 + (ɑ45 * ɑ136)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:
ɑ45x23 + ɑ88x22 + ɑ184x21 + ɑ251x20 + ɑ123x19 + ɑ88x18 + ɑ(284 % 255)x17 + ɑ168x16 + ɑ251x15 + ɑ(259 % 255)x14 + ɑ192x13 + ɑ69x12 + ɑ144x11 + ɑ195x10 + ɑ84x9 + ɑ(288 % 255)x8 + ɑ208x7 + ɑ181x6The result is:
ɑ45x23 + ɑ88x22 + ɑ184x21 + ɑ251x20 + ɑ123x19 + ɑ88x18 + ɑ29x17 + ɑ168x16 + ɑ251x15 + ɑ4x14 + ɑ192x13 + ɑ69x12 + ɑ144x11 + ɑ195x10 + ɑ84x9 + ɑ33x8 + ɑ208x7 + ɑ181x6
Now, convert this to integer notation:
193x23 + 254x22 + 149x21 + 216x20 + 197x19 + 254x18 + 48x17 + 252x16 + 216x15 + 16x14 + 130x13 + 47x12 + 168x11 + 100x10 + 107x9 + 39x8 + 81x7 + 49x6Step 3b: XOR the result with the result from step 2b
Use the result from step 2b to perform the next XOR.
(193 ⊕ 193)x23 + (201 ⊕ 254)x22 + (71 ⊕ 149)x21 + (43 ⊕ 216)x20 + (8 ⊕ 197)x19 + (231 ⊕ 254)x18 + (231 ⊕ 48)x17 + (222 ⊕ 252)x16 + (188 ⊕ 216)x15 + (99 ⊕ 16)x14 + (4 ⊕ 130)x13 + (174 ⊕ 47)x12 + (136 ⊕ 168)x11 + (220 ⊕ 100)x10 + (249 ⊕ 107)x9 + (33 ⊕ 39)x8 + (209 ⊕ 81)x7 + (0 ⊕ 49)x6
The result is:
0x23 + 55x22 + 210x21 + 243x20 + 205x19 + 25x18 + 215x17 + 34x16 + 100x15 + 115x14 + 134x13 + 129x12 + 32x11 + 184x10 + 146x9 + 6x8 + 128x7 + 49x6
Discard the lead 0 term to get:
55x22 + 210x21 + 243x20 + 205x19 + 25x18 + 215x17 + 34x16 + 100x15 + 115x14 + 134x13 + 129x12 + 32x11 + 184x10 + 146x9 + 6x8 + 128x7 + 49x6
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 55x22. Convert 55x22 to alpha notation. According to the log antilog table, for the integer value 55, the alpha exponent is 185. Therefore 55 = ɑ185. Multiply the generator polynomial by ɑ185:
(ɑ185 * ɑ0)x22 + (ɑ185 * ɑ43)x21 + (ɑ185 * ɑ139)x20 + (ɑ185 * ɑ206)x19 + (ɑ185 * ɑ78)x18 + (ɑ185 * ɑ43)x17 + (ɑ185 * ɑ239)x16 + (ɑ185 * ɑ123)x15 + (ɑ185 * ɑ206)x14 + (ɑ185 * ɑ214)x13 + (ɑ185 * ɑ147)x12 + (ɑ185 * ɑ24)x11 + (ɑ185 * ɑ99)x10 + (ɑ185 * ɑ150)x9 + (ɑ185 * ɑ39)x8 + (ɑ185 * ɑ243)x7 + (ɑ185 * ɑ163)x6 + (ɑ185 * ɑ136)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:
ɑ185x22 + ɑ228x21 + ɑ(324 % 255)x20 + ɑ(391 % 255)x19 + ɑ(263 % 255)x18 + ɑ228x17 + ɑ(424 % 255)x16 + ɑ(308 % 255)x15 + ɑ(391 % 255)x14 + ɑ(399 % 255)x13 + ɑ(332 % 255)x12 + ɑ209x11 + ɑ(284 % 255)x10 + ɑ(335 % 255)x9 + ɑ224x8 + ɑ(428 % 255)x7 + ɑ(348 % 255)x6 + ɑ(321 % 255)x5The result is:
ɑ185x22 + ɑ228x21 + ɑ69x20 + ɑ136x19 + ɑ8x18 + ɑ228x17 + ɑ169x16 + ɑ53x15 + ɑ136x14 + ɑ144x13 + ɑ77x12 + ɑ209x11 + ɑ29x10 + ɑ80x9 + ɑ224x8 + ɑ173x7 + ɑ93x6 + ɑ66x5
Now, convert this to integer notation:
55x22 + 61x21 + 47x20 + 79x19 + 29x18 + 61x17 + 229x16 + 40x15 + 79x14 + 168x13 + 60x12 + 162x11 + 48x10 + 253x9 + 18x8 + 246x7 + 182x6 + 97x5Step 4b: XOR the result with the result from step 3b
Use the result from step 3b to perform the next XOR.
(55 ⊕ 55)x22 + (210 ⊕ 61)x21 + (243 ⊕ 47)x20 + (205 ⊕ 79)x19 + (25 ⊕ 29)x18 + (215 ⊕ 61)x17 + (34 ⊕ 229)x16 + (100 ⊕ 40)x15 + (115 ⊕ 79)x14 + (134 ⊕ 168)x13 + (129 ⊕ 60)x12 + (32 ⊕ 162)x11 + (184 ⊕ 48)x10 + (146 ⊕ 253)x9 + (6 ⊕ 18)x8 + (128 ⊕ 246)x7 + (49 ⊕ 182)x6 + (0 ⊕ 97)x5
The result is:
0x22 + 239x21 + 220x20 + 130x19 + 4x18 + 234x17 + 199x16 + 76x15 + 60x14 + 46x13 + 189x12 + 130x11 + 136x10 + 111x9 + 20x8 + 118x7 + 135x6 + 97x5
Discard the lead 0 term to get:
239x21 + 220x20 + 130x19 + 4x18 + 234x17 + 199x16 + 76x15 + 60x14 + 46x13 + 189x12 + 130x11 + 136x10 + 111x9 + 20x8 + 118x7 + 135x6 + 97x5
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 239x21. Convert 239x21 to alpha notation. According to the log antilog table, for the integer value 239, the alpha exponent is 215. Therefore 239 = ɑ215. Multiply the generator polynomial by ɑ215:
(ɑ215 * ɑ0)x21 + (ɑ215 * ɑ43)x20 + (ɑ215 * ɑ139)x19 + (ɑ215 * ɑ206)x18 + (ɑ215 * ɑ78)x17 + (ɑ215 * ɑ43)x16 + (ɑ215 * ɑ239)x15 + (ɑ215 * ɑ123)x14 + (ɑ215 * ɑ206)x13 + (ɑ215 * ɑ214)x12 + (ɑ215 * ɑ147)x11 + (ɑ215 * ɑ24)x10 + (ɑ215 * ɑ99)x9 + (ɑ215 * ɑ150)x8 + (ɑ215 * ɑ39)x7 + (ɑ215 * ɑ243)x6 + (ɑ215 * ɑ163)x5 + (ɑ215 * ɑ136)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:
ɑ215x21 + ɑ(258 % 255)x20 + ɑ(354 % 255)x19 + ɑ(421 % 255)x18 + ɑ(293 % 255)x17 + ɑ(258 % 255)x16 + ɑ(454 % 255)x15 + ɑ(338 % 255)x14 + ɑ(421 % 255)x13 + ɑ(429 % 255)x12 + ɑ(362 % 255)x11 + ɑ239x10 + ɑ(314 % 255)x9 + ɑ(365 % 255)x8 + ɑ254x7 + ɑ(458 % 255)x6 + ɑ(378 % 255)x5 + ɑ(351 % 255)x4The result is:
ɑ215x21 + ɑ3x20 + ɑ99x19 + ɑ166x18 + ɑ38x17 + ɑ3x16 + ɑ199x15 + ɑ83x14 + ɑ166x13 + ɑ174x12 + ɑ107x11 + ɑ239x10 + ɑ59x9 + ɑ110x8 + ɑ254x7 + ɑ203x6 + ɑ123x5 + ɑ96x4
Now, convert this to integer notation:
239x21 + 8x20 + 134x19 + 63x18 + 148x17 + 8x16 + 14x15 + 187x14 + 63x13 + 241x12 + 104x11 + 22x10 + 210x9 + 103x8 + 142x7 + 224x6 + 197x5 + 217x4Step 5b: XOR the result with the result from step 4b
Use the result from step 4b to perform the next XOR.
(239 ⊕ 239)x21 + (220 ⊕ 8)x20 + (130 ⊕ 134)x19 + (4 ⊕ 63)x18 + (234 ⊕ 148)x17 + (199 ⊕ 8)x16 + (76 ⊕ 14)x15 + (60 ⊕ 187)x14 + (46 ⊕ 63)x13 + (189 ⊕ 241)x12 + (130 ⊕ 104)x11 + (136 ⊕ 22)x10 + (111 ⊕ 210)x9 + (20 ⊕ 103)x8 + (118 ⊕ 142)x7 + (135 ⊕ 224)x6 + (97 ⊕ 197)x5 + (0 ⊕ 217)x4
The result is:
0x21 + 212x20 + 4x19 + 59x18 + 126x17 + 207x16 + 66x15 + 135x14 + 17x13 + 76x12 + 234x11 + 158x10 + 189x9 + 115x8 + 248x7 + 103x6 + 164x5 + 217x4
Discard the lead 0 term to get:
212x20 + 4x19 + 59x18 + 126x17 + 207x16 + 66x15 + 135x14 + 17x13 + 76x12 + 234x11 + 158x10 + 189x9 + 115x8 + 248x7 + 103x6 + 164x5 + 217x4
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 212x20. Convert 212x20 to alpha notation. According to the log antilog table, for the integer value 212, the alpha exponent is 41. Therefore 212 = ɑ41. Multiply the generator polynomial by ɑ41:
(ɑ41 * ɑ0)x20 + (ɑ41 * ɑ43)x19 + (ɑ41 * ɑ139)x18 + (ɑ41 * ɑ206)x17 + (ɑ41 * ɑ78)x16 + (ɑ41 * ɑ43)x15 + (ɑ41 * ɑ239)x14 + (ɑ41 * ɑ123)x13 + (ɑ41 * ɑ206)x12 + (ɑ41 * ɑ214)x11 + (ɑ41 * ɑ147)x10 + (ɑ41 * ɑ24)x9 + (ɑ41 * ɑ99)x8 + (ɑ41 * ɑ150)x7 + (ɑ41 * ɑ39)x6 + (ɑ41 * ɑ243)x5 + (ɑ41 * ɑ163)x4 + (ɑ41 * ɑ136)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:
ɑ41x20 + ɑ84x19 + ɑ180x18 + ɑ247x17 + ɑ119x16 + ɑ84x15 + ɑ(280 % 255)x14 + ɑ164x13 + ɑ247x12 + ɑ255x11 + ɑ188x10 + ɑ65x9 + ɑ140x8 + ɑ191x7 + ɑ80x6 + ɑ(284 % 255)x5 + ɑ204x4 + ɑ177x3The result is:
ɑ41x20 + ɑ84x19 + ɑ180x18 + ɑ247x17 + ɑ119x16 + ɑ84x15 + ɑ25x14 + ɑ164x13 + ɑ247x12 + ɑ0x11 + ɑ188x10 + ɑ65x9 + ɑ140x8 + ɑ191x7 + ɑ80x6 + ɑ29x5 + ɑ204x4 + ɑ177x3
Now, convert this to integer notation:
212x20 + 107x19 + 150x18 + 131x17 + 147x16 + 107x15 + 3x14 + 198x13 + 131x12 + 1x11 + 165x10 + 190x9 + 132x8 + 65x7 + 253x6 + 48x5 + 221x4 + 219x3Step 6b: XOR the result with the result from step 5b
Use the result from step 5b to perform the next XOR.
(212 ⊕ 212)x20 + (4 ⊕ 107)x19 + (59 ⊕ 150)x18 + (126 ⊕ 131)x17 + (207 ⊕ 147)x16 + (66 ⊕ 107)x15 + (135 ⊕ 3)x14 + (17 ⊕ 198)x13 + (76 ⊕ 131)x12 + (234 ⊕ 1)x11 + (158 ⊕ 165)x10 + (189 ⊕ 190)x9 + (115 ⊕ 132)x8 + (248 ⊕ 65)x7 + (103 ⊕ 253)x6 + (164 ⊕ 48)x5 + (217 ⊕ 221)x4 + (0 ⊕ 219)x3
The result is:
0x20 + 111x19 + 173x18 + 253x17 + 92x16 + 41x15 + 132x14 + 215x13 + 207x12 + 235x11 + 59x10 + 3x9 + 247x8 + 185x7 + 154x6 + 148x5 + 4x4 + 219x3
Discard the lead 0 term to get:
111x19 + 173x18 + 253x17 + 92x16 + 41x15 + 132x14 + 215x13 + 207x12 + 235x11 + 59x10 + 3x9 + 247x8 + 185x7 + 154x6 + 148x5 + 4x4 + 219x3
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 111x19. Convert 111x19 to alpha notation. According to the log antilog table, for the integer value 111, the alpha exponent is 61. Therefore 111 = ɑ61. Multiply the generator polynomial by ɑ61:
(ɑ61 * ɑ0)x19 + (ɑ61 * ɑ43)x18 + (ɑ61 * ɑ139)x17 + (ɑ61 * ɑ206)x16 + (ɑ61 * ɑ78)x15 + (ɑ61 * ɑ43)x14 + (ɑ61 * ɑ239)x13 + (ɑ61 * ɑ123)x12 + (ɑ61 * ɑ206)x11 + (ɑ61 * ɑ214)x10 + (ɑ61 * ɑ147)x9 + (ɑ61 * ɑ24)x8 + (ɑ61 * ɑ99)x7 + (ɑ61 * ɑ150)x6 + (ɑ61 * ɑ39)x5 + (ɑ61 * ɑ243)x4 + (ɑ61 * ɑ163)x3 + (ɑ61 * ɑ136)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:
ɑ61x19 + ɑ104x18 + ɑ200x17 + ɑ(267 % 255)x16 + ɑ139x15 + ɑ104x14 + ɑ(300 % 255)x13 + ɑ184x12 + ɑ(267 % 255)x11 + ɑ(275 % 255)x10 + ɑ208x9 + ɑ85x8 + ɑ160x7 + ɑ211x6 + ɑ100x5 + ɑ(304 % 255)x4 + ɑ224x3 + ɑ197x2The result is:
ɑ61x19 + ɑ104x18 + ɑ200x17 + ɑ12x16 + ɑ139x15 + ɑ104x14 + ɑ45x13 + ɑ184x12 + ɑ12x11 + ɑ20x10 + ɑ208x9 + ɑ85x8 + ɑ160x7 + ɑ211x6 + ɑ100x5 + ɑ49x4 + ɑ224x3 + ɑ197x2
Now, convert this to integer notation:
111x19 + 13x18 + 28x17 + 205x16 + 66x15 + 13x14 + 193x13 + 149x12 + 205x11 + 180x10 + 81x9 + 214x8 + 230x7 + 178x6 + 17x5 + 140x4 + 18x3 + 141x2Step 7b: XOR the result with the result from step 6b
Use the result from step 6b to perform the next XOR.
(111 ⊕ 111)x19 + (173 ⊕ 13)x18 + (253 ⊕ 28)x17 + (92 ⊕ 205)x16 + (41 ⊕ 66)x15 + (132 ⊕ 13)x14 + (215 ⊕ 193)x13 + (207 ⊕ 149)x12 + (235 ⊕ 205)x11 + (59 ⊕ 180)x10 + (3 ⊕ 81)x9 + (247 ⊕ 214)x8 + (185 ⊕ 230)x7 + (154 ⊕ 178)x6 + (148 ⊕ 17)x5 + (4 ⊕ 140)x4 + (219 ⊕ 18)x3 + (0 ⊕ 141)x2
The result is:
0x19 + 160x18 + 225x17 + 145x16 + 107x15 + 137x14 + 22x13 + 90x12 + 38x11 + 143x10 + 82x9 + 33x8 + 95x7 + 40x6 + 133x5 + 136x4 + 201x3 + 141x2
Discard the lead 0 term to get:
160x18 + 225x17 + 145x16 + 107x15 + 137x14 + 22x13 + 90x12 + 38x11 + 143x10 + 82x9 + 33x8 + 95x7 + 40x6 + 133x5 + 136x4 + 201x3 + 141x2
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 160x18. Convert 160x18 to alpha notation. According to the log antilog table, for the integer value 160, the alpha exponent is 55. Therefore 160 = ɑ55. Multiply the generator polynomial by ɑ55:
(ɑ55 * ɑ0)x18 + (ɑ55 * ɑ43)x17 + (ɑ55 * ɑ139)x16 + (ɑ55 * ɑ206)x15 + (ɑ55 * ɑ78)x14 + (ɑ55 * ɑ43)x13 + (ɑ55 * ɑ239)x12 + (ɑ55 * ɑ123)x11 + (ɑ55 * ɑ206)x10 + (ɑ55 * ɑ214)x9 + (ɑ55 * ɑ147)x8 + (ɑ55 * ɑ24)x7 + (ɑ55 * ɑ99)x6 + (ɑ55 * ɑ150)x5 + (ɑ55 * ɑ39)x4 + (ɑ55 * ɑ243)x3 + (ɑ55 * ɑ163)x2 + (ɑ55 * ɑ136)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:
ɑ55x18 + ɑ98x17 + ɑ194x16 + ɑ(261 % 255)x15 + ɑ133x14 + ɑ98x13 + ɑ(294 % 255)x12 + ɑ178x11 + ɑ(261 % 255)x10 + ɑ(269 % 255)x9 + ɑ202x8 + ɑ79x7 + ɑ154x6 + ɑ205x5 + ɑ94x4 + ɑ(298 % 255)x3 + ɑ218x2 + ɑ191x1The result is:
ɑ55x18 + ɑ98x17 + ɑ194x16 + ɑ6x15 + ɑ133x14 + ɑ98x13 + ɑ39x12 + ɑ178x11 + ɑ6x10 + ɑ14x9 + ɑ202x8 + ɑ79x7 + ɑ154x6 + ɑ205x5 + ɑ94x4 + ɑ43x3 + ɑ218x2 + ɑ191x1
Now, convert this to integer notation:
160x18 + 67x17 + 50x16 + 64x15 + 109x14 + 67x13 + 53x12 + 171x11 + 64x10 + 19x9 + 112x8 + 240x7 + 57x6 + 167x5 + 113x4 + 119x3 + 43x2 + 65x1Step 8b: XOR the result with the result from step 7b
Use the result from step 7b to perform the next XOR.
(160 ⊕ 160)x18 + (225 ⊕ 67)x17 + (145 ⊕ 50)x16 + (107 ⊕ 64)x15 + (137 ⊕ 109)x14 + (22 ⊕ 67)x13 + (90 ⊕ 53)x12 + (38 ⊕ 171)x11 + (143 ⊕ 64)x10 + (82 ⊕ 19)x9 + (33 ⊕ 112)x8 + (95 ⊕ 240)x7 + (40 ⊕ 57)x6 + (133 ⊕ 167)x5 + (136 ⊕ 113)x4 + (201 ⊕ 119)x3 + (141 ⊕ 43)x2 + (0 ⊕ 65)x1
The result is:
0x18 + 162x17 + 163x16 + 43x15 + 228x14 + 85x13 + 111x12 + 141x11 + 207x10 + 65x9 + 81x8 + 175x7 + 17x6 + 34x5 + 249x4 + 190x3 + 166x2 + 65x1
Discard the lead 0 term to get:
162x17 + 163x16 + 43x15 + 228x14 + 85x13 + 111x12 + 141x11 + 207x10 + 65x9 + 81x8 + 175x7 + 17x6 + 34x5 + 249x4 + 190x3 + 166x2 + 65x1
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 162x17. Convert 162x17 to alpha notation. According to the log antilog table, for the integer value 162, the alpha exponent is 209. Therefore 162 = ɑ209. Multiply the generator polynomial by ɑ209:
(ɑ209 * ɑ0)x17 + (ɑ209 * ɑ43)x16 + (ɑ209 * ɑ139)x15 + (ɑ209 * ɑ206)x14 + (ɑ209 * ɑ78)x13 + (ɑ209 * ɑ43)x12 + (ɑ209 * ɑ239)x11 + (ɑ209 * ɑ123)x10 + (ɑ209 * ɑ206)x9 + (ɑ209 * ɑ214)x8 + (ɑ209 * ɑ147)x7 + (ɑ209 * ɑ24)x6 + (ɑ209 * ɑ99)x5 + (ɑ209 * ɑ150)x4 + (ɑ209 * ɑ39)x3 + (ɑ209 * ɑ243)x2 + (ɑ209 * ɑ163)x1 + (ɑ209 * ɑ136)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:
ɑ209x17 + ɑ252x16 + ɑ(348 % 255)x15 + ɑ(415 % 255)x14 + ɑ(287 % 255)x13 + ɑ252x12 + ɑ(448 % 255)x11 + ɑ(332 % 255)x10 + ɑ(415 % 255)x9 + ɑ(423 % 255)x8 + ɑ(356 % 255)x7 + ɑ233x6 + ɑ(308 % 255)x5 + ɑ(359 % 255)x4 + ɑ248x3 + ɑ(452 % 255)x2 + ɑ(372 % 255)x1 + ɑ(345 % 255)x0The result is:
ɑ209x17 + ɑ252x16 + ɑ93x15 + ɑ160x14 + ɑ32x13 + ɑ252x12 + ɑ193x11 + ɑ77x10 + ɑ160x9 + ɑ168x8 + ɑ101x7 + ɑ233x6 + ɑ53x5 + ɑ104x4 + ɑ248x3 + ɑ197x2 + ɑ117x1 + ɑ90x0
Now, convert this to integer notation:
162x17 + 173x16 + 182x15 + 230x14 + 157x13 + 173x12 + 25x11 + 60x10 + 230x9 + 252x8 + 34x7 + 243x6 + 40x5 + 13x4 + 27x3 + 141x2 + 237x1 + 223Step 9b: XOR the result with the result from step 8b
Use the result from step 8b to perform the next XOR.
(162 ⊕ 162)x17 + (163 ⊕ 173)x16 + (43 ⊕ 182)x15 + (228 ⊕ 230)x14 + (85 ⊕ 157)x13 + (111 ⊕ 173)x12 + (141 ⊕ 25)x11 + (207 ⊕ 60)x10 + (65 ⊕ 230)x9 + (81 ⊕ 252)x8 + (175 ⊕ 34)x7 + (17 ⊕ 243)x6 + (34 ⊕ 40)x5 + (249 ⊕ 13)x4 + (190 ⊕ 27)x3 + (166 ⊕ 141)x2 + (65 ⊕ 237)x1 + (0 ⊕ 223)x0
The result is:
0x17 + 14x16 + 157x15 + 2x14 + 200x13 + 194x12 + 148x11 + 243x10 + 167x9 + 173x8 + 141x7 + 226x6 + 10x5 + 244x4 + 165x3 + 43x2 + 172x1 + 223
Discard the lead 0 term to get:
14x16 + 157x15 + 2x14 + 200x13 + 194x12 + 148x11 + 243x10 + 167x9 + 173x8 + 141x7 + 226x6 + 10x5 + 244x4 + 165x3 + 43x2 + 172x1 + 223
Use the terms of the remainder as the error correction codewords
The division has been performed 9 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:
14 157 2 200 194 148 243 167 173 141 226 10 244 165 43 172 223