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:
67x14 + 85x13 + 70x12 + 134x11 + 87x10 + 38x9 + 85x8 + 194x7 + 119x6 + 50x5 + 6x4 + 18x3 + 6x2 + 103x1 + 38
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:
67x32 + 85x31 + 70x30 + 134x29 + 87x28 + 38x27 + 85x26 + 194x25 + 119x24 + 50x23 + 6x22 + 18x21 + 6x20 + 103x19 + 38x18
The lead term of the generator polynomial should also have the same exponent, so multiply by x14 to get
ɑ0x32 + ɑ215x31 + ɑ234x30 + ɑ158x29 + ɑ94x28 + ɑ184x27 + ɑ97x26 + ɑ118x25 + ɑ170x24 + ɑ79x23 + ɑ187x22 + ɑ152x21 + ɑ148x20 + ɑ252x19 + ɑ179x18 + ɑ5x17 + ɑ98x16 + ɑ96x15 + ɑ153x14
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 15 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 67x32. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 67x32 to alpha notation. According to the log antilog table, for the integer value 67, the alpha exponent is 98. Therefore 67 = ɑ98. Multiply the generator polynomial by ɑ98:
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:
ɑ98x32 + ɑ(313 % 255)x31 + ɑ(332 % 255)x30 + ɑ(256 % 255)x29 + ɑ192x28 + ɑ(282 % 255)x27 + ɑ195x26 + ɑ216x25 + ɑ(268 % 255)x24 + ɑ177x23 + ɑ(285 % 255)x22 + ɑ250x21 + ɑ246x20 + ɑ(350 % 255)x19 + ɑ(277 % 255)x18 + ɑ103x17 + ɑ196x16 + ɑ194x15 + ɑ251x14The result is:
ɑ98x32 + ɑ58x31 + ɑ77x30 + ɑ1x29 + ɑ192x28 + ɑ27x27 + ɑ195x26 + ɑ216x25 + ɑ13x24 + ɑ177x23 + ɑ30x22 + ɑ250x21 + ɑ246x20 + ɑ95x19 + ɑ22x18 + ɑ103x17 + ɑ196x16 + ɑ194x15 + ɑ251x14
Now, convert this to integer notation:
67x32 + 105x31 + 60x30 + 2x29 + 130x28 + 12x27 + 100x26 + 195x25 + 135x24 + 219x23 + 96x22 + 108x21 + 207x20 + 226x19 + 234x18 + 136x17 + 200x16 + 50x15 + 216x14Step 1b: XOR the result with the message polynomial
Since this is the first division step, XOR the result from 1a with the message polynomial.
(67 ⊕ 67)x32 + (85 ⊕ 105)x31 + (70 ⊕ 60)x30 + (134 ⊕ 2)x29 + (87 ⊕ 130)x28 + (38 ⊕ 12)x27 + (85 ⊕ 100)x26 + (194 ⊕ 195)x25 + (119 ⊕ 135)x24 + (50 ⊕ 219)x23 + (6 ⊕ 96)x22 + (18 ⊕ 108)x21 + (6 ⊕ 207)x20 + (103 ⊕ 226)x19 + (38 ⊕ 234)x18 + (0 ⊕ 136)x17 + (0 ⊕ 200)x16 + (0 ⊕ 50)x15 + (0 ⊕ 216)x14
The result is:
0x32 + 60x31 + 122x30 + 132x29 + 213x28 + 42x27 + 49x26 + 1x25 + 240x24 + 233x23 + 102x22 + 126x21 + 201x20 + 133x19 + 204x18 + 136x17 + 200x16 + 50x15 + 216x14
Discard the lead 0 term to get:
60x31 + 122x30 + 132x29 + 213x28 + 42x27 + 49x26 + 1x25 + 240x24 + 233x23 + 102x22 + 126x21 + 201x20 + 133x19 + 204x18 + 136x17 + 200x16 + 50x15 + 216x14
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 60x31. Convert 60x31 to alpha notation. According to the log antilog table, for the integer value 60, the alpha exponent is 77. Therefore 60 = ɑ77. Multiply the generator polynomial by ɑ77:
(ɑ77 * ɑ0)x31 + (ɑ77 * ɑ215)x30 + (ɑ77 * ɑ234)x29 + (ɑ77 * ɑ158)x28 + (ɑ77 * ɑ94)x27 + (ɑ77 * ɑ184)x26 + (ɑ77 * ɑ97)x25 + (ɑ77 * ɑ118)x24 + (ɑ77 * ɑ170)x23 + (ɑ77 * ɑ79)x22 + (ɑ77 * ɑ187)x21 + (ɑ77 * ɑ152)x20 + (ɑ77 * ɑ148)x19 + (ɑ77 * ɑ252)x18 + (ɑ77 * ɑ179)x17 + (ɑ77 * ɑ5)x16 + (ɑ77 * ɑ98)x15 + (ɑ77 * ɑ96)x14 + (ɑ77 * 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:
ɑ77x31 + ɑ(292 % 255)x30 + ɑ(311 % 255)x29 + ɑ235x28 + ɑ171x27 + ɑ(261 % 255)x26 + ɑ174x25 + ɑ195x24 + ɑ247x23 + ɑ156x22 + ɑ(264 % 255)x21 + ɑ229x20 + ɑ225x19 + ɑ(329 % 255)x18 + ɑ(256 % 255)x17 + ɑ82x16 + ɑ175x15 + ɑ173x14 + ɑ230x13The result is:
ɑ77x31 + ɑ37x30 + ɑ56x29 + ɑ235x28 + ɑ171x27 + ɑ6x26 + ɑ174x25 + ɑ195x24 + ɑ247x23 + ɑ156x22 + ɑ9x21 + ɑ229x20 + ɑ225x19 + ɑ74x18 + ɑ1x17 + ɑ82x16 + ɑ175x15 + ɑ173x14 + ɑ230x13
Now, convert this to integer notation:
60x31 + 74x30 + 93x29 + 235x28 + 179x27 + 64x26 + 241x25 + 100x24 + 131x23 + 228x22 + 58x21 + 122x20 + 36x19 + 137x18 + 2x17 + 211x16 + 255x15 + 246x14 + 244x13Step 2b: XOR the result with the result from step 1b
Use the result from step 1b to perform the next XOR.
(60 ⊕ 60)x31 + (122 ⊕ 74)x30 + (132 ⊕ 93)x29 + (213 ⊕ 235)x28 + (42 ⊕ 179)x27 + (49 ⊕ 64)x26 + (1 ⊕ 241)x25 + (240 ⊕ 100)x24 + (233 ⊕ 131)x23 + (102 ⊕ 228)x22 + (126 ⊕ 58)x21 + (201 ⊕ 122)x20 + (133 ⊕ 36)x19 + (204 ⊕ 137)x18 + (136 ⊕ 2)x17 + (200 ⊕ 211)x16 + (50 ⊕ 255)x15 + (216 ⊕ 246)x14 + (0 ⊕ 244)x13
The result is:
0x31 + 48x30 + 217x29 + 62x28 + 153x27 + 113x26 + 240x25 + 148x24 + 106x23 + 130x22 + 68x21 + 179x20 + 161x19 + 69x18 + 138x17 + 27x16 + 205x15 + 46x14 + 244x13
Discard the lead 0 term to get:
48x30 + 217x29 + 62x28 + 153x27 + 113x26 + 240x25 + 148x24 + 106x23 + 130x22 + 68x21 + 179x20 + 161x19 + 69x18 + 138x17 + 27x16 + 205x15 + 46x14 + 244x13
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 48x30. Convert 48x30 to alpha notation. According to the log antilog table, for the integer value 48, the alpha exponent is 29. Therefore 48 = ɑ29. Multiply the generator polynomial by ɑ29:
(ɑ29 * ɑ0)x30 + (ɑ29 * ɑ215)x29 + (ɑ29 * ɑ234)x28 + (ɑ29 * ɑ158)x27 + (ɑ29 * ɑ94)x26 + (ɑ29 * ɑ184)x25 + (ɑ29 * ɑ97)x24 + (ɑ29 * ɑ118)x23 + (ɑ29 * ɑ170)x22 + (ɑ29 * ɑ79)x21 + (ɑ29 * ɑ187)x20 + (ɑ29 * ɑ152)x19 + (ɑ29 * ɑ148)x18 + (ɑ29 * ɑ252)x17 + (ɑ29 * ɑ179)x16 + (ɑ29 * ɑ5)x15 + (ɑ29 * ɑ98)x14 + (ɑ29 * ɑ96)x13 + (ɑ29 * 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:
ɑ29x30 + ɑ244x29 + ɑ(263 % 255)x28 + ɑ187x27 + ɑ123x26 + ɑ213x25 + ɑ126x24 + ɑ147x23 + ɑ199x22 + ɑ108x21 + ɑ216x20 + ɑ181x19 + ɑ177x18 + ɑ(281 % 255)x17 + ɑ208x16 + ɑ34x15 + ɑ127x14 + ɑ125x13 + ɑ182x12The result is:
ɑ29x30 + ɑ244x29 + ɑ8x28 + ɑ187x27 + ɑ123x26 + ɑ213x25 + ɑ126x24 + ɑ147x23 + ɑ199x22 + ɑ108x21 + ɑ216x20 + ɑ181x19 + ɑ177x18 + ɑ26x17 + ɑ208x16 + ɑ34x15 + ɑ127x14 + ɑ125x13 + ɑ182x12
Now, convert this to integer notation:
48x30 + 250x29 + 29x28 + 220x27 + 197x26 + 242x25 + 102x24 + 41x23 + 14x22 + 208x21 + 195x20 + 49x19 + 219x18 + 6x17 + 81x16 + 78x15 + 204x14 + 51x13 + 98x12Step 3b: XOR the result with the result from step 2b
Use the result from step 2b to perform the next XOR.
(48 ⊕ 48)x30 + (217 ⊕ 250)x29 + (62 ⊕ 29)x28 + (153 ⊕ 220)x27 + (113 ⊕ 197)x26 + (240 ⊕ 242)x25 + (148 ⊕ 102)x24 + (106 ⊕ 41)x23 + (130 ⊕ 14)x22 + (68 ⊕ 208)x21 + (179 ⊕ 195)x20 + (161 ⊕ 49)x19 + (69 ⊕ 219)x18 + (138 ⊕ 6)x17 + (27 ⊕ 81)x16 + (205 ⊕ 78)x15 + (46 ⊕ 204)x14 + (244 ⊕ 51)x13 + (0 ⊕ 98)x12
The result is:
0x30 + 35x29 + 35x28 + 69x27 + 180x26 + 2x25 + 242x24 + 67x23 + 140x22 + 148x21 + 112x20 + 144x19 + 158x18 + 140x17 + 74x16 + 131x15 + 226x14 + 199x13 + 98x12
Discard the lead 0 term to get:
35x29 + 35x28 + 69x27 + 180x26 + 2x25 + 242x24 + 67x23 + 140x22 + 148x21 + 112x20 + 144x19 + 158x18 + 140x17 + 74x16 + 131x15 + 226x14 + 199x13 + 98x12
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 35x29. Convert 35x29 to alpha notation. According to the log antilog table, for the integer value 35, the alpha exponent is 47. Therefore 35 = ɑ47. Multiply the generator polynomial by ɑ47:
(ɑ47 * ɑ0)x29 + (ɑ47 * ɑ215)x28 + (ɑ47 * ɑ234)x27 + (ɑ47 * ɑ158)x26 + (ɑ47 * ɑ94)x25 + (ɑ47 * ɑ184)x24 + (ɑ47 * ɑ97)x23 + (ɑ47 * ɑ118)x22 + (ɑ47 * ɑ170)x21 + (ɑ47 * ɑ79)x20 + (ɑ47 * ɑ187)x19 + (ɑ47 * ɑ152)x18 + (ɑ47 * ɑ148)x17 + (ɑ47 * ɑ252)x16 + (ɑ47 * ɑ179)x15 + (ɑ47 * ɑ5)x14 + (ɑ47 * ɑ98)x13 + (ɑ47 * ɑ96)x12 + (ɑ47 * 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:
ɑ47x29 + ɑ(262 % 255)x28 + ɑ(281 % 255)x27 + ɑ205x26 + ɑ141x25 + ɑ231x24 + ɑ144x23 + ɑ165x22 + ɑ217x21 + ɑ126x20 + ɑ234x19 + ɑ199x18 + ɑ195x17 + ɑ(299 % 255)x16 + ɑ226x15 + ɑ52x14 + ɑ145x13 + ɑ143x12 + ɑ200x11The result is:
ɑ47x29 + ɑ7x28 + ɑ26x27 + ɑ205x26 + ɑ141x25 + ɑ231x24 + ɑ144x23 + ɑ165x22 + ɑ217x21 + ɑ126x20 + ɑ234x19 + ɑ199x18 + ɑ195x17 + ɑ44x16 + ɑ226x15 + ɑ52x14 + ɑ145x13 + ɑ143x12 + ɑ200x11
Now, convert this to integer notation:
35x29 + 128x28 + 6x27 + 167x26 + 21x25 + 245x24 + 168x23 + 145x22 + 155x21 + 102x20 + 251x19 + 14x18 + 100x17 + 238x16 + 72x15 + 20x14 + 77x13 + 84x12 + 28x11Step 4b: XOR the result with the result from step 3b
Use the result from step 3b to perform the next XOR.
(35 ⊕ 35)x29 + (35 ⊕ 128)x28 + (69 ⊕ 6)x27 + (180 ⊕ 167)x26 + (2 ⊕ 21)x25 + (242 ⊕ 245)x24 + (67 ⊕ 168)x23 + (140 ⊕ 145)x22 + (148 ⊕ 155)x21 + (112 ⊕ 102)x20 + (144 ⊕ 251)x19 + (158 ⊕ 14)x18 + (140 ⊕ 100)x17 + (74 ⊕ 238)x16 + (131 ⊕ 72)x15 + (226 ⊕ 20)x14 + (199 ⊕ 77)x13 + (98 ⊕ 84)x12 + (0 ⊕ 28)x11
The result is:
0x29 + 163x28 + 67x27 + 19x26 + 23x25 + 7x24 + 235x23 + 29x22 + 15x21 + 22x20 + 107x19 + 144x18 + 232x17 + 164x16 + 203x15 + 246x14 + 138x13 + 54x12 + 28x11
Discard the lead 0 term to get:
163x28 + 67x27 + 19x26 + 23x25 + 7x24 + 235x23 + 29x22 + 15x21 + 22x20 + 107x19 + 144x18 + 232x17 + 164x16 + 203x15 + 246x14 + 138x13 + 54x12 + 28x11
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 163x28. Convert 163x28 to alpha notation. According to the log antilog table, for the integer value 163, the alpha exponent is 91. Therefore 163 = ɑ91. Multiply the generator polynomial by ɑ91:
(ɑ91 * ɑ0)x28 + (ɑ91 * ɑ215)x27 + (ɑ91 * ɑ234)x26 + (ɑ91 * ɑ158)x25 + (ɑ91 * ɑ94)x24 + (ɑ91 * ɑ184)x23 + (ɑ91 * ɑ97)x22 + (ɑ91 * ɑ118)x21 + (ɑ91 * ɑ170)x20 + (ɑ91 * ɑ79)x19 + (ɑ91 * ɑ187)x18 + (ɑ91 * ɑ152)x17 + (ɑ91 * ɑ148)x16 + (ɑ91 * ɑ252)x15 + (ɑ91 * ɑ179)x14 + (ɑ91 * ɑ5)x13 + (ɑ91 * ɑ98)x12 + (ɑ91 * ɑ96)x11 + (ɑ91 * 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:
ɑ91x28 + ɑ(306 % 255)x27 + ɑ(325 % 255)x26 + ɑ249x25 + ɑ185x24 + ɑ(275 % 255)x23 + ɑ188x22 + ɑ209x21 + ɑ(261 % 255)x20 + ɑ170x19 + ɑ(278 % 255)x18 + ɑ243x17 + ɑ239x16 + ɑ(343 % 255)x15 + ɑ(270 % 255)x14 + ɑ96x13 + ɑ189x12 + ɑ187x11 + ɑ244x10The result is:
ɑ91x28 + ɑ51x27 + ɑ70x26 + ɑ249x25 + ɑ185x24 + ɑ20x23 + ɑ188x22 + ɑ209x21 + ɑ6x20 + ɑ170x19 + ɑ23x18 + ɑ243x17 + ɑ239x16 + ɑ88x15 + ɑ15x14 + ɑ96x13 + ɑ189x12 + ɑ187x11 + ɑ244x10
Now, convert this to integer notation:
163x28 + 10x27 + 94x26 + 54x25 + 55x24 + 180x23 + 165x22 + 162x21 + 64x20 + 215x19 + 201x18 + 125x17 + 22x16 + 254x15 + 38x14 + 217x13 + 87x12 + 220x11 + 250x10Step 5b: XOR the result with the result from step 4b
Use the result from step 4b to perform the next XOR.
(163 ⊕ 163)x28 + (67 ⊕ 10)x27 + (19 ⊕ 94)x26 + (23 ⊕ 54)x25 + (7 ⊕ 55)x24 + (235 ⊕ 180)x23 + (29 ⊕ 165)x22 + (15 ⊕ 162)x21 + (22 ⊕ 64)x20 + (107 ⊕ 215)x19 + (144 ⊕ 201)x18 + (232 ⊕ 125)x17 + (164 ⊕ 22)x16 + (203 ⊕ 254)x15 + (246 ⊕ 38)x14 + (138 ⊕ 217)x13 + (54 ⊕ 87)x12 + (28 ⊕ 220)x11 + (0 ⊕ 250)x10
The result is:
0x28 + 73x27 + 77x26 + 33x25 + 48x24 + 95x23 + 184x22 + 173x21 + 86x20 + 188x19 + 89x18 + 149x17 + 178x16 + 53x15 + 208x14 + 83x13 + 97x12 + 192x11 + 250x10
Discard the lead 0 term to get:
73x27 + 77x26 + 33x25 + 48x24 + 95x23 + 184x22 + 173x21 + 86x20 + 188x19 + 89x18 + 149x17 + 178x16 + 53x15 + 208x14 + 83x13 + 97x12 + 192x11 + 250x10
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 73x27. Convert 73x27 to alpha notation. According to the log antilog table, for the integer value 73, the alpha exponent is 152. Therefore 73 = ɑ152. Multiply the generator polynomial by ɑ152:
(ɑ152 * ɑ0)x27 + (ɑ152 * ɑ215)x26 + (ɑ152 * ɑ234)x25 + (ɑ152 * ɑ158)x24 + (ɑ152 * ɑ94)x23 + (ɑ152 * ɑ184)x22 + (ɑ152 * ɑ97)x21 + (ɑ152 * ɑ118)x20 + (ɑ152 * ɑ170)x19 + (ɑ152 * ɑ79)x18 + (ɑ152 * ɑ187)x17 + (ɑ152 * ɑ152)x16 + (ɑ152 * ɑ148)x15 + (ɑ152 * ɑ252)x14 + (ɑ152 * ɑ179)x13 + (ɑ152 * ɑ5)x12 + (ɑ152 * ɑ98)x11 + (ɑ152 * ɑ96)x10 + (ɑ152 * 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:
ɑ152x27 + ɑ(367 % 255)x26 + ɑ(386 % 255)x25 + ɑ(310 % 255)x24 + ɑ246x23 + ɑ(336 % 255)x22 + ɑ249x21 + ɑ(270 % 255)x20 + ɑ(322 % 255)x19 + ɑ231x18 + ɑ(339 % 255)x17 + ɑ(304 % 255)x16 + ɑ(300 % 255)x15 + ɑ(404 % 255)x14 + ɑ(331 % 255)x13 + ɑ157x12 + ɑ250x11 + ɑ248x10 + ɑ(305 % 255)x9The result is:
ɑ152x27 + ɑ112x26 + ɑ131x25 + ɑ55x24 + ɑ246x23 + ɑ81x22 + ɑ249x21 + ɑ15x20 + ɑ67x19 + ɑ231x18 + ɑ84x17 + ɑ49x16 + ɑ45x15 + ɑ149x14 + ɑ76x13 + ɑ157x12 + ɑ250x11 + ɑ248x10 + ɑ50x9
Now, convert this to integer notation:
73x27 + 129x26 + 92x25 + 160x24 + 207x23 + 231x22 + 54x21 + 38x20 + 194x19 + 245x18 + 107x17 + 140x16 + 193x15 + 164x14 + 30x13 + 213x12 + 108x11 + 27x10 + 5x9Step 6b: XOR the result with the result from step 5b
Use the result from step 5b to perform the next XOR.
(73 ⊕ 73)x27 + (77 ⊕ 129)x26 + (33 ⊕ 92)x25 + (48 ⊕ 160)x24 + (95 ⊕ 207)x23 + (184 ⊕ 231)x22 + (173 ⊕ 54)x21 + (86 ⊕ 38)x20 + (188 ⊕ 194)x19 + (89 ⊕ 245)x18 + (149 ⊕ 107)x17 + (178 ⊕ 140)x16 + (53 ⊕ 193)x15 + (208 ⊕ 164)x14 + (83 ⊕ 30)x13 + (97 ⊕ 213)x12 + (192 ⊕ 108)x11 + (250 ⊕ 27)x10 + (0 ⊕ 5)x9
The result is:
0x27 + 204x26 + 125x25 + 144x24 + 144x23 + 95x22 + 155x21 + 112x20 + 126x19 + 172x18 + 254x17 + 62x16 + 244x15 + 116x14 + 77x13 + 180x12 + 172x11 + 225x10 + 5x9
Discard the lead 0 term to get:
204x26 + 125x25 + 144x24 + 144x23 + 95x22 + 155x21 + 112x20 + 126x19 + 172x18 + 254x17 + 62x16 + 244x15 + 116x14 + 77x13 + 180x12 + 172x11 + 225x10 + 5x9
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 204x26. Convert 204x26 to alpha notation. According to the log antilog table, for the integer value 204, the alpha exponent is 127. Therefore 204 = ɑ127. Multiply the generator polynomial by ɑ127:
(ɑ127 * ɑ0)x26 + (ɑ127 * ɑ215)x25 + (ɑ127 * ɑ234)x24 + (ɑ127 * ɑ158)x23 + (ɑ127 * ɑ94)x22 + (ɑ127 * ɑ184)x21 + (ɑ127 * ɑ97)x20 + (ɑ127 * ɑ118)x19 + (ɑ127 * ɑ170)x18 + (ɑ127 * ɑ79)x17 + (ɑ127 * ɑ187)x16 + (ɑ127 * ɑ152)x15 + (ɑ127 * ɑ148)x14 + (ɑ127 * ɑ252)x13 + (ɑ127 * ɑ179)x12 + (ɑ127 * ɑ5)x11 + (ɑ127 * ɑ98)x10 + (ɑ127 * ɑ96)x9 + (ɑ127 * 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:
ɑ127x26 + ɑ(342 % 255)x25 + ɑ(361 % 255)x24 + ɑ(285 % 255)x23 + ɑ221x22 + ɑ(311 % 255)x21 + ɑ224x20 + ɑ245x19 + ɑ(297 % 255)x18 + ɑ206x17 + ɑ(314 % 255)x16 + ɑ(279 % 255)x15 + ɑ(275 % 255)x14 + ɑ(379 % 255)x13 + ɑ(306 % 255)x12 + ɑ132x11 + ɑ225x10 + ɑ223x9 + ɑ(280 % 255)x8The result is:
ɑ127x26 + ɑ87x25 + ɑ106x24 + ɑ30x23 + ɑ221x22 + ɑ56x21 + ɑ224x20 + ɑ245x19 + ɑ42x18 + ɑ206x17 + ɑ59x16 + ɑ24x15 + ɑ20x14 + ɑ124x13 + ɑ51x12 + ɑ132x11 + ɑ225x10 + ɑ223x9 + ɑ25x8
Now, convert this to integer notation:
204x26 + 127x25 + 52x24 + 96x23 + 69x22 + 93x21 + 18x20 + 233x19 + 181x18 + 83x17 + 210x16 + 143x15 + 180x14 + 151x13 + 10x12 + 184x11 + 36x10 + 9x9 + 3x8Step 7b: XOR the result with the result from step 6b
Use the result from step 6b to perform the next XOR.
(204 ⊕ 204)x26 + (125 ⊕ 127)x25 + (144 ⊕ 52)x24 + (144 ⊕ 96)x23 + (95 ⊕ 69)x22 + (155 ⊕ 93)x21 + (112 ⊕ 18)x20 + (126 ⊕ 233)x19 + (172 ⊕ 181)x18 + (254 ⊕ 83)x17 + (62 ⊕ 210)x16 + (244 ⊕ 143)x15 + (116 ⊕ 180)x14 + (77 ⊕ 151)x13 + (180 ⊕ 10)x12 + (172 ⊕ 184)x11 + (225 ⊕ 36)x10 + (5 ⊕ 9)x9 + (0 ⊕ 3)x8
The result is:
0x26 + 2x25 + 164x24 + 240x23 + 26x22 + 198x21 + 98x20 + 151x19 + 25x18 + 173x17 + 236x16 + 123x15 + 192x14 + 218x13 + 190x12 + 20x11 + 197x10 + 12x9 + 3x8
Discard the lead 0 term to get:
2x25 + 164x24 + 240x23 + 26x22 + 198x21 + 98x20 + 151x19 + 25x18 + 173x17 + 236x16 + 123x15 + 192x14 + 218x13 + 190x12 + 20x11 + 197x10 + 12x9 + 3x8
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 2x25. Convert 2x25 to alpha notation. According to the log antilog table, for the integer value 2, the alpha exponent is 1. Therefore 2 = ɑ1. Multiply the generator polynomial by ɑ1:
(ɑ1 * ɑ0)x25 + (ɑ1 * ɑ215)x24 + (ɑ1 * ɑ234)x23 + (ɑ1 * ɑ158)x22 + (ɑ1 * ɑ94)x21 + (ɑ1 * ɑ184)x20 + (ɑ1 * ɑ97)x19 + (ɑ1 * ɑ118)x18 + (ɑ1 * ɑ170)x17 + (ɑ1 * ɑ79)x16 + (ɑ1 * ɑ187)x15 + (ɑ1 * ɑ152)x14 + (ɑ1 * ɑ148)x13 + (ɑ1 * ɑ252)x12 + (ɑ1 * ɑ179)x11 + (ɑ1 * ɑ5)x10 + (ɑ1 * ɑ98)x9 + (ɑ1 * ɑ96)x8 + (ɑ1 * ɑ153)x7
The exponents of the alphas are added together. The result is:
ɑ1x25 + ɑ216x24 + ɑ235x23 + ɑ159x22 + ɑ95x21 + ɑ185x20 + ɑ98x19 + ɑ119x18 + ɑ171x17 + ɑ80x16 + ɑ188x15 + ɑ153x14 + ɑ149x13 + ɑ253x12 + ɑ180x11 + ɑ6x10 + ɑ99x9 + ɑ97x8 + ɑ154x7
Now, convert this to integer notation:
2x25 + 195x24 + 235x23 + 115x22 + 226x21 + 55x20 + 67x19 + 147x18 + 179x17 + 253x16 + 165x15 + 146x14 + 164x13 + 71x12 + 150x11 + 64x10 + 134x9 + 175x8 + 57x7Step 8b: XOR the result with the result from step 7b
Use the result from step 7b to perform the next XOR.
(2 ⊕ 2)x25 + (164 ⊕ 195)x24 + (240 ⊕ 235)x23 + (26 ⊕ 115)x22 + (198 ⊕ 226)x21 + (98 ⊕ 55)x20 + (151 ⊕ 67)x19 + (25 ⊕ 147)x18 + (173 ⊕ 179)x17 + (236 ⊕ 253)x16 + (123 ⊕ 165)x15 + (192 ⊕ 146)x14 + (218 ⊕ 164)x13 + (190 ⊕ 71)x12 + (20 ⊕ 150)x11 + (197 ⊕ 64)x10 + (12 ⊕ 134)x9 + (3 ⊕ 175)x8 + (0 ⊕ 57)x7
The result is:
0x25 + 103x24 + 27x23 + 105x22 + 36x21 + 85x20 + 212x19 + 138x18 + 30x17 + 17x16 + 222x15 + 82x14 + 126x13 + 249x12 + 130x11 + 133x10 + 138x9 + 172x8 + 57x7
Discard the lead 0 term to get:
103x24 + 27x23 + 105x22 + 36x21 + 85x20 + 212x19 + 138x18 + 30x17 + 17x16 + 222x15 + 82x14 + 126x13 + 249x12 + 130x11 + 133x10 + 138x9 + 172x8 + 57x7
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 103x24. Convert 103x24 to alpha notation. According to the log antilog table, for the integer value 103, the alpha exponent is 110. Therefore 103 = ɑ110. Multiply the generator polynomial by ɑ110:
(ɑ110 * ɑ0)x24 + (ɑ110 * ɑ215)x23 + (ɑ110 * ɑ234)x22 + (ɑ110 * ɑ158)x21 + (ɑ110 * ɑ94)x20 + (ɑ110 * ɑ184)x19 + (ɑ110 * ɑ97)x18 + (ɑ110 * ɑ118)x17 + (ɑ110 * ɑ170)x16 + (ɑ110 * ɑ79)x15 + (ɑ110 * ɑ187)x14 + (ɑ110 * ɑ152)x13 + (ɑ110 * ɑ148)x12 + (ɑ110 * ɑ252)x11 + (ɑ110 * ɑ179)x10 + (ɑ110 * ɑ5)x9 + (ɑ110 * ɑ98)x8 + (ɑ110 * ɑ96)x7 + (ɑ110 * 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:
ɑ110x24 + ɑ(325 % 255)x23 + ɑ(344 % 255)x22 + ɑ(268 % 255)x21 + ɑ204x20 + ɑ(294 % 255)x19 + ɑ207x18 + ɑ228x17 + ɑ(280 % 255)x16 + ɑ189x15 + ɑ(297 % 255)x14 + ɑ(262 % 255)x13 + ɑ(258 % 255)x12 + ɑ(362 % 255)x11 + ɑ(289 % 255)x10 + ɑ115x9 + ɑ208x8 + ɑ206x7 + ɑ(263 % 255)x6The result is:
ɑ110x24 + ɑ70x23 + ɑ89x22 + ɑ13x21 + ɑ204x20 + ɑ39x19 + ɑ207x18 + ɑ228x17 + ɑ25x16 + ɑ189x15 + ɑ42x14 + ɑ7x13 + ɑ3x12 + ɑ107x11 + ɑ34x10 + ɑ115x9 + ɑ208x8 + ɑ206x7 + ɑ8x6
Now, convert this to integer notation:
103x24 + 94x23 + 225x22 + 135x21 + 221x20 + 53x19 + 166x18 + 61x17 + 3x16 + 87x15 + 181x14 + 128x13 + 8x12 + 104x11 + 78x10 + 124x9 + 81x8 + 83x7 + 29x6Step 9b: XOR the result with the result from step 8b
Use the result from step 8b to perform the next XOR.
(103 ⊕ 103)x24 + (27 ⊕ 94)x23 + (105 ⊕ 225)x22 + (36 ⊕ 135)x21 + (85 ⊕ 221)x20 + (212 ⊕ 53)x19 + (138 ⊕ 166)x18 + (30 ⊕ 61)x17 + (17 ⊕ 3)x16 + (222 ⊕ 87)x15 + (82 ⊕ 181)x14 + (126 ⊕ 128)x13 + (249 ⊕ 8)x12 + (130 ⊕ 104)x11 + (133 ⊕ 78)x10 + (138 ⊕ 124)x9 + (172 ⊕ 81)x8 + (57 ⊕ 83)x7 + (0 ⊕ 29)x6
The result is:
0x24 + 69x23 + 136x22 + 163x21 + 136x20 + 225x19 + 44x18 + 35x17 + 18x16 + 137x15 + 231x14 + 254x13 + 241x12 + 234x11 + 203x10 + 246x9 + 253x8 + 106x7 + 29x6
Discard the lead 0 term to get:
69x23 + 136x22 + 163x21 + 136x20 + 225x19 + 44x18 + 35x17 + 18x16 + 137x15 + 231x14 + 254x13 + 241x12 + 234x11 + 203x10 + 246x9 + 253x8 + 106x7 + 29x6
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 69x23. Convert 69x23 to alpha notation. According to the log antilog table, for the integer value 69, the alpha exponent is 221. Therefore 69 = ɑ221. Multiply the generator polynomial by ɑ221:
(ɑ221 * ɑ0)x23 + (ɑ221 * ɑ215)x22 + (ɑ221 * ɑ234)x21 + (ɑ221 * ɑ158)x20 + (ɑ221 * ɑ94)x19 + (ɑ221 * ɑ184)x18 + (ɑ221 * ɑ97)x17 + (ɑ221 * ɑ118)x16 + (ɑ221 * ɑ170)x15 + (ɑ221 * ɑ79)x14 + (ɑ221 * ɑ187)x13 + (ɑ221 * ɑ152)x12 + (ɑ221 * ɑ148)x11 + (ɑ221 * ɑ252)x10 + (ɑ221 * ɑ179)x9 + (ɑ221 * ɑ5)x8 + (ɑ221 * ɑ98)x7 + (ɑ221 * ɑ96)x6 + (ɑ221 * 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:
ɑ221x23 + ɑ(436 % 255)x22 + ɑ(455 % 255)x21 + ɑ(379 % 255)x20 + ɑ(315 % 255)x19 + ɑ(405 % 255)x18 + ɑ(318 % 255)x17 + ɑ(339 % 255)x16 + ɑ(391 % 255)x15 + ɑ(300 % 255)x14 + ɑ(408 % 255)x13 + ɑ(373 % 255)x12 + ɑ(369 % 255)x11 + ɑ(473 % 255)x10 + ɑ(400 % 255)x9 + ɑ226x8 + ɑ(319 % 255)x7 + ɑ(317 % 255)x6 + ɑ(374 % 255)x5The result is:
ɑ221x23 + ɑ181x22 + ɑ200x21 + ɑ124x20 + ɑ60x19 + ɑ150x18 + ɑ63x17 + ɑ84x16 + ɑ136x15 + ɑ45x14 + ɑ153x13 + ɑ118x12 + ɑ114x11 + ɑ218x10 + ɑ145x9 + ɑ226x8 + ɑ64x7 + ɑ62x6 + ɑ119x5
Now, convert this to integer notation:
69x23 + 49x22 + 28x21 + 151x20 + 185x19 + 85x18 + 161x17 + 107x16 + 79x15 + 193x14 + 146x13 + 199x12 + 62x11 + 43x10 + 77x9 + 72x8 + 95x7 + 222x6 + 147x5Step 10b: XOR the result with the result from step 9b
Use the result from step 9b to perform the next XOR.
(69 ⊕ 69)x23 + (136 ⊕ 49)x22 + (163 ⊕ 28)x21 + (136 ⊕ 151)x20 + (225 ⊕ 185)x19 + (44 ⊕ 85)x18 + (35 ⊕ 161)x17 + (18 ⊕ 107)x16 + (137 ⊕ 79)x15 + (231 ⊕ 193)x14 + (254 ⊕ 146)x13 + (241 ⊕ 199)x12 + (234 ⊕ 62)x11 + (203 ⊕ 43)x10 + (246 ⊕ 77)x9 + (253 ⊕ 72)x8 + (106 ⊕ 95)x7 + (29 ⊕ 222)x6 + (0 ⊕ 147)x5
The result is:
0x23 + 185x22 + 191x21 + 31x20 + 88x19 + 121x18 + 130x17 + 121x16 + 198x15 + 38x14 + 108x13 + 54x12 + 212x11 + 224x10 + 187x9 + 181x8 + 53x7 + 195x6 + 147x5
Discard the lead 0 term to get:
185x22 + 191x21 + 31x20 + 88x19 + 121x18 + 130x17 + 121x16 + 198x15 + 38x14 + 108x13 + 54x12 + 212x11 + 224x10 + 187x9 + 181x8 + 53x7 + 195x6 + 147x5
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 185x22. Convert 185x22 to alpha notation. According to the log antilog table, for the integer value 185, the alpha exponent is 60. Therefore 185 = ɑ60. Multiply the generator polynomial by ɑ60:
(ɑ60 * ɑ0)x22 + (ɑ60 * ɑ215)x21 + (ɑ60 * ɑ234)x20 + (ɑ60 * ɑ158)x19 + (ɑ60 * ɑ94)x18 + (ɑ60 * ɑ184)x17 + (ɑ60 * ɑ97)x16 + (ɑ60 * ɑ118)x15 + (ɑ60 * ɑ170)x14 + (ɑ60 * ɑ79)x13 + (ɑ60 * ɑ187)x12 + (ɑ60 * ɑ152)x11 + (ɑ60 * ɑ148)x10 + (ɑ60 * ɑ252)x9 + (ɑ60 * ɑ179)x8 + (ɑ60 * ɑ5)x7 + (ɑ60 * ɑ98)x6 + (ɑ60 * ɑ96)x5 + (ɑ60 * 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:
ɑ60x22 + ɑ(275 % 255)x21 + ɑ(294 % 255)x20 + ɑ218x19 + ɑ154x18 + ɑ244x17 + ɑ157x16 + ɑ178x15 + ɑ230x14 + ɑ139x13 + ɑ247x12 + ɑ212x11 + ɑ208x10 + ɑ(312 % 255)x9 + ɑ239x8 + ɑ65x7 + ɑ158x6 + ɑ156x5 + ɑ213x4The result is:
ɑ60x22 + ɑ20x21 + ɑ39x20 + ɑ218x19 + ɑ154x18 + ɑ244x17 + ɑ157x16 + ɑ178x15 + ɑ230x14 + ɑ139x13 + ɑ247x12 + ɑ212x11 + ɑ208x10 + ɑ57x9 + ɑ239x8 + ɑ65x7 + ɑ158x6 + ɑ156x5 + ɑ213x4
Now, convert this to integer notation:
185x22 + 180x21 + 53x20 + 43x19 + 57x18 + 250x17 + 213x16 + 171x15 + 244x14 + 66x13 + 131x12 + 121x11 + 81x10 + 186x9 + 22x8 + 190x7 + 183x6 + 228x5 + 242x4Step 11b: XOR the result with the result from step 10b
Use the result from step 10b to perform the next XOR.
(185 ⊕ 185)x22 + (191 ⊕ 180)x21 + (31 ⊕ 53)x20 + (88 ⊕ 43)x19 + (121 ⊕ 57)x18 + (130 ⊕ 250)x17 + (121 ⊕ 213)x16 + (198 ⊕ 171)x15 + (38 ⊕ 244)x14 + (108 ⊕ 66)x13 + (54 ⊕ 131)x12 + (212 ⊕ 121)x11 + (224 ⊕ 81)x10 + (187 ⊕ 186)x9 + (181 ⊕ 22)x8 + (53 ⊕ 190)x7 + (195 ⊕ 183)x6 + (147 ⊕ 228)x5 + (0 ⊕ 242)x4
The result is:
0x22 + 11x21 + 42x20 + 115x19 + 64x18 + 120x17 + 172x16 + 109x15 + 210x14 + 46x13 + 181x12 + 173x11 + 177x10 + 1x9 + 163x8 + 139x7 + 116x6 + 119x5 + 242x4
Discard the lead 0 term to get:
11x21 + 42x20 + 115x19 + 64x18 + 120x17 + 172x16 + 109x15 + 210x14 + 46x13 + 181x12 + 173x11 + 177x10 + 1x9 + 163x8 + 139x7 + 116x6 + 119x5 + 242x4
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 11x21. Convert 11x21 to alpha notation. According to the log antilog table, for the integer value 11, the alpha exponent is 238. Therefore 11 = ɑ238. Multiply the generator polynomial by ɑ238:
(ɑ238 * ɑ0)x21 + (ɑ238 * ɑ215)x20 + (ɑ238 * ɑ234)x19 + (ɑ238 * ɑ158)x18 + (ɑ238 * ɑ94)x17 + (ɑ238 * ɑ184)x16 + (ɑ238 * ɑ97)x15 + (ɑ238 * ɑ118)x14 + (ɑ238 * ɑ170)x13 + (ɑ238 * ɑ79)x12 + (ɑ238 * ɑ187)x11 + (ɑ238 * ɑ152)x10 + (ɑ238 * ɑ148)x9 + (ɑ238 * ɑ252)x8 + (ɑ238 * ɑ179)x7 + (ɑ238 * ɑ5)x6 + (ɑ238 * ɑ98)x5 + (ɑ238 * ɑ96)x4 + (ɑ238 * 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:
ɑ238x21 + ɑ(453 % 255)x20 + ɑ(472 % 255)x19 + ɑ(396 % 255)x18 + ɑ(332 % 255)x17 + ɑ(422 % 255)x16 + ɑ(335 % 255)x15 + ɑ(356 % 255)x14 + ɑ(408 % 255)x13 + ɑ(317 % 255)x12 + ɑ(425 % 255)x11 + ɑ(390 % 255)x10 + ɑ(386 % 255)x9 + ɑ(490 % 255)x8 + ɑ(417 % 255)x7 + ɑ243x6 + ɑ(336 % 255)x5 + ɑ(334 % 255)x4 + ɑ(391 % 255)x3The result is:
ɑ238x21 + ɑ198x20 + ɑ217x19 + ɑ141x18 + ɑ77x17 + ɑ167x16 + ɑ80x15 + ɑ101x14 + ɑ153x13 + ɑ62x12 + ɑ170x11 + ɑ135x10 + ɑ131x9 + ɑ235x8 + ɑ162x7 + ɑ243x6 + ɑ81x5 + ɑ79x4 + ɑ136x3
Now, convert this to integer notation:
11x21 + 7x20 + 155x19 + 21x18 + 60x17 + 126x16 + 253x15 + 34x14 + 146x13 + 222x12 + 215x11 + 169x10 + 92x9 + 235x8 + 191x7 + 125x6 + 231x5 + 240x4 + 79x3Step 12b: XOR the result with the result from step 11b
Use the result from step 11b to perform the next XOR.
(11 ⊕ 11)x21 + (42 ⊕ 7)x20 + (115 ⊕ 155)x19 + (64 ⊕ 21)x18 + (120 ⊕ 60)x17 + (172 ⊕ 126)x16 + (109 ⊕ 253)x15 + (210 ⊕ 34)x14 + (46 ⊕ 146)x13 + (181 ⊕ 222)x12 + (173 ⊕ 215)x11 + (177 ⊕ 169)x10 + (1 ⊕ 92)x9 + (163 ⊕ 235)x8 + (139 ⊕ 191)x7 + (116 ⊕ 125)x6 + (119 ⊕ 231)x5 + (242 ⊕ 240)x4 + (0 ⊕ 79)x3
The result is:
0x21 + 45x20 + 232x19 + 85x18 + 68x17 + 210x16 + 144x15 + 240x14 + 188x13 + 107x12 + 122x11 + 24x10 + 93x9 + 72x8 + 52x7 + 9x6 + 144x5 + 2x4 + 79x3
Discard the lead 0 term to get:
45x20 + 232x19 + 85x18 + 68x17 + 210x16 + 144x15 + 240x14 + 188x13 + 107x12 + 122x11 + 24x10 + 93x9 + 72x8 + 52x7 + 9x6 + 144x5 + 2x4 + 79x3
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 45x20. Convert 45x20 to alpha notation. According to the log antilog table, for the integer value 45, the alpha exponent is 18. Therefore 45 = ɑ18. Multiply the generator polynomial by ɑ18:
(ɑ18 * ɑ0)x20 + (ɑ18 * ɑ215)x19 + (ɑ18 * ɑ234)x18 + (ɑ18 * ɑ158)x17 + (ɑ18 * ɑ94)x16 + (ɑ18 * ɑ184)x15 + (ɑ18 * ɑ97)x14 + (ɑ18 * ɑ118)x13 + (ɑ18 * ɑ170)x12 + (ɑ18 * ɑ79)x11 + (ɑ18 * ɑ187)x10 + (ɑ18 * ɑ152)x9 + (ɑ18 * ɑ148)x8 + (ɑ18 * ɑ252)x7 + (ɑ18 * ɑ179)x6 + (ɑ18 * ɑ5)x5 + (ɑ18 * ɑ98)x4 + (ɑ18 * ɑ96)x3 + (ɑ18 * 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:
ɑ18x20 + ɑ233x19 + ɑ252x18 + ɑ176x17 + ɑ112x16 + ɑ202x15 + ɑ115x14 + ɑ136x13 + ɑ188x12 + ɑ97x11 + ɑ205x10 + ɑ170x9 + ɑ166x8 + ɑ(270 % 255)x7 + ɑ197x6 + ɑ23x5 + ɑ116x4 + ɑ114x3 + ɑ171x2The result is:
ɑ18x20 + ɑ233x19 + ɑ252x18 + ɑ176x17 + ɑ112x16 + ɑ202x15 + ɑ115x14 + ɑ136x13 + ɑ188x12 + ɑ97x11 + ɑ205x10 + ɑ170x9 + ɑ166x8 + ɑ15x7 + ɑ197x6 + ɑ23x5 + ɑ116x4 + ɑ114x3 + ɑ171x2
Now, convert this to integer notation:
45x20 + 243x19 + 173x18 + 227x17 + 129x16 + 112x15 + 124x14 + 79x13 + 165x12 + 175x11 + 167x10 + 215x9 + 63x8 + 38x7 + 141x6 + 201x5 + 248x4 + 62x3 + 179x2Step 13b: XOR the result with the result from step 12b
Use the result from step 12b to perform the next XOR.
(45 ⊕ 45)x20 + (232 ⊕ 243)x19 + (85 ⊕ 173)x18 + (68 ⊕ 227)x17 + (210 ⊕ 129)x16 + (144 ⊕ 112)x15 + (240 ⊕ 124)x14 + (188 ⊕ 79)x13 + (107 ⊕ 165)x12 + (122 ⊕ 175)x11 + (24 ⊕ 167)x10 + (93 ⊕ 215)x9 + (72 ⊕ 63)x8 + (52 ⊕ 38)x7 + (9 ⊕ 141)x6 + (144 ⊕ 201)x5 + (2 ⊕ 248)x4 + (79 ⊕ 62)x3 + (0 ⊕ 179)x2
The result is:
0x20 + 27x19 + 248x18 + 167x17 + 83x16 + 224x15 + 140x14 + 243x13 + 206x12 + 213x11 + 191x10 + 138x9 + 119x8 + 18x7 + 132x6 + 89x5 + 250x4 + 113x3 + 179x2
Discard the lead 0 term to get:
27x19 + 248x18 + 167x17 + 83x16 + 224x15 + 140x14 + 243x13 + 206x12 + 213x11 + 191x10 + 138x9 + 119x8 + 18x7 + 132x6 + 89x5 + 250x4 + 113x3 + 179x2
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 27x19. Convert 27x19 to alpha notation. According to the log antilog table, for the integer value 27, the alpha exponent is 248. Therefore 27 = ɑ248. Multiply the generator polynomial by ɑ248:
(ɑ248 * ɑ0)x19 + (ɑ248 * ɑ215)x18 + (ɑ248 * ɑ234)x17 + (ɑ248 * ɑ158)x16 + (ɑ248 * ɑ94)x15 + (ɑ248 * ɑ184)x14 + (ɑ248 * ɑ97)x13 + (ɑ248 * ɑ118)x12 + (ɑ248 * ɑ170)x11 + (ɑ248 * ɑ79)x10 + (ɑ248 * ɑ187)x9 + (ɑ248 * ɑ152)x8 + (ɑ248 * ɑ148)x7 + (ɑ248 * ɑ252)x6 + (ɑ248 * ɑ179)x5 + (ɑ248 * ɑ5)x4 + (ɑ248 * ɑ98)x3 + (ɑ248 * ɑ96)x2 + (ɑ248 * 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:
ɑ248x19 + ɑ(463 % 255)x18 + ɑ(482 % 255)x17 + ɑ(406 % 255)x16 + ɑ(342 % 255)x15 + ɑ(432 % 255)x14 + ɑ(345 % 255)x13 + ɑ(366 % 255)x12 + ɑ(418 % 255)x11 + ɑ(327 % 255)x10 + ɑ(435 % 255)x9 + ɑ(400 % 255)x8 + ɑ(396 % 255)x7 + ɑ(500 % 255)x6 + ɑ(427 % 255)x5 + ɑ253x4 + ɑ(346 % 255)x3 + ɑ(344 % 255)x2 + ɑ(401 % 255)x1The result is:
ɑ248x19 + ɑ208x18 + ɑ227x17 + ɑ151x16 + ɑ87x15 + ɑ177x14 + ɑ90x13 + ɑ111x12 + ɑ163x11 + ɑ72x10 + ɑ180x9 + ɑ145x8 + ɑ141x7 + ɑ245x6 + ɑ172x5 + ɑ253x4 + ɑ91x3 + ɑ89x2 + ɑ146x1
Now, convert this to integer notation:
27x19 + 81x18 + 144x17 + 170x16 + 127x15 + 219x14 + 223x13 + 206x12 + 99x11 + 101x10 + 150x9 + 77x8 + 21x7 + 233x6 + 123x5 + 71x4 + 163x3 + 225x2 + 154x1Step 14b: XOR the result with the result from step 13b
Use the result from step 13b to perform the next XOR.
(27 ⊕ 27)x19 + (248 ⊕ 81)x18 + (167 ⊕ 144)x17 + (83 ⊕ 170)x16 + (224 ⊕ 127)x15 + (140 ⊕ 219)x14 + (243 ⊕ 223)x13 + (206 ⊕ 206)x12 + (213 ⊕ 99)x11 + (191 ⊕ 101)x10 + (138 ⊕ 150)x9 + (119 ⊕ 77)x8 + (18 ⊕ 21)x7 + (132 ⊕ 233)x6 + (89 ⊕ 123)x5 + (250 ⊕ 71)x4 + (113 ⊕ 163)x3 + (179 ⊕ 225)x2 + (0 ⊕ 154)x1
The result is:
0x19 + 169x18 + 55x17 + 249x16 + 159x15 + 87x14 + 44x13 + 0x12 + 182x11 + 218x10 + 28x9 + 58x8 + 7x7 + 109x6 + 34x5 + 189x4 + 210x3 + 82x2 + 154x1
Discard the lead 0 term to get:
169x18 + 55x17 + 249x16 + 159x15 + 87x14 + 44x13 + 0x12 + 182x11 + 218x10 + 28x9 + 58x8 + 7x7 + 109x6 + 34x5 + 189x4 + 210x3 + 82x2 + 154x1
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 169x18. Convert 169x18 to alpha notation. According to the log antilog table, for the integer value 169, the alpha exponent is 135. Therefore 169 = ɑ135. Multiply the generator polynomial by ɑ135:
(ɑ135 * ɑ0)x18 + (ɑ135 * ɑ215)x17 + (ɑ135 * ɑ234)x16 + (ɑ135 * ɑ158)x15 + (ɑ135 * ɑ94)x14 + (ɑ135 * ɑ184)x13 + (ɑ135 * ɑ97)x12 + (ɑ135 * ɑ118)x11 + (ɑ135 * ɑ170)x10 + (ɑ135 * ɑ79)x9 + (ɑ135 * ɑ187)x8 + (ɑ135 * ɑ152)x7 + (ɑ135 * ɑ148)x6 + (ɑ135 * ɑ252)x5 + (ɑ135 * ɑ179)x4 + (ɑ135 * ɑ5)x3 + (ɑ135 * ɑ98)x2 + (ɑ135 * ɑ96)x1 + (ɑ135 * 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:
ɑ135x18 + ɑ(350 % 255)x17 + ɑ(369 % 255)x16 + ɑ(293 % 255)x15 + ɑ229x14 + ɑ(319 % 255)x13 + ɑ232x12 + ɑ253x11 + ɑ(305 % 255)x10 + ɑ214x9 + ɑ(322 % 255)x8 + ɑ(287 % 255)x7 + ɑ(283 % 255)x6 + ɑ(387 % 255)x5 + ɑ(314 % 255)x4 + ɑ140x3 + ɑ233x2 + ɑ231x1 + ɑ(288 % 255)x0The result is:
ɑ135x18 + ɑ95x17 + ɑ114x16 + ɑ38x15 + ɑ229x14 + ɑ64x13 + ɑ232x12 + ɑ253x11 + ɑ50x10 + ɑ214x9 + ɑ67x8 + ɑ32x7 + ɑ28x6 + ɑ132x5 + ɑ59x4 + ɑ140x3 + ɑ233x2 + ɑ231x1 + ɑ33x0
Now, convert this to integer notation:
169x18 + 226x17 + 62x16 + 148x15 + 122x14 + 95x13 + 247x12 + 71x11 + 5x10 + 249x9 + 194x8 + 157x7 + 24x6 + 184x5 + 210x4 + 132x3 + 243x2 + 245x1 + 39Step 15b: XOR the result with the result from step 14b
Use the result from step 14b to perform the next XOR.
(169 ⊕ 169)x18 + (55 ⊕ 226)x17 + (249 ⊕ 62)x16 + (159 ⊕ 148)x15 + (87 ⊕ 122)x14 + (44 ⊕ 95)x13 + (0 ⊕ 247)x12 + (182 ⊕ 71)x11 + (218 ⊕ 5)x10 + (28 ⊕ 249)x9 + (58 ⊕ 194)x8 + (7 ⊕ 157)x7 + (109 ⊕ 24)x6 + (34 ⊕ 184)x5 + (189 ⊕ 210)x4 + (210 ⊕ 132)x3 + (82 ⊕ 243)x2 + (154 ⊕ 245)x1 + (0 ⊕ 39)x0
The result is:
0x18 + 213x17 + 199x16 + 11x15 + 45x14 + 115x13 + 247x12 + 241x11 + 223x10 + 229x9 + 248x8 + 154x7 + 117x6 + 154x5 + 111x4 + 86x3 + 161x2 + 111x1 + 39
Discard the lead 0 term to get:
213x17 + 199x16 + 11x15 + 45x14 + 115x13 + 247x12 + 241x11 + 223x10 + 229x9 + 248x8 + 154x7 + 117x6 + 154x5 + 111x4 + 86x3 + 161x2 + 111x1 + 39
Use the terms of the remainder as the error correction codewords
The division has been performed 15 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:
213 199 11 45 115 247 241 223 229 248 154 117 154 111 86 161 111 39