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:
182x15 + 230x14 + 247x13 + 119x12 + 50x11 + 7x10 + 118x9 + 134x8 + 87x7 + 38x6 + 82x5 + 6x4 + 134x3 + 151x2 + 50x1 + 7
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:
182x33 + 230x32 + 247x31 + 119x30 + 50x29 + 7x28 + 118x27 + 134x26 + 87x25 + 38x24 + 82x23 + 6x22 + 134x21 + 151x20 + 50x19 + 7x18
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 182x33. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 182x33 to alpha notation. According to the log antilog table, for the integer value 182, the alpha exponent is 93. Therefore 182 = ɑ93. Multiply the generator polynomial by ɑ93:
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:
ɑ93x33 + ɑ(308 % 255)x32 + ɑ(327 % 255)x31 + ɑ251x30 + ɑ187x29 + ɑ(277 % 255)x28 + ɑ190x27 + ɑ211x26 + ɑ(263 % 255)x25 + ɑ172x24 + ɑ(280 % 255)x23 + ɑ245x22 + ɑ241x21 + ɑ(345 % 255)x20 + ɑ(272 % 255)x19 + ɑ98x18 + ɑ191x17 + ɑ189x16 + ɑ246x15The result is:
ɑ93x33 + ɑ53x32 + ɑ72x31 + ɑ251x30 + ɑ187x29 + ɑ22x28 + ɑ190x27 + ɑ211x26 + ɑ8x25 + ɑ172x24 + ɑ25x23 + ɑ245x22 + ɑ241x21 + ɑ90x20 + ɑ17x19 + ɑ98x18 + ɑ191x17 + ɑ189x16 + ɑ246x15
Now, convert this to integer notation:
182x33 + 40x32 + 101x31 + 216x30 + 220x29 + 234x28 + 174x27 + 178x26 + 29x25 + 123x24 + 3x23 + 233x22 + 88x21 + 223x20 + 152x19 + 67x18 + 65x17 + 87x16 + 207x15Step 1b: XOR the result with the message polynomial
Since this is the first division step, XOR the result from 1a with the message polynomial.
(182 ⊕ 182)x33 + (230 ⊕ 40)x32 + (247 ⊕ 101)x31 + (119 ⊕ 216)x30 + (50 ⊕ 220)x29 + (7 ⊕ 234)x28 + (118 ⊕ 174)x27 + (134 ⊕ 178)x26 + (87 ⊕ 29)x25 + (38 ⊕ 123)x24 + (82 ⊕ 3)x23 + (6 ⊕ 233)x22 + (134 ⊕ 88)x21 + (151 ⊕ 223)x20 + (50 ⊕ 152)x19 + (7 ⊕ 67)x18 + (0 ⊕ 65)x17 + (0 ⊕ 87)x16 + (0 ⊕ 207)x15
The result is:
0x33 + 206x32 + 146x31 + 175x30 + 238x29 + 237x28 + 216x27 + 52x26 + 74x25 + 93x24 + 81x23 + 239x22 + 222x21 + 72x20 + 170x19 + 68x18 + 65x17 + 87x16 + 207x15
Discard the lead 0 term to get:
206x32 + 146x31 + 175x30 + 238x29 + 237x28 + 216x27 + 52x26 + 74x25 + 93x24 + 81x23 + 239x22 + 222x21 + 72x20 + 170x19 + 68x18 + 65x17 + 87x16 + 207x15
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 206x32. Convert 206x32 to alpha notation. According to the log antilog table, for the integer value 206, the alpha exponent is 111. Therefore 206 = ɑ111. Multiply the generator polynomial by ɑ111:
(ɑ111 * ɑ0)x32 + (ɑ111 * ɑ215)x31 + (ɑ111 * ɑ234)x30 + (ɑ111 * ɑ158)x29 + (ɑ111 * ɑ94)x28 + (ɑ111 * ɑ184)x27 + (ɑ111 * ɑ97)x26 + (ɑ111 * ɑ118)x25 + (ɑ111 * ɑ170)x24 + (ɑ111 * ɑ79)x23 + (ɑ111 * ɑ187)x22 + (ɑ111 * ɑ152)x21 + (ɑ111 * ɑ148)x20 + (ɑ111 * ɑ252)x19 + (ɑ111 * ɑ179)x18 + (ɑ111 * ɑ5)x17 + (ɑ111 * ɑ98)x16 + (ɑ111 * ɑ96)x15 + (ɑ111 * 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:
ɑ111x32 + ɑ(326 % 255)x31 + ɑ(345 % 255)x30 + ɑ(269 % 255)x29 + ɑ205x28 + ɑ(295 % 255)x27 + ɑ208x26 + ɑ229x25 + ɑ(281 % 255)x24 + ɑ190x23 + ɑ(298 % 255)x22 + ɑ(263 % 255)x21 + ɑ(259 % 255)x20 + ɑ(363 % 255)x19 + ɑ(290 % 255)x18 + ɑ116x17 + ɑ209x16 + ɑ207x15 + ɑ(264 % 255)x14The result is:
ɑ111x32 + ɑ71x31 + ɑ90x30 + ɑ14x29 + ɑ205x28 + ɑ40x27 + ɑ208x26 + ɑ229x25 + ɑ26x24 + ɑ190x23 + ɑ43x22 + ɑ8x21 + ɑ4x20 + ɑ108x19 + ɑ35x18 + ɑ116x17 + ɑ209x16 + ɑ207x15 + ɑ9x14
Now, convert this to integer notation:
206x32 + 188x31 + 223x30 + 19x29 + 167x28 + 106x27 + 81x26 + 122x25 + 6x24 + 174x23 + 119x22 + 29x21 + 16x20 + 208x19 + 156x18 + 248x17 + 162x16 + 166x15 + 58x14Step 2b: XOR the result with the result from step 1b
Use the result from step 1b to perform the next XOR.
(206 ⊕ 206)x32 + (146 ⊕ 188)x31 + (175 ⊕ 223)x30 + (238 ⊕ 19)x29 + (237 ⊕ 167)x28 + (216 ⊕ 106)x27 + (52 ⊕ 81)x26 + (74 ⊕ 122)x25 + (93 ⊕ 6)x24 + (81 ⊕ 174)x23 + (239 ⊕ 119)x22 + (222 ⊕ 29)x21 + (72 ⊕ 16)x20 + (170 ⊕ 208)x19 + (68 ⊕ 156)x18 + (65 ⊕ 248)x17 + (87 ⊕ 162)x16 + (207 ⊕ 166)x15 + (0 ⊕ 58)x14
The result is:
0x32 + 46x31 + 112x30 + 253x29 + 74x28 + 178x27 + 101x26 + 48x25 + 91x24 + 255x23 + 152x22 + 195x21 + 88x20 + 122x19 + 216x18 + 185x17 + 245x16 + 105x15 + 58x14
Discard the lead 0 term to get:
46x31 + 112x30 + 253x29 + 74x28 + 178x27 + 101x26 + 48x25 + 91x24 + 255x23 + 152x22 + 195x21 + 88x20 + 122x19 + 216x18 + 185x17 + 245x16 + 105x15 + 58x14
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 46x31. Convert 46x31 to alpha notation. According to the log antilog table, for the integer value 46, the alpha exponent is 130. Therefore 46 = ɑ130. Multiply the generator polynomial by ɑ130:
(ɑ130 * ɑ0)x31 + (ɑ130 * ɑ215)x30 + (ɑ130 * ɑ234)x29 + (ɑ130 * ɑ158)x28 + (ɑ130 * ɑ94)x27 + (ɑ130 * ɑ184)x26 + (ɑ130 * ɑ97)x25 + (ɑ130 * ɑ118)x24 + (ɑ130 * ɑ170)x23 + (ɑ130 * ɑ79)x22 + (ɑ130 * ɑ187)x21 + (ɑ130 * ɑ152)x20 + (ɑ130 * ɑ148)x19 + (ɑ130 * ɑ252)x18 + (ɑ130 * ɑ179)x17 + (ɑ130 * ɑ5)x16 + (ɑ130 * ɑ98)x15 + (ɑ130 * ɑ96)x14 + (ɑ130 * 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:
ɑ130x31 + ɑ(345 % 255)x30 + ɑ(364 % 255)x29 + ɑ(288 % 255)x28 + ɑ224x27 + ɑ(314 % 255)x26 + ɑ227x25 + ɑ248x24 + ɑ(300 % 255)x23 + ɑ209x22 + ɑ(317 % 255)x21 + ɑ(282 % 255)x20 + ɑ(278 % 255)x19 + ɑ(382 % 255)x18 + ɑ(309 % 255)x17 + ɑ135x16 + ɑ228x15 + ɑ226x14 + ɑ(283 % 255)x13The result is:
ɑ130x31 + ɑ90x30 + ɑ109x29 + ɑ33x28 + ɑ224x27 + ɑ59x26 + ɑ227x25 + ɑ248x24 + ɑ45x23 + ɑ209x22 + ɑ62x21 + ɑ27x20 + ɑ23x19 + ɑ127x18 + ɑ54x17 + ɑ135x16 + ɑ228x15 + ɑ226x14 + ɑ28x13
Now, convert this to integer notation:
46x31 + 223x30 + 189x29 + 39x28 + 18x27 + 210x26 + 144x25 + 27x24 + 193x23 + 162x22 + 222x21 + 12x20 + 201x19 + 204x18 + 80x17 + 169x16 + 61x15 + 72x14 + 24x13Step 3b: XOR the result with the result from step 2b
Use the result from step 2b to perform the next XOR.
(46 ⊕ 46)x31 + (112 ⊕ 223)x30 + (253 ⊕ 189)x29 + (74 ⊕ 39)x28 + (178 ⊕ 18)x27 + (101 ⊕ 210)x26 + (48 ⊕ 144)x25 + (91 ⊕ 27)x24 + (255 ⊕ 193)x23 + (152 ⊕ 162)x22 + (195 ⊕ 222)x21 + (88 ⊕ 12)x20 + (122 ⊕ 201)x19 + (216 ⊕ 204)x18 + (185 ⊕ 80)x17 + (245 ⊕ 169)x16 + (105 ⊕ 61)x15 + (58 ⊕ 72)x14 + (0 ⊕ 24)x13
The result is:
0x31 + 175x30 + 64x29 + 109x28 + 160x27 + 183x26 + 160x25 + 64x24 + 62x23 + 58x22 + 29x21 + 84x20 + 179x19 + 20x18 + 233x17 + 92x16 + 84x15 + 114x14 + 24x13
Discard the lead 0 term to get:
175x30 + 64x29 + 109x28 + 160x27 + 183x26 + 160x25 + 64x24 + 62x23 + 58x22 + 29x21 + 84x20 + 179x19 + 20x18 + 233x17 + 92x16 + 84x15 + 114x14 + 24x13
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 175x30. Convert 175x30 to alpha notation. According to the log antilog table, for the integer value 175, the alpha exponent is 97. Therefore 175 = ɑ97. Multiply the generator polynomial by ɑ97:
(ɑ97 * ɑ0)x30 + (ɑ97 * ɑ215)x29 + (ɑ97 * ɑ234)x28 + (ɑ97 * ɑ158)x27 + (ɑ97 * ɑ94)x26 + (ɑ97 * ɑ184)x25 + (ɑ97 * ɑ97)x24 + (ɑ97 * ɑ118)x23 + (ɑ97 * ɑ170)x22 + (ɑ97 * ɑ79)x21 + (ɑ97 * ɑ187)x20 + (ɑ97 * ɑ152)x19 + (ɑ97 * ɑ148)x18 + (ɑ97 * ɑ252)x17 + (ɑ97 * ɑ179)x16 + (ɑ97 * ɑ5)x15 + (ɑ97 * ɑ98)x14 + (ɑ97 * ɑ96)x13 + (ɑ97 * 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:
ɑ97x30 + ɑ(312 % 255)x29 + ɑ(331 % 255)x28 + ɑ255x27 + ɑ191x26 + ɑ(281 % 255)x25 + ɑ194x24 + ɑ215x23 + ɑ(267 % 255)x22 + ɑ176x21 + ɑ(284 % 255)x20 + ɑ249x19 + ɑ245x18 + ɑ(349 % 255)x17 + ɑ(276 % 255)x16 + ɑ102x15 + ɑ195x14 + ɑ193x13 + ɑ250x12The result is:
ɑ97x30 + ɑ57x29 + ɑ76x28 + ɑ0x27 + ɑ191x26 + ɑ26x25 + ɑ194x24 + ɑ215x23 + ɑ12x22 + ɑ176x21 + ɑ29x20 + ɑ249x19 + ɑ245x18 + ɑ94x17 + ɑ21x16 + ɑ102x15 + ɑ195x14 + ɑ193x13 + ɑ250x12
Now, convert this to integer notation:
175x30 + 186x29 + 30x28 + 1x27 + 65x26 + 6x25 + 50x24 + 239x23 + 205x22 + 227x21 + 48x20 + 54x19 + 233x18 + 113x17 + 117x16 + 68x15 + 100x14 + 25x13 + 108x12Step 4b: XOR the result with the result from step 3b
Use the result from step 3b to perform the next XOR.
(175 ⊕ 175)x30 + (64 ⊕ 186)x29 + (109 ⊕ 30)x28 + (160 ⊕ 1)x27 + (183 ⊕ 65)x26 + (160 ⊕ 6)x25 + (64 ⊕ 50)x24 + (62 ⊕ 239)x23 + (58 ⊕ 205)x22 + (29 ⊕ 227)x21 + (84 ⊕ 48)x20 + (179 ⊕ 54)x19 + (20 ⊕ 233)x18 + (233 ⊕ 113)x17 + (92 ⊕ 117)x16 + (84 ⊕ 68)x15 + (114 ⊕ 100)x14 + (24 ⊕ 25)x13 + (0 ⊕ 108)x12
The result is:
0x30 + 250x29 + 115x28 + 161x27 + 246x26 + 166x25 + 114x24 + 209x23 + 247x22 + 254x21 + 100x20 + 133x19 + 253x18 + 152x17 + 41x16 + 16x15 + 22x14 + 1x13 + 108x12
Discard the lead 0 term to get:
250x29 + 115x28 + 161x27 + 246x26 + 166x25 + 114x24 + 209x23 + 247x22 + 254x21 + 100x20 + 133x19 + 253x18 + 152x17 + 41x16 + 16x15 + 22x14 + 1x13 + 108x12
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 250x29. Convert 250x29 to alpha notation. According to the log antilog table, for the integer value 250, the alpha exponent is 244. Therefore 250 = ɑ244. Multiply the generator polynomial by ɑ244:
(ɑ244 * ɑ0)x29 + (ɑ244 * ɑ215)x28 + (ɑ244 * ɑ234)x27 + (ɑ244 * ɑ158)x26 + (ɑ244 * ɑ94)x25 + (ɑ244 * ɑ184)x24 + (ɑ244 * ɑ97)x23 + (ɑ244 * ɑ118)x22 + (ɑ244 * ɑ170)x21 + (ɑ244 * ɑ79)x20 + (ɑ244 * ɑ187)x19 + (ɑ244 * ɑ152)x18 + (ɑ244 * ɑ148)x17 + (ɑ244 * ɑ252)x16 + (ɑ244 * ɑ179)x15 + (ɑ244 * ɑ5)x14 + (ɑ244 * ɑ98)x13 + (ɑ244 * ɑ96)x12 + (ɑ244 * 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:
ɑ244x29 + ɑ(459 % 255)x28 + ɑ(478 % 255)x27 + ɑ(402 % 255)x26 + ɑ(338 % 255)x25 + ɑ(428 % 255)x24 + ɑ(341 % 255)x23 + ɑ(362 % 255)x22 + ɑ(414 % 255)x21 + ɑ(323 % 255)x20 + ɑ(431 % 255)x19 + ɑ(396 % 255)x18 + ɑ(392 % 255)x17 + ɑ(496 % 255)x16 + ɑ(423 % 255)x15 + ɑ249x14 + ɑ(342 % 255)x13 + ɑ(340 % 255)x12 + ɑ(397 % 255)x11The result is:
ɑ244x29 + ɑ204x28 + ɑ223x27 + ɑ147x26 + ɑ83x25 + ɑ173x24 + ɑ86x23 + ɑ107x22 + ɑ159x21 + ɑ68x20 + ɑ176x19 + ɑ141x18 + ɑ137x17 + ɑ241x16 + ɑ168x15 + ɑ249x14 + ɑ87x13 + ɑ85x12 + ɑ142x11
Now, convert this to integer notation:
250x29 + 221x28 + 9x27 + 41x26 + 187x25 + 246x24 + 177x23 + 104x22 + 115x21 + 153x20 + 227x19 + 21x18 + 158x17 + 88x16 + 252x15 + 54x14 + 127x13 + 214x12 + 42x11Step 5b: XOR the result with the result from step 4b
Use the result from step 4b to perform the next XOR.
(250 ⊕ 250)x29 + (115 ⊕ 221)x28 + (161 ⊕ 9)x27 + (246 ⊕ 41)x26 + (166 ⊕ 187)x25 + (114 ⊕ 246)x24 + (209 ⊕ 177)x23 + (247 ⊕ 104)x22 + (254 ⊕ 115)x21 + (100 ⊕ 153)x20 + (133 ⊕ 227)x19 + (253 ⊕ 21)x18 + (152 ⊕ 158)x17 + (41 ⊕ 88)x16 + (16 ⊕ 252)x15 + (22 ⊕ 54)x14 + (1 ⊕ 127)x13 + (108 ⊕ 214)x12 + (0 ⊕ 42)x11
The result is:
0x29 + 174x28 + 168x27 + 223x26 + 29x25 + 132x24 + 96x23 + 159x22 + 141x21 + 253x20 + 102x19 + 232x18 + 6x17 + 113x16 + 236x15 + 32x14 + 126x13 + 186x12 + 42x11
Discard the lead 0 term to get:
174x28 + 168x27 + 223x26 + 29x25 + 132x24 + 96x23 + 159x22 + 141x21 + 253x20 + 102x19 + 232x18 + 6x17 + 113x16 + 236x15 + 32x14 + 126x13 + 186x12 + 42x11
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 174x28. Convert 174x28 to alpha notation. According to the log antilog table, for the integer value 174, the alpha exponent is 190. Therefore 174 = ɑ190. Multiply the generator polynomial by ɑ190:
(ɑ190 * ɑ0)x28 + (ɑ190 * ɑ215)x27 + (ɑ190 * ɑ234)x26 + (ɑ190 * ɑ158)x25 + (ɑ190 * ɑ94)x24 + (ɑ190 * ɑ184)x23 + (ɑ190 * ɑ97)x22 + (ɑ190 * ɑ118)x21 + (ɑ190 * ɑ170)x20 + (ɑ190 * ɑ79)x19 + (ɑ190 * ɑ187)x18 + (ɑ190 * ɑ152)x17 + (ɑ190 * ɑ148)x16 + (ɑ190 * ɑ252)x15 + (ɑ190 * ɑ179)x14 + (ɑ190 * ɑ5)x13 + (ɑ190 * ɑ98)x12 + (ɑ190 * ɑ96)x11 + (ɑ190 * 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:
ɑ190x28 + ɑ(405 % 255)x27 + ɑ(424 % 255)x26 + ɑ(348 % 255)x25 + ɑ(284 % 255)x24 + ɑ(374 % 255)x23 + ɑ(287 % 255)x22 + ɑ(308 % 255)x21 + ɑ(360 % 255)x20 + ɑ(269 % 255)x19 + ɑ(377 % 255)x18 + ɑ(342 % 255)x17 + ɑ(338 % 255)x16 + ɑ(442 % 255)x15 + ɑ(369 % 255)x14 + ɑ195x13 + ɑ(288 % 255)x12 + ɑ(286 % 255)x11 + ɑ(343 % 255)x10The result is:
ɑ190x28 + ɑ150x27 + ɑ169x26 + ɑ93x25 + ɑ29x24 + ɑ119x23 + ɑ32x22 + ɑ53x21 + ɑ105x20 + ɑ14x19 + ɑ122x18 + ɑ87x17 + ɑ83x16 + ɑ187x15 + ɑ114x14 + ɑ195x13 + ɑ33x12 + ɑ31x11 + ɑ88x10
Now, convert this to integer notation:
174x28 + 85x27 + 229x26 + 182x25 + 48x24 + 147x23 + 157x22 + 40x21 + 26x20 + 19x19 + 236x18 + 127x17 + 187x16 + 220x15 + 62x14 + 100x13 + 39x12 + 192x11 + 254x10Step 6b: XOR the result with the result from step 5b
Use the result from step 5b to perform the next XOR.
(174 ⊕ 174)x28 + (168 ⊕ 85)x27 + (223 ⊕ 229)x26 + (29 ⊕ 182)x25 + (132 ⊕ 48)x24 + (96 ⊕ 147)x23 + (159 ⊕ 157)x22 + (141 ⊕ 40)x21 + (253 ⊕ 26)x20 + (102 ⊕ 19)x19 + (232 ⊕ 236)x18 + (6 ⊕ 127)x17 + (113 ⊕ 187)x16 + (236 ⊕ 220)x15 + (32 ⊕ 62)x14 + (126 ⊕ 100)x13 + (186 ⊕ 39)x12 + (42 ⊕ 192)x11 + (0 ⊕ 254)x10
The result is:
0x28 + 253x27 + 58x26 + 171x25 + 180x24 + 243x23 + 2x22 + 165x21 + 231x20 + 117x19 + 4x18 + 121x17 + 202x16 + 48x15 + 30x14 + 26x13 + 157x12 + 234x11 + 254x10
Discard the lead 0 term to get:
253x27 + 58x26 + 171x25 + 180x24 + 243x23 + 2x22 + 165x21 + 231x20 + 117x19 + 4x18 + 121x17 + 202x16 + 48x15 + 30x14 + 26x13 + 157x12 + 234x11 + 254x10
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 253x27. Convert 253x27 to alpha notation. According to the log antilog table, for the integer value 253, the alpha exponent is 80. Therefore 253 = ɑ80. Multiply the generator polynomial by ɑ80:
(ɑ80 * ɑ0)x27 + (ɑ80 * ɑ215)x26 + (ɑ80 * ɑ234)x25 + (ɑ80 * ɑ158)x24 + (ɑ80 * ɑ94)x23 + (ɑ80 * ɑ184)x22 + (ɑ80 * ɑ97)x21 + (ɑ80 * ɑ118)x20 + (ɑ80 * ɑ170)x19 + (ɑ80 * ɑ79)x18 + (ɑ80 * ɑ187)x17 + (ɑ80 * ɑ152)x16 + (ɑ80 * ɑ148)x15 + (ɑ80 * ɑ252)x14 + (ɑ80 * ɑ179)x13 + (ɑ80 * ɑ5)x12 + (ɑ80 * ɑ98)x11 + (ɑ80 * ɑ96)x10 + (ɑ80 * 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:
ɑ80x27 + ɑ(295 % 255)x26 + ɑ(314 % 255)x25 + ɑ238x24 + ɑ174x23 + ɑ(264 % 255)x22 + ɑ177x21 + ɑ198x20 + ɑ250x19 + ɑ159x18 + ɑ(267 % 255)x17 + ɑ232x16 + ɑ228x15 + ɑ(332 % 255)x14 + ɑ(259 % 255)x13 + ɑ85x12 + ɑ178x11 + ɑ176x10 + ɑ233x9The result is:
ɑ80x27 + ɑ40x26 + ɑ59x25 + ɑ238x24 + ɑ174x23 + ɑ9x22 + ɑ177x21 + ɑ198x20 + ɑ250x19 + ɑ159x18 + ɑ12x17 + ɑ232x16 + ɑ228x15 + ɑ77x14 + ɑ4x13 + ɑ85x12 + ɑ178x11 + ɑ176x10 + ɑ233x9
Now, convert this to integer notation:
253x27 + 106x26 + 210x25 + 11x24 + 241x23 + 58x22 + 219x21 + 7x20 + 108x19 + 115x18 + 205x17 + 247x16 + 61x15 + 60x14 + 16x13 + 214x12 + 171x11 + 227x10 + 243x9Step 7b: XOR the result with the result from step 6b
Use the result from step 6b to perform the next XOR.
(253 ⊕ 253)x27 + (58 ⊕ 106)x26 + (171 ⊕ 210)x25 + (180 ⊕ 11)x24 + (243 ⊕ 241)x23 + (2 ⊕ 58)x22 + (165 ⊕ 219)x21 + (231 ⊕ 7)x20 + (117 ⊕ 108)x19 + (4 ⊕ 115)x18 + (121 ⊕ 205)x17 + (202 ⊕ 247)x16 + (48 ⊕ 61)x15 + (30 ⊕ 60)x14 + (26 ⊕ 16)x13 + (157 ⊕ 214)x12 + (234 ⊕ 171)x11 + (254 ⊕ 227)x10 + (0 ⊕ 243)x9
The result is:
0x27 + 80x26 + 121x25 + 191x24 + 2x23 + 56x22 + 126x21 + 224x20 + 25x19 + 119x18 + 180x17 + 61x16 + 13x15 + 34x14 + 10x13 + 75x12 + 65x11 + 29x10 + 243x9
Discard the lead 0 term to get:
80x26 + 121x25 + 191x24 + 2x23 + 56x22 + 126x21 + 224x20 + 25x19 + 119x18 + 180x17 + 61x16 + 13x15 + 34x14 + 10x13 + 75x12 + 65x11 + 29x10 + 243x9
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 80x26. Convert 80x26 to alpha notation. According to the log antilog table, for the integer value 80, the alpha exponent is 54. Therefore 80 = ɑ54. Multiply the generator polynomial by ɑ54:
(ɑ54 * ɑ0)x26 + (ɑ54 * ɑ215)x25 + (ɑ54 * ɑ234)x24 + (ɑ54 * ɑ158)x23 + (ɑ54 * ɑ94)x22 + (ɑ54 * ɑ184)x21 + (ɑ54 * ɑ97)x20 + (ɑ54 * ɑ118)x19 + (ɑ54 * ɑ170)x18 + (ɑ54 * ɑ79)x17 + (ɑ54 * ɑ187)x16 + (ɑ54 * ɑ152)x15 + (ɑ54 * ɑ148)x14 + (ɑ54 * ɑ252)x13 + (ɑ54 * ɑ179)x12 + (ɑ54 * ɑ5)x11 + (ɑ54 * ɑ98)x10 + (ɑ54 * ɑ96)x9 + (ɑ54 * 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:
ɑ54x26 + ɑ(269 % 255)x25 + ɑ(288 % 255)x24 + ɑ212x23 + ɑ148x22 + ɑ238x21 + ɑ151x20 + ɑ172x19 + ɑ224x18 + ɑ133x17 + ɑ241x16 + ɑ206x15 + ɑ202x14 + ɑ(306 % 255)x13 + ɑ233x12 + ɑ59x11 + ɑ152x10 + ɑ150x9 + ɑ207x8The result is:
ɑ54x26 + ɑ14x25 + ɑ33x24 + ɑ212x23 + ɑ148x22 + ɑ238x21 + ɑ151x20 + ɑ172x19 + ɑ224x18 + ɑ133x17 + ɑ241x16 + ɑ206x15 + ɑ202x14 + ɑ51x13 + ɑ233x12 + ɑ59x11 + ɑ152x10 + ɑ150x9 + ɑ207x8
Now, convert this to integer notation:
80x26 + 19x25 + 39x24 + 121x23 + 82x22 + 11x21 + 170x20 + 123x19 + 18x18 + 109x17 + 88x16 + 83x15 + 112x14 + 10x13 + 243x12 + 210x11 + 73x10 + 85x9 + 166x8Step 8b: XOR the result with the result from step 7b
Use the result from step 7b to perform the next XOR.
(80 ⊕ 80)x26 + (121 ⊕ 19)x25 + (191 ⊕ 39)x24 + (2 ⊕ 121)x23 + (56 ⊕ 82)x22 + (126 ⊕ 11)x21 + (224 ⊕ 170)x20 + (25 ⊕ 123)x19 + (119 ⊕ 18)x18 + (180 ⊕ 109)x17 + (61 ⊕ 88)x16 + (13 ⊕ 83)x15 + (34 ⊕ 112)x14 + (10 ⊕ 10)x13 + (75 ⊕ 243)x12 + (65 ⊕ 210)x11 + (29 ⊕ 73)x10 + (243 ⊕ 85)x9 + (0 ⊕ 166)x8
The result is:
0x26 + 106x25 + 152x24 + 123x23 + 106x22 + 117x21 + 74x20 + 98x19 + 101x18 + 217x17 + 101x16 + 94x15 + 82x14 + 0x13 + 184x12 + 147x11 + 84x10 + 166x9 + 166x8
Discard the lead 0 term to get:
106x25 + 152x24 + 123x23 + 106x22 + 117x21 + 74x20 + 98x19 + 101x18 + 217x17 + 101x16 + 94x15 + 82x14 + 0x13 + 184x12 + 147x11 + 84x10 + 166x9 + 166x8
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 106x25. Convert 106x25 to alpha notation. According to the log antilog table, for the integer value 106, the alpha exponent is 40. Therefore 106 = ɑ40. Multiply the generator polynomial by ɑ40:
(ɑ40 * ɑ0)x25 + (ɑ40 * ɑ215)x24 + (ɑ40 * ɑ234)x23 + (ɑ40 * ɑ158)x22 + (ɑ40 * ɑ94)x21 + (ɑ40 * ɑ184)x20 + (ɑ40 * ɑ97)x19 + (ɑ40 * ɑ118)x18 + (ɑ40 * ɑ170)x17 + (ɑ40 * ɑ79)x16 + (ɑ40 * ɑ187)x15 + (ɑ40 * ɑ152)x14 + (ɑ40 * ɑ148)x13 + (ɑ40 * ɑ252)x12 + (ɑ40 * ɑ179)x11 + (ɑ40 * ɑ5)x10 + (ɑ40 * ɑ98)x9 + (ɑ40 * ɑ96)x8 + (ɑ40 * 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:
ɑ40x25 + ɑ255x24 + ɑ(274 % 255)x23 + ɑ198x22 + ɑ134x21 + ɑ224x20 + ɑ137x19 + ɑ158x18 + ɑ210x17 + ɑ119x16 + ɑ227x15 + ɑ192x14 + ɑ188x13 + ɑ(292 % 255)x12 + ɑ219x11 + ɑ45x10 + ɑ138x9 + ɑ136x8 + ɑ193x7The result is:
ɑ40x25 + ɑ0x24 + ɑ19x23 + ɑ198x22 + ɑ134x21 + ɑ224x20 + ɑ137x19 + ɑ158x18 + ɑ210x17 + ɑ119x16 + ɑ227x15 + ɑ192x14 + ɑ188x13 + ɑ37x12 + ɑ219x11 + ɑ45x10 + ɑ138x9 + ɑ136x8 + ɑ193x7
Now, convert this to integer notation:
106x25 + 1x24 + 90x23 + 7x22 + 218x21 + 18x20 + 158x19 + 183x18 + 89x17 + 147x16 + 144x15 + 130x14 + 165x13 + 74x12 + 86x11 + 193x10 + 33x9 + 79x8 + 25x7Step 9b: XOR the result with the result from step 8b
Use the result from step 8b to perform the next XOR.
(106 ⊕ 106)x25 + (152 ⊕ 1)x24 + (123 ⊕ 90)x23 + (106 ⊕ 7)x22 + (117 ⊕ 218)x21 + (74 ⊕ 18)x20 + (98 ⊕ 158)x19 + (101 ⊕ 183)x18 + (217 ⊕ 89)x17 + (101 ⊕ 147)x16 + (94 ⊕ 144)x15 + (82 ⊕ 130)x14 + (0 ⊕ 165)x13 + (184 ⊕ 74)x12 + (147 ⊕ 86)x11 + (84 ⊕ 193)x10 + (166 ⊕ 33)x9 + (166 ⊕ 79)x8 + (0 ⊕ 25)x7
The result is:
0x25 + 153x24 + 33x23 + 109x22 + 175x21 + 88x20 + 252x19 + 210x18 + 128x17 + 246x16 + 206x15 + 208x14 + 165x13 + 242x12 + 197x11 + 149x10 + 135x9 + 233x8 + 25x7
Discard the lead 0 term to get:
153x24 + 33x23 + 109x22 + 175x21 + 88x20 + 252x19 + 210x18 + 128x17 + 246x16 + 206x15 + 208x14 + 165x13 + 242x12 + 197x11 + 149x10 + 135x9 + 233x8 + 25x7
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 153x24. Convert 153x24 to alpha notation. According to the log antilog table, for the integer value 153, the alpha exponent is 68. Therefore 153 = ɑ68. Multiply the generator polynomial by ɑ68:
(ɑ68 * ɑ0)x24 + (ɑ68 * ɑ215)x23 + (ɑ68 * ɑ234)x22 + (ɑ68 * ɑ158)x21 + (ɑ68 * ɑ94)x20 + (ɑ68 * ɑ184)x19 + (ɑ68 * ɑ97)x18 + (ɑ68 * ɑ118)x17 + (ɑ68 * ɑ170)x16 + (ɑ68 * ɑ79)x15 + (ɑ68 * ɑ187)x14 + (ɑ68 * ɑ152)x13 + (ɑ68 * ɑ148)x12 + (ɑ68 * ɑ252)x11 + (ɑ68 * ɑ179)x10 + (ɑ68 * ɑ5)x9 + (ɑ68 * ɑ98)x8 + (ɑ68 * ɑ96)x7 + (ɑ68 * 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:
ɑ68x24 + ɑ(283 % 255)x23 + ɑ(302 % 255)x22 + ɑ226x21 + ɑ162x20 + ɑ252x19 + ɑ165x18 + ɑ186x17 + ɑ238x16 + ɑ147x15 + ɑ255x14 + ɑ220x13 + ɑ216x12 + ɑ(320 % 255)x11 + ɑ247x10 + ɑ73x9 + ɑ166x8 + ɑ164x7 + ɑ221x6The result is:
ɑ68x24 + ɑ28x23 + ɑ47x22 + ɑ226x21 + ɑ162x20 + ɑ252x19 + ɑ165x18 + ɑ186x17 + ɑ238x16 + ɑ147x15 + ɑ0x14 + ɑ220x13 + ɑ216x12 + ɑ65x11 + ɑ247x10 + ɑ73x9 + ɑ166x8 + ɑ164x7 + ɑ221x6
Now, convert this to integer notation:
153x24 + 24x23 + 35x22 + 72x21 + 191x20 + 173x19 + 145x18 + 110x17 + 11x16 + 41x15 + 1x14 + 172x13 + 195x12 + 190x11 + 131x10 + 202x9 + 63x8 + 198x7 + 69x6Step 10b: XOR the result with the result from step 9b
Use the result from step 9b to perform the next XOR.
(153 ⊕ 153)x24 + (33 ⊕ 24)x23 + (109 ⊕ 35)x22 + (175 ⊕ 72)x21 + (88 ⊕ 191)x20 + (252 ⊕ 173)x19 + (210 ⊕ 145)x18 + (128 ⊕ 110)x17 + (246 ⊕ 11)x16 + (206 ⊕ 41)x15 + (208 ⊕ 1)x14 + (165 ⊕ 172)x13 + (242 ⊕ 195)x12 + (197 ⊕ 190)x11 + (149 ⊕ 131)x10 + (135 ⊕ 202)x9 + (233 ⊕ 63)x8 + (25 ⊕ 198)x7 + (0 ⊕ 69)x6
The result is:
0x24 + 57x23 + 78x22 + 231x21 + 231x20 + 81x19 + 67x18 + 238x17 + 253x16 + 231x15 + 209x14 + 9x13 + 49x12 + 123x11 + 22x10 + 77x9 + 214x8 + 223x7 + 69x6
Discard the lead 0 term to get:
57x23 + 78x22 + 231x21 + 231x20 + 81x19 + 67x18 + 238x17 + 253x16 + 231x15 + 209x14 + 9x13 + 49x12 + 123x11 + 22x10 + 77x9 + 214x8 + 223x7 + 69x6
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 57x23. Convert 57x23 to alpha notation. According to the log antilog table, for the integer value 57, the alpha exponent is 154. Therefore 57 = ɑ154. Multiply the generator polynomial by ɑ154:
(ɑ154 * ɑ0)x23 + (ɑ154 * ɑ215)x22 + (ɑ154 * ɑ234)x21 + (ɑ154 * ɑ158)x20 + (ɑ154 * ɑ94)x19 + (ɑ154 * ɑ184)x18 + (ɑ154 * ɑ97)x17 + (ɑ154 * ɑ118)x16 + (ɑ154 * ɑ170)x15 + (ɑ154 * ɑ79)x14 + (ɑ154 * ɑ187)x13 + (ɑ154 * ɑ152)x12 + (ɑ154 * ɑ148)x11 + (ɑ154 * ɑ252)x10 + (ɑ154 * ɑ179)x9 + (ɑ154 * ɑ5)x8 + (ɑ154 * ɑ98)x7 + (ɑ154 * ɑ96)x6 + (ɑ154 * 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:
ɑ154x23 + ɑ(369 % 255)x22 + ɑ(388 % 255)x21 + ɑ(312 % 255)x20 + ɑ248x19 + ɑ(338 % 255)x18 + ɑ251x17 + ɑ(272 % 255)x16 + ɑ(324 % 255)x15 + ɑ233x14 + ɑ(341 % 255)x13 + ɑ(306 % 255)x12 + ɑ(302 % 255)x11 + ɑ(406 % 255)x10 + ɑ(333 % 255)x9 + ɑ159x8 + ɑ252x7 + ɑ250x6 + ɑ(307 % 255)x5The result is:
ɑ154x23 + ɑ114x22 + ɑ133x21 + ɑ57x20 + ɑ248x19 + ɑ83x18 + ɑ251x17 + ɑ17x16 + ɑ69x15 + ɑ233x14 + ɑ86x13 + ɑ51x12 + ɑ47x11 + ɑ151x10 + ɑ78x9 + ɑ159x8 + ɑ252x7 + ɑ250x6 + ɑ52x5
Now, convert this to integer notation:
57x23 + 62x22 + 109x21 + 186x20 + 27x19 + 187x18 + 216x17 + 152x16 + 47x15 + 243x14 + 177x13 + 10x12 + 35x11 + 170x10 + 120x9 + 115x8 + 173x7 + 108x6 + 20x5Step 11b: XOR the result with the result from step 10b
Use the result from step 10b to perform the next XOR.
(57 ⊕ 57)x23 + (78 ⊕ 62)x22 + (231 ⊕ 109)x21 + (231 ⊕ 186)x20 + (81 ⊕ 27)x19 + (67 ⊕ 187)x18 + (238 ⊕ 216)x17 + (253 ⊕ 152)x16 + (231 ⊕ 47)x15 + (209 ⊕ 243)x14 + (9 ⊕ 177)x13 + (49 ⊕ 10)x12 + (123 ⊕ 35)x11 + (22 ⊕ 170)x10 + (77 ⊕ 120)x9 + (214 ⊕ 115)x8 + (223 ⊕ 173)x7 + (69 ⊕ 108)x6 + (0 ⊕ 20)x5
The result is:
0x23 + 112x22 + 138x21 + 93x20 + 74x19 + 248x18 + 54x17 + 101x16 + 200x15 + 34x14 + 184x13 + 59x12 + 88x11 + 188x10 + 53x9 + 165x8 + 114x7 + 41x6 + 20x5
Discard the lead 0 term to get:
112x22 + 138x21 + 93x20 + 74x19 + 248x18 + 54x17 + 101x16 + 200x15 + 34x14 + 184x13 + 59x12 + 88x11 + 188x10 + 53x9 + 165x8 + 114x7 + 41x6 + 20x5
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 112x22. Convert 112x22 to alpha notation. According to the log antilog table, for the integer value 112, the alpha exponent is 202. Therefore 112 = ɑ202. Multiply the generator polynomial by ɑ202:
(ɑ202 * ɑ0)x22 + (ɑ202 * ɑ215)x21 + (ɑ202 * ɑ234)x20 + (ɑ202 * ɑ158)x19 + (ɑ202 * ɑ94)x18 + (ɑ202 * ɑ184)x17 + (ɑ202 * ɑ97)x16 + (ɑ202 * ɑ118)x15 + (ɑ202 * ɑ170)x14 + (ɑ202 * ɑ79)x13 + (ɑ202 * ɑ187)x12 + (ɑ202 * ɑ152)x11 + (ɑ202 * ɑ148)x10 + (ɑ202 * ɑ252)x9 + (ɑ202 * ɑ179)x8 + (ɑ202 * ɑ5)x7 + (ɑ202 * ɑ98)x6 + (ɑ202 * ɑ96)x5 + (ɑ202 * 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:
ɑ202x22 + ɑ(417 % 255)x21 + ɑ(436 % 255)x20 + ɑ(360 % 255)x19 + ɑ(296 % 255)x18 + ɑ(386 % 255)x17 + ɑ(299 % 255)x16 + ɑ(320 % 255)x15 + ɑ(372 % 255)x14 + ɑ(281 % 255)x13 + ɑ(389 % 255)x12 + ɑ(354 % 255)x11 + ɑ(350 % 255)x10 + ɑ(454 % 255)x9 + ɑ(381 % 255)x8 + ɑ207x7 + ɑ(300 % 255)x6 + ɑ(298 % 255)x5 + ɑ(355 % 255)x4The result is:
ɑ202x22 + ɑ162x21 + ɑ181x20 + ɑ105x19 + ɑ41x18 + ɑ131x17 + ɑ44x16 + ɑ65x15 + ɑ117x14 + ɑ26x13 + ɑ134x12 + ɑ99x11 + ɑ95x10 + ɑ199x9 + ɑ126x8 + ɑ207x7 + ɑ45x6 + ɑ43x5 + ɑ100x4
Now, convert this to integer notation:
112x22 + 191x21 + 49x20 + 26x19 + 212x18 + 92x17 + 238x16 + 190x15 + 237x14 + 6x13 + 218x12 + 134x11 + 226x10 + 14x9 + 102x8 + 166x7 + 193x6 + 119x5 + 17x4Step 12b: XOR the result with the result from step 11b
Use the result from step 11b to perform the next XOR.
(112 ⊕ 112)x22 + (138 ⊕ 191)x21 + (93 ⊕ 49)x20 + (74 ⊕ 26)x19 + (248 ⊕ 212)x18 + (54 ⊕ 92)x17 + (101 ⊕ 238)x16 + (200 ⊕ 190)x15 + (34 ⊕ 237)x14 + (184 ⊕ 6)x13 + (59 ⊕ 218)x12 + (88 ⊕ 134)x11 + (188 ⊕ 226)x10 + (53 ⊕ 14)x9 + (165 ⊕ 102)x8 + (114 ⊕ 166)x7 + (41 ⊕ 193)x6 + (20 ⊕ 119)x5 + (0 ⊕ 17)x4
The result is:
0x22 + 53x21 + 108x20 + 80x19 + 44x18 + 106x17 + 139x16 + 118x15 + 207x14 + 190x13 + 225x12 + 222x11 + 94x10 + 59x9 + 195x8 + 212x7 + 232x6 + 99x5 + 17x4
Discard the lead 0 term to get:
53x21 + 108x20 + 80x19 + 44x18 + 106x17 + 139x16 + 118x15 + 207x14 + 190x13 + 225x12 + 222x11 + 94x10 + 59x9 + 195x8 + 212x7 + 232x6 + 99x5 + 17x4
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 53x21. Convert 53x21 to alpha notation. According to the log antilog table, for the integer value 53, the alpha exponent is 39. Therefore 53 = ɑ39. Multiply the generator polynomial by ɑ39:
(ɑ39 * ɑ0)x21 + (ɑ39 * ɑ215)x20 + (ɑ39 * ɑ234)x19 + (ɑ39 * ɑ158)x18 + (ɑ39 * ɑ94)x17 + (ɑ39 * ɑ184)x16 + (ɑ39 * ɑ97)x15 + (ɑ39 * ɑ118)x14 + (ɑ39 * ɑ170)x13 + (ɑ39 * ɑ79)x12 + (ɑ39 * ɑ187)x11 + (ɑ39 * ɑ152)x10 + (ɑ39 * ɑ148)x9 + (ɑ39 * ɑ252)x8 + (ɑ39 * ɑ179)x7 + (ɑ39 * ɑ5)x6 + (ɑ39 * ɑ98)x5 + (ɑ39 * ɑ96)x4 + (ɑ39 * 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:
ɑ39x21 + ɑ254x20 + ɑ(273 % 255)x19 + ɑ197x18 + ɑ133x17 + ɑ223x16 + ɑ136x15 + ɑ157x14 + ɑ209x13 + ɑ118x12 + ɑ226x11 + ɑ191x10 + ɑ187x9 + ɑ(291 % 255)x8 + ɑ218x7 + ɑ44x6 + ɑ137x5 + ɑ135x4 + ɑ192x3The result is:
ɑ39x21 + ɑ254x20 + ɑ18x19 + ɑ197x18 + ɑ133x17 + ɑ223x16 + ɑ136x15 + ɑ157x14 + ɑ209x13 + ɑ118x12 + ɑ226x11 + ɑ191x10 + ɑ187x9 + ɑ36x8 + ɑ218x7 + ɑ44x6 + ɑ137x5 + ɑ135x4 + ɑ192x3
Now, convert this to integer notation:
53x21 + 142x20 + 45x19 + 141x18 + 109x17 + 9x16 + 79x15 + 213x14 + 162x13 + 199x12 + 72x11 + 65x10 + 220x9 + 37x8 + 43x7 + 238x6 + 158x5 + 169x4 + 130x3Step 13b: XOR the result with the result from step 12b
Use the result from step 12b to perform the next XOR.
(53 ⊕ 53)x21 + (108 ⊕ 142)x20 + (80 ⊕ 45)x19 + (44 ⊕ 141)x18 + (106 ⊕ 109)x17 + (139 ⊕ 9)x16 + (118 ⊕ 79)x15 + (207 ⊕ 213)x14 + (190 ⊕ 162)x13 + (225 ⊕ 199)x12 + (222 ⊕ 72)x11 + (94 ⊕ 65)x10 + (59 ⊕ 220)x9 + (195 ⊕ 37)x8 + (212 ⊕ 43)x7 + (232 ⊕ 238)x6 + (99 ⊕ 158)x5 + (17 ⊕ 169)x4 + (0 ⊕ 130)x3
The result is:
0x21 + 226x20 + 125x19 + 161x18 + 7x17 + 130x16 + 57x15 + 26x14 + 28x13 + 38x12 + 150x11 + 31x10 + 231x9 + 230x8 + 255x7 + 6x6 + 253x5 + 184x4 + 130x3
Discard the lead 0 term to get:
226x20 + 125x19 + 161x18 + 7x17 + 130x16 + 57x15 + 26x14 + 28x13 + 38x12 + 150x11 + 31x10 + 231x9 + 230x8 + 255x7 + 6x6 + 253x5 + 184x4 + 130x3
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 226x20. Convert 226x20 to alpha notation. According to the log antilog table, for the integer value 226, the alpha exponent is 95. Therefore 226 = ɑ95. Multiply the generator polynomial by ɑ95:
(ɑ95 * ɑ0)x20 + (ɑ95 * ɑ215)x19 + (ɑ95 * ɑ234)x18 + (ɑ95 * ɑ158)x17 + (ɑ95 * ɑ94)x16 + (ɑ95 * ɑ184)x15 + (ɑ95 * ɑ97)x14 + (ɑ95 * ɑ118)x13 + (ɑ95 * ɑ170)x12 + (ɑ95 * ɑ79)x11 + (ɑ95 * ɑ187)x10 + (ɑ95 * ɑ152)x9 + (ɑ95 * ɑ148)x8 + (ɑ95 * ɑ252)x7 + (ɑ95 * ɑ179)x6 + (ɑ95 * ɑ5)x5 + (ɑ95 * ɑ98)x4 + (ɑ95 * ɑ96)x3 + (ɑ95 * 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:
ɑ95x20 + ɑ(310 % 255)x19 + ɑ(329 % 255)x18 + ɑ253x17 + ɑ189x16 + ɑ(279 % 255)x15 + ɑ192x14 + ɑ213x13 + ɑ(265 % 255)x12 + ɑ174x11 + ɑ(282 % 255)x10 + ɑ247x9 + ɑ243x8 + ɑ(347 % 255)x7 + ɑ(274 % 255)x6 + ɑ100x5 + ɑ193x4 + ɑ191x3 + ɑ248x2The result is:
ɑ95x20 + ɑ55x19 + ɑ74x18 + ɑ253x17 + ɑ189x16 + ɑ24x15 + ɑ192x14 + ɑ213x13 + ɑ10x12 + ɑ174x11 + ɑ27x10 + ɑ247x9 + ɑ243x8 + ɑ92x7 + ɑ19x6 + ɑ100x5 + ɑ193x4 + ɑ191x3 + ɑ248x2
Now, convert this to integer notation:
226x20 + 160x19 + 137x18 + 71x17 + 87x16 + 143x15 + 130x14 + 242x13 + 116x12 + 241x11 + 12x10 + 131x9 + 125x8 + 91x7 + 90x6 + 17x5 + 25x4 + 65x3 + 27x2Step 14b: XOR the result with the result from step 13b
Use the result from step 13b to perform the next XOR.
(226 ⊕ 226)x20 + (125 ⊕ 160)x19 + (161 ⊕ 137)x18 + (7 ⊕ 71)x17 + (130 ⊕ 87)x16 + (57 ⊕ 143)x15 + (26 ⊕ 130)x14 + (28 ⊕ 242)x13 + (38 ⊕ 116)x12 + (150 ⊕ 241)x11 + (31 ⊕ 12)x10 + (231 ⊕ 131)x9 + (230 ⊕ 125)x8 + (255 ⊕ 91)x7 + (6 ⊕ 90)x6 + (253 ⊕ 17)x5 + (184 ⊕ 25)x4 + (130 ⊕ 65)x3 + (0 ⊕ 27)x2
The result is:
0x20 + 221x19 + 40x18 + 64x17 + 213x16 + 182x15 + 152x14 + 238x13 + 82x12 + 103x11 + 19x10 + 100x9 + 155x8 + 164x7 + 92x6 + 236x5 + 161x4 + 195x3 + 27x2
Discard the lead 0 term to get:
221x19 + 40x18 + 64x17 + 213x16 + 182x15 + 152x14 + 238x13 + 82x12 + 103x11 + 19x10 + 100x9 + 155x8 + 164x7 + 92x6 + 236x5 + 161x4 + 195x3 + 27x2
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 221x19. Convert 221x19 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)x19 + (ɑ204 * ɑ215)x18 + (ɑ204 * ɑ234)x17 + (ɑ204 * ɑ158)x16 + (ɑ204 * ɑ94)x15 + (ɑ204 * ɑ184)x14 + (ɑ204 * ɑ97)x13 + (ɑ204 * ɑ118)x12 + (ɑ204 * ɑ170)x11 + (ɑ204 * ɑ79)x10 + (ɑ204 * ɑ187)x9 + (ɑ204 * ɑ152)x8 + (ɑ204 * ɑ148)x7 + (ɑ204 * ɑ252)x6 + (ɑ204 * ɑ179)x5 + (ɑ204 * ɑ5)x4 + (ɑ204 * ɑ98)x3 + (ɑ204 * ɑ96)x2 + (ɑ204 * 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:
ɑ204x19 + ɑ(419 % 255)x18 + ɑ(438 % 255)x17 + ɑ(362 % 255)x16 + ɑ(298 % 255)x15 + ɑ(388 % 255)x14 + ɑ(301 % 255)x13 + ɑ(322 % 255)x12 + ɑ(374 % 255)x11 + ɑ(283 % 255)x10 + ɑ(391 % 255)x9 + ɑ(356 % 255)x8 + ɑ(352 % 255)x7 + ɑ(456 % 255)x6 + ɑ(383 % 255)x5 + ɑ209x4 + ɑ(302 % 255)x3 + ɑ(300 % 255)x2 + ɑ(357 % 255)x1The result is:
ɑ204x19 + ɑ164x18 + ɑ183x17 + ɑ107x16 + ɑ43x15 + ɑ133x14 + ɑ46x13 + ɑ67x12 + ɑ119x11 + ɑ28x10 + ɑ136x9 + ɑ101x8 + ɑ97x7 + ɑ201x6 + ɑ128x5 + ɑ209x4 + ɑ47x3 + ɑ45x2 + ɑ102x1
Now, convert this to integer notation:
221x19 + 198x18 + 196x17 + 104x16 + 119x15 + 109x14 + 159x13 + 194x12 + 147x11 + 24x10 + 79x9 + 34x8 + 175x7 + 56x6 + 133x5 + 162x4 + 35x3 + 193x2 + 68x1Step 15b: XOR the result with the result from step 14b
Use the result from step 14b to perform the next XOR.
(221 ⊕ 221)x19 + (40 ⊕ 198)x18 + (64 ⊕ 196)x17 + (213 ⊕ 104)x16 + (182 ⊕ 119)x15 + (152 ⊕ 109)x14 + (238 ⊕ 159)x13 + (82 ⊕ 194)x12 + (103 ⊕ 147)x11 + (19 ⊕ 24)x10 + (100 ⊕ 79)x9 + (155 ⊕ 34)x8 + (164 ⊕ 175)x7 + (92 ⊕ 56)x6 + (236 ⊕ 133)x5 + (161 ⊕ 162)x4 + (195 ⊕ 35)x3 + (27 ⊕ 193)x2 + (0 ⊕ 68)x1
The result is:
0x19 + 238x18 + 132x17 + 189x16 + 193x15 + 245x14 + 113x13 + 144x12 + 244x11 + 11x10 + 43x9 + 185x8 + 11x7 + 100x6 + 105x5 + 3x4 + 224x3 + 218x2 + 68x1
Discard the lead 0 term to get:
238x18 + 132x17 + 189x16 + 193x15 + 245x14 + 113x13 + 144x12 + 244x11 + 11x10 + 43x9 + 185x8 + 11x7 + 100x6 + 105x5 + 3x4 + 224x3 + 218x2 + 68x1
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 238x18. Convert 238x18 to alpha notation. According to the log antilog table, for the integer value 238, the alpha exponent is 44. Therefore 238 = ɑ44. Multiply the generator polynomial by ɑ44:
(ɑ44 * ɑ0)x18 + (ɑ44 * ɑ215)x17 + (ɑ44 * ɑ234)x16 + (ɑ44 * ɑ158)x15 + (ɑ44 * ɑ94)x14 + (ɑ44 * ɑ184)x13 + (ɑ44 * ɑ97)x12 + (ɑ44 * ɑ118)x11 + (ɑ44 * ɑ170)x10 + (ɑ44 * ɑ79)x9 + (ɑ44 * ɑ187)x8 + (ɑ44 * ɑ152)x7 + (ɑ44 * ɑ148)x6 + (ɑ44 * ɑ252)x5 + (ɑ44 * ɑ179)x4 + (ɑ44 * ɑ5)x3 + (ɑ44 * ɑ98)x2 + (ɑ44 * ɑ96)x1 + (ɑ44 * 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:
ɑ44x18 + ɑ(259 % 255)x17 + ɑ(278 % 255)x16 + ɑ202x15 + ɑ138x14 + ɑ228x13 + ɑ141x12 + ɑ162x11 + ɑ214x10 + ɑ123x9 + ɑ231x8 + ɑ196x7 + ɑ192x6 + ɑ(296 % 255)x5 + ɑ223x4 + ɑ49x3 + ɑ142x2 + ɑ140x1 + ɑ197x0The result is:
ɑ44x18 + ɑ4x17 + ɑ23x16 + ɑ202x15 + ɑ138x14 + ɑ228x13 + ɑ141x12 + ɑ162x11 + ɑ214x10 + ɑ123x9 + ɑ231x8 + ɑ196x7 + ɑ192x6 + ɑ41x5 + ɑ223x4 + ɑ49x3 + ɑ142x2 + ɑ140x1 + ɑ197x0
Now, convert this to integer notation:
238x18 + 16x17 + 201x16 + 112x15 + 33x14 + 61x13 + 21x12 + 191x11 + 249x10 + 197x9 + 245x8 + 200x7 + 130x6 + 212x5 + 9x4 + 140x3 + 42x2 + 132x1 + 141Step 16b: XOR the result with the result from step 15b
Use the result from step 15b to perform the next XOR.
(238 ⊕ 238)x18 + (132 ⊕ 16)x17 + (189 ⊕ 201)x16 + (193 ⊕ 112)x15 + (245 ⊕ 33)x14 + (113 ⊕ 61)x13 + (144 ⊕ 21)x12 + (244 ⊕ 191)x11 + (11 ⊕ 249)x10 + (43 ⊕ 197)x9 + (185 ⊕ 245)x8 + (11 ⊕ 200)x7 + (100 ⊕ 130)x6 + (105 ⊕ 212)x5 + (3 ⊕ 9)x4 + (224 ⊕ 140)x3 + (218 ⊕ 42)x2 + (68 ⊕ 132)x1 + (0 ⊕ 141)x0
The result is:
0x18 + 148x17 + 116x16 + 177x15 + 212x14 + 76x13 + 133x12 + 75x11 + 242x10 + 238x9 + 76x8 + 195x7 + 230x6 + 189x5 + 10x4 + 108x3 + 240x2 + 192x1 + 141
Discard the lead 0 term to get:
148x17 + 116x16 + 177x15 + 212x14 + 76x13 + 133x12 + 75x11 + 242x10 + 238x9 + 76x8 + 195x7 + 230x6 + 189x5 + 10x4 + 108x3 + 240x2 + 192x1 + 141
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:
148 116 177 212 76 133 75 242 238 76 195 230 189 10 108 240 192 141