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.

Enter the coefficients of the message polynomial, separated by commas (no spaces):
(example: 32,91,11,120,209,114,220,77,67,64,236,17,236 )

Enter the desired number of ECC blocks:
(example: 13 )

Perform Division

Polynomial Division Steps

The first step to the division is to prepare the message polynomial for the division. The full message polynomial is:

70x¹⁵ + 247x¹⁴ + 118x¹³ + 86x¹² + 194x¹¹ + 6x¹⁰ + 151x⁹ + 50x⁸ + 16x⁷ + 236x⁶ + 17x⁵ + 236x⁴ + 17x³ + 236x² + 17x¹ + 236

To make sure that the exponent of the lead term doesn't become too small during the division, multiply the message polynomial by xⁿ where n is the number of error correction codewords that are needed. In this case n is 18, for 18 error correction codewords, so multiply the message polynomial by x¹⁸, which gives us:

70x³³ + 247x³² + 118x³¹ + 86x³⁰ + 194x²⁹ + 6x²⁸ + 151x²⁷ + 50x²⁶ + 16x²⁵ + 236x²⁴ + 17x²³ + 236x²² + 17x²¹ + 236x²⁰ + 17x¹⁹ + 236x¹⁸

The lead term of the generator polynomial should also have the same exponent, so multiply by x¹⁵ to get

α⁰x³³ + α²¹⁵x³² + α²³⁴x³¹ + α¹⁵⁸x³⁰ + α⁹⁴x²⁹ + α¹⁸⁴x²⁸ + α⁹⁷x²⁷ + α¹¹⁸x²⁶ + α¹⁷⁰x²⁵ + α⁷⁹x²⁴ + α¹⁸⁷x²³ + α¹⁵²x²² + α¹⁴⁸x²¹ + α²⁵²x²⁰ + α¹⁷⁹x¹⁹ + α⁵x¹⁸ + α⁹⁸x¹⁷ + α⁹⁶x¹⁶ + α¹⁵³x¹⁵

Now it is possible to perform the repeated division steps. The number of steps in the division must equal the number of terms in the message polynomial. In this case, the division will take 16 steps to complete. This will result in a remainder that has 18 terms. These terms will be the 18 error correction codewords that are required.

Step 1a: Multiply the Generator Polynomial by the Lead Term of the Message Polynomial

The first step is to multiply the generator polynomial by the lead term of the message polynomial. The lead term in this case is 70x³³. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 70x³³ to alpha notation. According to the log antilog table, for the integer value 70, the alpha exponent is 48. Therefore 70 = α⁴⁸. Multiply the generator polynomial by α⁴⁸:

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:

α⁴⁸x³³ + α^{(263 % 255)}x³² + α^{(282 % 255)}x³¹ + α²⁰⁶x³⁰ + α¹⁴²x²⁹ + α²³²x²⁸ + α¹⁴⁵x²⁷ + α¹⁶⁶x²⁶ + α²¹⁸x²⁵ + α¹²⁷x²⁴ + α²³⁵x²³ + α²⁰⁰x²² + α¹⁹⁶x²¹ + α^{(300 % 255)}x²⁰ + α²²⁷x¹⁹ + α⁵³x¹⁸ + α¹⁴⁶x¹⁷ + α¹⁴⁴x¹⁶ + α²⁰¹x¹⁵

The result is:

α⁴⁸x³³ + α⁸x³² + α²⁷x³¹ + α²⁰⁶x³⁰ + α¹⁴²x²⁹ + α²³²x²⁸ + α¹⁴⁵x²⁷ + α¹⁶⁶x²⁶ + α²¹⁸x²⁵ + α¹²⁷x²⁴ + α²³⁵x²³ + α²⁰⁰x²² + α¹⁹⁶x²¹ + α⁴⁵x²⁰ + α²²⁷x¹⁹ + α⁵³x¹⁸ + α¹⁴⁶x¹⁷ + α¹⁴⁴x¹⁶ + α²⁰¹x¹⁵

Now, convert this to integer notation:

70x³³ + 29x³² + 12x³¹ + 83x³⁰ + 42x²⁹ + 247x²⁸ + 77x²⁷ + 63x²⁶ + 43x²⁵ + 204x²⁴ + 235x²³ + 28x²² + 200x²¹ + 193x²⁰ + 144x¹⁹ + 40x¹⁸ + 154x¹⁷ + 168x¹⁶ + 56x¹⁵

Step 1b: XOR the result with the message polynomial

Since this is the first division step, XOR the result from 1a with the message polynomial.

(70 ⊕ 70)x³³ + (247 ⊕ 29)x³² + (118 ⊕ 12)x³¹ + (86 ⊕ 83)x³⁰ + (194 ⊕ 42)x²⁹ + (6 ⊕ 247)x²⁸ + (151 ⊕ 77)x²⁷ + (50 ⊕ 63)x²⁶ + (16 ⊕ 43)x²⁵ + (236 ⊕ 204)x²⁴ + (17 ⊕ 235)x²³ + (236 ⊕ 28)x²² + (17 ⊕ 200)x²¹ + (236 ⊕ 193)x²⁰ + (17 ⊕ 144)x¹⁹ + (236 ⊕ 40)x¹⁸ + (0 ⊕ 154)x¹⁷ + (0 ⊕ 168)x¹⁶ + (0 ⊕ 56)x¹⁵

The result is:

0x³³ + 234x³² + 122x³¹ + 5x³⁰ + 232x²⁹ + 241x²⁸ + 218x²⁷ + 13x²⁶ + 59x²⁵ + 32x²⁴ + 250x²³ + 240x²² + 217x²¹ + 45x²⁰ + 129x¹⁹ + 196x¹⁸ + 154x¹⁷ + 168x¹⁶ + 56x¹⁵

Discard the lead 0 term to get:

234x³² + 122x³¹ + 5x³⁰ + 232x²⁹ + 241x²⁸ + 218x²⁷ + 13x²⁶ + 59x²⁵ + 32x²⁴ + 250x²³ + 240x²² + 217x²¹ + 45x²⁰ + 129x¹⁹ + 196x¹⁸ + 154x¹⁷ + 168x¹⁶ + 56x¹⁵

Step 2a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 234x³². Convert 234x³² to alpha notation. According to the log antilog table, for the integer value 234, the alpha exponent is 22. Therefore 234 = α²². Multiply the generator polynomial by α²²:

(α²² * α⁰)x³² + (α²² * α²¹⁵)x³¹ + (α²² * α²³⁴)x³⁰ + (α²² * α¹⁵⁸)x²⁹ + (α²² * α⁹⁴)x²⁸ + (α²² * α¹⁸⁴)x²⁷ + (α²² * α⁹⁷)x²⁶ + (α²² * α¹¹⁸)x²⁵ + (α²² * α¹⁷⁰)x²⁴ + (α²² * α⁷⁹)x²³ + (α²² * α¹⁸⁷)x²² + (α²² * α¹⁵²)x²¹ + (α²² * α¹⁴⁸)x²⁰ + (α²² * α²⁵²)x¹⁹ + (α²² * α¹⁷⁹)x¹⁸ + (α²² * α⁵)x¹⁷ + (α²² * α⁹⁸)x¹⁶ + (α²² * α⁹⁶)x¹⁵ + (α²² * α¹⁵³)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:

α²²x³² + α²³⁷x³¹ + α^{(256 % 255)}x³⁰ + α¹⁸⁰x²⁹ + α¹¹⁶x²⁸ + α²⁰⁶x²⁷ + α¹¹⁹x²⁶ + α¹⁴⁰x²⁵ + α¹⁹²x²⁴ + α¹⁰¹x²³ + α²⁰⁹x²² + α¹⁷⁴x²¹ + α¹⁷⁰x²⁰ + α^{(274 % 255)}x¹⁹ + α²⁰¹x¹⁸ + α²⁷x¹⁷ + α¹²⁰x¹⁶ + α¹¹⁸x¹⁵ + α¹⁷⁵x¹⁴

The result is:

α²²x³² + α²³⁷x³¹ + α¹x³⁰ + α¹⁸⁰x²⁹ + α¹¹⁶x²⁸ + α²⁰⁶x²⁷ + α¹¹⁹x²⁶ + α¹⁴⁰x²⁵ + α¹⁹²x²⁴ + α¹⁰¹x²³ + α²⁰⁹x²² + α¹⁷⁴x²¹ + α¹⁷⁰x²⁰ + α¹⁹x¹⁹ + α²⁰¹x¹⁸ + α²⁷x¹⁷ + α¹²⁰x¹⁶ + α¹¹⁸x¹⁵ + α¹⁷⁵x¹⁴

Now, convert this to integer notation:

234x³² + 139x³¹ + 2x³⁰ + 150x²⁹ + 248x²⁸ + 83x²⁷ + 147x²⁶ + 132x²⁵ + 130x²⁴ + 34x²³ + 162x²² + 241x²¹ + 215x²⁰ + 90x¹⁹ + 56x¹⁸ + 12x¹⁷ + 59x¹⁶ + 199x¹⁵ + 255x¹⁴

Step 2b: XOR the result with the result from step 1b

Use the result from step 1b to perform the next XOR.

(234 ⊕ 234)x³² + (122 ⊕ 139)x³¹ + (5 ⊕ 2)x³⁰ + (232 ⊕ 150)x²⁹ + (241 ⊕ 248)x²⁸ + (218 ⊕ 83)x²⁷ + (13 ⊕ 147)x²⁶ + (59 ⊕ 132)x²⁵ + (32 ⊕ 130)x²⁴ + (250 ⊕ 34)x²³ + (240 ⊕ 162)x²² + (217 ⊕ 241)x²¹ + (45 ⊕ 215)x²⁰ + (129 ⊕ 90)x¹⁹ + (196 ⊕ 56)x¹⁸ + (154 ⊕ 12)x¹⁷ + (168 ⊕ 59)x¹⁶ + (56 ⊕ 199)x¹⁵ + (0 ⊕ 255)x¹⁴

The result is:

0x³² + 241x³¹ + 7x³⁰ + 126x²⁹ + 9x²⁸ + 137x²⁷ + 158x²⁶ + 191x²⁵ + 162x²⁴ + 216x²³ + 82x²² + 40x²¹ + 250x²⁰ + 219x¹⁹ + 252x¹⁸ + 150x¹⁷ + 147x¹⁶ + 255x¹⁵ + 255x¹⁴

Discard the lead 0 term to get:

241x³¹ + 7x³⁰ + 126x²⁹ + 9x²⁸ + 137x²⁷ + 158x²⁶ + 191x²⁵ + 162x²⁴ + 216x²³ + 82x²² + 40x²¹ + 250x²⁰ + 219x¹⁹ + 252x¹⁸ + 150x¹⁷ + 147x¹⁶ + 255x¹⁵ + 255x¹⁴

Step 3a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 241x³¹. Convert 241x³¹ to alpha notation. According to the log antilog table, for the integer value 241, the alpha exponent is 174. Therefore 241 = α¹⁷⁴. Multiply the generator polynomial by α¹⁷⁴:

(α¹⁷⁴ * α⁰)x³¹ + (α¹⁷⁴ * α²¹⁵)x³⁰ + (α¹⁷⁴ * α²³⁴)x²⁹ + (α¹⁷⁴ * α¹⁵⁸)x²⁸ + (α¹⁷⁴ * α⁹⁴)x²⁷ + (α¹⁷⁴ * α¹⁸⁴)x²⁶ + (α¹⁷⁴ * α⁹⁷)x²⁵ + (α¹⁷⁴ * α¹¹⁸)x²⁴ + (α¹⁷⁴ * α¹⁷⁰)x²³ + (α¹⁷⁴ * α⁷⁹)x²² + (α¹⁷⁴ * α¹⁸⁷)x²¹ + (α¹⁷⁴ * α¹⁵²)x²⁰ + (α¹⁷⁴ * α¹⁴⁸)x¹⁹ + (α¹⁷⁴ * α²⁵²)x¹⁸ + (α¹⁷⁴ * α¹⁷⁹)x¹⁷ + (α¹⁷⁴ * α⁵)x¹⁶ + (α¹⁷⁴ * α⁹⁸)x¹⁵ + (α¹⁷⁴ * α⁹⁶)x¹⁴ + (α¹⁷⁴ * α¹⁵³)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:

α¹⁷⁴x³¹ + α^{(389 % 255)}x³⁰ + α^{(408 % 255)}x²⁹ + α^{(332 % 255)}x²⁸ + α^{(268 % 255)}x²⁷ + α^{(358 % 255)}x²⁶ + α^{(271 % 255)}x²⁵ + α^{(292 % 255)}x²⁴ + α^{(344 % 255)}x²³ + α²⁵³x²² + α^{(361 % 255)}x²¹ + α^{(326 % 255)}x²⁰ + α^{(322 % 255)}x¹⁹ + α^{(426 % 255)}x¹⁸ + α^{(353 % 255)}x¹⁷ + α¹⁷⁹x¹⁶ + α^{(272 % 255)}x¹⁵ + α^{(270 % 255)}x¹⁴ + α^{(327 % 255)}x¹³

The result is:

α¹⁷⁴x³¹ + α¹³⁴x³⁰ + α¹⁵³x²⁹ + α⁷⁷x²⁸ + α¹³x²⁷ + α¹⁰³x²⁶ + α¹⁶x²⁵ + α³⁷x²⁴ + α⁸⁹x²³ + α²⁵³x²² + α¹⁰⁶x²¹ + α⁷¹x²⁰ + α⁶⁷x¹⁹ + α¹⁷¹x¹⁸ + α⁹⁸x¹⁷ + α¹⁷⁹x¹⁶ + α¹⁷x¹⁵ + α¹⁵x¹⁴ + α⁷²x¹³

Now, convert this to integer notation:

241x³¹ + 218x³⁰ + 146x²⁹ + 60x²⁸ + 135x²⁷ + 136x²⁶ + 76x²⁵ + 74x²⁴ + 225x²³ + 71x²² + 52x²¹ + 188x²⁰ + 194x¹⁹ + 179x¹⁸ + 67x¹⁷ + 75x¹⁶ + 152x¹⁵ + 38x¹⁴ + 101x¹³

Step 3b: XOR the result with the result from step 2b

Use the result from step 2b to perform the next XOR.

(241 ⊕ 241)x³¹ + (7 ⊕ 218)x³⁰ + (126 ⊕ 146)x²⁹ + (9 ⊕ 60)x²⁸ + (137 ⊕ 135)x²⁷ + (158 ⊕ 136)x²⁶ + (191 ⊕ 76)x²⁵ + (162 ⊕ 74)x²⁴ + (216 ⊕ 225)x²³ + (82 ⊕ 71)x²² + (40 ⊕ 52)x²¹ + (250 ⊕ 188)x²⁰ + (219 ⊕ 194)x¹⁹ + (252 ⊕ 179)x¹⁸ + (150 ⊕ 67)x¹⁷ + (147 ⊕ 75)x¹⁶ + (255 ⊕ 152)x¹⁵ + (255 ⊕ 38)x¹⁴ + (0 ⊕ 101)x¹³

The result is:

0x³¹ + 221x³⁰ + 236x²⁹ + 53x²⁸ + 14x²⁷ + 22x²⁶ + 243x²⁵ + 232x²⁴ + 57x²³ + 21x²² + 28x²¹ + 70x²⁰ + 25x¹⁹ + 79x¹⁸ + 213x¹⁷ + 216x¹⁶ + 103x¹⁵ + 217x¹⁴ + 101x¹³

Discard the lead 0 term to get:

221x³⁰ + 236x²⁹ + 53x²⁸ + 14x²⁷ + 22x²⁶ + 243x²⁵ + 232x²⁴ + 57x²³ + 21x²² + 28x²¹ + 70x²⁰ + 25x¹⁹ + 79x¹⁸ + 213x¹⁷ + 216x¹⁶ + 103x¹⁵ + 217x¹⁴ + 101x¹³

Step 4a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 221x³⁰. Convert 221x³⁰ to alpha notation. According to the log antilog table, for the integer value 221, the alpha exponent is 204. Therefore 221 = α²⁰⁴. Multiply the generator polynomial by α²⁰⁴:

(α²⁰⁴ * α⁰)x³⁰ + (α²⁰⁴ * α²¹⁵)x²⁹ + (α²⁰⁴ * α²³⁴)x²⁸ + (α²⁰⁴ * α¹⁵⁸)x²⁷ + (α²⁰⁴ * α⁹⁴)x²⁶ + (α²⁰⁴ * α¹⁸⁴)x²⁵ + (α²⁰⁴ * α⁹⁷)x²⁴ + (α²⁰⁴ * α¹¹⁸)x²³ + (α²⁰⁴ * α¹⁷⁰)x²² + (α²⁰⁴ * α⁷⁹)x²¹ + (α²⁰⁴ * α¹⁸⁷)x²⁰ + (α²⁰⁴ * α¹⁵²)x¹⁹ + (α²⁰⁴ * α¹⁴⁸)x¹⁸ + (α²⁰⁴ * α²⁵²)x¹⁷ + (α²⁰⁴ * α¹⁷⁹)x¹⁶ + (α²⁰⁴ * α⁵)x¹⁵ + (α²⁰⁴ * α⁹⁸)x¹⁴ + (α²⁰⁴ * α⁹⁶)x¹³ + (α²⁰⁴ * α¹⁵³)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:

α²⁰⁴x³⁰ + α^{(419 % 255)}x²⁹ + α^{(438 % 255)}x²⁸ + α^{(362 % 255)}x²⁷ + α^{(298 % 255)}x²⁶ + α^{(388 % 255)}x²⁵ + α^{(301 % 255)}x²⁴ + α^{(322 % 255)}x²³ + α^{(374 % 255)}x²² + α^{(283 % 255)}x²¹ + α^{(391 % 255)}x²⁰ + α^{(356 % 255)}x¹⁹ + α^{(352 % 255)}x¹⁸ + α^{(456 % 255)}x¹⁷ + α^{(383 % 255)}x¹⁶ + α²⁰⁹x¹⁵ + α^{(302 % 255)}x¹⁴ + α^{(300 % 255)}x¹³ + α^{(357 % 255)}x¹²

The result is:

α²⁰⁴x³⁰ + α¹⁶⁴x²⁹ + α¹⁸³x²⁸ + α¹⁰⁷x²⁷ + α⁴³x²⁶ + α¹³³x²⁵ + α⁴⁶x²⁴ + α⁶⁷x²³ + α¹¹⁹x²² + α²⁸x²¹ + α¹³⁶x²⁰ + α¹⁰¹x¹⁹ + α⁹⁷x¹⁸ + α²⁰¹x¹⁷ + α¹²⁸x¹⁶ + α²⁰⁹x¹⁵ + α⁴⁷x¹⁴ + α⁴⁵x¹³ + α¹⁰²x¹²

Now, convert this to integer notation:

221x³⁰ + 198x²⁹ + 196x²⁸ + 104x²⁷ + 119x²⁶ + 109x²⁵ + 159x²⁴ + 194x²³ + 147x²² + 24x²¹ + 79x²⁰ + 34x¹⁹ + 175x¹⁸ + 56x¹⁷ + 133x¹⁶ + 162x¹⁵ + 35x¹⁴ + 193x¹³ + 68x¹²

Step 4b: XOR the result with the result from step 3b

Use the result from step 3b to perform the next XOR.

(221 ⊕ 221)x³⁰ + (236 ⊕ 198)x²⁹ + (53 ⊕ 196)x²⁸ + (14 ⊕ 104)x²⁷ + (22 ⊕ 119)x²⁶ + (243 ⊕ 109)x²⁵ + (232 ⊕ 159)x²⁴ + (57 ⊕ 194)x²³ + (21 ⊕ 147)x²² + (28 ⊕ 24)x²¹ + (70 ⊕ 79)x²⁰ + (25 ⊕ 34)x¹⁹ + (79 ⊕ 175)x¹⁸ + (213 ⊕ 56)x¹⁷ + (216 ⊕ 133)x¹⁶ + (103 ⊕ 162)x¹⁵ + (217 ⊕ 35)x¹⁴ + (101 ⊕ 193)x¹³ + (0 ⊕ 68)x¹²

The result is:

0x³⁰ + 42x²⁹ + 241x²⁸ + 102x²⁷ + 97x²⁶ + 158x²⁵ + 119x²⁴ + 251x²³ + 134x²² + 4x²¹ + 9x²⁰ + 59x¹⁹ + 224x¹⁸ + 237x¹⁷ + 93x¹⁶ + 197x¹⁵ + 250x¹⁴ + 164x¹³ + 68x¹²

Discard the lead 0 term to get:

42x²⁹ + 241x²⁸ + 102x²⁷ + 97x²⁶ + 158x²⁵ + 119x²⁴ + 251x²³ + 134x²² + 4x²¹ + 9x²⁰ + 59x¹⁹ + 224x¹⁸ + 237x¹⁷ + 93x¹⁶ + 197x¹⁵ + 250x¹⁴ + 164x¹³ + 68x¹²

Step 5a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 42x²⁹. Convert 42x²⁹ to alpha notation. According to the log antilog table, for the integer value 42, the alpha exponent is 142. Therefore 42 = α¹⁴². Multiply the generator polynomial by α¹⁴²:

(α¹⁴² * α⁰)x²⁹ + (α¹⁴² * α²¹⁵)x²⁸ + (α¹⁴² * α²³⁴)x²⁷ + (α¹⁴² * α¹⁵⁸)x²⁶ + (α¹⁴² * α⁹⁴)x²⁵ + (α¹⁴² * α¹⁸⁴)x²⁴ + (α¹⁴² * α⁹⁷)x²³ + (α¹⁴² * α¹¹⁸)x²² + (α¹⁴² * α¹⁷⁰)x²¹ + (α¹⁴² * α⁷⁹)x²⁰ + (α¹⁴² * α¹⁸⁷)x¹⁹ + (α¹⁴² * α¹⁵²)x¹⁸ + (α¹⁴² * α¹⁴⁸)x¹⁷ + (α¹⁴² * α²⁵²)x¹⁶ + (α¹⁴² * α¹⁷⁹)x¹⁵ + (α¹⁴² * α⁵)x¹⁴ + (α¹⁴² * α⁹⁸)x¹³ + (α¹⁴² * α⁹⁶)x¹² + (α¹⁴² * α¹⁵³)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:

α¹⁴²x²⁹ + α^{(357 % 255)}x²⁸ + α^{(376 % 255)}x²⁷ + α^{(300 % 255)}x²⁶ + α²³⁶x²⁵ + α^{(326 % 255)}x²⁴ + α²³⁹x²³ + α^{(260 % 255)}x²² + α^{(312 % 255)}x²¹ + α²²¹x²⁰ + α^{(329 % 255)}x¹⁹ + α^{(294 % 255)}x¹⁸ + α^{(290 % 255)}x¹⁷ + α^{(394 % 255)}x¹⁶ + α^{(321 % 255)}x¹⁵ + α¹⁴⁷x¹⁴ + α²⁴⁰x¹³ + α²³⁸x¹² + α^{(295 % 255)}x¹¹

The result is:

α¹⁴²x²⁹ + α¹⁰²x²⁸ + α¹²¹x²⁷ + α⁴⁵x²⁶ + α²³⁶x²⁵ + α⁷¹x²⁴ + α²³⁹x²³ + α⁵x²² + α⁵⁷x²¹ + α²²¹x²⁰ + α⁷⁴x¹⁹ + α³⁹x¹⁸ + α³⁵x¹⁷ + α¹³⁹x¹⁶ + α⁶⁶x¹⁵ + α¹⁴⁷x¹⁴ + α²⁴⁰x¹³ + α²³⁸x¹² + α⁴⁰x¹¹

Now, convert this to integer notation:

42x²⁹ + 68x²⁸ + 118x²⁷ + 193x²⁶ + 203x²⁵ + 188x²⁴ + 22x²³ + 32x²² + 186x²¹ + 69x²⁰ + 137x¹⁹ + 53x¹⁸ + 156x¹⁷ + 66x¹⁶ + 97x¹⁵ + 41x¹⁴ + 44x¹³ + 11x¹² + 106x¹¹

Step 5b: XOR the result with the result from step 4b

Use the result from step 4b to perform the next XOR.

(42 ⊕ 42)x²⁹ + (241 ⊕ 68)x²⁸ + (102 ⊕ 118)x²⁷ + (97 ⊕ 193)x²⁶ + (158 ⊕ 203)x²⁵ + (119 ⊕ 188)x²⁴ + (251 ⊕ 22)x²³ + (134 ⊕ 32)x²² + (4 ⊕ 186)x²¹ + (9 ⊕ 69)x²⁰ + (59 ⊕ 137)x¹⁹ + (224 ⊕ 53)x¹⁸ + (237 ⊕ 156)x¹⁷ + (93 ⊕ 66)x¹⁶ + (197 ⊕ 97)x¹⁵ + (250 ⊕ 41)x¹⁴ + (164 ⊕ 44)x¹³ + (68 ⊕ 11)x¹² + (0 ⊕ 106)x¹¹

The result is:

0x²⁹ + 181x²⁸ + 16x²⁷ + 160x²⁶ + 85x²⁵ + 203x²⁴ + 237x²³ + 166x²² + 190x²¹ + 76x²⁰ + 178x¹⁹ + 213x¹⁸ + 113x¹⁷ + 31x¹⁶ + 164x¹⁵ + 211x¹⁴ + 136x¹³ + 79x¹² + 106x¹¹

Discard the lead 0 term to get:

181x²⁸ + 16x²⁷ + 160x²⁶ + 85x²⁵ + 203x²⁴ + 237x²³ + 166x²² + 190x²¹ + 76x²⁰ + 178x¹⁹ + 213x¹⁸ + 113x¹⁷ + 31x¹⁶ + 164x¹⁵ + 211x¹⁴ + 136x¹³ + 79x¹² + 106x¹¹

Step 6a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 181x²⁸. Convert 181x²⁸ to alpha notation. According to the log antilog table, for the integer value 181, the alpha exponent is 42. Therefore 181 = α⁴². Multiply the generator polynomial by α⁴²:

(α⁴² * α⁰)x²⁸ + (α⁴² * α²¹⁵)x²⁷ + (α⁴² * α²³⁴)x²⁶ + (α⁴² * α¹⁵⁸)x²⁵ + (α⁴² * α⁹⁴)x²⁴ + (α⁴² * α¹⁸⁴)x²³ + (α⁴² * α⁹⁷)x²² + (α⁴² * α¹¹⁸)x²¹ + (α⁴² * α¹⁷⁰)x²⁰ + (α⁴² * α⁷⁹)x¹⁹ + (α⁴² * α¹⁸⁷)x¹⁸ + (α⁴² * α¹⁵²)x¹⁷ + (α⁴² * α¹⁴⁸)x¹⁶ + (α⁴² * α²⁵²)x¹⁵ + (α⁴² * α¹⁷⁹)x¹⁴ + (α⁴² * α⁵)x¹³ + (α⁴² * α⁹⁸)x¹² + (α⁴² * α⁹⁶)x¹¹ + (α⁴² * α¹⁵³)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:

α⁴²x²⁸ + α^{(257 % 255)}x²⁷ + α^{(276 % 255)}x²⁶ + α²⁰⁰x²⁵ + α¹³⁶x²⁴ + α²²⁶x²³ + α¹³⁹x²² + α¹⁶⁰x²¹ + α²¹²x²⁰ + α¹²¹x¹⁹ + α²²⁹x¹⁸ + α¹⁹⁴x¹⁷ + α¹⁹⁰x¹⁶ + α^{(294 % 255)}x¹⁵ + α²²¹x¹⁴ + α⁴⁷x¹³ + α¹⁴⁰x¹² + α¹³⁸x¹¹ + α¹⁹⁵x¹⁰

The result is:

α⁴²x²⁸ + α²x²⁷ + α²¹x²⁶ + α²⁰⁰x²⁵ + α¹³⁶x²⁴ + α²²⁶x²³ + α¹³⁹x²² + α¹⁶⁰x²¹ + α²¹²x²⁰ + α¹²¹x¹⁹ + α²²⁹x¹⁸ + α¹⁹⁴x¹⁷ + α¹⁹⁰x¹⁶ + α³⁹x¹⁵ + α²²¹x¹⁴ + α⁴⁷x¹³ + α¹⁴⁰x¹² + α¹³⁸x¹¹ + α¹⁹⁵x¹⁰

Now, convert this to integer notation:

181x²⁸ + 4x²⁷ + 117x²⁶ + 28x²⁵ + 79x²⁴ + 72x²³ + 66x²² + 230x²¹ + 121x²⁰ + 118x¹⁹ + 122x¹⁸ + 50x¹⁷ + 174x¹⁶ + 53x¹⁵ + 69x¹⁴ + 35x¹³ + 132x¹² + 33x¹¹ + 100x¹⁰

Step 6b: XOR the result with the result from step 5b

Use the result from step 5b to perform the next XOR.

(181 ⊕ 181)x²⁸ + (16 ⊕ 4)x²⁷ + (160 ⊕ 117)x²⁶ + (85 ⊕ 28)x²⁵ + (203 ⊕ 79)x²⁴ + (237 ⊕ 72)x²³ + (166 ⊕ 66)x²² + (190 ⊕ 230)x²¹ + (76 ⊕ 121)x²⁰ + (178 ⊕ 118)x¹⁹ + (213 ⊕ 122)x¹⁸ + (113 ⊕ 50)x¹⁷ + (31 ⊕ 174)x¹⁶ + (164 ⊕ 53)x¹⁵ + (211 ⊕ 69)x¹⁴ + (136 ⊕ 35)x¹³ + (79 ⊕ 132)x¹² + (106 ⊕ 33)x¹¹ + (0 ⊕ 100)x¹⁰

The result is:

0x²⁸ + 20x²⁷ + 213x²⁶ + 73x²⁵ + 132x²⁴ + 165x²³ + 228x²² + 88x²¹ + 53x²⁰ + 196x¹⁹ + 175x¹⁸ + 67x¹⁷ + 177x¹⁶ + 145x¹⁵ + 150x¹⁴ + 171x¹³ + 203x¹² + 75x¹¹ + 100x¹⁰

Discard the lead 0 term to get:

20x²⁷ + 213x²⁶ + 73x²⁵ + 132x²⁴ + 165x²³ + 228x²² + 88x²¹ + 53x²⁰ + 196x¹⁹ + 175x¹⁸ + 67x¹⁷ + 177x¹⁶ + 145x¹⁵ + 150x¹⁴ + 171x¹³ + 203x¹² + 75x¹¹ + 100x¹⁰

Step 7a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 20x²⁷. Convert 20x²⁷ to alpha notation. According to the log antilog table, for the integer value 20, the alpha exponent is 52. Therefore 20 = α⁵². Multiply the generator polynomial by α⁵²:

(α⁵² * α⁰)x²⁷ + (α⁵² * α²¹⁵)x²⁶ + (α⁵² * α²³⁴)x²⁵ + (α⁵² * α¹⁵⁸)x²⁴ + (α⁵² * α⁹⁴)x²³ + (α⁵² * α¹⁸⁴)x²² + (α⁵² * α⁹⁷)x²¹ + (α⁵² * α¹¹⁸)x²⁰ + (α⁵² * α¹⁷⁰)x¹⁹ + (α⁵² * α⁷⁹)x¹⁸ + (α⁵² * α¹⁸⁷)x¹⁷ + (α⁵² * α¹⁵²)x¹⁶ + (α⁵² * α¹⁴⁸)x¹⁵ + (α⁵² * α²⁵²)x¹⁴ + (α⁵² * α¹⁷⁹)x¹³ + (α⁵² * α⁵)x¹² + (α⁵² * α⁹⁸)x¹¹ + (α⁵² * α⁹⁶)x¹⁰ + (α⁵² * α¹⁵³)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:

α⁵²x²⁷ + α^{(267 % 255)}x²⁶ + α^{(286 % 255)}x²⁵ + α²¹⁰x²⁴ + α¹⁴⁶x²³ + α²³⁶x²² + α¹⁴⁹x²¹ + α¹⁷⁰x²⁰ + α²²²x¹⁹ + α¹³¹x¹⁸ + α²³⁹x¹⁷ + α²⁰⁴x¹⁶ + α²⁰⁰x¹⁵ + α^{(304 % 255)}x¹⁴ + α²³¹x¹³ + α⁵⁷x¹² + α¹⁵⁰x¹¹ + α¹⁴⁸x¹⁰ + α²⁰⁵x⁹

The result is:

α⁵²x²⁷ + α¹²x²⁶ + α³¹x²⁵ + α²¹⁰x²⁴ + α¹⁴⁶x²³ + α²³⁶x²² + α¹⁴⁹x²¹ + α¹⁷⁰x²⁰ + α²²²x¹⁹ + α¹³¹x¹⁸ + α²³⁹x¹⁷ + α²⁰⁴x¹⁶ + α²⁰⁰x¹⁵ + α⁴⁹x¹⁴ + α²³¹x¹³ + α⁵⁷x¹² + α¹⁵⁰x¹¹ + α¹⁴⁸x¹⁰ + α²⁰⁵x⁹

Now, convert this to integer notation:

20x²⁷ + 205x²⁶ + 192x²⁵ + 89x²⁴ + 154x²³ + 203x²² + 164x²¹ + 215x²⁰ + 138x¹⁹ + 92x¹⁸ + 22x¹⁷ + 221x¹⁶ + 28x¹⁵ + 140x¹⁴ + 245x¹³ + 186x¹² + 85x¹¹ + 82x¹⁰ + 167x⁹

Step 7b: XOR the result with the result from step 6b

Use the result from step 6b to perform the next XOR.

(20 ⊕ 20)x²⁷ + (213 ⊕ 205)x²⁶ + (73 ⊕ 192)x²⁵ + (132 ⊕ 89)x²⁴ + (165 ⊕ 154)x²³ + (228 ⊕ 203)x²² + (88 ⊕ 164)x²¹ + (53 ⊕ 215)x²⁰ + (196 ⊕ 138)x¹⁹ + (175 ⊕ 92)x¹⁸ + (67 ⊕ 22)x¹⁷ + (177 ⊕ 221)x¹⁶ + (145 ⊕ 28)x¹⁵ + (150 ⊕ 140)x¹⁴ + (171 ⊕ 245)x¹³ + (203 ⊕ 186)x¹² + (75 ⊕ 85)x¹¹ + (100 ⊕ 82)x¹⁰ + (0 ⊕ 167)x⁹

The result is:

0x²⁷ + 24x²⁶ + 137x²⁵ + 221x²⁴ + 63x²³ + 47x²² + 252x²¹ + 226x²⁰ + 78x¹⁹ + 243x¹⁸ + 85x¹⁷ + 108x¹⁶ + 141x¹⁵ + 26x¹⁴ + 94x¹³ + 113x¹² + 30x¹¹ + 54x¹⁰ + 167x⁹

Discard the lead 0 term to get:

24x²⁶ + 137x²⁵ + 221x²⁴ + 63x²³ + 47x²² + 252x²¹ + 226x²⁰ + 78x¹⁹ + 243x¹⁸ + 85x¹⁷ + 108x¹⁶ + 141x¹⁵ + 26x¹⁴ + 94x¹³ + 113x¹² + 30x¹¹ + 54x¹⁰ + 167x⁹

Step 8a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 24x²⁶. Convert 24x²⁶ to alpha notation. According to the log antilog table, for the integer value 24, the alpha exponent is 28. Therefore 24 = α²⁸. Multiply the generator polynomial by α²⁸:

(α²⁸ * α⁰)x²⁶ + (α²⁸ * α²¹⁵)x²⁵ + (α²⁸ * α²³⁴)x²⁴ + (α²⁸ * α¹⁵⁸)x²³ + (α²⁸ * α⁹⁴)x²² + (α²⁸ * α¹⁸⁴)x²¹ + (α²⁸ * α⁹⁷)x²⁰ + (α²⁸ * α¹¹⁸)x¹⁹ + (α²⁸ * α¹⁷⁰)x¹⁸ + (α²⁸ * α⁷⁹)x¹⁷ + (α²⁸ * α¹⁸⁷)x¹⁶ + (α²⁸ * α¹⁵²)x¹⁵ + (α²⁸ * α¹⁴⁸)x¹⁴ + (α²⁸ * α²⁵²)x¹³ + (α²⁸ * α¹⁷⁹)x¹² + (α²⁸ * α⁵)x¹¹ + (α²⁸ * α⁹⁸)x¹⁰ + (α²⁸ * α⁹⁶)x⁹ + (α²⁸ * α¹⁵³)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:

α²⁸x²⁶ + α²⁴³x²⁵ + α^{(262 % 255)}x²⁴ + α¹⁸⁶x²³ + α¹²²x²² + α²¹²x²¹ + α¹²⁵x²⁰ + α¹⁴⁶x¹⁹ + α¹⁹⁸x¹⁸ + α¹⁰⁷x¹⁷ + α²¹⁵x¹⁶ + α¹⁸⁰x¹⁵ + α¹⁷⁶x¹⁴ + α^{(280 % 255)}x¹³ + α²⁰⁷x¹² + α³³x¹¹ + α¹²⁶x¹⁰ + α¹²⁴x⁹ + α¹⁸¹x⁸

The result is:

α²⁸x²⁶ + α²⁴³x²⁵ + α⁷x²⁴ + α¹⁸⁶x²³ + α¹²²x²² + α²¹²x²¹ + α¹²⁵x²⁰ + α¹⁴⁶x¹⁹ + α¹⁹⁸x¹⁸ + α¹⁰⁷x¹⁷ + α²¹⁵x¹⁶ + α¹⁸⁰x¹⁵ + α¹⁷⁶x¹⁴ + α²⁵x¹³ + α²⁰⁷x¹² + α³³x¹¹ + α¹²⁶x¹⁰ + α¹²⁴x⁹ + α¹⁸¹x⁸

Now, convert this to integer notation:

24x²⁶ + 125x²⁵ + 128x²⁴ + 110x²³ + 236x²² + 121x²¹ + 51x²⁰ + 154x¹⁹ + 7x¹⁸ + 104x¹⁷ + 239x¹⁶ + 150x¹⁵ + 227x¹⁴ + 3x¹³ + 166x¹² + 39x¹¹ + 102x¹⁰ + 151x⁹ + 49x⁸

Step 8b: XOR the result with the result from step 7b

Use the result from step 7b to perform the next XOR.

(24 ⊕ 24)x²⁶ + (137 ⊕ 125)x²⁵ + (221 ⊕ 128)x²⁴ + (63 ⊕ 110)x²³ + (47 ⊕ 236)x²² + (252 ⊕ 121)x²¹ + (226 ⊕ 51)x²⁰ + (78 ⊕ 154)x¹⁹ + (243 ⊕ 7)x¹⁸ + (85 ⊕ 104)x¹⁷ + (108 ⊕ 239)x¹⁶ + (141 ⊕ 150)x¹⁵ + (26 ⊕ 227)x¹⁴ + (94 ⊕ 3)x¹³ + (113 ⊕ 166)x¹² + (30 ⊕ 39)x¹¹ + (54 ⊕ 102)x¹⁰ + (167 ⊕ 151)x⁹ + (0 ⊕ 49)x⁸

The result is:

0x²⁶ + 244x²⁵ + 93x²⁴ + 81x²³ + 195x²² + 133x²¹ + 209x²⁰ + 212x¹⁹ + 244x¹⁸ + 61x¹⁷ + 131x¹⁶ + 27x¹⁵ + 249x¹⁴ + 93x¹³ + 215x¹² + 57x¹¹ + 80x¹⁰ + 48x⁹ + 49x⁸

Discard the lead 0 term to get:

244x²⁵ + 93x²⁴ + 81x²³ + 195x²² + 133x²¹ + 209x²⁰ + 212x¹⁹ + 244x¹⁸ + 61x¹⁷ + 131x¹⁶ + 27x¹⁵ + 249x¹⁴ + 93x¹³ + 215x¹² + 57x¹¹ + 80x¹⁰ + 48x⁹ + 49x⁸

Step 9a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 244x²⁵. Convert 244x²⁵ to alpha notation. According to the log antilog table, for the integer value 244, the alpha exponent is 230. Therefore 244 = α²³⁰. Multiply the generator polynomial by α²³⁰:

(α²³⁰ * α⁰)x²⁵ + (α²³⁰ * α²¹⁵)x²⁴ + (α²³⁰ * α²³⁴)x²³ + (α²³⁰ * α¹⁵⁸)x²² + (α²³⁰ * α⁹⁴)x²¹ + (α²³⁰ * α¹⁸⁴)x²⁰ + (α²³⁰ * α⁹⁷)x¹⁹ + (α²³⁰ * α¹¹⁸)x¹⁸ + (α²³⁰ * α¹⁷⁰)x¹⁷ + (α²³⁰ * α⁷⁹)x¹⁶ + (α²³⁰ * α¹⁸⁷)x¹⁵ + (α²³⁰ * α¹⁵²)x¹⁴ + (α²³⁰ * α¹⁴⁸)x¹³ + (α²³⁰ * α²⁵²)x¹² + (α²³⁰ * α¹⁷⁹)x¹¹ + (α²³⁰ * α⁵)x¹⁰ + (α²³⁰ * α⁹⁸)x⁹ + (α²³⁰ * α⁹⁶)x⁸ + (α²³⁰ * α¹⁵³)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:

α²³⁰x²⁵ + α^{(445 % 255)}x²⁴ + α^{(464 % 255)}x²³ + α^{(388 % 255)}x²² + α^{(324 % 255)}x²¹ + α^{(414 % 255)}x²⁰ + α^{(327 % 255)}x¹⁹ + α^{(348 % 255)}x¹⁸ + α^{(400 % 255)}x¹⁷ + α^{(309 % 255)}x¹⁶ + α^{(417 % 255)}x¹⁵ + α^{(382 % 255)}x¹⁴ + α^{(378 % 255)}x¹³ + α^{(482 % 255)}x¹² + α^{(409 % 255)}x¹¹ + α²³⁵x¹⁰ + α^{(328 % 255)}x⁹ + α^{(326 % 255)}x⁸ + α^{(383 % 255)}x⁷

The result is:

α²³⁰x²⁵ + α¹⁹⁰x²⁴ + α²⁰⁹x²³ + α¹³³x²² + α⁶⁹x²¹ + α¹⁵⁹x²⁰ + α⁷²x¹⁹ + α⁹³x¹⁸ + α¹⁴⁵x¹⁷ + α⁵⁴x¹⁶ + α¹⁶²x¹⁵ + α¹²⁷x¹⁴ + α¹²³x¹³ + α²²⁷x¹² + α¹⁵⁴x¹¹ + α²³⁵x¹⁰ + α⁷³x⁹ + α⁷¹x⁸ + α¹²⁸x⁷

Now, convert this to integer notation:

244x²⁵ + 174x²⁴ + 162x²³ + 109x²² + 47x²¹ + 115x²⁰ + 101x¹⁹ + 182x¹⁸ + 77x¹⁷ + 80x¹⁶ + 191x¹⁵ + 204x¹⁴ + 197x¹³ + 144x¹² + 57x¹¹ + 235x¹⁰ + 202x⁹ + 188x⁸ + 133x⁷

Step 9b: XOR the result with the result from step 8b

Use the result from step 8b to perform the next XOR.

(244 ⊕ 244)x²⁵ + (93 ⊕ 174)x²⁴ + (81 ⊕ 162)x²³ + (195 ⊕ 109)x²² + (133 ⊕ 47)x²¹ + (209 ⊕ 115)x²⁰ + (212 ⊕ 101)x¹⁹ + (244 ⊕ 182)x¹⁸ + (61 ⊕ 77)x¹⁷ + (131 ⊕ 80)x¹⁶ + (27 ⊕ 191)x¹⁵ + (249 ⊕ 204)x¹⁴ + (93 ⊕ 197)x¹³ + (215 ⊕ 144)x¹² + (57 ⊕ 57)x¹¹ + (80 ⊕ 235)x¹⁰ + (48 ⊕ 202)x⁹ + (49 ⊕ 188)x⁸ + (0 ⊕ 133)x⁷

The result is:

0x²⁵ + 243x²⁴ + 243x²³ + 174x²² + 170x²¹ + 162x²⁰ + 177x¹⁹ + 66x¹⁸ + 112x¹⁷ + 211x¹⁶ + 164x¹⁵ + 53x¹⁴ + 152x¹³ + 71x¹² + 0x¹¹ + 187x¹⁰ + 250x⁹ + 141x⁸ + 133x⁷

Discard the lead 0 term to get:

243x²⁴ + 243x²³ + 174x²² + 170x²¹ + 162x²⁰ + 177x¹⁹ + 66x¹⁸ + 112x¹⁷ + 211x¹⁶ + 164x¹⁵ + 53x¹⁴ + 152x¹³ + 71x¹² + 0x¹¹ + 187x¹⁰ + 250x⁹ + 141x⁸ + 133x⁷

Step 10a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 243x²⁴. Convert 243x²⁴ to alpha notation. According to the log antilog table, for the integer value 243, the alpha exponent is 233. Therefore 243 = α²³³. Multiply the generator polynomial by α²³³:

(α²³³ * α⁰)x²⁴ + (α²³³ * α²¹⁵)x²³ + (α²³³ * α²³⁴)x²² + (α²³³ * α¹⁵⁸)x²¹ + (α²³³ * α⁹⁴)x²⁰ + (α²³³ * α¹⁸⁴)x¹⁹ + (α²³³ * α⁹⁷)x¹⁸ + (α²³³ * α¹¹⁸)x¹⁷ + (α²³³ * α¹⁷⁰)x¹⁶ + (α²³³ * α⁷⁹)x¹⁵ + (α²³³ * α¹⁸⁷)x¹⁴ + (α²³³ * α¹⁵²)x¹³ + (α²³³ * α¹⁴⁸)x¹² + (α²³³ * α²⁵²)x¹¹ + (α²³³ * α¹⁷⁹)x¹⁰ + (α²³³ * α⁵)x⁹ + (α²³³ * α⁹⁸)x⁸ + (α²³³ * α⁹⁶)x⁷ + (α²³³ * α¹⁵³)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:

α²³³x²⁴ + α^{(448 % 255)}x²³ + α^{(467 % 255)}x²² + α^{(391 % 255)}x²¹ + α^{(327 % 255)}x²⁰ + α^{(417 % 255)}x¹⁹ + α^{(330 % 255)}x¹⁸ + α^{(351 % 255)}x¹⁷ + α^{(403 % 255)}x¹⁶ + α^{(312 % 255)}x¹⁵ + α^{(420 % 255)}x¹⁴ + α^{(385 % 255)}x¹³ + α^{(381 % 255)}x¹² + α^{(485 % 255)}x¹¹ + α^{(412 % 255)}x¹⁰ + α²³⁸x⁹ + α^{(331 % 255)}x⁸ + α^{(329 % 255)}x⁷ + α^{(386 % 255)}x⁶

The result is:

α²³³x²⁴ + α¹⁹³x²³ + α²¹²x²² + α¹³⁶x²¹ + α⁷²x²⁰ + α¹⁶²x¹⁹ + α⁷⁵x¹⁸ + α⁹⁶x¹⁷ + α¹⁴⁸x¹⁶ + α⁵⁷x¹⁵ + α¹⁶⁵x¹⁴ + α¹³⁰x¹³ + α¹²⁶x¹² + α²³⁰x¹¹ + α¹⁵⁷x¹⁰ + α²³⁸x⁹ + α⁷⁶x⁸ + α⁷⁴x⁷ + α¹³¹x⁶

Now, convert this to integer notation:

243x²⁴ + 25x²³ + 121x²² + 79x²¹ + 101x²⁰ + 191x¹⁹ + 15x¹⁸ + 217x¹⁷ + 82x¹⁶ + 186x¹⁵ + 145x¹⁴ + 46x¹³ + 102x¹² + 244x¹¹ + 213x¹⁰ + 11x⁹ + 30x⁸ + 137x⁷ + 92x⁶

Step 10b: XOR the result with the result from step 9b

Use the result from step 9b to perform the next XOR.

(243 ⊕ 243)x²⁴ + (243 ⊕ 25)x²³ + (174 ⊕ 121)x²² + (170 ⊕ 79)x²¹ + (162 ⊕ 101)x²⁰ + (177 ⊕ 191)x¹⁹ + (66 ⊕ 15)x¹⁸ + (112 ⊕ 217)x¹⁷ + (211 ⊕ 82)x¹⁶ + (164 ⊕ 186)x¹⁵ + (53 ⊕ 145)x¹⁴ + (152 ⊕ 46)x¹³ + (71 ⊕ 102)x¹² + (0 ⊕ 244)x¹¹ + (187 ⊕ 213)x¹⁰ + (250 ⊕ 11)x⁹ + (141 ⊕ 30)x⁸ + (133 ⊕ 137)x⁷ + (0 ⊕ 92)x⁶

The result is:

0x²⁴ + 234x²³ + 215x²² + 229x²¹ + 199x²⁰ + 14x¹⁹ + 77x¹⁸ + 169x¹⁷ + 129x¹⁶ + 30x¹⁵ + 164x¹⁴ + 182x¹³ + 33x¹² + 244x¹¹ + 110x¹⁰ + 241x⁹ + 147x⁸ + 12x⁷ + 92x⁶

Discard the lead 0 term to get:

234x²³ + 215x²² + 229x²¹ + 199x²⁰ + 14x¹⁹ + 77x¹⁸ + 169x¹⁷ + 129x¹⁶ + 30x¹⁵ + 164x¹⁴ + 182x¹³ + 33x¹² + 244x¹¹ + 110x¹⁰ + 241x⁹ + 147x⁸ + 12x⁷ + 92x⁶

Step 11a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 234x²³. Convert 234x²³ to alpha notation. According to the log antilog table, for the integer value 234, the alpha exponent is 22. Therefore 234 = α²². Multiply the generator polynomial by α²²:

(α²² * α⁰)x²³ + (α²² * α²¹⁵)x²² + (α²² * α²³⁴)x²¹ + (α²² * α¹⁵⁸)x²⁰ + (α²² * α⁹⁴)x¹⁹ + (α²² * α¹⁸⁴)x¹⁸ + (α²² * α⁹⁷)x¹⁷ + (α²² * α¹¹⁸)x¹⁶ + (α²² * α¹⁷⁰)x¹⁵ + (α²² * α⁷⁹)x¹⁴ + (α²² * α¹⁸⁷)x¹³ + (α²² * α¹⁵²)x¹² + (α²² * α¹⁴⁸)x¹¹ + (α²² * α²⁵²)x¹⁰ + (α²² * α¹⁷⁹)x⁹ + (α²² * α⁵)x⁸ + (α²² * α⁹⁸)x⁷ + (α²² * α⁹⁶)x⁶ + (α²² * α¹⁵³)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:

α²²x²³ + α²³⁷x²² + α^{(256 % 255)}x²¹ + α¹⁸⁰x²⁰ + α¹¹⁶x¹⁹ + α²⁰⁶x¹⁸ + α¹¹⁹x¹⁷ + α¹⁴⁰x¹⁶ + α¹⁹²x¹⁵ + α¹⁰¹x¹⁴ + α²⁰⁹x¹³ + α¹⁷⁴x¹² + α¹⁷⁰x¹¹ + α^{(274 % 255)}x¹⁰ + α²⁰¹x⁹ + α²⁷x⁸ + α¹²⁰x⁷ + α¹¹⁸x⁶ + α¹⁷⁵x⁵

The result is:

α²²x²³ + α²³⁷x²² + α¹x²¹ + α¹⁸⁰x²⁰ + α¹¹⁶x¹⁹ + α²⁰⁶x¹⁸ + α¹¹⁹x¹⁷ + α¹⁴⁰x¹⁶ + α¹⁹²x¹⁵ + α¹⁰¹x¹⁴ + α²⁰⁹x¹³ + α¹⁷⁴x¹² + α¹⁷⁰x¹¹ + α¹⁹x¹⁰ + α²⁰¹x⁹ + α²⁷x⁸ + α¹²⁰x⁷ + α¹¹⁸x⁶ + α¹⁷⁵x⁵

Now, convert this to integer notation:

234x²³ + 139x²² + 2x²¹ + 150x²⁰ + 248x¹⁹ + 83x¹⁸ + 147x¹⁷ + 132x¹⁶ + 130x¹⁵ + 34x¹⁴ + 162x¹³ + 241x¹² + 215x¹¹ + 90x¹⁰ + 56x⁹ + 12x⁸ + 59x⁷ + 199x⁶ + 255x⁵

Step 11b: XOR the result with the result from step 10b

Use the result from step 10b to perform the next XOR.

(234 ⊕ 234)x²³ + (215 ⊕ 139)x²² + (229 ⊕ 2)x²¹ + (199 ⊕ 150)x²⁰ + (14 ⊕ 248)x¹⁹ + (77 ⊕ 83)x¹⁸ + (169 ⊕ 147)x¹⁷ + (129 ⊕ 132)x¹⁶ + (30 ⊕ 130)x¹⁵ + (164 ⊕ 34)x¹⁴ + (182 ⊕ 162)x¹³ + (33 ⊕ 241)x¹² + (244 ⊕ 215)x¹¹ + (110 ⊕ 90)x¹⁰ + (241 ⊕ 56)x⁹ + (147 ⊕ 12)x⁸ + (12 ⊕ 59)x⁷ + (92 ⊕ 199)x⁶ + (0 ⊕ 255)x⁵

The result is:

0x²³ + 92x²² + 231x²¹ + 81x²⁰ + 246x¹⁹ + 30x¹⁸ + 58x¹⁷ + 5x¹⁶ + 156x¹⁵ + 134x¹⁴ + 20x¹³ + 208x¹² + 35x¹¹ + 52x¹⁰ + 201x⁹ + 159x⁸ + 55x⁷ + 155x⁶ + 255x⁵

Discard the lead 0 term to get:

92x²² + 231x²¹ + 81x²⁰ + 246x¹⁹ + 30x¹⁸ + 58x¹⁷ + 5x¹⁶ + 156x¹⁵ + 134x¹⁴ + 20x¹³ + 208x¹² + 35x¹¹ + 52x¹⁰ + 201x⁹ + 159x⁸ + 55x⁷ + 155x⁶ + 255x⁵

Step 12a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 92x²². Convert 92x²² to alpha notation. According to the log antilog table, for the integer value 92, the alpha exponent is 131. Therefore 92 = α¹³¹. Multiply the generator polynomial by α¹³¹:

(α¹³¹ * α⁰)x²² + (α¹³¹ * α²¹⁵)x²¹ + (α¹³¹ * α²³⁴)x²⁰ + (α¹³¹ * α¹⁵⁸)x¹⁹ + (α¹³¹ * α⁹⁴)x¹⁸ + (α¹³¹ * α¹⁸⁴)x¹⁷ + (α¹³¹ * α⁹⁷)x¹⁶ + (α¹³¹ * α¹¹⁸)x¹⁵ + (α¹³¹ * α¹⁷⁰)x¹⁴ + (α¹³¹ * α⁷⁹)x¹³ + (α¹³¹ * α¹⁸⁷)x¹² + (α¹³¹ * α¹⁵²)x¹¹ + (α¹³¹ * α¹⁴⁸)x¹⁰ + (α¹³¹ * α²⁵²)x⁹ + (α¹³¹ * α¹⁷⁹)x⁸ + (α¹³¹ * α⁵)x⁷ + (α¹³¹ * α⁹⁸)x⁶ + (α¹³¹ * α⁹⁶)x⁵ + (α¹³¹ * α¹⁵³)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:

α¹³¹x²² + α^{(346 % 255)}x²¹ + α^{(365 % 255)}x²⁰ + α^{(289 % 255)}x¹⁹ + α²²⁵x¹⁸ + α^{(315 % 255)}x¹⁷ + α²²⁸x¹⁶ + α²⁴⁹x¹⁵ + α^{(301 % 255)}x¹⁴ + α²¹⁰x¹³ + α^{(318 % 255)}x¹² + α^{(283 % 255)}x¹¹ + α^{(279 % 255)}x¹⁰ + α^{(383 % 255)}x⁹ + α^{(310 % 255)}x⁸ + α¹³⁶x⁷ + α²²⁹x⁶ + α²²⁷x⁵ + α^{(284 % 255)}x⁴

The result is:

α¹³¹x²² + α⁹¹x²¹ + α¹¹⁰x²⁰ + α³⁴x¹⁹ + α²²⁵x¹⁸ + α⁶⁰x¹⁷ + α²²⁸x¹⁶ + α²⁴⁹x¹⁵ + α⁴⁶x¹⁴ + α²¹⁰x¹³ + α⁶³x¹² + α²⁸x¹¹ + α²⁴x¹⁰ + α¹²⁸x⁹ + α⁵⁵x⁸ + α¹³⁶x⁷ + α²²⁹x⁶ + α²²⁷x⁵ + α²⁹x⁴

Now, convert this to integer notation:

92x²² + 163x²¹ + 103x²⁰ + 78x¹⁹ + 36x¹⁸ + 185x¹⁷ + 61x¹⁶ + 54x¹⁵ + 159x¹⁴ + 89x¹³ + 161x¹² + 24x¹¹ + 143x¹⁰ + 133x⁹ + 160x⁸ + 79x⁷ + 122x⁶ + 144x⁵ + 48x⁴

Step 12b: XOR the result with the result from step 11b

Use the result from step 11b to perform the next XOR.

(92 ⊕ 92)x²² + (231 ⊕ 163)x²¹ + (81 ⊕ 103)x²⁰ + (246 ⊕ 78)x¹⁹ + (30 ⊕ 36)x¹⁸ + (58 ⊕ 185)x¹⁷ + (5 ⊕ 61)x¹⁶ + (156 ⊕ 54)x¹⁵ + (134 ⊕ 159)x¹⁴ + (20 ⊕ 89)x¹³ + (208 ⊕ 161)x¹² + (35 ⊕ 24)x¹¹ + (52 ⊕ 143)x¹⁰ + (201 ⊕ 133)x⁹ + (159 ⊕ 160)x⁸ + (55 ⊕ 79)x⁷ + (155 ⊕ 122)x⁶ + (255 ⊕ 144)x⁵ + (0 ⊕ 48)x⁴

The result is:

0x²² + 68x²¹ + 54x²⁰ + 184x¹⁹ + 58x¹⁸ + 131x¹⁷ + 56x¹⁶ + 170x¹⁵ + 25x¹⁴ + 77x¹³ + 113x¹² + 59x¹¹ + 187x¹⁰ + 76x⁹ + 63x⁸ + 120x⁷ + 225x⁶ + 111x⁵ + 48x⁴

Discard the lead 0 term to get:

68x²¹ + 54x²⁰ + 184x¹⁹ + 58x¹⁸ + 131x¹⁷ + 56x¹⁶ + 170x¹⁵ + 25x¹⁴ + 77x¹³ + 113x¹² + 59x¹¹ + 187x¹⁰ + 76x⁹ + 63x⁸ + 120x⁷ + 225x⁶ + 111x⁵ + 48x⁴

Step 13a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 68x²¹. Convert 68x²¹ to alpha notation. According to the log antilog table, for the integer value 68, the alpha exponent is 102. Therefore 68 = α¹⁰². Multiply the generator polynomial by α¹⁰²:

(α¹⁰² * α⁰)x²¹ + (α¹⁰² * α²¹⁵)x²⁰ + (α¹⁰² * α²³⁴)x¹⁹ + (α¹⁰² * α¹⁵⁸)x¹⁸ + (α¹⁰² * α⁹⁴)x¹⁷ + (α¹⁰² * α¹⁸⁴)x¹⁶ + (α¹⁰² * α⁹⁷)x¹⁵ + (α¹⁰² * α¹¹⁸)x¹⁴ + (α¹⁰² * α¹⁷⁰)x¹³ + (α¹⁰² * α⁷⁹)x¹² + (α¹⁰² * α¹⁸⁷)x¹¹ + (α¹⁰² * α¹⁵²)x¹⁰ + (α¹⁰² * α¹⁴⁸)x⁹ + (α¹⁰² * α²⁵²)x⁸ + (α¹⁰² * α¹⁷⁹)x⁷ + (α¹⁰² * α⁵)x⁶ + (α¹⁰² * α⁹⁸)x⁵ + (α¹⁰² * α⁹⁶)x⁴ + (α¹⁰² * α¹⁵³)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:

α¹⁰²x²¹ + α^{(317 % 255)}x²⁰ + α^{(336 % 255)}x¹⁹ + α^{(260 % 255)}x¹⁸ + α¹⁹⁶x¹⁷ + α^{(286 % 255)}x¹⁶ + α¹⁹⁹x¹⁵ + α²²⁰x¹⁴ + α^{(272 % 255)}x¹³ + α¹⁸¹x¹² + α^{(289 % 255)}x¹¹ + α²⁵⁴x¹⁰ + α²⁵⁰x⁹ + α^{(354 % 255)}x⁸ + α^{(281 % 255)}x⁷ + α¹⁰⁷x⁶ + α²⁰⁰x⁵ + α¹⁹⁸x⁴ + α²⁵⁵x³

The result is:

α¹⁰²x²¹ + α⁶²x²⁰ + α⁸¹x¹⁹ + α⁵x¹⁸ + α¹⁹⁶x¹⁷ + α³¹x¹⁶ + α¹⁹⁹x¹⁵ + α²²⁰x¹⁴ + α¹⁷x¹³ + α¹⁸¹x¹² + α³⁴x¹¹ + α²⁵⁴x¹⁰ + α²⁵⁰x⁹ + α⁹⁹x⁸ + α²⁶x⁷ + α¹⁰⁷x⁶ + α²⁰⁰x⁵ + α¹⁹⁸x⁴ + α⁰x³

Now, convert this to integer notation:

68x²¹ + 222x²⁰ + 231x¹⁹ + 32x¹⁸ + 200x¹⁷ + 192x¹⁶ + 14x¹⁵ + 172x¹⁴ + 152x¹³ + 49x¹² + 78x¹¹ + 142x¹⁰ + 108x⁹ + 134x⁸ + 6x⁷ + 104x⁶ + 28x⁵ + 7x⁴ + 1x³

Step 13b: XOR the result with the result from step 12b

Use the result from step 12b to perform the next XOR.

(68 ⊕ 68)x²¹ + (54 ⊕ 222)x²⁰ + (184 ⊕ 231)x¹⁹ + (58 ⊕ 32)x¹⁸ + (131 ⊕ 200)x¹⁷ + (56 ⊕ 192)x¹⁶ + (170 ⊕ 14)x¹⁵ + (25 ⊕ 172)x¹⁴ + (77 ⊕ 152)x¹³ + (113 ⊕ 49)x¹² + (59 ⊕ 78)x¹¹ + (187 ⊕ 142)x¹⁰ + (76 ⊕ 108)x⁹ + (63 ⊕ 134)x⁸ + (120 ⊕ 6)x⁷ + (225 ⊕ 104)x⁶ + (111 ⊕ 28)x⁵ + (48 ⊕ 7)x⁴ + (0 ⊕ 1)x³

The result is:

0x²¹ + 232x²⁰ + 95x¹⁹ + 26x¹⁸ + 75x¹⁷ + 248x¹⁶ + 164x¹⁵ + 181x¹⁴ + 213x¹³ + 64x¹² + 117x¹¹ + 53x¹⁰ + 32x⁹ + 185x⁸ + 126x⁷ + 137x⁶ + 115x⁵ + 55x⁴ + 1x³

Discard the lead 0 term to get:

232x²⁰ + 95x¹⁹ + 26x¹⁸ + 75x¹⁷ + 248x¹⁶ + 164x¹⁵ + 181x¹⁴ + 213x¹³ + 64x¹² + 117x¹¹ + 53x¹⁰ + 32x⁹ + 185x⁸ + 126x⁷ + 137x⁶ + 115x⁵ + 55x⁴ + 1x³

Step 14a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 232x²⁰. Convert 232x²⁰ to alpha notation. According to the log antilog table, for the integer value 232, the alpha exponent is 11. Therefore 232 = α¹¹. Multiply the generator polynomial by α¹¹:

(α¹¹ * α⁰)x²⁰ + (α¹¹ * α²¹⁵)x¹⁹ + (α¹¹ * α²³⁴)x¹⁸ + (α¹¹ * α¹⁵⁸)x¹⁷ + (α¹¹ * α⁹⁴)x¹⁶ + (α¹¹ * α¹⁸⁴)x¹⁵ + (α¹¹ * α⁹⁷)x¹⁴ + (α¹¹ * α¹¹⁸)x¹³ + (α¹¹ * α¹⁷⁰)x¹² + (α¹¹ * α⁷⁹)x¹¹ + (α¹¹ * α¹⁸⁷)x¹⁰ + (α¹¹ * α¹⁵²)x⁹ + (α¹¹ * α¹⁴⁸)x⁸ + (α¹¹ * α²⁵²)x⁷ + (α¹¹ * α¹⁷⁹)x⁶ + (α¹¹ * α⁵)x⁵ + (α¹¹ * α⁹⁸)x⁴ + (α¹¹ * α⁹⁶)x³ + (α¹¹ * α¹⁵³)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:

α¹¹x²⁰ + α²²⁶x¹⁹ + α²⁴⁵x¹⁸ + α¹⁶⁹x¹⁷ + α¹⁰⁵x¹⁶ + α¹⁹⁵x¹⁵ + α¹⁰⁸x¹⁴ + α¹²⁹x¹³ + α¹⁸¹x¹² + α⁹⁰x¹¹ + α¹⁹⁸x¹⁰ + α¹⁶³x⁹ + α¹⁵⁹x⁸ + α^{(263 % 255)}x⁷ + α¹⁹⁰x⁶ + α¹⁶x⁵ + α¹⁰⁹x⁴ + α¹⁰⁷x³ + α¹⁶⁴x²

The result is:

α¹¹x²⁰ + α²²⁶x¹⁹ + α²⁴⁵x¹⁸ + α¹⁶⁹x¹⁷ + α¹⁰⁵x¹⁶ + α¹⁹⁵x¹⁵ + α¹⁰⁸x¹⁴ + α¹²⁹x¹³ + α¹⁸¹x¹² + α⁹⁰x¹¹ + α¹⁹⁸x¹⁰ + α¹⁶³x⁹ + α¹⁵⁹x⁸ + α⁸x⁷ + α¹⁹⁰x⁶ + α¹⁶x⁵ + α¹⁰⁹x⁴ + α¹⁰⁷x³ + α¹⁶⁴x²

Now, convert this to integer notation:

232x²⁰ + 72x¹⁹ + 233x¹⁸ + 229x¹⁷ + 26x¹⁶ + 100x¹⁵ + 208x¹⁴ + 23x¹³ + 49x¹² + 223x¹¹ + 7x¹⁰ + 99x⁹ + 115x⁸ + 29x⁷ + 174x⁶ + 76x⁵ + 189x⁴ + 104x³ + 198x²

Step 14b: XOR the result with the result from step 13b

Use the result from step 13b to perform the next XOR.

(232 ⊕ 232)x²⁰ + (95 ⊕ 72)x¹⁹ + (26 ⊕ 233)x¹⁸ + (75 ⊕ 229)x¹⁷ + (248 ⊕ 26)x¹⁶ + (164 ⊕ 100)x¹⁵ + (181 ⊕ 208)x¹⁴ + (213 ⊕ 23)x¹³ + (64 ⊕ 49)x¹² + (117 ⊕ 223)x¹¹ + (53 ⊕ 7)x¹⁰ + (32 ⊕ 99)x⁹ + (185 ⊕ 115)x⁸ + (126 ⊕ 29)x⁷ + (137 ⊕ 174)x⁶ + (115 ⊕ 76)x⁵ + (55 ⊕ 189)x⁴ + (1 ⊕ 104)x³ + (0 ⊕ 198)x²

The result is:

0x²⁰ + 23x¹⁹ + 243x¹⁸ + 174x¹⁷ + 226x¹⁶ + 192x¹⁵ + 101x¹⁴ + 194x¹³ + 113x¹² + 170x¹¹ + 50x¹⁰ + 67x⁹ + 202x⁸ + 99x⁷ + 39x⁶ + 63x⁵ + 138x⁴ + 105x³ + 198x²

Discard the lead 0 term to get:

23x¹⁹ + 243x¹⁸ + 174x¹⁷ + 226x¹⁶ + 192x¹⁵ + 101x¹⁴ + 194x¹³ + 113x¹² + 170x¹¹ + 50x¹⁰ + 67x⁹ + 202x⁸ + 99x⁷ + 39x⁶ + 63x⁵ + 138x⁴ + 105x³ + 198x²

Step 15a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 23x¹⁹. Convert 23x¹⁹ to alpha notation. According to the log antilog table, for the integer value 23, the alpha exponent is 129. Therefore 23 = α¹²⁹. Multiply the generator polynomial by α¹²⁹:

(α¹²⁹ * α⁰)x¹⁹ + (α¹²⁹ * α²¹⁵)x¹⁸ + (α¹²⁹ * α²³⁴)x¹⁷ + (α¹²⁹ * α¹⁵⁸)x¹⁶ + (α¹²⁹ * α⁹⁴)x¹⁵ + (α¹²⁹ * α¹⁸⁴)x¹⁴ + (α¹²⁹ * α⁹⁷)x¹³ + (α¹²⁹ * α¹¹⁸)x¹² + (α¹²⁹ * α¹⁷⁰)x¹¹ + (α¹²⁹ * α⁷⁹)x¹⁰ + (α¹²⁹ * α¹⁸⁷)x⁹ + (α¹²⁹ * α¹⁵²)x⁸ + (α¹²⁹ * α¹⁴⁸)x⁷ + (α¹²⁹ * α²⁵²)x⁶ + (α¹²⁹ * α¹⁷⁹)x⁵ + (α¹²⁹ * α⁵)x⁴ + (α¹²⁹ * α⁹⁸)x³ + (α¹²⁹ * α⁹⁶)x² + (α¹²⁹ * α¹⁵³)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:

α¹²⁹x¹⁹ + α^{(344 % 255)}x¹⁸ + α^{(363 % 255)}x¹⁷ + α^{(287 % 255)}x¹⁶ + α²²³x¹⁵ + α^{(313 % 255)}x¹⁴ + α²²⁶x¹³ + α²⁴⁷x¹² + α^{(299 % 255)}x¹¹ + α²⁰⁸x¹⁰ + α^{(316 % 255)}x⁹ + α^{(281 % 255)}x⁸ + α^{(277 % 255)}x⁷ + α^{(381 % 255)}x⁶ + α^{(308 % 255)}x⁵ + α¹³⁴x⁴ + α²²⁷x³ + α²²⁵x² + α^{(282 % 255)}x¹

The result is:

α¹²⁹x¹⁹ + α⁸⁹x¹⁸ + α¹⁰⁸x¹⁷ + α³²x¹⁶ + α²²³x¹⁵ + α⁵⁸x¹⁴ + α²²⁶x¹³ + α²⁴⁷x¹² + α⁴⁴x¹¹ + α²⁰⁸x¹⁰ + α⁶¹x⁹ + α²⁶x⁸ + α²²x⁷ + α¹²⁶x⁶ + α⁵³x⁵ + α¹³⁴x⁴ + α²²⁷x³ + α²²⁵x² + α²⁷x¹

Now, convert this to integer notation:

23x¹⁹ + 225x¹⁸ + 208x¹⁷ + 157x¹⁶ + 9x¹⁵ + 105x¹⁴ + 72x¹³ + 131x¹² + 238x¹¹ + 81x¹⁰ + 111x⁹ + 6x⁸ + 234x⁷ + 102x⁶ + 40x⁵ + 218x⁴ + 144x³ + 36x² + 12x¹

Step 15b: XOR the result with the result from step 14b

Use the result from step 14b to perform the next XOR.

(23 ⊕ 23)x¹⁹ + (243 ⊕ 225)x¹⁸ + (174 ⊕ 208)x¹⁷ + (226 ⊕ 157)x¹⁶ + (192 ⊕ 9)x¹⁵ + (101 ⊕ 105)x¹⁴ + (194 ⊕ 72)x¹³ + (113 ⊕ 131)x¹² + (170 ⊕ 238)x¹¹ + (50 ⊕ 81)x¹⁰ + (67 ⊕ 111)x⁹ + (202 ⊕ 6)x⁸ + (99 ⊕ 234)x⁷ + (39 ⊕ 102)x⁶ + (63 ⊕ 40)x⁵ + (138 ⊕ 218)x⁴ + (105 ⊕ 144)x³ + (198 ⊕ 36)x² + (0 ⊕ 12)x¹

The result is:

0x¹⁹ + 18x¹⁸ + 126x¹⁷ + 127x¹⁶ + 201x¹⁵ + 12x¹⁴ + 138x¹³ + 242x¹² + 68x¹¹ + 99x¹⁰ + 44x⁹ + 204x⁸ + 137x⁷ + 65x⁶ + 23x⁵ + 80x⁴ + 249x³ + 226x² + 12x¹

Discard the lead 0 term to get:

18x¹⁸ + 126x¹⁷ + 127x¹⁶ + 201x¹⁵ + 12x¹⁴ + 138x¹³ + 242x¹² + 68x¹¹ + 99x¹⁰ + 44x⁹ + 204x⁸ + 137x⁷ + 65x⁶ + 23x⁵ + 80x⁴ + 249x³ + 226x² + 12x¹

Step 16a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 18x¹⁸. Convert 18x¹⁸ to alpha notation. According to the log antilog table, for the integer value 18, the alpha exponent is 224. Therefore 18 = α²²⁴. Multiply the generator polynomial by α²²⁴:

(α²²⁴ * α⁰)x¹⁸ + (α²²⁴ * α²¹⁵)x¹⁷ + (α²²⁴ * α²³⁴)x¹⁶ + (α²²⁴ * α¹⁵⁸)x¹⁵ + (α²²⁴ * α⁹⁴)x¹⁴ + (α²²⁴ * α¹⁸⁴)x¹³ + (α²²⁴ * α⁹⁷)x¹² + (α²²⁴ * α¹¹⁸)x¹¹ + (α²²⁴ * α¹⁷⁰)x¹⁰ + (α²²⁴ * α⁷⁹)x⁹ + (α²²⁴ * α¹⁸⁷)x⁸ + (α²²⁴ * α¹⁵²)x⁷ + (α²²⁴ * α¹⁴⁸)x⁶ + (α²²⁴ * α²⁵²)x⁵ + (α²²⁴ * α¹⁷⁹)x⁴ + (α²²⁴ * α⁵)x³ + (α²²⁴ * α⁹⁸)x² + (α²²⁴ * α⁹⁶)x¹ + (α²²⁴ * α¹⁵³)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:

α²²⁴x¹⁸ + α^{(439 % 255)}x¹⁷ + α^{(458 % 255)}x¹⁶ + α^{(382 % 255)}x¹⁵ + α^{(318 % 255)}x¹⁴ + α^{(408 % 255)}x¹³ + α^{(321 % 255)}x¹² + α^{(342 % 255)}x¹¹ + α^{(394 % 255)}x¹⁰ + α^{(303 % 255)}x⁹ + α^{(411 % 255)}x⁸ + α^{(376 % 255)}x⁷ + α^{(372 % 255)}x⁶ + α^{(476 % 255)}x⁵ + α^{(403 % 255)}x⁴ + α²²⁹x³ + α^{(322 % 255)}x² + α^{(320 % 255)}x¹ + α^{(377 % 255)}x⁰

The result is:

α²²⁴x¹⁸ + α¹⁸⁴x¹⁷ + α²⁰³x¹⁶ + α¹²⁷x¹⁵ + α⁶³x¹⁴ + α¹⁵³x¹³ + α⁶⁶x¹² + α⁸⁷x¹¹ + α¹³⁹x¹⁰ + α⁴⁸x⁹ + α¹⁵⁶x⁸ + α¹²¹x⁷ + α¹¹⁷x⁶ + α²²¹x⁵ + α¹⁴⁸x⁴ + α²²⁹x³ + α⁶⁷x² + α⁶⁵x¹ + α¹²²x⁰

Now, convert this to integer notation:

18x¹⁸ + 149x¹⁷ + 224x¹⁶ + 204x¹⁵ + 161x¹⁴ + 146x¹³ + 97x¹² + 127x¹¹ + 66x¹⁰ + 70x⁹ + 228x⁸ + 118x⁷ + 237x⁶ + 69x⁵ + 82x⁴ + 122x³ + 194x² + 190x¹ + 236

Step 16b: XOR the result with the result from step 15b

Use the result from step 15b to perform the next XOR.

(18 ⊕ 18)x¹⁸ + (126 ⊕ 149)x¹⁷ + (127 ⊕ 224)x¹⁶ + (201 ⊕ 204)x¹⁵ + (12 ⊕ 161)x¹⁴ + (138 ⊕ 146)x¹³ + (242 ⊕ 97)x¹² + (68 ⊕ 127)x¹¹ + (99 ⊕ 66)x¹⁰ + (44 ⊕ 70)x⁹ + (204 ⊕ 228)x⁸ + (137 ⊕ 118)x⁷ + (65 ⊕ 237)x⁶ + (23 ⊕ 69)x⁵ + (80 ⊕ 82)x⁴ + (249 ⊕ 122)x³ + (226 ⊕ 194)x² + (12 ⊕ 190)x¹ + (0 ⊕ 236)x⁰

The result is:

0x¹⁸ + 235x¹⁷ + 159x¹⁶ + 5x¹⁵ + 173x¹⁴ + 24x¹³ + 147x¹² + 59x¹¹ + 33x¹⁰ + 106x⁹ + 40x⁸ + 255x⁷ + 172x⁶ + 82x⁵ + 2x⁴ + 131x³ + 32x² + 178x¹ + 236

Discard the lead 0 term to get:

235x¹⁷ + 159x¹⁶ + 5x¹⁵ + 173x¹⁴ + 24x¹³ + 147x¹² + 59x¹¹ + 33x¹⁰ + 106x⁹ + 40x⁸ + 255x⁷ + 172x⁶ + 82x⁵ + 2x⁴ + 131x³ + 32x² + 178x¹ + 236

Use the terms of the remainder as the error correction codewords

The division has been performed 16 times, which is the number of terms in the message polynomial. This means that the division is complete and the terms of the above polynomial are the error correction codewords to use for the original message polynomial:

235  159  5  173  24  147  59  33  106  40  255  172  82  2  131  32  178  236