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:

64x³⁰ + 2x²⁹ + 149x²⁸ + 2x²⁷ + 6x²⁶ + 2x²⁵ + 22x²⁴ + 2x²³ + 118x²² + 2x²¹ + 86x²⁰ + 2x¹⁹ + 68x¹⁸ + 2x¹⁷ + 247x¹⁶ + 2x¹⁵ + 87x¹⁴ + 2x¹³ + 66x¹² + 2x¹¹ + 16x¹⁰ + 2x⁹ + 236x⁸ + 2x⁷ + 17x⁶ + 2x⁵ + 236x⁴ + 2x³ + 17x² + 2x¹ + 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 10, for 10 error correction codewords, so multiply the message polynomial by x¹⁰, which gives us:

64x⁴⁰ + 2x³⁹ + 149x³⁸ + 2x³⁷ + 6x³⁶ + 2x³⁵ + 22x³⁴ + 2x³³ + 118x³² + 2x³¹ + 86x³⁰ + 2x²⁹ + 68x²⁸ + 2x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

The lead term of the generator polynomial should also have the same exponent, so multiply by x³⁰ to get

α⁰x⁴⁰ + α²⁵¹x³⁹ + α⁶⁷x³⁸ + α⁴⁶x³⁷ + α⁶¹x³⁶ + α¹¹⁸x³⁵ + α⁷⁰x³⁴ + α⁶⁴x³³ + α⁹⁴x³² + α³²x³¹ + α⁴⁵x³⁰

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 31 steps to complete. This will result in a remainder that has 10 terms. These terms will be the 10 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 64x⁴⁰. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 64x⁴⁰ to alpha notation. According to the log antilog table, for the integer value 64, the alpha exponent is 6. Therefore 64 = α⁶. Multiply the generator polynomial by α⁶:

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:

α⁶x⁴⁰ + α^{(257 % 255)}x³⁹ + α⁷³x³⁸ + α⁵²x³⁷ + α⁶⁷x³⁶ + α¹²⁴x³⁵ + α⁷⁶x³⁴ + α⁷⁰x³³ + α¹⁰⁰x³² + α³⁸x³¹ + α⁵¹x³⁰

The result is:

α⁶x⁴⁰ + α²x³⁹ + α⁷³x³⁸ + α⁵²x³⁷ + α⁶⁷x³⁶ + α¹²⁴x³⁵ + α⁷⁶x³⁴ + α⁷⁰x³³ + α¹⁰⁰x³² + α³⁸x³¹ + α⁵¹x³⁰

Now, convert this to integer notation:

64x⁴⁰ + 4x³⁹ + 202x³⁸ + 20x³⁷ + 194x³⁶ + 151x³⁵ + 30x³⁴ + 94x³³ + 17x³² + 148x³¹ + 10x³⁰

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.

(64 ⊕ 64)x⁴⁰ + (2 ⊕ 4)x³⁹ + (149 ⊕ 202)x³⁸ + (2 ⊕ 20)x³⁷ + (6 ⊕ 194)x³⁶ + (2 ⊕ 151)x³⁵ + (22 ⊕ 30)x³⁴ + (2 ⊕ 94)x³³ + (118 ⊕ 17)x³² + (2 ⊕ 148)x³¹ + (86 ⊕ 10)x³⁰ + (2 ⊕ 0)x²⁹ + (68 ⊕ 0)x²⁸ + (2 ⊕ 0)x²⁷ + (247 ⊕ 0)x²⁶ + (2 ⊕ 0)x²⁵ + (87 ⊕ 0)x²⁴ + (2 ⊕ 0)x²³ + (66 ⊕ 0)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x⁴⁰ + 6x³⁹ + 95x³⁸ + 22x³⁷ + 196x³⁶ + 149x³⁵ + 8x³⁴ + 92x³³ + 103x³² + 150x³¹ + 92x³⁰ + 2x²⁹ + 68x²⁸ + 2x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

6x³⁹ + 95x³⁸ + 22x³⁷ + 196x³⁶ + 149x³⁵ + 8x³⁴ + 92x³³ + 103x³² + 150x³¹ + 92x³⁰ + 2x²⁹ + 68x²⁸ + 2x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 6x³⁹. Convert 6x³⁹ to alpha notation. According to the log antilog table, for the integer value 6, the alpha exponent is 26. Therefore 6 = α²⁶. Multiply the generator polynomial by α²⁶:

(α²⁶ * α⁰)x³⁹ + (α²⁶ * α²⁵¹)x³⁸ + (α²⁶ * α⁶⁷)x³⁷ + (α²⁶ * α⁴⁶)x³⁶ + (α²⁶ * α⁶¹)x³⁵ + (α²⁶ * α¹¹⁸)x³⁴ + (α²⁶ * α⁷⁰)x³³ + (α²⁶ * α⁶⁴)x³² + (α²⁶ * α⁹⁴)x³¹ + (α²⁶ * α³²)x³⁰ + (α²⁶ * α⁴⁵)x²⁹

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:

α²⁶x³⁹ + α^{(277 % 255)}x³⁸ + α⁹³x³⁷ + α⁷²x³⁶ + α⁸⁷x³⁵ + α¹⁴⁴x³⁴ + α⁹⁶x³³ + α⁹⁰x³² + α¹²⁰x³¹ + α⁵⁸x³⁰ + α⁷¹x²⁹

The result is:

α²⁶x³⁹ + α²²x³⁸ + α⁹³x³⁷ + α⁷²x³⁶ + α⁸⁷x³⁵ + α¹⁴⁴x³⁴ + α⁹⁶x³³ + α⁹⁰x³² + α¹²⁰x³¹ + α⁵⁸x³⁰ + α⁷¹x²⁹

Now, convert this to integer notation:

6x³⁹ + 234x³⁸ + 182x³⁷ + 101x³⁶ + 127x³⁵ + 168x³⁴ + 217x³³ + 223x³² + 59x³¹ + 105x³⁰ + 188x²⁹

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

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

(6 ⊕ 6)x³⁹ + (95 ⊕ 234)x³⁸ + (22 ⊕ 182)x³⁷ + (196 ⊕ 101)x³⁶ + (149 ⊕ 127)x³⁵ + (8 ⊕ 168)x³⁴ + (92 ⊕ 217)x³³ + (103 ⊕ 223)x³² + (150 ⊕ 59)x³¹ + (92 ⊕ 105)x³⁰ + (2 ⊕ 188)x²⁹ + (68 ⊕ 0)x²⁸ + (2 ⊕ 0)x²⁷ + (247 ⊕ 0)x²⁶ + (2 ⊕ 0)x²⁵ + (87 ⊕ 0)x²⁴ + (2 ⊕ 0)x²³ + (66 ⊕ 0)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³⁹ + 181x³⁸ + 160x³⁷ + 161x³⁶ + 234x³⁵ + 160x³⁴ + 133x³³ + 184x³² + 173x³¹ + 53x³⁰ + 190x²⁹ + 68x²⁸ + 2x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

181x³⁸ + 160x³⁷ + 161x³⁶ + 234x³⁵ + 160x³⁴ + 133x³³ + 184x³² + 173x³¹ + 53x³⁰ + 190x²⁹ + 68x²⁸ + 2x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 181x³⁸. Convert 181x³⁸ 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)x³⁸ + (α⁴² * α²⁵¹)x³⁷ + (α⁴² * α⁶⁷)x³⁶ + (α⁴² * α⁴⁶)x³⁵ + (α⁴² * α⁶¹)x³⁴ + (α⁴² * α¹¹⁸)x³³ + (α⁴² * α⁷⁰)x³² + (α⁴² * α⁶⁴)x³¹ + (α⁴² * α⁹⁴)x³⁰ + (α⁴² * α³²)x²⁹ + (α⁴² * α⁴⁵)x²⁸

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:

α⁴²x³⁸ + α^{(293 % 255)}x³⁷ + α¹⁰⁹x³⁶ + α⁸⁸x³⁵ + α¹⁰³x³⁴ + α¹⁶⁰x³³ + α¹¹²x³² + α¹⁰⁶x³¹ + α¹³⁶x³⁰ + α⁷⁴x²⁹ + α⁸⁷x²⁸

The result is:

α⁴²x³⁸ + α³⁸x³⁷ + α¹⁰⁹x³⁶ + α⁸⁸x³⁵ + α¹⁰³x³⁴ + α¹⁶⁰x³³ + α¹¹²x³² + α¹⁰⁶x³¹ + α¹³⁶x³⁰ + α⁷⁴x²⁹ + α⁸⁷x²⁸

Now, convert this to integer notation:

181x³⁸ + 148x³⁷ + 189x³⁶ + 254x³⁵ + 136x³⁴ + 230x³³ + 129x³² + 52x³¹ + 79x³⁰ + 137x²⁹ + 127x²⁸

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

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

(181 ⊕ 181)x³⁸ + (160 ⊕ 148)x³⁷ + (161 ⊕ 189)x³⁶ + (234 ⊕ 254)x³⁵ + (160 ⊕ 136)x³⁴ + (133 ⊕ 230)x³³ + (184 ⊕ 129)x³² + (173 ⊕ 52)x³¹ + (53 ⊕ 79)x³⁰ + (190 ⊕ 137)x²⁹ + (68 ⊕ 127)x²⁸ + (2 ⊕ 0)x²⁷ + (247 ⊕ 0)x²⁶ + (2 ⊕ 0)x²⁵ + (87 ⊕ 0)x²⁴ + (2 ⊕ 0)x²³ + (66 ⊕ 0)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³⁸ + 52x³⁷ + 28x³⁶ + 20x³⁵ + 40x³⁴ + 99x³³ + 57x³² + 153x³¹ + 122x³⁰ + 55x²⁹ + 59x²⁸ + 2x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

52x³⁷ + 28x³⁶ + 20x³⁵ + 40x³⁴ + 99x³³ + 57x³² + 153x³¹ + 122x³⁰ + 55x²⁹ + 59x²⁸ + 2x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 52x³⁷. Convert 52x³⁷ to alpha notation. According to the log antilog table, for the integer value 52, the alpha exponent is 106. Therefore 52 = α¹⁰⁶. Multiply the generator polynomial by α¹⁰⁶:

(α¹⁰⁶ * α⁰)x³⁷ + (α¹⁰⁶ * α²⁵¹)x³⁶ + (α¹⁰⁶ * α⁶⁷)x³⁵ + (α¹⁰⁶ * α⁴⁶)x³⁴ + (α¹⁰⁶ * α⁶¹)x³³ + (α¹⁰⁶ * α¹¹⁸)x³² + (α¹⁰⁶ * α⁷⁰)x³¹ + (α¹⁰⁶ * α⁶⁴)x³⁰ + (α¹⁰⁶ * α⁹⁴)x²⁹ + (α¹⁰⁶ * α³²)x²⁸ + (α¹⁰⁶ * α⁴⁵)x²⁷

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:

α¹⁰⁶x³⁷ + α^{(357 % 255)}x³⁶ + α¹⁷³x³⁵ + α¹⁵²x³⁴ + α¹⁶⁷x³³ + α²²⁴x³² + α¹⁷⁶x³¹ + α¹⁷⁰x³⁰ + α²⁰⁰x²⁹ + α¹³⁸x²⁸ + α¹⁵¹x²⁷

The result is:

α¹⁰⁶x³⁷ + α¹⁰²x³⁶ + α¹⁷³x³⁵ + α¹⁵²x³⁴ + α¹⁶⁷x³³ + α²²⁴x³² + α¹⁷⁶x³¹ + α¹⁷⁰x³⁰ + α²⁰⁰x²⁹ + α¹³⁸x²⁸ + α¹⁵¹x²⁷

Now, convert this to integer notation:

52x³⁷ + 68x³⁶ + 246x³⁵ + 73x³⁴ + 126x³³ + 18x³² + 227x³¹ + 215x³⁰ + 28x²⁹ + 33x²⁸ + 170x²⁷

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

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

(52 ⊕ 52)x³⁷ + (28 ⊕ 68)x³⁶ + (20 ⊕ 246)x³⁵ + (40 ⊕ 73)x³⁴ + (99 ⊕ 126)x³³ + (57 ⊕ 18)x³² + (153 ⊕ 227)x³¹ + (122 ⊕ 215)x³⁰ + (55 ⊕ 28)x²⁹ + (59 ⊕ 33)x²⁸ + (2 ⊕ 170)x²⁷ + (247 ⊕ 0)x²⁶ + (2 ⊕ 0)x²⁵ + (87 ⊕ 0)x²⁴ + (2 ⊕ 0)x²³ + (66 ⊕ 0)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³⁷ + 88x³⁶ + 226x³⁵ + 97x³⁴ + 29x³³ + 43x³² + 122x³¹ + 173x³⁰ + 43x²⁹ + 26x²⁸ + 168x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

88x³⁶ + 226x³⁵ + 97x³⁴ + 29x³³ + 43x³² + 122x³¹ + 173x³⁰ + 43x²⁹ + 26x²⁸ + 168x²⁷ + 247x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 88x³⁶. Convert 88x³⁶ to alpha notation. According to the log antilog table, for the integer value 88, the alpha exponent is 241. Therefore 88 = α²⁴¹. Multiply the generator polynomial by α²⁴¹:

(α²⁴¹ * α⁰)x³⁶ + (α²⁴¹ * α²⁵¹)x³⁵ + (α²⁴¹ * α⁶⁷)x³⁴ + (α²⁴¹ * α⁴⁶)x³³ + (α²⁴¹ * α⁶¹)x³² + (α²⁴¹ * α¹¹⁸)x³¹ + (α²⁴¹ * α⁷⁰)x³⁰ + (α²⁴¹ * α⁶⁴)x²⁹ + (α²⁴¹ * α⁹⁴)x²⁸ + (α²⁴¹ * α³²)x²⁷ + (α²⁴¹ * α⁴⁵)x²⁶

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:

α²⁴¹x³⁶ + α^{(492 % 255)}x³⁵ + α^{(308 % 255)}x³⁴ + α^{(287 % 255)}x³³ + α^{(302 % 255)}x³² + α^{(359 % 255)}x³¹ + α^{(311 % 255)}x³⁰ + α^{(305 % 255)}x²⁹ + α^{(335 % 255)}x²⁸ + α^{(273 % 255)}x²⁷ + α^{(286 % 255)}x²⁶

The result is:

α²⁴¹x³⁶ + α²³⁷x³⁵ + α⁵³x³⁴ + α³²x³³ + α⁴⁷x³² + α¹⁰⁴x³¹ + α⁵⁶x³⁰ + α⁵⁰x²⁹ + α⁸⁰x²⁸ + α¹⁸x²⁷ + α³¹x²⁶

Now, convert this to integer notation:

88x³⁶ + 139x³⁵ + 40x³⁴ + 157x³³ + 35x³² + 13x³¹ + 93x³⁰ + 5x²⁹ + 253x²⁸ + 45x²⁷ + 192x²⁶

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

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

(88 ⊕ 88)x³⁶ + (226 ⊕ 139)x³⁵ + (97 ⊕ 40)x³⁴ + (29 ⊕ 157)x³³ + (43 ⊕ 35)x³² + (122 ⊕ 13)x³¹ + (173 ⊕ 93)x³⁰ + (43 ⊕ 5)x²⁹ + (26 ⊕ 253)x²⁸ + (168 ⊕ 45)x²⁷ + (247 ⊕ 192)x²⁶ + (2 ⊕ 0)x²⁵ + (87 ⊕ 0)x²⁴ + (2 ⊕ 0)x²³ + (66 ⊕ 0)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³⁶ + 105x³⁵ + 73x³⁴ + 128x³³ + 8x³² + 119x³¹ + 240x³⁰ + 46x²⁹ + 231x²⁸ + 133x²⁷ + 55x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

105x³⁵ + 73x³⁴ + 128x³³ + 8x³² + 119x³¹ + 240x³⁰ + 46x²⁹ + 231x²⁸ + 133x²⁷ + 55x²⁶ + 2x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 105x³⁵. Convert 105x³⁵ to alpha notation. According to the log antilog table, for the integer value 105, the alpha exponent is 58. Therefore 105 = α⁵⁸. Multiply the generator polynomial by α⁵⁸:

(α⁵⁸ * α⁰)x³⁵ + (α⁵⁸ * α²⁵¹)x³⁴ + (α⁵⁸ * α⁶⁷)x³³ + (α⁵⁸ * α⁴⁶)x³² + (α⁵⁸ * α⁶¹)x³¹ + (α⁵⁸ * α¹¹⁸)x³⁰ + (α⁵⁸ * α⁷⁰)x²⁹ + (α⁵⁸ * α⁶⁴)x²⁸ + (α⁵⁸ * α⁹⁴)x²⁷ + (α⁵⁸ * α³²)x²⁶ + (α⁵⁸ * α⁴⁵)x²⁵

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:

α⁵⁸x³⁵ + α^{(309 % 255)}x³⁴ + α¹²⁵x³³ + α¹⁰⁴x³² + α¹¹⁹x³¹ + α¹⁷⁶x³⁰ + α¹²⁸x²⁹ + α¹²²x²⁸ + α¹⁵²x²⁷ + α⁹⁰x²⁶ + α¹⁰³x²⁵

The result is:

α⁵⁸x³⁵ + α⁵⁴x³⁴ + α¹²⁵x³³ + α¹⁰⁴x³² + α¹¹⁹x³¹ + α¹⁷⁶x³⁰ + α¹²⁸x²⁹ + α¹²²x²⁸ + α¹⁵²x²⁷ + α⁹⁰x²⁶ + α¹⁰³x²⁵

Now, convert this to integer notation:

105x³⁵ + 80x³⁴ + 51x³³ + 13x³² + 147x³¹ + 227x³⁰ + 133x²⁹ + 236x²⁸ + 73x²⁷ + 223x²⁶ + 136x²⁵

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

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

(105 ⊕ 105)x³⁵ + (73 ⊕ 80)x³⁴ + (128 ⊕ 51)x³³ + (8 ⊕ 13)x³² + (119 ⊕ 147)x³¹ + (240 ⊕ 227)x³⁰ + (46 ⊕ 133)x²⁹ + (231 ⊕ 236)x²⁸ + (133 ⊕ 73)x²⁷ + (55 ⊕ 223)x²⁶ + (2 ⊕ 136)x²⁵ + (87 ⊕ 0)x²⁴ + (2 ⊕ 0)x²³ + (66 ⊕ 0)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³⁵ + 25x³⁴ + 179x³³ + 5x³² + 228x³¹ + 19x³⁰ + 171x²⁹ + 11x²⁸ + 204x²⁷ + 232x²⁶ + 138x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

25x³⁴ + 179x³³ + 5x³² + 228x³¹ + 19x³⁰ + 171x²⁹ + 11x²⁸ + 204x²⁷ + 232x²⁶ + 138x²⁵ + 87x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 25x³⁴. Convert 25x³⁴ to alpha notation. According to the log antilog table, for the integer value 25, the alpha exponent is 193. Therefore 25 = α¹⁹³. Multiply the generator polynomial by α¹⁹³:

(α¹⁹³ * α⁰)x³⁴ + (α¹⁹³ * α²⁵¹)x³³ + (α¹⁹³ * α⁶⁷)x³² + (α¹⁹³ * α⁴⁶)x³¹ + (α¹⁹³ * α⁶¹)x³⁰ + (α¹⁹³ * α¹¹⁸)x²⁹ + (α¹⁹³ * α⁷⁰)x²⁸ + (α¹⁹³ * α⁶⁴)x²⁷ + (α¹⁹³ * α⁹⁴)x²⁶ + (α¹⁹³ * α³²)x²⁵ + (α¹⁹³ * α⁴⁵)x²⁴

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:

α¹⁹³x³⁴ + α^{(444 % 255)}x³³ + α^{(260 % 255)}x³² + α²³⁹x³¹ + α²⁵⁴x³⁰ + α^{(311 % 255)}x²⁹ + α^{(263 % 255)}x²⁸ + α^{(257 % 255)}x²⁷ + α^{(287 % 255)}x²⁶ + α²²⁵x²⁵ + α²³⁸x²⁴

The result is:

α¹⁹³x³⁴ + α¹⁸⁹x³³ + α⁵x³² + α²³⁹x³¹ + α²⁵⁴x³⁰ + α⁵⁶x²⁹ + α⁸x²⁸ + α²x²⁷ + α³²x²⁶ + α²²⁵x²⁵ + α²³⁸x²⁴

Now, convert this to integer notation:

25x³⁴ + 87x³³ + 32x³² + 22x³¹ + 142x³⁰ + 93x²⁹ + 29x²⁸ + 4x²⁷ + 157x²⁶ + 36x²⁵ + 11x²⁴

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

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

(25 ⊕ 25)x³⁴ + (179 ⊕ 87)x³³ + (5 ⊕ 32)x³² + (228 ⊕ 22)x³¹ + (19 ⊕ 142)x³⁰ + (171 ⊕ 93)x²⁹ + (11 ⊕ 29)x²⁸ + (204 ⊕ 4)x²⁷ + (232 ⊕ 157)x²⁶ + (138 ⊕ 36)x²⁵ + (87 ⊕ 11)x²⁴ + (2 ⊕ 0)x²³ + (66 ⊕ 0)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³⁴ + 228x³³ + 37x³² + 242x³¹ + 157x³⁰ + 246x²⁹ + 22x²⁸ + 200x²⁷ + 117x²⁶ + 174x²⁵ + 92x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

228x³³ + 37x³² + 242x³¹ + 157x³⁰ + 246x²⁹ + 22x²⁸ + 200x²⁷ + 117x²⁶ + 174x²⁵ + 92x²⁴ + 2x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 228x³³. Convert 228x³³ to alpha notation. According to the log antilog table, for the integer value 228, the alpha exponent is 156. Therefore 228 = α¹⁵⁶. Multiply the generator polynomial by α¹⁵⁶:

(α¹⁵⁶ * α⁰)x³³ + (α¹⁵⁶ * α²⁵¹)x³² + (α¹⁵⁶ * α⁶⁷)x³¹ + (α¹⁵⁶ * α⁴⁶)x³⁰ + (α¹⁵⁶ * α⁶¹)x²⁹ + (α¹⁵⁶ * α¹¹⁸)x²⁸ + (α¹⁵⁶ * α⁷⁰)x²⁷ + (α¹⁵⁶ * α⁶⁴)x²⁶ + (α¹⁵⁶ * α⁹⁴)x²⁵ + (α¹⁵⁶ * α³²)x²⁴ + (α¹⁵⁶ * α⁴⁵)x²³

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:

α¹⁵⁶x³³ + α^{(407 % 255)}x³² + α²²³x³¹ + α²⁰²x³⁰ + α²¹⁷x²⁹ + α^{(274 % 255)}x²⁸ + α²²⁶x²⁷ + α²²⁰x²⁶ + α²⁵⁰x²⁵ + α¹⁸⁸x²⁴ + α²⁰¹x²³

The result is:

α¹⁵⁶x³³ + α¹⁵²x³² + α²²³x³¹ + α²⁰²x³⁰ + α²¹⁷x²⁹ + α¹⁹x²⁸ + α²²⁶x²⁷ + α²²⁰x²⁶ + α²⁵⁰x²⁵ + α¹⁸⁸x²⁴ + α²⁰¹x²³

Now, convert this to integer notation:

228x³³ + 73x³² + 9x³¹ + 112x³⁰ + 155x²⁹ + 90x²⁸ + 72x²⁷ + 172x²⁶ + 108x²⁵ + 165x²⁴ + 56x²³

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

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

(228 ⊕ 228)x³³ + (37 ⊕ 73)x³² + (242 ⊕ 9)x³¹ + (157 ⊕ 112)x³⁰ + (246 ⊕ 155)x²⁹ + (22 ⊕ 90)x²⁸ + (200 ⊕ 72)x²⁷ + (117 ⊕ 172)x²⁶ + (174 ⊕ 108)x²⁵ + (92 ⊕ 165)x²⁴ + (2 ⊕ 56)x²³ + (66 ⊕ 0)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³³ + 108x³² + 251x³¹ + 237x³⁰ + 109x²⁹ + 76x²⁸ + 128x²⁷ + 217x²⁶ + 194x²⁵ + 249x²⁴ + 58x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

108x³² + 251x³¹ + 237x³⁰ + 109x²⁹ + 76x²⁸ + 128x²⁷ + 217x²⁶ + 194x²⁵ + 249x²⁴ + 58x²³ + 66x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 108x³². Convert 108x³² to alpha notation. According to the log antilog table, for the integer value 108, the alpha exponent is 250. Therefore 108 = α²⁵⁰. Multiply the generator polynomial by α²⁵⁰:

(α²⁵⁰ * α⁰)x³² + (α²⁵⁰ * α²⁵¹)x³¹ + (α²⁵⁰ * α⁶⁷)x³⁰ + (α²⁵⁰ * α⁴⁶)x²⁹ + (α²⁵⁰ * α⁶¹)x²⁸ + (α²⁵⁰ * α¹¹⁸)x²⁷ + (α²⁵⁰ * α⁷⁰)x²⁶ + (α²⁵⁰ * α⁶⁴)x²⁵ + (α²⁵⁰ * α⁹⁴)x²⁴ + (α²⁵⁰ * α³²)x²³ + (α²⁵⁰ * α⁴⁵)x²²

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:

α²⁵⁰x³² + α^{(501 % 255)}x³¹ + α^{(317 % 255)}x³⁰ + α^{(296 % 255)}x²⁹ + α^{(311 % 255)}x²⁸ + α^{(368 % 255)}x²⁷ + α^{(320 % 255)}x²⁶ + α^{(314 % 255)}x²⁵ + α^{(344 % 255)}x²⁴ + α^{(282 % 255)}x²³ + α^{(295 % 255)}x²²

The result is:

α²⁵⁰x³² + α²⁴⁶x³¹ + α⁶²x³⁰ + α⁴¹x²⁹ + α⁵⁶x²⁸ + α¹¹³x²⁷ + α⁶⁵x²⁶ + α⁵⁹x²⁵ + α⁸⁹x²⁴ + α²⁷x²³ + α⁴⁰x²²

Now, convert this to integer notation:

108x³² + 207x³¹ + 222x³⁰ + 212x²⁹ + 93x²⁸ + 31x²⁷ + 190x²⁶ + 210x²⁵ + 225x²⁴ + 12x²³ + 106x²²

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

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

(108 ⊕ 108)x³² + (251 ⊕ 207)x³¹ + (237 ⊕ 222)x³⁰ + (109 ⊕ 212)x²⁹ + (76 ⊕ 93)x²⁸ + (128 ⊕ 31)x²⁷ + (217 ⊕ 190)x²⁶ + (194 ⊕ 210)x²⁵ + (249 ⊕ 225)x²⁴ + (58 ⊕ 12)x²³ + (66 ⊕ 106)x²² + (2 ⊕ 0)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³² + 52x³¹ + 51x³⁰ + 185x²⁹ + 17x²⁸ + 159x²⁷ + 103x²⁶ + 16x²⁵ + 24x²⁴ + 54x²³ + 40x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

52x³¹ + 51x³⁰ + 185x²⁹ + 17x²⁸ + 159x²⁷ + 103x²⁶ + 16x²⁵ + 24x²⁴ + 54x²³ + 40x²² + 2x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 52x³¹. Convert 52x³¹ to alpha notation. According to the log antilog table, for the integer value 52, the alpha exponent is 106. Therefore 52 = α¹⁰⁶. Multiply the generator polynomial by α¹⁰⁶:

(α¹⁰⁶ * α⁰)x³¹ + (α¹⁰⁶ * α²⁵¹)x³⁰ + (α¹⁰⁶ * α⁶⁷)x²⁹ + (α¹⁰⁶ * α⁴⁶)x²⁸ + (α¹⁰⁶ * α⁶¹)x²⁷ + (α¹⁰⁶ * α¹¹⁸)x²⁶ + (α¹⁰⁶ * α⁷⁰)x²⁵ + (α¹⁰⁶ * α⁶⁴)x²⁴ + (α¹⁰⁶ * α⁹⁴)x²³ + (α¹⁰⁶ * α³²)x²² + (α¹⁰⁶ * α⁴⁵)x²¹

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:

α¹⁰⁶x³¹ + α^{(357 % 255)}x³⁰ + α¹⁷³x²⁹ + α¹⁵²x²⁸ + α¹⁶⁷x²⁷ + α²²⁴x²⁶ + α¹⁷⁶x²⁵ + α¹⁷⁰x²⁴ + α²⁰⁰x²³ + α¹³⁸x²² + α¹⁵¹x²¹

The result is:

α¹⁰⁶x³¹ + α¹⁰²x³⁰ + α¹⁷³x²⁹ + α¹⁵²x²⁸ + α¹⁶⁷x²⁷ + α²²⁴x²⁶ + α¹⁷⁶x²⁵ + α¹⁷⁰x²⁴ + α²⁰⁰x²³ + α¹³⁸x²² + α¹⁵¹x²¹

Now, convert this to integer notation:

52x³¹ + 68x³⁰ + 246x²⁹ + 73x²⁸ + 126x²⁷ + 18x²⁶ + 227x²⁵ + 215x²⁴ + 28x²³ + 33x²² + 170x²¹

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

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

(52 ⊕ 52)x³¹ + (51 ⊕ 68)x³⁰ + (185 ⊕ 246)x²⁹ + (17 ⊕ 73)x²⁸ + (159 ⊕ 126)x²⁷ + (103 ⊕ 18)x²⁶ + (16 ⊕ 227)x²⁵ + (24 ⊕ 215)x²⁴ + (54 ⊕ 28)x²³ + (40 ⊕ 33)x²² + (2 ⊕ 170)x²¹ + (16 ⊕ 0)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³¹ + 119x³⁰ + 79x²⁹ + 88x²⁸ + 225x²⁷ + 117x²⁶ + 243x²⁵ + 207x²⁴ + 42x²³ + 9x²² + 168x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

119x³⁰ + 79x²⁹ + 88x²⁸ + 225x²⁷ + 117x²⁶ + 243x²⁵ + 207x²⁴ + 42x²³ + 9x²² + 168x²¹ + 16x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 119x³⁰. Convert 119x³⁰ to alpha notation. According to the log antilog table, for the integer value 119, the alpha exponent is 43. Therefore 119 = α⁴³. Multiply the generator polynomial by α⁴³:

(α⁴³ * α⁰)x³⁰ + (α⁴³ * α²⁵¹)x²⁹ + (α⁴³ * α⁶⁷)x²⁸ + (α⁴³ * α⁴⁶)x²⁷ + (α⁴³ * α⁶¹)x²⁶ + (α⁴³ * α¹¹⁸)x²⁵ + (α⁴³ * α⁷⁰)x²⁴ + (α⁴³ * α⁶⁴)x²³ + (α⁴³ * α⁹⁴)x²² + (α⁴³ * α³²)x²¹ + (α⁴³ * α⁴⁵)x²⁰

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:

α⁴³x³⁰ + α^{(294 % 255)}x²⁹ + α¹¹⁰x²⁸ + α⁸⁹x²⁷ + α¹⁰⁴x²⁶ + α¹⁶¹x²⁵ + α¹¹³x²⁴ + α¹⁰⁷x²³ + α¹³⁷x²² + α⁷⁵x²¹ + α⁸⁸x²⁰

The result is:

α⁴³x³⁰ + α³⁹x²⁹ + α¹¹⁰x²⁸ + α⁸⁹x²⁷ + α¹⁰⁴x²⁶ + α¹⁶¹x²⁵ + α¹¹³x²⁴ + α¹⁰⁷x²³ + α¹³⁷x²² + α⁷⁵x²¹ + α⁸⁸x²⁰

Now, convert this to integer notation:

119x³⁰ + 53x²⁹ + 103x²⁸ + 225x²⁷ + 13x²⁶ + 209x²⁵ + 31x²⁴ + 104x²³ + 158x²² + 15x²¹ + 254x²⁰

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

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

(119 ⊕ 119)x³⁰ + (79 ⊕ 53)x²⁹ + (88 ⊕ 103)x²⁸ + (225 ⊕ 225)x²⁷ + (117 ⊕ 13)x²⁶ + (243 ⊕ 209)x²⁵ + (207 ⊕ 31)x²⁴ + (42 ⊕ 104)x²³ + (9 ⊕ 158)x²² + (168 ⊕ 15)x²¹ + (16 ⊕ 254)x²⁰ + (2 ⊕ 0)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x³⁰ + 122x²⁹ + 63x²⁸ + 0x²⁷ + 120x²⁶ + 34x²⁵ + 208x²⁴ + 66x²³ + 151x²² + 167x²¹ + 238x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

122x²⁹ + 63x²⁸ + 0x²⁷ + 120x²⁶ + 34x²⁵ + 208x²⁴ + 66x²³ + 151x²² + 167x²¹ + 238x²⁰ + 2x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 122x²⁹. Convert 122x²⁹ to alpha notation. According to the log antilog table, for the integer value 122, the alpha exponent is 229. Therefore 122 = α²²⁹. Multiply the generator polynomial by α²²⁹:

(α²²⁹ * α⁰)x²⁹ + (α²²⁹ * α²⁵¹)x²⁸ + (α²²⁹ * α⁶⁷)x²⁷ + (α²²⁹ * α⁴⁶)x²⁶ + (α²²⁹ * α⁶¹)x²⁵ + (α²²⁹ * α¹¹⁸)x²⁴ + (α²²⁹ * α⁷⁰)x²³ + (α²²⁹ * α⁶⁴)x²² + (α²²⁹ * α⁹⁴)x²¹ + (α²²⁹ * α³²)x²⁰ + (α²²⁹ * α⁴⁵)x¹⁹

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:

α²²⁹x²⁹ + α^{(480 % 255)}x²⁸ + α^{(296 % 255)}x²⁷ + α^{(275 % 255)}x²⁶ + α^{(290 % 255)}x²⁵ + α^{(347 % 255)}x²⁴ + α^{(299 % 255)}x²³ + α^{(293 % 255)}x²² + α^{(323 % 255)}x²¹ + α^{(261 % 255)}x²⁰ + α^{(274 % 255)}x¹⁹

The result is:

α²²⁹x²⁹ + α²²⁵x²⁸ + α⁴¹x²⁷ + α²⁰x²⁶ + α³⁵x²⁵ + α⁹²x²⁴ + α⁴⁴x²³ + α³⁸x²² + α⁶⁸x²¹ + α⁶x²⁰ + α¹⁹x¹⁹

Now, convert this to integer notation:

122x²⁹ + 36x²⁸ + 212x²⁷ + 180x²⁶ + 156x²⁵ + 91x²⁴ + 238x²³ + 148x²² + 153x²¹ + 64x²⁰ + 90x¹⁹

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

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

(122 ⊕ 122)x²⁹ + (63 ⊕ 36)x²⁸ + (0 ⊕ 212)x²⁷ + (120 ⊕ 180)x²⁶ + (34 ⊕ 156)x²⁵ + (208 ⊕ 91)x²⁴ + (66 ⊕ 238)x²³ + (151 ⊕ 148)x²² + (167 ⊕ 153)x²¹ + (238 ⊕ 64)x²⁰ + (2 ⊕ 90)x¹⁹ + (236 ⊕ 0)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²⁹ + 27x²⁸ + 212x²⁷ + 204x²⁶ + 190x²⁵ + 139x²⁴ + 172x²³ + 3x²² + 62x²¹ + 174x²⁰ + 88x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

27x²⁸ + 212x²⁷ + 204x²⁶ + 190x²⁵ + 139x²⁴ + 172x²³ + 3x²² + 62x²¹ + 174x²⁰ + 88x¹⁹ + 236x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 27x²⁸. Convert 27x²⁸ to alpha notation. According to the log antilog table, for the integer value 27, the alpha exponent is 248. Therefore 27 = α²⁴⁸. Multiply the generator polynomial by α²⁴⁸:

(α²⁴⁸ * α⁰)x²⁸ + (α²⁴⁸ * α²⁵¹)x²⁷ + (α²⁴⁸ * α⁶⁷)x²⁶ + (α²⁴⁸ * α⁴⁶)x²⁵ + (α²⁴⁸ * α⁶¹)x²⁴ + (α²⁴⁸ * α¹¹⁸)x²³ + (α²⁴⁸ * α⁷⁰)x²² + (α²⁴⁸ * α⁶⁴)x²¹ + (α²⁴⁸ * α⁹⁴)x²⁰ + (α²⁴⁸ * α³²)x¹⁹ + (α²⁴⁸ * α⁴⁵)x¹⁸

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:

α²⁴⁸x²⁸ + α^{(499 % 255)}x²⁷ + α^{(315 % 255)}x²⁶ + α^{(294 % 255)}x²⁵ + α^{(309 % 255)}x²⁴ + α^{(366 % 255)}x²³ + α^{(318 % 255)}x²² + α^{(312 % 255)}x²¹ + α^{(342 % 255)}x²⁰ + α^{(280 % 255)}x¹⁹ + α^{(293 % 255)}x¹⁸

The result is:

α²⁴⁸x²⁸ + α²⁴⁴x²⁷ + α⁶⁰x²⁶ + α³⁹x²⁵ + α⁵⁴x²⁴ + α¹¹¹x²³ + α⁶³x²² + α⁵⁷x²¹ + α⁸⁷x²⁰ + α²⁵x¹⁹ + α³⁸x¹⁸

Now, convert this to integer notation:

27x²⁸ + 250x²⁷ + 185x²⁶ + 53x²⁵ + 80x²⁴ + 206x²³ + 161x²² + 186x²¹ + 127x²⁰ + 3x¹⁹ + 148x¹⁸

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

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

(27 ⊕ 27)x²⁸ + (212 ⊕ 250)x²⁷ + (204 ⊕ 185)x²⁶ + (190 ⊕ 53)x²⁵ + (139 ⊕ 80)x²⁴ + (172 ⊕ 206)x²³ + (3 ⊕ 161)x²² + (62 ⊕ 186)x²¹ + (174 ⊕ 127)x²⁰ + (88 ⊕ 3)x¹⁹ + (236 ⊕ 148)x¹⁸ + (2 ⊕ 0)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²⁸ + 46x²⁷ + 117x²⁶ + 139x²⁵ + 219x²⁴ + 98x²³ + 162x²² + 132x²¹ + 209x²⁰ + 91x¹⁹ + 120x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

46x²⁷ + 117x²⁶ + 139x²⁵ + 219x²⁴ + 98x²³ + 162x²² + 132x²¹ + 209x²⁰ + 91x¹⁹ + 120x¹⁸ + 2x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 46x²⁷. Convert 46x²⁷ to alpha notation. According to the log antilog table, for the integer value 46, the alpha exponent is 130. Therefore 46 = α¹³⁰. Multiply the generator polynomial by α¹³⁰:

(α¹³⁰ * α⁰)x²⁷ + (α¹³⁰ * α²⁵¹)x²⁶ + (α¹³⁰ * α⁶⁷)x²⁵ + (α¹³⁰ * α⁴⁶)x²⁴ + (α¹³⁰ * α⁶¹)x²³ + (α¹³⁰ * α¹¹⁸)x²² + (α¹³⁰ * α⁷⁰)x²¹ + (α¹³⁰ * α⁶⁴)x²⁰ + (α¹³⁰ * α⁹⁴)x¹⁹ + (α¹³⁰ * α³²)x¹⁸ + (α¹³⁰ * α⁴⁵)x¹⁷

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:

α¹³⁰x²⁷ + α^{(381 % 255)}x²⁶ + α¹⁹⁷x²⁵ + α¹⁷⁶x²⁴ + α¹⁹¹x²³ + α²⁴⁸x²² + α²⁰⁰x²¹ + α¹⁹⁴x²⁰ + α²²⁴x¹⁹ + α¹⁶²x¹⁸ + α¹⁷⁵x¹⁷

The result is:

α¹³⁰x²⁷ + α¹²⁶x²⁶ + α¹⁹⁷x²⁵ + α¹⁷⁶x²⁴ + α¹⁹¹x²³ + α²⁴⁸x²² + α²⁰⁰x²¹ + α¹⁹⁴x²⁰ + α²²⁴x¹⁹ + α¹⁶²x¹⁸ + α¹⁷⁵x¹⁷

Now, convert this to integer notation:

46x²⁷ + 102x²⁶ + 141x²⁵ + 227x²⁴ + 65x²³ + 27x²² + 28x²¹ + 50x²⁰ + 18x¹⁹ + 191x¹⁸ + 255x¹⁷

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

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

(46 ⊕ 46)x²⁷ + (117 ⊕ 102)x²⁶ + (139 ⊕ 141)x²⁵ + (219 ⊕ 227)x²⁴ + (98 ⊕ 65)x²³ + (162 ⊕ 27)x²² + (132 ⊕ 28)x²¹ + (209 ⊕ 50)x²⁰ + (91 ⊕ 18)x¹⁹ + (120 ⊕ 191)x¹⁸ + (2 ⊕ 255)x¹⁷ + (17 ⊕ 0)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²⁷ + 19x²⁶ + 6x²⁵ + 56x²⁴ + 35x²³ + 185x²² + 152x²¹ + 227x²⁰ + 73x¹⁹ + 199x¹⁸ + 253x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

19x²⁶ + 6x²⁵ + 56x²⁴ + 35x²³ + 185x²² + 152x²¹ + 227x²⁰ + 73x¹⁹ + 199x¹⁸ + 253x¹⁷ + 17x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 19x²⁶. Convert 19x²⁶ to alpha notation. According to the log antilog table, for the integer value 19, the alpha exponent is 14. Therefore 19 = α¹⁴. Multiply the generator polynomial by α¹⁴:

(α¹⁴ * α⁰)x²⁶ + (α¹⁴ * α²⁵¹)x²⁵ + (α¹⁴ * α⁶⁷)x²⁴ + (α¹⁴ * α⁴⁶)x²³ + (α¹⁴ * α⁶¹)x²² + (α¹⁴ * α¹¹⁸)x²¹ + (α¹⁴ * α⁷⁰)x²⁰ + (α¹⁴ * α⁶⁴)x¹⁹ + (α¹⁴ * α⁹⁴)x¹⁸ + (α¹⁴ * α³²)x¹⁷ + (α¹⁴ * α⁴⁵)x¹⁶

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:

α¹⁴x²⁶ + α^{(265 % 255)}x²⁵ + α⁸¹x²⁴ + α⁶⁰x²³ + α⁷⁵x²² + α¹³²x²¹ + α⁸⁴x²⁰ + α⁷⁸x¹⁹ + α¹⁰⁸x¹⁸ + α⁴⁶x¹⁷ + α⁵⁹x¹⁶

The result is:

α¹⁴x²⁶ + α¹⁰x²⁵ + α⁸¹x²⁴ + α⁶⁰x²³ + α⁷⁵x²² + α¹³²x²¹ + α⁸⁴x²⁰ + α⁷⁸x¹⁹ + α¹⁰⁸x¹⁸ + α⁴⁶x¹⁷ + α⁵⁹x¹⁶

Now, convert this to integer notation:

19x²⁶ + 116x²⁵ + 231x²⁴ + 185x²³ + 15x²² + 184x²¹ + 107x²⁰ + 120x¹⁹ + 208x¹⁸ + 159x¹⁷ + 210x¹⁶

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

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

(19 ⊕ 19)x²⁶ + (6 ⊕ 116)x²⁵ + (56 ⊕ 231)x²⁴ + (35 ⊕ 185)x²³ + (185 ⊕ 15)x²² + (152 ⊕ 184)x²¹ + (227 ⊕ 107)x²⁰ + (73 ⊕ 120)x¹⁹ + (199 ⊕ 208)x¹⁸ + (253 ⊕ 159)x¹⁷ + (17 ⊕ 210)x¹⁶ + (2 ⊕ 0)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²⁶ + 114x²⁵ + 223x²⁴ + 154x²³ + 182x²² + 32x²¹ + 136x²⁰ + 49x¹⁹ + 23x¹⁸ + 98x¹⁷ + 195x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

114x²⁵ + 223x²⁴ + 154x²³ + 182x²² + 32x²¹ + 136x²⁰ + 49x¹⁹ + 23x¹⁸ + 98x¹⁷ + 195x¹⁶ + 2x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

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 114x²⁵. Convert 114x²⁵ to alpha notation. According to the log antilog table, for the integer value 114, the alpha exponent is 155. Therefore 114 = α¹⁵⁵. Multiply the generator polynomial by α¹⁵⁵:

(α¹⁵⁵ * α⁰)x²⁵ + (α¹⁵⁵ * α²⁵¹)x²⁴ + (α¹⁵⁵ * α⁶⁷)x²³ + (α¹⁵⁵ * α⁴⁶)x²² + (α¹⁵⁵ * α⁶¹)x²¹ + (α¹⁵⁵ * α¹¹⁸)x²⁰ + (α¹⁵⁵ * α⁷⁰)x¹⁹ + (α¹⁵⁵ * α⁶⁴)x¹⁸ + (α¹⁵⁵ * α⁹⁴)x¹⁷ + (α¹⁵⁵ * α³²)x¹⁶ + (α¹⁵⁵ * α⁴⁵)x¹⁵

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:

α¹⁵⁵x²⁵ + α^{(406 % 255)}x²⁴ + α²²²x²³ + α²⁰¹x²² + α²¹⁶x²¹ + α^{(273 % 255)}x²⁰ + α²²⁵x¹⁹ + α²¹⁹x¹⁸ + α²⁴⁹x¹⁷ + α¹⁸⁷x¹⁶ + α²⁰⁰x¹⁵

The result is:

α¹⁵⁵x²⁵ + α¹⁵¹x²⁴ + α²²²x²³ + α²⁰¹x²² + α²¹⁶x²¹ + α¹⁸x²⁰ + α²²⁵x¹⁹ + α²¹⁹x¹⁸ + α²⁴⁹x¹⁷ + α¹⁸⁷x¹⁶ + α²⁰⁰x¹⁵

Now, convert this to integer notation:

114x²⁵ + 170x²⁴ + 138x²³ + 56x²² + 195x²¹ + 45x²⁰ + 36x¹⁹ + 86x¹⁸ + 54x¹⁷ + 220x¹⁶ + 28x¹⁵

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

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

(114 ⊕ 114)x²⁵ + (223 ⊕ 170)x²⁴ + (154 ⊕ 138)x²³ + (182 ⊕ 56)x²² + (32 ⊕ 195)x²¹ + (136 ⊕ 45)x²⁰ + (49 ⊕ 36)x¹⁹ + (23 ⊕ 86)x¹⁸ + (98 ⊕ 54)x¹⁷ + (195 ⊕ 220)x¹⁶ + (2 ⊕ 28)x¹⁵ + (236 ⊕ 0)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²⁵ + 117x²⁴ + 16x²³ + 142x²² + 227x²¹ + 165x²⁰ + 21x¹⁹ + 65x¹⁸ + 84x¹⁷ + 31x¹⁶ + 30x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

117x²⁴ + 16x²³ + 142x²² + 227x²¹ + 165x²⁰ + 21x¹⁹ + 65x¹⁸ + 84x¹⁷ + 31x¹⁶ + 30x¹⁵ + 236x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Step 17a: 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 117x²⁴. Convert 117x²⁴ to alpha notation. According to the log antilog table, for the integer value 117, the alpha exponent is 21. Therefore 117 = α²¹. Multiply the generator polynomial by α²¹:

(α²¹ * α⁰)x²⁴ + (α²¹ * α²⁵¹)x²³ + (α²¹ * α⁶⁷)x²² + (α²¹ * α⁴⁶)x²¹ + (α²¹ * α⁶¹)x²⁰ + (α²¹ * α¹¹⁸)x¹⁹ + (α²¹ * α⁷⁰)x¹⁸ + (α²¹ * α⁶⁴)x¹⁷ + (α²¹ * α⁹⁴)x¹⁶ + (α²¹ * α³²)x¹⁵ + (α²¹ * α⁴⁵)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:

α²¹x²⁴ + α^{(272 % 255)}x²³ + α⁸⁸x²² + α⁶⁷x²¹ + α⁸²x²⁰ + α¹³⁹x¹⁹ + α⁹¹x¹⁸ + α⁸⁵x¹⁷ + α¹¹⁵x¹⁶ + α⁵³x¹⁵ + α⁶⁶x¹⁴

The result is:

α²¹x²⁴ + α¹⁷x²³ + α⁸⁸x²² + α⁶⁷x²¹ + α⁸²x²⁰ + α¹³⁹x¹⁹ + α⁹¹x¹⁸ + α⁸⁵x¹⁷ + α¹¹⁵x¹⁶ + α⁵³x¹⁵ + α⁶⁶x¹⁴

Now, convert this to integer notation:

117x²⁴ + 152x²³ + 254x²² + 194x²¹ + 211x²⁰ + 66x¹⁹ + 163x¹⁸ + 214x¹⁷ + 124x¹⁶ + 40x¹⁵ + 97x¹⁴

Step 17b: XOR the result with the result from step 16b

Use the result from step 16b to perform the next XOR.

(117 ⊕ 117)x²⁴ + (16 ⊕ 152)x²³ + (142 ⊕ 254)x²² + (227 ⊕ 194)x²¹ + (165 ⊕ 211)x²⁰ + (21 ⊕ 66)x¹⁹ + (65 ⊕ 163)x¹⁸ + (84 ⊕ 214)x¹⁷ + (31 ⊕ 124)x¹⁶ + (30 ⊕ 40)x¹⁵ + (236 ⊕ 97)x¹⁴ + (2 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²⁴ + 136x²³ + 112x²² + 33x²¹ + 118x²⁰ + 87x¹⁹ + 226x¹⁸ + 130x¹⁷ + 99x¹⁶ + 54x¹⁵ + 141x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

136x²³ + 112x²² + 33x²¹ + 118x²⁰ + 87x¹⁹ + 226x¹⁸ + 130x¹⁷ + 99x¹⁶ + 54x¹⁵ + 141x¹⁴ + 2x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Step 18a: 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 136x²³. Convert 136x²³ to alpha notation. According to the log antilog table, for the integer value 136, the alpha exponent is 103. Therefore 136 = α¹⁰³. Multiply the generator polynomial by α¹⁰³:

(α¹⁰³ * α⁰)x²³ + (α¹⁰³ * α²⁵¹)x²² + (α¹⁰³ * α⁶⁷)x²¹ + (α¹⁰³ * α⁴⁶)x²⁰ + (α¹⁰³ * α⁶¹)x¹⁹ + (α¹⁰³ * α¹¹⁸)x¹⁸ + (α¹⁰³ * α⁷⁰)x¹⁷ + (α¹⁰³ * α⁶⁴)x¹⁶ + (α¹⁰³ * α⁹⁴)x¹⁵ + (α¹⁰³ * α³²)x¹⁴ + (α¹⁰³ * α⁴⁵)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:

α¹⁰³x²³ + α^{(354 % 255)}x²² + α¹⁷⁰x²¹ + α¹⁴⁹x²⁰ + α¹⁶⁴x¹⁹ + α²²¹x¹⁸ + α¹⁷³x¹⁷ + α¹⁶⁷x¹⁶ + α¹⁹⁷x¹⁵ + α¹³⁵x¹⁴ + α¹⁴⁸x¹³

The result is:

α¹⁰³x²³ + α⁹⁹x²² + α¹⁷⁰x²¹ + α¹⁴⁹x²⁰ + α¹⁶⁴x¹⁹ + α²²¹x¹⁸ + α¹⁷³x¹⁷ + α¹⁶⁷x¹⁶ + α¹⁹⁷x¹⁵ + α¹³⁵x¹⁴ + α¹⁴⁸x¹³

Now, convert this to integer notation:

136x²³ + 134x²² + 215x²¹ + 164x²⁰ + 198x¹⁹ + 69x¹⁸ + 246x¹⁷ + 126x¹⁶ + 141x¹⁵ + 169x¹⁴ + 82x¹³

Step 18b: XOR the result with the result from step 17b

Use the result from step 17b to perform the next XOR.

(136 ⊕ 136)x²³ + (112 ⊕ 134)x²² + (33 ⊕ 215)x²¹ + (118 ⊕ 164)x²⁰ + (87 ⊕ 198)x¹⁹ + (226 ⊕ 69)x¹⁸ + (130 ⊕ 246)x¹⁷ + (99 ⊕ 126)x¹⁶ + (54 ⊕ 141)x¹⁵ + (141 ⊕ 169)x¹⁴ + (2 ⊕ 82)x¹³ + (17 ⊕ 0)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²³ + 246x²² + 246x²¹ + 210x²⁰ + 145x¹⁹ + 167x¹⁸ + 116x¹⁷ + 29x¹⁶ + 187x¹⁵ + 36x¹⁴ + 80x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

246x²² + 246x²¹ + 210x²⁰ + 145x¹⁹ + 167x¹⁸ + 116x¹⁷ + 29x¹⁶ + 187x¹⁵ + 36x¹⁴ + 80x¹³ + 17x¹² + 2x¹¹ + 236x¹⁰

Step 19a: 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 246x²². Convert 246x²² to alpha notation. According to the log antilog table, for the integer value 246, the alpha exponent is 173. Therefore 246 = α¹⁷³. Multiply the generator polynomial by α¹⁷³:

(α¹⁷³ * α⁰)x²² + (α¹⁷³ * α²⁵¹)x²¹ + (α¹⁷³ * α⁶⁷)x²⁰ + (α¹⁷³ * α⁴⁶)x¹⁹ + (α¹⁷³ * α⁶¹)x¹⁸ + (α¹⁷³ * α¹¹⁸)x¹⁷ + (α¹⁷³ * α⁷⁰)x¹⁶ + (α¹⁷³ * α⁶⁴)x¹⁵ + (α¹⁷³ * α⁹⁴)x¹⁴ + (α¹⁷³ * α³²)x¹³ + (α¹⁷³ * α⁴⁵)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:

α¹⁷³x²² + α^{(424 % 255)}x²¹ + α²⁴⁰x²⁰ + α²¹⁹x¹⁹ + α²³⁴x¹⁸ + α^{(291 % 255)}x¹⁷ + α²⁴³x¹⁶ + α²³⁷x¹⁵ + α^{(267 % 255)}x¹⁴ + α²⁰⁵x¹³ + α²¹⁸x¹²

The result is:

α¹⁷³x²² + α¹⁶⁹x²¹ + α²⁴⁰x²⁰ + α²¹⁹x¹⁹ + α²³⁴x¹⁸ + α³⁶x¹⁷ + α²⁴³x¹⁶ + α²³⁷x¹⁵ + α¹²x¹⁴ + α²⁰⁵x¹³ + α²¹⁸x¹²

Now, convert this to integer notation:

246x²² + 229x²¹ + 44x²⁰ + 86x¹⁹ + 251x¹⁸ + 37x¹⁷ + 125x¹⁶ + 139x¹⁵ + 205x¹⁴ + 167x¹³ + 43x¹²

Step 19b: XOR the result with the result from step 18b

Use the result from step 18b to perform the next XOR.

(246 ⊕ 246)x²² + (246 ⊕ 229)x²¹ + (210 ⊕ 44)x²⁰ + (145 ⊕ 86)x¹⁹ + (167 ⊕ 251)x¹⁸ + (116 ⊕ 37)x¹⁷ + (29 ⊕ 125)x¹⁶ + (187 ⊕ 139)x¹⁵ + (36 ⊕ 205)x¹⁴ + (80 ⊕ 167)x¹³ + (17 ⊕ 43)x¹² + (2 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²² + 19x²¹ + 254x²⁰ + 199x¹⁹ + 92x¹⁸ + 81x¹⁷ + 96x¹⁶ + 48x¹⁵ + 233x¹⁴ + 247x¹³ + 58x¹² + 2x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

19x²¹ + 254x²⁰ + 199x¹⁹ + 92x¹⁸ + 81x¹⁷ + 96x¹⁶ + 48x¹⁵ + 233x¹⁴ + 247x¹³ + 58x¹² + 2x¹¹ + 236x¹⁰

Step 20a: 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 19x²¹. Convert 19x²¹ to alpha notation. According to the log antilog table, for the integer value 19, the alpha exponent is 14. Therefore 19 = α¹⁴. Multiply the generator polynomial by α¹⁴:

(α¹⁴ * α⁰)x²¹ + (α¹⁴ * α²⁵¹)x²⁰ + (α¹⁴ * α⁶⁷)x¹⁹ + (α¹⁴ * α⁴⁶)x¹⁸ + (α¹⁴ * α⁶¹)x¹⁷ + (α¹⁴ * α¹¹⁸)x¹⁶ + (α¹⁴ * α⁷⁰)x¹⁵ + (α¹⁴ * α⁶⁴)x¹⁴ + (α¹⁴ * α⁹⁴)x¹³ + (α¹⁴ * α³²)x¹² + (α¹⁴ * α⁴⁵)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:

α¹⁴x²¹ + α^{(265 % 255)}x²⁰ + α⁸¹x¹⁹ + α⁶⁰x¹⁸ + α⁷⁵x¹⁷ + α¹³²x¹⁶ + α⁸⁴x¹⁵ + α⁷⁸x¹⁴ + α¹⁰⁸x¹³ + α⁴⁶x¹² + α⁵⁹x¹¹

The result is:

α¹⁴x²¹ + α¹⁰x²⁰ + α⁸¹x¹⁹ + α⁶⁰x¹⁸ + α⁷⁵x¹⁷ + α¹³²x¹⁶ + α⁸⁴x¹⁵ + α⁷⁸x¹⁴ + α¹⁰⁸x¹³ + α⁴⁶x¹² + α⁵⁹x¹¹

Now, convert this to integer notation:

19x²¹ + 116x²⁰ + 231x¹⁹ + 185x¹⁸ + 15x¹⁷ + 184x¹⁶ + 107x¹⁵ + 120x¹⁴ + 208x¹³ + 159x¹² + 210x¹¹

Step 20b: XOR the result with the result from step 19b

Use the result from step 19b to perform the next XOR.

(19 ⊕ 19)x²¹ + (254 ⊕ 116)x²⁰ + (199 ⊕ 231)x¹⁹ + (92 ⊕ 185)x¹⁸ + (81 ⊕ 15)x¹⁷ + (96 ⊕ 184)x¹⁶ + (48 ⊕ 107)x¹⁵ + (233 ⊕ 120)x¹⁴ + (247 ⊕ 208)x¹³ + (58 ⊕ 159)x¹² + (2 ⊕ 210)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²¹ + 138x²⁰ + 32x¹⁹ + 229x¹⁸ + 94x¹⁷ + 216x¹⁶ + 91x¹⁵ + 145x¹⁴ + 39x¹³ + 165x¹² + 208x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

138x²⁰ + 32x¹⁹ + 229x¹⁸ + 94x¹⁷ + 216x¹⁶ + 91x¹⁵ + 145x¹⁴ + 39x¹³ + 165x¹² + 208x¹¹ + 236x¹⁰

Step 21a: 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 138x²⁰. Convert 138x²⁰ to alpha notation. According to the log antilog table, for the integer value 138, the alpha exponent is 222. Therefore 138 = α²²². Multiply the generator polynomial by α²²²:

(α²²² * α⁰)x²⁰ + (α²²² * α²⁵¹)x¹⁹ + (α²²² * α⁶⁷)x¹⁸ + (α²²² * α⁴⁶)x¹⁷ + (α²²² * α⁶¹)x¹⁶ + (α²²² * α¹¹⁸)x¹⁵ + (α²²² * α⁷⁰)x¹⁴ + (α²²² * α⁶⁴)x¹³ + (α²²² * α⁹⁴)x¹² + (α²²² * α³²)x¹¹ + (α²²² * α⁴⁵)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:

α²²²x²⁰ + α^{(473 % 255)}x¹⁹ + α^{(289 % 255)}x¹⁸ + α^{(268 % 255)}x¹⁷ + α^{(283 % 255)}x¹⁶ + α^{(340 % 255)}x¹⁵ + α^{(292 % 255)}x¹⁴ + α^{(286 % 255)}x¹³ + α^{(316 % 255)}x¹² + α²⁵⁴x¹¹ + α^{(267 % 255)}x¹⁰

The result is:

α²²²x²⁰ + α²¹⁸x¹⁹ + α³⁴x¹⁸ + α¹³x¹⁷ + α²⁸x¹⁶ + α⁸⁵x¹⁵ + α³⁷x¹⁴ + α³¹x¹³ + α⁶¹x¹² + α²⁵⁴x¹¹ + α¹²x¹⁰

Now, convert this to integer notation:

138x²⁰ + 43x¹⁹ + 78x¹⁸ + 135x¹⁷ + 24x¹⁶ + 214x¹⁵ + 74x¹⁴ + 192x¹³ + 111x¹² + 142x¹¹ + 205x¹⁰

Step 21b: XOR the result with the result from step 20b

Use the result from step 20b to perform the next XOR.

(138 ⊕ 138)x²⁰ + (32 ⊕ 43)x¹⁹ + (229 ⊕ 78)x¹⁸ + (94 ⊕ 135)x¹⁷ + (216 ⊕ 24)x¹⁶ + (91 ⊕ 214)x¹⁵ + (145 ⊕ 74)x¹⁴ + (39 ⊕ 192)x¹³ + (165 ⊕ 111)x¹² + (208 ⊕ 142)x¹¹ + (236 ⊕ 205)x¹⁰

The result is:

0x²⁰ + 11x¹⁹ + 171x¹⁸ + 217x¹⁷ + 192x¹⁶ + 141x¹⁵ + 219x¹⁴ + 231x¹³ + 202x¹² + 94x¹¹ + 33x¹⁰

Discard the lead 0 term to get:

11x¹⁹ + 171x¹⁸ + 217x¹⁷ + 192x¹⁶ + 141x¹⁵ + 219x¹⁴ + 231x¹³ + 202x¹² + 94x¹¹ + 33x¹⁰

Step 22a: 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 11x¹⁹. Convert 11x¹⁹ to alpha notation. According to the log antilog table, for the integer value 11, the alpha exponent is 238. Therefore 11 = α²³⁸. Multiply the generator polynomial by α²³⁸:

(α²³⁸ * α⁰)x¹⁹ + (α²³⁸ * α²⁵¹)x¹⁸ + (α²³⁸ * α⁶⁷)x¹⁷ + (α²³⁸ * α⁴⁶)x¹⁶ + (α²³⁸ * α⁶¹)x¹⁵ + (α²³⁸ * α¹¹⁸)x¹⁴ + (α²³⁸ * α⁷⁰)x¹³ + (α²³⁸ * α⁶⁴)x¹² + (α²³⁸ * α⁹⁴)x¹¹ + (α²³⁸ * α³²)x¹⁰ + (α²³⁸ * α⁴⁵)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:

α²³⁸x¹⁹ + α^{(489 % 255)}x¹⁸ + α^{(305 % 255)}x¹⁷ + α^{(284 % 255)}x¹⁶ + α^{(299 % 255)}x¹⁵ + α^{(356 % 255)}x¹⁴ + α^{(308 % 255)}x¹³ + α^{(302 % 255)}x¹² + α^{(332 % 255)}x¹¹ + α^{(270 % 255)}x¹⁰ + α^{(283 % 255)}x⁹

The result is:

α²³⁸x¹⁹ + α²³⁴x¹⁸ + α⁵⁰x¹⁷ + α²⁹x¹⁶ + α⁴⁴x¹⁵ + α¹⁰¹x¹⁴ + α⁵³x¹³ + α⁴⁷x¹² + α⁷⁷x¹¹ + α¹⁵x¹⁰ + α²⁸x⁹

Now, convert this to integer notation:

11x¹⁹ + 251x¹⁸ + 5x¹⁷ + 48x¹⁶ + 238x¹⁵ + 34x¹⁴ + 40x¹³ + 35x¹² + 60x¹¹ + 38x¹⁰ + 24x⁹

Step 22b: XOR the result with the result from step 21b

Use the result from step 21b to perform the next XOR.

(11 ⊕ 11)x¹⁹ + (171 ⊕ 251)x¹⁸ + (217 ⊕ 5)x¹⁷ + (192 ⊕ 48)x¹⁶ + (141 ⊕ 238)x¹⁵ + (219 ⊕ 34)x¹⁴ + (231 ⊕ 40)x¹³ + (202 ⊕ 35)x¹² + (94 ⊕ 60)x¹¹ + (33 ⊕ 38)x¹⁰ + (0 ⊕ 24)x⁹

The result is:

0x¹⁹ + 80x¹⁸ + 220x¹⁷ + 240x¹⁶ + 99x¹⁵ + 249x¹⁴ + 207x¹³ + 233x¹² + 98x¹¹ + 7x¹⁰ + 24x⁹

Discard the lead 0 term to get:

80x¹⁸ + 220x¹⁷ + 240x¹⁶ + 99x¹⁵ + 249x¹⁴ + 207x¹³ + 233x¹² + 98x¹¹ + 7x¹⁰ + 24x⁹

Step 23a: 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 80x¹⁸. Convert 80x¹⁸ to alpha notation. According to the log antilog table, for the integer value 80, the alpha exponent is 54. Therefore 80 = α⁵⁴. Multiply the generator polynomial by α⁵⁴:

(α⁵⁴ * α⁰)x¹⁸ + (α⁵⁴ * α²⁵¹)x¹⁷ + (α⁵⁴ * α⁶⁷)x¹⁶ + (α⁵⁴ * α⁴⁶)x¹⁵ + (α⁵⁴ * α⁶¹)x¹⁴ + (α⁵⁴ * α¹¹⁸)x¹³ + (α⁵⁴ * α⁷⁰)x¹² + (α⁵⁴ * α⁶⁴)x¹¹ + (α⁵⁴ * α⁹⁴)x¹⁰ + (α⁵⁴ * α³²)x⁹ + (α⁵⁴ * α⁴⁵)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:

α⁵⁴x¹⁸ + α^{(305 % 255)}x¹⁷ + α¹²¹x¹⁶ + α¹⁰⁰x¹⁵ + α¹¹⁵x¹⁴ + α¹⁷²x¹³ + α¹²⁴x¹² + α¹¹⁸x¹¹ + α¹⁴⁸x¹⁰ + α⁸⁶x⁹ + α⁹⁹x⁸

The result is:

α⁵⁴x¹⁸ + α⁵⁰x¹⁷ + α¹²¹x¹⁶ + α¹⁰⁰x¹⁵ + α¹¹⁵x¹⁴ + α¹⁷²x¹³ + α¹²⁴x¹² + α¹¹⁸x¹¹ + α¹⁴⁸x¹⁰ + α⁸⁶x⁹ + α⁹⁹x⁸

Now, convert this to integer notation:

80x¹⁸ + 5x¹⁷ + 118x¹⁶ + 17x¹⁵ + 124x¹⁴ + 123x¹³ + 151x¹² + 199x¹¹ + 82x¹⁰ + 177x⁹ + 134x⁸

Step 23b: XOR the result with the result from step 22b

Use the result from step 22b to perform the next XOR.

(80 ⊕ 80)x¹⁸ + (220 ⊕ 5)x¹⁷ + (240 ⊕ 118)x¹⁶ + (99 ⊕ 17)x¹⁵ + (249 ⊕ 124)x¹⁴ + (207 ⊕ 123)x¹³ + (233 ⊕ 151)x¹² + (98 ⊕ 199)x¹¹ + (7 ⊕ 82)x¹⁰ + (24 ⊕ 177)x⁹ + (0 ⊕ 134)x⁸

The result is:

0x¹⁸ + 217x¹⁷ + 134x¹⁶ + 114x¹⁵ + 133x¹⁴ + 180x¹³ + 126x¹² + 165x¹¹ + 85x¹⁰ + 169x⁹ + 134x⁸

Discard the lead 0 term to get:

217x¹⁷ + 134x¹⁶ + 114x¹⁵ + 133x¹⁴ + 180x¹³ + 126x¹² + 165x¹¹ + 85x¹⁰ + 169x⁹ + 134x⁸

Step 24a: 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 217x¹⁷. Convert 217x¹⁷ to alpha notation. According to the log antilog table, for the integer value 217, the alpha exponent is 96. Therefore 217 = α⁹⁶. Multiply the generator polynomial by α⁹⁶:

(α⁹⁶ * α⁰)x¹⁷ + (α⁹⁶ * α²⁵¹)x¹⁶ + (α⁹⁶ * α⁶⁷)x¹⁵ + (α⁹⁶ * α⁴⁶)x¹⁴ + (α⁹⁶ * α⁶¹)x¹³ + (α⁹⁶ * α¹¹⁸)x¹² + (α⁹⁶ * α⁷⁰)x¹¹ + (α⁹⁶ * α⁶⁴)x¹⁰ + (α⁹⁶ * α⁹⁴)x⁹ + (α⁹⁶ * α³²)x⁸ + (α⁹⁶ * α⁴⁵)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:

α⁹⁶x¹⁷ + α^{(347 % 255)}x¹⁶ + α¹⁶³x¹⁵ + α¹⁴²x¹⁴ + α¹⁵⁷x¹³ + α²¹⁴x¹² + α¹⁶⁶x¹¹ + α¹⁶⁰x¹⁰ + α¹⁹⁰x⁹ + α¹²⁸x⁸ + α¹⁴¹x⁷

The result is:

α⁹⁶x¹⁷ + α⁹²x¹⁶ + α¹⁶³x¹⁵ + α¹⁴²x¹⁴ + α¹⁵⁷x¹³ + α²¹⁴x¹² + α¹⁶⁶x¹¹ + α¹⁶⁰x¹⁰ + α¹⁹⁰x⁹ + α¹²⁸x⁸ + α¹⁴¹x⁷

Now, convert this to integer notation:

217x¹⁷ + 91x¹⁶ + 99x¹⁵ + 42x¹⁴ + 213x¹³ + 249x¹² + 63x¹¹ + 230x¹⁰ + 174x⁹ + 133x⁸ + 21x⁷

Step 24b: XOR the result with the result from step 23b

Use the result from step 23b to perform the next XOR.

(217 ⊕ 217)x¹⁷ + (134 ⊕ 91)x¹⁶ + (114 ⊕ 99)x¹⁵ + (133 ⊕ 42)x¹⁴ + (180 ⊕ 213)x¹³ + (126 ⊕ 249)x¹² + (165 ⊕ 63)x¹¹ + (85 ⊕ 230)x¹⁰ + (169 ⊕ 174)x⁹ + (134 ⊕ 133)x⁸ + (0 ⊕ 21)x⁷

The result is:

0x¹⁷ + 221x¹⁶ + 17x¹⁵ + 175x¹⁴ + 97x¹³ + 135x¹² + 154x¹¹ + 179x¹⁰ + 7x⁹ + 3x⁸ + 21x⁷

Discard the lead 0 term to get:

221x¹⁶ + 17x¹⁵ + 175x¹⁴ + 97x¹³ + 135x¹² + 154x¹¹ + 179x¹⁰ + 7x⁹ + 3x⁸ + 21x⁷

Step 25a: 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 221x¹⁶. Convert 221x¹⁶ 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)x¹⁶ + (α²⁰⁴ * α²⁵¹)x¹⁵ + (α²⁰⁴ * α⁶⁷)x¹⁴ + (α²⁰⁴ * α⁴⁶)x¹³ + (α²⁰⁴ * α⁶¹)x¹² + (α²⁰⁴ * α¹¹⁸)x¹¹ + (α²⁰⁴ * α⁷⁰)x¹⁰ + (α²⁰⁴ * α⁶⁴)x⁹ + (α²⁰⁴ * α⁹⁴)x⁸ + (α²⁰⁴ * α³²)x⁷ + (α²⁰⁴ * α⁴⁵)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:

α²⁰⁴x¹⁶ + α^{(455 % 255)}x¹⁵ + α^{(271 % 255)}x¹⁴ + α²⁵⁰x¹³ + α^{(265 % 255)}x¹² + α^{(322 % 255)}x¹¹ + α^{(274 % 255)}x¹⁰ + α^{(268 % 255)}x⁹ + α^{(298 % 255)}x⁸ + α²³⁶x⁷ + α²⁴⁹x⁶

The result is:

α²⁰⁴x¹⁶ + α²⁰⁰x¹⁵ + α¹⁶x¹⁴ + α²⁵⁰x¹³ + α¹⁰x¹² + α⁶⁷x¹¹ + α¹⁹x¹⁰ + α¹³x⁹ + α⁴³x⁸ + α²³⁶x⁷ + α²⁴⁹x⁶

Now, convert this to integer notation:

221x¹⁶ + 28x¹⁵ + 76x¹⁴ + 108x¹³ + 116x¹² + 194x¹¹ + 90x¹⁰ + 135x⁹ + 119x⁸ + 203x⁷ + 54x⁶

Step 25b: XOR the result with the result from step 24b

Use the result from step 24b to perform the next XOR.

(221 ⊕ 221)x¹⁶ + (17 ⊕ 28)x¹⁵ + (175 ⊕ 76)x¹⁴ + (97 ⊕ 108)x¹³ + (135 ⊕ 116)x¹² + (154 ⊕ 194)x¹¹ + (179 ⊕ 90)x¹⁰ + (7 ⊕ 135)x⁹ + (3 ⊕ 119)x⁸ + (21 ⊕ 203)x⁷ + (0 ⊕ 54)x⁶

The result is:

0x¹⁶ + 13x¹⁵ + 227x¹⁴ + 13x¹³ + 243x¹² + 88x¹¹ + 233x¹⁰ + 128x⁹ + 116x⁸ + 222x⁷ + 54x⁶

Discard the lead 0 term to get:

13x¹⁵ + 227x¹⁴ + 13x¹³ + 243x¹² + 88x¹¹ + 233x¹⁰ + 128x⁹ + 116x⁸ + 222x⁷ + 54x⁶

Step 26a: 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 13x¹⁵. Convert 13x¹⁵ to alpha notation. According to the log antilog table, for the integer value 13, the alpha exponent is 104. Therefore 13 = α¹⁰⁴. Multiply the generator polynomial by α¹⁰⁴:

(α¹⁰⁴ * α⁰)x¹⁵ + (α¹⁰⁴ * α²⁵¹)x¹⁴ + (α¹⁰⁴ * α⁶⁷)x¹³ + (α¹⁰⁴ * α⁴⁶)x¹² + (α¹⁰⁴ * α⁶¹)x¹¹ + (α¹⁰⁴ * α¹¹⁸)x¹⁰ + (α¹⁰⁴ * α⁷⁰)x⁹ + (α¹⁰⁴ * α⁶⁴)x⁸ + (α¹⁰⁴ * α⁹⁴)x⁷ + (α¹⁰⁴ * α³²)x⁶ + (α¹⁰⁴ * α⁴⁵)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:

α¹⁰⁴x¹⁵ + α^{(355 % 255)}x¹⁴ + α¹⁷¹x¹³ + α¹⁵⁰x¹² + α¹⁶⁵x¹¹ + α²²²x¹⁰ + α¹⁷⁴x⁹ + α¹⁶⁸x⁸ + α¹⁹⁸x⁷ + α¹³⁶x⁶ + α¹⁴⁹x⁵

The result is:

α¹⁰⁴x¹⁵ + α¹⁰⁰x¹⁴ + α¹⁷¹x¹³ + α¹⁵⁰x¹² + α¹⁶⁵x¹¹ + α²²²x¹⁰ + α¹⁷⁴x⁹ + α¹⁶⁸x⁸ + α¹⁹⁸x⁷ + α¹³⁶x⁶ + α¹⁴⁹x⁵

Now, convert this to integer notation:

13x¹⁵ + 17x¹⁴ + 179x¹³ + 85x¹² + 145x¹¹ + 138x¹⁰ + 241x⁹ + 252x⁸ + 7x⁷ + 79x⁶ + 164x⁵

Step 26b: XOR the result with the result from step 25b

Use the result from step 25b to perform the next XOR.

(13 ⊕ 13)x¹⁵ + (227 ⊕ 17)x¹⁴ + (13 ⊕ 179)x¹³ + (243 ⊕ 85)x¹² + (88 ⊕ 145)x¹¹ + (233 ⊕ 138)x¹⁰ + (128 ⊕ 241)x⁹ + (116 ⊕ 252)x⁸ + (222 ⊕ 7)x⁷ + (54 ⊕ 79)x⁶ + (0 ⊕ 164)x⁵

The result is:

0x¹⁵ + 242x¹⁴ + 190x¹³ + 166x¹² + 201x¹¹ + 99x¹⁰ + 113x⁹ + 136x⁸ + 217x⁷ + 121x⁶ + 164x⁵

Discard the lead 0 term to get:

242x¹⁴ + 190x¹³ + 166x¹² + 201x¹¹ + 99x¹⁰ + 113x⁹ + 136x⁸ + 217x⁷ + 121x⁶ + 164x⁵

Step 27a: 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 242x¹⁴. Convert 242x¹⁴ to alpha notation. According to the log antilog table, for the integer value 242, the alpha exponent is 213. Therefore 242 = α²¹³. Multiply the generator polynomial by α²¹³:

(α²¹³ * α⁰)x¹⁴ + (α²¹³ * α²⁵¹)x¹³ + (α²¹³ * α⁶⁷)x¹² + (α²¹³ * α⁴⁶)x¹¹ + (α²¹³ * α⁶¹)x¹⁰ + (α²¹³ * α¹¹⁸)x⁹ + (α²¹³ * α⁷⁰)x⁸ + (α²¹³ * α⁶⁴)x⁷ + (α²¹³ * α⁹⁴)x⁶ + (α²¹³ * α³²)x⁵ + (α²¹³ * α⁴⁵)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:

α²¹³x¹⁴ + α^{(464 % 255)}x¹³ + α^{(280 % 255)}x¹² + α^{(259 % 255)}x¹¹ + α^{(274 % 255)}x¹⁰ + α^{(331 % 255)}x⁹ + α^{(283 % 255)}x⁸ + α^{(277 % 255)}x⁷ + α^{(307 % 255)}x⁶ + α²⁴⁵x⁵ + α^{(258 % 255)}x⁴

The result is:

α²¹³x¹⁴ + α²⁰⁹x¹³ + α²⁵x¹² + α⁴x¹¹ + α¹⁹x¹⁰ + α⁷⁶x⁹ + α²⁸x⁸ + α²²x⁷ + α⁵²x⁶ + α²⁴⁵x⁵ + α³x⁴

Now, convert this to integer notation:

242x¹⁴ + 162x¹³ + 3x¹² + 16x¹¹ + 90x¹⁰ + 30x⁹ + 24x⁸ + 234x⁷ + 20x⁶ + 233x⁵ + 8x⁴

Step 27b: XOR the result with the result from step 26b

Use the result from step 26b to perform the next XOR.

(242 ⊕ 242)x¹⁴ + (190 ⊕ 162)x¹³ + (166 ⊕ 3)x¹² + (201 ⊕ 16)x¹¹ + (99 ⊕ 90)x¹⁰ + (113 ⊕ 30)x⁹ + (136 ⊕ 24)x⁸ + (217 ⊕ 234)x⁷ + (121 ⊕ 20)x⁶ + (164 ⊕ 233)x⁵ + (0 ⊕ 8)x⁴

The result is:

0x¹⁴ + 28x¹³ + 165x¹² + 217x¹¹ + 57x¹⁰ + 111x⁹ + 144x⁸ + 51x⁷ + 109x⁶ + 77x⁵ + 8x⁴

Discard the lead 0 term to get:

28x¹³ + 165x¹² + 217x¹¹ + 57x¹⁰ + 111x⁹ + 144x⁸ + 51x⁷ + 109x⁶ + 77x⁵ + 8x⁴

Step 28a: 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 28x¹³. Convert 28x¹³ to alpha notation. According to the log antilog table, for the integer value 28, the alpha exponent is 200. Therefore 28 = α²⁰⁰. Multiply the generator polynomial by α²⁰⁰:

(α²⁰⁰ * α⁰)x¹³ + (α²⁰⁰ * α²⁵¹)x¹² + (α²⁰⁰ * α⁶⁷)x¹¹ + (α²⁰⁰ * α⁴⁶)x¹⁰ + (α²⁰⁰ * α⁶¹)x⁹ + (α²⁰⁰ * α¹¹⁸)x⁸ + (α²⁰⁰ * α⁷⁰)x⁷ + (α²⁰⁰ * α⁶⁴)x⁶ + (α²⁰⁰ * α⁹⁴)x⁵ + (α²⁰⁰ * α³²)x⁴ + (α²⁰⁰ * α⁴⁵)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:

α²⁰⁰x¹³ + α^{(451 % 255)}x¹² + α^{(267 % 255)}x¹¹ + α²⁴⁶x¹⁰ + α^{(261 % 255)}x⁹ + α^{(318 % 255)}x⁸ + α^{(270 % 255)}x⁷ + α^{(264 % 255)}x⁶ + α^{(294 % 255)}x⁵ + α²³²x⁴ + α²⁴⁵x³

The result is:

α²⁰⁰x¹³ + α¹⁹⁶x¹² + α¹²x¹¹ + α²⁴⁶x¹⁰ + α⁶x⁹ + α⁶³x⁸ + α¹⁵x⁷ + α⁹x⁶ + α³⁹x⁵ + α²³²x⁴ + α²⁴⁵x³

Now, convert this to integer notation:

28x¹³ + 200x¹² + 205x¹¹ + 207x¹⁰ + 64x⁹ + 161x⁸ + 38x⁷ + 58x⁶ + 53x⁵ + 247x⁴ + 233x³

Step 28b: XOR the result with the result from step 27b

Use the result from step 27b to perform the next XOR.

(28 ⊕ 28)x¹³ + (165 ⊕ 200)x¹² + (217 ⊕ 205)x¹¹ + (57 ⊕ 207)x¹⁰ + (111 ⊕ 64)x⁹ + (144 ⊕ 161)x⁸ + (51 ⊕ 38)x⁷ + (109 ⊕ 58)x⁶ + (77 ⊕ 53)x⁵ + (8 ⊕ 247)x⁴ + (0 ⊕ 233)x³

The result is:

0x¹³ + 109x¹² + 20x¹¹ + 246x¹⁰ + 47x⁹ + 49x⁸ + 21x⁷ + 87x⁶ + 120x⁵ + 255x⁴ + 233x³

Discard the lead 0 term to get:

109x¹² + 20x¹¹ + 246x¹⁰ + 47x⁹ + 49x⁸ + 21x⁷ + 87x⁶ + 120x⁵ + 255x⁴ + 233x³

Step 29a: 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 109x¹². Convert 109x¹² to alpha notation. According to the log antilog table, for the integer value 109, the alpha exponent is 133. Therefore 109 = α¹³³. Multiply the generator polynomial by α¹³³:

(α¹³³ * α⁰)x¹² + (α¹³³ * α²⁵¹)x¹¹ + (α¹³³ * α⁶⁷)x¹⁰ + (α¹³³ * α⁴⁶)x⁹ + (α¹³³ * α⁶¹)x⁸ + (α¹³³ * α¹¹⁸)x⁷ + (α¹³³ * α⁷⁰)x⁶ + (α¹³³ * α⁶⁴)x⁵ + (α¹³³ * α⁹⁴)x⁴ + (α¹³³ * α³²)x³ + (α¹³³ * α⁴⁵)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:

α¹³³x¹² + α^{(384 % 255)}x¹¹ + α²⁰⁰x¹⁰ + α¹⁷⁹x⁹ + α¹⁹⁴x⁸ + α²⁵¹x⁷ + α²⁰³x⁶ + α¹⁹⁷x⁵ + α²²⁷x⁴ + α¹⁶⁵x³ + α¹⁷⁸x²

The result is:

α¹³³x¹² + α¹²⁹x¹¹ + α²⁰⁰x¹⁰ + α¹⁷⁹x⁹ + α¹⁹⁴x⁸ + α²⁵¹x⁷ + α²⁰³x⁶ + α¹⁹⁷x⁵ + α²²⁷x⁴ + α¹⁶⁵x³ + α¹⁷⁸x²

Now, convert this to integer notation:

109x¹² + 23x¹¹ + 28x¹⁰ + 75x⁹ + 50x⁸ + 216x⁷ + 224x⁶ + 141x⁵ + 144x⁴ + 145x³ + 171x²

Step 29b: XOR the result with the result from step 28b

Use the result from step 28b to perform the next XOR.

(109 ⊕ 109)x¹² + (20 ⊕ 23)x¹¹ + (246 ⊕ 28)x¹⁰ + (47 ⊕ 75)x⁹ + (49 ⊕ 50)x⁸ + (21 ⊕ 216)x⁷ + (87 ⊕ 224)x⁶ + (120 ⊕ 141)x⁵ + (255 ⊕ 144)x⁴ + (233 ⊕ 145)x³ + (0 ⊕ 171)x²

The result is:

0x¹² + 3x¹¹ + 234x¹⁰ + 100x⁹ + 3x⁸ + 205x⁷ + 183x⁶ + 245x⁵ + 111x⁴ + 120x³ + 171x²

Discard the lead 0 term to get:

3x¹¹ + 234x¹⁰ + 100x⁹ + 3x⁸ + 205x⁷ + 183x⁶ + 245x⁵ + 111x⁴ + 120x³ + 171x²

Step 30a: 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 3x¹¹. Convert 3x¹¹ to alpha notation. According to the log antilog table, for the integer value 3, the alpha exponent is 25. Therefore 3 = α²⁵. Multiply the generator polynomial by α²⁵:

(α²⁵ * α⁰)x¹¹ + (α²⁵ * α²⁵¹)x¹⁰ + (α²⁵ * α⁶⁷)x⁹ + (α²⁵ * α⁴⁶)x⁸ + (α²⁵ * α⁶¹)x⁷ + (α²⁵ * α¹¹⁸)x⁶ + (α²⁵ * α⁷⁰)x⁵ + (α²⁵ * α⁶⁴)x⁴ + (α²⁵ * α⁹⁴)x³ + (α²⁵ * α³²)x² + (α²⁵ * α⁴⁵)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:

α²⁵x¹¹ + α^{(276 % 255)}x¹⁰ + α⁹²x⁹ + α⁷¹x⁸ + α⁸⁶x⁷ + α¹⁴³x⁶ + α⁹⁵x⁵ + α⁸⁹x⁴ + α¹¹⁹x³ + α⁵⁷x² + α⁷⁰x¹

The result is:

α²⁵x¹¹ + α²¹x¹⁰ + α⁹²x⁹ + α⁷¹x⁸ + α⁸⁶x⁷ + α¹⁴³x⁶ + α⁹⁵x⁵ + α⁸⁹x⁴ + α¹¹⁹x³ + α⁵⁷x² + α⁷⁰x¹

Now, convert this to integer notation:

3x¹¹ + 117x¹⁰ + 91x⁹ + 188x⁸ + 177x⁷ + 84x⁶ + 226x⁵ + 225x⁴ + 147x³ + 186x² + 94x¹

Step 30b: XOR the result with the result from step 29b

Use the result from step 29b to perform the next XOR.

(3 ⊕ 3)x¹¹ + (234 ⊕ 117)x¹⁰ + (100 ⊕ 91)x⁹ + (3 ⊕ 188)x⁸ + (205 ⊕ 177)x⁷ + (183 ⊕ 84)x⁶ + (245 ⊕ 226)x⁵ + (111 ⊕ 225)x⁴ + (120 ⊕ 147)x³ + (171 ⊕ 186)x² + (0 ⊕ 94)x¹

The result is:

0x¹¹ + 159x¹⁰ + 63x⁹ + 191x⁸ + 124x⁷ + 227x⁶ + 23x⁵ + 142x⁴ + 235x³ + 17x² + 94x¹

Discard the lead 0 term to get:

159x¹⁰ + 63x⁹ + 191x⁸ + 124x⁷ + 227x⁶ + 23x⁵ + 142x⁴ + 235x³ + 17x² + 94x¹

Step 31a: 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 159x¹⁰. Convert 159x¹⁰ to alpha notation. According to the log antilog table, for the integer value 159, the alpha exponent is 46. Therefore 159 = α⁴⁶. Multiply the generator polynomial by α⁴⁶:

(α⁴⁶ * α⁰)x¹⁰ + (α⁴⁶ * α²⁵¹)x⁹ + (α⁴⁶ * α⁶⁷)x⁸ + (α⁴⁶ * α⁴⁶)x⁷ + (α⁴⁶ * α⁶¹)x⁶ + (α⁴⁶ * α¹¹⁸)x⁵ + (α⁴⁶ * α⁷⁰)x⁴ + (α⁴⁶ * α⁶⁴)x³ + (α⁴⁶ * α⁹⁴)x² + (α⁴⁶ * α³²)x¹ + (α⁴⁶ * α⁴⁵)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:

α⁴⁶x¹⁰ + α^{(297 % 255)}x⁹ + α¹¹³x⁸ + α⁹²x⁷ + α¹⁰⁷x⁶ + α¹⁶⁴x⁵ + α¹¹⁶x⁴ + α¹¹⁰x³ + α¹⁴⁰x² + α⁷⁸x¹ + α⁹¹x⁰

The result is:

α⁴⁶x¹⁰ + α⁴²x⁹ + α¹¹³x⁸ + α⁹²x⁷ + α¹⁰⁷x⁶ + α¹⁶⁴x⁵ + α¹¹⁶x⁴ + α¹¹⁰x³ + α¹⁴⁰x² + α⁷⁸x¹ + α⁹¹x⁰

Now, convert this to integer notation:

159x¹⁰ + 181x⁹ + 31x⁸ + 91x⁷ + 104x⁶ + 198x⁵ + 248x⁴ + 103x³ + 132x² + 120x¹ + 163

Step 31b: XOR the result with the result from step 30b

Use the result from step 30b to perform the next XOR.

(159 ⊕ 159)x¹⁰ + (63 ⊕ 181)x⁹ + (191 ⊕ 31)x⁸ + (124 ⊕ 91)x⁷ + (227 ⊕ 104)x⁶ + (23 ⊕ 198)x⁵ + (142 ⊕ 248)x⁴ + (235 ⊕ 103)x³ + (17 ⊕ 132)x² + (94 ⊕ 120)x¹ + (0 ⊕ 163)x⁰

The result is:

0x¹⁰ + 138x⁹ + 160x⁸ + 39x⁷ + 139x⁶ + 209x⁵ + 118x⁴ + 140x³ + 149x² + 38x¹ + 163

Discard the lead 0 term to get:

138x⁹ + 160x⁸ + 39x⁷ + 139x⁶ + 209x⁵ + 118x⁴ + 140x³ + 149x² + 38x¹ + 163

Use the terms of the remainder as the error correction codewords

The division has been performed 31 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:

138  160  39  139  209  118  140  149  38  163