Error correction codewords allow QR code readers to detect and correct errors in QR codes. This page explains how to create these error correction codewords after encoding the data.

Step 1: Break Data Codewords into Blocks if Necessary

Before generating error correction codewords, it may be necessary to break up the data codewords into smaller blocks if the QR code is larger than version 2. As an example, if creating a 5-Q QR code, the error correction table says that a 5-Q code has 62 data codewords (strings of 8-bit binary numbers). For example:

(codeword #1) 01000011
(codeword #2) 01010101
(codeword #3) 01000110
(codeword #4) 10000110
(codeword #5) 01010111
(codeword #6) 00100110
(codeword #7) 01010101
(codeword #8) 11000010
(codeword #9) 01110111
(codeword #10) 00110010
(codeword #11) 00000110
(codeword #12) 00010010
(codeword #13) 00000110
(codeword #14) 01100111
(codeword #15) 00100110
(codeword #16) 11110110
(codeword #17) 11110110
(codeword #18) 01000010
(codeword #19) 00000111
(codeword #20) 01110110
(codeword #21) 10000110
(codeword #22) 11110010
(codeword #23) 00000111
(codeword #24) 00100110
(codeword #25) 01010110
(codeword #26) 00010110
(codeword #27) 11000110
(codeword #28) 11000111
(codeword #29) 10010010
(codeword #30) 00000110
(codeword #31) 10110110
(codeword #32) 11100110
(codeword #33) 11110111
(codeword #34) 01110111
(codeword #35) 00110010
(codeword #36) 00000111
(codeword #37) 01110110
(codeword #38) 10000110
(codeword #39) 01010111
(codeword #40) 00100110
(codeword #41) 01010010
(codeword #42) 00000110
(codeword #43) 10000110
(codeword #44) 10010111
(codeword #45) 00110010
(codeword #46) 00000111
(codeword #47) 01000110
(codeword #48) 11110111
(codeword #49) 01110110
(codeword #50) 01010110
(codeword #51) 11000010
(codeword #52) 00000110
(codeword #53) 10010111
(codeword #54) 00110010
(codeword #55) 11100000
(codeword #56) 11101100
(codeword #57) 00010001
(codeword #58) 11101100
(codeword #59) 00010001
(codeword #60) 11101100
(codeword #61) 00010001
(codeword #62) 11101100

The error correction table mentions "group 1" and "group 2", as well as "number of blocks." This means that the data codewords must be split up into up to two groups, and within each group, the data codewords may be further broken up into blocks. The data codewords are broken up sequentially (i.e. starting with codeword 1, then codeword 2, and so on.)

For a 5-Q code, it says that there are two groups, the first of which should be broken up into 2 blocks that each contain 15 data codewords, and the second of which should be broken up into 2 blocks that each contain 16 data codewords. Notice that 15+15+16+16 = 62, which is the total number of data codewords. Below, are the codewords from above, broken up into the correct groups and blocks to illustrate how to do this step.

In the table, notice that the blocks in group 1 consist of 15 data codewords, and the blocks in group 2 consist of 16 data codewords as specified in the error correction table. Note also that they are broken up sequentially. That is, they are in the same order as they were before being separated into blocks.

The error correction table also says that for a 5-Q code, there are 18 error correction codewords per block. In this example, there are four blocks, so there will be four sets of 18 error correction codewords, for a total of 72 error correction codewords.

As shown in the error correction table, the smaller QR codes do not require the data codewords to be broken up at all, so, for a 1-M code (for example) all 16 data codewords would be used as a single block, and only 10 error correction codewords would need to be generated.

Before continuing the 5-Q example, it is necessary to explain the underlying steps of Reed-Solomon Error correction.

Step 2: Understand Polynomial Long Division

The error correction codewords will be generated using a method called Reed-Solomon Error correction. Part of the process is to perform polynomial long division. That is, dividing one polynomial by another polynomial. This tutorial assumes an understanding of how to do standard long division (i.e. with numbers rather than polyomials) by hand. For more information, read wikiHow's article about how to do long division.

Polynomial long division is slightly more complicated than standard long division. To help clarify the process, here is a simple example of polynomial long division.

This example show 3x² + x - 1 (the dividend) divided by x + 1 (the divisor). These polynomials are first put into a tableau (see Wikipedia's article on long division for an explanation of this notation).

In each step of the long division, multiply x + 1 by something to make its first term equal that of the polynomial at the bottom of the tableau.

For the first step, 3x² + x - 1 is at the bottom of the tableau, so multiply x + 1 by 3x. This results in 3x² + 3x, which has the same first term as that of the polynomial at the bottom of the tableau. Here is the updated tableau.

Now subtract 3x² + 3x from 3x² + x - 1. This results in -2x - 1 (because x - 3x is (1 - 3)x, or -2x). The - 1 at the end is carried over from the original polynomial. Here is the updated tableau.

Now, once again, multiply x + 1 by a term that will result in a polynomial whose first term is the same as that of the bottom polynomial. In this case, do -2 (x + 1) to get -2x - 2. Here is the updated tableau.

And now, once again, subtract the two polynomials. The -2x cancels out, but this leaves -1 - 2, which equals 1, as shown in the final tableau below.

The remainder is 1. Technically it is 1x⁰. It would be necessary to multiply (x + 1) by 1/x to make the first term the same, but since this is a fraction, it means that no further division is possible. The 1 at the bottom of the tableau is the remainder.

Overall, the steps of polynomial long division are:

1. Find the appropriate term to multiply the divisor by. The result of the multiplication should have the same first term as the the dividend (in the first multiplication step) or remainder (in all subsequent multiplication steps).
2. Subtract the result from the dividend (in the first multiplication step) or remainder (in all subsequent multiplication steps).
3. Repeat steps 1 and 2 until it is no longer possible to multiply by an integer, or in other words, it would be necessary to multiply by a fraction. The number at the bottom of the tableau is the remainder.

The polynomial long division necessary for Reed-Solomon Error Correction is simpler in some ways than this example, because it will not be necessary to deal with the exponents of the polynomial terms.

Step 3: Understand The Galois Field

As mentioned in the previous section, to generate error correction codewords, this process uses a method called Reed-Solomon Error correction. Along with polynomial long division, this method uses a Galois field, which is essentially a restricted set of numbers, as well as some mathematical operations that create numbers that are still in that set.

The QR Code standard says to use bit-wise modulo 2 arithmetic and byte-wise modulo 100011101 arithmetic. This means using Galois Field 2⁸, or in other words Galois Field 256, sometimes written as GF(256).

The numbers in GF(256) will all be in the range of 0 through 255 (inclusive). Notice that this is the same range of numbers that can be represented with an eight-bit byte (the largest possible eight-bit byte is 11111111, which equals 255).

This means that all of the mathematical operations in GF(256) will result in numbers that can be represented as eight-bit bytes.

Step 4: Understand Galois Field Arithmetic

As mentioned previously, GF(256) contains the numbers 0 through 255 inclusive. Mathematical operations in GF(256) are cyclical in nature, meaning that if a mathematical operation is performed within GF(256) that results in a number larger than 255, it will be necessary to use the modulo operation to get a number that is still in the Galois Field.

In the Galois Field, negative numbers have the same value as positive numbers, so -n = n. In other words, always use the absolute value of the numbers involved in Galois Field arithmetic.

This means that addition and subtraction within the Galois Field are the same thing. Addition and subtraction in the Galois Field are performed by doing the addition or subtraction as normal, but then performing the modulo operation. And since we are using bit-wise modulo 2 arithmetic (as mentioned in the QR code specification), this is the same as performing the XOR operation. For example:
1 + 1 = 2 % 2 = 0
or
1 ^ 1 = 0

and

0 + 1 = 1 % 2 = 1
or
0 ^ 1 = 1

For the purposes of encoding a QR code, all addition and subtraction in GF(256) is performed by XORing the two numbers together.

Step 5: Generate Powers of 2 using Byte-Wise Modulo 100011101

All of the numbers in GF(256) can be represented as a power of 2. Specifically, all numbers in GF(256) can be represented as 2ⁿ, where 0 8 would seem to be too big for the Galois Field since it is equal to 256.

The powers of 2 from 0 to 8 are:
2⁰ = 1
2¹ = 2
2² = 4
2³ = 8
2⁴ = 16
2⁵ = 32
2⁶ = 64
2⁷ = 128
2⁸ = 256

The QR code specification says to use byte-wise modulo 100011101 arithmetic (where 100011101 is a binary number that is equivalent to 285 in decimal). This means that when a number is 256 or larger, it should be XORed with 285. This leads to unexpected values for 2⁸ and larger.

Note that when continuing on to 2⁹, do not take its usual value of 512 and XOR with 285 (which would result in too large a number anyway). Instead, since 2⁹ = 2⁸ * 2, use the value of 2⁸ that was calculated in the previous step.

In other words:
2⁹ = 2⁸ * 2 = 29 * 2 = 58

Continue using the previous power of 2 to create the next powers of 2:
2¹⁰ = 2⁹ * 2 = 58 * 2 = 116
2¹¹ = 2¹⁰ * 2 = 116 * 2 = 232

Whenever a value greater than or equal to 256 is obtained, once again XOR with 285:
2¹² = 2¹¹ * 2 = 232 * 2 = 464 ^ 285 = 205

In general, the values are always equal to 2 times the previous power, and if that value is 256 or greater, it is XORed with 285.

Using this procedure, all numbers in GF(256) can be represented with 2ⁿ, where n is a number in the range 0

Step 6: Understand Multiplication with Logs and Antilogs

Because all of the values can be represented with 2ⁿ as explained above, it is possible to logs and antilogs to simplify multiplication in GF(256). (Incidentally, this is what slide rules do, although they are not restricted to GF(256)).

The simplification is possible because, in general (i.e. not just in Galois Fields), one can multiply two numbers p and q with the following operation:
b^{(log_bp + log_bq)}

In this case, base 2 is in use, so the above operation becomes
2^{(log₂p + log₂q)}
And the log₂x operation tells us how many times to multiply 2 to get an answer of x. In other words, it determines n where 2ⁿ = x.

This provides a shortcut for multiplying numbers in GF(256). When multiplying two numbers that have the same base, such as 2² * 2⁸, this is equivalent to adding the two exponents together, like so:

2² * 2⁸ = 2^{(2 + 8)} = 2¹⁰

This means that since all GF(256) values have been expressed as powers of two (explained in the previous section), all multiplication in GF(256) can be performed as addition of base-2 exponents.

For example, to multiply 16 * 32, this is the same as multiplying 2⁴ by 2⁵. As the previous paragraph explained, this is the same as 2^{(4 + 5)} or 2⁹. The value of 2⁹ was already calculated, therefore in GF(256) with byte-wise modulo 285, 16 * 32 is 2⁹, which was determined in the previous section to be equal to 58.

When adding exponents, if the exponent becomes greater than or equal to 256, simply apply modulo 255. In other words:

2¹⁷⁰ * 2¹⁶⁴ = 2^(170+164) = 2³³⁴ → 2^{334 % 255} = 2⁷⁹.

Therefore, all that is needed to perform multiplication in GF(256) is to generate all the powers of 2. These values have been calculated already and can be found in the log antilog table. The table uses alpha notation, where ɑ = 2. The QR Code specification also uses alpha notation in Annex A.

Step 7: Understanding The Generator Polynomial

We have made a lot of progress toward generating error correction codewords, but we are not quite there yet. The next step is understanding generator polynomials.

As mentioned earlier, error correction coding uses polynomial long division. To do that, two polynomials are needed. The first polynomial to use is called the message polynomial. The message polynomial uses the data codewords from the data encoding step as its coefficients. For example, if the data codewords, converted to integers, were 25, 218, and 35, the message polynomial would be 25x² + 218x + 35. In practice, real message polynomials for standard QR codes are much longer, but this is just an example.

The message polynomial will be divided by a generator polynomial. The generator polynomial is a polynomial that is created by multiplying
(x - ɑ⁰) ... (x - ɑ^n-1)
where n is the number of error correction codewords that must be generated (see the error correction table). As mentioned in the previous section, ɑ (alpha) is equal to 2.

It is possible to programmatically create all the generator polynomials that might be needed. During the process, this tutorial will make use of alpha notation and the log antilog multiplication that was described earlier on this page.

The QR code specification lists generator polynomials in Annex A, starting with 2 and ending with 68. Although standard QR codes will always require more than 2 error correction codewords per block, this page will show how to calculate the generator polynomial for 2 error correction codewords, because it illustrates the process of calculating all the rest of the generator polynomials as well.

Generator Polynomial for 2 Error Correction Codewords

First, multiply (x - ɑ⁰) and (x - ɑ¹).
Since the coefficient of x is 1, and since ɑ⁰ = 1, this can be written as
(ɑ⁰x - ɑ⁰) * (ɑ⁰x - ɑ¹)

Multiply each term of the first part by each term of the second part to get this:
(ɑ⁰x¹ * ɑ⁰x¹) + (ɑ⁰x⁰ * ɑ⁰x¹) + (ɑ⁰x¹ * ɑ¹x⁰) + (ɑ⁰x⁰ * ɑ¹x⁰)

Notice that exponent addition can be used here to perform the multiplications:
(ɑ⁽⁰⁺⁰⁾x⁽¹⁺¹⁾) + (ɑ⁽⁰⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺¹⁾x⁽¹⁺⁰⁾) + (ɑ⁽⁰⁺¹⁾x⁽⁰⁺⁰⁾)

The result:
ɑ⁰x² + ɑ⁰x¹ + ɑ¹x¹ + ɑ¹x⁰

Now, combine like terms. There are two x¹ terms, so add them together.
ɑ⁰x² + (ɑ⁰+ɑ¹)x¹ + ɑ¹x⁰

Remember that addition in GF(256) is performed by XOR. Convert the alphas to their integer counterparts using the log antilog table, then perform the XOR.

x² + (1^2)x¹ + 2x⁰
x² + 3x¹ + 2x⁰

Back in alpha notation, this is
ɑ⁰x² + ɑ²⁵x¹ + ɑ¹x⁰
This is the generator polynomial for 2 error correction codewords.

Generator Polynomial for 3 Error Correction Codewords

All the rest of the generator polynomials can be created the same way, using the generator polynomial from each previous step. Start with the previous generator polynomial and multiply it by the next factor, in this case (x - ɑ²)

(ɑ⁰x² + ɑ²⁵x¹ + ɑ¹x⁰) * (ɑ⁰x¹ + ɑ²x⁰)

(ɑ⁰x² * ɑ⁰x¹) + (ɑ²⁵x¹ * ɑ⁰x¹) + (ɑ¹x⁰ * ɑ⁰x¹) + (ɑ⁰x² * ɑ²x⁰) + (ɑ²⁵x¹ * ɑ²x⁰) + (ɑ¹x⁰ * ɑ²x⁰)

To multiply, add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽²⁺¹⁾) + (ɑ⁽²⁵⁺⁰⁾x⁽¹⁺¹⁾) + (ɑ⁽¹⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺²⁾x⁽²⁺⁰⁾) + (ɑ⁽²⁵⁺²⁾x⁽¹⁺⁰⁾) + (ɑ⁽¹⁺²⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x³ + ɑ²⁵x² + ɑ¹x¹ + ɑ²x² + ɑ²⁷x¹ + ɑ³x⁰

Now combine like terms.
ɑ⁰x³ + (ɑ²⁵ ^ ɑ²)x² + (ɑ¹ ^ ɑ²⁷)x¹ + ɑ³x⁰

ɑ⁰x³ + (3 ^ 4)x² + (2 ^ 12)x¹ + ɑ³x⁰
ɑ⁰x³ + 7x² + 14x¹ + ɑ³x⁰
ɑ⁰x³ + ɑ¹⁹⁸x² + ɑ¹⁹⁹x¹ + ɑ³x⁰

This is the generator polynomial for 3 error correction codewords.

When the Exponents are Greater than or Equal to 256

When performing multiplication by adding exponents together, sometimes this may result in an exponent that is greater than or equal to 256. In this case, apply modulo 255 BEFORE combining like terms. For example
ɑ²⁵⁷x⁴ = ɑ^{(257 % 255)}x⁴ = ɑ²x⁴

Other Generator Polynomials

In general, it is possible to create all the generator polynomials g(x) as illustrated below:
(g(x) for 2 error correction codewords) * (x - ɑ²) = (g(x) for 3 error correction codewords)
(g(x) for 3 error correction codewords) * (x - ɑ³) = (g(x) for 4 error correction codewords)
(g(x) for 4 error correction codewords) * (x - ɑ⁴) = (g(x) for 5 error correction codewords)
...
(g(x) for n-1 error correction codewords) * (x - ɑ^n-1) = (g(x) for n error correction codewords)

To obtain a specific generator polynomial without having to calculate it programmatically, the generator polynomial tool can create it.

Step 8: Generating Error Correction Codewords

Now it is finally time to start generating error correction codewords. This part uses the data codewords from the HELLO WORLD example on the alphanumeric mode encoding page. These codewords will be used as the coefficients of the message polynomial. This example uses error correction level M, and version 1, to create a 1-M code.

The Message Polynomial

The data encoding step resulted in the following data codewords for HELLO WORLD as a 1-M code.
00100000 01011011 00001011 01111000 11010001 01110010 11011100 01001101 01000011 01000000 11101100 00010001 11101100 00010001 11101100 00010001

Convert those binary numbers into decimal:
32, 91, 11, 120, 209, 114, 220, 77, 67, 64, 236, 17, 236, 17, 236, 17

These numbers are the coefficients of the message polynomial. In other words:
32x¹⁵ + 91x¹⁴ + 11x¹³ ... and so on

The Generator Polynomial

First, get the generator polynomial. Since this is a 1-M code, the error correction table says to create 10 error correction codewords. Therefore, use the following generator polynomial:
x¹⁰ + ɑ²⁵¹x⁹ + ɑ⁶⁷x⁸ + ɑ⁴⁶x⁷ + ɑ⁶¹x⁶ + ɑ¹¹⁸x⁵ + ɑ⁷⁰x⁴ + ɑ⁶⁴x³ + ɑ⁹⁴x² + ɑ³²x + ɑ⁴⁵

(Refer to the generator polynomial tool and the discussion about generator polynomials earlier on this page.)

The Division Steps

Now it is time to divide the message polynomial by the generator polynomial. This is done in much the same way as the polynomial long division shown earlier on the page. Here are the division steps, updated to take Galois Field arithmetic into account:

1. Find the appropriate term to multiply the generator polynomial by. The result of the multiplication should have the same first term as the the message polynomial (in the first multiplication step) or remainder (in all subsequent multiplication steps).
2. XOR the result with the message polynomial (in the first multiplication step) or remainder (in all subsequent multiplication steps).
3. Perform these steps n times, where n is the number of data codewords.

Notice the differences between these steps and those of normal polynomial long division. Instead of subtracting after the multiplication step, we perform an XOR (which, in GF(256), is the same thing).

More importantly, after dividing the two polynomials, there will be a remainder. The coefficients of this remainder are the error correction codewords.

The division will be illustrated step-by-step in the next section.

Step 9: Divide the Message Polynomial by the Generator Polynomial

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

32x¹⁵ + 91x¹⁴ + 11x¹³ + 120x¹² + 209x¹¹ + 114x¹⁰ + 220x⁹ + 77x⁸ + 67x⁷ + 64x⁶ + 236x⁵ + 17x⁴ + 236x³ + 17x² + 236x¹ + 17

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:

32x²⁵ + 91x²⁴ + 11x²³ + 120x²² + 209x²¹ + 114x²⁰ + 220x¹⁹ + 77x¹⁸ + 67x¹⁷ + 64x¹⁶ + 236x¹⁵ + 17x¹⁴ + 236x¹³ + 17x¹² + 236x¹¹ + 17x¹⁰

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

ɑ⁰x²⁵ + ɑ²⁵¹x²⁴ + ɑ⁶⁷x²³ + ɑ⁴⁶x²² + ɑ⁶¹x²¹ + ɑ¹¹⁸x²⁰ + ɑ⁷⁰x¹⁹ + ɑ⁶⁴x¹⁸ + ɑ⁹⁴x¹⁷ + ɑ³²x¹⁶ + ɑ⁴⁵x¹⁵

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 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 32x²⁵. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 32x²⁵ to alpha notation. According to the log antilog table, for the integer value 32, the alpha exponent is 5. Therefore 32 = ɑ⁵. Multiply the generator polynomial by ɑ⁵:

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:

The result is:

ɑ⁵x²⁵ + ɑ¹x²⁴ + ɑ⁷²x²³ + ɑ⁵¹x²² + ɑ⁶⁶x²¹ + ɑ¹²³x²⁰ + ɑ⁷⁵x¹⁹ + ɑ⁶⁹x¹⁸ + ɑ⁹⁹x¹⁷ + ɑ³⁷x¹⁶ + ɑ⁵⁰x¹⁵

Now, convert this to integer notation:

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.

(32 ⊕ 32)x²⁵ + (91 ⊕ 2)x²⁴ + (11 ⊕ 101)x²³ + (120 ⊕ 10)x²² + (209 ⊕ 97)x²¹ + (114 ⊕ 197)x²⁰ + (220 ⊕ 15)x¹⁹ + (77 ⊕ 47)x¹⁸ + (67 ⊕ 134)x¹⁷ + (64 ⊕ 74)x¹⁶ + (236 ⊕ 5)x¹⁵ + (17 ⊕ 0)x¹⁴ + (236 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (236 ⊕ 0)x¹¹ + (17 ⊕ 0)x¹⁰

The result is:

0x²⁵ + 89x²⁴ + 110x²³ + 114x²² + 176x²¹ + 183x²⁰ + 211x¹⁹ + 98x¹⁸ + 197x¹⁷ + 10x¹⁶ + 233x¹⁵ + 17x¹⁴ + 236x¹³ + 17x¹² + 236x¹¹ + 17x¹⁰

Discard the lead 0 term to get:

89x²⁴ + 110x²³ + 114x²² + 176x²¹ + 183x²⁰ + 211x¹⁹ + 98x¹⁸ + 197x¹⁷ + 10x¹⁶ + 233x¹⁵ + 17x¹⁴ + 236x¹³ + 17x¹² + 236x¹¹ + 17x¹⁰

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 89x²⁴. Convert 89x²⁴ to alpha notation. According to the log antilog table, for the integer value 89, the alpha exponent is 210. Therefore 89 = ɑ²¹⁰. Multiply the generator polynomial by ɑ²¹⁰:

(ɑ²¹⁰ * ɑ⁰)x²⁴ + (ɑ²¹⁰ * ɑ²⁵¹)x²³ + (ɑ²¹⁰ * ɑ⁶⁷)x²² + (ɑ²¹⁰ * ɑ⁴⁶)x²¹ + (ɑ²¹⁰ * ɑ⁶¹)x²⁰ + (ɑ²¹⁰ * ɑ¹¹⁸)x¹⁹ + (ɑ²¹⁰ * ɑ⁷⁰)x¹⁸ + (ɑ²¹⁰ * ɑ⁶⁴)x¹⁷ + (ɑ²¹⁰ * ɑ⁹⁴)x¹⁶ + (ɑ²¹⁰ * ɑ³²)x¹⁵ + (ɑ²¹⁰ * ɑ⁴⁵)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:

The result is:

ɑ²¹⁰x²⁴ + ɑ²⁰⁶x²³ + ɑ²²x²² + ɑ¹x²¹ + ɑ¹⁶x²⁰ + ɑ⁷³x¹⁹ + ɑ²⁵x¹⁸ + ɑ¹⁹x¹⁷ + ɑ⁴⁹x¹⁶ + ɑ²⁴²x¹⁵ + ɑ⁰x¹⁴

Now, convert this to integer notation:

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

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

(89 ⊕ 89)x²⁴ + (110 ⊕ 83)x²³ + (114 ⊕ 234)x²² + (176 ⊕ 2)x²¹ + (183 ⊕ 76)x²⁰ + (211 ⊕ 202)x¹⁹ + (98 ⊕ 3)x¹⁸ + (197 ⊕ 90)x¹⁷ + (10 ⊕ 140)x¹⁶ + (233 ⊕ 176)x¹⁵ + (17 ⊕ 1)x¹⁴ + (236 ⊕ 0)x¹³ + (17 ⊕ 0)x¹² + (236 ⊕ 0)x¹¹ + (17 ⊕ 0)x¹⁰

The result is:

0x²⁴ + 61x²³ + 152x²² + 178x²¹ + 251x²⁰ + 25x¹⁹ + 97x¹⁸ + 159x¹⁷ + 134x¹⁶ + 89x¹⁵ + 16x¹⁴ + 236x¹³ + 17x¹² + 236x¹¹ + 17x¹⁰

Discard the lead 0 term to get:

61x²³ + 152x²² + 178x²¹ + 251x²⁰ + 25x¹⁹ + 97x¹⁸ + 159x¹⁷ + 134x¹⁶ + 89x¹⁵ + 16x¹⁴ + 236x¹³ + 17x¹² + 236x¹¹ + 17x¹⁰

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 61x²³. Convert 61x²³ to alpha notation. According to the log antilog table, for the integer value 61, the alpha exponent is 228. Therefore 61 = ɑ²²⁸. Multiply the generator polynomial by ɑ²²⁸:

(ɑ²²⁸ * ɑ⁰)x²³ + (ɑ²²⁸ * ɑ²⁵¹)x²² + (ɑ²²⁸ * ɑ⁶⁷)x²¹ + (ɑ²²⁸ * ɑ⁴⁶)x²⁰ + (ɑ²²⁸ * ɑ⁶¹)x¹⁹ + (ɑ²²⁸ * ɑ¹¹⁸)x¹⁸ + (ɑ²²⁸ * ɑ⁷⁰)x¹⁷ + (ɑ²²⁸ * ɑ⁶⁴)x¹⁶ + (ɑ²²⁸ * ɑ⁹⁴)x¹⁵ + (ɑ²²⁸ * ɑ³²)x¹⁴ + (ɑ²²⁸ * ɑ⁴⁵)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:

The result is:

ɑ²²⁸x²³ + ɑ²²⁴x²² + ɑ⁴⁰x²¹ + ɑ¹⁹x²⁰ + ɑ³⁴x¹⁹ + ɑ⁹¹x¹⁸ + ɑ⁴³x¹⁷ + ɑ³⁷x¹⁶ + ɑ⁶⁷x¹⁵ + ɑ⁵x¹⁴ + ɑ¹⁸x¹³

Now, convert this to integer notation:

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

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

(61 ⊕ 61)x²³ + (152 ⊕ 18)x²² + (178 ⊕ 106)x²¹ + (251 ⊕ 90)x²⁰ + (25 ⊕ 78)x¹⁹ + (97 ⊕ 163)x¹⁸ + (159 ⊕ 119)x¹⁷ + (134 ⊕ 74)x¹⁶ + (89 ⊕ 194)x¹⁵ + (16 ⊕ 32)x¹⁴ + (236 ⊕ 45)x¹³ + (17 ⊕ 0)x¹² + (236 ⊕ 0)x¹¹ + (17 ⊕ 0)x¹⁰

The result is:

0x²³ + 138x²² + 216x²¹ + 161x²⁰ + 87x¹⁹ + 194x¹⁸ + 232x¹⁷ + 204x¹⁶ + 155x¹⁵ + 48x¹⁴ + 193x¹³ + 17x¹² + 236x¹¹ + 17x¹⁰

Discard the lead 0 term to get:

138x²² + 216x²¹ + 161x²⁰ + 87x¹⁹ + 194x¹⁸ + 232x¹⁷ + 204x¹⁶ + 155x¹⁵ + 48x¹⁴ + 193x¹³ + 17x¹² + 236x¹¹ + 17x¹⁰

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 138x²². Convert 138x²² to alpha notation. According to the log antilog table, for the integer value 138, the alpha exponent is 222. Therefore 138 = ɑ²²². Multiply the generator polynomial by ɑ²²²:

(ɑ²²² * ɑ⁰)x²² + (ɑ²²² * ɑ²⁵¹)x²¹ + (ɑ²²² * ɑ⁶⁷)x²⁰ + (ɑ²²² * ɑ⁴⁶)x¹⁹ + (ɑ²²² * ɑ⁶¹)x¹⁸ + (ɑ²²² * ɑ¹¹⁸)x¹⁷ + (ɑ²²² * ɑ⁷⁰)x¹⁶ + (ɑ²²² * ɑ⁶⁴)x¹⁵ + (ɑ²²² * ɑ⁹⁴)x¹⁴ + (ɑ²²² * ɑ³²)x¹³ + (ɑ²²² * ɑ⁴⁵)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:

The result is:

ɑ²²²x²² + ɑ²¹⁸x²¹ + ɑ³⁴x²⁰ + ɑ¹³x¹⁹ + ɑ²⁸x¹⁸ + ɑ⁸⁵x¹⁷ + ɑ³⁷x¹⁶ + ɑ³¹x¹⁵ + ɑ⁶¹x¹⁴ + ɑ²⁵⁴x¹³ + ɑ¹²x¹²

Now, convert this to integer notation:

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

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

(138 ⊕ 138)x²² + (216 ⊕ 43)x²¹ + (161 ⊕ 78)x²⁰ + (87 ⊕ 135)x¹⁹ + (194 ⊕ 24)x¹⁸ + (232 ⊕ 214)x¹⁷ + (204 ⊕ 74)x¹⁶ + (155 ⊕ 192)x¹⁵ + (48 ⊕ 111)x¹⁴ + (193 ⊕ 142)x¹³ + (17 ⊕ 205)x¹² + (236 ⊕ 0)x¹¹ + (17 ⊕ 0)x¹⁰

The result is:

0x²² + 243x²¹ + 239x²⁰ + 208x¹⁹ + 218x¹⁸ + 62x¹⁷ + 134x¹⁶ + 91x¹⁵ + 95x¹⁴ + 79x¹³ + 220x¹² + 236x¹¹ + 17x¹⁰

Discard the lead 0 term to get:

243x²¹ + 239x²⁰ + 208x¹⁹ + 218x¹⁸ + 62x¹⁷ + 134x¹⁶ + 91x¹⁵ + 95x¹⁴ + 79x¹³ + 220x¹² + 236x¹¹ + 17x¹⁰

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 243x²¹. Convert 243x²¹ to alpha notation. According to the log antilog table, for the integer value 243, the alpha exponent is 233. Therefore 243 = ɑ²³³. Multiply the generator polynomial by ɑ²³³:

(ɑ²³³ * ɑ⁰)x²¹ + (ɑ²³³ * ɑ²⁵¹)x²⁰ + (ɑ²³³ * ɑ⁶⁷)x¹⁹ + (ɑ²³³ * ɑ⁴⁶)x¹⁸ + (ɑ²³³ * ɑ⁶¹)x¹⁷ + (ɑ²³³ * ɑ¹¹⁸)x¹⁶ + (ɑ²³³ * ɑ⁷⁰)x¹⁵ + (ɑ²³³ * ɑ⁶⁴)x¹⁴ + (ɑ²³³ * ɑ⁹⁴)x¹³ + (ɑ²³³ * ɑ³²)x¹² + (ɑ²³³ * ɑ⁴⁵)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:

The result is:

ɑ²³³x²¹ + ɑ²²⁹x²⁰ + ɑ⁴⁵x¹⁹ + ɑ²⁴x¹⁸ + ɑ³⁹x¹⁷ + ɑ⁹⁶x¹⁶ + ɑ⁴⁸x¹⁵ + ɑ⁴²x¹⁴ + ɑ⁷²x¹³ + ɑ¹⁰x¹² + ɑ²³x¹¹

Now, convert this to integer notation:

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

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

(243 ⊕ 243)x²¹ + (239 ⊕ 122)x²⁰ + (208 ⊕ 193)x¹⁹ + (218 ⊕ 143)x¹⁸ + (62 ⊕ 53)x¹⁷ + (134 ⊕ 217)x¹⁶ + (91 ⊕ 70)x¹⁵ + (95 ⊕ 181)x¹⁴ + (79 ⊕ 101)x¹³ + (220 ⊕ 116)x¹² + (236 ⊕ 201)x¹¹ + (17 ⊕ 0)x¹⁰

The result is:

0x²¹ + 149x²⁰ + 17x¹⁹ + 85x¹⁸ + 11x¹⁷ + 95x¹⁶ + 29x¹⁵ + 234x¹⁴ + 42x¹³ + 168x¹² + 37x¹¹ + 17x¹⁰

Discard the lead 0 term to get:

149x²⁰ + 17x¹⁹ + 85x¹⁸ + 11x¹⁷ + 95x¹⁶ + 29x¹⁵ + 234x¹⁴ + 42x¹³ + 168x¹² + 37x¹¹ + 17x¹⁰

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 149x²⁰. Convert 149x²⁰ to alpha notation. According to the log antilog table, for the integer value 149, the alpha exponent is 184. Therefore 149 = ɑ¹⁸⁴. Multiply the generator polynomial by ɑ¹⁸⁴:

(ɑ¹⁸⁴ * ɑ⁰)x²⁰ + (ɑ¹⁸⁴ * ɑ²⁵¹)x¹⁹ + (ɑ¹⁸⁴ * ɑ⁶⁷)x¹⁸ + (ɑ¹⁸⁴ * ɑ⁴⁶)x¹⁷ + (ɑ¹⁸⁴ * ɑ⁶¹)x¹⁶ + (ɑ¹⁸⁴ * ɑ¹¹⁸)x¹⁵ + (ɑ¹⁸⁴ * ɑ⁷⁰)x¹⁴ + (ɑ¹⁸⁴ * ɑ⁶⁴)x¹³ + (ɑ¹⁸⁴ * ɑ⁹⁴)x¹² + (ɑ¹⁸⁴ * ɑ³²)x¹¹ + (ɑ¹⁸⁴ * ɑ⁴⁵)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:

The result is:

ɑ¹⁸⁴x²⁰ + ɑ¹⁸⁰x¹⁹ + ɑ²⁵¹x¹⁸ + ɑ²³⁰x¹⁷ + ɑ²⁴⁵x¹⁶ + ɑ⁴⁷x¹⁵ + ɑ²⁵⁴x¹⁴ + ɑ²⁴⁸x¹³ + ɑ²³x¹² + ɑ²¹⁶x¹¹ + ɑ²²⁹x¹⁰

Now, convert this to integer notation:

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

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

(149 ⊕ 149)x²⁰ + (17 ⊕ 150)x¹⁹ + (85 ⊕ 216)x¹⁸ + (11 ⊕ 244)x¹⁷ + (95 ⊕ 233)x¹⁶ + (29 ⊕ 35)x¹⁵ + (234 ⊕ 142)x¹⁴ + (42 ⊕ 27)x¹³ + (168 ⊕ 201)x¹² + (37 ⊕ 195)x¹¹ + (17 ⊕ 122)x¹⁰

The result is:

0x²⁰ + 135x¹⁹ + 141x¹⁸ + 255x¹⁷ + 182x¹⁶ + 62x¹⁵ + 100x¹⁴ + 49x¹³ + 97x¹² + 230x¹¹ + 107x¹⁰

Discard the lead 0 term to get:

135x¹⁹ + 141x¹⁸ + 255x¹⁷ + 182x¹⁶ + 62x¹⁵ + 100x¹⁴ + 49x¹³ + 97x¹² + 230x¹¹ + 107x¹⁰

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 135x¹⁹. Convert 135x¹⁹ to alpha notation. According to the log antilog table, for the integer value 135, the alpha exponent is 13. Therefore 135 = ɑ¹³. Multiply the generator polynomial by ɑ¹³:

(ɑ¹³ * ɑ⁰)x¹⁹ + (ɑ¹³ * ɑ²⁵¹)x¹⁸ + (ɑ¹³ * ɑ⁶⁷)x¹⁷ + (ɑ¹³ * ɑ⁴⁶)x¹⁶ + (ɑ¹³ * ɑ⁶¹)x¹⁵ + (ɑ¹³ * ɑ¹¹⁸)x¹⁴ + (ɑ¹³ * ɑ⁷⁰)x¹³ + (ɑ¹³ * ɑ⁶⁴)x¹² + (ɑ¹³ * ɑ⁹⁴)x¹¹ + (ɑ¹³ * ɑ³²)x¹⁰ + (ɑ¹³ * ɑ⁴⁵)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:

The result is:

ɑ¹³x¹⁹ + ɑ⁹x¹⁸ + ɑ⁸⁰x¹⁷ + ɑ⁵⁹x¹⁶ + ɑ⁷⁴x¹⁵ + ɑ¹³¹x¹⁴ + ɑ⁸³x¹³ + ɑ⁷⁷x¹² + ɑ¹⁰⁷x¹¹ + ɑ⁴⁵x¹⁰ + ɑ⁵⁸x⁹

Now, convert this to integer notation:

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

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

(135 ⊕ 135)x¹⁹ + (141 ⊕ 58)x¹⁸ + (255 ⊕ 253)x¹⁷ + (182 ⊕ 210)x¹⁶ + (62 ⊕ 137)x¹⁵ + (100 ⊕ 92)x¹⁴ + (49 ⊕ 187)x¹³ + (97 ⊕ 60)x¹² + (230 ⊕ 104)x¹¹ + (107 ⊕ 193)x¹⁰ + (0 ⊕ 105)x⁹

The result is:

0x¹⁹ + 183x¹⁸ + 2x¹⁷ + 100x¹⁶ + 183x¹⁵ + 56x¹⁴ + 138x¹³ + 93x¹² + 142x¹¹ + 170x¹⁰ + 105x⁹

Discard the lead 0 term to get:

183x¹⁸ + 2x¹⁷ + 100x¹⁶ + 183x¹⁵ + 56x¹⁴ + 138x¹³ + 93x¹² + 142x¹¹ + 170x¹⁰ + 105x⁹

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 183x¹⁸. Convert 183x¹⁸ to alpha notation. According to the log antilog table, for the integer value 183, the alpha exponent is 158. Therefore 183 = ɑ¹⁵⁸. Multiply the generator polynomial by ɑ¹⁵⁸:

(ɑ¹⁵⁸ * ɑ⁰)x¹⁸ + (ɑ¹⁵⁸ * ɑ²⁵¹)x¹⁷ + (ɑ¹⁵⁸ * ɑ⁶⁷)x¹⁶ + (ɑ¹⁵⁸ * ɑ⁴⁶)x¹⁵ + (ɑ¹⁵⁸ * ɑ⁶¹)x¹⁴ + (ɑ¹⁵⁸ * ɑ¹¹⁸)x¹³ + (ɑ¹⁵⁸ * ɑ⁷⁰)x¹² + (ɑ¹⁵⁸ * ɑ⁶⁴)x¹¹ + (ɑ¹⁵⁸ * ɑ⁹⁴)x¹⁰ + (ɑ¹⁵⁸ * ɑ³²)x⁹ + (ɑ¹⁵⁸ * ɑ⁴⁵)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:

The result is:

ɑ¹⁵⁸x¹⁸ + ɑ¹⁵⁴x¹⁷ + ɑ²²⁵x¹⁶ + ɑ²⁰⁴x¹⁵ + ɑ²¹⁹x¹⁴ + ɑ²¹x¹³ + ɑ²²⁸x¹² + ɑ²²²x¹¹ + ɑ²⁵²x¹⁰ + ɑ¹⁹⁰x⁹ + ɑ²⁰³x⁸

Now, convert this to integer notation:

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

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

(183 ⊕ 183)x¹⁸ + (2 ⊕ 57)x¹⁷ + (100 ⊕ 36)x¹⁶ + (183 ⊕ 221)x¹⁵ + (56 ⊕ 86)x¹⁴ + (138 ⊕ 117)x¹³ + (93 ⊕ 61)x¹² + (142 ⊕ 138)x¹¹ + (170 ⊕ 173)x¹⁰ + (105 ⊕ 174)x⁹ + (0 ⊕ 224)x⁸

The result is:

0x¹⁸ + 59x¹⁷ + 64x¹⁶ + 106x¹⁵ + 110x¹⁴ + 255x¹³ + 96x¹² + 4x¹¹ + 7x¹⁰ + 199x⁹ + 224x⁸

Discard the lead 0 term to get:

59x¹⁷ + 64x¹⁶ + 106x¹⁵ + 110x¹⁴ + 255x¹³ + 96x¹² + 4x¹¹ + 7x¹⁰ + 199x⁹ + 224x⁸

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 59x¹⁷. Convert 59x¹⁷ to alpha notation. According to the log antilog table, for the integer value 59, the alpha exponent is 120. Therefore 59 = ɑ¹²⁰. Multiply the generator polynomial by ɑ¹²⁰:

(ɑ¹²⁰ * ɑ⁰)x¹⁷ + (ɑ¹²⁰ * ɑ²⁵¹)x¹⁶ + (ɑ¹²⁰ * ɑ⁶⁷)x¹⁵ + (ɑ¹²⁰ * ɑ⁴⁶)x¹⁴ + (ɑ¹²⁰ * ɑ⁶¹)x¹³ + (ɑ¹²⁰ * ɑ¹¹⁸)x¹² + (ɑ¹²⁰ * ɑ⁷⁰)x¹¹ + (ɑ¹²⁰ * ɑ⁶⁴)x¹⁰ + (ɑ¹²⁰ * ɑ⁹⁴)x⁹ + (ɑ¹²⁰ * ɑ³²)x⁸ + (ɑ¹²⁰ * ɑ⁴⁵)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:

The result is:

ɑ¹²⁰x¹⁷ + ɑ¹¹⁶x¹⁶ + ɑ¹⁸⁷x¹⁵ + ɑ¹⁶⁶x¹⁴ + ɑ¹⁸¹x¹³ + ɑ²³⁸x¹² + ɑ¹⁹⁰x¹¹ + ɑ¹⁸⁴x¹⁰ + ɑ²¹⁴x⁹ + ɑ¹⁵²x⁸ + ɑ¹⁶⁵x⁷

Now, convert this to integer notation:

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

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

(59 ⊕ 59)x¹⁷ + (64 ⊕ 248)x¹⁶ + (106 ⊕ 220)x¹⁵ + (110 ⊕ 63)x¹⁴ + (255 ⊕ 49)x¹³ + (96 ⊕ 11)x¹² + (4 ⊕ 174)x¹¹ + (7 ⊕ 149)x¹⁰ + (199 ⊕ 249)x⁹ + (224 ⊕ 73)x⁸ + (0 ⊕ 145)x⁷

The result is:

0x¹⁷ + 184x¹⁶ + 182x¹⁵ + 81x¹⁴ + 206x¹³ + 107x¹² + 170x¹¹ + 146x¹⁰ + 62x⁹ + 169x⁸ + 145x⁷

Discard the lead 0 term to get:

184x¹⁶ + 182x¹⁵ + 81x¹⁴ + 206x¹³ + 107x¹² + 170x¹¹ + 146x¹⁰ + 62x⁹ + 169x⁸ + 145x⁷

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 184x¹⁶. Convert 184x¹⁶ to alpha notation. According to the log antilog table, for the integer value 184, the alpha exponent is 132. Therefore 184 = ɑ¹³². Multiply the generator polynomial by ɑ¹³²:

(ɑ¹³² * ɑ⁰)x¹⁶ + (ɑ¹³² * ɑ²⁵¹)x¹⁵ + (ɑ¹³² * ɑ⁶⁷)x¹⁴ + (ɑ¹³² * ɑ⁴⁶)x¹³ + (ɑ¹³² * ɑ⁶¹)x¹² + (ɑ¹³² * ɑ¹¹⁸)x¹¹ + (ɑ¹³² * ɑ⁷⁰)x¹⁰ + (ɑ¹³² * ɑ⁶⁴)x⁹ + (ɑ¹³² * ɑ⁹⁴)x⁸ + (ɑ¹³² * ɑ³²)x⁷ + (ɑ¹³² * ɑ⁴⁵)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:

The result is:

ɑ¹³²x¹⁶ + ɑ¹²⁸x¹⁵ + ɑ¹⁹⁹x¹⁴ + ɑ¹⁷⁸x¹³ + ɑ¹⁹³x¹² + ɑ²⁵⁰x¹¹ + ɑ²⁰²x¹⁰ + ɑ¹⁹⁶x⁹ + ɑ²²⁶x⁸ + ɑ¹⁶⁴x⁷ + ɑ¹⁷⁷x⁶

Now, convert this to integer notation:

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

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

(184 ⊕ 184)x¹⁶ + (182 ⊕ 133)x¹⁵ + (81 ⊕ 14)x¹⁴ + (206 ⊕ 171)x¹³ + (107 ⊕ 25)x¹² + (170 ⊕ 108)x¹¹ + (146 ⊕ 112)x¹⁰ + (62 ⊕ 200)x⁹ + (169 ⊕ 72)x⁸ + (145 ⊕ 198)x⁷ + (0 ⊕ 219)x⁶

The result is:

0x¹⁶ + 51x¹⁵ + 95x¹⁴ + 101x¹³ + 114x¹² + 198x¹¹ + 226x¹⁰ + 246x⁹ + 225x⁸ + 87x⁷ + 219x⁶

Discard the lead 0 term to get:

51x¹⁵ + 95x¹⁴ + 101x¹³ + 114x¹² + 198x¹¹ + 226x¹⁰ + 246x⁹ + 225x⁸ + 87x⁷ + 219x⁶

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 51x¹⁵. Convert 51x¹⁵ to alpha notation. According to the log antilog table, for the integer value 51, the alpha exponent is 125. Therefore 51 = ɑ¹²⁵. Multiply the generator polynomial by ɑ¹²⁵:

(ɑ¹²⁵ * ɑ⁰)x¹⁵ + (ɑ¹²⁵ * ɑ²⁵¹)x¹⁴ + (ɑ¹²⁵ * ɑ⁶⁷)x¹³ + (ɑ¹²⁵ * ɑ⁴⁶)x¹² + (ɑ¹²⁵ * ɑ⁶¹)x¹¹ + (ɑ¹²⁵ * ɑ¹¹⁸)x¹⁰ + (ɑ¹²⁵ * ɑ⁷⁰)x⁹ + (ɑ¹²⁵ * ɑ⁶⁴)x⁸ + (ɑ¹²⁵ * ɑ⁹⁴)x⁷ + (ɑ¹²⁵ * ɑ³²)x⁶ + (ɑ¹²⁵ * ɑ⁴⁵)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:

The result is:

ɑ¹²⁵x¹⁵ + ɑ¹²¹x¹⁴ + ɑ¹⁹²x¹³ + ɑ¹⁷¹x¹² + ɑ¹⁸⁶x¹¹ + ɑ²⁴³x¹⁰ + ɑ¹⁹⁵x⁹ + ɑ¹⁸⁹x⁸ + ɑ²¹⁹x⁷ + ɑ¹⁵⁷x⁶ + ɑ¹⁷⁰x⁵

Now, convert this to integer notation:

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

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

(51 ⊕ 51)x¹⁵ + (95 ⊕ 118)x¹⁴ + (101 ⊕ 130)x¹³ + (114 ⊕ 179)x¹² + (198 ⊕ 110)x¹¹ + (226 ⊕ 125)x¹⁰ + (246 ⊕ 100)x⁹ + (225 ⊕ 87)x⁸ + (87 ⊕ 86)x⁷ + (219 ⊕ 213)x⁶ + (0 ⊕ 215)x⁵

The result is:

0x¹⁵ + 41x¹⁴ + 231x¹³ + 193x¹² + 168x¹¹ + 159x¹⁰ + 146x⁹ + 182x⁸ + 1x⁷ + 14x⁶ + 215x⁵

Discard the lead 0 term to get:

41x¹⁴ + 231x¹³ + 193x¹² + 168x¹¹ + 159x¹⁰ + 146x⁹ + 182x⁸ + 1x⁷ + 14x⁶ + 215x⁵

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 41x¹⁴. Convert 41x¹⁴ to alpha notation. According to the log antilog table, for the integer value 41, the alpha exponent is 147. Therefore 41 = ɑ¹⁴⁷. Multiply the generator polynomial by ɑ¹⁴⁷:

(ɑ¹⁴⁷ * ɑ⁰)x¹⁴ + (ɑ¹⁴⁷ * ɑ²⁵¹)x¹³ + (ɑ¹⁴⁷ * ɑ⁶⁷)x¹² + (ɑ¹⁴⁷ * ɑ⁴⁶)x¹¹ + (ɑ¹⁴⁷ * ɑ⁶¹)x¹⁰ + (ɑ¹⁴⁷ * ɑ¹¹⁸)x⁹ + (ɑ¹⁴⁷ * ɑ⁷⁰)x⁸ + (ɑ¹⁴⁷ * ɑ⁶⁴)x⁷ + (ɑ¹⁴⁷ * ɑ⁹⁴)x⁶ + (ɑ¹⁴⁷ * ɑ³²)x⁵ + (ɑ¹⁴⁷ * ɑ⁴⁵)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:

The result is:

ɑ¹⁴⁷x¹⁴ + ɑ¹⁴³x¹³ + ɑ²¹⁴x¹² + ɑ¹⁹³x¹¹ + ɑ²⁰⁸x¹⁰ + ɑ¹⁰x⁹ + ɑ²¹⁷x⁸ + ɑ²¹¹x⁷ + ɑ²⁴¹x⁶ + ɑ¹⁷⁹x⁵ + ɑ¹⁹²x⁴

Now, convert this to integer notation:

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

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

(41 ⊕ 41)x¹⁴ + (231 ⊕ 84)x¹³ + (193 ⊕ 249)x¹² + (168 ⊕ 25)x¹¹ + (159 ⊕ 81)x¹⁰ + (146 ⊕ 116)x⁹ + (182 ⊕ 155)x⁸ + (1 ⊕ 178)x⁷ + (14 ⊕ 88)x⁶ + (215 ⊕ 75)x⁵ + (0 ⊕ 130)x⁴

The result is:

0x¹⁴ + 179x¹³ + 56x¹² + 177x¹¹ + 206x¹⁰ + 230x⁹ + 45x⁸ + 179x⁷ + 86x⁶ + 156x⁵ + 130x⁴

Discard the lead 0 term to get:

179x¹³ + 56x¹² + 177x¹¹ + 206x¹⁰ + 230x⁹ + 45x⁸ + 179x⁷ + 86x⁶ + 156x⁵ + 130x⁴

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 179x¹³. Convert 179x¹³ to alpha notation. According to the log antilog table, for the integer value 179, the alpha exponent is 171. Therefore 179 = ɑ¹⁷¹. Multiply the generator polynomial by ɑ¹⁷¹:

(ɑ¹⁷¹ * ɑ⁰)x¹³ + (ɑ¹⁷¹ * ɑ²⁵¹)x¹² + (ɑ¹⁷¹ * ɑ⁶⁷)x¹¹ + (ɑ¹⁷¹ * ɑ⁴⁶)x¹⁰ + (ɑ¹⁷¹ * ɑ⁶¹)x⁹ + (ɑ¹⁷¹ * ɑ¹¹⁸)x⁸ + (ɑ¹⁷¹ * ɑ⁷⁰)x⁷ + (ɑ¹⁷¹ * ɑ⁶⁴)x⁶ + (ɑ¹⁷¹ * ɑ⁹⁴)x⁵ + (ɑ¹⁷¹ * ɑ³²)x⁴ + (ɑ¹⁷¹ * ɑ⁴⁵)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:

The result is:

ɑ¹⁷¹x¹³ + ɑ¹⁶⁷x¹² + ɑ²³⁸x¹¹ + ɑ²¹⁷x¹⁰ + ɑ²³²x⁹ + ɑ³⁴x⁸ + ɑ²⁴¹x⁷ + ɑ²³⁵x⁶ + ɑ¹⁰x⁵ + ɑ²⁰³x⁴ + ɑ²¹⁶x³

Now, convert this to integer notation:

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

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

(179 ⊕ 179)x¹³ + (56 ⊕ 126)x¹² + (177 ⊕ 11)x¹¹ + (206 ⊕ 155)x¹⁰ + (230 ⊕ 247)x⁹ + (45 ⊕ 78)x⁸ + (179 ⊕ 88)x⁷ + (86 ⊕ 235)x⁶ + (156 ⊕ 116)x⁵ + (130 ⊕ 224)x⁴ + (0 ⊕ 195)x³

The result is:

0x¹³ + 70x¹² + 186x¹¹ + 85x¹⁰ + 17x⁹ + 99x⁸ + 235x⁷ + 189x⁶ + 232x⁵ + 98x⁴ + 195x³

Discard the lead 0 term to get:

70x¹² + 186x¹¹ + 85x¹⁰ + 17x⁹ + 99x⁸ + 235x⁷ + 189x⁶ + 232x⁵ + 98x⁴ + 195x³

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 70x¹². Convert 70x¹² to alpha notation. According to the log antilog table, for the integer value 70, the alpha exponent is 48. Therefore 70 = ɑ⁴⁸. Multiply the generator polynomial by ɑ⁴⁸:

(ɑ⁴⁸ * ɑ⁰)x¹² + (ɑ⁴⁸ * ɑ²⁵¹)x¹¹ + (ɑ⁴⁸ * ɑ⁶⁷)x¹⁰ + (ɑ⁴⁸ * ɑ⁴⁶)x⁹ + (ɑ⁴⁸ * ɑ⁶¹)x⁸ + (ɑ⁴⁸ * ɑ¹¹⁸)x⁷ + (ɑ⁴⁸ * ɑ⁷⁰)x⁶ + (ɑ⁴⁸ * ɑ⁶⁴)x⁵ + (ɑ⁴⁸ * ɑ⁹⁴)x⁴ + (ɑ⁴⁸ * ɑ³²)x³ + (ɑ⁴⁸ * ɑ⁴⁵)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:

The result is:

ɑ⁴⁸x¹² + ɑ⁴⁴x¹¹ + ɑ¹¹⁵x¹⁰ + ɑ⁹⁴x⁹ + ɑ¹⁰⁹x⁸ + ɑ¹⁶⁶x⁷ + ɑ¹¹⁸x⁶ + ɑ¹¹²x⁵ + ɑ¹⁴²x⁴ + ɑ⁸⁰x³ + ɑ⁹³x²

Now, convert this to integer notation:

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

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

(70 ⊕ 70)x¹² + (186 ⊕ 238)x¹¹ + (85 ⊕ 124)x¹⁰ + (17 ⊕ 113)x⁹ + (99 ⊕ 189)x⁸ + (235 ⊕ 63)x⁷ + (189 ⊕ 199)x⁶ + (232 ⊕ 129)x⁵ + (98 ⊕ 42)x⁴ + (195 ⊕ 253)x³ + (0 ⊕ 182)x²

The result is:

0x¹² + 84x¹¹ + 41x¹⁰ + 96x⁹ + 222x⁸ + 212x⁷ + 122x⁶ + 105x⁵ + 72x⁴ + 62x³ + 182x²

Discard the lead 0 term to get:

84x¹¹ + 41x¹⁰ + 96x⁹ + 222x⁸ + 212x⁷ + 122x⁶ + 105x⁵ + 72x⁴ + 62x³ + 182x²

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 84x¹¹. Convert 84x¹¹ to alpha notation. According to the log antilog table, for the integer value 84, the alpha exponent is 143. Therefore 84 = ɑ¹⁴³. Multiply the generator polynomial by ɑ¹⁴³:

(ɑ¹⁴³ * ɑ⁰)x¹¹ + (ɑ¹⁴³ * ɑ²⁵¹)x¹⁰ + (ɑ¹⁴³ * ɑ⁶⁷)x⁹ + (ɑ¹⁴³ * ɑ⁴⁶)x⁸ + (ɑ¹⁴³ * ɑ⁶¹)x⁷ + (ɑ¹⁴³ * ɑ¹¹⁸)x⁶ + (ɑ¹⁴³ * ɑ⁷⁰)x⁵ + (ɑ¹⁴³ * ɑ⁶⁴)x⁴ + (ɑ¹⁴³ * ɑ⁹⁴)x³ + (ɑ¹⁴³ * ɑ³²)x² + (ɑ¹⁴³ * ɑ⁴⁵)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:

The result is:

ɑ¹⁴³x¹¹ + ɑ¹³⁹x¹⁰ + ɑ²¹⁰x⁹ + ɑ¹⁸⁹x⁸ + ɑ²⁰⁴x⁷ + ɑ⁶x⁶ + ɑ²¹³x⁵ + ɑ²⁰⁷x⁴ + ɑ²³⁷x³ + ɑ¹⁷⁵x² + ɑ¹⁸⁸x¹

Now, convert this to integer notation:

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

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

(84 ⊕ 84)x¹¹ + (41 ⊕ 66)x¹⁰ + (96 ⊕ 89)x⁹ + (222 ⊕ 87)x⁸ + (212 ⊕ 221)x⁷ + (122 ⊕ 64)x⁶ + (105 ⊕ 242)x⁵ + (72 ⊕ 166)x⁴ + (62 ⊕ 139)x³ + (182 ⊕ 255)x² + (0 ⊕ 165)x¹

The result is:

0x¹¹ + 107x¹⁰ + 57x⁹ + 137x⁸ + 9x⁷ + 58x⁶ + 155x⁵ + 238x⁴ + 181x³ + 73x² + 165x¹

Discard the lead 0 term to get:

107x¹⁰ + 57x⁹ + 137x⁸ + 9x⁷ + 58x⁶ + 155x⁵ + 238x⁴ + 181x³ + 73x² + 165x¹

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 107x¹⁰. Convert 107x¹⁰ to alpha notation. According to the log antilog table, for the integer value 107, the alpha exponent is 84. Therefore 107 = ɑ⁸⁴. Multiply the generator polynomial by ɑ⁸⁴:

(ɑ⁸⁴ * ɑ⁰)x¹⁰ + (ɑ⁸⁴ * ɑ²⁵¹)x⁹ + (ɑ⁸⁴ * ɑ⁶⁷)x⁸ + (ɑ⁸⁴ * ɑ⁴⁶)x⁷ + (ɑ⁸⁴ * ɑ⁶¹)x⁶ + (ɑ⁸⁴ * ɑ¹¹⁸)x⁵ + (ɑ⁸⁴ * ɑ⁷⁰)x⁴ + (ɑ⁸⁴ * ɑ⁶⁴)x³ + (ɑ⁸⁴ * ɑ⁹⁴)x² + (ɑ⁸⁴ * ɑ³²)x¹ + (ɑ⁸⁴ * ɑ⁴⁵)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:

The result is:

ɑ⁸⁴x¹⁰ + ɑ⁸⁰x⁹ + ɑ¹⁵¹x⁸ + ɑ¹³⁰x⁷ + ɑ¹⁴⁵x⁶ + ɑ²⁰²x⁵ + ɑ¹⁵⁴x⁴ + ɑ¹⁴⁸x³ + ɑ¹⁷⁸x² + ɑ¹¹⁶x¹ + ɑ¹²⁹x⁰

Now, convert this to integer notation:

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

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

(107 ⊕ 107)x¹⁰ + (57 ⊕ 253)x⁹ + (137 ⊕ 170)x⁸ + (9 ⊕ 46)x⁷ + (58 ⊕ 77)x⁶ + (155 ⊕ 112)x⁵ + (238 ⊕ 57)x⁴ + (181 ⊕ 82)x³ + (73 ⊕ 171)x² + (165 ⊕ 248)x¹ + (0 ⊕ 23)x⁰

The result is:

0x¹⁰ + 196x⁹ + 35x⁸ + 39x⁷ + 119x⁶ + 235x⁵ + 215x⁴ + 231x³ + 226x² + 93x¹ + 23

Discard the lead 0 term to get:

196x⁹ + 35x⁸ + 39x⁷ + 119x⁶ + 235x⁵ + 215x⁴ + 231x³ + 226x² + 93x¹ + 23

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:

196  35  39  119  235  215  231  226  93  23

Next: Structure Final Message

The data codewords and error correction codewords have now been generated. For larger QR codes, the QR code specification requires that the codewords be interleaved according to a particular pattern. The next section, structure final message explains the interleaving process.