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cyclotomic polynomial, which is of degree in .NET Generator QR Code 2d barcode in .NET cyclotomic polynomial, which is of degree

cyclotomic polynomial, which is of degree use vs .net denso qr bar code maker todevelop qr code 2d barcode with .net gs1 databar <p(21 - 1 QR Code ISO/IEC18004 for .NET ) = 127 - 1 = 2 3 3 7. (since 127 i s prime). So if we imagine that it"s exactly the product of the primitive septies, then there should be 2 . 3 .

3 = 18 of them. How many irreducible septies are there mod 2 It turns out that there are. 21 - 21 126 7 =-;;-=18. 16 . Primitive Polynomials So all irred ucible septics mod 2 are primitive. An octic polynomial in F2[X) is primitive, by definition, if it is irreducible and divides the (28 - l)th = 255 th = (3.5.

17)th cyclotomic polynomial, which is of degree. c,o(28 - 1). = (3 -. 1)(5 - 1)(7 - 1). = 24 6 = 48. So if we ima QR-Code for .NET gine that it"s exactly the product of the primitive octics, then there should be 48/8 = 6 of them. How many irreducible octics are there mod 2 It turns out that there are.

So only 6 of the 30 irreducible octics are primitive. 16.3 Testing for primitivity For large de gree d, it is not good to test for "primitivity of polynomials in F q[x) by actually computing the (pd - l)th cyclotomic polynomial, since the degree of this grows exponentially in d, and too much memory would be needed to do this! There is a better approach expressible as an algorithm, which uses hardly any memory. Theorem: Let P be an irreducible polynomial of degree n in Fq[x). Let N = pn _ 1.

Then P is primitive if and only if. xN/r 1 mod P for every pr qr barcode for .NET ime number r dividing N. Remark: As usual, the way we test whether a polynomial f is 1 mod P is to divide-with-remainder f by P, and see whether or not the remainder is 1.

Remark: Of course, we should use the fast exponentiation algorithm here. Remark: The fast exponentiation algorithm inF 2 [x) runs especially fast, since all that is necessary to square such a polynomial is to double the exponents (of terms which actually occur). For example, in F 2 [x),.

(x 5 + x4 + X + 1)2 = x lO + x 8 + x 2 + 1 + x 9 is pri mitive, while the nonie 1 + x + x 9 is irreducible but not primitive. In both cases, we"ll take the irreducibility for granted. Also, for brevity, let"s express polynomials in F2[X) as arrays of non-negative integers, where an integer i occurs if and only if Xi occurs in the polynomial.

(Of course this trick is special to the case that the coefficients are in F 2 .) For example, x 3 + x + 1 would be denoted by [0,1,3). First, note that 29 -1 = 7 73.

Example: Let"s verify that the nonic (degree nine polynomial) 1 + X4 16.3 Testing for primitivity so (by Lagrange"s theorem, etc.) it might be that = 1 mod irred nonic X 73. = 1 mod irred nonic if the nonic is not primitive. In the first case, x 7 is already reduced modulo any nonic polynomial, so it cannot be that x 7 = 1 modulo any nonie. Next, compute x 73 mod 1 + x4 + x 9 via the Fast Modular Exponentiation Algorithm: initiate (X, E, Y) = (x, 73,1) and at each step in the algorithm we list the triple of values (X, E, Y) at that point.

When E = 0 the algorithm terminates and the value of Y is the desired x 73 modulo the polynomial. [1] [1] [2] [4] [8] [8] [2,6,7] [0,3,5,7J [1,4,6J [1,4,6J Thus,. x 73. 73 72 36 18 9 8 4 2 1 0. [0] [1] [1] [1] [1] [0,4] [0,4] [0,4J [0,4J [4,6,8J + x 6 + x 8 mod 1 + x4 + x 9 whieh is not visual .net QR 1 mod 1 + X4 + x 9 Just to check, let"s verify that x 511 = 1 mod 1 + X4 + x 9 Of course, taking advantage of the power-of-2 situation, it would be smarter to verify that X 512 = x mod 1 + X4 + x 9 . We again use the Fast Modular Exponentiation Algorithm, displaying the values of (X, EY) at successive steps in the execution: [IJ [2J [4] [8J [2,6,7] [0,3,5,7] [1,4,6] [2,3,7,8] [0,2,5,7] [1] [1] So that.

X 512. 512 256 128 64 32 16. [0] [0] [0]. [OJ [0] [OJ visual .net qr bidimensional barcode . [OJ [OJ [OJ [0] [1].

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