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8 Μαΐ 2024 · Use modulo-2 binary division to divide binary data by the key and store remainder of division. Perform modulo-2 division again and if the remainder is 0, then there are no errors. In this article we will focus only on finding the remainder i.e. check word and the code word.
20 Οκτ 2023 · Modulo calculator finds a mod b, the remainder when a is divided by b. The modulo operation returns the remainder in division of 2 positive or negative numbers or decimals.
We must then calculate the required remainder from a modulo-2 divide and add this to the data, in order that the remainder will be zero when we perform the divide. To take a simple example, we have 32, and make it divisible by 9, we add a ‘0’ to make ‘320’, and now divide by 9, to give 35 remainder 4.
For the general case of linear block codes, all the n digits of X are formed by linear combinations (modulo-2 additions) of k message bits. A special case, where x0 = m0, x1 = m1, x2 = m2....xk-1 = mk-1 and the remaining digits from xk+1 to xn are linear combinations of m0, m1, m2, ..... mk-1 is known as a systematic code.
Modulo 2 arithmetic is performed digit by digit on binary numbers. Each digit is considered independently from its neighbours. Numbers are not carried or borrowed. Modulo 2 addition/subtraction is performed using an exclusive OR (xor) operation on the corresponding binary digits of each operand.
So the polynomial $R(x)$ must be the negative of the remainder, when $M(x)x^5$ is divided by $P(x)$. Because modulo 2 we have that $-0=0$ and $-1=1$ we can forget about that 'negative'. Let's calculate! The product $R_0:=M(x)x^5=x^{12}+x^{11}+x^{10}+x^6+x^5$ is of degree twelve. In other words it has a total of $n=13$ coefficients.
Theorem Let m ≥ 2 be an integer and a a number in the range 1 ≤ a ≤ m − 1 (i.e. a standard rep. of a number modulo m). Then a has a multiplicative inverse modulo m if a and m are relatively prime.
